Free probability and combinatorics
Preliminary version
c
Michael Anshelevich 2012
April 24, 2015
Preface
These notes were used in a topics course Free probability and combinatorics taught at Texas A&M University in the fall of 2012. The course was based on Nica and Speicher’s textbook [NS06], however there
were sufficient differences to warrant separate notes. I tried to minimize the list of references, but many
more are included in the body of the text. Thanks to the students in the course, and to March Boedihardjo,
for numerous comments and corrections; further corrections or comments are welcome. Eventually I will
probably replace hand-drawn figures by typeset ones.
Contents
1
2
3
4
5
Noncommutative probability spaces and distributions.
3
1.1
Non-commutative probability spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.2
Distributions. Weak law of large numbers. . . . . . . . . . . . . . . . . . . . . . . . . .
6
Functional analysis background.
9
2.1
C∗ - and von Neumann algebras. Spectral theorem. . . . . . . . . . . . . . . . . . . . . .
9
2.2
Gelfand-Naimark-Segal construction. . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
2.3
Other topics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14
Free independence.
19
3.1
Independence. Free independence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
3.2
Free products. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22
Set partition lattices.
29
4.1
All, non-crossing, interval partitions. Enumeration. . . . . . . . . . . . . . . . . . . . .
29
4.2
Incidence algebra. Multiplicative functions. . . . . . . . . . . . . . . . . . . . . . . . .
33
Free cumulant machinery.
38
5.1
Free cumulants. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38
5.2
Free independence and free cumulants. . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
5.3
Free convolution, distributions, and limit theorems. . . . . . . . . . . . . . . . . . . . .
49
1
6
Lattice paths and Fock space models.
54
6.1
Dyck paths and the full Fock space. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
54
6.2
Łukasiewicz paths and Voiculescu’s operator model. . . . . . . . . . . . . . . . . . . .
62
6.3
Motzkin paths and Schürmann’s operator model. Freely infinitely divisible distributions.
63
6.4
Motzkin paths and orthogonal polynomials. . . . . . . . . . . . . . . . . . . . . . . . .
69
6.5
q-deformed free probability. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
74
7
Free Lévy processes.
77
8
Free multiplicative convolution and the R-transform.
81
9
Belinschi-Nica evolution.
85
10 ∗-distribution of a non-self-adjoint element
93
11 Combinatorics of random matrices.
97
11.1 Gaussian random matrices. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
97
11.2 Map enumeration. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
11.3 Partitions and permutations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
11.4 Asymptotic free independence for Gaussian random matrices. . . . . . . . . . . . . . . 112
11.5 Asymptotic free independence for unitary random matrices. . . . . . . . . . . . . . . . . 115
12 Operator-valued free probability.
121
13 A very brief survey of analytic results.
128
13.1 Complex analytic methods. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
13.2 Free entropy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
13.3 Operator algebras. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
2
Chapter 1
Noncommutative probability spaces and
distributions.
See Lectures 1, 4, 8 of [NS06].
1.1
Non-commutative probability spaces.
Definition 1.1. An (algebraic) (non-commutative) probability space is a pair (A, ϕ), where A is a unital
∗-algebra and ϕ is a state on A. That is:
• A is an algebra over C, with operations za + wb, ab for z, w ∈ C, a, b ∈ A. Unital: 1 ∈ A.
• ∗ is an anti-linear involution, (za)∗ = za∗ , (ab)∗ = b∗ a∗ .
• ϕ : A → C is a linear functional. Self-adjoint: ϕ [a∗ ] = ϕ [a]. Unital: ϕ [1] = 1. Positive:
ϕ [a∗ a] ≥ 0.
Definition 1.2. a ∈ A is symmetric if a = a∗ . A symmetric a ∈ (A, ϕ) is a (n.c.) random variable.
Examples of commutative probability spaces.
Example 1.3. Let (X, Σ, P ) be a measure space, P a probability measure. Take
Z
∞
∗
A = L (X, P ), f = f , E(f ) =
f dP.
X
Then E (usually called the expectation) is a state, in particular positive and unital. Thus (A, E) is a
(commutative) probability space. Note: f = f ∗ means f real-valued.
3
A related construction is
A = L∞− (X, P ) =
\
Lp (X, P ),
p≥1
the space of complex-valued random variables all of whose moments are finite.
Example 1.4. Let X be a compact topological space (e.g. X = [0, 1]), and µ a Borel probability measure
on X. Then for
Z
f dµ,
A = C(X), ϕ [f ] =
X
(A, ϕ) is again a commutative probability space.
Example 1.5. Let A = C[x] (polynomials in x with complex coefficients), x∗ = x. Then any state ϕ on
C[x] gives a commutative probability space. Does such a state always come from a measure?
Examples of non-commutative probability spaces.
Example 1.6. Let x1 , x2 , . . . , xd be non-commuting indeterminates. Let
A = Chx1 , x2 , . . . , xd i = Chxi
be polynomials in d non-commuting variables, with the involution
x∗i = xi ,
(xu(1) xu(2) . . . xu(n) )∗ = xu(n) . . . xu(2) xu(1) .
Then any state ϕ on Chxi gives a non-commutative probability space. These do not come from measures.
One example: for z = (z1 , z2 , . . . , zd ) ∈ Rd ,
δz (f (x1 , x2 , . . . , xd )) = f (z1 , z2 , . . . , zd )
is a state (check!). Other examples?
Example 1.7. A = Mn (C), the n × n matrices over C, with the involution (A∗ )i,j = Aji and
n
1 X
1
ϕ [A] =
Aii =
Tr(A) = tr(A),
N i=1
N
the normalized trace of A. Note that indeed, tr(A∗ ) = tr(A) and tr(A∗ A) ≥ 0.
Definition 1.8. Let ϕ be a state on A.
a. ϕ is tracial, or a trace, if for any a, b ∈ A,
ϕ [ab] = ϕ [ba] .
Note that A is, in general, not commutative.
4
b. ϕ is faithful if ϕ [a∗ a] = 0 only if a = 0.
Example 1.9. For a probability space (X, Σ, P ), let
A = Mn (C) ⊗ L∞− (X, P ) ' Mn (L∞− (X, P )).
These are random matrices = matrix-valued random variables = matrices with random entries. Take
Z
ϕ [A] = (tr ⊗ E)(A) =
tr(A) dP.
X
Example 1.10. Let H be a Hilbert space, and A a ∗-subalgebra of B(H), the algebra of bounded linear
operators on H. a∗ is the adjoint operator to a. If ξ ∈ H is a unit vector, then
ϕ [a] = haξ, ξi
is a state. Why unital:
ϕ [1] = hξ, ξi = kξk2 = 1.
Why self-adjoint:
ϕ [a∗ ] = ha∗ ξ, ξi = hξ, aξi = haξ, ξi.
Why positive:
ϕ [a∗ a] = ha∗ aξ, ξi = haξ, aξi = kaξk2 ≥ 0.
Typically not tracial or faithful.
Example 1.11 (Group algebra). Let Γ be a discrete group (finite, Zn , Fn , etc.).
C[Γ] = functions Γ → C of finite support
= finite linear combinations of elements of Γ with C coefficients.
(f : x 7→ f (x))
↔
X
f (x)x.
x∈Γ
This is a vector space. It is an algebra with multiplication
!
!
X
X
X
f (x)x
g(y)y =
(f (x)g(y))z,
x∈Γ
y∈Γ
z∈Γ
z=xy
in other words
(f g)(z) =
X
f (x)g(y) =
z=xy
X
x∈Γ
the convolution multiplication. The involution is
f ∗ (x) = f (x−1 ).
5
f (x)g(x−1 z),
Check that indeed, (f g)∗ = g ∗ f ∗ .
So C[Γ] is a unital ∗-algebra, with the unit δe , where for x ∈ Γ
(
1, y = x,
δx (y) =
0, y 6= x
and e is the unit of Γ.
Moreover, define
τ [f ] = f (e).
Exercise 1.12. Prove that τ is a faithful, tracial state, called the von Neumann trace.
Later: other versions of group algebras.
1.2
Distributions. Weak law of large numbers.
Definition 1.13. Let (A, ϕ) be an n.c. probability space.
a. Let a ∈ (A, ϕ) be symmetric. The distribution of a is the linear functional
ϕa : C[x] → C,
ϕa [p(x)] = ϕ [p(a)] .
Note that ϕa is a state on C[x].
The sequence of numbers
{mn [ϕa ] = ϕa [xn ] = ϕ [an ] , n = 0, 1, 2, . . .}
are the moments of a. In particular m0 [ϕa ] = 1 and m1 [ϕa ] = ϕ [a] is the mean of a.
b. More generally, let a1 , a2 , . . . , ad ∈ (A, ϕ) be symmetric. Their joint distribution is the state
ϕa1 ,...,ad : Chx1 , . . . , xd i → C,
ϕa1 ,...,ad [p(x1 , . . . , xd )] = p(a1 , . . . , ad ).
The numbers
ϕ au(1) au(2) . . . au(n) : n ≥ 0, 1 ≤ u(i) ≤ d
are the joint moments of a1 , . . . , ad .
Denote by D(d) the space of all joint distributions of d-tuples of symmetric random variables, which
is the space of all states on Chx1 , . . . , xd i.
6
c. We say that
(N )
(N )
(a1 , . . . , ad ) → (a1 , . . . , ad )
in moments (or, for d > 1, in distribution) if for each p,
ϕ(a(N ) ,...,a(N ) ) [p] → ϕ(a1 ,...,ad ) [p]
1
d
as N → ∞.
Remark 1.14. Each µ ∈ D(d) can be realized as a distribution of a d-tuple of random variables, namely
(x1 , x2 , . . . , xd ) ⊂ (Chxi, µ).
Definition 1.15. a1 , a2 , . . . , ad ∈ (A, ϕ) are singleton independent if
ϕ au(1) au(2) . . . au(n) = 0
whenever all ai are centered (that is, ϕ [ai ] = 0) and some index in ~u appears only once.
Proposition 1.16 (Weak law of large numbers). Suppose {an : n ∈ N} ⊂ (A, ϕ) are singleton independent, identically distributed (that is, all ϕai are the same), and uniformly bounded, in the sense that for a
fixed C and all ~u,
ϕ au(1) au(2) . . . au(n) ≤ C n .
Denote
1
(a1 + a2 + . . . + an ).
n
Then sn → ϕ [a1 ] in moments (sn converges to the mean, a scalar).
sn =
Proof. Note first that
1
(a1 − ϕ [a1 ]) + . . . + (an − ϕ [an ]) = sn − ϕ [a1 ] .
n
So without loss of generality, may assume ϕ [a1 ] = 0, and we need to show that sn → 0 in moments.
n
n
X
k
1 1 X
k
ϕ sn = k ϕ (a1 + . . . + an ) = k
...
ϕ au(1) au(2) . . . au(k) .
n
n
u(1)=1
u(k)=1
How many non-zero terms?
Denote B(k) the number of partitions of k points into disjoint non-empty subsets. Don’t need the exact
value, but see Chapter 4. For each partition
π = (B1 , B2 , . . . , Br ),
B1 ∪ B2 ∪ . . . Br = {1, . . . , k} ,
Bi ∩ Bj = ∅ for i 6= j.
How many k-tuples (u(1), u(2), . . . , u(k)) such that
u(i) = u(j)
⇔
π
i∼j
( i.e. i, j in the same block of π)?
7
n(n − 1)(n − 2) . . . (n − r + 1) ≤ nr ≤ nk/2
because of the singleton condition. Each term in the sum bounded by
ϕ au(1) au(2) . . . au(k) ≤ C k .
Thus
k ϕ xn ≤ 1 B(k)nk/2 C k → 0
nk
as n → ∞ for any k > 0.
Remark 1.17. For centered independent random variables, showed
sn =
Expect a non-trivial limit for
a1 + . . . + an
→ 0.
n
a1 + . . . + an
√
.
n
What limit depends on the notion of independence used. See Lecture 8 of [NS06] for more.
8
Chapter 2
Functional analysis background.
See Lectures 3, 7 of [NS06].
2.1
C∗- and von Neumann algebras. Spectral theorem.
Definition 2.1. An (abstract) C ∗ -algebra is a Banach ∗-algebra with an extra axiom
ka∗ ak = kak2 .
A C ∗ -probability space is a pair (A, ϕ), where A is a C ∗ -algebra and ϕ is a state on it continuous in the
norm topology. (It follows from Corollary 2.13(c) that continuity is actually automatic).
Example 2.2. If X is a compact Hausdorff space, the algebra of continuous functions on it C(X) is a C ∗ algebra (with the uniform norm). States on C(X) come from integration with respect to Borel probability
measures.
Theorem 2.3 (Gelfand-Naimark theorem). Any unital, commutative C ∗ -algebra is of the form in the
preceding example.
Example 2.4. If A is a ∗-subalgebra of B(H) closed in the norm topology, then A is a C ∗ -algebra. If
ξ ∈ H is a unit vector, then ϕ = h·ξ, ξi is a state on A.
Theorem 2.5. Any abstract C ∗ -algebra is a concrete C ∗ -algebra, that is, it has a representation as in the
preceding example.
Definition 2.6. A ∗-subalgebra A ⊂ B(H) is a von Neumann algebra (or a W ∗ -algebra) if A is closed in
the weak operator topology. A W ∗ -probability space is a pair (A, ϕ), where A is a W ∗ -algebra and ϕ is
a normal state (continuous in the ultraweak operator topology).
9
Example 2.7. Let X be a compact Hausdorff space, and µ be a Borel probability measure on X. Then
(C(X), µ) is a C ∗ -probability space. But C(X) ⊂ B(L2 (X, µ)) is not WOT-closed. However, L∞ (X, µ)
is so closed, and therefore is a von Neumann algebra. In fact, the WOT on L∞ (X, µ) is the weak-∗
topology on L∞ = (L1 )∗ . Normal states on L∞ (X, µ) come from integration with respect to f dµ,
f ∈ L1 (X, µ).
Definition 2.8. For a ∈ B(H), its spectrum is the set
σ(a) = {z ∈ C : (z − a) is not invertible} .
The spectrum is always a compact, non-empty subset of C. The spectral radius of a is
r = sup {|z| : z ∈ σ(a)} .
We always have r(a) ≤ kak.
An operator a ∈ B(H) is positive (written a ≥ 0) if a = a∗ and σ(a) ⊂ [0, ∞).
Proposition 2.9. Let A be a C ∗ -algebra and a ∈ A.
a. kak2 = r(a∗ a). Thus norm (topology) is determined by the spectrum (algebra).
b. a ≥ 0 if and only if for some b ∈ A, a = b∗ b. (In fact, may take b ≥ 0 and a = b2 ).
Definition 2.10. In an (algebraic) n.c. probability space A, say a ≥ 0 if ∃b1 , . . . , bk such that
a=
k
X
b∗i bi .
i=1
Remark 2.11. Note that in the algebra C[x1 , x2 , . . . , xd ] of multivariate polynomials in commuting variables, we can have p(x) ≥ 0 for all x without being able to write p as a sum of squares.
Theorem 2.12 (Spectral theorem and continuous functional calculus, bounded symmetric case). Let a =
a∗ ∈ B(H). Then the C ∗ -algebra generated by a
C ∗ (a) ' C(σ(a))
(isometric C ∗ -isomorphism). Moreover, this defines a map f 7→ f (a) ∈ B(H), so that f (a) is a welldefined operator for any continuous f .
Corollary 2.13. Let (A, ϕ) be a C ∗ -probability space.
a. For any a = a∗ ∈ A, the operator kak − a is positive.
b. For any a = a∗ , b ∈ A, ϕ [b∗ ab] ≤ kak ϕ [b∗ b] (why real?).
10
c. For any a ∈ A, |ϕ [a]| ≤ kak.
d. Assume ϕ is faithful. Then for any a ∈ A, kak = limn→∞ (ϕ [(a∗ a)n ])1/2n . Thus the norm can be
computed by using moments.
Proof. For (a), using the identification in the spectral theorem, in C(σ(a)), a corresponds to the identity
function (f (x) = x), kak = kf ku , and kak − a corresponds to kf ku − f , which is positive.
For (b), using part (a) we can write kak − a = c∗ c, from which
ϕ [b∗ (kak − a)b] = ϕ [b∗ c∗ cb] ≥ 0.
If a is symmetric, (c) follows from (b) (with b = 1). In general, Cauchy-Schwartz inequality below
implies
p
p
|ϕ [a]| ≤ ϕ [a∗ a] ϕ [1∗ 1] ≤ ka∗ ak = kak .
Finally, applying the spectral theorem to a symmetric element a∗ a, the last statement follows from that
fact that for a finite measure µ, limn→∞ kf k2n,µ = kf k∞,µ and, if µ has full support, kf k∞,µ = kf ku .
2.2
Gelfand-Naimark-Segal construction.
Remark 2.14 (GNS construction I.). Let (A, ϕ) be an n.c. probability space. Let V = A as a vector
space. For a, b ∈ V , define
ha, biϕ = ϕ [b∗ a] .
This is a possibly degenerate inner product, i.e. may have kakϕ = 0 for a 6= 0 in V . In fact, inner product
non-degenerate if and only if ϕ faithful.
For a ∈ A, define λ(a) ∈ L(V ) (linear, not necessarily bounded operators) by
λ(a)b = ab.
Note that
hλ(a∗ )b, ciϕ = ϕ [c∗ a∗ b] = ϕ [(ac)∗ b] = hb, λ(a)ciϕ ,
so with respect to this inner produce, the adjoint of the operator λ(a) is λ(a∗ ). We conclude that
{λ(a) : a ∈ A} is a ∗-representation of A on V . Moreover,
ϕ [a] = hλ(a)1, 1iϕ .
Even if h·, ·iϕ is degenerate, we still have the Cauchy-Schwartz inequality
2
ha, biϕ ≤ ha, aiϕ hb, biϕ ,
11
in other words |ϕ [b∗ a]|2 ≤ ϕ [a∗ a] ϕ [b∗ b]. Let
n
o
N = a ∈ A : ϕ [a∗ a] = kak2ϕ = 0 .
If a, b ∈ N so are their linear combinations. So N is a subspace of V , on V /N , the inner product is
non-degenerate, and induces a norm k·kϕ . Let
k·k
H = (V /N )
which is a Hilbert space. Denote H (or perhaps V ) by L2 (A, ϕ). We have a natural map a 7→ â = a + N
of A → Â ⊂ L2 (A, ϕ) with dense range.
Moreover, for any a ∈ A and b ∈ N
kabk4ϕ = ϕ [b∗ a∗ ab]2 ≤ kb∗ a∗ aa∗ abkϕ kb∗ bkϕ = 0.
So N is a left ideal in A. For a ∈ A,
b
λ(a)b̂ = ab + aN = ab.
Thus have a (not necessarily faithful) ∗-representation of A on the Hilbert space H, by a priori unbounded
operators with a common dense domain  = V /N .
Corollary 2.15. Each µ ∈ D(d) can be realized as the joint distribution of (possibly unbounded) operators on a Hilbert space with respect to a vector state.
Proof. µ is the joint distribution of the operators λ(x1 ), . . . , λ(xd ) in the GNS representation of
(Chx1 , . . . , xd i, µ)
with respect to the GNS state.
Would like to say
(A, ϕ) ' (A ⊂ B(H), ϕ = h·ξ, ξi).
Implies A is a C ∗ -algebra! So have to assume this.
Remark 2.16 (GNS construction II.). Let (A, ϕ) be an C ∗ -probability space. Use the notation from the
preceding remark.
To show: each kλ(a)k < ∞, in fact kλ(a)k ≤ kak. Indeed,
2
2
2
2
2 λ(a)b̂ = kab + N kϕ ≤ kabkϕ = ϕ [b∗ a∗ ab] ≤ ka∗ ak ϕ [b∗ b] = kak b̂ .
ϕ
ϕ
Finally,
ϕ [a] = λ(a)1̂, 1̂ ϕ .
12
Corollary 2.17. If (A, ϕ) is a C ∗ -probability space and ϕ is faithful, may assume A ⊂ B(H), ϕ = h·ξ, ξi.
In addition, ξ is cyclic, that is
k·k
Aξ = H.
w
Moreover, in this case (A , h·ξ, ξi) is a W ∗ -probability space.
How is the preceding statement related to Theorem 2.5? Does it give a complete proof for it?
The following result is inspired by Proposition 1.2 in [PV10].
Proposition 2.18. Let µ be a state on Chx1 , . . . , xd i such that for a fixed C and all ~u,
µ xu(1) xu(2) . . . xu(n) ≤ C n .
Then µ can be realized as a joint distribution of a d-tuple of bounded operators on a Hilbert space.
Proof. The construction of the Hilbert space H is as in the GNS construction I. We need to show that
the representation is by bounded operators. It suffices to show that each λ(xi ) is bounded. Define the
“non-commutative ball of radius C” to be the space of formal power series
(
)
X
X
BC hxi =
α~u x~u : α~u ∈ C,
|α~u | C |~u| < ∞ .
~
u
~
u
It is easily seen to be a ∗-algebra, to which µ extends via
#
"
X
X
µ
e
α~u x~u =
α~u µ [x~u ] .
~
u
~
u
Define gi to be the power series expansion of 1 −
and so lies in BC hxi. Since
x2i
4C 2
gi2 = 1 −
1/2
. This series has radius of convergence 2C,
1 2
x,
4C 2 i
for f ∈ Â,
0≤
µ
e [f ∗ gi∗ gi f ]
1
1 2
xi f = µ [f ∗ f ] −
µ [(xi f )∗ (xi f )] .
=µ f 1−
2
2
4C
4C
∗
It follows that
kxi f kµ ≤ 2C kf kµ ,
so each operator λ(xi ) is bounded.
Remark 2.19. Any µ ∈ D(d) as in the previous proposition produces, via the GNS construction, a C ∗ algebra and a von Neumann algebra. Thus, at least in principle, one can study von Neumann algebras by
studying such joint distributions. See also Theorem 4.11 of [NS06].
13
2.3
Other topics.
Remark 2.20 (Group algebras).
(
L2 (C[Γ], τ ) = L2 (Γ) =
)
f : Γ → C,
X
|f (x)|2 < ∞ .
x∈Γ
Γ acts on L2 (Γ) on the left by
(λ(y)g)(x) = (δy g)(x) = g(y −1 x).
C[Γ] acts on L2 (Γ) on the left by
(λ(f )g)(x) = (f g)(x) =
X
f (y)g(y −1 x).
y∈Γ
The reduced group C ∗ -algebra is
Cr∗ (Γ) = C[Γ]
k·k
⊂ B(L2 (Γ))
(there is also a full C ∗ -algebra C ∗ (Γ)). The group von Neumann algebra is
weak
L(Γ) = W ∗ (Γ) = C[Γ]
⊂ B(L2 (Γ)).
The vector state τ = h·δe , δe i is the extension of the von Neumann trace, which is still faithful and tracial
on Cr∗ (Γ) and L(Γ).
Remark 2.21 (The isomorphism problem). It is easy to show that
F2 6' F3
and not too hard that
C[F2 ] 6' C[F3 ].
Using K-theory, Pimsner and Voiculescu showed that
Cr∗ (F2 ) 6' Cr∗ (F3 ).
The question of whether
?
L(F2 ) ' L(F3 )
is open.
14
Remark 2.22 (Distributions and moments). Let a be a symmetric, bounded operator. C ∗ (a) ' C(σ(a)),
σ(a) compact. So (continuous) states on C ∗ (a) correspond to Borel probability measures on σ(a). In
particular, if a is a symmetric, bounded operator in (A, ϕ), there exists a Borel probability measure on R
such that for any f ∈ C(σ(a)),
Z
ϕ [f (a)] =
f dµa .
µa is the distribution of a (with respect to ϕ). Note that µa is supported on σ(a), so in particular (for a
bounded operator) compactly supported. Note also that for a polynomial p,
Z
ϕa [p(x)] = p(x) dµa (x).
By the Weierstrass theorem and continuity of ϕ, ϕa determines µa . On the other hand, if ϕ is faithful on
C ∗ (a), for a = a∗
kak = sup {|z| : z ∈ supp(µa )} .
Exercise 2.23. Let A ∈ (Mn (C), tr) be a Hermitian matrix, with real eigenvalues λ1 , . . . , λn . Compute
the algebraic distribution ϕA and the analytic distribution µA of A with respect to tr.
Remark 2.24 (Generating functions).
Z
n
xn dµa (x)
mn = ϕ [a ] =
are the moments of a. Always take m0 = 1. For a formal indeterminate z,
∞
X
M (z) =
mn z n
n=0
is the (formal) moment generating function of a (or of ϕa , or of µa ). If a is bounded, then in fact
Z
1
dµa (z)
M (z) =
R 1 − xz
for z ∈ C, |z| ≤ kak−1 . The (formal) Cauchy transform is
Gµ (z) =
∞
X
mn
n=0
1
z n+1
1
= M (1/z).
z
But in fact, the Cauchy transform is defined by
Z
Gµ (z) =
R
1
dµ(x)
z−x
for any finite measure µ on R, as an analytic map Gµ : C+ → C− .
The measure can be recovered from its Cauchy transform via the Stieltjes inversion formula
dµ(x) = −
1
lim =Gµ (x + iy) dx.
π y→0+
15
Example 2.25. Let Γ = F1 = Z, with a single generator x. In (C[Z], τ ), have the generating element
u = δx . Note that
u∗ = δx−1 = u−1 ,
so u is a unitary. Moreover,
τ [un ] = δn=0 ,
n ∈ Z,
which by definition says that u is a Haar unitary.
What is the distribution of the symmetric operator u + u∗ ? Moments:
τ (u + u−1 )n =?
Number of walks with n steps, starting and ending at zero, with steps of length 1 to the right or to the left.
2n
−1 2n+1
−1 2n
τ (u + u )
= 0, τ (u + u )
=
.
n
Cauchy transform of the distribution is
∞ ∞
∞
X
2n −2n−1 X (2n)! −2n−1 X 2n 1 · 3 · . . . · (2n − 1) −2n−1
G(z) =
z
=
z
=
z
n
n!n!
n!
n=0
n=0
n=0
∞
n
X
1
3
2n − 1
4
=
...
z −2n−1
n!
2
2
2
n=0
∞
X
−1/2
(−4)n
1
3
2n − 1
1
−
−
... −
z −2n−1 = z −1 1 − 4z −2
=√
.
=
2
n!
2
2
2
z −4
n=0
Therefore the distribution is
dµ(x) = −
1
1
1
1
lim+ = p
dx = √
dx.
2
π y→0
π 4 − x2
(x + iy) − 4
This is the arcsine distribution. We will see this example again.
Exercise 2.26. Verify Corollary 2.13(d) for the operator u + u∗ in the preceding example. You might want
to use Remark 2.22,
Unbounded operators.
Remark 2.27 (Commutative setting). Let (X, Σ, P ) be a probability space.
A = L∞ (X, µ).
Ae are all the measurable functions, which form an algebra. Moreover,
f ∈ Ae
⇔
∀g ∈ Cb (C, C), g ◦ f ∈ A.
16
Remark 2.28. On a Hilbert space H, an unbounded operator a is defined only on a (dense) subspace
D(a). Since may have D(a) ∩ D(b) = {0}, cannot in general define a + b.
For a von Neumann algebra A, an unbounded, self-adjoint operator a is affiliated to A, if f (a) ∈ A for
e if in its polar decomposition
all bounded continuous f . A general operator T is affiliated to A, T ∈ A,
T = ua, the partial isometry u ∈ A and the positive operator a is affiliated to A.
If (A, ϕ) is a W ∗ -probability space with ϕ a faithful, tracial state (so that A is a finite von Neumann
algebra), then Ae form an algebra.
Moment problem.
If a is an unbounded, self-adjoint operator, there exists a probability measure µa on R, not necessarily
compactly supported, such that
Z
f dµa
ϕ [f (a)] =
R
R
for f ∈ Cb (R). Moments ϕ [an ] = R xn dµa (x) need not be finite, so ϕa may be undefined. Also, can
have non-uniqueness in the moment problem: µ 6= ν with
Z
Z
p(x) dµ = p(x) dν ∀p ∈ C[x].
R
Definition 2.29. ak → a in distribution if µak → µa weakly, that is
Z
Z
f dµak → f dµ ∀f ∈ Cb (R).
Proposition 2.30. If all µak , µa compactly supported (in particular, if ak , a are bounded), then
ak → a in moments
⇔
ak → a in distribution.
More generally, if ak → a in moments and µa is determined by its moments, then ak → a is distribution.
Tensor products.
Remark 2.31 (Vector spaces). Let V1 , . . . , Vn be vector spaces. Their algebraic tensor product is
( k
)
X (i)
(i)
V1 ⊗ V2 ⊗ . . . ⊗ Vn =
a1 ⊗ . . . ⊗ an(i) , k ≥ 0, aj ∈ Vj /linearity relations
i=1
a1 ⊗ . . . ⊗ (za + wb) ⊗ . . . ⊗ an = za1 ⊗ . . . ⊗ a ⊗ . . . ⊗ an + wa1 ⊗ . . . ⊗ b ⊗ . . . ⊗ an .
Thus for example a ⊗ +a ⊗ c = a ⊗ (b + c), but a ⊗ b + c ⊗ d cannot in general be simplified. Note also
that
a ⊗ b 6= b ⊗ a.
17
Remark 2.32 (Algebras). Let A1 , . . . , An be algebras. Their algebraic tensor product A1 ⊗ . . . ⊗ An has
the vector space structure from the preceding remark and algebra structure
(a1 ⊗ . . . ⊗ an )(b1 ⊗ . . . ⊗ bn ) = a1 b1 ⊗ a2 b2 ⊗ . . . ⊗ an bn .
If the algebras are unital, we have natural embeddings
Ai ,→
n
O
Aj
j=1
via
ai 7→ 1 ⊗ . . . ⊗ ai ⊗ . . . ⊗ 1.
Note that the images of Ai , Aj commute:
(a ⊗ 1)(1 ⊗ b) = a ⊗ b = (1 ⊗ b)(a ⊗ 1).
This is a universal object for this property.
Also have tensor products of Hilbert spaces, C ∗ , von Neumann algebras. Require the taking of closures,
which in some cases are not unique.
Remark 2.33 (Probability spaces). Let (A1 , ϕ1 ), . . . , (An , ϕn ) be n.c. probability spaces. The tensor
product state
n
O
ϕ=
ϕi
i=1
on
Nn
i=1
Ai is defined via the linear extension of
ϕ [a1 ⊗ . . . ⊗ an ] = ϕ1 [a1 ]ϕ2 [a2 ] . . . ϕn [an ].
18
Chapter 3
Free independence.
See Lectures 5, 6 of [NS06].
3.1
Independence. Free independence.
Definition 3.1. Subalgebras A1 , A2 , . . . , An ⊂ (A, ϕ) are independent (with respect to ϕ) if they commute and for any ai ∈ Ai , 1 ≤ i ≤ n
ϕ [a1 a2 . . . an ] = ϕ [a1 ] ϕ [a2 ] . . . ϕ [an ] .
Elements a1 , a2 , . . . , an are independent if the ∗-subalgebras they generate are independent.
Note that we do not assume that each Ai is itself commutative.
Example 3.2. If
(A, ϕ) =
n
O
(Ai , ϕi ),
i=1
then A1 , . . . , An considered as subalgebras of A are independent. For this reason, sometimes call independence tensor independence.
Remark 3.3. If a1 , . . . , an are independent (and in particular commute), their individual distributions
ϕa1 , . . . , ϕan completely determine their joint distribution ϕa1 ,...,an . Indeed, commutativity implies that all
the joint moments can be brought into the form
h
i
h
i h
i
u(1) u(2)
u(1)
u(2)
ϕ a1 a2 . . . anu(n) = ϕ a1
ϕ a2
. . . ϕ au(n)
.
n
This is not true for singleton independence: if ϕ [a1 ] = ϕ [a2 ] = 0, then
ϕ [a1 a2 a1 ] = 0,
19
but
ϕ [a1 a2 a1 a2 ]
is not determined.
Want an independence-type rule for joint distributions of non-commuting variables.
Definition 3.4 (Voiculescu). Let (A, ϕ) be an n.c. probability space.
a. Subalgebras A1 , A2 , . . . , Ak ⊂ (A, ϕ) are freely independent (or free) with respect to ϕ if whenever
u(1) 6= u(2), u(2) 6= u(3), u(3) 6= u(4), . . . ,
ai ∈ Au(i) , ϕ [ai ] = 0, then
ϕ [a1 a2 . . . an ] = 0.
b. Elements a1 , a2 , . . . , ak are freely independent if the ∗-subalgebras they generate are freely independent.
Proposition 3.5. Let Fn be the free group with generators x1 , x2 , . . . , xn , and consider the n.c. probability space (C[Fn ], τ ). Then with respect to τ , the Haar unitaries λ(x1 ), λ(x2 ), . . . , λ(xn ) are freely
independent.
Proof. We want to show that
"
τ
n
Y
Pj
#
λ(xu(j) ), λ(x−1
=0
u(j) )
j=1
whenever
τ Pj (λ(x), λ(x−1 )) = 0
(3.1)
u(1) 6= u(2) 6= u(3) 6= . . . 6= u(n).
(3.2)
and
Note first that
Pj (λ(x), λ(x−1 )) =
X
(j)
αk λ(xk ),
k∈Z
(j)
and so equation (3.1) implies that each α0 = 0. Moreover,
n
Y
X
(1)
(n)
k(1)
k(n)
Pj λ(xu(j) ), λ(x−1
)
=
αk(1) . . . αk(n) λ(xu(1) . . . xu(n) ),
u(j)
j=1
~k
where all k(i) 6= 0. But then by (3.2), the word
k(1)
k(n)
xu(1) . . . xu(n)
is reduced, has no cancellations, and in particular never equals e. It follows that τ applied to any of the
terms in the sum is zero.
20
Remark 3.6. Pairwise free does not imply free. For example, in (C[F2 ], τ ), let
a = λ(x1 ),
b = λ(x2 ),
c = λ(x1 x2 ) = ab.
Then the elements in each pair {a, b}, {a, c}, {b, c} are free, but the triple {a, b, c} is not free.
Remark 3.7. If A1 , . . . , An ⊂ (A, ϕ) ⊂ B(H) are star-subalgebras which are free, then
C ∗ (A1 ), . . . , C ∗ (An ) are free, and W ∗ (A1 ), . . . , W ∗ (An ) are free.
Claim 3.8. Suppose know ϕa1 , . . . , ϕan and a1 , . . . , an are free. Then ϕa1 ,...,an is determined.
Example 3.9. Let a, b ∈ (A, ϕ) be free. How to compute ϕ [abab]?
Write a◦ = a − ϕ [a]. Note
ϕ (a◦ )2 = ϕ a2 − 2aϕ [a] + ϕ [a]2 = ϕ a2 − ϕ [a]2 .
Then
ϕ [abab] = ϕ [(a◦ + ϕ [a])(b◦ + ϕ [b])(a◦ + ϕ [a])(b◦ + ϕ [b])] .
Using freeness and linearity, this reduces to
ϕ [abab] = ϕ [a] ϕ [b] ϕ [a] ϕ [b] + ϕ [a]2 ϕ (b◦ )2 + ϕ [b]2 ϕ (a◦ )2
= ϕ [a] ϕ [b] ϕ [a] ϕ [b] + ϕ [a]2 (ϕ b2 − ϕ [b]2 ) + ϕ [b]2 (ϕ a2 − ϕ [a]2 )
= ϕ a2 ϕ [b]2 + ϕ [a]2 ϕ b2 − ϕ [a]2 ϕ [b]2 .
Moral: not a good way to compute.
Proof of Claim. Inside A, the ∗-algebra generated by a1 , . . . , an is
Alg∗ (a1 , a2 , . . . , an )

= C ⊕ Span 
∞
[
[
b1 b2 . . . bk : bi ∈ Alg∗

au(i) , ϕ [bi ] = 0  .
k=1 u(1)6=u(2)6=...6=u(k)
ϕ is zero on the second component, and is determined by ϕ [1] = 1 on the first.
Exercise 3.10. Exercise 5.25 in [NS06]. In this exercise we prove that free independence behaves well
under successive
decompositions
and thus is associative. Let {Ai : i ∈ I} be unital ∗-subalgebras of
n
o
(j)
(A, ϕ), and Bi : j ∈ J(i) be unital ∗-subalgebras of Ai . Then we have the following.
n
o
(j)
a. If {Ai : i ∈ I} are freely independent in (A, ϕ) and for each i ∈ I, Bi : j ∈ J(i) are freely
n
o
(j)
independent in (Ai , ϕ|Ai ), then all Bi : i ∈ I, j ∈ J(i) are freely independent in (A, ϕ).
21
o
(j)
Bi : i ∈ I, j ∈ J(i) are freely independent in (A, ϕ) and if, for each i ∈ I, Ai is the
n
o
(j)
∗-algebra generated by Bi : j ∈ J(i) , then {Ai : i ∈ I} are freely independent in (A, ϕ).
b. If all
n
Exercise 3.11. Let (A, τ ) be a tracial n.c. probability space, a1 , a2 , . . . , an free, and u a unitary. Then
{u∗ ai u, 1 ≤ i ≤ n} are free. Is the conclusion still true if τ is not assumed to be a trace?
Exercise 3.12. Exercise 5.24 in [NS06]. Let (A, ϕ) be a n.c. probability space. Consider a unital ∗subalgebra B ⊂ A and a Haar unitary u ∈ A freely independent from B. Show that then also B and u∗ Bu
are free.
3.2
Free products.
Groups.
Γ1 , Γ2 , . . . , Γn groups with units ei ∈ Γi .
S
Any word x1 x2 . . . xk with all xi ∈ ni=1 Γi can be reduced by identifying
x1 x2 . . . xi−1 ej xi . . . xk = x1 x2 . . . xi−1 xi . . . xk
and
x1 x2 . . . xi−1 yzxi . . . xk = x1 x2 . . . xi−1 (yz)xi . . . xk
if y, z ∈ Γi for the same i. In this way any word can be reduced to a unique (!) reduced word: unit e or
x1 x2 . . . xk , xi ∈ Γu(i) \ eu(i) , u(1) 6= u(2) . . . 6= u(k).
Define the free product of groups
(
Γ = Γ1 ∗ Γ2 ∗ . . . ∗ Γn = ∗ni=1 Γi =
reduced words in elements of
n
[
i=1
with unit e, inverse
−1 −1
(x1 x2 . . . xk )−1 = x−1
k . . . x2 x 1
and product
(x1 x2 . . . xk )(y1 y2 . . . yl ) = reduced version of (x1 . . . xk y1 . . . yl ).
22
)
Γi
Reduced free product of n.c. probability spaces.
Let (A1 , ϕ1 ), . . . , (An , ϕn ) be n.c. probability spaces. Denote
A◦i = {a ∈ Ai : ϕi [a] = 0} ,
W∅ = C = C1, and for u(1) 6= u(2) . . . 6= u(k), define the vector space
W~u = A◦u(1) ⊗ A◦u(2) ⊗ . . . ⊗ A◦u(k) .
Let
A=
∞
M
k=0
M
W~u = C1 ⊕
|~
u|=k
u(1)6=u(2)...6=u(k)
∞
M
M
W~u
k=1 u(1)6=u(2)...6=u(k)
By abuse of notation we will write a1 ⊗ a2 ⊗ . . . ⊗ ak as a1 a2 . . . ak . Define
(a1 a2 . . . ak )∗ = a∗k . . . a∗2 a∗1 .
On A, we define multiplication as follows. First,
z(a1 a2 . . . ak ) = (za1 )a2 . . . ak .
Next, let ai ∈ A◦u(i) , bi ∈ A◦v(i) , where ~u and ~v are alternating. If u(1) 6= v(1),
(ak . . . a2 a1 )(b1 b2 . . . bl ) = ak . . . a2 a1 b1 b2 . . . bl .
In general,
(ak . . . a2 a1 )(b1 b2 . . . bl ) = ak . . . a2 (a1 b1 )◦ b2 . . . bl + ϕu(1) [a1 b1 ] ak . . . a3 (a2 b2 )◦ b3 . . . bl + . . .
+ ϕu(1) [a1 b1 ] ϕu(2) [a2 b2 ] . . . ϕu(j−1) [aj−1 bj−1 ] ak . . . aj bj . . . bl ,
where j ≤ min(k, l) is the first index (if it exists) such that u(j) 6= v(j).
On A, define the free product state
ϕ = ∗ni=1 ϕi
by: ϕ [1] = 1 and on an alternating word in centered elements, ϕ [a1 a2 . . . ak ] = 0. We will prove later ϕ
is positive; other properties are clear.
Remark 3.13. This is a reduced free product, since we identify the units of all Ai with the unit 1 ∈ A.
Thus this is a product with amalgamation over C. There is also a full free product construction.
Remark 3.14. If we identify
Ai ↔ C ⊕ A◦i ⊂ A,
a ↔ ϕ [a] ⊕ (a − ϕ [a])
then ϕi = ϕ|Ai and A1 , . . . , An are freely independent with respect to ϕ.
23
Remark 3.15.
∗ni=1 C[Γi ] = C [∗ni=1 Γi ] .
Proposition 3.16. If each ϕi is tracial, so is their free product ϕ.
Corollary 3.17. Free product of commutative algebras is tracial. In particular, free product of onedimensional distributions is tracial.
Proof of the Proposition. Let ai ∈ A◦u(i) , bi ∈ A◦v(i) , u(1) 6= u(2) 6= . . . 6= u(k), v(1) 6= v(2) 6= . . . 6=
v(l). Suppose
u(1) = v(1), u(2) = v(2), . . . u(j) = v(j), u(j + 1) 6= v(j + 1).
Then
ϕ [ak . . . a1 b1 . . . bl ] = ϕ [ak . . . a2 (a1 b1 )◦ b2 . . . bl ] + ϕu(1) [a1 b1 ] ϕ [ak . . . a3 (a2 b2 )◦ b3 . . . bl ] + . . .
+ ϕu(1) [a1 b1 ] ϕu(2) [a2 b2 ] . . . ϕu(j−1) [aj−1 bj−1 ] ϕ [ak . . . aj bj . . . bl ]
This is zero unless j = k = l, in which case, since each ϕi is tracial,
ϕ [ak . . . a1 b1 . . . bk ] = ϕu(1) [a1 b1 ] ϕu(2) [a2 b2 ] . . . ϕu(j) [aj bj ]
= ϕu(1) [b1 a1 ] ϕu(2) [b2 a2 ] . . . ϕu(j) [bj aj ] = ϕ [b1 . . . bk ak . . . a1 ] .
This implies that ϕ has the trace property for general ai , bi (why?).
Remark 3.18 (Reduced free product of Hilbert spaces with distinguished vectors). Given Hilbert spaces
with distinguished vectors (H1 , ξ1 )ni=1 , define their reduced free product
(H, ξ) = ∗ni=1 (Hi , ξi )
as follows. Denote Hi◦ = Hi Cξi . Then
Halg = Cξ ⊕
∞
M
k=1
M
◦
◦
◦
H~u = Hu(1)
⊗ Hu(2)
⊗ . . . ⊗ Hu(k)
.
|~
u|=k
u(1)6=u(2)...6=u(k)
H is the completion of Halg with respect to the inner product for which H~u ⊥ H~v for ~u 6= ~v , and on each
H~u we use the usual tensor inner product
hf1 ⊗ f2 ⊗ . . . ⊗ fk , g1 ⊗ g2 ⊗ . . . ⊗ gk i =
k
Y
hfi , gi iu(i)
i=1
◦
for fi , gi ∈ Hu(i)
.
If (Hi , ξi ) = L2 (Ai , ϕ1 ), then, at least in the faithful case,
(H, ξ) = L2 (∗ni=1 (Ai , ϕi )) .
More generally, if each Ai is represented on Hi with ϕi = h·ξi , ξi i, then can represent ∗ni=1 (Ai , ϕi ) on H
so that ϕ = h·ξ, ξi. This implies in particular that ϕ is positive.
By using this representation and taking appropriate closures, can define reduced free products of C ∗ and
W ∗ -probability spaces.
24
Positivity.
How to prove positivity of a joint distribution? Represent it as a joint distribution of symmetric operators
on a Hilbert space.
How to prove positivity of the free product state? One way is to use the preceding remark. We will use a
different, more algebraic proof.
Remark 3.19. A matrix A ∈ Mn (C) is positive if and only if one of the following equivalent conditions
holds:
a. A = A∗ and σ(A) ⊂ [0, ∞) (non-negative eigenvalues).
b. A = B ∗ B for some B (may take B = B ∗ ).
c. for all z = (z1 , z2 , . . . , zn )t ∈ Cn ,
hAz, zi =
n
X
z i Aij zj ≥ 0.
i,j=1
Proposition 3.20. A linear functional ϕ on A is positive if and only if for all n and all
a1 , a2 , . . . , an ∈ A, the (numerical) matrix [ϕ(a∗i aj )]ni,j=1 is positive.
Proof. ⇐ clear by taking n = 1. For the converse, let a1 , . . . , an ∈ A and z1 , . . . zn ∈ C. Then since ϕ is
positive,
#
" n
!∗ n
n
X
X
X
0≤ϕ
zj aj =
z i ϕ [a∗i aj ] zj .
zi ai
i=1
i,j=1
j=1
By the preceding remark, the matrix [ϕ(a∗i aj )]ni,j=1 is positive.
Definition 3.21. Let A be a ∗-algebra. Then Mn (A) = Mn (C) ⊗ A is also a ∗-algebra, and so has a
notion of positivity. If T : A → B, define
Tn : Mn (A) → Mn (B) Tn ([aij ]ni,j=1 ) = [T (aij )]ni,j=1 .
We say that T is completely positive if each Tn is positive.
Remark 3.22. Usually defined only for C ∗ -algebras. Even in that case, positive does not imply completely positive.
Compare with Section 3.5 of [Spe98].
Proposition 3.23. If (A, ϕ) is an n.c. probability space, so that ϕ is positive, then each ϕn : Mn (A) →
Mn (C) is positive, so that ϕ is completely positive.
25
P
∗
Proof. Let A ∈ Mn (A) be positive. By definition, that means A = N
i=1 Bi Bi . It suffices to show that
∗
∗
each ϕn [Bi Bi ] is positive. So without loss of generality, assume that A = B B. That is
Aij =
n
X
b∗ki bkj ,
buv ∈ A.
k=1
Then
[ϕn (A)]ni,j=1
=
n
X
[ϕ(b∗ki bkj )]ni,j=1
k=1
and for each k, this matrix is positive.
Definition 3.24. For A, B, C ∈ Mn (C), C is the Schur product of A and B if
Cij = Aij Bij .
Proposition 3.25. If A, B are positive, so is their Schur product.
Proof. Let A = D∗ D, Aij =
n
X
i,j=1
z i Cij zj =
Pn
k=1
n
X
Dki Dkj . Then
z i Dki Dkj Bij zj =
i,j,k=1
n
X
n
X
k=1
i,j=1
!
(Dki zi )Bij (Dkj zj )
≥ 0,
since B is positive. Therefore C is positive.
Theorem 3.26. Let (Ai , ϕi )ni=1 be n.c. probability spaces and (A, ϕ) = ∗ni=1 (Ai , ϕi ). Then ϕ is positive.
If each ϕi is faithful, so is ϕ.
Proof. Recall the representation
A=
∞
M
k=1
M
W~u
|~
u|=k
u(1)6=u(2)...6=u(k)
L
and ϕ [ξη] = 0 unless ξ, η ∈ W~u for the same ~u. Thus for ξ = ~u ξ~u ,
X
ϕ [ξ ∗ ξ] =
ϕ [ξ~u∗ ξ~u ] .
~
u
So it suffices to show that ϕ [ξ ∗ ξ] ≥ 0 for ξ ∈ W~u . In its turn,
ξ=
r
X
(j) (j)
(j)
a1 a2 . . . ak ,
j=1
26
(j)
ai ∈ A◦u(i) .
Then
"
∗
ϕ [ξ ξ] = ϕ
r X
(i)
ak
∗
...
(i)
a1
i=1
=
r
X
ϕu(1)
r
∗ X
#
(j)
a1
(j)
. . . ak
j=1
h
(i)
(j)
(a1 )∗ a1
i
h
i
(i) ∗ (j)
. . . ϕu(k) (ak ) ak .
i,j=1
h
ir
(i) ∗ (j)
Each matrix ϕu(s) (as ) as
is positive, therefore so is their Schur product.
i,j=1
The proof of faithfulness is similar, see Proposition 6.14 in [NS06].
Remark 3.27 (Free convolutions). Let µ, ν be two distributions (probability measures on R). Free products allow us to construct two symmetric, freely independent variables a, b in some tracial C ∗ -probability
space (A, ϕ) such that µ = µa , ν = µb (why?). Know the distribution of a + b is a probability measure on
R, depends only on µ, ν and not on a, b. Thus define the additive free convolution µ ν by µ ν = µa+b .
How to compute?
Similarly, define the multiplicative free convolution µ ν = ϕab . Note that ab is not symmetric, so in
general µ ν is not positive and does not correspond to a measure on R. Also note that since ϕ is tracial,
is commutative.
Suppose that µ, ν are supported in [0, ∞). Then we may choose a, b to be positive. In that case, a1/2 ba1/2
is also positive, and since ϕ is trace,
ϕab = ϕa1/2 ba1/2 .
In this case µ ν may be identified with a probability measure on [0, ∞).
Now suppose instead that µ, ν are supported on the unit circle T = {z ∈ C : |z| = 1}. Then we may
choose a, b to be unitary. Then ab is also unitary, so in this case µ ν may be identified with a probability
measure on T.
Proposition 3.28 (Compare with Exercise 3.11). Let (A, τ ) be a n.c. probability space, a1 , a2 , . . . , an
free, and u a unitary free from them. Then {u∗ ai u, 1 ≤ i ≤ n} are free.
Proof. The key point is that even if τ is not tracial, if a and u are freely independent then
τ [u∗ au] = τ [u∗ τ [a] u] + τ [u∗ a◦ u]
= τ [a] τ [u∗ u] + τ [(u∗ )◦ a◦ u◦ ] + τ [u∗ ] τ [a◦ u◦ ] + τ [(u∗ )◦ a◦ ] τ [u] + τ [u∗ ] τ [a◦ ] τ [u]
= τ [a] .
In particular, τ [a] = 0 if and only if τ [u∗ au] = 0. The rest of the argument proceeds as in the solution to
the exercise.
By a similar argument, we can also get (how?) the following useful weakening of the hypothesis in the
definition of free independence:
27
Proposition 3.29. Let (A, ϕ) be an n.c. probability space. Subalgebras A1 , A2 , . . . , Ak ⊂ (A, ϕ) are
freely independent with respect to ϕ if and only if whenever
u(1) 6= u(2) 6= u(3) 6= . . . 6= u(n),
ai ∈ Au(i) for 1 ≤ i ≤ n, and ϕ [ai ] = 0 for 2 ≤ i ≤ n − 1, then
ϕ [a1 a2 . . . an ] = 0.
28
Chapter 4
Set partition lattices.
See Lectures 9, 10 of [NS06]. For general combinatorial background, see [Sta97]. The general theory
of incidence algebras, Möbius inversion etc. goes back to the series On the foundations of combinatorial
theory I–X by Gian-Carlo Rota and various co-authors.
4.1
All, non-crossing, interval partitions. Enumeration.
Let S be a finite ordered set. π = {B1 , B2 , . . . , Bk } is a (set) partition of S, π ∈ P(S) if
B1 ∪ B2 ∪ . . . ∪ Bk = S,
Bi ∩ Bj = ∅ for i 6= j,
Bi 6= ∅.
{Bi } are the blocks or classes of π. We also write
π
i∼j
if i and j are in the same block of π. Denote |π| the number of blocks of π.
Usually take S = [n] = {1, 2, . . . , n}. Write P([n]) = P(n).
Partitions are partially ordered by reverse refinement: if
π = {B1 , . . . , Bk } ,
σ = {C1 , . . . , Cr }
then
π≤σ
⇔
∀i ∃j : Bi ⊂ Cj .
For example, if π = {(1, 3, 5)(2)(4, 6)} and σ = {(1, 3, 5)(2, 4, 6)}, then π ≤ σ. The largest partition is
1̂ = 1̂n = {(1, 2, . . . , n)}
and the smallest one
0̂ = 0̂n = {(1)(2) . . . (n)} .
29
30
Example 4.1. P(3) has 5 elements: one largest, one smallest, and three which are not comparable to each
other.
Partitions form a lattice: for any π, σ ∈ P(S), there exists a largest partition τ such that τ ≤ π, τ ≤ σ,
called the meet and denoted π ∧ σ; and the smallest partition τ such that π ≤ τ , σ ≤ τ , called the join and
denoted π ∨ σ. Clearly
π∧σ
π
σ
i ∼ j ⇔ i ∼ j and i ∼ j;
π ∨ σ can be thought of as the equivalence relation on S generated by π and σ. For example,
if π = {(1, 3, 5)(2, 4, 6)} and σ = {(1, 2, 3)(4, 5, 6)}, then π ∧ σ = {(1, 3)(2)(4, 6)(5)}.
If π = {(1, 2)(3, 4)(4, 6)(7)} and σ = {(1)(2, 3)(4, 5)(6, 7)}, then π ∨ σ = 1̂7 .
Definition 4.2. A partition π ∈ P(S) of an ordered set S is a non-crossing partition if
π
π
i ∼ j, k ∼ l, i < k < j < l
⇒
π
π
π
i ∼ k ∼ j ∼ l.
See Figure 4. Denote non-crossing partitions by NC(S), NC(n).
Note that 0̂, 1̂ ∈ NC(S). With the same operations ∧ and ∨ as above, NC(S) is a lattice. If π, σ ∈ NC(S)
they have the same π ∧σ in P(S) and in NC(S). However, for π = {(1, 3)(2)(4)} and σ = {(1)(2, 4)(3)},
π ∨ σ = {(1, 3)(2, 4)} in P(4) but π ∨ σ = 1̂4 in NC(4).
Definition 4.3. π ∈ NC(S) ⊂ P(S) is an interval partition if its classes are intervals. See Figure 4. They
also form a lattice. Denote them by Int(S), Int(n).
Exercise 4.4. Denote
S(n) = {subsets of [n]} .
Then S(n) is a lattice (by reverse inclusion), and
Int(n) ' S(n − 1).
For this reason, interval partitions are also called Boolean partitions.
Will study NC(n), but also P(n), Int(n), in a lot more detail.
Enumeration I.
Definition 4.5.
B(n) = |P(n)| = Bell number.
S(n, k) = |{π ∈ P(n) : |π| = k}| = Stirling number of the second kind.
No formula.
31
Lemma 4.6. The Bell numbers satisfy a recursion relation
B(n + 1) =
n X
n
i=0
i
B(n − i),
n ≥ 1,
where by convention B(0) = 1. Consequently, their exponential generating function is
∞
X
1
z
F (z) =
B(n)z n = ee −1 .
n!
n=0
Remark 4.7. ee
z −1
grows faster than exponentially, so B(n) grows faster than any an .
Proof of the Lemma. If π ∈ P(n + 1) and the block containing 1 contains i + 1 elements, there are ni
ways to choose the remaining i elements of the block, and the partition induced by π on the complement
of this block is arbitrary. This gives the recursion. Differentiating the generating function term-by-term,
!
∞
∞
n
X
X
X
1
1
n!
B(n + 1)z n =
B(n − i) z n
F 0 (z) =
n!
n!
i!(n
−
i)!
n=0
n=0
i=0
=
∞
∞
X
1 iX 1
z
B(j)z j = ez F (z).
i! j=0 j!
i=0
Solving this differential equation with the initial condition F (0) = 1, we get the result.
Exercise 4.8. Derive a recursion for the Stirling numbers, and use it to show that their generating function
is
∞ X
n
X
1
z
S(n, k)z n wk = ew(e −1) .
F (z, w) = 1 +
n!
n=1 k=1
Definition 4.9.
cn = |NC(n)| = Catalan number.
Lemma 4.10. The Catalan numbers satisfy recursion relations
cn =
n
X
i=1
X
cu(1) cu(2) . . . cu(i) ,
u(1),u(2)...,u(i)≥0
u(1)+u(2)+...+u(i)=n−i
where by convention c0 = 1, and
cn =
n−1
X
ck cn−k−1 .
k=0
32
n ≥ 1,
Their ordinary generating function is
F (z) =
∞
X
n
cn z =
1−
n=0
√
1 − 4z
.
2z
Consequently,
1
2n
1
2n + 1
cn =
=
.
n+1 n
2n + 1
n
Proof. If the first block of π ∈ NC(n) is
(1, 2 + u(1), 3 + u(1) + u(2), . . . , i + u(1) + . . . + u(i)),
we can complete it to a non-crossing partition by choosing arbitrary non-crossing partitions on each of
the i intervals in the complement of this block. This proves the first recursion. Next,
F (z) = 1 +
∞
X
n
cn z = 1 +
=1+
i=1
X
cu(1) cu(2) . . . cu(i) z u(1) z u(2) . . . z u(i) z i
n=1 i=1
n=1
∞
X
∞ X
n
X
z
i
∞
X
u(1),u(2)...,u(i)≥0
u(1)+u(2)+...+u(i)=n−i
∞
X
u(1)
cu(1) z
...
cu(i) z u(i) = 1
u(1)=0
+
∞
X
zF (z) =
i=1
u(i)=0
1
.
1 − zF (z)
Thus
zF (z)2 − F (z) + 1 = 0,
and
√
1 − 4z
.
2z
It is easy to see that the second recursion leads to the same quadratic equation, and so holds as well.
Finally, using the binomial series,
F (z) =
1−
∞ ∞
∞
X
1 X 1/2
(2k − 2)! k−1 X (2n)!
k
F (z) = −
(−4z) =
z
=
zn,
2z k=1 k
(k
−
1)!k!
n!(n
+
1)!
n=0
k=1
which leads to the formula for cn .
4.2
Incidence algebra. Multiplicative functions.
Remark 4.11 (Incidence algebra). For now, L is a general (locally finite) lattice. Let
I = functions on L
33
and
I2 = functions on {(σ, π) : σ, π ∈ L, σ ≤ π} .
Then I2 is the incidence algebra under convolution
X
(f ∗ g)(σ, π) =
f (σ, τ )g(τ, π).
σ≤τ ≤π
Moreover, I2 acts on I by convolution: if f ∈ I and g ∈ I2 , then f ∗ g ∈ I via
X
(f ∗ g)(π) =
f (τ )g(τ, π).
τ ≤π
Thus may identify f (σ) = f (0̂, σ).
It is easy to see that I2 has the identity element
(
1,
δ(σ, π) =
0,
σ=π
σ 6= π.
Another important element is
ζ(σ, π) = 1
for all σ ≤ π. In particular, for f ∈ I,
(f ∗ ζ)(π) =
X
f (σ).
σ≤π
Möbius inversion.
Definition 4.12. The Möbius function µ ∈ I2 is defined by
ζ ∗ µ = µ ∗ ζ = δ.
In our context, µ exists by general results. We will compute it later.
Proposition 4.13 (Partial Möbius inversion). Let f, g ∈ I. Suppose for all π ∈ L,
X
f (π) =
g(σ).
σ≤π
a. Möbius inversion: for all π ∈ L,
g(π) =
X
σ≤π
34
f (σ)µ(σ, π).
b. Partial Möbius inversion: for any ω, π ∈ L,
X
X
g(τ ) =
f (σ)µ(σ, π).
ω∨τ =π
ω≤σ≤π
Note two particular cases.
Proof.
f (π) =
X
X
g(σ) =
0̂≤σ≤π
g(σ)ζ(σ, π) = (g ∗ ζ)(π).
0̂≤σ≤π
Therefore
g(π) = (g ∗ ζ ∗ µ)(π) = (f ∗ µ)(π) =
X
f (σ)µ(σ, π).
σ≤π
More generally,
X
X X
f (σ)µ(σ, π) =
g(τ )µ(σ, π) =
ω≤σ≤π τ ≤σ
ω≤σ≤π
=
X
X
τ ∈L
g(τ )δ(ω ∨ τ, π) =
τ ∈L
X
X
g(τ )
(ω∨τ )≤σ≤π
g(τ ).
ω∨τ =π
Multiplicative functions.
f (σ, π) or even f (π) too complicated. Want multiplicative functions.
Remark 4.14. In a lattice L, denote
[σ, π] = {τ ∈ L : σ ≤ τ ≤ π} .
In P(S),
[0̂, π] '
Y
P(Vi ) '
Vi ∈π
Y
P(|Vi |)
Vi ∈π
and [π, 1̂] ' P(|π|). More generally, if σ ≤ π,
[σ, π] '
Y
P(|σ|Vi |).
Vi ∈π
In NC(S),
[0̂, π] '
Y
NC(Vi ) '
Vi ∈π
Y
NC(|Vi |).
Vi ∈π
But what is [π, 1̂]? For example,
[{(1, 6)(2, 5)(3, 4)} , 1̂6 ] 6' NC(3).
35
ζ(ω ∨ τ, σ)µ(σ, π)
Definition 4.15. Let π ∈ NC(n). Its (right) Kreweras complement K[π] ∈ NC(n) is the largest partition
in NC({1̄, 2̄, . . . , n̄}) such that
π ∪ K[π] ∈ NC({1, 1̄, 2, 2̄, . . . , n, n̄}).
Similarly, have the left Kreweras complement. See Figure 4.
Lemma 4.16. Let K be the Kreweras complement on NC(n).
a. K −1 is the left Kreweras complement.
b. K is an anti-isomorphism of lattices.
c. |K[π]| + |π| = n + 1.
Proof.
a. Picture.
b. σ ≤ π if and only if K[σ] ≥ K[π].
c. Any element of NC(n) can be included in a maximal chain, all of which look like
0̂n σn−1 . . . σ3 σ2 1̂n ,
where |σk | = k. K maps it to a maximal chain
1̂n K[σn−1 ] . . . K[σ3 ] K[σ2 ] 0̂n
with |K[σk ]| = n − k + 1.
Corollary 4.17. In NC(n),
[π, 1̂] ' [0̂, K[π]] '
Y
NC(Vi )
Vi ∈K[π]
and more generally
[σ, π] '
Y
[σ|Vi , 1̂Vi ] '
Vi ∈π
Y
Y
NC(Uj ).
Vi ∈π Uj ∈K[σ|Vi ]
Exercise 4.18. Identify [σ, π] in Int(n).
Definition 4.19. Given a family of lattices {Li : i ∈ N}, a function f on
{(σ, π) : σ ≤ π, σ, π ∈ Li , i ∈ N}
36
is multiplicative if there exist fixed {fi : i ∈ N} such that whenever
[σ, π] '
n
Y
Lki i ,
i=1
then
f (σ, π) =
n
Y
fiki .
i=1
Conversely, given any {fi : i ∈ N}, can use this formula to define a multiplicative function f .
Example 4.20. ζ(σ, π) = 1. So ζ is multiplicative, with ζn = 1 for all n ∈ N.
δ(σ, π) = δσ=π . Note that [π, π] ' L1 . So δ is multiplicative, with δ1 = 1, δn = 0 for n ≥ 2.
Remark 4.21. By general theory, the Möbius function is also always multiplicative. In fact, in general
µL×L0 = µL µL0 .
37
Chapter 5
Free cumulant machinery.
See Lectures 8, 11, 12, 16 of [NS06].
5.1
Free cumulants.
Definition 5.1. Let (A, ϕ) be an n.c. probability space. For each n, define the moment functional
M :A
{z. . . × A} → C
| ×A×
n
by
M [a1 , a2 , . . . , an ] = ϕ [a1 a2 . . . an ] .
For each n, M is a multilinear functional. Combine these into a multiplicative function
Y
Mπ [a1 , a2 , . . . , an ] =
M [ai : i ∈ V ]
V ∈π
for π ∈ NC(n).
Example 5.2.
M{(1,4)(2,3)} [a1 , a2 , a3 , a4 ] = M [a1 , a4 ]M [a2 , a3 ] = ϕ [a1 a4 ] ϕ [a2 a3 ] .
Definition 5.3. In the context of the preceding definition, define the partitioned free cumulants
Rπ [a1 , a2 , . . . , an ]
of {a1 , . . . , an } via
M = R ∗ ζ,
R = M ∗ µ.
38
That is,
Mπ [a1 , a2 , . . . , an ] =
X
Rσ [a1 , a2 , . . . , an ]
σ≤π
and
Rπ [a1 , a2 , . . . , an ] =
X
Mσ [a1 , a2 , . . . , an ]µ(σ, π).
σ≤π
In particular, for π = 1̂, we have the moment-cumulant formulas
X
R[a1 , a2 , . . . , an ] =
Mσ [a1 , a2 , . . . , an ]µ(σ, 1̂),
σ∈NC(n)
and
M [a1 , a2 , . . . , an ] =
X
Rσ [a1 , a2 , . . . , an ],
σ∈NC(n)
where
Rσ [a1 , a2 , . . . , an ] =
Y
R[ai : i ∈ V ].
V ∈σ
Example 5.4.
R[a] = M [a] = ϕ [a]
is the mean of a.
M [a, b] = R[a, b] + R[a]R[b],
so
R[a, b] = ϕ [ab] − ϕ [a] ϕ [b]
is the covariance of a and b.
M [a, b, c] = R[a, b, c] + R[a]R[b, c] + R[a, c]R[b] + R[a, b]R[c] + R[a]R[b]R[c].
Notation 5.5.
R[a1 , . . . , an ] = Rϕ [a1 , . . . , an ]
is a multilinear functional on (A, ϕ). If µ ∈ D(d) is the joint distribution of (a1 , . . . , ad ), then x1 , . . . , xd ∈
Chx1 , . . . , xd i also have joint distribution µ with respect to µ. So can consider
Rµ [xu(1) , xu(2) , . . . , xu(n) ].
These satisfy
ϕ au(1) au(2) . . . au(n) = µ xu(1) xu(2) . . . xu(n) =
X
Rπµ [xu(1) , xu(2) , . . . , xu(n) ].
π∈NC(n)
For µ ∈ D(d), can define Rµ ∈ Chx1 , . . . , xd i∗ (not necessarily positive) via
Rµ xu(1) xu(2) . . . xu(n) = Rµ [xu(1) , xu(2) , . . . , xu(n) ].
39
In one variable, we have the moments
mn [µ] = µ [xn ] = ϕ [an ]
and the moment-cumulant formula
mn [µ] =
X
rπ [µ] =
X
Y
r|V | [µ].
π∈NC(n) V ∈π
π∈NC(n)
Here
rn [µ] = Rµ [xn ] = Rµ [x, x, . . . , x].
Finally, we have the free cumulant generating function (combinatorial R-transform)
µ
R (z) =
∞
X
rn [µ]z n
n=1
and more generally for µ ∈ D(d)
µ
R (z1 , z2 , . . . , zd ) =
∞
X
d
X
Rµ [xu(1) xu(2) . . . xu(n) ]zu(1) zu(2) . . . zu(n)
n=1 u(1),...,u(n)=1
for non-commuting indeterminates z1 , z2 , . . . , zd .
Proposition 5.6. The moment and free cumulant generating functions are related by
M (z) = 1 + R(zM (z))
(recall that M has a constant term while R does not), and more generally
M (z1 , . . . , zd ) = 1 + R z1 M (z), . . . , zd M (z)
= 1 + R M (z)z1 , . . . , M (z)zd ,
where
M (z1 , . . . , zd ) = 1 +
∞ X
X
n=1 |~
u|=n
40
M [x~u ]z~u .
Proof. Very similar to the enumeration argument.
X
M [xu(1) , xu(2) , . . . , xu(n) ]z~u =
Rπ [x~u ]z~u
π∈NC(n)
=
n
X
k=1
X
S⊂[n]
S={1=v(1),v(2),...,v(k)}
v(k+1)=n+1
R[xu(v(i)) : i ∈ S]
k
Y


zu(v(j))
j=1
=
n
X
k=1
X
Rπj [xu(v(i)) : i ∈ Vj ]
πj ∈NC(Vj =[v(j)+1,v(j+1)−1])
X
R[xu(v(i)) : i ∈ S]
k
Y
Y
i∈Vj


zu(v(j)) M [xu(v(j)+1) , . . . , xu(v(j+1)−1) ]
j=1
S⊂[n]
S={1=v(1),v(2),...,v(k)}
v(k+1)=n+1
M (z) = 1 +
R[xw~ ]zw(1) M (z)zw(2) M (z) . . . zw(k) M (z)
k=1 |w|=k
~
= 1 + R z1 M (z), z2 M (z), . . . , zd M (z) .
The second identity is obtained by starting with the block containing n.
Remark 5.7. Classical (tensor) cumulants are similarly defined via
X
mn [µ] =
kπ [µ]
π∈P(n)
and Boolean cumulants by
M [a1 , . . . , an ] =
X
Bπ [a1 , . . . , an ].
π∈Int(n)
For the exponential moment generating function
µ
Mexp
(z)
Y
i∈Vj
Therefore
∞ X
X
zu(v(i)) 
∞
X
1
=
mn [µ]z n ,
n!
n=0
can show as before that the cumulant generating function is
∞
X
1
`(z) =
kn [µ]z n = log Mexp (z).
n!
n=1
41
zu(v(i))  .
Exercise 5.8. For the Boolean cumulant generating function
B(z) =
∞ X
X
B[xu(1) , . . . , xu(n) ]z~u ,
n=1 |~
u|=n
show that
1 − B(z) = M (z)−1
(inverse with respect to multiplication).
Enumeration II.
Proposition 5.9. On NC(n), the Möbius function is given by
µn = (−1)n−1 cn−1 ,
where cn is the Catalan number.
Proof. Recall µ ∗ ζ = δ. Thus
X
Y
µ|V |
π∈NC(n) V ∈π
(
1,
=
µ(0̂, π)ζ(π, 1̂) = δ(0̂n , 1̂n ) =
0,
π∈NC(n)
X
n = 1,
n ≥ 2.
It follows that {µn : n ∈ N} are
P∞the freen cumulants of a “distribution” (not positive) with m1 = 1, mn = 0
for n ≥ 2. Thus for F (z) = n=1 µn z , we have
1 + z = 1 + F (z(1 + z)).
If w = z 2 + z, then z =
√
−1+ 1+4w
2
F (w) =
and
∞
X
n=1
n
µn w =
−1 +
√
∞
1 + 4w X
=
(−1)n cn−1 wn
2
n=1
as before.
Exercise 5.10. Use Remark 5.7 and Exercise 5.8 to compute the Möbius functions on P(n) and Int(n).
Definition 5.11. A multi-chain in a lattice is a linearly ordered subset
0̂ ≤ σ1 ≤ σ2 ≤ . . . ≤ σp ≤ 1̂,
possibly with repetitions.
42
Proposition 5.12. The number M (n, p) of multi-chains of length p in NC(n) is the Fuss-Catalan number
1 (p + 1)n
.
n n−1
Proof. Note that
X
|N C(n)| =
1 = (ζ ∗ ζ)(0̂n , 1̂n ).
0̂n ≤σ1 ≤1̂n
Similarly,
X
M (n, p) =
0̂≤σ1 ≤σ2 ≤...≤σp ≤1̂
1 = (ζ ∗ ζ ∗ . . . ∗ ζ )(0̂n , 1̂n ) = (ζ ∗(p+1) )n .
{z
}
|
P∞
p+1
If Fp (z) = 1 + n=1 (ζ ∗(p+1) )n z n , then F0 (z) =
that Fp satisfies z(Fp (z))p+1 = Fp (z) − 1, since
1
1−z
and Fp (z) = Fp−1 (zFp (z)). It follows by induction
Fp (z) − 1 = Fp−1 (zFp (z)) − 1 = (zFp (z))(Fp−1 (zFp (z)))p = z(Fp (z))p+1 .
It is known that the coefficients in the expansion of the solution of this equation are the Fuss-Catalan
numbers.
Proposition 5.13. The number of non-crossing partitions in NC(n) with ki blocks of size i is the multinomial coefficient
n!
P
.
N (n; ki , i ∈ N) = Q
i∈N ki )!
i∈N ki !(n + 1 −
For the proof, see Lecture 9 from [NS06], or an argument in [Spe98] using the Lagrange inversion formula.
Distributions.
Recall: Stieltjes inversion formula
dµ(x) = −
1
lim =Gµ (x + iy) dx,
π y→0+
where
∞
X
1
1
G(z) = M (1/z) =
mn n+1
z
z
n=0
is the Cauchy transform. Define the analytic R-transform
∞
X
1
R(z) = R(z) =
rn z n−1 = r1 + r2 z + r3 z 2 + . . . .
z
n=1
43
Exercise 5.14. Use Proposition 5.6 to show that (as formal power series)
1
1
G
+ R(z) = z, i.e. R(z) = G−1 (z) − ,
z
z
inverse with respect to composition.
Example 5.15. Let r1 [µt ] = t, rn [µt ] = 0 for n 6= 1. Then mn [µt ] = tn and µt = δt .
Lemma 5.16. Denote NC2 (n) the non-crossing pair partitions. Then
|NC2 (2n)| = cn = |N C(n)| .
Proof. By considering the n choices of elements in [2n] to which 1 can be paired by a non-crossing
partitions, it is easy to see that |NC2 (2n)| satisfy the second recursion for the Catalan numbers.
Remark 5.17. A direct bijection NC(n) and NC2 (2n) is provided by the map
π 7→ K[π ∪ K[π]].
See Exercise 9.42 in [NS06].
Example 5.18. Let r2 [µt ] = t, rn [µt ] = 0 for n 6= 2. Then
m2n [µt ] = tn |NC2 (2n)| = tn cn
and m2n+1 [µt ] = 0. Therefore
√
1 − 4tz 2
,
2tz 2
√
z − z 2 − 4t
Gµt (z) =
,
2t
µt
M (z) =
1−
and so
1 √
4t − x2 1[−2√t,2√t] .
2πt
These are the semicircular distributions, which we will denote by S(0, t). The law for t = 1 is the Wigner
semicircle law. Figure omitted. We will see that it is a free analog of the Gaussian (normal) distribution.
A random variable with a semicircular distribution is a semicircular element.
dµt (x) =
Example 5.19. Let rn [µt ] = t for all n. Thus its R-transform is
R(z) =
∞
X
tz n−1 =
n=1
Then
t
.
1−z
1
1 + (t − 1)z
+ R(z) =
,
z
z − z2
44
and so
Gµt (z) =
z+1−t−
p
(z − (t + 1))2 − 4t
.
2z
From this one can deduce that
p
4t − (x − (t + 1))2
+ max(1 − t, 0)δ0 ,
2x
√
√
the distribution being supported on the interval [( t − 1)2 , ( t + 1)2 ] and has an atom at 0 for 0 < t < 1.
These are the free Poisson distributions, also called the Marchenko-Pastur distributions, which we will
denote by πt . Of particular interest is the free Poisson distribution with parameter t = 1,
r
1 4−x
dµ(x) =
2π
x
1
dµ(x) =
2π
on [0, 4]. Note that
mn [π1 ] = |NC(n)| = cn = m2n [S(0, 1)].
Thus if s has semicircular distribution, s2 has free Poisson distribution. Thus “free χ2 = free Poisson”.
Remark 5.20. Note that in the previous example,
X
mn [µt ] =
π∈NC(n)
t
|π|
=
n
X
tk |{π ∈ NC(n) : |π| = k}| .
k=1
Since we know the generating function of these moments, from this we may deduce that
1 n
n
|{π ∈ NC(n) : |π| = k}| = N (n, k) =
n k
k−1
are the Narayana numbers.
5.2
Free independence and free cumulants.
Free cumulants of products.
Let a1 , a2 , . . . , an ∈ (A, ϕ). Group these into consecutive products: choose
1 ≤ v(1) < v(2) < . . . < v(k) = n,
and let
A1 = a1 . . . av(1) ,
A2 = av(1)+1 . . . av(2)
45
up to
Ak = av(k−1)+1 . . . an .
In other words, for Vj = [v(j − 1) + 1, v(j)] (where as usual v(0) = 0), Aj =
Q
i∈Vj
ai . Clearly
a1 . . . an = A1 . . . Ak .
So
M [A1 , A2 , . . . , Ak ] = ϕ [A1 A2 . . . Ak ] = M [a1 , a2 , . . . , an ].
What about R[A1 , A2 , . . . , Ak ] in terms of Rπ [a1 , a2 , . . . , an ]?
Let
σ = {V1 , V2 , . . . , Vk } ∈ Int(n).
Theorem 5.21.
X
R[A1 , A2 , . . . , Ak ] =
Rπ [a1 , a2 , . . . , an ].
π∈NC(n)
π∨σ=1̂n
Here π ∨ σ = 1̂n means any two A’s are eventually connected via π. More generally,
X
Rτ [A1 , A2 , . . . , Ak ] =
Rπ [a1 , a2 , . . . , an ].
π∈NC(n)
π∨σ=τ̂
Example 5.22.
R[a1 a2 , a3 ] = R{(1,2,3)} + R{(1,3)(2)} + R{(1)(2,3)} = R[a1 , a2 , a3 ] + R[a1 , a3 ]R[a2 ] + R[a1 ]R[a2 , a3 ].
Remark 5.23. Idea of proof:
X
R[A1 , A2 , . . . , Ak ] =
Mπ [A1 , . . . , Ak ]µ(π, 1̂)
π∈NC(k)
and each of these π induces a partition of [n].
Given a fixed σ = {V1 , V2 , . . . , Vk } ∈ Int(n), define the map
NC(k) → NC(n),
τ 7→ τ̂
by
τ = {U1 , U2 , . . . , Uj } 7→

[

i∈U1
Vi ,
[
Vi , . . . ,
i∈U2
[
i∈Uj
Vi


.

For example, if σ = {(1)(2, 3, 4)(5, 6)} ∈ Int(6) and τ = {(1, 2)(3)} ∈ NC(3), then
τ̂ = {(1, 2, 3, 4)(5, 6)} ∈ NC(6).
If instead τ = {(1, 3)(2)} then τ̂ = {(1, 5, 6)(2, 3, 4)}.
Then
46
a. τ̂ ∈ NC(n).
b. The map τ 7→ τ̂ is injective, with 0̂k 7→ σ, 1̂k 7→ 1̂n .
c. τ 7→ τ̂ preserves the partial order and its image is
{π ∈ NC(n) : π ≥ σ} = [σ, 1̂n ].
Thus [0̂k , 1̂k ] ' [σ, 1̂n ]. In particular, µ(σ, π) = µ(σ̂, π̂), where these are two Möbius functions on
different lattices.
Proof of the Theorem.
Rτ [A1 , A2 , . . . , Ak ] =
X
Mπ [A1 , . . . , Ak ]µ(π, τ ) =
π∈NC(k)
π≤τ
X
Mπ̂ [a1 , a2 , . . . , an ]µ(π̂, τ̂ ).
π∈NC(k)
π≤τ
Denoting ω = π̂ and using partial Möbius inversion (Proposition 4.13), the expression continues as
X
X
=
Mω [a1 , a2 , . . . , an ]µ(ω, τ̂ ) =
Rπ [a1 , a2 , . . . , an ].
ω∈NC(n)
σ≤ω≤τ̂
π∈NC(n)
π∨σ=τ̂
Free independence as vanishing mixed free cumulants.
Theorem 5.24. Let A1 , A2 , . . . , Ak ⊂ (A, ϕ). They are freely independent with respect to ϕ if and only
if “mixed free cumulants vanish”, that is, for all u(1), u(2), . . . , u(n), ai ∈ Au(i) ,
R[a1 , a2 , . . . , an ] 6= 0
⇒
u(1) = u(2) = . . . = u(n).
Remark 5.25. No centered or alternating condition.
Remark 5.26. Any block of maximal depth in a non-crossing partition is an interval block.
Proof of the easy direction. Suppose all the mixed free cumulants are zero. Let u(1) 6= u(2) 6= . . . 6=
u(n), ai ∈ Ai , ϕ [ai ] = 0. Write
X
ϕ [a1 a2 . . . an ] =
Rπ [a1 , a2 , . . . , an ].
π∈NC(n)
By the preceding remark, Rπ has a factor
R[ai , ai+1 , . . . , aj ].
If i < j, this is a mixed free cumulant, and so is zero. If i = j, then R[ai ] = ϕ [ai ], and so also is zero. It
follows that each term in the sum is zero.
47
Lemma 5.27. If n ≥ 2 and any of a1 , a2 , . . . , an is a scalar, then
R[a1 , a2 , . . . , an ] = 0.
Proof. Without loss of generality, an = 1. The proof is by induction on n. For n = 2,
R[a, 1] = ϕ [a1] − ϕ [a] ϕ [1] = 0.
Assuming the result for k < n,
ϕ [a1 a2 . . . an−1 1] = R[a1 , a2 , . . . , an−1 , 1] +
X
Rπ [a1 , a2 , . . . , an−1 , 1]
π6=1̂n
X
= R[a1 , a2 , . . . , an−1 , 1] +
Rπ [a1 , a2 , . . . , an−1 ]R[1],
π∈NC(n−1)
where we have used the induction hypothesis in the second step. On the other hand, this expression is also
equal to
X
ϕ [a1 a2 . . . an−1 ] =
Rπ [a1 , a2 , . . . , an−1 ].
π∈NC(n−1)
Since R[1] = 1, it follows that R[a1 , a2 , . . . , an−1 , 1] = 0.
Proof of the hard direction of the theorem. Suppose A1 , . . . , Ak are freely independent and ai ∈ Au(i) .
First write
R[a1 , . . . , an ] = R[a◦1 + ϕ [a1 ] , . . . , a◦n + ϕ [an ]].
Since R is multilinear, and using the lemma, we may assume that ai = a◦i , ϕ [ai ] = 0. The rest of the
proof is by induction on n. For n = 2, mixed cumulants condition says u(1) 6= u(2), so by freeness
R[a1 , a2 ] = ϕ [a1 a2 ] − ϕ [a1 ] ϕ [a2 ] = 0.
For general n, if u(1) 6= u(2) 6= . . . 6= u(n), then
R[a1 , . . . , an ] = ϕ [a1 . . . an ] −
X
Rπ [a1 , . . . , an ] = 0
π6=1̂n
since the first term is zero by freeness, and the second by the induction hypothesis. Finally, if not all but
some of the u(i)’s are different, we may group them so that
u(1) = . . . = u(v(1)),
u(v(1) + 1) = . . . = u(v(2))
u(v(j − 1) + 1) = . . . = u(n),
48
with u(v(1)) 6= u(v(2)) 6= u(v(3)) 6= . . . 6= u(n). Write
A1 = a1 . . . , av(1) , . . . , Aj = av(j−1) . . . an ,
with j < n. By the induction hypothesis, R[A1 , A2 , . . . , Aj ] = 0. But by Theorem 5.21, this expression
also equals
X
Rπ [a1 , a2 , . . . , an ]
π∨σ=1̂n
for a certain σ ∈ Int(n), σ 6= 1̂n . Take π 6= 1̂n . By the induction hypothesis, in any non-zero term in
the sum, all the ai ’s in the same block of π have to belong to the same subalgebra. Also by construction,
all the ai ’s in the same block of σ belong to the same subalgebra. But then the condition π ∨ σ = 1̂n
implies that all the ai ’s belong to the same subalgebra, contradicting the assumption. It follows that all
Rπ [a1 , a2 , . . . , an ] = 0 for π 6= 1̂n , and so also for π = 1̂n .
5.3
Free convolution, distributions, and limit theorems.
Corollary 5.28. The free cumulant functional is a linearizing transform for the additive free convolution,
in the sense that
Rµν = Rµ + Rν .
Proof.
(µ ν) xu(1) , xu(2) , . . . , xu(n) = ϕ Xu(1) + Yu(1) , . . . , Xu(n) + Yu(n) ,
where {Xi }di=1 , {Yi }di=1 ⊂ (A, ϕ) are free, µ = ϕX1 ,...,Xd , and ν = ϕY1 ,...,Yd . Thus
Rµν [xu(1) , . . . , xu(n) ] = Rϕ [Xu(1) + Yu(1) , . . . , Xu(n) + Yu(n) ]
X
=
Rϕ [Xu(1) or Yu(1) , . . . , Xu(n) or Yu(n) ]
= Rϕ [Xu(1) , . . . , Xu(n) ] + Rϕ [Yu(1) . . . . , Yu(n) ]
= Rµ [xu(1) , . . . , xu(n) ] + Rν [xu(1) , . . . , xu(n) ],
where we used freeness in the middle step.
Note that the same property holds for the free cumulant generating function and for the R-transform.
Remark 5.29. Random variables are classically independent if and only if their mixed classical cumulants
are zero. The cumulant generating function satisfies
`µ∗ν = `µ + `ν ,
where µ∗ν is the classical convolution, which plays for independence the same role as the free convolution
plays for free independence.
49
Exercise 5.30. Let A1 , A2 , . . . , Ak ⊂ (A, ϕ) be subalgebras not containing the unit of A. They are
Boolean independent if for any u(1) 6= u(2) 6= . . . 6= u(n) and ai ∈ Au(i) ,
ϕ [a1 a2 . . . an ] = ϕ [a1 ] ϕ [a2 ] . . . ϕ [an ] .
a. Show that this definition might be inconsistent if the subalgebras were allowed to contain the unit.
b. Show that Boolean independence is equivalent to the vanishing of mixed Boolean cumulants.
Part (b) implies in particular (you don’t need to prove it, the argument is exactly the same as above) that
B µ]ν = B µ + B ν . Here ] is the Boolean convolution on D(d): if a, b have distributions µ, ν, respectively,
and are Boolean independent, then the distribution of a + b is µ ] ν.
Example 5.31 (Symmetric random walk on a free group). Recall the setting from Example 2.25: let Fn
be the free group with generators x1 , . . . , xn . In (C[Fn ], τ ), have the generating elements ui = δxi , all of
which are Haar unitaries. Consider the element
1
−1
−1
(u1 + u−1
X=
1 ) + (u2 + u2 ) + . . . + (un + un ) .
2n
Note that
mk [X] = probability of the return to e after k steps in a simple random walk on Fn starting at e.
So want to find the distribution of X. Know that each ui + u∗i has the arcsine distribution, and they are
freely independent. Thus
1
Gui +u∗i (z) = √
z2 − 4
and
r
1
1
Rui +u∗i (z) = 4 + 2 − .
z
z
So
s
! r
1
1
n
n2 n
RX (z) =
n 4+
−
=
1
+
−
2n
(z/2n)2 z/2n
z2
z
and
r
n2 n − 1
−
.
z2
z
Then
p
−(n − 1)z + n2 z 2 − (2n − 1)
GX (z) =
z2 − 1
and
p
1 (2n − 1) − n2 x2
dµX (x) =
dx.
π
1 − x2
These are the Kesten-McKay distributions.
1
+ RX (z) =
z
1+
50
Exercise 5.32. Let
1
1
µ = δ−a + δa
2
2
be a symmetric Bernoulli distribution. Compute µ µ. Conclude that a free convolution of two discrete
distributions may be continuous.
Limit theorems.
Theorem 5.33 (Free central limit theorem). Let {Xn : n ∈ X} ⊂ (A, ϕ) be freely independent, identically
distributed, mean zero, variance t
ϕ [Xi ] = 0, ϕ Xi2 = t.
Then
Sn =
X1 + . . . + Xn
√
n
converges in distribution to S(0, t), the semicircular distribution with mean 0 and variance t.
Equivalently, denoting by Dα the dilation operator
Z
Z
f (αx) dµ(x)
f (x) d(Dα µ)(x) =
R
R
and µ = µXi ,
D1/√n (µ µ . . . µ) → S(0, t).
|
{z
}
n
Proof. Note first that by definition mn [Dα µ] = αn mn [µ], from which it follows that rn [Dα µ] = αn rn [µ].
Since S(0, t) is compactly supported, to prove convergence in distribution it suffices to prove convergence
in moments,
Z
k
ϕ Sn →
xk dS(0, t).
R
By the moment-cumulant formula, it is in fact enough to prove the convergence of free cumulants, that is
rk [Sn ] → rk [S(0, t)] = tδk=2 .
Indeed,
r1 [Sn ] =
nr1 [X]
√
= 0,
n
r2 [Sn ] =
nr2 [X]
= t,
n
and
rk [Sn ] =
nrk [X]
→ 0,
nk/2
51
k > 2.
Remark 5.34. Can replace the “identically distributed” condition with a “uniformly bounded moments”
condition. See [NS06] for an extensive discussion.
Theorem 5.35 (Poisson limit theorem).
n
t
t
1−
δ0 + δ1
→ πt ,
n
n
the free Poisson distribution with parameter t.
Again, the same conclusion can be obtained under weaker assumptions.
Remark 5.36. If µX = (1 − α)δ0 + αδ1 , then mn [X] = α for all n ≥ 1 (m0 [X] = 1). Thus X 2 and X
have the same distribution. May take
X 2 = X = X ∗,
so that X is an (orthogonal) projection, ϕ [X] = α.
Proof. Again, it suffices to show that
"
n #
t
t
t
t
δ0 + δ1 = rk
δ0 + δ1
1−
→ rk [πt ] = t.
nrk 1 −
n
n
n
n
Denote µn = 1 −
t
n
δ0 + nt δ1 . Note that mk [µn ] = nt . So by the moment-cumulant formula,
t
1
rk [µn ] = + o
,
n
n
and
nrk [µn ] = t + o(1) → t
as n → ∞.
Remark 5.37 (Generalized Poisson limit theorem). Let ν ∈ D(d) be a state on Chx1 , . . . , xd i. Denote
t
t
µn =
1−
δ0 + ν ∈ D(d).
n
n
Then for any |~u| = k ≥ 1,
M µn [x~u ] =
t
ν[x~u ].
n
Therefore the same argument as above gives
R
µn
n
µn
[x~u ] = nR [x~u ] = n
t
ν[x~u ] + o
n
52
1
→ tν[x~u ].
n
We have thus proved the existence of the limit
lim
n→∞
t
1−
n
t
δ0 + ν
n
n
= πν,t ,
which is called the free compound Poisson distribution. In particular, πt = πδ1 ,t . Note that since a limit
of states is a state, we have proved that for any ν ∈ D(d) there exists πν,t ∈ D(d) such that
Rπν,t = tν.
We will later extend this result.
53
Chapter 6
Lattice paths and Fock space models.
See Lectures 2, 7, 8, 9, 13, 21 of [NS06]. For the background on lattice paths, see [Fla80], [Vie84],
[Vie85].
6.1
Dyck paths and the full Fock space.
Lattice paths.
Piece-wise linear paths with vertices in Z × Z.
Each step: one unit to the right, {−1, 0} ∪ N units up.
(1, n) rising or NE step.
(1, 0) flat or E step.
(1, −1) falling of SE step.
Start at (0, 0) at height 0. Never go below the x-axis.
In most cases, end on the x axis at height 0.
Combinatorics: move to the right. Fock spaces: flip, move to the left.
Paths needed for enumeration: attach to each step a weight or a label w(step) = a symbol, number,
operator, etc.
Y
w(path) =
w(step).
Often look at
X
paths
w(path) =
XY
paths
54
w(step).
55
56
Definition 6.1. Dyck paths start and end at height 0, and consist of steps: (1, 1) (rise), (1, −1) (fall).
Incomplete Dyck paths: same conditions except ending at the height ≥ 0.
Lemma 6.2. Dyck paths with 2n steps are in a bijective correspondence with NC2 (2n).
Proof. See Figure 6. Rising step corresponds to an opener, falling step to a closer in a partitions. Key
point: a Dyck path corresponds to a unique non-crossing partition.
Corollary 6.3. |Dyck paths(2n)| = cn , the Catalan number. Moreover, if we put on the steps the weights
w(rise) = 1, w(fall) = t, then
X
w(path) = tn cn = m2n [S(0, t)].
Dyck paths(2n)
In fact, much more is true.
Full Fock space.
Let HR be a real Hilbert space, H = HR ⊗R C its complexification. The algebraic full Fock space is the
vector space
∞
M
H⊗n = CΩ ⊕ H ⊕ (H ⊗ H) ⊕ (H ⊗ H ⊗ H) ⊕ . . . .
Falg (H) =
n=0
Here Ω is called the vacuum vector. The full Fock space is the completion of Falg (H) with respect to the
norm coming from the inner product
hf1 ⊗ f2 ⊗ . . . ⊗ fn , g1 ⊗ g2 ⊗ . . . ⊗ gk i = δn=k
n
Y
hfi , gi iH .
i=1
Let f ∈ HR . Define the (free) creation operator `(f ) and (free) annihilation operator `∗ (f ) on Falg (H)
by
`(f )(g1 ⊗ g2 ⊗ . . . ⊗ gn ) = f ⊗ g1 ⊗ g2 ⊗ . . . ⊗ gn ,
`(f )Ω = f ,
`∗ (f )(g1 ⊗ g2 ⊗ . . . ⊗ gn ) = hg1 , f i g2 ⊗ . . . ⊗ gn ,
`∗ (f )(g) = hg, f i Ω,
`∗ (f )Ω = 0.
Remark 6.4. Frequently denote the creation operator by a∗ or a+ , annihilation operator by a or a− .
Lemma 6.5. `∗ (f ) is the adjoint of `(f ).
57
Proof.
h`(f )(g1 ⊗ . . . ⊗ gn , h0 ⊗ h1 ⊗ . . . ⊗ hn i = hf ⊗ g1 ⊗ . . . ⊗ gn ), h0 ⊗ h1 ⊗ . . . ⊗ hn i
n
Y
= hf, h0 i
hgi , hi i
i=1
D
= g1 ⊗ . . . ⊗ gn , hf, h0 ih1 ⊗ . . . ⊗ hn
E
= hg1 ⊗ . . . ⊗ gn , `∗ (f )(h0 ⊗ h1 ⊗ . . . ⊗ hn )i .
Therefore for each f ∈ HR ,
X(f ) = `(f ) + `∗ (f )
is a symmetric operator. Consider the n.c. probability space
(A = Alg∗ (X(f ) : f ∈ HR ) , ϕ = h·Ω, Ωi).
ϕ is the vacuum expectation.
One can check (using the commutant) that ϕ faithful. It follows from Corollary 6.8 below that, because
f ∈ HR , ϕ is tracial.
Proposition 6.6. Using right-to-left Dyck paths,
X
X(f1 )X(f2 ) . . . X(fn )Ω =
w(path)fu(1) ⊗ fu(2) ⊗ . . . ⊗ fu(k) ,
incomplete Dyck paths(n)
where the weights are as follows: w(rise) = 1; if the i’th step is a fall matched with a rise on step j, then
w(fall) = hfi , fj i, and the unmatched rises are in positions u(1) < u(2) < . . . < u(k).
Example 6.7.
X(f1 )Ω = f1 ,
X(f1 )X(f2 )Ω = X(f1 )f2 = f1 ⊗ f2 + hf1 , f2 i Ω
See Figure 6.
Note: since all fi ∈ HR , hfi , fj i = hfj , fi i.
Proof. By induction,
X
X(f0 )
w(path)fu(1) ⊗ fu(2) ⊗ . . . ⊗ fu(k)
incomplete Dyck paths(n)
X
=
w(path)f0 ⊗ fu(1) ⊗ fu(2) ⊗ . . . ⊗ fu(k)
incomplete Dyck paths(n)
+
X
w(path) fu(1) , f0 fu(2) ⊗ . . . ⊗ fu(k)
incomplete Dyck paths(n)
=
X
w(path)fu(1) ⊗ fu(2) ⊗ . . . ⊗ fu(k) .
incomplete Dyck paths(n+1)
58
A different argument was given in class.
Corollary 6.8 (Free Wick formula).
a.
ϕ [X(f1 )X(f2 ) . . . X(fn )] =
X
k
Y
fu(i) , fv(i)
X
w(path) =
i=1
π∈NC2 (n)
π={(u(1),v(1))...(u(k),v(k))}
u(i)<v(i)
Dyck paths(n)
(again, order in the inner products irrelevant).
b.
Rϕ [X(f1 ), X(f2 ), . . . , X(fn )] = δn=2 hf1 , f2 i .
c. Each X(f ) has distribution S(0, kf k2 ).
d. If f1 , f2 , . . . , fn are mutually orthogonal, then {X(f1 ), X(f2 ), . . . , X(fn )} are freely independent.
Call {X(f ) : f ∈ HR } a semicircular system, and any
{X(fi ) : i ∈ S, fi ⊥ fj for i 6= j}
a free semicircular system.
Definition 6.9. Let a = (a1 , a2 , . . . , ad ) ∈ (A, ϕ) be random variables.
Mean of a = vector (ϕ [a1 ] , . . . , ϕ [ad ])t .
h
id
Covariance of a = matrix ϕ [ai aj ] − ϕ [ai ] ϕ [aj ] = Rϕ [ai , aj ]
.
i,j=1
Random variables are uncorrelated if their covariance is the zero matrix.
Corollary 6.10. If elements in a semicircular system are uncorrelated, they are freely independent.
Warning: semicircular variables may be uncorrelated without being freely independent.
Theorem 6.11 (Multivariate free CLT). Let {ai,n : i = 1, . . . , d, n ∈ N} ⊂ (A, ϕ) be random variables
such that the d-tuples (a1,n , . . . , ad,n ) are, for n ∈ N, freely independent, identically distributed, have
mean 0 and covariance Q. Then
!
N
N
X
X
1
√
a1,n , . . .
ad,n → (s1 , . . . , sd )
N n=1
n=1
in distribution, where (s1 , . . . , sd ) are a semicircular system with covariance Q.
59
Meaning of incomplete Dyck paths? Let f1 , . . . , fn ∈ HR . Define the Wick product W (f1 , f2 , . . . , fn ) ∈
B(F(H)) by W (∅) = 1 and the recursion
X(f )W (f1 , f2 , . . . , fn ) = W (f, f1 , f2 , . . . , fn ) + hf1 , f i W (f2 , . . . , fn ),
For example, W (f ) = X(f )W (∅) = X(f ),
W (f1 , f2 ) = X(f1 )X(f2 ) − hf2 , f1 i .
Clearly W (f1 , . . . , fn ) is a polynomial in X(f1 ), . . . , X(fn ).
Lemma 6.12.
W (f1 , . . . , fn )Ω = f1 ⊗ . . . ⊗ fn ,
and it is a unique operator in Alg∗ (X(f ) : f ∈ HR ) with this property.
Proof. Use the recursion. The second statement follows from the fact (which we do not prove) that the
vector Ω is separating for the algebra, i.e. the only operator in it with AΩ = 0 is A = 0.
Fix f ∈ HR with kf k2 = t, and denote X(f ) = X. Then W (f, . . . , f ) = Un (X, t) is a polynomial in X.
It satisfies the recursion
XUn (X, t) = Un+1 (X, t) + tUn−1 (X, t),
U0 (X, t) = 1.
{Un } are Chebyshev polynomials of the second kind (with a slightly unusual normalization). Satisfy a
3-term recursion, so by general theory (Theorem 6.43) are orthogonal with respect to some distribution.
Lemma 6.13. {Un : n ∈ N} are the monic orthogonal polynomials with respect to µ = S(0, t), that is
Z
Un (x, t)Uk (x, t) dµ(x) = 0, n 6= k.
R
Proof.
Z
Un (x, t)Uk (x, t) dµ(x) = hUn (X, t)Uk (X, t)Ω, Ωi
R
= hUk (X, t)Ω, Un (X, t)Ωi = f ⊗k , f
⊗n
(
0, k 6= n,
= n
t , n = k.
Corollary 6.14.
Xn =
X
t#matched steps U#unmatched steps (X, t)
incomplete Dyck paths(n)
n
n
= Un (X, t) +
c1 tUn−2 (X, t) +
c2 t2 Un−4 (X, t) + . . . .
2
4
60
Proof. Combine Proposition 6.6 with the definition of the Wick products.
What about general W (f1 , f2 , . . . , fn )?
Lemma 6.15. Let f1 , . . . , fd be mutually orthogonal and ~u = (u(1), . . . , u(n)), u(i) ∈ [d], be such that
~u = (v(1), . . . , v(1), v(2), . . . , v(2), . . . , v(r), . . . , v(r))
|
{z
} |
{z
}
|
{z
}
k(1)
k(2)
k(r)
with v(1) 6= v(2) 6= . . . 6= v(r) (neighbors different). Then
W (fu(1) , fu(2) , . . . fu(n) ) =
r
Y
2 Uk(i) X(fv(i) ), fv(i) .
i=1
It is natural to call these multivariate Chebyshev polynomials.
Proof. Using Lemma 6.12, it suffices to show that if f ⊥ g1 then
Uk (X(f ), kf k2 )g1 ⊗ . . . ⊗ gn = f ⊗k ⊗ g1 ⊗ . . . ⊗ gn .
Indeed, by induction, for k ≥ 1
Uk+1 (X, kf k2 )g1 ⊗ . . . ⊗ gn = XUk X(f ), kf k2 − kf k2 Uk−1 X(f ), kf k2 g1 ⊗ . . . ⊗ gn
= X(f )f ⊗k ⊗ g1 ⊗ . . . ⊗ gn − kf k2 f ⊗(k−1) ⊗ g1 ⊗ . . . ⊗ gn
= f ⊗(k+1) ⊗ g1 ⊗ . . . ⊗ gn .
Exercise 6.16. Let HR = Rn be a real Hilbert space, H = Cn its complexification, and
Γ0 (H) = W ∗ (X(f ) : f ∈ HR ).
Prove that
Γ0 (H) ' L(Fdim H ).
You may use without proof the analytic facts L(Z) ' L∞ (T) (proved via Fourier transforms) and, for
a = a∗ , W ∗ (a) ' L∞ (σ(a)). To be completely precise, you would also need a von Neumann algebra
version of Proposition 6.6 and Theorem 7.9 from [NS06], namely that the von Neumann algebra generated
by freely independent subalgebras is (isomorphic to) their free product.
Exercise 6.17. Exercise 8.26 in [NS06]. Fill in the details in the following argument.
For f ∈ HR , show that the operator X(f ) = `(f ) + `∗ (f ) on F(H) has, with respect to the vacuum
expectation on B(F(H)), the same distribution as the operator
f ⊕ f ⊕ ... ⊕ f
f ⊕ f ⊕ ... ⊕ f
f ⊕ f ⊕ ... ⊕ f
∗
√
√
√
=`
+`
X
N
N
N
on F(HN ), with respect to the vacuum expectation on B(F(HN )). Here HN = |H ⊕ .{z
. . ⊕ H}. Show that
N
the second operator is a sum of N freely independent operators,
and apply
the free central limit theorem
f ⊕f√⊕...⊕f to conclude that X(f ) is a semicircular element. Hint: what is ?
N
61
6.2
Łukasiewicz paths and Voiculescu’s operator model.
Lattice paths and Fock spaces II.
Definition 6.18. Łukasiewicz paths are lattice paths starting and ending at height 0, with rising steps
(1, k), k ∈ N, flat steps (1, 0), and falling steps (1, −1). We will always use weight 1 for the falling steps
and weight rk+1 for a rising step (1, k), including weight r1 for a flat step (1, 0).
Exercise 6.19. Use the idea in Figure 6 to show that Łukasiewicz paths with n steps are in a bijection
with NC(n).
Proposition 6.20 (Voiculescu’s operator model). Let µ ∈ D(1), with free cumulants {rk : k ∈ N}. Let f
be a unit vector, and ` = `(f ) a free creation operator on a full Fock space. Note that the only relation
between ` and `∗ is `∗ ` = 1. Let
Xµ = ` +
∞
X
rk (`∗ )k−1 = ` + r1 + r2 `∗ + r3 (`∗ )2 + . . .
k=1
be a formal power series. Then
Xµn Ω, Ω = mn [µ],
where this expression depends only on finitely many terms from the series, and so is well defined. In other
words, µ is the distribution of Xµ .
Proof.
*
Xµn Ω, Ω =
`+
∞
X
!n
rk (`∗ )k−1
+
Ω, Ω
=
k=1
X
w(path) =
Łukasiewicz paths(n)
X
Y
r|V | .
π∈NC(n) V ∈π
Proposition 6.21. If f1 , . . . , fk are mutually orthogonal, then {Alg∗ (`(fi )) : 1 ≤ i ≤ k} are freely independent with respect to the vacuum expectation.
Proof. See Proposition 7.15 in [NS06], or imitate the proof of Proposition 3.5.
Proposition 6.22. Let f1 ⊥ f2 be unit vectors, `1 = `(f1 ) and `2 = `(f2 ), so that the relation between
them is
`∗i `j = δi=j
Let
Xµ = `1 +
∞
X
rk [µ](`∗1 )k−1 ,
k=1
Xν = `2 +
∞
X
k=1
62
rk [ν](`∗2 )k−1 .
Then Xµ and Xν are freely independent, and Xµ + Xν has the same distribution as
Y = l1 +
∞
X
(rk [µ] + rk [ν])(l1∗ )k−1 .
k=1
This gives an alternative proof of the fact that
rk [µ ν] = rk [µ] + rk [ν].
Proof. Free independence follows from the preceding proposition. Let
Z = Xµ + Xν = (`1 + `2 ) +
∞
X
rk [µ](`∗1 )k−1
k=1
and note that
Y = `1 +
∞
X
rk [µ](`∗1 )k−1
+
∞
X
rk [ν](`∗2 )k−1
k=1
+
k=1
∞
X
rk [ν](`∗1 )k−1 .
k=1
We verify that for all n,
hZ n Ω, Ωi = hY n Ω, Ωi ,
since in a Łukasiewicz path, a step of one type (1 or 2) can only be matched with a step of the same
type.
Can do this in the multivariate case. See [Haa97] for an alternative argument.
6.3
Motzkin paths and Schürmann’s operator model. Freely infinitely divisible distributions.
Lattice paths and Fock spaces III. See [GSS92].
Definition 6.23. A Motzkin path is a lattice path starting and ending at height 0, with rising steps (1, 1),
flat steps (1, 0), and falling steps (1, −1).
Remark 6.24. In this section, rising and falling steps have weight 1, while flat steps at positive height
have weight 2. Equivalently, flat steps come with two possible labels, “s” and “m”, where flat steps at
height 0 can also have the s label.
Definition 6.25. Each step in a Motzkin path has a height: a rising step from k to k + 1, flat step from k
to k, falling step from k to k − 1 all have height k.
For a partition π, for each k, the number of open blocks is
|{V ∈ π : min(V ) < k ≤ max(V )}| .
See the figure for the next lemma for an example.
63
Lemma 6.26. Motzkin paths with n steps with labels as in the preceding remark are in a bijection with
NC(n). Under this bijection, the height of a step in the path equals the number of open blocks at that
location in the partition.
Proof. See Figure 6. Openers of blocks correspond to rising steps, closers to falling steps, middle elements to flat steps labeled with m, singletons to flat steps labeled with s.
Remark 6.27. Under this bijection, Dyck paths correspond to NC2 (n), Motzkin paths with no flat steps
at height 0 to non-crossing partitions with no singletons, and unlabeled Motzkin paths to NC1,2 (n), noncrossing partitions whose blocks have at most 2 elements.
Remark 6.28. Let HR be a real Hilbert space, H its complexification, F(H) its full Fock space, and
ϕ = h·Ω, Ωi the vacuum expectation. Let
λ1 , λ2 , . . . , λd ∈ R,
ξ1 , ξ2 , . . . , ξd ∈ HR ,
and
T1 , T2 , . . . , Td ∈ L(HR )
be symmetric, possibly unbounded, linear operators defined on a common domain D ⊂ HR containing
{ξi : 1 ≤ i ≤ d} and invariant under all Ti . That is, for any ~u,
Tu(1) Tu(2) . . . Tu(n−1) ξu(n)
is defined. On the Fock space, we have the free creation and annihilation operators `(ξi ), `∗ (ξi ). Define
Λ(Ti )(f1 ⊗ f2 ⊗ . . . ⊗ fn ) = (Ti f1 ) ⊗ f2 ⊗ . . . ⊗ fn ,
Λ(Ti )Ω = 0, and
Xi = `(ξi ) + `∗ (ξi ) + Λ(Ti ) + λi .
It is easy to check that Λ(T ) is symmetric if T is. So each Xi is symmetric. What is the joint distribution
of (X1 , . . . , Xd ) with respect to ϕ?
Lemma 6.29.
Rϕ [Xi ] = λi ,
Rϕ [Xi , Xj ] = hξi , ξj i ,
Rϕ [Xu(1) , Xu(2) , . . . , Xu(n) ] = ξu(1) , Tu(2) . . . Tu(n−1) ξu(n) .
Proof.
ϕ Xu(1) Xu(2) . . . , Xu(n) = Xu(1) Xu(2) . . . , Xu(n) Ω, Ω =
X
Motzkin paths(n)
64
hw(path)Ω, Ωi ,
where the rising step in position i has weight `∗ (ξu(i) ), the falling step `(ξu(i) ), a flat “m” step weight Tu(i) ,
and a flat “s” step weight λu(i) . For example, the term corresponding to the path/partition in Figure 6 is
(moving right-to-left)
hξ12 , T10 ξ4 i λ11 hξ9 , ξ5 i hξ8 , T7 ξ6 i λ3 hξ2 , ξ1 i .
Which joint distributions arise in this way?
Definition 6.30. A distribution µ ∈ D(d) is freely (or )-infinitely divisible if for all n, µ(1/n) exists,
that is, there exists νn ∈ D(d) such that νnn = µ. Note: positivity is crucial.
Proposition 6.31. The following are equivalent.
a. µ is -infinitely divisible.
b. For all t ≥ 0, all k ∈ N, and |~u| = k, t R~uµ = R~uµt for some µt ∈ D(d).
c. There exists a free convolution semigroup
{µt : t ≥ 0}
µt µs = µt+s ,
µ1 = µ.
Proof. µ is -infinitely divisible if and only if there exists νn with n1 R~uµ = R~uνn . Denote µ1/n = νn , and
for p/q ∈ Q+ , µp/q = νqp ∈ D(d) satisfies
p
µ
R~up/q = R~uµt .
q
By continuity, can extend this to real t.
Example 6.32. Recall that rn [S(0, t)] = tδn=2 and rn [πt ] = t. It follows that {S(0, t) : t ≥ 0} and
{πt : t ≥ 0} form free convolution semigroups, and in particular are -infinitely divisible. More generally, for any ν ∈ D and any t ≥ 0, the corresponding free compound Poisson distribution µν,t is
-infinitely divisible.
See Lecture 13 of [NS06] for a discussion of general limit theorems.
Definition 6.33. Denote
C0 hx1 , . . . , xd i = {P ∈ Chx1 , . . . , xd i with no constant term.}
A self-adjoint linear functional ρ on Chxi is conditionally positive if it is positive on C0 hxi.
Example 6.34. On C[x], let ρ [xn ] = δn=2 . Then
ρ (x2 − 1)2 = −2,
so ρ is not positive.
65
Lemma 6.35. Every P ∈ C0 hx1 , . . . , xd i is of the form
P =
d
X
i=1
λ i xi +
d
X
xi Pij (x)xj ,
Pij ∈ Chx1 , . . . , xd i.
i,j=1
If ψ ∈ D(d), then ρ defined arbitrarily on 1 and xi , and
" d
#
d
X
X
ρ
xi Pij xj =
ψ [Pii ] ,
i,j=1
i=1
is conditionally positive. If d = 1, the converse also holds. That is, any conditionally positive functional
on C[x] is of the form ρ [xn ] = ψ [xn−2 ] for some positive ψ and n ≥ 2.
Proof. For P ∈ C0 hx1 , . . . , xd i,
#
" d
#
" d
!∗ d
d
X
X
X
X
∗
∗
ψ [Pi∗ Pi ] ≥ 0.
ρ [P P ] = ρ
P j xi = ρ
x i Pi Pj x j =
Pi x i
i=1
j=1
i,j=1
i=1
Conversely, if d = 1, ρ is conditionally positive, and ψ is defined by ψ [xn ] = ρ [xn+2 ], then
ψ [P ∗ P ] = ρ [xP ∗ P x] ≥ 0.
Remark 6.36 (Schürmann’s construction). Let ρ on C0 hx1 , . . . , xd i be conditionally positive. GNS construction for (C0 hx1 , . . . , xd i, ρ) produces a Hilbert space
H = (C0 hx1 , . . . , xd i/N )
with the inner product
D
E
P̂ , Q̂ = ρ [Q∗ P ] .
Note that C0 hx1 , . . . , xd i is not unital, so there is no natural state vector for ρ. However, we can define,
for i = 1, 2, . . . , d
λi = ρ[xi ] ∈ R, ξi = x̂i ∈ H, Ti = λ(xi ) ∈ L(H).
Then
ρ [xi ] = λi ,
ρ [xi xj ] = hξi , ξj i ,
ρ xu(1) xu(2) . . . xu(n) = ξu(1) , Tu(2) . . . Tu(n−1) ξu(n) , .
Note also that the conditions from Remark 6.28 are satisfied. Therefore that construction give a state
µ ∈ D(d) with
Rµ [xu(1) , xu(2) , . . . , xu(n) ] = Rϕ [Xu(1) , Xu(2) , . . . , Xu(n) ] = ρ xu(1) xu(2) . . . xu(n) .
66
Proposition 6.37 (Schoenberg correspondence). For distributions with finite moments, µ is -infinitely
divisible if and only if Rµ is a conditionally positive functional.
Proof. ⇒. µ is part of a free convolution semigroup µt , where
X
µt xu(1) xu(2) . . . xu(n) =
t|π| Rπµ [xu(1) , xu(2) , . . . , xu(n) ].
π∈NC(n)
Then
d R [xu(1) xu(2) . . . xu(n) ] = R [xu(1) , xu(2) , . . . , xu(n) ] = µt xu(1) xu(2) . . . xu(n) .
dt t=0
µ
µ
On the other hand, if P ∗ P has no constant term
d µt [P ∗ P ] − µ0 [P ∗ P ]
∗
≥0
µt [P P ] = lim+
t→0
dt t=0
t
since µ0 [P ∗ P ] = δ0 [P ∗ P ] = (P ∗ P )(0) = 0.
For the converse direction, suppose that Rµ is conditionally positive. Then tRµ is also conditionally
positive, and so by the Fock space constructions above, tRµ = Rµt for some µt ∈ D(d).
Exercise 6.38. Let (A, ψ) be an n.c. probability space, and suppose for simplicity that ψ is faithful. On
the full Fock space F(L2 (A, ψ)), for f ∈ A symmetric, define the creation and annihilation operators
`(f ), `∗ (f ) as before, and
Λ(f )(g1 ⊗ g2 ⊗ . . . ⊗ gn ) = (f g1 ) ⊗ g2 ⊗ . . . ⊗ gn ,
Λ(f )Ω = 0. Then each Λ(f ), and also
X(f ) = `(f ) + `∗ (f ) + Λ(f ) + ψ [f ] ,
is symmetric. Compute
hX(f1 )X(f2 ) . . . X(fn )Ω, Ωi .
Deduce that each X(f ) has a free compound Poisson distribution, and if fi fj = 0 for i 6= j, then X(fi )
are freely independent with respect to the vacuum expectation.
Remark 6.39 (Classically infinitely divisible distributions). Let HR and the algebraic Fock space Falg (H)
be as before. On this space, define the pre-inner product
hf1 ⊗ . . . ⊗ fn , g1 ⊗ . . . ⊗ gk is = δn=k
X
n
Y
fi , gα(i) .
α∈Sym(n) i=1
For example
hf1 ⊗ f2 , g1 ⊗ g2 is = hf1 , g1 i hf2 , g2 i + hf1 , g2 i hf2 , g1 i .
67
This is a degenerate inner product, since kf ⊗ g − g ⊗ f ks = 0. Quotient out by tensors of norm 0, get
the symmetric Fock space Fs (H), on which f ⊗s g = g ⊗s f . On this space, define the creation operators
a+ (f )(g1 ⊗ . . . ⊗ gn ) = f ⊗ g1 ⊗ . . . ⊗ gn ,
annihilation operators
−
a (f )(g1 ⊗ . . . ⊗ gn ) =
n
X
hgi , f i g1 ⊗ . . . ⊗ ĝj ⊗ . . . ⊗ gn ,
j=1
and preservation operators
e )(g1 ⊗ . . . ⊗ gn ) =
Λ(T
n
X
g1 ⊗ . . . ⊗ (T gj ) ⊗ . . . ⊗ gn
j=1
Again,
e i ) + λi
Xi = a+ (ξi ) + a− (ξi ) + Λ(T
are symmetric (with respect to the new inner product). However, for {ξi , Ti , λi } as in Remark 6.28, these
Xi ’s commute. Moreover, if µ ∈ D(d) is their joint distribution, its classical cumulants are precisely
k µ [xu(1) , . . . , xu(n) ] = ξu(1) , Tu(2) . . . Tu(n−1) ξu(n)
Proposition 6.40. µ ∈ D(d) is ∗-infinitely divisible if and only if µ comes from (λi , ξi , Ti )di=1 on a symmetric Fock space, if and only if k µ is a conditionally positive functional.
Corollary 6.41. The Bercovici-Pata bijection is the bijection
BP : {∗ − infinitely divisible distributions}
↔
{ − infinitely divisible distributions} ,
where
k µ = RBP[µ] .
This is a bijection since both sets {k µ : µ ∗ −ID} and {Rµ : µ −ID} consist precisely of all conditionally positive functionals. Moreover, this map is a homomorphism with respect to the convolution
operations,
BP[µ ∗ ν] = BP[µ] BP[ν],
and homeomorphism with respect to convergence in moments.
This homomorphism property does not extend to D(d), even for d = 1.
Exercise 6.42 (Boolean infinite divisibility). Let H and (λi , ξi , Ti )di=1 be as before. The Boolean Fock
space is simply CΩ ⊕ H, with Boolean creation operators
a+ (f )Ω = f,
68
a+ (f )g = 0,
and annihilation operators
a− (f )Ω = 0,
a− (f )g = hg, f i Ω.
Extend T ∈ L(H) to the Boolean Fock space by T Ω = 0. Finally, denote PΩ the projection onto Ω,
PΩ Ω = Ω, PΩ g = 0. Define
Xi = a+ (ξi ) + a− (ξi ) + Ti + λi PΩ .
a. Show that the joint distribution of (X1 , . . . , Xd ) with respect to the vacuum expectation is ]infinitely divisible.
Hint: in the free case, the argument goes as follows. Via the Schoenberg correspondence, free
infinite divisibility is equivalent to the conditional positivity of the free cumulant functional. This
in turn is equivalent to a representation on a free Fock space coming from the data (λi , ξi , Ti )di=1 :
one direction is Schürmann’s construction, the other follows from the existence of the Fock space
constructions for all re-scalings of this data by positive t.
b. Let µ ∈ D(d) be arbitrary. Using the GNS construction, we can represent the n.c. probability space
(Chx1 , . . . , xd i, µ) on a Hilbert space K with µ = h·Ω, Ωi. Let H = K CΩ,
λi = hλ(xi )Ω, Ωi ∈ C,
ξi = (λ(xi ) − λi PΩ )Ω ∈ K,
Ti = λ(xi ) − a+ (ξi ) − a− (ξi ) − λi PΩ ∈ L(K).
Show that these satisfy all the conditions in Remark 6.28 and in part (a) (on H, not on K). Conclude
that any distribution is infinitely divisible in the Boolean sense.
6.4
Motzkin paths and orthogonal polynomials.
Lattice paths IV.
Let {β0 , β1 , β2 , . . .} ⊂ R and {γ1 , γ2 , γ3 , . . .} ⊂ [0, ∞). Consider Motzkin paths with the following
weights: rising steps have weight 1, flat steps at height n have weight βn , falling steps at height n have
weight γn . Also fix γ0 = 1. See Figure 6.
Let H = C. Then each C⊗n ' C, so by denoting
{z. . . ⊗ 1} = en ,
|1 ⊗ 1 ⊗
n
e0 = Ω, we identify
F(C) = Span (en : n ≥ 0).
69
Choose the inner product so that {en : n ≥ 0} form an orthogonal basis, with ken k2 = γn γn−1 . . . γ1 ,
ke0 k = 1. Define
a+ en = en+1 ,
a− en = γn en−1 ,
a0 e n = β n e n .
a− e0 = 0,
Then (a+ )∗ = a− , a0 is symmetric, and
X = a+ + a− + a0
is also symmetric, with some distribution µ,
Z
n
hX e0 , e0 i =
xn dµ(x).
R
Theorem 6.43 (Darboux, Chebyshev, Stieltjes, Viennot, Flajolet, etc.).
a. See Figure 6.
m2 [µ] = β02 + γ1 ,
m1 [µ] = β0 ,
m3 [µ] = β03 + 2β0 γ1 + β1 γ1 ,
and in general
X
mn [µ] =
w(path).
Motzkin paths(n)
b. The moment generating function of µ has a continued fraction expansion
M (z) =
∞
X
1
mn [µ]z n =
n=0
γ1 z 2
1 − β0 z −
1 − β1 z −
γ2 z 2
1 − β2 z −
γ3 z 2
1 − ...
c. Any measure µ all of whose moments are finite arises in this way. The corresponding pair of
sequences
β0 β1 β2 . . .
J(µ) =
γ1 γ2 γ3 . . .
are the Jacobi parameters of µ.
d. Let {Pn (x) : n ≥ 0} be the monic orthogonal polynomials with respect to µ, that is Pn (x) = xn +. . .
and
Z
Pn (x)Pk (x) dµ(x) = 0, n 6= k
R
70
Then these polynomials satisfy the three-term recursion
xPn (x) = Pn+1 (x) + βn Pn (x) + γn Pn−1 (x),
where by convention P0 (x) = 1, P−1 = 0.
Proof.
a. Clear from the Fock space model.
b. Figure omitted. We will show by induction that the continued fraction up to γn expands to the sum
over Motzkin paths of height at most n, plus higher order terms. Indeed, taking an extra level in the
continued fraction corresponds to replacing each γn with γn times the sum over Motzkin paths of
height at most 1, with labels βn , γn+1 . But inserting such paths between each rise at height n and
the corresponding fall is exactly how one goes from a path of height n to a path of height n + 1.
c. It suffices to show how to compute the Jacobi parameters from the moments. Indeed,
γ1 γ2 . . . γn = m2n − Poly(γ1 , . . . , γn−1 , β0 , . . . , βn−1 ),
γ1 γ2 . . . γn βn = m2n+1 − Poly(γ1 , . . . , γn , β0 , . . . , βn−1 ),
d. Note that by assumption,
hPn (X)e0 , Pk (X)e0 i = 0,
n 6= k
and
Pn (X)e0 ∈ Span (ei : 0 ≤ i ≤ n)
It follows that for each n, {Pi (X)e0 : 0 ≤ i ≤ n} is a basis for Span (ei : 0 ≤ i ≤ n). Moreover,
Pn (X)e0 − en ∈ Span (ei : 0 ≤ i ≤ n − 1), which implies that in fact Pn (X)e0 = en . But
XPn (x)e0 = Xen = en+1 + βn en + γn en−1 = Pn+1 (X)e0 + βn Pn (X)e0 + γn Pn−1 (X)e0 .
Example 6.44. Let βn = 0 and γn = t for all n. Thus
0 0 0 ...
J(µt ) =
t t t ...
Then
mn [µt ] =
X
tn/2 ,
Dyck paths(n)
and µt = S(0, t). The orthogonal polynomials are the Chebyshev polynomials of the second kind.
71
Example 6.45. Let βn = βt and γn = γt for all n. Thus
βt βt βt . . .
J(µt ) =
γt γt γt . . .
Then
mn [µt ] =
X
X
(γt)#falling steps (βt)#flat steps =
Rπ [µt ],
π∈NC1,2 (n)
Motzkin paths(n)
where r1 [µt ] = βt, r2 [µt ] = γt, rn [µt ] = 0 for n > 2. Thus µt = S(βt, γt) = S(β, γ)t .
Example 6.46. Let β0 = βt, βn = βt + b, γn = γt, so that
βt βt + b βt + b . . .
J(µt ) =
γt
γt
γt
...
Then
mn [µt ] =
X
X
(γt)#falling steps (βt)#singleton flat steps b#middle flat steps =
labeled Motzkin paths
Rπ [µt ],
π∈NC(n)
where r1 [µt ] = βt, rn [µt ] = γbn−2 t. Thus µt = µt , where r1 [µ] = β, rn [µ] = γbn−2 . µ is a shifted,
scaled free Poisson distribution:
(
a + r1 [ν], n = 1,
rn [Dα ν δa ] =
αn rn [ν],
n > 1.
Definition 6.47. Let β, b, c ∈ R and γ ≥ 0. The free Meixner distributions {µt } with these parameters
are determined by their Jacobi parameters
βt βt + b βt + b . . .
J(µt ) =
γt γt + c γt + c . . .
Thus either c ≥ 0, or c < 0 and t ≥ −c/γ.
Proposition 6.48. Whenever µs , µt are positive, then µs µt = µs+t . In particular, if c ≥ 0, then µt is
-infinitely divisible. If c < 0, µ = µ−c/γ is not -infinitely divisible, but µt is positive for all t ≥ 1.
Proof. See Figure 6. Take a Motzkin path with labels βt, βt + b, γt, γt + c. Separate it into Motzkin
paths with labels βt on flat steps, b on flat steps at height at least 1, γt on falling steps, c on falling steps
at height at least 1. Form a new labeled Motzkin path by
βt-flat 7→ s-flat ,
γt-fall 7→ fall,
matching rise 7→ rise,
b or c steps 7→ m-flat steps.
72
This corresponds to a π ∈ NC(n). On each non-singleton class of π, have a further (b, c)-Motzkin path,
starting with a γt-rise. Thus
X
Y
X
mn [µt ] =
(βt)|Sing(π)|
(γt)
w(path)
=
π∈NC(n)
V ∈π,|V |≥2
X
Y
(βt)|Sing(π)|
(b,c)−Motzkin paths on |V |−2
(γt)m|V |−2 (S(b, c))
V ∈π,|V |≥2
π∈NC(n)
and
r1 [µt ] = βt,
rn [µt ] = γtmn−2 (S(b, c)).
Free Meixner distributions include the semicircular, free Poisson, arcsine, Kesten-McKay, Bernoulli and
free binomial distributions.
Remark 6.49. Let β = b = 0, γ = 1, c = −1. Thus
0
0
0
...
J(µt ) =
t t − 1 t − 1 ...
For t = 1,
0 0 0 ...
J(µt ) =
1 0 0 ...
so that
1
1 − z2
z
1
1
1
= 2
=
−
.
z −1
2 z−1 z+1
M µ (z) =
and
Gµ (z) =
1
1
z 1 − z −2
So
1
µ = (δ−1 + δ1 )
2
is the symmetric Bernoulli distribution. Easy to see: µ∗t is defined only for t ∈ N. But µt is defined for
all t ≥ 1! Compare with the Bercovici-Pata bijection.
Exercise 6.50.
a. If
β0 β1 β2 . . .
J(µ) =
,
γ1 γ2 γ3 . . .
show that its Boolean cumulants are B1 [µ] = β0 , Bn+2 [µ] = γ1 mn [ν], where
β1 β2 . . .
J(ν) =
γ2 γ3 . . .
The map µ 7→ ν is the coefficient stripping.
73
b. Conclude that
]t
J(µ ) =
6.5
β0 t β1 β2 . . .
,
γ1 t γ2 γ3 . . .
q-deformed free probability.
See [BS91], [Ans01].
Definition 6.51. Let q be a number or a symbol. The q-integers are
[n]q = 1 + q + q 2 + . . . + q n−1 =
1 − qn
1−q
The notion goes back to Euler (≈ 1750), Rogers (1894), Ramanujan (1913).
Example 6.52. [n]1 = n,
(
0,
[n]0 =
1,
n = 0,
n ≥ 1.
Consider Dyck paths with weights 1 on rising steps, and weight [n]q = 1 + q + . . . + q n−1 on a falling
step at height n.
From general theory, if
mn [µ(x|q)] =
X
w(path),
Dyck paths
then µ(x|q) is the orthogonality measure for the polynomials {Pn (x|q)} which are defined via the recursion
xPn (x|q) = Pn+1 (x|q) + [n]q Pn−1 (x|q).
These are the continuous (Rogers) q-Hermite polynomials. For q = 1, they are the Hermite polynomials.
For q = 0, they are the Chebyshev polynomials of the 2nd kind.
Have explicit formulas for µ(x|q), Pn (x|q). µ(x|q) is the q-Gaussian distribution.
Equivalently,
mn [µ(x|q)] =
X
w(path),
labeled Dyck paths
where each falling step at height n may have one of the labels 1, q, . . . , q n−1 . Using this idea, on Falg (H),
define a+ (f ) = `(f ) but
−
a (f )(g1 ⊗ . . . ⊗ gn ) =
n
X
q i−1 hfi , gi g1 ⊗ . . . ⊗ ĝi ⊗ . . . ⊗ gn .
i=1
Note that on the full Fock space F(H), these are not adjoints.
74
Definition 6.53. On the algebraic full Fock space, define the q-inner product
hf1 ⊗ . . . ⊗ fn , g1 ⊗ . . . ⊗ gk iq = δn=k
X
α∈Sym(n)
q
inv(α)
n
Y
fi , gα(i) ,
i=1
where inv (α) is the number of inversions in the permutation α,
inv (α) = |{(i, j) : 1 ≤ i < j ≤ n, α(i) > α(j)}|
Theorem 6.54 (Bożejko, Speicher 1991; Zagier 1992). The inner product h·, ·iq is positive definite for
−1 < q < 1. It is positive semi-definite for q = −1, 1.
k·kq
Proposition 6.55. Let Fq (H) = Falg (H)
. On this space, (a+ (f ))∗ = a− (f ), and so
X(f ) = a+ (f ) + a− (f )
is symmetric.
Proposition 6.56 (Wick formula). Let f be a unit vector. The distribution of X(f ) is the q-Gaussian
distribution. Its moments are
X
X
w(path) =
q cr(π) ,
X(f )2n Ω, Ω =
labeled Dyck paths
π∈P 2 (2n)
where for a general π ∈ P(n), cr (π) is the number of reduced crossings of the partition π,
n
o
π
π
cr (π) = (i < j < k < l) : i ∼ k, j ∼ l, both consecutive in their blocks. Example 6.57. cr ({(1, 3, 6)(2, 4)(5, 7)}) = 3. See Figure 6.
Proof. See Figure 6. To show: a bijection between P 2 (2n) and labeled Dyck paths with 2n steps. Openers
correspond to rising steps, closers to falling steps. Note that for each i, the number of open blocks before
i is equal to the height of the path on step i. Thus if on the i’th step, we close the j’th open block, that
step is labeled by q j−1 .
Corollary 6.58.
X
q inv(α) = [n]q ! = [n]q [n − 1]q . . . [1]q .
α∈Sym(n)
Proof. For kf k = 1,
X
q inv(α) = f ⊗n , f ⊗n q = (a− (f ))n f ⊗n , Ω q .
α∈Sym(n)
75
Definition 6.59. For a measure µ, its q-cumulants are defined via
Y (q)
X
r|V | .
mn [µ] =
q cr(π)
π∈P(n)
(0)
V ∈π
(1)
Note: rn = rn , rn = kn .
Theorem 6.60. If ρ is conditionally positive, there exists a positive measure µ such that
(q)
rn+2 [µ] = ρ [xn ] .
Denote all measures arising in this way by IDq .
Corollary 6.61. If define µ q ν by
rn(q) [µ q ν] = rn(q) [µ] + r(q) [ν],
then q is a well-defined operation on IDq .
Question 6.62. Is q a well-defined operation on D(1)?
Remark 6.63. As in Exercise 6.16, let
Γq (H) = W ∗ (X(f ) : f ∈ HR ),
where X(f ) are now operators defined in Proposition 6.55. It is a spectacular result of (Guionnet,
Shlyakhtenko 2012), based on previous work (Dabrowski 2010), that Γq (H) ' L(Fdim H ) for sufficiently
small q (depending on dim H).
76
Chapter 7
Free Lévy processes.
This chapter was omitted in the course, so what follows is only an outline. See [GSS92].
Definition 7.1. In (A, ϕ), a family of random variables {X(t) : t ≥ 0} is a free Lévy process if
a. It has free increments: for all 0 < t1 , < t2 < . . . < tn ,
{X(t1 ), X(t2 ) − X(t1 ), . . . , X(tn ) − X(tn−1 )}
are free.
b. It has stationary increments: for s < t,
µX(t)−X(s) = µX(t−s) .
Assuming X(0) ∼ δ0 and denoting µX(t) = µt , it follows that {µt : t ≥ 0} form a -semigroup.
c. The semigroup of distributions is continuous at t = 0.
Remark 7.2. Can construct such processes as injective limits of free products. More explicitly, start with
a -semigroup {µt : t ≥ 0}. Let ρ = Rµ , Rµt = tρ. Let H be the Hilbert space and (λ, ξ, T ) the triple
from Remark 6.28. Let K = H ⊗ L2 [0, ∞). For I ⊂ [0, ∞) a finite half-open interval, denote
λI = λ |I| ,
ξI = ξ ⊗ 1I ,
TI (ζ ⊗ f ) = (T ζ) ⊗ (1I f ).
On F(H ⊗ L2 [0, ∞)), let
X(I) = l(ξI ) + l∗ (ξI ) + Λ(TI ) + λ |I| .
Then whenever I1 , I2 , . . . , In are disjoint, X(I1 ), X(I2 ), . . . , X(In ) are free, and µX(I) = µ|I| . So get
X(t) = X(1[0,t) ) : t ≥ 0 .
More generally, can construct {X(f ) : f ∈ L2 [0, ∞) ∩ L∞ [0, ∞)}. For the free Brownian motion, only
need f ∈ L2 [0, ∞).
77
Markov property.
See [Bia98].
For free Lévy processes, in particular for the free Brownian motion, have free stochastic calculus etc.
Here we only show that such a process has the Markov property.
Definition 7.3. Let (A, ϕ) be a C ∗ -probability space, B ⊂ A a C ∗ -subalgebra. The map ϕ [·|B] : A → B
is a conditional expectation if it is self-adjoint and for any a ∈ A and b1 , b2 ∈ B,
ϕ [b1 ab2 |B] = b1 ϕ [a|B] b2 .
It does not always exist. If ϕ is a trace, can choose a conditional expectation so that moreover ϕ [ϕ [a|B]] =
ϕ [a]. Then
ϕ [ϕ [a|B] b] = ϕ [ϕ [ab|B]] = ϕ [ab] .
Thus ϕ [a|B] ∈ B and hb, ϕ [a|B]iϕ = hb, aiϕ . For ϕ faithful, ϕ-preserving conditional expectation is
unique. Again, it may not exist.
Definition 7.4. A process {X(t) : t ≥ 0} is a Markov process if
ϕ [f (X(t))|C ∗ (X(r), r ≤ s)] ∈ C ∗ (X(s)).
Theorem 7.5.
a. A free Lévy process is a Markov process.
b. For t ≥ s,
∗
1
1
C (X(r) : r ≤ s) =
ϕ
.
µ
t
1 − X(t)z + R (z)
1 − X(s)z + Rµs (z)
That is, t 7→
1
1−X(t)z+Rµt (z)
is a martingale. Need to be more precise on what type of object z is.
There are deep conceptual reasons for the Markov property of a process with freely independent increments discovered by Voiculescu.
Free Appell polynomials.
See [Ans04].
Definition 7.6. The linear map
∂xi : Chx1 , . . . , xd i → Chx1 , . . . , xd i ⊗ Chx1 , . . . , xd i,
78
defined by
∂xi (xu(1) xu(2) . . . xu(n) ) =
X
xu(1) . . . xu(j−1) ⊗ xu(j+1) . . . xu(n)
u(j)=i
is the free difference quotient. It is so named because in one variable,
n
∂x x =
n−1
X
i
x ⊗x
n−i−1
n−1
X
↔
i=0
xi y n−i−1 =
i=0
and so
∂x p(x) =
xn − y n
,
x−y
p(x) − p(y)
.
x−y
Definition 7.7. In an n.c. probability space (A, ϕ), define the free Appell polynomials to be multilinear
maps A(X1 , X2 , . . . , Xn ) from A × A × . . . × A to A such that
a. Each A(X1 , X2 , . . . , Xn ) is a polynomial in X1 , . . . , Xn .
b. A(∅) = 1 and
∂Xi A(X1 , X2 , . . . , Xn ) = A(X1 , . . . , Xi−1 ) ⊗ A(Xi+1 , . . . , Xn ).
c. ϕ [A(X1 , X2 , . . . , Xn )] = 0 for all n ≥ 1.
RFor A = C[x], these are ordinary polynomials An (x), such that ∂x An (x) =
An (x) dµ(x) = 0 for n ≥ 1.
P
Ai (x) ⊗ An−i−1 (x) and
Theorem 7.8.
a.
∞
X
An (x)z n =
n=0
1
,
1 − xz + Rµ (z)
and a similar result holds in the multivariate case.
b.
X0 A(X1 , X2 , . . . , Xn ) = A(X0 , X1 , . . . , Xn ) +
n
X
R[X0 , X1 , . . . , Xj ]A(Xj+1 , . . . , Xn ).
j=1
c. If A1 , . . . , Ad ⊂ A are free,
X1,1 , X1,2 , . . . , X1,u(1) ∈ Av(1) ,
...,
Xk,1 , Xk,2 , . . . , Xk,u(k) ∈ Av(k)
for v(1) 6= v(2) 6= . . ., then
A(X1,1 , . . . , X1,u(1) , X2,1 , . . . , X2,u(2) , . . . , Xk,1 , . . . , Xku(k) )
= A(X1,1 , . . . , X1,u(1) )A(X2,1 , . . . , X2,u(2) ) . . . A(Xk,1 , . . . , Xku(k) ).
79
Proof. For (a), if F (x, z) =
An (x)z n , then µ[F ] = 1 and
X
∂x F (x, z) =
Ai ⊗ An−1−i z n = zF ⊗ F.
P
n
n,i
1
These conditions uniquely determine F (x, z) = 1−xz+R
µ (z) . The proof of part (b) uses a similar technique.
For part (c), we show that the factored polynomials also satisfy the initial conditions and the recursion
from part (b).
Proposition 7.9. In a tracial C ∗ -probability space (A, ϕ), if B ⊂ A, Y ∈ B, and X is free from B, then


ϕ A(X + Y, X + Y, . . . , X + Y ) |B  = A(Y, Y, . . . , Y ).
|
{z
}
n
Proof. ϕ is tracial, so the corresponding conditional expectation exists. For any b ∈ B,
hX
i
ϕ [A(X + Y, . . . , X + Y )b] = ϕ
A(X or Y, . . . , X or Y )b
Using the preceding theorem, freeness, and the fact that the free Appell polynomials are centered, we
continue the expression as
X
=
ϕ [A(X, . . . , X)A(Y, . . . , Y )A(X, . . . , X) . . . A(X or Y )b] = ϕ [A(Y, Y, . . . , Y )b] .
The result follows from the second characterization of conditional expectations.
Proof of Theorem 7.5. We are interested in ϕ [P (X(t))|C ∗ (X(r), r ≤ s)]. Since they span the space, it
suffices to show this for P = An . In that case,
ϕ [An (X(t))|C ∗ (X(r), r ≤ s)] = ϕ [An (X(s) + (X(t) − X(s)))|C ∗ (X(r), r ≤ s)] = An (X(s))
by the preceding proposition and free independence of increments.
80
Chapter 8
Free multiplicative convolution and the
R-transform.
See Lectures 12, 14 of [NS06]. In the preceding chapters, we tried to emphasize the similarity and parallels
between classical and free probability, through the use of all/non-crossing partitions, symmetric/full Fock
space, and infinitely divisible distributions. By contrast, the phenomena described in this and the following
chapters are special to free probability.
Remark 8.1. Recall: if a, b are free, µab = µa µb , the free multiplicative convolution. Note that if
{a1 , . . . , an } and {b1 , . . . , bn } are free,
X
ϕ [a1 b1 a2 b2 . . . an bn ] =
Rπ [a1 , b1 , a2 , b2 , . . . , an , bn ]
π∈NC(2n)
=
X
Rσ [a1 , a2 , . . . , an ]
X
Rσ [a1 , a2 , . . . , an ]
X
X
Rτ [b1 , b2 , . . . , bn ]
τ ∈NC(n)
τ ≤K[σ]
σ∈NC(n)
=
Rτ [b1 , b2 , . . . , bn ]
τ ∈NC(n)
σ∪τ ∈NC(2n)
σ∈NC(n)
=
X
Rσ [a1 , a2 , . . . , an ]MK[σ] [b1 , b2 , . . . , bn ].
σ∈NC(n)
In particular, mn [µab ] =
P
σ∈NC(n)
Rσ [a]MK[σ] [b].
Compound Poisson realization.
Recall that if s is a semicircular variable, then s2 has the free Poisson distribution, all of whose free
cumulants are 1.
81
Lemma 8.2. Let a1 , . . . , ak be self-adjoint, orthogonal random variables in a n.c. probability space
(A, τ ), in the sense that ai aj = 0 for i 6= j. Let s be a semicircular variable free from {ai : 1 ≤ i ≤ k}.
Then
{sa1 s, sa2 s, . . . , sak s}
are freely independent and have free compound Poisson distributions.
See Example 12.19 in [NS06] for the proof in the non-tracial case.
Proof in the tracial case.
M [sau(1) s, sau(2) s, . . . , sau(n) s] = τ sau(1) ssau(2) s . . . sau(n) s = τ au(1) s2 au(2) s2 . . . au(n) s2
X
=
Mπ [au(1) , au(2) , . . . , au(n) ]RK[π] [s2 , s2 , . . . , s2 ]
π∈NC(n)
=
X
Mπ [au(1) , au(2) , . . . , au(n) ].
π∈NC(n)
It follows that
R[sau(1) s, sau(2) s, . . . , sau(n) s] = M [au(1) , au(2) , . . . , au(n) ] = τ au(1) au(2) . . . au(n) = 0
unless all u(1) = u(2) = . . . = u(n). Thus {sa1 s, sa2 s, . . . , sak s} are free. Moreover,
rn [sai s] = R[sai s, sai s, . . . , sai s] = M [ai , ai , . . . , ai ] = mn [ai ],
so sai s has a free compound Poisson distribution.
Example 8.3. Let A = C ∗ (s) ∗ L∞ [0, 1], with the free product state τ = ϕ ∗
R1
0
· dx. Then the map
t 7→ s1[0,t) s
is a free Poisson process for t ∈ [0, 1]. Indeed, for I ∩ J = ∅, 1I 1J = 0, so the process has freely
independent increments, and
rk [s1I s] = mk [1I ] = |I|
so s1I s has distribution π|I| .
Free compression.
Proposition 8.4. Let a1 , . . . , ad be in a n.c. probability space (A, τ ). Choose p freely independent from
them, where p is a projection, p = p2 = p∗ . Denote α = τ [p]. Define a new (compressed) n.c. probability
space
e τe) = (pAp, 1 τ |pAp ).
(A,
α
Then
1
Rτe[pau(1) p, pau(2) p, . . . , pau(n) p] = Rτ [α au(1) , α au(2) , . . . , α au(n) ].
α
82
See Theorem 14.10 in [NS06] for the proof in the non-tracial case.
Proof in the tracial case.
1 τ
M [au(1) p, au(2) p, . . . , au(n) p]
α
1 X
=
Rπτ [au(1) , au(2) , . . . , au(n) ]MK[π] [p, p, . . . , p]
α
M τe[pau(1) p, pau(2) p, . . . , pau(n) p] =
π∈NC(n)
1
=
α
=
X
Rπτ [au(1) , au(2) , . . . , au(n) ]αn+1−|π|
π∈NC(n)
X 1 |π|
Rπτ [α au(1) , α au(2) , . . . , α au(n) ].
α
π∈NC(n)
The conclusion follows.
Corollary 8.5. For arbitrary µ ∈ D(d) and arbitrary t ≥ 1, the expression µt is well defined, in the
sense that there exists µt ∈ D(d) with Rµt = tRµ .
Proof. Choose a1 , . . . , ad and p as in the proposition so that τa1 ,...,ad = Dt µ and τ [p] =
(pAp, τe), (pa1 p, pa2 p, . . . , pad p) has distribution µt .
1
.
t
Then in
Corollary 8.6. The Bercovici-Pata bijection ID∗ → ID cannot be extended to general measures.
Proof. Such an extension T would in particular have the property that T −1 [T [µ]t ] = µ∗t . However,
taking µ to be the Bernoulli distribution, the right-hand-side is defined only for integer t, while the lefthand-side is defined for any t ≥ 1.
Exercise 8.7. Let p, q ∈ (A, τ ) be freely independent projections in a (for convenience) tracial n.c.
probability space, with τ [p] = α, τ [q] = β. Note that the distributions of pq, pqp, and qpq are all the
same. Use the results above, and either the R-transforms or an extension of Remark 6.49 to compute
the distribution of pq. The computation of the moment generating function or the Cauchy transform is
sufficient, you do not need to find the actual measure. See Example 13.5 for the answer.
Outline of the argument: for p as above and any a free from it, in the notation of Proposition 8.4
e
= (µταa )(1/α) .
µτpap
On the other hand,
e
µpap = µτpap = (1 − α)δ0 + αµτpap
.
Thus
µpap = (1 − α)δ0 + αµ(1/α)
αa
83
and
1
Gpap (z) = (1 − α) + αGµ(1/α)
(z).
αa
z
Finally,
1
Rαa (z) = Ra (αz).
α
, and finally Gpqp .
, then Gµ(1/α)
Using these results, we compute Gq , then Rq , then Rµ(1/α)
αq
αq
Rµαa
(1/α) (z) =
See Lecture 14 of [NS06] for related results about compression by matrix units, very useful in applications
of free probability to the study of operator algebras.
84
Chapter 9
Belinschi-Nica evolution.
See [BN09].
Remark 9.1. Recall that
µ -infinitely divisible
⇔
Rµ = ρ is conditionally positive,
µ ] -infinitely divisible
⇔
B µ = ρ is conditionally positive,
while
and any µ is ]-infinitely divisible. So we can define a map
B : D(d) → ID
via RB[µ] = B µ . This is the Boolean Bercovici-Pata bijection.
Relation between free and Boolean cumulants.
Recall
M [a1 , a2 , . . . , an ] =
X
Rπ [a1 , a2 , . . . , an ]
π∈NC(n)
while
M [a1 , a2 , . . . , an ] =
X
Bπ [a1 , a2 , . . . , an ]
π∈Int(n)
Corollary 9.2. Denote
n
o
π
f
NC(n)
= π ∈ NC(n) : 1 ∼ n .
Then
B[a1 , a2 , . . . , an ] =
X
g
π∈NC(n)
85
Rπ [a1 , a2 , . . . , an ]
86
Proof. Figure omitted. Group the blocks of π ∈ NC(n) according to the smallest σ ∈ Int(n) such that
π ≤ σ. Then the restriction of π to each block of σ connects the minimum and the maximum of that
block.
Example 9.3. See Figure 9.
B[a1 , a2 , a3 , a4 ] = R[a1 , a2 , a3 , a4 ] + R[a1 , a2 , a4 ]R[a3 ] + R[a1 , a3 , a4 ]R[a2 ]
+ R[a1 , a4 ]R[a2 , a3 ] + R[a1 , a4 ]R[a2 ]R[a3 ].
Corollary 9.4.
B B[µ] [xu(1) , xu(2) , . . . , xu(n) ] =
X
RπB[µ] [xu(1) , xu(2) , . . . , xu(n) ] =
g
π∈NC(n)
X
Bπµ [xu(1) , xu(2) , . . . , xu(n) ]
g
π∈NC(n)
Definition 9.5. On NC(n), define a new partial order:
σπ
⇔
σ
σ ≤ π and ∀ V ∈ π, min(V ) ∼ max(V ).
Example 9.6.
f
NC(n)
= σ ∈ NC(n) : σ 1̂n .
f ). For another example, see Figure 9.
Note also that if σ π and V ∈ π, then σ|V ∈ NC(V
f
Definition 9.7. Denote by Out(σ) the outer blocks of σ. If σ ∈ NC(n),
it has a unique outer block o(σ).
Inn(σ) are the inner blocks of σ.
Proposition 9.8.
a. For π ∈ NC(n),
{σ ∈ NC(n) : σ π} ' {σ ∈ NC(n) : π0 ≤ σ ≤ π} = [π0 , π] ⊂ NC(n).
b. For σ ∈ NC(n),
{π ∈ NC(n) : σ π} ' {S ⊂ σ : Out(σ) ⊂ S} .
The isomorphism is implemented by the map
π = (V1 , V2 , . . . , Vk ) 7→ {o(σ|V1 ), o(σ|V2 ), . . . , o(σ|Vk )} .
f
In particular, for σ ∈ NC(n),
{π ∈ NC(n) : σ π} ' {S ⊂ σ : o(σ) ∈ S}.
Proof. See Figure 9 for an illustration. For part (a), let π = (V1 , V2 , . . . , Vk ) and
π0 = {(min(V1 ), max(V1 )), (min(V2 ), max(V2 )), . . . , singletons} .
π0 ∈ NC1,2 (n). Then π0 π and σ π if and only if π0 ≤ σ ≤ π.
For part (b), we need to show that the map is a bijection. We construct the inverse map as follows. Let
Out(σ) ⊂ S ⊂ σ. For V ∈ σ, choose U ∈ S with the largest depth such that V ≤ U (see the illustration).
Since S contains the outer blocks, U exists. In π, adjoin V to U . Get π ∈ NC(n), σ ≤ π, σ π, each
block of π contains a unique element of S.
87
Corollary 9.9.
X
t|π|−|Out(σ)| s|σ|−|π| = (t + s)|Inn(σ)| .
π∈NC(n)
σπ
f
In particular, if σ ∈ NC(n),
then
X
t|π|−1 s|σ|−|π| = (t + s)|σ|−1 .
π∈NC(n)
σπ
Proof.
X
X
t|π|−|Out(σ)| s|σ|−|π| =
π∈NC(n)
σπ
t|S|−|Out(σ)| s|σ|−|S|
Out(σ)⊂S⊂σ
|Inn(σ)| =
|Inn(σ)| i |σ|−i
ts
= (t + s)|Inn(σ)| .
i
X
i=0
Bt transformation.
Definition 9.10. For t ∈ R, define the linear map (Belinschi-Nica remarkable transformation)
Bt : D(d) → Chx1 , . . . , xd i∗
by
X
B Bt [µ] [xu(1) , xu(2) , . . . , xu(n) ] =
t|π|−1 Bπµ [xu(1) , xu(2) , . . . , xu(n) ].
g
π∈NC(n)
Remark 9.11. B1 = B, while B0 is the identity transformation.
Proposition 9.12. For t ≥ 0,
Bt [µ] = µ(1+t)
1
] 1+t
.
In particular, the Boolean Bercovici-Pata bijection is given by the formula B[µ] = µ2
] 21
Corollary 9.13. For t ≥ 0, Bt maps D(d) to itself, and so preserves positivity.
Proof of the Proposition.
B Bt [µ]
](1+t)
= (1 + t)B Bt [µ]
(by definition of ])
X
= (1 + t)
t|π|−1 Bπµ
(by definition of Bt )
g
π∈NC(n)
= (1 + t)
X
g
π∈NC(n)
t|π|−1
Y
X
V ∈π σi ∈NC(V
g i)
88
Rσµi
(by Corollary 9.2).
.
Let
σ = σ1 ∪ σ2 ∪ . . . ∪ σ|π| .
f
Then σ ≤ π, σ π, and σ ∈ NC(n).
So we can continue the expression above as
X X
= (1 + t)
t|π|−1 Rσµ
σπ
g
σ∈NC(n)
X
= (1 + t)
(1 + t)|σ|−1 Rσµ
(by Corollary 9.9)
g
σ∈NC(n)
X
=
(1 + t)|σ| Rσµ
g
σ∈NC(n)
X
=
Rσµ
(1+t)
(by definition of )
g
σ∈NC(n)
(1+t)
= Bµ
(by Corollary 9.2).
We conclude that
Bt [µ]](1+t) = µ(1+t) .
Proposition 9.14. {Bt : t ∈ R} form a group,
Bt ◦ Bs = Bt+s ,
with the identity element B0 and B−1
t = B−t . In particular, {Bt : t ≥ 0} form a semigroup.
Remark 9.15. It follows that
µ
1
(1+s) ] 1+s
1
(1+t) ] 1+t
= µ(1+t+s)
1
] 1+t+s
,
so there is a commutation relation between the Boolean and free convolution powers.
Proof of the proposition.
B (Bt ◦Bs )[µ] =
X
t|π|−1 BπBs [µ]
(by definition of Bt )
g
π∈NC(n)
=
X
t|π|−1
g
π∈NC(n)
=
X X
Y
X
s|σi |−1 Bσµi
(by definition of Bs )
Vi ∈π σi ∈NC(V
g i)
t|π|−1 s|σ|−|π| Bσµ
(by definition of )
σπ
g
σ∈NC(n)
=
X
(t + s)|σ|−1 Bσµ
(by Corollary 9.9)
g
σ∈NC(n)
= B Bt+s [µ]
(by definition of Bt+s ).
89
Exercise 9.16. Prove that we could also have defined the same Bt in terms of the free cumulants via
X
RBt [µ] [xu(1) , xu(2) , . . . , xu(n) ] =
t|π|−1 Rπµ [xu(1) , xu(2) , . . . , xu(n) ].
g
π∈NC(n)
(Combine Remark 9.1, Definition 9.10, Remark 9.11, and Proposition 9.14.)
Φ transformation.
Definition 9.17. Define the linear map
Φ : D(d) → Chx1 , . . . , xd i∗
by
B
Φ[µ]
(z1 , z2 , . . . , zd ) =
d
X
zi M µ (z1 , z2 , . . . , zd )zi .
i=1
Equivalently, B
Φ[µ]
[xi ] = 0, and
B Φ[µ] [xu(1) , xu(2) , . . . , xu(n) ] = δu(1)=u(n) M µ [xu(2) , . . . , xu(n−1) ].
Remark 9.18. Recall that in one variable,
1
M µ (z) =
γ0 z 2
1 − β0 z −
1 − β1 z −
Thus
B
Φ[µ]
γ1 z 2
γ2 z 2
1 − β2 z −
1 − ...
z2
(z) =
γ0 z 2
1 − β0 z −
1 − β1 z −
γ1 z 2
1 − β2 z −
γ2 z 2
1 − ...
and so
M Φ[µ] (z) =
1
=
1 − B Φ[µ] (z)
1
z2
1−
γ0 z 2
1 − β0 z −
1 − β1 z −
γ1 z 2
1 − β2 z −
90
γ2 z 2
1 − ...
In other words,
0, β0 β1 β2 . . .
J(Φ[µ]) =
1, γ1 γ2 γ3 . . .
It follows that in this case, the coefficient stripping operation of Exercise 6.50 is the left inverse of Φ.
Exercise 9.19. Prove, from the definition of Φ, that for µ ∈ D(d), the functional B Φ[µ] is conditionally
positive. Conclude that Φ preserves positivity.
Evolution equation.
Theorem 9.20. For any ρ ∈ D(d) and any t ≥ 0,
Bt [Φ[ρ]] = Φ[ρ γt ],
where γt = S(0, t) for d = 1, and in general is the distribution of a free semicircular system with
covariance tI.
Proof. To keep the notation short, we will denote
RU = R[xu(j) : j ∈ U ].
B Bt [Φ[ρ]] =
X
t|π|−1 BπΦ[ρ] [xu(1) , . . . , xu(n) ]
(by definition of Bt )
g
π∈NC(n)
=
X
t|π|−1
=
t|π|−1
=
g
π∈NC(n)
(by definition of Φ)
Y
γ1
R{min(V
MVρi \{min(Vi ),max(Vi )}
i ),max(Vi )}
Vi ∈π
g
π∈NC(n)
X
δu(min(Vi ))=u(max(Vi )) MVρi \{min(Vi ),max(Vi )}
Vi ∈π
g
π∈NC(n)
X
Y
t|π|−1
Y
X
γ1
Ro(σ
i)
Vi ∈π σi ∈NC(V
g i)
(by the moment-cumulant formula and because
Y
RUρ
U ∈Inn(σi )
γ1
Ro(σ
i)
91
(
δu(min(Vi )=u(max(Vi )) , |o(σi )| = 2,
=
)
0,
otherwise
=
X
g
π∈NC(n)
=
Y
t−1
X
Vi ∈π σi ∈NC(V
g i)
X X
Y
t−1
=
X
X
t−1
=
X
=
X
i
RUρ
)
(by definition of )
U ∈Inn(σ|Vi )
Y
RUγt
Y
RUρ
(by Proposition 9.8)
U 6∈S
U ∈Inn(σ)
Y
γt
t−1 Ro(σ)
RUργt
(by definition of )
U ∈Inn(σ)
g
σ∈NC(n)
=
(because Rγt = tRγ1 )
Y γ
ρ
t
RU + RU
γt
t−1 Ro(σ)
g
σ∈NC(n)
X
Y
U ∈S
g
o(σ)∈S⊂σ
σ∈NC(n)
RUρ
U ∈Inn(σi )
γt
Ro(σ|
V
Vi ∈π
σπ
g
σ∈NC(n)
Y
γt
Ro(σ
i)
γ1
Ro(σ)
g
σ∈NC(n)
= δu(1)=u(n) M
Y
RUργt
U ∈Inn(σ)
ργt
[xu(2) , . . . , xu(n−1) ] = B Φ[ργt ] .
Remark 9.21 (A 2012). Using
Bt [ρ] = (Bt−1 ◦ B)[ρ] = B[ρ]t
] 1t
,
the identity Bt [Φ[ρ]] = Φ[ρ γt ] implies
B[Φ[ρ]]t = Φ[ρ γt ]]t .
Denote µ = B[Φ[ρ]]. Then µ is -infinitely divisible, and
µt = µt = Φ[ρ γt ]]t .
What is ρ? In the single-variable case,
ρ
.
RnB[Φ[ρ]] = BnΦ[ρ] = Mn−2
Thus ρ is the generator of the -semigroup {µt : t ≥ 0}. This calculation shows that any free convolution
semigroup (with finite variance), once stripped, gives a semicircular evolution.
92
Chapter 10
∗-distribution of a non-self-adjoint element
See Lectures 1, 2, 11 of [NS06].
Definition 10.1. For a ∈ (A, ϕ), its ∗-distribution is the joint distribution of (a, a∗ ), a state on Chx, x∗ i.
Notation 10.2. Let
ε(i) ∈ {∗, ·}
~ε = (ε(1), . . . , ε(n)),
Denote
P 2 (~ε) = {π ∈ P 2 (n) : each block of π connects a ∗ and a ·} .
See Figure 10.
Example 10.3. Let u be a Haar unitary, so that
ϕ[un ] = δn=0 ,
n ∈ Z.
Write
u~ε = uε(1) uε(2) . . . uε(n) ,
so that each term is either u∗ or u. Then the ∗-distribution is of u is
(
~ε
|{i:ε(i)=·}|−|{i:ε(i)=∗}| ~ε
1, |{i : ε(i) = ·}| = |{i : ε(i) = ∗}| ,
µ x =ϕ u =ϕ u
=
0, otherwise.
More pictorially, we can identify ~ε with a lattice path, with ∗ corresponding to a rising step (1, 1), and · to
a falling step (1, −1). See Figure 10. Note that the paths start at height zero, but are not necessarily Dyck
paths. In this language,
(
~ε
1, path ends at height 0,
µ x =
0, otherwise.
What are their free cumulants? See Lecture 15 of [NS06].
93
94
Example 10.4. Let ` be a free creation operator, `∗ ` = 1. The ∗-distribution of ` is
(
(
~ε
~ε
1, can reduce the word `~ε to 1,
1, ~ε corresponds to a Dyck path,
µ` x = ϕ ` =
=
0, otherwise.
0, otherwise.
See Figure 10. It follows that the joint free cumulants of `, `∗ are R[`∗ , `] = 1, with the rest all equal to
zero.
Definition 10.5. An operator c is a standard circular element if c =
standard semicircular.
√1 (s1
2
+ is2 ), where s1 , s2 are free
Example 10.6. Let c be a standard circular element. What is its ∗-distribution? Clearly its free cumulants
are
1
R[c, c∗ ] = R[c∗ , c] = ϕ [(s1 + is2 )(s1 − is2 )] = 1,
2
with the rest all equal to zero. What are the ∗-moments? Using Notation 10.2,
µc x~ε = |NC2 (~ε)| =?
Proposition 10.7. c∗ c has a free Poisson distribution.
Note: this is not obvious, as
i
1
c∗ c = (s21 + s22 ) + (s1 s2 − s2 s1 ).
2
2
Proof. To show:
ϕ [(c∗ c)n ] = s2n ,
where s is standard semicircular. Indeed, using the non-crossing property (see Figure 10),
ϕ [(c∗ c)n ] = |NC2 (∗, ·, ∗, ·, . . . , ∗, ·)| = |NC2 (2n)| = ϕ s2n .
Remark 10.8. In fact (Oravecz 2001, Kemp, Speicher 2007, Schumacher, Yan 2012),
∗ p p n
ϕ [((c ) c ) ] = NC2 (∗, . . . , ∗, ·, . . . , ·, . . . , ∗, . . . , ∗, ·, . . . , ·) = Fuss-Catalan number
| {z } | {z }
| {z } | {z } p
p
p
p
from Proposition 5.12. What about more general ~ε?
Exercise 10.9. Let c1 , . . . , cp are freely independent circular elements. Use the idea from Proposition 10.7
and the preceding remark to compute
n ϕ c∗p . . . c∗2 c∗1 c1 c2 . . . cp
.
95
Example 10.10. Let f1 ⊥ f2 be unit vectors and X(f1 ), X(f2 ) corresponding q-Gaussian elements from
Proposition 6.55.
1
Y = √ (X(f1 ) + iX(f2 ))
2
is a q-circular element. Using the q-Wick formula (Proposition 6.56),
X
ϕ Y ~ε =
q cr(π) .
π∈P 2 (~
ε)
Know very little about them.
Question 10.11. Compute
1/n

kY k2 = kY ∗ Y k = lim ϕ [(Y ∗ Y )n ]1/n = lim 
n→∞
X
n→∞
π∈P 2 (∗,·,∗,·,...,∗,·)
96
q cr(π) 
.
Chapter 11
Combinatorics of random matrices.
See Lectures 22, 23 of [NS06], Chapter 7 of [Gui09].
11.1
Gaussian random matrices.
Remark 11.1 (Random matrices). Recall from Example 1.9: let (Ω, P ) be a probability space, and
\
L∞− (Ω, P ) =
Lp (Ω, P )
p≥1
the algebra of random variables all of whose moments are finite. On this algebra we have the expectation
functional
Z
Z
E[f ] =
f dP =
x dµf (x),
Ω
R
where µf is the distribution of f . For each N , we have the algebra of N × N random matrices
MN (C) ⊗ L∞− (Ω, P ) ' MN (L∞− (Ω, P )).
We can identify A ∈ MN (L∞− (Ω, P )) with a N 2 -tuple of random variables, so it has a N 2 -dimensional
distribution. But also: for
N
1 X
1
tr[A] =
Tr[A] =
Aii
N
N i=1
the normalized trace (note: itself a random variable)
(MN (C) ⊗ L∞− (Ω, P ), E ⊗ tr) = (MN (L∞− (Ω, P )), E ◦ tr)
is an n.c. probability space, so A also has a 1-dimensional distribution as an element in this space.
97
98
99
Definition 11.2 (Gaussian random variables). X has the Gaussian (or normal) distribution N (0, t) with
mean zero and variance t if
1 −x2 /2t
dµX (x) = √
e
dx.
2πt
(X1 , . . . , Xd ) are jointly Gaussian with mean 0 and (positive definite) covariance matrix Q = (Qij )di,j=1 if
!
d
X
1
xi (Q−1 )ij xj /2 dx1 . . . dxd .
dµX1 ,X2 ,...,Xd (x1 , . . . , xd ) = exp −
Z
i,j=1
Remark 11.3. Z is the normalization constant, in this case
Z
exp −
Z=
Rd
d
X
!
xi (Q−1 )ij xj /2
dx1 . . . dxd .
i,j=1
We will use the same notation for other normalization constants.
In particular, X1 , . . . , Xd are independent Gaussian if all Qij = 0 for i 6= j (in the jointly Gaussian case,
uncorrelated implies independent).
Exercise 11.4. Show that the (classical) cumulant generating function of a jointly Gaussian distribution
is
d
1X
`(z1 , . . . , zd ) =
zi Qij zj .
2 i,j=1
Hint: denote x = (x1 , . . . , xd )t , and the same for y, z. Re-write the jointly Gaussian density as
dµ(x) =
1 −xt Q−1 x/2
e
dx.
Z
Recall from Remark 5.7 that
µ
Z
` (z) = log
t
ex z dµ(x)
R
(why?). Now use the change of variables x = y + Qz.
Corollary 11.5 (Wick formula). If X1 , . . . , Xd are jointly Gaussian,
X Y E Xu(1) Xu(2) . . . Xu(n) =
E Xu(i) Xu(j) ,
π∈P 2 (n) (i,j)∈π
where P 2 (n) are the pair partitions. Equivalently, their classical cumulants are
(
0,
n 6= 2,
k[Xu(1) , . . . , Xu(n) ] =
E[Xu(1) Xu(2) ], n = 2.
100
Remark 11.6 (Complex Gaussians). A complex-valued random variable Y is complex Gaussian if <Y ,
=Y are independent (centered) Gaussian. We will only consider <Y ∼ N (0, a), =Y ∼ N (0, a), with the
same variance. Note that
E[Y ] = 0
and
E[Y 2 ] = E[(<Y )2 − (=Y )2 + 2i(<Y )(=Y )] = a − a = 0.
However,
E[Y Ȳ ] = E[(<Y )2 + (=Y )2 ] = 2a.
Moreover, if Y1 , . . . , Yd are independent complex Gaussian (i.e. their real and imaginary parts are independent), by multi-linearity of classical cumulants all of their joint cumulants are zero except k[Yi , Yi∗ ] =
k[Yi∗ , Yi ] = 2ai .
Corollary 11.7 (Wick formula). Using Notation 10.2, for independent complex Gaussian Y1 , . . . , Yd ,
i
h
X Y h ε(i) ε(j) i
ε(n)
ε(1)
E Yu(i) Yu(j) .
E Yu(1) . . . Yu(n) =
π∈P 2 (~
ε) (i,j)∈π
Remark 11.8. The distribution of Y as a C ' R2 -valued random variable is
1 −(x2 +y2 )/2a
e
dx dy.
Z
Definition 11.9. A random N × N matrix X is a GUE matrix (Gaussian unitary ensemble) if
a. X is self-adjoint, Xij = X̄ji , each Xii real.
b. {Xii , <Xij , =Xij } for 1 ≤ i ≤ N , i < j are independent Gaussian.
1
c. Xii ∼ N (0, N1 ), <Xij , =Xij ∼ N (0, 2N
). In other words,
E [Xij Xkl ] =
Note that
1
δi=l δj=k .
N
N
1 X
E ◦ tr[X ] =
E[Xij Xji ] = 1.
N i,j=1
2
Remark 11.10. We can identify the space of self-adjoint N × N matrices with
2
{(xii : 1 ≤ 1 ≤ N, <xij , =xij : 1 ≤ i < j ≤ N )} ' RN .
Since
N
Y
i=1
2
e−xii /(2/N )
Y
e−((<xij )
2 +(=x )2 )/(1/N )
ij
i<j
101
= e−
P
i,j |xij |
2
/(2/N )
= e−N Tr[X
2 ]/2
,
2
the RN distribution of a GUE matrix is
N
Y
1 −N Tr[X 2 ]/2 Y
1
2
= e
dxii
d<xij d=xij = e−N Tr[X ]/2 dX.
Z
Z
i=1
i<j
Exercise 11.11.
a. Show that a d-dimensional Gaussian distribution dµ(x) with the identity covariance matrix (so that
its components are independent and have variance 1) is rotationally invariant, in the sense that
dµ(x) = dµ(Ox)
for any orthogonal d × d matrix O. (Maxwell’s theorem asserts that up to rescaling, these are the
only rotationally invariant distributions with independent components).
b. Show that a GUE matrix is unitarily invariant, in the sense that for any N × N unitary matrix U ,
if X is an N × N GUE matrix, so is U XU ∗ . (You can use the definition, or Remark 11.10). See
Remark 11.48.
Example 11.12. We would like to compute
i
1 hX
4
E
E ◦ tr[X ] =
Xi(1)i(2) Xi(2)i(3) Xi(3)i(4) Xi(4)i(1) .
N
Apply the Wick formula:
(
1
, if i(j) = i(k + 1), i(j + 1) = i(k),
E Xi(j)i(j+1) Xi(k)i(k+1) = N
0, otherwise.
Draw a ribbon (fat) 4-star (with a marked edge). See Figure 11. Note that these pictures are dual to those
used in topology. The rule above says we match the two edges with the correct orientation. Thus
X
1
4
E ◦ tr[X ] =
E Xi(1)i(2) Xi(2)i(3) E Xi(3)i(4) Xi(4)i(1) δi(1)=i(3)
N
i(1),i(2),i(3),i(4)
+ E Xi(1)i(2) Xi(4)i(1) E Xi(2)i(3) Xi(3)i(4) δi(2)=i(4)
+ E Xi(1)i(2) Xi(3)i(4) E Xi(2)i(3) Xi(4)i(1) δi(1)=i(4) δi(2)=i(3) δi(2)=i(1)
=
1
1 1
(N 3 + N 3 + N ) = 1 + 1 + 2 .
2
NN
N
What about a general moment E ◦ tr[X 2n ]? It corresponds to a ribbon 2n-star (with a marked edge). See
Figure 11.
In the diagram, identify edges pairwise, in an orientation-preserving way, get a closed surface with a graph
drawn on it. # of vertices = 1. # of edges = n. # of faces (loops) = # free parameters. So for each
π ∈ P 2 (2n), get a term
1 # faces 1
N
= N #F −#E−1 .
n
N
N
102
Remark 11.13. The Euler characteristic of a connected surface is
χ = #F − #E + #V = 2 − 2g,
where g is the genus of the surface. Thus in the preceding example,
N #F −#E−1 = N χ−2 =
1
.
N 2g
Proposition 11.14. For X GUE,
1
X
2n
E ◦ tr[X ] =
π∈P 2 (2n)
N 2g(π)
n
X
1
=
|{π ∈ P 2 (2n) of genus g}| .
N 2g
g=0
Definition 11.15. A map is a connected (ribbon) graph embedded in an (orientable) compact surface in
which all edges are matched in an orientation-preserving way, and each face (loop) is homeomorphic to a
disk. The genus of a map is the genus of a surface into which it is embedded.
Remark 11.16. Note that
Maps with one vertex and n edges
↔
P 2 (2n).
Thus we proved: for X GUE,
n
X
1
|Mg (2n)| .
E ◦ tr[X ] =
N 2g
g=0
2n
In particular, for large N
E ◦ tr[X 2n ] ≈ |M0 (2n)| = #(planar maps) = |NC2 (2n)| = cn = ϕ s2n ,
where s is standard semicircular.
Definition 11.17. (XN )∞
N =1 have asymptotic semicircular distribution if µXN → µs as N → ∞. Thus
XN ∈ (MN (L∞− ), E ◦ tr)
→
s ∈ (A, ϕ)
in distribution. Compare with Definition 1.13. Equivalently, mn (X) → mn (s).
Corollary 11.18. Let X be a GUE matrix. Then
E ◦ tr[X 2n ] → cn = ϕ s2n ,
and so the distribution of a GUE matrix is asymptotically semicircular.
103
Remark 11.19 (Wigner’s theorem). Much stronger statements hold. First, in fact
tr[X 2n ] → ϕ[s2n ] almost surely.
Second, the assumption that X is GUE can be weakened to assuming it is a Wigner matrix: a self-adjoint
matrix for which
N
1 X 2
=0
E [Xij ] = 0,
lim
−
1
N
E
|X
|
ij
N →∞ N 2
i,j=1
and for each n,
n i
h√
sup sup E N Xij < ∞.
N 1≤i,j≤N
P
Third, recall from Exercise 2.23 that for a self-adjoint matrix, its distribution is N1 N
i=1 δλi , where {λi }
are its (real) eigenvalues. Similarly, we may identify the distribution of a random matrix as
µ̂N =
1
(δλ1 + . . . + δλN ) ,
N
the random spectral distribution of X, which is a random measure. Then in fact, for a Wigner matrix, as
N → ∞,
1√
4 − x2 dx,
µ̂N →
2π
the Wigner semicircle law.
The results go back to the work of (Wigner 1958) and (Dyson 1962) on modeling a heavy nucleus, and are
a (weak) example of universality, which is more properly exhibited in the behavior of eigenvalue spacings
rather than the eigenvalues themselves.
Exercise 11.20. Let (A, ϕ) be a n.c. probability space. Let
{si : 1 ≤ i ≤ N } ⊂ A,
{cij : 1 ≤ i < j ≤ N } ⊂ A
be standard semicircular, respectively, standard circular, all free among themselves. Let X ∈ MN (A)
have entries
1
1
1
Xij = √ cij , Xji = √ c∗ij for i < j.
Xii = √ si ,
N
N
N
Prove that for any N , the distribution of XN ∈ (MN (A), ϕ ◦ tr) is (exactly, not asymptotically) standard
semicircular. (Adapt Corollary 11.7 and the argument in Example 11.12 to this setting.)
11.2
Map enumeration.
Change the point of view: think of the matrix integrals as generating functions for maps.
104
Remark 11.21. We know that
n
X
1
E ◦ tr[X ] =
|{maps with 1 vertex, n edges, of genus g}| .
N 2g
g=0
2n
What about
" k
#
Y
E N Tr(X n(1) )N Tr(X n(2) ) . . . N Tr(X n(k) ) = E
N Tr(X n(i) )
i=1
= Nk
X
E Xi(1)i(2) . . . Xi(n(1))i(1) Xi(n(1)+1)i(n(1)+2) . . . Xi(n(1)+n(2))i(n(1)+1) . . . ?
k stars, of types n(1), . . . , n(k), pair off all edges in orientation-preserving ways. See Figure 11 for a
single term from the expansion of E [N Tr(X 4 )N T r(X 3 )N T r(X 3 )]. Get k vertices, with the number of
faces again equal to the number of free parameters. Apply the Wick theorem, each term in the expression
above is of the form
1
N k #E N #F = N #V −#E+#F .
N
In general, if a map has c components,
2c −
c
X
gi = 2c − 2g = #V − #E + #F.
i=1
Thus
"
E
k
Y
#
N Tr(X
n(i)
) =
∞ X
∞
X
g=0 c=1
i=1
1
N 2g−2c
|Gg,c (n(i), 1 ≤ i ≤ k)| ,
where
Gg,c (n(i), 1 ≤ i ≤ k) = {union of c maps built out of stars of type n(1), . . . , n(k), of total genus g}
What is the largest power of N ? Unclear.
Remark 11.22. Next,
E exp
k
X
i=1
!
ti N Tr[X i ]
∞
X
1
=E
n!
n=0
=E
∞
X
n=0
=E
1
n!
∞
X
k
X
!n
ti N Tr[X i ]
i=1
n(1)+...+n(k)=n
X
n=0 n(1)+...+n(k)=n
=
∞ X
∞
X
g=0 c=1
1
N 2g−2c
Y
k
n
n(i)
ti (N Tr[X i ])n(i)
n(1), n(2), . . . , n(k) i=1
X
k
n(i)
Y
ti
(N Tr[X i ])n(i)
n(i)!
i=1
∞
X
n(1),...,n(k)=0
k
n(i)
Y
ti
|Gg,c ((i, n(i)), 1 ≤ i ≤ k)| .
n(i)!
i=1
Note: powers of N may be positive or negative; not very nice.
105
Remark 11.23 (Connected maps). A general combinatorial principle:
X
X
Fdiagram =
diagrams
Fπ .
π∈{partitions of the components of the diagram}
This is the relation between moments and classical cumulants, and we know from Remark 5.7 that these
are also related via
!
∞
∞
X
X
1
1
n
n
mn z = exp
kn z .
n!
n!
n=0
n=1
It follows that the generating function for connected maps is
!
k
∞
k
n(i)
X
X
X
Y
1
ti
i
log E exp
ti N Tr[X ] =
|Mg ((i, n(i)), 1 ≤ i ≤ k)| .
N 2g−2
n(i)!
g=0
i=1
i=1
n(1),...,n(k)≥0
n6=(0,0,...,0)
where
Mg ((i, n(i)), 1 ≤ i ≤ k) = {maps of genus g built out of n(i) stars of type i, 1 ≤ i ≤ k} .
Corollary 11.24. The generating function for connected maps is
!
k
∞
k
n(i)
X
X
X
Y
1
1
ti
i
log
E
exp
t
N
Tr[X
]
=
|Mg ((i, n(i)), 1 ≤ i ≤ k)| .
i
2g
N2
N
n(i)!
g=0
i=1
i=1
n(1),...,n(k)≥0
n6=(0,0,...,0)
Remark 11.25 (Further results).
a. One can look at maps where every edge has a color. These arise by gluing stars where only edges
of the same color are glued. Their generating functions are matrix integrals in several independent
GUE matrices, since in the Wick formula, only matrices of the same color can be paired.
b. The expansions above are purely formal. Proofs that these are actual asymptotic expansions are
very hard, see (Ercolani, McLaughlin 2003), (Guionnet, Maurel-Segala 2006, 2007). Their results
apply to N12 log E exp (N Tr[V (X)]), where V is a convex potential.
c. Physicists use the term “planar sector of quantum field theory” or “planar approximation” for the
leading term in the expansion.
d. Note that since X is GUE,
E exp
n
X
!
ti N Tr[X i ]
Z
=
"
exp N Tr −X 2 /2 +
i=1
n
X
i=1
106
#!
ti X i
dX = Z,
where
"
#!
n
X
1
exp N Tr −X 2 /2 +
ti X i
dX
Z
i=1
is a probability measure. Z is called the partition function. It can be thought of as a normalization
constant, but N12 log Z can also be thought of as a generating function for maps.
e. Sometimes the matrix integrals can be computed directly, thus giving formulas for generating functions. Namely, for the matrix model above and sufficiently nice V , the asymptotic distribution µ is
a minimizer of
Z
Z
log |x − y| dµ(x) dµ(y) − V (x) dµ(x)
(11.1)
In particular, it satisfies the Schwinger-Dyson equation
Z
dµ(x)
= V 0 (y),
y ∈ supp(µ).
2
y−x
(11.2)
This has generalizations to colored models and multi-variate equations, using cyclic derivatives etc.
Exercise 11.26. Sketch the arguments below; I am not asking for complete technical details.
a. Explain why equation 11.1 implies that
Z
2 log |x − y| dµ(x) − V (y) = C,
y ∈ supp(µ),
where C is a constant independent of y. Note that differentiating this expression we get equation 11.2. Hint: Consider deformations µ + εν for a signed measure ν.
b. Show that for the quadratic potential V (x) = x2 /2, the semicircular distribution is a solution of
equationR11.2. Hint: use the real part of the Cauchy transform. Note that care is required in even
defining dµ(x)
for y real.
y−x
Example 11.27. Let X be GUE, and D(1) , . . . , D(n) (for each N ) fixed (non-random) matrices.
E ◦ tr XD(1) . . . XD(n) =?
Consider
E ◦ tr XD(1) XD(2) XD(3) XD(4) .
See Figure 11. According to the Wick formula, this expression is again a sum of three terms. The first
one is
1 E Xi(1)i(2) Di(2)i(2) Xi(2)i(1) Di(1)i(3) Xi(3)i(4) Di(4)i(4) Xi(4)i(3) Di(3)i(1)
N
X
X
1 1 X
D
Di(1)i(3) Di(3)i(1) = tr[D(1) ] tr[D(3) ] tr[D(2) D(4) ].
=
D
i(2)i(2)
i(4)i(4)
2
NN
i(2)
i(4)
i(1),i(3)
107
Similarly, the second term is tr[D(2) ] tr[D(4) ] tr[D(1) D(3) ]. In the third term,
1 1
E Xi(1)i(2) Di(2)i(3) Xi(3)i(4) Di(4)i(2) Xi(2)i(1) Di(1)i(4) Xi(4)i(3) Di(3)i(1) = 2 tr D(1) D(4) D(3) D(2) .
N
N
Note the order!
11.3
Partitions and permutations.
Remark 11.28. Denote by Sym(n) the permutations of the set [n]. We will use the cycle notation
1 2 3 4 5 6 7
↔ (1374)(26)(5).
3 6 7 1 5 2 4
This suggests a natural embedding
P : P(n) ,→ Sym(n),
π 7→ Pπ ,
where a block of a partition is mapped to a cycle of a permutation, in increasing order.
For a partition π, had for example
Y
Rπ [a1 , a2 , . . . , an ] =
R[ai : i ∈ V ].
V ∈π
For a permutation α and matrices, can write
trα [A1 , A2 , . . . , An ] =
Y
tr Au(1) Au(2) . . . Au(k) .
V a cycle in α
V =(u(1),u(2),...,u(k))
Note that this is well-defined because tr is cyclically symmetric.
Denote by SNC (n) the image of NC(n) in Sym(n) under the embedding above. We would like to find a
more intrinsic definition of this subset.
Remark 11.29. Recall: a transposition is a 2-cycle (i, j). Sym(n) is generated by transpositions. In fact,
(u(1)u(2) . . . u(k)) = (u(1)u(2)) · (u(2)u(3)) · . . . · (u(k − 2)u(k − 1)) · (u(k − 1)u(k)).
Lemma 11.30. Let α ∈ Sym(n), and σ = (ab) be a transposition. Then
(
#cyc(α) + 1, if a, b are in the same cycle of α,
#cyc(ασ) =
#cyc(α) − 1, if a, b are in different cycles of α.
108
Proof. See Figure 11. The result follows from the observation that
(u(1) . . . u(i) . . . u(j) . . . u(k)) · (u(i)u(j)) =(u(1) . . . u(i − 1)u(i)u(j + 1) . . . u(k))
(u(i + 1) . . . u(j − 1)u(j))
and
(u(1) . . . u(k))(v(1) . . . v(m))·(u(i)v(j)) = (u(1) . . . u(i)v(j +1) . . . v(m)v(1) . . . v(j)u(i+1) . . . u(k)).
Definition 11.31. For α ∈ Sym(n), denote
|α| = min {k : α = σ1 σ2 . . . σk for σi transpositions} ,
|e| = 0, where e is the identity element in Sym(n). Also denote
γ = (1, 2, . . . , n) ∈ Sym(n)
the long cycle.
Lemma 11.32. Let α, β ∈ Sym(n).
a. d(α, β) = |α−1 β| is a metric on Sym(n). (In fact, it is a distance in a certain Cayley graph of
Sym(n).)
b. |α| = n − #cyc(α).
c. |αβ| = |βα|.
Proof. For part (a), d(α, α) = |e| = 0, since α−1 ν = (α−1 β)(β −1 ν),
−1 −1 −1 α ν ≤ α β + β ν ,
and
α−1 β = σ1 σ2 . . . σk
⇔
β −1 α = σk . . . σ2 σ1 .
X
(|V | − 1) ≥ |α| .
For part (b), by Remark 11.29,
n − #cyc(α) =
V a cycle in α
On the other hand, since #cyc(e) = n, by Lemma 11.30
#cyc(σ1 . . . σk ) ≥ n − k,
and so #cyc(α) ≥ n − |α|. The result follows. Part (c) is equivalent to proving that in general, |α−1 βα| =
|β|, which follows from part (b), since conjugation preserves the cycle structure.
109
Theorem 11.33. (Biane 1997)
a. α ∈ SNC (n) if and only if
|α| + α−1 γ = n − 1.
Since d(e, γ) = n − 1, this says that
d(e, α) + d(α, γ) = d(e, γ),
so α lies on a geodesic from e to γ.
b. For α, β ∈ SNC (n), denote α ≤ β if
|α| + α−1 β + β −1 γ = n − 1.
Equivalently, α and β lie on the same geodesic from e to γ, with α closer to e. Then the map
NC(n) → SNC (n) is a lattice isomorphism.
Proof. Suppose |α| + |α−1 γ| = n − 1, so that
α = σ1 σ2 . . . σk ,
α−1 γ = σk+1 . . . σn−1 .
Then γ = σ1 σ2 . . . σn−1 . Recall that e has n cycles, γ has 1 cycle, and each multiplication by σj changes
the number of cycles by 1. It follows that
#cyc(σ1 . . . σj ) = n − j.
Therefore going from σ1 . . . σj to
σ1 . . . σj−1 = (σ1 . . . σj )σj ,
σj cuts one cycle into two, which by the proof of Lemma 11.30 happens in a non-crossing way. The
converse is similar. For part (b), we note that α ≤ β if and only if we can write
α = σ1 . . . σj ,
β = σ1 . . . σj . . . σk ,
γ = σ1 . . . σj . . . σk . . . σn−1 .
Lemma 11.34. Under the identification NC(n) ↔ SNC (n),
K[π]
↔
Pπ−1 γ.
Sketch of proof. See Figure 11. The proof is by induction. For α = σ1 = (i, j), i < j,
α−1 γ = (1 . . . (i − 1)j . . . n)(i . . . (j − 1)) = K[α].
We want to build up to the case of
α = σ1 . . . σk ,
α−1 γ = σk+1 . . . σn−1 .
On the induction step, multiplying ασj connects two cycles of α, while σj α−1 γ cuts a cycle into two.
110
Remark 11.35. Biane’s embedding above allows us to define noncrossing partitions for other Coxeter
groups, for example noncrossing partitions of type B.
Notation 11.36. For α ∈ Sym(n),

Y
trα [X1 , . . . , Xn ] =
tr 
cycles of α

Y
Xi  .
i∈ cycle
For example,
tr(152)(34) [X1 , X2 , X3 , X4 , X5 ] = tr [X1 X5 X2 ] tr [X3 X4 ] .
Note: this makes sense since tr is cyclically symmetric.
Theorem 11.37. Let X be a GUE matrix, and D(1) , . . . , D(2n) be fixed matrices.
a.
1
X
E ◦ tr[X 2n ] =
π∈P 2 (2n)
N n−#cyc(πγ)+1
b.
E ◦ tr XD(1) XD(2) . . . XD(2n) =
1
X
π∈P 2 (2n)
N n−#cyc(πγ)+1
.
trπγ D(1) , D(2) , . . . , D(2n) .
Proof. For (a), using Example 11.12, we only need to show that
#cyc(πγ) = #faces in the map corresponding to π = #free parameters.
We are computing
E Xi(1)i(2) Xi(2)i(3) . . . Xi(2n)i(1) .
The term in the Wick sum corresponding to π pairs together
Xi(j)i(j+1) , Xi(π(j))i(π(j)+1) .
In other words,
i(j) = i(π(j) + 1) = i(γπ(j)),
i(j + 1) = i(γ(j)) = i(π(j)).
Thus indeed,
#free parameters = #cyc(γπ) = #cyc(πγ).
For part (b), we write
X E ◦ tr XD(1) XD(2) . . . D(2n) =
E Xi(1)u(1) Du(1)i(2) Xi(2)u(2) Du(2)i(3) . . . Xi(2n)u(2n) Du(2n)i(1) .
111
From the Wick formula i(j) = u(π(j)), u(j) = i(π(j)). So get
X
π∈P 2 (2n)
=
1
N n+1
X
X
π∈P 2 (2n)
=
X
π∈P 2 (2n)
=
(2)
(2n)
i(j)=u(π(j))
X
π∈P 2 (2n)
=
(1)
Du(1)i(2) Du(2)i(3) . . . Du(2n)i(1)
1
X
N n+1
1
1
N n+1
π∈P 2 (2n)
(2n)
(1)
(πγ)2 (1)
(πγ(1))
Du(1)u(πγ(1)) Du(πγ(1))u(πγπγ(1)) Du((πγ)2 (1))u((πγ)3 (1)) . . .
u(j)
Trπγ D(1) , D(2) , . . . , D(2n)
1
X
(2)
u(j)
X
N n+1
(1)
Du(1)u(π(2)) Du(2)u(π(3)) . . . Du(2n)u(π(1))
N n−#cyc(πγ)+1
trπγ D(1) , D(2) , . . . , D(2n) .
Remark 11.38. Note that the leading term in the expansion corresponds to g = 0, but also to π ∈
NC2 (2n). Equivalently,
n − #cyc(πγ) + 1 = 2n − #cyc(πγ) − n + 1 = |πγ| − n + 1 = π −1 γ + |π| − 2n + 1 = 0
if and only if π ∈ NC2 (2n) ⊂ Sym(2n).
11.4
Asymptotic free independence for Gaussian random matrices.
(N )
(N )
(N )
Definition 11.39. Recall from Definition 1.13: let a1 , a2 , . . . , ak
∈ (AN , ϕN ).
a. We say that
(N )
(N )
(a1 , . . . , ak ) → (a1 , . . . , ak ) ⊂ (A, ϕ)
in distribution if for each ~u,
h
i
(N ) (N )
(N )
ϕN au(1) au(2) . . . au(n) → ϕ au(1) au(2) . . . au(n)
as N → ∞. Equivalently,
h
i
(N )
(N )
(N )
ϕN
R
au(1) , au(2) , . . . , au(n) → Rϕ au(1) , au(2) , . . . , au(n) .
(N )
(N )
(N )
In this case we say that (a1 , a2 , . . . , ak ) have the asymptotic distribution
µa1 ,a2 ,...,ak .
112
(N )
(N )
(N )
b. a1 , a2 , . . . , ak
are asymptotically free if (a1 , . . . , ak ) ⊂ (A, ϕ) are free. Equivalently,
h
i
(N )
(N )
(N ) N →∞
RϕN au(1) , au(2) , . . . , au(n) −→ 0
unless all u(1) = u(2) = . . . = u(n).
Note that limits of tracial joint distributions are tracial.
Remark 11.40. Let X1 , . . . , Xk be entry-wise independent GUE. Say π ∈ P(n) is consistent with
~u = (u(1), u(2), . . . , u(n)),
if
π
i∼j
⇒
1 ≤ u(i) ≤ k
u(i) = u(j).
See Figure11. Denote such π by P ~u (n). Then as in Theorem 11.37,
1
X
E ◦ tr Xu(1) Xu(2) . . . Xu(2n) =
π∈P 2,~u (2n)
N n−#cyc(πγ)+1
.
and
E ◦ tr Xu(1) D(1) Xu(2) D(2) . . . Xu(2n) D(2n) =
X
π∈P 2,~u (2n)
1
N n−#cyc(πγ)+1
trπγ D(1) , D(2) , . . . , D(2n) .
Theorem 11.41. Let X1 , . . . , Xp be entry-wise independent GUE matrices, and D1 , . . . , Dq be nonrandom matrices with an asymptotic (tracial) joint distribution µd1 ,...,dq .
a. X1 , . . . , Xp are asymptotically freely independent. Consequently, in distribution
(X1 , . . . , Xp ) → (s1 , . . . , sp ),
where (s1 , . . . , sp ) is a free semicircular system.
b. More generally, in distribution
(X1 , . . . , Xp , D1 , . . . , Dq ) → (s1 , . . . , sp , d1 , . . . , dq ),
where s1 , . . . , sp are free semicircular elements freely independent from d1 , . . . , dq .
Proof. Taking the limit N → ∞ in the preceding remark and using Remark 11.38,
X
E ◦ tr Xu(1) Xu(2) . . . Xu(n) →
1,
π∈NC2,~u (n)
113
which implies that for all π ∈ NC(n)
(
1, π ∈ NC2 (n) ~u-consistent,
Rπ Xu(1) , Xu(2) , . . . , Xu(n) →
= Rπ su(1) , su(2) , . . . , su(n) .
0, otherwise.
Similarly, taking D(i) = fi (D1 , . . . , Dq ) in the preceding remark,
lim E ◦ tr Xu(1) f1 (D1 , . . . , Dq )Xu(2) f2 (D1 , . . . , Dq ) . . . Xu(n) fn (D1 , . . . , Dq )
N →∞
X
trπγ [f1 (D1 , . . . , Dq ), f2 (D1 , . . . , Dq ), . . . , fn (D1 , . . . , Dq )]
= lim
N →∞
π∈NC2,~u (n)
=
X
ϕπγ [f1 (d1 , . . . , dq ), f2 (d1 , . . . , dq ), . . . , fn (d1 , . . . , dq )]
π∈NC2,~u (n)
=
X
ϕ
MK[π]
[f1 (d1 , . . . , dq ), f2 (d1 , . . . , dq ), . . . , fn (d1 , . . . , dq )]
π∈NC2,~u (n)
=
X
ϕ
Rπϕ su(1) , su(2) , . . . , su(k) MK[π]
[f1 (d1 , . . . , dq ), f2 (d1 , . . . , dq ), . . . , fn (d1 , . . . , dq )]
π∈NC(n)
= ϕ su(1) f1 (d1 , . . . , dq )su(2) f2 (d1 , . . . , dq ) . . . su(n) fn (d1 , . . . , dq ) ,
where we have used Lemma 11.34, properties of free cumulants of semicircular elements, and the results
from Remark 8.1.
Remark 11.42 (Other Gaussian ensembles).
a. GOE (Gaussian orthogonal ensemble): X real symmetric Gaussian. Counts maps on non-oriented
surfaces. The leading term still planar, since any identification of two edges with the opposite
orientation leads to non-zero genus.
b. GSE (Gaussian symplectic ensemble): X certain quaternionic matrices. These all have a common
generalization to β-ensembles, where the real case corresponds to β = 1, complex to β = 2, and
quaternionic to β = 4.
c. Ginibre ensemble: real Gaussian matrices, with no symmetry, all entries independent. Note that
this is typically not a normal matrix. It still has N complex eigenvalues, and its spectral distribution
is a measure on C. Asymptotically,
N
1 X
1
δλi → circular law 1|z|≤1 on C.
N i=1
π
Old result for Gaussian matrices; for Wigner-type matrices, proved by various authors, culminating
in the proof by (Tao, Vu 2010) under optimal conditions. In distribution,
(X, X ∗ ) → (c, c∗ ),
where c is a circular operator, from Definition 10.5. Note that a circular operator is not normal, so
its distribution cannot come from a measure; but its Brown measure is a (rescaled) circular law.
114
Exercise 11.43. The oldest random matrix model (Wishart 1928) arose in statistics, and its relatives
appear in numerous applications, from cell phone communications to medicine. Let (y1 , . . . yK )t be jointly
Gaussian, with the K × K covariance matrix Q. For j = 1, 2, . . . , N (with N ≥ K) let (Yij )K
i=1 be
independent samples from this Gaussian distribution, and Y = (Yij ) a K × N random matrix, whose
columns are independent and identically distributed. Then the N × N matrix N1 Y t Y is called a Wishart
matrix. Note that the K × K matrix
N
1
1 X
t
(Y Y )ij =
Yik Yjk
N
N k=1
is the sample covariance matrix for this distribution, and in particular E N1 (Y Y t ) = Q, and that the
distributions of Y t Y and Y Y t are closely related, although they live in different spaces.
a. Suppose that Q is the identity matrix, so that all entries of Y are independent. Show that
1 t
Y Y = X t P X,
N
where X is a N ×N random matrix all of whose entries are independent N (0, N1 ) random variables,
and P a diagonal N × N matrix with K ones and N − K zeros on the diagonal.
b. Let now both N and K go to infinity, so that K/N → t, 0 ≤ t ≤ 1. According to the preceding
remark, (X, X t ) → (c, c∗ ). Use a variation of Theorem 11.41 and a combination of Proposition 10.7
with Lemma 8.2 to conclude that the asymptotic distribution of a Wishart matrix with identity
covariance is πt , the free Poisson distribution, which in this context is called the Marchenko-Pastur
distribution.
A similar argument for general Q shows that the limiting distribution of a general Wishart matrix is
free compound Poisson. An alternative approach of rectangular free probability (which corresponds to
taking limits of rectangular matrices directly) was developed by Benaych-Georges. Compare also with
Exercise 22.18 in [NS06].
11.5
Asymptotic free independence for unitary random matrices.
Remark 11.44 (Unitary random matrices). UN is the group of unitary N × N complex matrices. On UN ,
have the Haar measure, which is right and left invariant. In particular,
Z
Z
Z
∗
f (U ) dU =
f (V U V ) dU =
V f (U )V ∗ dU.
UN
UN
UN
A random N × N unitary matrix whose distribution is the Haar measure is a Haar unitary matrix. Note
that for any λ ∈ C, |λ| = 1, if U is a random unitary matrix whose distribution is the Haar measure, then
Z
Z
Z
n
n
n
U dU =
(λU ) d(λU ) = λ
U n dU,
UN
UN
UN
115
and so
Z
U n dU = δn=0 ,
n ∈ Z.
UN
In particular, in the n.c. probability space (MN (L∞− (Ω, P )), E ◦ tr), a Haar unitary matrix is a Haar
unitary.
In physics, Haar unitary matrices are called CUE (circular unitary ensemble) matrices.
Theorem 11.45. Let U1 , . . . , Up be independent N × N Haar unitary matrices, and (D1 , . . . , Dq ) nonrandom matrices converging in ∗-distribution to (d1 , . . . , dq ) ⊂ (A, ϕ). Then in ∗-distribution,
(U1 , . . . , Up , D1 , . . . , Dq ) → (u1 , . . . , up , d1 , . . . , dq ),
where u1 , . . . , up are free Haar unitaries free from {d1 , . . . , dq }.
Remark 11.46. Recall from Exercise 3.12, that if u1 , . . . , up , d0 , d1 , . . . , dp are as above, then
d0 , u1 d1 u∗1 , . . . , up dp u∗p
are free.
Corollary 11.47. Let AN , BN be N × N (non-random) matrices such that AN and BN converge in
distribution as N → ∞. Let UN be an N × N Haar unitary matrix. Then UN AN UN∗ and BN are
asymptotically free.
Remark 11.48.
a. Let X be an N × N random matrix, with matrix-valued distribution PX . We say that PX (or X) is
unitarily invariant if for any (fixed) V ∈ UN
Z
Z
f (X) dPX = f (V XV ∗ )dPX .
b. For example, if X is a GUE matrix,
dPX =
1 −N Tr[X 2 ]/2
e
dX
Z
and Tr[(V XV ∗ )2 ] = Tr[X 2 ], so this is a unitarily invariant ensemble (compare with GOE, GSE).
More generally, matrix models Z1 e−T r[V (X)] dX are unitarily invariant.
c. If X is any random matrix, and U is a Haar unitary matrix, then U ∗ XU is unitarily invariant (by
invariance of the Haar measure). Thus the corollary says “independent unitarily invariant matrices
are asymptotically free”.
116
d. In particular, asymptotic freeness for unitarily invariant matrices implies asymptotic freeness for
Gaussian matrices (Theorem 11.41). Conversely, if X is GUE, its polar decomposition is X =
U (X ∗ X)1/2 , where U ia a Haar unitary matrix. So we could use the Gaussian result to prove the
unitary result. We will use combinatorics instead.
e. GUE are the only matrices which are at once Wigner matrices and unitarily invariant.
Remark 11.49. Let
{λ1 , . . . , λN } , {ρ1 , . . . , ρN } ⊂ R
be eigenvalue lists, and A, B be self-adjoint N × N matrices with these eigenvalues. What are the eigenvalues of A + B?
(i) If A, B commute, they have a common basis of eigenvectors, and
eigenvalues of (A + B) = eigenvalues of A + eigenvalues of B.
(ii) In general, eigenvalues of A + B depend on A, B and not just on {λi }, {ρj }. They satisfy the Horn
inequalities.
(iii) If A, B are in a “general position”: so that B is fixed (without loss of generality, diagonal) and A is
random,


λ1

 ∗
..
A=U
U ,
.
λN
U a Haar unitary matrix. Then if
N
1 X
µA =
δλ ,
N i=1 i
N
1 X
µB =
δρ ,
N i=1 i
then µA+B ≈ µA µB for large N . Thus µA+B (asymptotically) depends only on {λi }, {ρj }.
Remark 11.50. We will not prove Theorem 11.45. To prove Corollary 11.47, we need to understand
expressions of the form
E ◦ tr U A(1) U ∗ B (1) U A(2) U ∗ B (2) . . . ,
where A(i) = Au(i) , B (i) = B v(i) . What is the analog of the Wick formula?
Remark 11.51. Let U = (Uij )N
i,j=1 be an N × N Haar unitary matrix. For N ≥ n,
E Ui(1)j(1) . . . Ui(n)j(n) Ūu(1)v(1) . . . Ūu(n)v(n)
X
=
δi(β(1))=u(1) . . . δi(β(n))=u(n) δj(α(1))=v(1) . . . δj(α(n))=v(n) Wg(N, βα−1 ).
α,β∈Sym(n)
117
Here for β ∈ Sym(n) and N ≥ n
Wg(N, β) = E U11 . . . Unn Ū1β(1) . . . Ūnβ(n)
is the Weingarten function.
Example 11.52.
E U11 U12 Ū22 Ū12 = 0,
E U11 U12 Ū12 Ū11 = Wg(N, (12)) + Wg(N, e),
where for the first term α = e, β = (12), for the second α = β = (12).
Remark 11.53. There is no “useful” formula for Wg. Can write it in terms of characters. Can write it in
terms of an inverse of an explicit matrix. It is known that Wg(N, α) only depends on the conjugacy class
(cycle structure) of α, and
1
1
.
Wg(N, α) = µ(α) 2n−#cyc(α) + O
N
N 2n−#cyc(α)+2
Moreover, for α = Pσ , β = Pπ ∈ SNC (n), µ(β) = µ(0̂n , π) and more generally
µ(α−1 β) = µ(Pσ−1 Pπ ) = µ(σ, π),
the Möbius function on NC(n).
118
Remark 11.54. Thus we can now compute
E ◦ tr U A(1) U ∗ B (1) U A(2) U ∗ B (2) . . . U A(n) U ∗ B (n)
i
1 X h
(1)
(1)
(n)
(n)
=
E Ui(1)j(1) Aj(1)v(1) Ūu(1)v(1) Bu(1)i(2) . . . Ui(n)j(n) Aj(n)v(n) Ūu(n)v(n) Bu(n)i(1)
N
~i,~j
~
u,~v
=
1 X (1)
(1)
(2)
(2)
(n)
(n)
Aj(1)v(1) Bu(1)i(2) Aj(2)v(2) Bu(2)i(3) . . . Aj(n)v(n) Bu(n)i(1)
N
~i,~j
~
u,~v
E Ui(1)j(1) Ūu(1)v(1) Ui(2)j(2) Ūu(2)v(2) . . . Ui(n)j(n) Ūu(n)v(n)
1 X (1)
(1)
(2)
(2)
(n)
(n)
Aj(1)v(1) Bu(1)i(2) Aj(2)v(2) Bu(2)i(3) . . . Aj(n)v(n) Bu(n)i(1)
=
N
~i,~j
~
u,~v
X
δi(β(1))=u(1) . . . δi(β(n))=u(n) δj(α(1))=v(1) . . . δj(α(n))=v(n) Wg(N, βα−1 )
α,β∈Sym(n)
=
=
=
1
N
1
N
1
N
X
Wg(N, βα−1 )
(n)
(1)
Aj(1)j(α(1)) . . . Aj(n)j(α(n))
X
(1)
(n)
Bu(1)u(β −1 γ(1)) Bu(n)u(β −1 γ(n))
~
u
~j
α,β∈Sym(n)
X
X
Wg(N, βα−1 ) Trα A(1) , . . . , A(n) Trβ −1 γ B (1) , . . . , B (n)
α,β∈Sym(n)
X
Wg(N, βα−1 )N #cyc(α)+#cyc(β
−1 γ)
trα A(1) , . . . , A(n) trβ −1 γ B (1) , . . . , B (n) .
α,β∈Sym(n)
Using the Weingarten function asymptotics, the degree of
1
N
in the leading term is
(2n − #cyc(α−1 β)) + 1 − #cyc(α) − #cyc(β −1 γ)
= 2n + 1 − (n − α−1 β ) − (n − |α|) − (n − β −1 γ )
= |α| + α−1 β + β −1 γ − (n − 1).
Recall that |α| + |α−1 β| + |β −1 γ| ≥ n − 1, with equality only if α = Pσ , β = Pπ , α ≤ β, or equivalently
σ ≤ π. Thus as N → ∞, the leading term in the expression above is
X
(1)
(1)
1
(n)
(n)
−1
µ(Pσ Pπ ) trσ A , . . . , A
trK[π] B , . . . , B
+O
.
N2
σ,π∈NC(n)
σ≤π
Corollary 11.55. Let (A(1) , . . . , A(n) ), (B (1) , . . . , B (n) ) be N × N (random or not) matrices such that
(A(1) , . . . , A(n) ) → (a1 , . . . , an ) ∈ (A, ϕ),
(B (1) , . . . , B (n) ) → (b1 , . . . , bn ) ∈ (A, ϕ),
119
with tracial state ϕ, and UN be an N × N Haar unitary matrix. Then as N → ∞,
X
E ◦ tr U A(1) U ∗ B (1) . . . U A(n) U ∗ B (n) →
µ(Pσ−1 Pπ )ϕσ [a1 , . . . , an ] ϕK[π] [b1 , . . . , bn ] .
σ,π∈NC(n)
σ≤π
Sketch of the proof of Corollary 11.47. Say AN → a, BN → b in distribution. To prove asymptotic free
independence of UN AN UN∗ and BN , we want to show that for any fj , gj with ϕ[fj (a)] = ϕ[gj (b)] = 0,
the expression (see text)
E ◦ tr [f1 (UN AN UN∗ )g1 (BN ) . . . gn−1 (BN )fn (UN AN UN∗ )]
= E ◦ tr [UN f1 (AN )UN∗ g1 (BN ) . . . gn−1 (BN )UN fn (AN )UN∗ ]
goes to zero. This expression asymptotically equals
X
µ(Pσ−1 Pπ )ϕσ [f1 (a), . . . , fn (a)] ϕK[π] [g1 (b), . . . , gn−1 (b), 1] .
σ,π∈NC(n)
σ≤π
Using Exercise 11.56 below, we conclude that each term in the sum above involves either ϕ[fi (a)] or
ϕ[fj (b)], and so equals zero.
Exercise 11.56.
a. For any σ ∈ NC(n), the partition σ ∪ K[σ] ∈ NC(2n) has exactly two outer blocks.
b. For any outer block of a non-crossing partition, there is an interval block covered by it.
c. If σ ∈ N C(n) has no singletons, and σ ≤ π, then K[π] has at least 2 singletons.
120
Chapter 12
Operator-valued free probability.
This chapter was omitted in the course, so what follows is only an outline. See [Spe98], [NSS02].
B-valued probability spaces.
Definition 12.1. Let A be a ∗-algebra, and B ⊂ A a C ∗ -algebra. In particular, A is a B-bimodule,
B · A ⊂ A,
A · B ⊂ A.
Assume that we have a conditional expectation E : A → B. That is, E is a positive B-bimodule map,
a ≥ 0 ⇒ E[a] ≥ 0 in B,
E [b1 ab2 ] = b1 E[a]b2 .
Then (A, E) is a B-valued n.c. probability space.
Example 12.2. (X, Σ, P ) a probability space, Ω ⊂ Σ a sub-σ-algebra. Take A = Σ-measurable functions, B = Ω-measurable functions. Then the Radon-Nikodym theorem gives a conditional expectation
E : A → B.
Example 12.3. B a C ∗ -algebra. Let
A = Bhxi = C − Span (b0 xb1 x . . . , xbn : n ≥ 0) /(linearity relations) ' B ⊕ B ⊗C 2 ⊕ B ⊗C 3 ⊕ . . . .
Example 12.4. (C, ϕ) an n.c. probability space. Then
(A = Mn (C), tr ◦ϕ)
is also an n.c. probability space. Let
B = Mn (C),
E : A → B,
A 7→ (ϕ[Aij ])ni,j=1 .
In particular, may have
(A = Mnk (C) = Mn (C) ⊗ Mk (C), B = Mn (C), E = Id ⊗ tr = partial trace = block-trace).
121
Remark 12.5 (Motivation).
(i) Many scalar-valued constructions work.
(ii) Free products with amalgamation.
(iii) Random matrices with independent blocks, random band matrices.
Definition 12.6. X ∈ (A, B, E). A (B-valued) distribution of X is a completely positive (c.p.) functional
µX : Bhxi → B,
µX [b0 xb1 . . . xbn ] = E [b0 Xb1 . . . Xbn ] = b0 E [Xb1 . . . X] bn .
Same for joint distributions.
Remark 12.7 (Generating functions). Moment-generating function: for b ∈ B,
M X (b) = E (1 − bX)−1 = µX (1 − bx)−1 = 1 + b E[X] + b E[XbX] + b E[XbXbX] + . . . .
Note: involves only the symmetric moments E[XbXb . . . bX], not general moments
E[Xb1 Xb2 . . . bn−1 X].
To get full information from generating functions, need fully matricial constructions (as studied by
Voiculescu 2004, 2010, Belinschi, Popa, Vinnikov 2010, 2011, 2012, Helton, Klep, McCullough 2009,
2010, 2011). No analog of Stieltjes inversion.
Functional analysis background
B-valued spectral theory not well developed.
Remark 12.8 (GNS construction). (A, B, E). Can define a B-valued inner product on A,
ha1 , a2 i = E[a∗2 a1 ].
Note that
hb1 a1 b2 , a2 b3 i = E [b∗3 a∗2 b1 a1 b2 ] = b∗3 E [(b∗1 a2 )∗ a1 ] b2 = b∗3 ha1 , b∗1 a2 i b2 .
Remark 12.9 (Tensor products with amalgamation). V1 , . . . , Vn B-bimodules, i.e. have commuting actions
B × Vi → Vi , Vi × B → Vi .
Then
V1 ⊗B V2 ⊗B . . . ⊗B Vn =
( k
X
)
(i)
ξ1
⊗ . . . ⊗ ξn(i) :
(i)
ξj
∈ Vj
/(B − linearity),
i=1
i.e.
ξ1 ⊗B . . . ⊗B (b1 ξb2 + b3 ηb4 ) ⊗B . . . ⊗B ξn
= ξ1 ⊗B . . . ⊗B ξk−1 b1 ⊗B ξ ⊗B b2 ξk+1 ⊗B . . . ⊗B ξn + ξ1 ⊗B . . . ⊗B ξk−1 b3 ⊗B η ⊗B b4 ξk+1 ⊗B . . . ⊗B ξn .
s
122
Definition 12.10. Subalgebras A1 , . . . , An ⊂ (A, B, E) are freely independent (with amalgamation) over
B if whenever u(1) 6= u(2) 6= . . . 6= u(k), ai ∈ Au(i) , E[ai ] = 0 (in B), then
E [a1 . . . ak ] = 0.
Remark 12.11 (Free product with amalgamation). (A, B, E) = ∗ni=1 (Ai , B, Ei ). Denote
A◦i = {a ∈ Ai : Ei [a] = 0} .
Let W∅ = B,
W~u = A◦u(1) ⊗B A◦u(2) ⊗B . . . ⊗B A◦u(k)
Then
A=
∞
M
k=0
M
W~u ,
|~
u|=k
u(1)6=u(2)6=...6=u(k)
and E[1] = 1, E[a1 a2 . . . ak ] = 0.
Complete positivity of the free product.
Lemma 12.12. Let B be a unital C ∗ -algebra. A matrix (Bij )ni,j=1 ∈ Mn (B) is positive (as an element in
the C ∗ -algebra) if and only if
n
X
∀ b1 , b2 , . . . , bn ∈ B,
b∗i Bij bj ≥ 0.
i,j=1
Proposition 12.13. Let B be a unital C ∗ -algebra and E : A → B be a B-bimodule map. E is positive if
and only if it is completely positive.
Proof. One direction is clear. Assume E is a positive B-bimodule map. Let A ∈ Mn (A). Without loss of
generality A = B ∗ B, B ∈ Mn (A), i.e.
Aij =
n
X
b∗ki bkj ,
buv ∈ A.
k=1
Then using notation from Definition 3.21,
(En [A])ni,j=1 =
n
X
(E[b∗ki bkj ])ni,j=1 ,
k=1
so ∀ b1 , . . . , bn ∈ B,
X
b∗i E[b∗ki bkj ]bj =
n
X
k=1
#
n
n
X
X
E
(bki bi )∗
(bkj bj ) ≥ 0 in B.
"
i=1
123
j=1
Theorem 12.14. Free product of completely positive maps is completely positive.
The proof is the same as in Theorem 3.26.
Operator-valued free cumulants.
Let X1 , . . . , Xn ∈ (A, B, E). Their moment functional is a C-multilinear functional
M [b0 X1 b1 , X2 b2 , . . . , Xn bn ] = b0 M [X1 , b1 X2 , . . . , bn−1 Xn ]bn = E[b0 X1 b1 X2 b2 . . . Xn bn ].
Define the free cumulant functionals
R[b0 X1 b1 , X2 b2 , . . . , Xn bn ] = b0 R[X1 , b1 X2 , . . . , bn−1 Xn ]bn
by
M [b0 X1 b1 , X2 b2 , . . . , Xn bn ] =
X
Rπ [b0 X1 b1 , X2 b2 , . . . , Xn bn ].
π∈NC(n)
Here Rπ uses the nesting structure on NC(n). In particular, this will not work for general partitions.
Example 12.15. Figure omitted. Let π = {(16)(235)(4)(78)}. Then
Rπ [b0 X1 b1 , X2 b2 , X3 b3 , X4 b4 , X5 b5 , X6 b6 , X7 b7 , X8 b8 ]
= b0 R[X1 b1 R[X2 b2 , X3 b3 R[X4 ]b4 X5 ]b5 X6 ]b6 R[X7 b7 , X8 ]b8 .
Theorem 12.16. Free independence is equivalent to the vanishing of mixed free cumulants.
Definition 12.17. Free convolution. Free cumulants satisfy Rµν = Rµ + Rν .
Operator-valued semicircular distributions.
B-valued free cumulants take values in B. In particular,
R2X [b] = E[XbX] = variance
is a completely positive map R2X : B → B. Thus: let η : B → B be a completely positive map. The
functional µ : Bhxi → B defined by
(
η(b1 ), n = 2,
Rµ [x, b1 x, b2 x, . . . , bn−1 x] =
0,
n 6= 2
is the B-valued semicircular distribution with variance η.
124
Proposition 12.18. µ is positive.
Proof I. Free central limit theorem.
Proof II. Fock space (module) realization.
Example 12.19. The moments of a semicircular distribution are
E[b0 ] = b0 ,
E[b0 Xb1 ] = 0,
E[b0 Xb1 Xb2 ] = b0 η(b1 )b2 ,
E[b0 Xb1 Xb2 Xb3 Xb4 ] = b0 η(b1 )b2 η(b3 )b4 + b0 η(b1 η(b2 )b3 )b4 ,
etc.
The full Fock module construction.
Let M be a B-bimodule. A B-valued inner product is a map
h·, ·i : M × M → B
such that
hb1 f b2 , gb3 i = b∗3 hf, b∗1 gi b2
(compare with Remark 12.8),
hf, gi∗ = hg, f i ,
hf, f i ≥ 0 in B.
The full Fock module F(M) is the B-bimodule
F(M) =
∞
M
M⊗B k = B ⊕ M ⊕ (M ⊗B M) ⊕ (M ⊗B M ⊗B M) ⊕ . . . ,
k=0
with the inner product
hf1 ⊗ . . . ⊗ fn , g1 ⊗ . . . ⊗ gk i = δn=k hgk hgk−1 h. . . hg1 , f1 i , f2 i . . . , fk−1 i , fk i
Remark 12.20. If the inner product on M is positive, it is completely positive, which implies that the
inner product on F(M) is positive.
125
Now specialize: Let M = principal B-bimodule, M = BξB, ξ a symbol. Then
F ' B ⊕ BξB ⊕ BξBξB ⊕ . . . ' Bhξi.
A fixed completely positive map η : B → B gives an inner product on M:
hb1 ξb2 , b3 ξb4 i = b∗4 η(b∗3 b1 )b2 .
On F(M), get the creation operator
`∗ (b1 ξb2 ξ . . . bn−1 ξbn ) = ξb1 ξb2 ξ . . . bn−1 ξbn
and the annihilation operator
`(b1 ξb2 ξ . . . bn−1 ξbn ) = η(b1 )b2 ξ . . . bn−1 ξbn ,
`(b1 ) = η(b1 )1 ∈ B. If we define X = `∗ + `, then with respect to the conditional expectation h·1, 1i, X
has the B-valued semicircular distribution with variance η.
Convolution powers.
Definition 12.21. µ : Bhxi → B is -infinitely divisible if for all t ≥ 0, tRµ = Rµt for some positive µt .
In that case write µt = µt . But remember: Rµt ∈ B. So can ask: if η : B → B is completely positive, is
there a µη such that
Rµη = η(Rµ )?
Theorem 12.22.
a. For any c.p. µ and c.p. η, µ]η is well defined.
b. If µ is -infinitely divisible (i.e. µt ≥ 0), then automatically µη ≥ 0 for any c.p. η.
c. For any c.p. µ and c.p. η, µ(1+η) ≥ 0.
Proof. (a), (b) proved using a Fock module construction. For (c), Method I (A, Belinschi, Fevrier, Nica
2011) is to define the map Bη combinatorially as in Definition 9.10, prove that it is c.p., and define
µ(1+η) = (Bη [µ])](1+η) .
Method II (Shlyakhtenko 2012) is to use an operator-valued Voiculescu’s operator model from Proposition 6.20.
126
Relation between different operator-valued cumulants.
If D ⊂ B ⊂ A, what is the relation between D and B-valued free cumulants? Between D and B-valued
freeness?
Theorem 12.23. Let 1 ∈ D ⊂ B ⊂ A, (A, B, E), (B, D, E) n.c. probability spaces. Let a1 , . . . , ak ∈ A.
Suppose that for all d1 , . . . , dn−1 ∈ D,
RB au(1) d1 , au(2) d2 , . . . , au(n) ∈ D.
Then
RB au(1) d1 , au(2) d2 , . . . , au(n) = RD au(1) d1 , au(2) d2 , . . . , au(n)
Theorem 12.24. Let 1 ∈ D ⊂ B, N ⊂ A, (A, B, E), (B, D, E) n.c. probability spaces. Suppose E :
B → D is faithful, i.e.
∀b2 ∈ B, E[b1 b2 ] = 0 ⇒ b1 = 0.
Then the following are equivalent.
a. B and N are free with amalgamation over D.
b. For any n1 , . . . , nk ∈ N and b1 , . . . , bk−1 ∈ B,
RB [n1 b1 , n2 b2 , . . . , nk−1 bk−1 , nk ] = RD [n1 E[b1 ], . . . , nk−1 E[bk−1 ], nk ]
Remark 12.25 (Applications). Suppose X, Y ∈ (A, ϕ) are not free. How to compute µX+Y ? Sometimes
have (A, B, E), ϕ = ϕ ◦ E, and X, Y are free with respect to E. Then can compute
µBX+Y : Bhxi → B
by µBX+Y = µBX µBY . Then can also compute
µX+Y [xn ] = ϕ µBX+Y [xn ] ∈ C.
127
Chapter 13
A very brief survey of analytic results.
See [VDN92], [Voi02].
Another combinatorial topic we did not cover is the connection between free probability and asymptotic
representation theory of the symmetric and unitary groups, starting with the work of Biane (1995, 1998)
and continued by Féray, Lassalle, Śniady and many others.
13.1
Complex analytic methods.
Let µ be an arbitrary probability measure on R. Its Cauchy transform is
Z
1
Gµ (z) =
dµ(x).
R z −x
Lemma 13.1. Gµ is an analytic function
Gµ : C+ → C− ∪ R,
where C+ = {z ∈ C : =z > 0}.
Proof.
Z
1
=Gµ (z) =
=
dµ(x) =
z−x
R
Z
R
−=z
dµ(x).
|z − x|2
Remark 13.2. Gµ is invertible in a Stolz angle
z ∈ C+ : |z| > α, =z > β |<z| .
G−1
µ is defined on
z ∈ C+ : |z| < α, =z > β |<z| .
128
Figure omitted. On such a domain, define the R-transform by
1
Rµ (z) = G−1
µ (z) − .
z
Theorem 13.3. For µ, ν arbitrary probability measures, can define µ ν, such that
Rµ + Rν = Rµν
on a domain as above. Key point: µ ν is again positive.
Exercise 13.4. Let {γt : t ≥ 0} be the free convolution semigroup of semicircular distributions; recall
that Rγt (z) = tz. Let µ be an arbitrary probability measure, and denote
G(z, t) = Gµγt (z).
Prove Voiculescu’s evolution equation:
∂t G(z, t) + G(z, t) ∂z G(z, t) = 0.
What is the corresponding result for a general free convolution semigroup {νt : t ≥ 0}?
Multiplicative free convolution
Let µ, ν be probability measure on R+ , X, Y freely independent, unbounded operators with distributions
µ, ν. Define the multiplicative free convolution by
µ ν = µX 1/2 Y X 1/2 = µY 1/2 XY 1/2 .
Or: Let µ, ν be probability measure on T, X, Y freely independent, unitary operators with distributions
µ, ν. Define
µ ν = µXY .
How to compute? Define
Z
ψµ (z) =
R+ or T
xz
dµ(x) = Mµ (z) − 1,
1 − xz
χµ (z) = ψµ−1 (z) (on a domain), and the S-transform
Sµ (z) =
1+z
χµ (z).
z
Then
Sµν (z) = Sµ (z)Sν (z).
But note: constructed a multivariate R-transform. No good multivariate S-transform, although there are
attempts.
129
Example 13.5. Let
µ = (1 − α)δ0 + αδ1 ,
ν = (1 − β)δ0 + βδ1 .
Recall: if p, q are projections, ϕ[p] = α, ϕ[q] = β, in Proposition 8.4 computed the distribution of pqp
using R-transforms. Using S-transforms:
ψµ (z) = α
z
,
1−z
Thus
χµ (z) =
z
,
α+z
Sµ (z) =
1+z
.
α+z
(1 + z)2
Sµν (z) =
,
(α + z)(β + z)
z(1 + z)
,
(α + z)(β + z)
p
1 − z(α + β) + (az − 1)(bz − 1)
ψµν (z) =
,
2(z − 1)
χµν (z) =
where
a, b = α + β − 2αβ ±
Then G(z) =
1
z
1+ψ
1
z
p
4αβ(1 − α)(1 − β).
gives
p
1 z − (α + β) + (z − a)(z − b)
,
Gµν (z) = +
z
2(1 − z)z
and can find µ ν using Stieltjes inversion.
Free infinite divisibility
Recall: a free convolution semigroup is a family of probability measures {µt : t ≥ 0} such that
Rµt = tRµ .
Recall: if the moments of µt are finite, have
r1 [µt ] = tλ̃,
rn [µt ] = tmn−2 [ρ],
in other words
"
Rµt (z) = t λ̃ + ρ
∞
X
#!
x
n−2 n−1
z
Z
= t λ̃ +
R
n=2
z
dρ(x) ,
1 − xz
where ρ is a finite positive measure. This formula still holds for general µ with finite variance (Kolmogorov representation). In general, have a free Lévy-Khinchin representation
Z
z+x
Rµt (z) = t λ +
dρ(x) .
R 1 − xz
130
It follows that
n
o
ID = νλ,ρ : λ ∈ R, ρ finite Borel measure .
Can also define ν∗λ,ρ via its Fourier transform
Z
x2 + 1
ixθ
Fν∗λ,ρ (θ) = exp iλθ + (e − 1 − ixθ)
dρ(x) .
x2
R
There are precisely all the classically infinitely divisible distributions. But moreover,
Theorem 13.6 (Bercovici, Pata 1999). Let {µn : n ∈ N} be probability measures, and k1 < k2 < . . .
integers. The following are equivalent:
a. µn ∗ µn ∗ . . . ∗ µn → ν∗λ,ρ weakly.
{z
}
|
kn
b. µn µn . . . µn → νλ,ρ weakly.
{z
}
|
kn
Recall however, that there does not exist a bijection T such that
T (µ ∗ ν) = T (µ) T (ν).
The theorem above has many generalizations, to ], , non-identically distributed case, etc.
An alternative description of -infinitely divisible distributions is that they are precisely those µ whose
R-transforms Rµ extend to an analytic function C+ → C+ .
Stable distributions
µ is -stable if µ µ is µ up to a shift and rescaling. For example, for µ semicircular,
µ µ = D√2 µ.
Have a complete description. Again, -stable distributions are in a bijection with ∗-stable distributions,
and their domains of attraction coincide.
Except for Gaussian/semicircular, stable distributions do not have finite moments.
Example 13.7. Define µ by Rµ (z) = α, α ∈ C. Note µ µ = D2 µ, so µ is stable.
G−1
µ (w) =
1
+ α,
w
Gµ (z) =
1
.
z−α
If α is real, then µ = δα . But if α is purely imaginary, α = ia, then
=
1
a − =z
=
z − ia
(<z)2 + (=z − a)2
131
and
1
a
dx.
2
π x + a2
These are the Cauchy distributions. They have a very special property that
dµ(x) =
µ µ = µ ∗ µ;
in particular they are both and ∗-stable.
13.2
Free entropy.
The description in this section is very brief and imprecise, and is meant to give only a very vague idea of
the field. It should not be used as a reference.
Microstates free entropy
Remark 13.8 (Bolzmann entropy). A macrostate of the system (say, ideal gas) is described by some
macro parameters, such as temperature, pressure, or volume. A microstate of the system is the complete
description of the positions and velocities of all molecules in the system. Entropy of a macrostate measures
the amount of microstates compatible with it.
Remark 13.9. In this remark we give a heuristic derivation of the classical entropy formula. Let µ be a
measure with density f , and let F be its cdf, i.e. the primitive of f . Assume that µ is supported in an
interval [a, b]. Fix n, and for 0 ≤ i ≤ n let ai = F −1 (i/n). Now look at all the atomic measures
1
(δx + δx2 + . . . + δxn )
n 1
with ai−1 ≤ xi ≤ ai .
On the one hand, both the k-th moment of this measure and of µ lie
" n−1
#
n
1X k 1X k
a ,
a .
n i=0 i n i=1 i
Therefore the difference between them is at most (1/n)(akn − ak0 ) = (1/n)(bk − ak ). So for a fixed k, this
can be made arbitrarily small by choosing sufficiently big n.
On the other hand, the volume of the set of these atomic measures is
n
n Y
Y
i−1
i
−1
−1
Vn =
(ai − ai−1 ) =
F
−F
n
n
i=1
i=1
132
and so its logarithm is
n
X
i=1
Now
F
−1
i
i−1
−1
−1
log F
−F
n
n
i
i−1
1 −1 0 i
−1
−F
≈ (F )
,
n
n
n
n
so the above expression is approximately
log Vn ≈
n
X
i=1
log((F
i
))
− n log n.
n
−1 0
Therefore as n goes to infinity,
Z
Z
Z
1
−1 0
−1
log(Vn ) + log(n) → log((F ) (x))dx = − log(f (F (x)))dx = − log(f (x))dF (x).
n
See (Voiculescu 1993) for the corresponding heuristics in the single-variable free entropy case.
Remark 13.10 (Gaussian
Now look at all the atomic measures with the second moment
Pn maximization).
2
equal
to 1, that is, i=1 xi = n. The volume of this set is related to the area of the sphere of radius
√
n in Rn , An = 2π n/2 Γ(n/2)−1 n(n−1)/2 . The corresponding normalized logarithmic volume converges
to 12 log(2π) + 12 , which is precisely the entropy of the standard gaussian distribution. Thus (1) almost
every such atomic measure is an approximant to the gaussian distribution, and (2) any other distribution
with variance 1 has entropy smaller than the Gaussian. Compare with the properties of the semicircular
distribution below.
Definition 13.11. Given X1 , . . . , Xn ∈ (A, τ ) in a tracial n.c. probability space. What is the volume of
the set of N × N self-adjoint matrices whose joint distribution is close to µX1 ,...,Xn ? Define the matricial
microstates
Γ(X1 , . . . , Xn ; m, N, ε)
= A1 , . . . , An ∈ MNsa×N (C) : τ Xu(1) . . . Xu(p) − tr Au(1) . . . Au(p) < ε for p ≤ m
(cutoff omitted for simplicity). The free entropy of (X1 , . . . , Xn ) with respect to τ is
1
n
log vol Γ(X1 , . . . , Xn ; m, N, ε) + log N .
χ(X1 , . . . Xn ) = inf inf lim sup
m∈N ε>0 N →∞
N2
2
Remark 13.12. Is Γ(X1 , . . . , Xn ; m, N, ε) always non-empty? The question is equivalent to the Connes
embedding problem.
Remark 13.13. If could replace the lim supN →∞ by limN →∞ , many results would follow.
Remark 13.14 (Properties of χ).
133
a. (One-variable formula) If µ = µX ,
ZZ
3 1
χ(X) =
log |s − t| dµ(s) dµ(t) + + log 2π,
4 2
minus logarithmic energy.
b. χ(X1 , . . . , Xn ) is either finite or −∞.
c. (Subadditivity)
χ(X1 , . . . , Xm+n ) ≤ χ(X1 , . . . , Xm ) + χ(Xm+1 , . . . , Xm+n ).
d. (Free independence) Let χ(Xj ) > −∞, 1 ≤ j ≤ n. Then
χ(X1 , . . . , Xn ) = χ(X1 ) + . . . + χ(Xn )
if and only if {X1 , . . . , Xn } are freely independent.
Remark: Know this only for individual variables, not for subsets.
e. Normalize τ [Xj2 ] = 1, 1 ≤ j ≤ n. Then χ(X1 , . . . , Xn ) is maximal if and only if X1 , . . . , Xn are
free semicircular.
Free entropy dimension
Definition 13.15. (X1 , . . . , Xn ) ∈ (A, τ ). S1 , . . . , Sn free semicircular free from {X1 , . . . , Xn }. The free
entropy dimension is
δ(X1 , . . . , Xn ) = n + lim sup
ε↓0
χ(X1 + εS1 , . . . , Xn + εSn )
.
|log ε|
Many related versions.
Question 13.16. Is χ or δ a von Neumann algebra invariant? That is, if A = W ∗ (X1 , . . . , Xn ) =
W ∗ (Y1 , . . . , Yk ), is χ(X1 , . . . , Xn ) = χ(Y1 , . . . , Yk )? Would solve the isomorphism problem. Know (a
version of) δ is an algebra invariant.
Remark 13.17 (Extra properties of free dimension).
a. δ(free semicircular n-tuple) = n.
b. For any set of generators, δ(matrix algebra) = 1 −
1
.
n2
Question 13.18. If δ(X1 , . . . , Xn ) = n, does it follow that W ∗ (X1 , . . . , Xn ) ' L(Fn )?
Answer: no (Nate Brown 2005).
134
Remark 13.19 (Major applications).
a. Absence of Cartan subalgebras (Voiculescu 1996): if M has a Cartan subalgebra, then δ(M) ≤ 1.
So L(Fn ) does not come from a measurable equivalent relation.
b. Primality (Ge 1998): if M ' M1 ⊗M2 for infinite-dimensional M1 , M2 , then δ(M) ≤ 1. So L(Fn )
is prime.
Extensions by Popa and his school.
Non-microstates free entropy
Recall from Definition 7.6: the free difference quotient
X
∂Xi (Xu(1) . . . Xu(k) ) =
Xu(1) . . . Xu(j−1) ⊗ Xu(j+1) . . . Xu(k) .
u(j)=i
For X1 , . . . , Xn ∈ (A, τ ), define their conjugate variables ξ1 , . . . , ξn by
(τ ⊗ τ ) [∂Xi P (X1 , . . . , Xn )] = τ [ξi P (X1 , . . . , Xn )]
(Schwinger-Dyson equations). Typically ξi is not a polynomial. Sometimes, ξi ∈ L2 (A, τ ) exists. Then
define the free Fisher information
∗
Φ (X1 , . . . , Xn ) =
n
X
kξi k22
=
i=1
n
X
τ [ξi∗ ξi ]
i=1
(actually, a relative version of it).
Remark 13.20 (Properties).
a. Formula in the single-variable case.
b. Superadditivity.
c. Free additivity for subsets.
d. Minimum achieved if {Xi } free semicircular.
Define the non-microstates free entropy
Z 1 ∞
n
n
∗
∗
1/2
1/2
χ (X1 , . . . , Xn ) =
− Φ (X1 + t S1 , . . . , Xn + t Sn ) dt + log 2πe,
2 0
1+t
2
where Si are free semicircular.
Properties similar to χ. In particular χ(X) = χ∗ (X).
Question 13.21. Is χ = χ∗ ?
Theorem 13.22 (Biane, Capitaine, Guionnet 2003). χ ≤ χ∗ .
135
13.3
Operator algebras.
The description in this section is very brief and imprecise, and is meant to give only a very vague idea of
the field. It should not be used as a reference.
The von Neumann algebras of all countable, amenable groups with the infinite conjugacy class property
are the same, namely the hyperfinite II1 -factor. Similarly,
W ∗ (M2 (C), tr)⊗∞ ' W ∗ (M3 (C), tr)⊗∞ .
?
This is why a positive answer to the isomorphism problem for L(F2 ) ' L(F3 ) is not unreasonable.
Exercise 11.20 allows us to combine free semicircular and circular operators into a single semicircular
matrix, while the results from Lecture 14 of [NS06] allow us to cut up a semicircular element by a family
of matrix units.
Definition 13.23. If M is a a II1 -factor, p ∈ M a projection with τ [p] = α, the amplification of M is
Mα = pMp.
Mα depends only on α and not on p. Can do this also for larger α.
Theorem 13.24. If N ≥ 2, n ≥ 2, L(Fn ) ' MN (C) ⊗ L(FN 2 (n−1)+1 ), so that
(L(Fn ))1/N ' L(FN 2 (n−1)+1 ).
Idea of proof: realize L(Fn ) inside MN (C) ⊗ (A, ϕ) for some A.
1/2
Theorem 13.25. More generally if p, q ≥ 2 and λ = p−1
, then
q−1
(L(Fp ))λ ' L(Fq ).
Using this idea, Dykema and Rădulescu independently defined interpolated free group factors:
{L(Ft ) : t ∈ (1, ∞)} ,
L(Ft ) ∗ L(Fs ) ' L(Ft+s ).
Isomorphism problem: not solved. But know: either all L(Ft ), 1 < t ≤ ∞ are the same, or they are all
different.
Theorem 13.26 (Dykema). For n, k ≥ 2,
Mn (C) ∗ Mk (C) ' L(Fs ),
where (1 − n12 ) + (1 − k12 ) = s. More generally, formulas for free products of finite dimensional and even
hyperfinite algebras. Even with amalgamation.
Much, much more to be said.
136
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