FREE PROBABILITY
FOR
COMBINATORIALISTS
Richard P. Stanley
Department of Mathematics
M.I.T. 2−375
Cambridge, MA 02139
rstan@math.mit.edu
http://www−math.mit.edu/~rstan
Transparencies available at:
http://www−math.mit.edu/~rstan/trans.html
1
D. Voiulesu
R. Speiher
A. Nia
P. Biane
2
Classial (ommutative) probability: bi; i 2 I random (omplex)
variables in a ommutative C -algebra B
with 1
fbi : i 2 I g are independent if
E (bri11 brinn ) = E (bri11 ) E (brinn ) (s:t:a:)
8i1; : : : ; in all distint.
Equivalently, let Bi; i 2 I be unital subalgebras, with
Bi; i 2 I are
E (1)
= 1. Then
if
independent
E (b1 bn) = 0
whenever E (bj ) = 0 (1 j n) and
bj 2 Bij with i1; : : : ; in all distint.
3
Free (nonommutative) probability: A = C -algebra with 1; regard
elements as random variables.
Ai; i 2 I : unital subalgebras
' : A ! C linear (expetation)
'(1) = 1
The Ai's are free if '(a1 an) = 0
whenever '(aj ) = 0 (1 j n) and
aj 2 Aij with ij 6= ij +1 (1 j < n).
Note.
not
Does
redue to lassial
ase when A is ommutative. E.g., independent a; b need not satisfy '(aba) =
0.
4
Let B; C be free, b; bi
mal omputation gives:
'(b)
'(b11b22)
=
=
2 B, et.
For-
'(b)'()
'(b1b2)'(1)'(2)
+'(b1)'(b2)'(12)
'(b1)'(b2)'(1)'(2);
et.
5
MOMENTS AND
CUMULANTS
Classial ase. Let x; y be inde-
pendent random variables from the probability distributions (measures) x; y ,
say with ompat support. Given the
E (xn); E (yn), we want
moments
Let
E ((x + y)n):
X
tn
F(t) =
E (x ) ;
n
!
n0
essentially the
.
n
Fourier transform of
6
onvolution: Classial onvolution
dened by
x+y = x y
Then
F (t) = F(t)F (t)
log F (t) = log F(t) + log F (t):
(log F linearizes )
mn = E (xn) and
X tn
X tn
F(t) =
mn = exp n ;
n!
n!
n1
n0
Hene if
then n is a
n(x + y)
mn
umulant and
=
=
n(x)X
+ n(y )
fB1;:::;Brg2n
7
#B1 #Br :
a; b be free. Dene
a+b = a+
2b
the free onvolution of a and b.
Now let
CauhyZ transform:
G(z )
a
=
=
z
d(a)
z X
a
1
+
ordinary
1
n
'(an)z
n
1 2 C [ 1=z ℄ ;
the
generating funtion for
the moments mn = '(an).
8
R-transform R(z) by
1
R(z) + z = G(z)h 1i
Dene the
(ompositional inverse), so
G R(z ) +
1
z
= z:
Theorem. R2+ = R + R
Set R(z ) =
1
z
+
X
kn(a)z
n
1
kn(a + b) = kn(a) + kn(b)
when a; b are free. Call kn(a) a
umulant of .
1, so
n
9
free
1
12
2
11
10
3
9
4
8
5
6
7
NCn: lattie of nonrossing partitions
of f1; 2; : : : ; ng.
10
1234
123
124
12
13
12−34 14−23
134
234
23
24
34
14
NC4
11
#NCn
#(max. hains)
(^0; ^1)
'(a
n
2n
1
= Cn =
n+1 n
= nn 2
= ( 1)n 1Cn 1
X
) = mn =
fB1;:::;Brg2NCn
where ki = ki(a).
12
k#B1 k#Br ;
2
3
k13(a b ca 2 c 5 )
µab = µa X µ b
etc.
13
RANDOM MATRICES
Let A be a random n n hermitian
matrix, hosen from the
(
). Let
Gaussian unitary ensemble GUE
1 > > n
be the spetrum of
Choose
w
A.
2 Sn uniformly, and let
(1; : : : ; n) be the shape of w under
the RSK algorithm.
Theorem. i and i have the same
distribution (after resaling) as
1.
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n!
Let A; B; C be n n hermitian matries with eigenvalues = (1; : : : ; n),
= (1; : : : ; n) and = (1; : : : ; n).
Horn's onjeture. Charateriza-
tion of (; ; ) when A + B = C .
ss =
X
s
: Littlewood-Rihardson
Saturation onjeture.
k
=
6
0
)
6= 0
k;k
oeÆient
Klyahko. Saturation ) Horn.
Knutson-Tao. Saturation onje-
ture is true.
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A = n n hermitian matrix
spetrum : 1; : : : ; n
A = \natural" measure on n n
hermitian matries A0 with
spe(A) = spe(A0)
n
X
1
Æj
A =
n j =1
(mass 1=n at eah eigenvalue).
An; Bn : n n hermitian (s.t.a.)
Let 1; 2 be probability measures with
ompat support on R suh that
An ! 1; Bn ! 2
(weakly).
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Theorem. Choose A0n and Bn0 from
An , Bn . Then
+ 2
An+Bn ! 1 2
(weakly, in probability) as n ! 1.
0
0
I.e., if we know spe(A) and spe(B )
(n large) then we an bet with a good
hane to win, that
A+B A 2
+ B :
Thus spe(A) and spe(B ) determine
spe(A + B ) almost surely as n ! 1.
Equivalently, A0n and Bn0 beome free
in the limit n ! 1 with respet to
'() =
1
n
17
tr():
ASYMPTOTIC
REPRESENTATION
THEORY OF Sn
Fix
` k.
Let
` n,
= irred. har. of Sn indexed by () := (1n k );
value at w 2 Sn of yle type (; 1n k ).
Asymptotis of
large?
()
18
for
xed,
n
1
λ
x1 y1 x 2
19
y2 x3 y3 x4
ωλ
ω
j!(u) !(v)j ju vj; !(u) = juj for juj >> 0
1
(u) = 2 (!(u) juj) (ompat support)
20
Z
G! (z) := z exp x z 0(x)dx
R
Y
(z yi)
G! = Y
(z xi)
1 X
G! (z ) = +
a
(! )z n 1
n
z
1
1
1
n
K! (z) = G! (z)h 1i
1X
n
C
(
!
)
z
=
n
z
1
1
C1(!) = 0 8!; C2(!) = jj
an(!) is a moment and Cn(!) a free
n
umulant of some probability measure
m! .
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Consequenes.
Theorem. Let n ` n and
!^n(u) = n 1=2!n(n1=2u)
(so diagram of is resaled to have
area 1). Supppose that
!^ n ! !
n
(so C2(! ) = 1). Then (; 1n
1
0
Ỳ
= Ci+1(! )A n(k
`)=2
i=1
where
f
n
= (1; : : : ; `) ` k
n n
= (1 )
= # SYT of shape
22
k)
=f
1+O
n :
n
1 n
;
Theorem (asymptoti Littlewood-Rih-
ardson rule). Let
!^ n ! !; !^ n ! !0:
Almost all shapes (ounting multipliity) appearing in sn s n are \near" a
shape n suh that
where
!^ n ! ! 2
+ ! 0;
Cn(! 2
+ ! 0) = Cn(! ) + Cn(! 0).
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