Sharp Bounds for Sums Associated
to a Graph of Matrices
James Mingo
(joint work with Roland Speicher)
Queen’s University at Kingston
Brazilian Operator Algebras Symposium
Florianópolis, 2011
[arXiv:0909.4277]
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Graph Sums
I
G = (V, E) is a directed graph, e ∈ E runs from s(e) to t(e)
I
T : E −→ MN (C), e 7→ Te = (tij ) is an oriented assignment
of matrices to edges (flip edge = transpose matrix)
tij
I
(e)
S(G, T) =
X
Y
i
(e)
tit(e) is(e)
j
where the sum runs over all
i:V→[N] e∈E
functions i : V → [N] = {1, 2, 3, . . . , N}; ik = i(k)
Examples
i
T
T
S(G, T) = Tr(T) =
i
X
i
tii
S(G, T) =
j
X
tij
i,j
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More Examples
j
T1
i
T1
T3
S(G, T) =
j
k
T2
X
i
l
(1) (2) (3)
tij tjk tjl
S(G, T) =
i,j,k,l
T3
X
k
(1) (2) (3)
tij tjk tki
i,j,k
T4
i4
i3
T3
S( G, T) =
N
X
(1) (2) (3) (4) (5) (6)
ti1 i2 ti3 i2 ti3 i4 ti4 i4 ti5 i3 ti2 i5
T2
i1
T2
T 1 i2
T5
i1 ,i2 ,...,i8 =1
T6
i5
T11
i8
T12 i7
T10 T8
T7
(7) (8) (9) (10) (11) (12)
× ti6 i5 ti6 i5 ti6 i6 ti7 i5 ti8 i7 ti8 i7
i6
T9
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the question (inspired by Bai & Silverstein (2006) (1) )
S(G, T) =
X
Y
(e)
tit(e) is(e)
i:V→[N] e∈E
I
given G = (V, E) find r(G) such that for all N and all
T : E → MN (C) we have
|S(G, T)| 6 Nr(G)
Y
kTe k
e∈E
where kTk is the operator norm of T ∈ MN (C)
I
note that r(G) > 1
I
answer: r(G) depends on the number of leaves of a tree (or
union of trees) associated with G
(1)
but with a different answer, also Yin & Krishnaiah (1983), and Bai (1999)
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motivation
I
Let {AN }N be a sequence of symmetric random matrices
with i. i. d. centred entries and TN,1 , . . . TN,k a sequence of
deterministic matrices (i.e. “just a matrix”)
I
we want to find the leading order terms of
E Tr(T1 AT2 A · · · ATk−1 ATk )
I
this leads to finding which for partitions π of [2k] the sum
N
X
(1)
(k)
ti1 i2 · · · ti2k−1 i2k
i1 ,...,ik =1
(subject to the constraint that ir = is whenever r and s are in
the same block of π) will contribute to the leading order
I
this can easily be written as a graph sum of a graph Gπ
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some notation, so we can say what r(G) is
I
let G = (V, E) be a directed graph
I
a cutting edge is one whose removal disconnects the
connected component which contains it
I
a subgraph is two-edge connected if it cannot be
disconnected by removing a single edge
I
a two-edge connected component is a two-edge connected
subgraph not properly contained in another two-edge
connected subgraph
I
F(G) is the graph whose vertices are the two-edge
connected components of G and whose edges are the
cutting edges of G
I
F(G) is a union of trees (= a forest)
I
a leaf of a tree (or forest) is a vertex with only one incident
edge
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The Theorem
I
let (G = (V, E) be a directed graph with F(G) the forest of
two-edge connected components
I
if G is connected r(G) = max{1, `/2} where ` is the number
of leaves of F(G)
I
in general, r(G) is the sum of the contributions of each
connected component
T
i4
i3
T2
T1
j
i
T3
r = 3/2
i1
T1
T 1 i2
T2
T5
T6
i5
T11
l
i
T3
T2
j
k
4
T3
r=1
k
i8
T12 i7
T10 T8
T7
i6
T9
r = 3/2
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Input-Output Graphs
I
I
I
I
I
I
I
I
G = (V, E), a directed graph is a input-output graph if we
can write V as a disjoint union: Vin ∪ Vout ∪ Vint , with
Vin = input vertices;
Vout = output vertices; and
Vint = internal vertices = everything else, and
G contains no directed cycles;
each v ∈ Vint is on a directed path from a vertex in Vin to a
vertex in Vout ;
each input vertex has only outgoing edges; and
each output vertex has only incoming edges.
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The Operator TG when G = (V, E) is input-output
I
I
I
for each v ∈ V, let Hv = CN , ξi = [0, . . . , 1, . . . , 0] ∈ Hv , √
and ξ(v) = ξ1 + · · · + ξN = [1, . . . , 1, . . . , 1] ∈ Hv , kξk2 = N
O
a basis element of
Hv is specified by a function
v∈Vin
j : Vin → [N] and the corresponding vector is
O
ξjv
v∈Vin
I
define TG :
O
Hv −→
v∈Vin
D O
w∈Vout
ξjw , TG
O
Hv by
v∈Vout
O
ξiv
E
v∈Vint
=
X Y
k:V→[N] e∈E
hξkt(e) , Te ξks(e) i
(e)
(=tk
t(e) ks(e)
)
where the sum runs over all k : V → [N] such that k|V = i and
in
k|Vout = j
O
O (w)
ξ(v)
S(G, T) =
ξ , TG
w∈Vout
v∈Vin
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S(G, T) =
O
w∈Vout
O
ξ(w) , TG
ξ(v)
v∈Vin
thus |S(G, T)| 6 N(#(Vin )+#(Vout ))/2 kTG k
I
moreover we can write TG = Lr Tr Lr−1 · · · T1 L0 where the
Li ’s are partial isometries and each Ti is a tensor product of
some Te ’s
L0 T1
L1 T2 L2 T3 L3
T4 L4
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conclusion of method
I
suppose our graph is
where
L∗ L"
L : H → H ⊗ H is given by L(ξi ) = ξi ⊗ ξi and
L 0 : H → H ⊗ H ⊗ H is given by L 0 (ξi ) = ξi ⊗ ξi ⊗ ξi
Theorem Let G be a graph. By splitting vertices and adding I,
flipping edges we can modify a graph to obtain G 0 so that
I
the graph sums are the same;
I
the forests of two edge connected components are the same
I
the products of the norms kTe k are the same
such that
I
G 0 is a input-output graph
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I
think of our graph as an algorithm, where we are feeding
input vectors into the input vertices and then operate them
through the graph, each edge doing some calculation, and
each vertex acting like a logic gate, doing some
compatibility checks.
I
the main problem is the timing of the various operations,
in particular, how long one has to wait at a vertex, before
applying an operator on an outgoing edge.
I
in algorithmic terms, it is clear that one has to wait until all
the input information is processed; i.e. one has to wait for
information to arrive along the longest path from an input
vertex to the given vertex.
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converting a two-edge connected graph to an
input-output graph
I
let v and w be distinct vertices, declare v to be input and w
to be output;
I
choose a simple path from v to w (no loops or backtracks),
this makes a subgraph G1 which is an input-output
subgraph;
I
we form an increasing sequence of subgraphs
G1 ⊂ G2 ⊂ · · · ⊂ Gk by adding paths: if e is an edge with a
vertex x in Gk and the other vertex z not in Gk , then there is
a path from z to Gk without passing through x, say
returning to Gk at y; orient this path connecting x to y and
add to Gk to get Gk+1 so that Gk+1 has no oriented cycles;
I
this works unless x = y, in which case we split a vertex.
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vertex splitting
I
sometimes to make a nice input-out graph we need to split
a vertex
(3)
T3
i3
T2
(2)
(4)
ti2 i3
ti4 i2
T5
i2
I
i4
i4
T4
T1
i1
i3 ti3 i4
i5 i1 (1)
ti1 i2
i2
!
δi2 i!2 i2
(5)
ti2 i5
i5
so we split the vertex and add an edge; on the new edge
we put the identity matrix; this forces the indices at either
end to be equal
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optimality and edge collapsing
How do we know that
we can always find T : E → MN (C) such
Y
r(G)
that |S(G, T)| = N
kTe k?
e
I
on each edge which is not a cutting edge place I; this
collapses each two-edge connected component to a point;
resulting graph os a tree (or forest)
I
on each edge which is not incident to a leaf place the
identity matrix; flip vertices so that graph now looks like:
V
V
V
t
V
I
√
let V(ξ1 ) = ( N)−1 ξ, V(ξi ) = 0 for i > 1
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