RANDOM MATRICES AND FREE PROBABILITY

JACOB CAMPBELL

These are notes for a graduate topics course at the University of Virginia in Spring 2026. The main aim of the course is to learn about recent developments in strong convergence and its applications. See the table of contents for more details.

Contents

Contents
 1.  GUE and genus expansion
 1.1.  Moments of gaussians
 1.2.  Self-adjoint gaussian matrix
 1.3.  Exercises
 2.  Semicircle law and noncommutative probability spaces
 2.1.  Which pairings survive in the limit?
 2.2.  Moment problem
 2.3.  Noncommutative probability spaces
 2.4.  Exercises
 3.  Free independence
 3.1.  Motivation and definition
 3.2.  What does freeness actually do?
 3.3.  Free central limit theorem
 3.4.  Exercises
 4.  Asymptotic freeness of multiple GUEs
 4.1.  Free semicircular families
 4.2.  Asymptotic freeness of GUEs
 4.3.  Exercises
References

1. GUE and genus expansion

1.1. Moments of gaussians. Recall the following distribution:

Definition 1.1. A random variable X has a standard gaussian distribution if it has density

fX(x) = 1 2πex2 2

for x . This is the case of mean 0 and variance 1.

Actually, we care about all the moments:

Proposition 1.2. If X is standard gaussian, then

𝔼(X2k) = (2k 1)!! and 𝔼(X2k1) = 0

for k 1.

Proof. Exercise.

Definition 1.3 (Set partitions). A partition of the set [n] := {1,,n} is a collection {V 1,,V l} of blocks, which are disjoint subsets of [n], with [n] = i=1lV i. The set of partitions of [n] will be denoted by P(n). A pair partition is a partition whose blocks all have size 2; the set of pair partitions of [n] will be denoted by P2(n). Note that P2(n) = when n is odd.

Let ϕ : P(n) Sn be the map which turns blocks into cycles, with the usual order inherited from [n]. This is obviously injective, and the range of P2(n) is the set of permutations in Sn with order 2 and no fixed points. We will refer interchangeably to P2(n) and its image in Sn; notice that in the case of P2(n), the arbitrary choice of order in the definition of ϕ does not actually matter.

Example 1.4.

P2(4) = {{{1,2},{3,4}},{{1,3},{2,4}},{{1,4},{2,3}}}

Proposition 1.5. We have |P2(2k)| = (2k 1)!! for k 1.

Idea of proof. When you choose what gets paired with 1, you have 2k 1 choices. Next, you have 2k 3 choices. This goes on to give you (2k 1)!! choices in total. Exercise: formalize this idea.

Theorem 1.6 (Wick formula). Let X1,,Xn be gaussian with covariance matrix Σ. Then for 𝐢 : [k] [n], we have

𝔼(X𝐢(1)X𝐢(k)) = πP2(k) (r,s)π𝔼(X𝐢(r)X𝐢(s)).

Proof. Exercise.

Definition 1.7 (Complex gaussian). A standard complex gaussian variable Z is obtained by letting X and Y be independent standard real gaussians and letting Z = 1 2(X + 𝑖𝑌 ). This has mean 0 and variance 1:

𝔼(Z) = 1 2(𝔼(X) + 𝑖𝔼(Y )) = 0

and

𝔼(ZZ¯) = 1 2𝔼((X + 𝑖𝑌 )(X 𝑖𝑌 )) = 1 2(𝔼(X2) + 𝔼(Y 2)) = 1.

Proposition 1.8. Let Z be a standard complex gaussian variable. Then

𝔼(ZmZ¯n) = { m!if m = n 0 otherwise

for m,n 0.

Proof. Exercise.

Remark 1.9. By multilinearity, the structure of the Wick formula is retained when one replaces some of the real gaussians with complex gaussians and/or their adjoints. We will use this observation freely.

1.2. Self-adjoint gaussian matrix.

Definition 1.10 (Gaussian unitary ensemble). Let A = (a𝑖𝑗)1i,jN be a matrix where

The point of this definition is that A is self-adjoint with independent gaussian entries, except as required by the self-adjointness. The choice of variance 1 N is for normalization and will play an important role when we make N .

Definition 1.11. Let (Ω,F,P) be a probability space and for 1 i,j N, let a𝑖𝑗 : Ω be a random variable. Assume that A = (a𝑖𝑗)i,j is self-adjoint. This random matrix produces a random probability measure

νA := 1 N i=1Nδ λi

on , where λ1,,λN are the eigenvalues of A. This is called the empirical spectral distribution (or ESD) of A.

The average eigenvalue distribution of A, denoted by μA, is simply the mean of νA: define μA by

f(x)dμA(x) = 𝔼 (f(x)dνA(x))

for measurable f. This gives us a nice way to access the moments:

xmdμ A(x) = 𝔼 (xmdν A(x)) = 𝔼 ( 1 N i=1Nλ im) = 𝔼tr N(Am).

Theorem 1.12 (Wigner’s semicircle law). Let A be the random matrix from Definition 1.10. Then μA μ weakly as N , where 𝑑𝜇 = f𝑑𝑥 with

f(x) = { 1 2π4 x2if x [2,2] 0 otherwise .

Remark 1.13. The mode of convergence in Theorem 1.12 can be strengthened. Namely, one can show νA μ weakly in probability or weakly almost surely. This could be a good starting point for your project.

In this class and next, we will prove Theorem 1.12 using moments. The moments of the average eigenvalue distribution turn out to be encoded by a combinatorially rich sequence of polynomials – related to topological genus of certain surfaces – evaluated at 1 N. We will come back to the latter connection later on; for now, we will only be concerned with the leading order.

Notation 1.14. Let γn = (1,,n). We will use the notation #() for the number of disjoint cycles in a permutation (including singletons/fixed points).

Theorem 1.15 (Genus expansion). We have

𝔼(trN(Am)) = πP2(m) ( 1 N )m 2 +1#(γmπ)

for m 1. When m is odd, the formula above should be read as an empty sum, i.e. as 0.

Remark 1.16. For π P2(2k), there are at least two ways to interpret the exponent k + 1 #(γ2kπ). It is 2gπ, where gπ is defined in either of the following two ways:

  1. gπ is the genus of the surface obtained from a polygon with 2k sides by gluing the sides together in pairs according to π;
  2. gπ is the smallest possible genus for which the following can be done: put the elements of [2k] on a circle clockwise, make that circle the boundary of a surface with genus gπ, and draw π on the surface without crossings.

This is where the phrase “genus expansion” comes from.

Proof of Theorem 1.15. We have

𝔼(trN(Am)) = 1 N 𝐢:[m][N]𝔼(a𝐢(1)𝐢(2)a𝐢(m)𝐢(1)) = 1 N 𝐢:[m][N] πP2(m) (r,s)π𝔼(a𝐢(r)𝐢(r+1)a𝐢(s)𝐢(s+1)) = 1 N 𝐢:[m][N] πP2(m) (r,s)πδ𝐢(r)=𝐢(s+1)δ𝐢(r+1)=𝐢(s) 1 N = 1 N 𝐢:[m][N] πP2(m) 𝐢=𝐢γmπ ( 1 N )m 2 = ( 1 N )m 2 +1 πP2(m)|{𝐢 : [m] [N] : 𝐢 = 𝐢 γm π}| = ( 1 N )m 2 +1 πP2(m)N#(γmπ) = πP2(m) ( 1 N )k+1#(γmπ)

since a map 𝐢 : [m] [N] with 𝐢 = 𝐢 γm π amounts to a choice of label from [N] for each cycle in γmπ.

1.3. Exercises.

Exercise 1.17. In this exercise, you will show that the moments of a standard gaussian variable count pair partitions.

  1. Let X be a standard gaussian variable. Prove that

    𝔼(X2k) = (2k 1)!! and 𝔼(X2k1) = 0

    for all k 1. Use integration by parts to find a recursion.

  2. Prove that |P2(2k)| = (2k 1)!! by putting P2(2k) in bijection with a set of cardinality (2k 1)|P2(2k 2)|.

Exercise 1.18. In this exercise, you will prove the Wick formula: if X1,,Xn are gaussian with mean 0 and covariance matrix Σ, then

𝔼(X𝐢(1)X𝐢(k)) = πP2(k) (r,s)π𝔼(X𝐢(r)X𝐢(s))

for all 𝐢 : [k] [n].

  1. Assume Σ is diagonal and prove the claim.
  2. Prove the claim in general by diagonalizing Σ and using the multilinearity of the claimed formula.

Exercise 1.19 ([3, Exercise 1.6]). Let Z be a standard complex gaussian variable. Show that the moments are

𝔼(ZmZ¯n) = { m!if m = n 0 otherwise .

Hint: start by showing that

𝔼(ZmZ¯n) = 1 π2(t1 + it2)m(t 1 it2)ne(t12+t 22) dt1dt1

and then switch to polar coordinates to show that

𝔼(ZmZ¯n) = 1 π02π0rm+n+1e𝑖𝜃(mn)er2 𝑑𝑟𝑑𝜃.

Use this to prove the claim.

2. Semicircle law and noncommutative probability spaces

Recall the theorem from last class:

Theorem 1.15 (Genus expansion). We have

𝔼(trN(Am)) = πP2(m) ( 1 N )m 2 +1#(γmπ)

for m 1. When m is odd, the formula above should be read as an empty sum, i.e. as 0.

What happens when we send N ? Clearly, for the RHS of Theorem 1.15 to converge, the exponents have to be non-negative; if this is indeed the case, the only summands that will survive are the ones with m 2 + 1 #(𝛾𝜋) = 0.

Definition 2.1. A partition π P(n) is said to be noncrossing if the following situation never occurs: there are some blocks V,W π with V W, and some a,c V and b,d W with a /mo> b /mo> c /mo> d. The set of noncrossing partitions of [n] is denoted by 𝑁𝐶(n).

Theorem 2.2 ([1]). We have #(𝛾𝜋) k + 1 for all π P2(2k). Moreover, we have equality if and only if π is noncrossing.

We will come back to this in a moment. First, let’s use it to find the limiting moments of μA.

Notation 2.3. Write NC2(m) for the non-crossing pair partitions of [m]. When m is odd, NC2(m) is of course empty.

Proposition 2.4. We have

|NC2(2k)| = 1 k + 1( 2k k)  and |NC2(2k 1)| = 0

for all k 1.

Proof. Exercise.

Definition 2.5. The k-th Catalan number is Cat(k) := 1 k+1( 2k k) .

Corollary 2.6. We have

lim N𝔼(trN(A2k)) = Cat(k) and lim N𝔼(trN(A2k1)) = 0

for k 1.

2.1. Which pairings survive in the limit?. Recall the core combinatorial theorem:

Theorem 2.2 ([1]). We have #(𝛾𝜋) k + 1 for all π P2(2k). Moreover, we have equality if and only if π is noncrossing.

To see the inequality, we can make a simple observation about permutations:

Lemma 2.7. Let α Sn and let τ = (i,j) be a transposition. Then

#(𝛼𝜏) = { #(α) + 1if i and j are in the same cycle of α #(α) 1 if i and j are in different cycles of α .

Remark 2.8. The permutation γ2kπ can be thought of in terms of a building process: each pair in π is a transposition, which bumps the cycle count up or down by 1. We start with the one cycle of γ2k, and the maximal situation is that each of the k pairs in π increases the number of cycles. This makes #(γ2kπ) k + 1.

For example, let k = 3 and π = (1,4)(2,3)(5,6). Then we start at (1,2,3,4,5,6) with 1 cycle. Next, we multiply by (1,4):

(1,2,3,4,5,6)(1,4) = (1,5,6)(2,3,4)

which has two cycles. Next,

(1,2,3,4,5,6)(1,4)(2,3) = (1,5,6)(2,4)(3)

which has three cycles. Finally,

(1,2,3,4,5,6)(1,4)(2,3)(5,6) = (1,5)(2,4)(3)(6)

which has four cycles. This is the maximal situation: each pair split a cycle in two.

On the other hand, now consider π = (1,5)(2,3)(4,6). Then we again start at (1,2,3,4,5,6) with 1 cycle. Then

(1,2,3,4,5,6)(1,5) = (1,6)(2,3,4,5)

has two cycles, as before. Next,

(1,2,3,4,5,6)(1,5)(2,3) = (1,6)(2,4,5)(3)

which is another splitting step (notice we haven’t arrived at the crossing yet) giving us three cycles. The last step is different:

(1,2,3,4,5,6)(1,5)(2,3)(4,6) = (1,6,5,2,4)(3)

which is back down to two cycles. This is the generic situation: we can have both splits and merges.

Analyzing the case of equality can probably be done directly, but it is really more insightful to put all this in a proper algebraic framework. Let’s work with general partitions and permutations, with size n; the results we need for pairings will fall out as special cases.

Definition 2.9. For α Sn, let |α| be the length of α: the minimal number of transpositions needed to factor α. For α,β Sn, write d(α,β) := |α1β|.

Proposition 2.10. Notation as above. Then

  1. |α| = n #(α) for all α Sn;
  2. (Sn,d) is a metric space.

Proof. For (1), use Lemma 2.7. (2) is a very direct and straightforward verification of axioms.

Proposition 2.11 ([1]). Let ϕ : P(n) Sn be the injective map which turns blocks into cycles. Let

S𝑁𝐶(γn) := {α Sn : d(e,α) + d(α,γn) = d(e,γn)}.

Then S𝑁𝐶(γn) is the range of ϕ|𝑁𝐶(n).

Proof. Finicky induction but elementary. See [1, Theorem 1].

Proof of Theorem 2.2. In (S2k,d), for π P2(2k), the triangle inequality gives

d(e,γ2k) d(e,π) + d(π,γ2k).

Unpacking notation, this translates to

2k 1 (2k k) + (2k #(γ2kπ))

which simplifies to #(γ2kπ) k + 1. The case of equality, i.e.

d(e,γ2k) = d(e,π) + d(π,γ2k),

corresponds to π being noncrossing due to Proposition 2.11.

2.2. Moment problem.

Question 2.12. We just showed that the moments of the average eigenvalue distribution of a GUE converge to the Catalan numbers. How do we know this is a moment sequence, and how do we know this determines the weak limit of the average eigenvalue distribution?

The Catalan numbers appear as the moments of a particular distribution:

Proposition 2.13. Let 𝑑𝜇 = f(x)𝑑𝑥 where

f(x) = { 1 2π4 x2if x [2,2] 0 otherwise .

Then

x2k𝑑𝜇(x) = 1 k + 1( 2k k)  and x2k1𝑑𝜇(x) = 0

for all k 1.

Proof. Exercise.

Okay, so we can guess that maybe the average eigenvalue distribution of a GUE converges to the semicircle distribution. But we just found the moments match – a priori this could just be a funny coincidence. So we can refocus Question 2.12:

Question 2.12. If (μn)n1 is a sequence in Prob() and μ Prob() has

lim nxmdμ n(x) =xm𝑑𝜇(x)

for all m 1, then does μn converge to μ in some meaningful sense at the level of measures?

The answer to Question  is a qualified “yes”:

Theorem 2.14. Let μ be a probability measure on with compact support.

  1. If ν is another probability measure on with

    xm𝑑𝜇(x) =xm𝑑𝜈(x)

    for all m 1, then μ = ν.

  2. If (μn)n1 is a sequence of probability measures on with

    lim nxmdμ n(x) =x,𝑑𝜇(x)

    for all m 1, then μn μ weakly as n .

Proof. Big exercise. See [2, Theorems 30.1 &30.2].

Proof of Theorem 1.12. Let μN be the average eigenvalue distribution of an N × N GUE, and let μ be the semicircle distribution with radius 2. We have already proved that

lim Nxmdμ N(x) =xm𝑑𝜇(x)

for all m 1. Since μ has compact support, Theorem 2.14 shows that μA μ weakly as N .

2.3. Noncommutative probability spaces.

Definition 2.15. A -algebra is a unital associative algebra over with an operation 𝒜 𝒜 : aa which is conjugate-linear and has (a) = a and (𝑎𝑏) = ba for all a,b 𝒜. An element a 𝒜 is said to be positive if a = bb for some b 𝒜.

Definition 2.16. A -probability space is a pair (𝒜,φ) where 𝒜 is a -algebra and φ is a linear functional on 𝒜 with φ(aa) 0 for all a 𝒜, and φ(1) = 1.

Remark 2.17. The point of Definition 2.16 is to view the elements of 𝒜 as “random variables” and the linear functional φ as the “expectation”. This abstraction becomes useful when one needs to consider noncommutative objects as random variables.

Example 2.18 (Commutative case). Let (Ω,F,P) be a (classical) probability space, i.e. a measure space with total measure 1. Let 𝒜 := L(Ω,F,P) be the algebra of essentially bounded functions, and let φ be the linear functional defined by

φ(f) =Ωf(ω)𝑑𝑃(ω)

for f 𝒜. This is a -probability space, where the -operation is complex conjugation – in fact, it has a lot of analytic structure that we will put aside for now.

Example 2.19 (Finite moments). Major objection to Example 2.18: many random variables we care about are not bounded. We can set up a similar example that will actually be more relevant for us: let 𝒜 = L(Ω,F,P) where L(Ω,F,P) := p1Lp(Ω,F,P), i.e. the algebra of random variables with all moments finite. (Use Hölder’s inequality to show 𝒜 is a -algebra: once to show Lp Lq for p q, and once to show 𝑓𝑔 L1 for f,g 𝒜.) The expectation φ is defined in the same way, by integrating functions against P. This is again a -probability space.

Example 2.20 (Scalar matrices). Let 𝒜 = MN be the algebra of N × N matrices and let φ(A) = 1 NTr(A). This is a -probability space which is not commutative.

Example 2.21 (Random matrices). Let 𝒜 = MN(L(Ω,F,P)) be the -algebra of matrices with entries that are RVs with finite moments, and let φ be the expected trace: φ(A) = 𝔼tr(A). For example, our GUE random matrix lives here.

So far, we haven’t seen anything new. The real reason for making this abstraction is to include genuinely noncommutative situations – where there is no hope of identifying an underlying classical probability space – and view them as essentially probabilistic in nature. The best example comes from groups and their group rings.

Example 2.22 (Group algebra). Let G be a group and let 𝒜 := [G]. The expectation is defined by

φ ( gGagg) = ae.

This is a -probability space.

Next class, we will use group algebras and group-theoretical freeness to motivate Voiculescu’s highly influential notion of free independence. Then, we will use it to construct a concrete operator model for the asymptotics of multiple GUEs.

2.4. Exercises.

Exercise 2.23. In this exercise, you will enumerate the noncrossing pairings and show that they are counted by the moments of the semicircle distribution. (1) and (2) are classic textbook combinatorics. (3) is a somewhat involved calculus problem.

  1. Find a recursion for |NC2(2k)|. Think back to the enumeration of P2(2k).
  2. Show that Cat(k) := 1 k+1( 2k k) satisfy the same recursion that you found in (1). To do this, first show that with C(z) := k=0Cat(k)zk, we have C(z) = 114z 2z , and derive the functional equation C(z) = 1 + 𝑧𝐶(z)2. Recover the recursion from this functional equation.
  3. Let 𝑑𝜇 = f(x)𝑑𝑥 where

    f(x) = { 1 2π1 4xif x [2,2] 0 otherwise .

    Show that

    x2k𝑑𝜇(x) = Cat(k) and x2k1𝑑𝜇(x)

    for k 1. To do this, make the substitution x = 2cos𝜃. You can use the following identity without proof:

    0π cos2m𝜃𝑑𝜃 = (2m 1)!! (2m)!! π.

Exercise 2.24. In this exercise, you will fill in the details of why convergence in moments implies weak convergence in the case of the semicircular law. In both parts, let μ be a probability measure on with compact support. This is part genuine exercise, part “book report”. Both parts require some knowledge of measure theory.

  1. Let ν be a probability measure on with

    xm𝑑𝜇(x) =xm𝑑𝜈(x)

    for all m 1. Prove that μ = ν. Note: there is no assumption that ν has compact support! Be careful and think about how to get around this issue.

  2. Let (μn)n1 be a sequence of probability measures on with

    lim nxmdμ n(x) =xm𝑑𝜇(x)

    for all m 1. Prove that μn μ weakly. This is standard textbook material (Billingsley P& Theorems 30.1 and 30.2). What I’m looking for is a concise summary of the argument that addresses its various subtleties: for example, how exactly is (1) being used, how do you establish tightness, etc. You don’t need to reproduce proofs of basic classical theorems, like Markov inequality or Helly selection theorem/Prokhorov’s theorem.

3. Free independence

3.1. Motivation and definition.

Definition 3.1 (Group-theoretical freeness). Let G be a group and let {Gi : i I} be a family of subgroups. The subgroups are free if for all k 1, for all i1,,ik I with ijij+1, and for all gj Gij with gje, we have g1gke.

How does this look in terms of the -probability space ([G],φ)? Well, the condition that

gjeg1gke translates to φ(gj) = 0φ(g1gk) = 0.

Definition 3.2 (Freeness of subalgebras). Let (𝒜,φ) be a -probability space, and let {𝒜i : i I} be a family of unital -subalgebras. The family {𝒜i : i I} is said to be freely independent (or free) if for all k 1, for all i1,,ik I with ijij+1, and for all aj 𝒜ij with φ(aj) = 0, we have φ(a1ak) = 0.

Proposition 3.3. Let G be a group and let {Gi : i I} be a family of subgroups. Then the subgroups are free in the group-theoretical sense if and only if the subalgebras [Gi] of [G] are freely independent with respect to φ.

Proof. Exercise. I think this is an essential one.

Definition 3.4 (Freeness of variables). Let (𝒜,φ) be a -probability space and let {ai : i I} be a family in 𝒜. For i I, let 𝒜i be the -subalgebra of 𝒜 generated by ai. The family {ai : i I} is said to be freely independent (or free) if the family {𝒜i : i I} is freely independent.

3.2. What does freeness actually do?. To understand what freeness means, let’s look at some examples. This section is entirely based on [4, Lecture 5].

Example 3.5. Suppose that 𝒜 and B are freely independent subalgebras of some ambient algebra. Let’s look at some words with mixtures between 𝒜 and B.

Length two: for a 𝒜 and b B, freeness says

0 = φ((a φ(a))(b φ(b))) = φ(𝑎𝑏) φ(a)φ(b)

so φ(𝑎𝑏) = φ(a)φ(b).

Length three: for a1,a2 𝒜 and b B, freeness says

φ((a φ(a))(b φ(b))(a φ(a))) = 0.

Expanding and using the length two factorization, this says

0 = φ(a1ba2) φ(a2)φ(a1b) φ(b)φ(a1a2) + φ(a1)φ(a2)φ(b) φ(a1)φ(ba2) + φ(a1)φ(a2)φ(b) + φ(a1)φ(b)φ(a2) φ(a1)φ(a2)φ(b) = φ(a1ba2) φ(a2)φ(a1)φ(b) φ(b)φ(a1a2) + φ(a1)φ(a2)φ(b) φ(a1)φ(b)φ(a2) + φ(a1)φ(a2)φ(b) + φ(a1)φ(b)φ(a2) φ(a1)φ(a2)φ(b) = φ(a1ba2) φ(a1a2)φ(b)

so φ(a1ba2) = φ(a1a2)φ(b).

IMPORTANT: this factorization pattern does not continue. The same kind of calculation, with a1,a2 𝒜 and b1,b2 B, shows that

φ(a1b1a2b2) = φ(a1a2)φ(b1)φ(b2) + φ(a1)φ(a2)φ(b1b2) φ(a1)φ(b1)φ(a2)φ(b2).

The point of all this is that mixed moments only depend on mixed moments from the same subalgebra, via some complicated pattern.

Proposition 3.6. Let (𝒜,φ) be a -probability space and let {𝒜i : i I} be freely independent -subalgebras. Let B be the -subalgebra of 𝒜 generated by {𝒜i : i I}. Then φ|B is uniquely determined by {φ|𝒜i : i I}.

Proof. See [4, Lemma 5.13].

3.3. Free central limit theorem. Similar to the last section, this one is entirely based on [4, Lecture 8]. From now on, fix a -probability space (𝒜,φ) and a sequence (an)n1 in 𝒜 with an = an, φ(an) = 0, and φ(an2) = 1 for all n 1. Also assume all the an have the same moment sequence and they are freely independent.

Theorem 3.7 ([5]). We have

lim Nφ ((a1 + + aN N )m) = { Cat(k)if m = 2k 0 otherwise

for m 1.

The first step is to simply unpack the LHS:

φ ((a1 + + aN N )m) = 1 Nm2 𝐢:[m][N]φ(a𝐢(1)a𝐢(m)).

Notation 3.8. For 𝐢 : [m] , let ker(𝐢) be the partition of [m] defined by letting p and q be in the same block if and only if 𝐢(p) = 𝐢(q).

Lemma 3.9. If 𝐢,𝐣 : [m] have ker(𝐢) = ker(𝐣), then

φ(a𝐢(1)a𝐢(m)) = φ(a𝐣(1)a𝐣(m)).

Idea of proof. This follows from the assumption that the an all have the same distribution, plus the fact that free independence means mixed moments are determined by the individual distributions.

For example, suppose the common partition is {{1,3},{2,4}}, and 𝐢 = (1,2,1,2) and 𝐣 = (5,3,5,3). Then, using the computations from Example 3.5, we have

φ(a1a2a1a2) = φ(a12)φ(a 2)2 + φ(a 1)2φ(a 22) φ(a 1)2φ(a 2)2

and

φ(a5a3a5a3) = φ(a52)φ(a 3)2 + φ(a 5)2φ(a 32) φ(a 5)2φ(a 3)2.

Since we assume the an all have the same moment sequence, these are the same.

Notation 3.10. In light of Lemma 3.9, for π P(m), write φ(π) for the common value of φ(a𝐢(1)a𝐢(m)) where 𝐢 : [m] [N] has ker(𝐢) = π. Let AπN be the number of maps 𝐢 : [m] [N] with ker(𝐢) = π.

So we have

φ ((a1 + + aN N )m) = 1 Nm2 πP(m)AπNφ(π)

and the task is to see what happens to AπN as N . We can immediately dispense with any π that has a singleton block:

Lemma 3.11. If π P(m) and there is some V π with |V | = 1, then φ(π) = 0.

Proof. Suppose that 𝐢(j) = r and {j} π. Then

φ(π) = φ(a𝐢(1)a𝐢(j1)ara𝐢(j+1)a𝐢(m)) = φ(a𝐢(1)a𝐢(j1)a𝐢(j+1)a𝐢(m))φ(ar)

using freeness and one of the computations in Example 3.5. We have φ(ar) = 0, so φ(π) = 0.

Now, we need to understand the limit of AπN where π P(m) has |V | 2 for all V π. This is straightforward: a map 𝐢 : [m] [N] with π = ker(𝐢) amounts to a choice of label from [N] for each block of π, where we do not allow duplicate labels. There are

N(N 1)(N |π| + 1)

choices, so we have

φ ((a1 + + aN N )m) = πP(m) |V |2V π N(N 1)(N |π| + 1) Nm2 φ(π).

The condition on π makes |π| m2, and when N , the fraction involving N goes to 1 if |π| = m2, and goes to 0 otherwise. The only way we can have |π| = 1 is if π P2(m). So we have

lim Nφ ((a1 + + aN N )m) = πP2(m)φ(π).

Of course, if m is odd, this is 0.

Now let us compute φ(π) for π P2(2k). Pick 𝐢 : [m] with π = ker(𝐢). If 𝐢(j)𝐢(j + 1) for all 1 j m 1, then φ(π) = 0 by freeness; otherwise, there is some 1 j m 1 with 𝐢(j) = 𝐣(j + 1). Using the computations from Example 3.5, we have

φ(π) = φ(a𝐢(1)a𝐢(j1)(a𝐢(j)a𝐢(j+1))a𝐣(j+2)a𝐢(m)) = φ(a𝐢(1)a𝐢(j1)a𝐢(j+2)a𝐢(m))φ(a𝐢(j)a𝐢(j+1)) = φ(a𝐢(1)a𝐢(j1)a𝐢(j+2)a𝐢(m)).

The pairings where we can keep doing this k times are exactly the ones that are noncrossing, so we have

φ(π) = { 1if π NC2(2k) 0if π P2(2k) NC2(2k) .

This shows that

lim Nφ ((a1 + + aN N )m) = { Cat(k)if m = 2k 0 otherwise

which is the claim of Theorem 3.7.

3.4. Exercises.

Exercise 3.12. Let G be a group and let 𝒜 = [G]. Let φ be the linear functional on 𝒜 defined by

φ ( gGagg) = ae.

I would say (3) is essential for understanding the concept of free independence. I strongly suggest you do it, even if you don’t hand it in.

  1. Prove that φ is positive by directly using the definition above.
  2. Find a Hilbert space H, a -homomorphism π : 𝒜 B(H), and a vector ξ H such that φ(a) = π(a)ξ,ξ for all a 𝒜. Use the fact that such a thing exists to prove that φ is positive.
  3. Let {Gi : i I} be a family of subgroups of G, and for i I, let 𝒜i := [Gi]. Prove that {Gi : I} are free in the group-theoretical sense if and only if {𝒜i : i I} are freely independent.

4. Asymptotic freeness of multiple GUEs

4.1. Free semicircular families. The example of freeness in the last section will become relevant when we talk about random unitary matrices, but for now, we are looking for some noncommutative data that captures the asymptotics of multiple GUEs. The answer is that when N , they become freely independent families of semicircular variables.

Definition 4.1 (Semicircular family). Let (𝒜,φ) be a -probability space. A free semicircular family in (𝒜,φ) consists of some elements s1,,sr 𝒜 such that

Theorem 4.2. For each r 1, there is a -probability space (𝒜,φ) and some elements s1,,sr 𝒜 such that

  1. {s1,,sr} is a free semicircular family, and
  2. 𝒜 is generated by {s1,,sr}.

In this section, we will construct a concrete instance of a free semicircular family using creation and annihilation operators on the so-called full Fock space. This model will allow for very concrete calculations involving certain lattice paths which are in obvious bijection with NC2.

Definition 4.3 (Full Fock space). Fix a Hilbert space H. The full Fock space on H is the Hilbert space

F(H) := Ω n=1Hn.

Here, the inner product of Hn is given on elementary tensors by

ξ1 ξn,η1 ηn = ξ1,η1ξn,ηn

and the direct sum is the set of sequences (ξn)n0 where ξ0 Ω and ξn Hn for n 1, with n=1ξn2 /mo> . The inner product is given by

(ξn)n0,(ηn)n0 = n=0ξ n,ηn.

The -algebra B(F(H)) carries a canonical vacuum state:

φ(T) = TΩ,Ω.

Then (B(F(H))),φ) is a -probability space.

Proposition 4.4. For ξ H, there is an operator c(ξ) B(F(H)) defined by c(ξ)Ω = ξ and

c(ξ)ξ1 ξn = ξ ξ1 ξn.

The adjoint is given by

c(ξ)Ω = 0c(ξ)ξ 1 = ξ1,ξΩ,

and

c(ξ)ξ 1 ξn = ξ1,ξξ2 ξn

and has the following useful property:

c(ξ)c(η) = η,ξ 1.

Specifically, c(ξ) is a scalar multiple of an isometry, and c(ξ) = ξ.

Proof. Exercise.

Proposition 4.5 (Semicircular). Let H be a Hilbert space and let ξ H. Then c(ξ) + c(ξ)B(F(H)) is a semicircular element with radius 2ξ.

The combinatorics of noncrossing pairings are directly built into this model, but in a slightly different form.

Definition 4.6. A Dyck path with m steps is a path in 0 × 0 which

Note that this definition includes the requirement that the path never goes below the x-axis. Let D(m) be the set of Dyck paths with m steps; when m is odd, D(m) = .

Lemma 4.7. There is a canonical bijection NC2(m) D(m).

Proof of Proposition 4.5. To clean up notation, let a = c(ξ). We will compute moments:

φ((a + a)m) = 𝜖1,,𝜖m{1,}φ(a𝜖1 a𝜖m ).

The key observation is that φ(a𝜖1a𝜖m) is just an indicator of a certain combinatorial property of the sequence (𝜖1,,𝜖m): the sequence determines a lattice path with NE and SE steps, and the indicator detects whether or not it is a Dyck path.

To see this, for 𝜖1,,𝜖m {1,}, let

λj = { 1 if 𝜖mj+1 = 1 1if 𝜖mj+1 =

for 1 j m. Then, if we read the expression ξ0 as Ω, we have

a𝜖1 a𝜖m Ω = { ξ(λ1++λm)if λ 1 + + λk 01 k m 0 otherwise

and

φ(a𝜖1 a𝜖m ) = a𝜖1 a𝜖m Ω,Ω = { ξ(λ1++λm),Ωif λ 1 + + λk 01 k m 0 otherwise

The inner product in the last line is 1 if λ1 + + λm = 0, and 0 otherwise. The condition λ1 + + λk 0 for all 1 k m means the path doesn’t go below the x-axis, and the condition that λ1 + + λm = 0 means it does return to the x-axis, i.e. it’s a Dyck path. So φ((a + a)m) is the number of Dyck paths with m steps. By Lemma 4.7, this shows that a + a is semicircular.

Proposition 4.8 (Freeness). Let H be a Hilbert space and let H1,,Hr be mutually orthogonal subspaces of H. For 1 i r, let 𝒜i be the -subalgebra of B(F(H)) generated by {c(ξ) : ξ Hi}. Then 𝒜1,,𝒜r are freely independent.

Proof. Let 1 i1,,ik r with ijij+1, and pick Tj 𝒜ij with φ(Tj) = 0. Assume without loss of generality that

Tj = c(ξ1j)c(ξ mjj)c(η 1j)c(η njj)

with mj + nj 1 and all vectors in Hij. The reason we can do this, and ignore contributions from 1, is that any non-zero contribution from 1 will make φ non-zero.

If there is some 1 j k 1 with nj0 and mj+10, then the product TjTj+1 includes the product c(ηnjj)c(ξ1j+1) = ξ1j+1,ηnjj1. Since Hj Hj+1, this is 0 and then φ(T1Tk) = 0.

Otherwise, for all 1 j k 1, we have either nj = 0 or mj+1 = 0. Then

T1Tk = c(ξ1)c(ξm)c(η1)c(η n)

for some ξ1,,ξm,η1,,ηn H with m + n 1, and

φ(T1Tk) = c(ξ1)c(ξm)c(η1)c(η n)Ω,Ω = c(η1)c(η n)Ω,c(ξ m)c(ξ 1)Ω.

Since at least one of m or n is 1, one of the two arguments in the inner product is of the form c(ξ)Ω = 0, so φ(T1Tk) = 0.

Proof of Theorem 4.2. Let {e1,,er} be the standard orthonormal basis of r, and let sj = c(ej) B(F(r)) for 1 j r. Let 𝒜 be the -algebra in B(F(r)) generated by {s1,,sr}, and let φ(T) = TΩ,Ω for T 𝒜.

Remark 4.9. Theorem 4.2 is a bit artificially weak, because I don’t want to assume any background in operator algebras. If we let 𝒜 be the C-subalgebra of B(F(r)) generated by {c(e1) + c(e1),,c(er) + c(er)} and let φ(T) = TΩ,Ω for T 𝒜, then

  1. Ω is a cyclic vector for 𝒜, meaning that {TΩ : T 𝒜} is dense in F(H);
  2. φ is a faithful trace;
  3. for any C-algebra B with a faithful state ψ, if B is generated by a free semicircular family of size r, then there is a state-preserving -isomorphism 𝒜 B.

This is a very nice object and in light of (3), it is referred to as the semicircular C-algebra with r generators. See [4, Lecture 7] for details.

4.2. Asymptotic freeness of GUEs.

Theorem 4.10 ([6]). Let {A1,,Ar} be an independent family of N × N GUEs, and let {s1,,sr}be a free semicircular family. Then (A1,,Ar) converges in distribution to (s1,,sr), in the following sense: for all m 1 and 𝐢 : [m] [r], we have

lim N𝔼tr(A𝐢(1)A𝐢(m)) = φ(s𝐢(1)s𝐢(m)).

This theorem follows from an easy multivariate generalization of the genus expansion:

Theorem 1.15. Let m 1 and 𝐢 : [m] [r]. Then

𝔼(trN(A𝐢(1)A𝐢(m))) = πP2(m) πker(𝐢) ( 1 N )m 2 +1#(γmπ).

When m is odd, the formula should be read as an empty sum, i.e. as 0.

Proof of Theorem 4.10. When N in the formula above, the only summands that survive are the ones with m 2 + 1 #(γmπ) = 0. These are the ones with π noncrossing, so

𝔼(trN(A𝐢(1)A𝐢(m))) = |{π NC2(m) : π ker(𝐢)}| + O(N2)

and we need to show that

φ(s𝐢(1)s𝐢(m)) = |{π NC2(m) : π ker(𝐢)}|.

This is similar to Proposition 4.5: in the proof there, the words a𝜖1a𝜖m which contribute to the count |NC2(m)| correspond to Dyck paths. There is a similar constraint here, but we also need pairs (r,s) of up and down steps to have 𝐢(r) = 𝐢(s). In terms of NC2(m), this amounts to π ker(𝐢).

Here is an example of how this works: consider the path which makes two up steps and then two down steps, or in other words the noncrossing pairing {{1,4},{2,3}}. This corresponds to

a𝐢(1)a 𝐢(2)a 𝐢(3)a𝐢(4)Ω = a𝐢(1)a 𝐢(2)(e 𝐢(3) e𝐢(4)) = a𝐢(1)e 𝐢(3),e𝐢(2)e𝐢(4) = e𝐢(3),e𝐢(2)e𝐢(1),e𝐢(4)Ω

so we have the additional requirement that 𝐢(2) = 𝐢(3) and 𝐢(1) = 𝐢(4), i.e. the pairing is compatible with 𝐢.

4.3. Exercises.

Exercise 4.11. Let H be a Hilbert space, let F(H) be its full Fock space, and let F0(H) be the dense subspace of finite linear combinations of elementary tensors. In this exercise, you will fill in the details of the construction of creation and annihilation operators on F(H).

  1. For ξ H, define c(ξ) : F0(H) F(H) by letting

    c(ξ)(ξ1 ξn) = ξ ξ1 ξn

    and extending by linearity. Prove that c(ξ) is bounded and extends to an element c(ξ) B(H) with c(ξ) = ξ. Try to do it in a way that implies c(ξ) is a scalar multiple of an isometry.

  2. Prove that

    c(ξ)Ω = 0c(ξ)ξ 1 = ξ1,ξΩ,

    and

    c(ξ)(ξ 1 ξn) = ξ1,ξξ2 ξn

    for n 2.

  3. Prove that c(ξ)c(η) = η,ξ 1.

Exercise 4.12. A C-probability space is a -probability space (𝒜,φ) where 𝒜 is equipped with a norm, with respect to which it is complete, which has the properties

This might be familiar to you as a C-algebra equipped with a state. As usual in this course, we consider all algebras and related structures to be unital. This exercise requires some knowledge of functional analysis beyond the prerequisites.

  1. Let (𝒜,φ) be a C-probability space and let a 𝒜 be normal (meaning that aa = aa). Prove that there is a unique compactly supported measure on such that

    φ(apaq) =zpz¯q𝑑𝜇(z)

    for all p,q 0. This is called the -distribution, or simply the distribution, of a.

  2. Let (𝒜,φ) be a C-probability space and let {𝒜i : i I} be a freely independent family of -subalgebras of 𝒜. For i I, let Bi be the norm-closure of 𝒜i. Prove that {Bi : i I} are freely independent.

References

1.
Philippe Biane, Some properties of crossings and partitions, Discrete Math. 175 (1997), no. 1-3, 41–53.
2.
Patrick Billingsley, Probability and measure, third ed., Wiley Series in Probability and Mathematical Statistics, John Wiley &Sons, Inc., New York, 1995.
3.
James A. Mingo and Roland Speicher, Free probability and random matrices, Fields Institute Monographs, vol. 35, Springer, New York; Fields Institute for Research in Mathematical Sciences, Toronto, ON, 2017.
4.
Alexandru Nica and Roland Speicher, Lectures on the combinatorics of free probability, London Mathematical Society Lecture Note Series, vol. 335, Cambridge University Press, Cambridge, 2006.
5.
Dan Voiculescu, Symmetries of some reduced free product C-algebras, Operator algebras and their connections with topology and ergodic theory (Buşteni, 1983), Lecture Notes in Math., vol. 1132, Springer, Berlin, 1985, pp. 556–588.
6.
_________ , Limit laws for random matrices and free products, Invent. Math. 104 (1991), no. 1, 201–220.