Free Probability Notes

Free probability and combinatorics Preliminary version

Michael Anshelevich c 2012

July 17, 2018 Preface

These notes were used in a topics course Free probability and combinatorics taught at Texas A&M Uni- versity 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 Noncommutative probability spaces and distributions. 3 1.1 Non-commutative probability spaces...... 3 1.2 Distributions. Weak law of large numbers...... 6

2 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

3 Free independence. 19 3.1 Independence. Free independence...... 19 3.2 Free products...... 22

4 Set partition lattices. 29 4.1 All, non-crossing, interval partitions. Enumeration...... 29 4.2 Incidence algebra. Multiplicative functions...... 33

5 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 Schurmann’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 Levy´ 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 X on . Then for Z A = C(X), ϕ [f] = f dµ, 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, . . . , xdi = Chxi be polynomials in d non-commuting variables, with the involution

∗ ∗ xi = 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. d One example: for z = (z1, z2, . . . , zd) ∈ R ,

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] = A = Tr(A) = tr(A), N ii N i=1 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.

n Example 1.11 (Group algebra). Let Γ be a discrete group (ﬁnite, Z , Fn, etc.).

C[Γ] = functions Γ → C of finite support = finite linear combinations of elements of Γ with C coefficients. X (f : x 7→ f(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 X X (fg)(z) = f(x)g(y) = f(x)g(x−1z), z=xy x∈Γ the convolution multiplication. The involution is

f ∗(x) = f(x−1).

5 Check that indeed, (fg)∗ = 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

n n {mn[ϕa] = ϕa [x ] = ϕ [a ] , 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, . . . , xdi → C, ϕa1,...,ad [p(x1, . . . , xd)] = ϕ [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, . . . , xdi.

6 c. We say that (N) (N) (a1 , . . . , ad ) → (a1, . . . , ad) in moments (or, for d > 1, in distribution) if for each p,

ϕ (N) (N) [p] → ϕ(a1,...,ad) [p] (a1 ,...,ad ) 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, n ϕ au(1)au(2) . . . au(n) ≤ C . Denote 1 s = (a + a + ... + a ). n n 1 2 n

Then sn → ϕ [a1] in moments (sn converges to the mean, a scalar).

Proof. Note ﬁrst that 1 (a − ϕ [a ]) + ... + (a − ϕ [a ]) = s − ϕ [a ] . n 1 1 n n n 1

So without loss of generality, may assume ϕ [a1] = 0, and we need to show that sn → 0 in moments.

n n 1 1 X X ϕ sk = ϕ (a + ... + a )k = ... ϕ a a . . . a . n nk 1 n nk u(1) u(2) u(k) 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) ⇔ i ∼π 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

k ϕ au(1)au(2) . . . au(k) ≤ C .

Thus k 1 k/2 k ϕ x ≤ B(k)n C → 0 n nk as n → ∞ for any k > 0.

Remark 1.17. For centered independent random variables, showed a + ... + a s = 1 n → 0. n n Expect a non-trivial limit for a + ... + a 1 √ n . 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

k X ∗ a = bi 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 well- deﬁned 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.

∗ n 1/2n d. Assume ϕ is faithful. Then for any a ∈ A, kak = limn→∞ (ϕ [(a a) ]) . 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 = kfku, and kak − a corresponds to kfku − 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 |ϕ [a]| ≤ pϕ [a∗a] ϕ [1∗1] ≤ pka∗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→∞ kfk2n,µ = kfk∞,µ and, if µ has full support, kfk∞,µ = kfku.

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, bi ≤ ha, ai hb, bi , ϕ ϕ ϕ

11 in other words |ϕ [b∗a]|2 ≤ ϕ [a∗a] ϕ [b∗b]. Let

n ∗ 2 o N = a ∈ A : ϕ [a a] = kakϕ = 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→ aˆ = a + N of A → Aˆ ⊂ L2(A, ϕ) with dense range. Moreover, for any a ∈ A and b ∈ N

2 ∗ ∗ ∗ kabkϕ = ϕ [b a ab] ≤ ka abkϕ kbkϕ = 0. So N is a left ideal in A. For a ∈ A,

λ(a)ˆb = ab + aN = ab.b Thus have a (not necessarily faithful) ∗-representation of A on the Hilbert space H, by a priori unbounded operators with a common dense domain Aˆ = 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, . . . , xdi, 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 ˆ 2 2 ∗ ∗ ∗ ∗ 2 ˆ λ(a)b = kab + N kϕ ≤ kabkϕ = ϕ [b a ab] ≤ ka ak ϕ [b b] = kak b . ϕ ϕ Finally, ˆ ˆ ϕ [a] = λ(a)1, 1 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, . . . , xdi such that for a ﬁxed C and all ~u, n µ xu(1)xu(2) . . . xu(n) ≤ C . 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 |~u| BC hxi = α~ux~u : α~u ∈ C, |α~u| C < ∞ . ~u ~u It is easily seen to be a ∗-algebra, to which µ extends via " # X X µe α~ux~u = α~uµ [x~u] . ~u ~u

2 1/2 xi Deﬁne gi to be the power series expansion of 1 − 4C2 . This series has radius of convergence 2C, and so lies in BC hxi. Since 1 g2 = 1 − x2, i 4C2 i for f ∈ Aˆ, 1 1 0 ≤ µ [f ∗g∗g f] = µ f ∗ 1 − x2 f = µ [f ∗f] − µ [(x f)∗(x f)] . e i i 4C2 i 4C2 i i It follows that

kxifkµ ≤ 2C kfkµ , 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). ( ) 2 2 X 2 L (C[Γ], τ) = L (Γ) = f :Γ → C, |f(x)| < ∞ . x∈Γ

Γ acts on L2(Γ) on the left by −1 (λ(y)g)(x) = (δyg)(x) = g(y x). C[Γ] acts on L2(Γ) on the left by X (λ(f)g)(x) = (fg)(x) = f(y)g(y−1x). y∈Γ

The reduced group C∗-algebra is

∗ k·k 2 Cr (Γ) = C[Γ] ⊂ B(L (Γ))

(there is also a full C∗-algebra C∗(Γ)). The group von Neumann algebra is

∗ weak 2 L(Γ) = W (Γ) = C[Γ] ⊂ B(L (Γ)).

The vector state τ = h·δe, δei 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.

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

14 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 n mn = ϕ [a ] = x dµa(x) are the moments of a. Always take m0 = 1. For a formal indeterminate z,

∞ X n M(z) = mnz 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 M(z) = dµa(x) R 1 − xz −1 for z ∈ C, |z| ≤ kak . The (formal) Cauchy transform is ∞ X 1 1 G (z) = m = M(1/z). µ n zn+1 z n=0 But in fact, the Cauchy transform is deﬁned by Z 1 Gµ(z) = dµ(x) R 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 1 dµ(x) = − 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 ∗ −1 u = δx−1 = u , so u is a unitary. Moreover, n τ [u ] = 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 τ (u + u−1)2n+1 = 0, τ (u + u−1)2n = . n Cauchy transform of the distribution is

∞ ∞ ∞ X 2n X (2n)! X 2n1 · 3 · ... · (2n − 1) G(z) = z−2n−1 = z−2n−1 = z−2n−1 n n!n! n! n=0 n=0 n=0 ∞ X 4n 1 3 2n − 1 = ... z−2n−1 n! 2 2 2 n=0 ∞ n X (−4) 1 3 2n − 1 −1/2 1 = − − ... − z−2n−1 = z−1 1 − 4z−2 = √ . n! 2 2 2 2 n=0 z − 4 Therefore the distribution is 1 1 1 1 dµ(x) = − lim =p dx = √ dx. π y→0+ (x + iy)2 − 4 π 4 − x2 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 all bounded continuous f. A general operator T is affiliated to A, T ∈ Ae, if in its polar decomposition 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 ϕ [f(a)] = f dµa R n R n for f ∈ Cb( ). Moments ϕ [a ] = x dµa(x) need not be ﬁnite, so ϕa may be undeﬁned. Also, can R R 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) (i) V1 ⊗ V2 ⊗ ... ⊗ Vn = a1 ⊗ ... ⊗ an , k ≥ 0, aj ∈ Vj /linearity relations i=1

a1 ⊗ ... ⊗ (za + wb) ⊗ ... ⊗ an = za1 ⊗ ... ⊗ a ⊗ ... ⊗ an + wa1 ⊗ ... ⊗ b ⊗ ... ⊗ an. Thus for example a ⊗ b + 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) = a1b1 ⊗ a2b2 ⊗ ... ⊗ anbn.

If the algebras are unital, we have natural embeddings

n O Ai ,→ 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 Nn on 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 [a1a2 . . . 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 n O (A, ϕ) = (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 u(1) u(2) u(n)i h u(1)i h u(2)i u(n) ϕ a1 a2 . . . an = ϕ a1 ϕ a2 . . . ϕ an .

This is not true for singleton independence: if ϕ [a1] = ϕ [a2] = 0, then

ϕ [a1a2a1] = 0,

19 but ϕ [a1a2a1a2] 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 [a1a2 . . . 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 " n # Y −1 τ Pj λ(xu(j)), λ(xu(j)) = 0 j=1 whenever −1 τ Pj(λ(x), λ(x )) = 0 (3.1) and u(1) 6= u(2) 6= u(3) 6= ... 6= u(n). (3.2) Note ﬁrst that −1 X (j) k Pj(λ(x), λ(x )) = αk λ(x ), k∈Z (j) and so equation (3.1) implies that each α0 = 0. Moreover, n Y −1 X (1) (n) k(1) k(n) Pj λ(xu(j)), λ(xu(j)) = αk(1) . . . αk(n)λ(xu(1) . . . xu(n)), 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(x1x2) = 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  b1b2 . . . 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 (j) o (A, ϕ), and Bi : j ∈ J(i) be unital ∗-subalgebras of Ai. Then we have the following.

n (j) o a. If {Ai : i ∈ I} are freely independent in (A, ϕ) and for each i ∈ I, Bi : j ∈ J(i) are freely n (j) o independent in (Ai, ϕ|Ai ), then all Bi : i ∈ I, j ∈ J(i) are freely independent in (A, ϕ).

21 n (j) o b. If all Bi : i ∈ I, j ∈ J(i) are freely independent in (A, ϕ) and if, for each i ∈ I, Ai is the n (j) o ∗-algebra generated by Bi : j ∈ J(i) , then {Ai : i ∈ I} are freely independent in (A, ϕ).

Exercise 3.11. Let (A, τ) be a tracial n.c. probability space, a1, a2, . . . , an free, and u a unitary. Then ∗ {u aiu, 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. Sn Any word x1x2 . . . xk with all xi ∈ i=1 Γi can be reduced by identifying

x1x2 . . . xi−1ejxi . . . xk = x1x2 . . . xi−1xi . . . xk and x1x2 . . . xi−1yzxi . . . xk = x1x2 . . . 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 x1x2 . . . xk, xi ∈ Γu(i) \ eu(i) , u(1) 6= u(2) ... 6= u(k).

Deﬁne the free product of groups

( n ) n [ Γ = Γ1 ∗ Γ2 ∗ ... ∗ Γn = ∗i=1Γi = reduced words in elements of Γi i=1 with unit e, inverse −1 −1 −1 −1 (x1x2 . . . xk) = xk . . . x2 x1 and product (x1x2 . . . xk)(y1y2 . . . yl) = reduced version of (x1 . . . xky1 . . . yl).

22 Reduced free product of n.c. probability spaces.

Let (A1, ϕ1),..., (An, ϕn) be n.c. probability spaces. Denote

◦ Ai = {a ∈ Ai : ϕi [a] = 0} ,

W∅ = C = C1, and for u(1) 6= u(2) ... 6= u(k), deﬁne the vector space

◦ ◦ ◦ W~u = Au(1) ⊗ Au(2) ⊗ ... ⊗ Au(k). Let ∞ ∞ M M M M A = W~u = C1 ⊕ W~u k=0 |~u|=k k=1 u(1)6=u(2)...6=u(k) u(1)6=u(2)...6=u(k)

By abuse of notation we will write a1 ⊗ a2 ⊗ ... ⊗ ak as a1a2 . . . ak. Deﬁne

∗ ∗ ∗ ∗ (a1a2 . . . ak) = ak . . . a2a1. On A, we deﬁne multiplication as follows. First,

z(a1a2 . . . ak) = (za1)a2 . . . ak.

◦ ◦ Next, let ai ∈ Au(i), bi ∈ Av(i), where ~u and ~v are alternating. If u(1) 6= v(1),

(ak . . . a2a1)(b1b2 . . . bl) = ak . . . a2a1b1b2 . . . bl.

In general,

◦ ◦ (ak . . . a2a1)(b1b2 . . . bl) = ak . . . a2(a1b1) b2 . . . bl + ϕu(1) [a1b1] ak . . . a3(a2b2) b3 . . . bl + ...

+ ϕu(1) [a1b1] ϕu(2) [a2b2] . . . ϕu(j−1) [aj−1bj−1] ak . . . ajbj . . . 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 n ϕ = ∗i=1ϕi by: ϕ [1] = 1 and on an alternating word in centered elements, ϕ [a1a2 . . . 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 ⊕ Ai ⊂ 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. n n ∗i=1C[Γi] = C [∗i=1Γi] .

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 one- dimensional distributions is tracial.

◦ ◦ Proof of the Proposition. Let ai ∈ Au(i), bi ∈ Av(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 . . . a1b1 . . . bl] = ϕ [ak . . . a2(a1b1) b2 . . . bl] + ϕu(1) [a1b1] ϕ [ak . . . a3(a2b2) b3 . . . bl] + ...

+ ϕu(1) [a1b1] ϕu(2) [a2b2] . . . ϕu(j−1) [aj−1bj−1] ϕ [ak . . . ajbj . . . bl]

This is zero unless j = k = l, in which case, since each ϕi is tracial,

ϕ [ak . . . a1b1 . . . bk] = ϕu(1) [a1b1] ϕu(2) [a2b2] . . . ϕu(j) [ajbj]

= ϕu(1) [b1a1] ϕu(2) [b2a2] . . . ϕu(j) [bjaj] = ϕ [b1 . . . bkak . . . 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 n with distinguished vectors (H1, ξ1)i=1, deﬁne their reduced free product n (H, ξ) = ∗i=1(Hi, ξi) ◦ as follows. Denote Hi = Hi Cξi. Then ∞ M M ◦ ◦ ◦ Halg = Cξ ⊕ H~u = Hu(1) ⊗ Hu(2) ⊗ ... ⊗ Hu(k) . k=1 |~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 k Y hf1 ⊗ f2 ⊗ ... ⊗ fk, g1 ⊗ g2 ⊗ ... ⊗ gki = hfi, giiu(i) i=1 ◦ for fi, gi ∈ Hu(i). 2 If (Hi, ξi) = L (Ai, ϕ1), then, at least in the faithful case, 2 n (H, ξ) = L (∗i=1(Ai, ϕi)) . n More generally, if each Ai is represented on Hi with ϕi = h·ξi, ξii, then can represent ∗i=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∗).

t n c. for all z = (z1, z2, . . . , zn) ∈ C ,

n X hAz, zi = ziAijzj ≥ 0. i,j=1 Proposition 3.20. A linear functional ϕ on A is positive if and only if for all n and all ∗ n a1, a2, . . . , an ∈ A, the (numerical) matrix [ϕ(ai aj)]i,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 ziai zjaj = ziϕ [ai aj] zj. i=1 j=1 i,j=1 ∗ n By the preceding remark, the matrix [ϕ(ai aj)]i,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

n n Tn : Mn(A) → Mn(B) Tn([aij]i,j=1) = [T (aij)]i,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 PN ∗ Proof. Let A ∈ Mn(A) be positive. By deﬁnition, that means A = 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

n X ∗ Aij = bkibkj, buv ∈ A. k=1 Then n n X ∗ n [ϕn(A)]i,j=1 = [ϕ(bkibkj)]i,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 = AijBij.

Proposition 3.25. If A, B are positive, so is their Schur product.

∗ Pn Proof. Let A = D D, Aij = k=1 DkiDkj. Then

n n n n ! X X X X ziCijzj = ziDkiDkjBijzj = (Dkizi)Bij(Dkjzj) ≥ 0, i,j=1 i,j,k=1 k=1 i,j=1 since B is positive. Therefore C is positive.

n n Theorem 3.26. Let (Ai, ϕi)i=1 be n.c. probability spaces and (A, ϕ) = ∗i=1(Ai, ϕi). Then ϕ is positive. If each ϕi is faithful, so is ϕ.

Proof. Recall the representation ∞ M M A = W~u k=1 |~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~uξ~u] . ~u

∗ So it sufﬁces to show that ϕ [ξ ξ] ≥ 0 for ξ ∈ W~u. In its turn,

r X (j) (j) (j) (j) ◦ ξ = a1 a2 . . . ak , ai ∈ Au(i). j=1

26 Then " r r # ∗ ∗ ∗ X (i) (i) X (j) (j) ϕ [ξ ξ] = ϕ ak ... a1 a1 . . . ak i=1 j=1 r X h (i) ∗ (j)i h (i) ∗ (j)i = ϕu(1) (a1 ) a1 . . . ϕu(k) (ak ) ak . i,j=1

r h (i) ∗ (j)i 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, a1/2ba1/2 is also positive, and since ϕ is trace,

ϕab = ϕa1/2ba1/2 . 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 aiu, 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 [a1a2 . . . 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, Mobius¨ 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 i ∼π 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

1ˆ = 1ˆn = {(1, 2, . . . , n)} and the smallest one 0ˆ = 0ˆn = {(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 i π∼∧σ j ⇔ i ∼π j and i ∼σ 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)(5, 6)(7)} and σ = {(1)(2, 3)(4, 5)(6, 7)}, then π ∨ σ = 1ˆ7. Deﬁnition 4.2. A partition π ∈ P(S) of an ordered set S is a non-crossing partition if

i ∼π j, k ∼π l, i < k < j < l ⇒ i ∼π k ∼π j ∼π l.

See Figure 4. Denote non-crossing partitions by NC(S), NC(n).

Note that 0ˆ, 1ˆ ∈ 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 = 1ˆ4 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

n X n B(n + 1) = B(n − i), n ≥ 1, i i=0 where by convention B(0) = 1. Consequently, their exponential generating function is

∞ X 1 z F (z) = B(n)zn = ee −1. n! n=0

Remark 4.7. eez−1 grows faster than exponentially, so B(n) grows faster than any an.

n Proof of the Lemma. If π ∈ P(n + 1) and the block containing 1 contains i + 1 elements, there are i 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 1 X 1 X n! F 0(z) = B(n + 1)zn = B(n − i) zn n! n! i!(n − i)! n=0 n=0 i=0 ∞ ∞ X 1 X 1 = zi B(j)zj = ezF (z). i! j! i=0 j=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 ∞ n X X 1 z F (z, w) = 1 + S(n, k)znwk = ew(e −1). n! n=1 k=1 Deﬁnition 4.9. cn = |NC(n)| = Catalan number. Lemma 4.10. The Catalan numbers satisfy recursion relations

n X X cn = cu(1)cu(2) . . . cu(i), n ≥ 1, i=1 u(1),u(2)...,u(i)≥0 u(1)+u(2)+...+u(i)=n−i where by convention c0 = 1, and n−1 X cn = ckcn−k−1. k=0

32 Their ordinary generating function is

∞ √ X 1 − 1 − 4z F (z) = c zn = . n 2z n=0 Consequently, 1 2n 1 2n + 1 c = = . n 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,

∞ ∞ n X n X X X u(1) u(2) u(i) i F (z) = 1 + cnz = 1 + cu(1)cu(2) . . . cu(i)z z . . . z z n=1 n=1 i=1 u(1),u(2)...,u(i)≥0 u(1)+u(2)+...+u(i)=n−i ∞ ∞ ∞ ∞ X X X X 1 = 1 + zi c zu(1) ... c zu(i) = 1 + zF (z) = . u(1) u(i) 1 − zF (z) i=1 u(1)=0 u(i)=0 i=1

Thus zF (z)2 − F (z) + 1 = 0, and √ 1 − 1 − 4z F (z) = . 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,

∞ ∞ ∞ 1 X 1/2 X (2k − 2)! X (2n)! F (z) = − (−4z)k = zk−1 = zn, 2z k (k − 1)!k! n!(n + 1)! k=1 k=1 n=0 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(0ˆ, σ).

It is easy to see that I2 has the identity element ( 1, σ = π δ(σ, π) = 0, σ 6= π.

Another important element is ζ(σ, π) = 1 for all σ ≤ π. In particular, for f ∈ I, X (f ∗ ζ)(π) = f(σ). σ≤π

Mobius¨ inversion.

Deﬁnition 4.12. The Mobius¨ 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 Mobius¨ inversion). Let f, g ∈ I. Suppose for all π ∈ L, X f(π) = g(σ). σ≤π

a. Mobius¨ inversion: for all π ∈ L, X g(π) = f(σ)µ(σ, π). σ≤π

34 b. Partial Mobius¨ 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. X X f(π) = g(σ) = g(σ)ζ(σ, π) = (g ∗ ζ)(π). 0ˆ≤σ≤π 0ˆ≤σ≤π Therefore X g(π) = (g ∗ ζ ∗ µ)(π) = (f ∗ µ)(π) = f(σ)µ(σ, π). σ≤π More generally, X X X X X f(σ)µ(σ, π) = g(τ)µ(σ, π) = g(τ) ζ(ω ∨ τ, σ)µ(σ, π) ω≤σ≤π ω≤σ≤π τ≤σ τ∈L (ω∨τ)≤σ≤π X X = g(τ)δ(ω ∨ τ, π) = g(τ). τ∈L 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), Y Y [0ˆ, π] ' P(Vi) ' P(|Vi|)

Vi∈π Vi∈π and [π, 1]ˆ 'P(|π|). More generally, if σ ≤ π, Y [σ, π] ' P(|σ|Vi |).

Vi∈π In NC(S), Y Y [0ˆ, π] ' NC(Vi) ' NC(|Vi|).

Vi∈π Vi∈π But what is [π, 1]ˆ ? For example,

[{(1, 6)(2, 5)(3, 4)} , 1ˆ6] 6' NC(3).

35 Deﬁnition 4.15. Let π ∈ NC(n). Its (right) Kreweras complement K[π] ∈ NC(n) is the largest partition in NC({1¯, 2¯,..., n¯}) such that

π ∪ K[π] ∈ NC({1, 1¯, 2, 2¯, . . . , n, n¯}).

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 ˆ ˆ 0n σn−1 ... σ3 σ2 1n,

where |σk| = k. K maps it to a maximal chain ˆ ˆ 1n K[σn−1] ... K[σ3] K[σ2] 0n

with |K[σk]| = n − k + 1. Corollary 4.17. In NC(n), Y [π, 1]ˆ ' [0ˆ,K[π]] ' NC(Vi)

Vi∈K[π] and more generally Y ˆ Y Y [σ, π] ' [σ|Vi , 1Vi ] ' NC(Uj). V ∈π V ∈π i i 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

n Y ki [σ, π] ' Li , i=1 then n Y ki f(σ, π) = fi . 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 Mobius¨ function is also always multiplicative. In fact, in general

µL×L0 = µLµL0 .

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 × A × ... × A → C | {z } n by M[a1, a2, . . . , an] = ϕ [a1a2 . . . 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 [a1a4] ϕ [a2a3] .

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, X Mπ[a1, a2, . . . , an] = Rσ[a1, a2, . . . , an] σ≤π and X Rπ[a1, a2, . . . , an] = Mσ[a1, a2, . . . , an]µ(σ, π). σ≤π In particular, for π = 1ˆ, we have the moment-cumulant formulas X R[a1, a2, . . . , an] = Mσ[a1, a2, . . . , an]µ(σ, 1)ˆ , σ∈NC(n) and X M[a1, a2, . . . , an] = Rσ[a1, a2, . . . , an], σ∈NC(n) where Y Rσ[a1, a2, . . . , an] = 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. ϕ R[a1, . . . , an] = R [a1, . . . , an] is a multilinear functional on (A, ϕ). If µ ∈ D(d) is the joint distribution of (a1, . . . , ad), then x1, . . . , xd ∈ Chx1, . . . , xdi also have joint distribution µ with respect to µ. So can consider µ R [xu(1), xu(2), . . . , xu(n)]. These satisfy

X µ ϕ au(1)au(2) . . . au(n) = µ xu(1)xu(2) . . . xu(n) = Rπ[xu(1), xu(2), . . . , xu(n)]. π∈NC(n)

µ ∗ For µ ∈ D(d), can deﬁne R ∈ Chx1, . . . , xdi (not necessarily positive) via µ µ R xu(1)xu(2) . . . xu(n) = R [xu(1), xu(2), . . . , xu(n)].

39 In one variable, we have the moments

n n mn[µ] = µ [x ] = ϕ [a ] and the moment-cumulant formula X X Y mn[µ] = rπ[µ] = r|V |[µ]. π∈NC(n) π∈NC(n) V ∈π

Here µ n µ rn[µ] = R [x ] = R [x, x, . . . , x]. Finally, we have the free cumulant generating function (combinatorial R-transform)

∞ µ X n R (z) = rn[µ]z n=1 and more generally for µ ∈ D(d)

∞ d µ X X µ R (z1, z2, . . . , zd) = R [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 z1M(z), . . . , zdM(z) = 1 + R M(z)z1,...,M(z)zd , where ∞ X X M(z1, . . . , zd) = 1 + M[x~u]z~u. n=1 |~u|=n

40 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 X = k=1 S⊂[n] S={1=v(1),v(2),...,v(k)} v(k+1)=n+1

k   Y X Y R[xu(v(j)) : 1 ≤ j ≤ k] zu(v(j)) Rπj [xu(i) : i ∈ Vj] zu(i)

j=1 πj ∈NC(Vj =[v(j)+1,v(j+1)−1]) i∈Vj

n k   X X Y Y = R[xu(v(j)) : 1 ≤ j ≤ k] zu(v(j))M[xu(v(j)+1), . . . , xu(v(j+1)−1)] zu(i) . k=1 S⊂[n] j=1 i∈Vj S={1=v(1),v(2),...,v(k)} v(k+1)=n+1

Therefore

∞ X X M(z) = 1 + R[x ~w]zw(1)M(z)zw(2)M(z) . . . zw(k)M(z) k=1 | ~w|=k = 1 + R z1M(z), z2M(z), . . . , zdM(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 X M[a1, . . . , an] = Bπ[a1, . . . , an]. π∈Int(n) For the exponential moment generating function

∞ X 1 M µ (z) = m [µ]zn, exp n! n n=0 can show as before that the cumulant generating function is

∞ X 1 `(z) = k [µ]zn = log M (z). n! n exp n=1

41 Exercise 5.8. For the Boolean cumulant generating function

∞ X X B(z) = 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 Mobius¨ function is given by

n−1 µn = (−1) cn−1, where cn is the Catalan number.

Proof. Recall µ ∗ ζ = δ. Thus ( X Y X 1, n = 1, µ|V | = µ(0ˆ, π)ζ(π, 1)ˆ = δ(0ˆn, 1ˆn) = 0, n ≥ 2. π∈NC(n) V ∈π π∈NC(n)

It follows that {µn : n ∈ N} are the free cumulants of a “distribution” (not positive) with m1 = 1, mn = 0 P∞ n for n ≥ 2. Thus for F (z) = n=1 µnz , we have 1 + z = 1 + F (z(1 + z)). √ 2 −1+ 1+4w If w = z + z, then z = 2 and

∞ √ ∞ X −1 + 1 + 4w X F (w) = µ wn = = (−1)nc wn n 2 n−1 n=1 n=1 as before.

Exercise 5.10. Use Remark 5.7 and Exercise 5.8 to compute the Mobius¨ functions on P(n) and Int(n).

Deﬁnition 5.11. A multi-chain in a lattice is a linearly ordered subset

0ˆ ≤ σ1 ≤ σ2 ≤ ... ≤ σp ≤ 1ˆ, 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 |NC(n)| = 1 = (ζ ∗ ζ)(0ˆn, 1ˆn).

0ˆn≤σ1≤1ˆn Similarly, X ∗(p+1) M(n, p) = 1 = (ζ ∗ ζ ∗ ... ∗ ζ)(0ˆn, 1ˆn) = (ζ )n. | {z } 0ˆ≤σ1≤σ2≤...≤σp≤1ˆ p+1 P∞ ∗(p+1) n 1 If Fp(z) = 1 + n=1(ζ )nz , then F0(z) = 1−z and Fp(z) = Fp−1(zFp(z)). It follows by induction p+1 that Fp satisﬁes z(Fp(z)) = Fp(z) − 1, since

p p+1 Fp(z) − 1 = Fp−1(zFp(z)) − 1 = (zFp(z))(Fp−1(zFp(z))) = z(Fp(z)) .

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 multino- mial coefﬁcient n! N(n; k , i ∈ ) = . i N Q k !(n + 1 − P k )! i∈N i i∈N i

For the proof, see Lecture 9 from [NS06], or an argument in [Spe98] using the Lagrange inversion formula.

Distributions.

Recall: Stieltjes inversion formula 1 dµ(x) = − lim =Gµ(x + iy) dx, π y→0+ where ∞ 1 X 1 G(z) = M(1/z) = m z n zn+1 n=0 is the Cauchy transform. Deﬁne the analytic R-transform

∞ 1 X R(z) = R(z) = r zn−1 = r + r z + r z2 + .... z n 1 2 3 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.

n Example 5.15. Let r1[µt] = t, rn[µt] = 0 for n 6= 1. Then mn[µt] = t and µt = δt.

Lemma 5.16. Denote NC2(n) the non-crossing pair partitions. Then

|NC2(2n)| = cn = |NC(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

n n m2n[µt] = t |NC2(2n)| = t cn and m2n+1[µt] = 0. Therefore √ 1 − 1 − 4tz2 M µt (z) = , 2tz2 √ z − z2 − 4t G (z) = , µt 2t and so 1 √ dµ (x) = 4t − x21 √ √ . t 2πt [−2 t,2 t] 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.

Example 5.19. Let rn[µt] = t for all n. Thus its R-transform is

∞ X t R(z) = tzn−1 = . 1 − z n=1 Then 1 1 + (t − 1)z + R(z) = , z z − z2

44 and so z + 1 − t − p(z − (t + 1))2 − 4t G (z) = . µt 2z From this one can deduce that 1 p4t − (x − (t + 1))2 dµ(x) = + max(1 − t, 0)δ , 2π 2x 0 √ √ 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,

1 r4 − x dµ(x) = 2π x 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,

n X |π| X k mn[µt] = t = t |{π ∈ NC(n): |π| = k}| . π∈NC(n) 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 V = [v(j − 1) + 1, v(j)] (where as usual v(0) = 0), A = Q a . Clearly j j i∈Vj i

a1 . . . an = A1 ...Ak. So M[A1,A2,...,Ak] = ϕ [A1A2 ...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=1ˆn

Here π ∨ σ = 1ˆn 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=ˆτ Example 5.22.

R[a1a2, 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, 1)ˆ 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→ τˆ by    [ [ [  τ = {U1,U2,...,Uj} 7→ Vi, Vi,..., Vi .

i∈U1 i∈U2 i∈Uj  For example, if σ = {(1)(2, 3, 4)(5, 6)} ∈ Int(6) and τ = {(1, 2)(3)} ∈ NC(3), then τˆ = {(1, 2, 3, 4)(5, 6)} ∈ NC(6). If instead τ = {(1, 3)(2)} then τˆ = {(1, 5, 6)(2, 3, 4)}. Then

46 a. τˆ ∈ NC(n).

b. The map τ 7→ τˆ is injective, with 0ˆk 7→ σ, 1ˆk 7→ 1ˆn. c. τ 7→ τˆ preserves the partial order and its image is

{π ∈ NC(n): π ≥ σ} = [σ, 1ˆn].

Thus [0ˆk, 1ˆk] ' [σ, 1ˆn]. In particular, µ(σ, π) = µ(ˆσ, πˆ), where these are two Mobius¨ functions on different lattices.

Proof of the Theorem. X X Rτ [A1,A2,...,Ak] = Mπ[A1,...,Ak]µ(π, τ) = Mπˆ[a1, a2, . . . , an]µ(ˆπ, τˆ). π∈NC(k) π∈NC(k) π≤τ π≤τ Denoting ω =π ˆ and using partial Mobius¨ inversion (Proposition 4.13), the expression continues as X X = Mω[a1, a2, . . . , an]µ(ω, τˆ) = Rπ[a1, a2, . . . , an]. ω∈NC(n) π∈NC(n) σ≤ω≤τˆ π∨σ=ˆτ

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 [a1a2 . . . 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, X ϕ [a1a2 . . . an−11] = R[a1, a2, . . . , an−1, 1] + Rπ[a1, a2, . . . , an−1, 1]

π6=1ˆn 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 [a1a2 . . . 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[a1 + ϕ [a1] , . . . , an + ϕ [an]]. ◦ Since R is multilinear, and using the lemma, we may assume that ai = ai , 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 [a1a2] − ϕ [a1] ϕ [a2] = 0.

For general n, if u(1) 6= u(2) 6= ... 6= u(n), then X R[a1, . . . , an] = ϕ [a1 . . . an] − Rπ[a1, . . . , an] = 0

π6=1ˆn 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=1ˆn for a certain σ ∈ Int(n), σ 6= 1ˆn. Take π 6= 1ˆn. 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 = 1ˆn 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= 1ˆn, and so also for π = 1ˆn.

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) , d d where {Xi}i=1 , {Yi}i=1 ⊂ (A, ϕ) are free, µ = ϕX1,...,Xd , and ν = ϕY1,...,Yd . Thus

µν ϕ R [xu(1), . . . , xu(n)] = R [Xu(1) + Yu(1),...,Xu(n) + Yu(n)] X ϕ = R [Xu(1) or Yu(1),...,Xu(n) or Yu(n)] ϕ ϕ = R [Xu(1),...,Xu(n)] + R [Yu(1)....,Yu(n)] µ ν = R [xu(1), . . . , xu(n)] + R [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 [a1a2 . . . an] = ϕ [a1] ϕ [a2] . . . ϕ [an] .

a. Show that this deﬁnition gives a degenerate structure 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 Bµ]ν = Bµ +Bν. 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 X = (u + u−1) + (u + u−1) + ... + (u + u−1) . 2n 1 1 2 2 n n 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 + ui has the arcsine distribution, and they are freely independent. Thus 1 Gu +u∗ (z) = √ i i z2 − 4 and r 1 1 Ru +u∗ (z) = 4 + − . i i z2 z So s ! r 1 1 n n2 n R (z) = n 4 + − = 1 + − X 2n (z/2n)2 z/2n z2 z and r 1 n2 n − 1 + R (z) = 1 + − . z X z2 z Then −(n − 1)z + pn2z2 − (2n − 1) G (z) = X z2 − 1 and 1 p(2n − 1) − n2x2 dµ (x) = dx. X π 1 − x2 These are the Kesten-McKay distributions.

50 Exercise 5.32. Let 1 1 µ = δ + δ 2 −a 2 a 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 2 ϕ [Xi] = 0, ϕ Xi = t. Then X + ... + X S = 1 √ n n 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(x) d(Dαµ)(x) = f(αx) dµ(x) R R and µ = µXi , √ D1/ n(µ µ ... µ) → S(0, t). | {z } n

n n Proof. Note first that by definition mn[Dαµ] = α mn[µ], from which it follows that rn[Dαµ] = α rn[µ]. Since S(0, t) is compactly supported, to prove convergence in distribution it suffices to prove convergence in moments, Z k k ϕ Sn → x 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, nr [X] r [S ] = √1 = 0, 1 n n nr [X] r [S ] = 2 = t, 2 n n and nr [X] r [S ] = k → 0, k > 2. k n nk/2

51 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).

t t n 1 − δ + δ → π , n 0 n 1 t the free Poisson distribution with parameter t.

Again, the same conclusion can be obtained under weaker assumptions.

2 Remark 5.36. If µX = (1 − α)δ0 + αδ1, then mn[X] = α for all n ≥ 1 (m0[X] = 1). Thus X and X have the same distribution. May take X2 = X = X∗, so that X is an (orthogonal) projection, ϕ [X] = α.

Proof. Again, it sufﬁces to show that " # t t t t n nr 1 − δ + δ = r 1 − δ + δ → r [π ] = t. k n 0 n 1 k n 0 n 1 k t

t t t Denote µn = 1 − n δ0 + n δ1. Note that mk[µn] = n . So by the moment-cumulant formula, t 1 r [µ ] = + o , k n 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, . . . , xdi. Denote t t µ = 1 − δ + ν ∈ D(d). n n 0 n

Then for any |~u| = k ≥ 1, t M µn [x ] = ν[x ]. ~u n ~u Therefore the same argument as above gives n t 1 Rµn [x ] = nRµn [x ] = n ν[x ] + o → tν[x ]. ~u ~u n ~u n ~u

52 We have thus proved the existence of the limit

t t n lim 1 − δ0 + ν = πν,t, n→∞ n n

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 X Y w(path) = w(step). paths paths 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.

54 55 56 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 n w(path) = t 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 ⊗n Falg(H) = H = CΩ ⊕ H ⊕ (H ⊗ H) ⊕ (H ⊗ H ⊗ 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

n Y hf1 ⊗ f2 ⊗ ... ⊗ fn, g1 ⊗ g2 ⊗ ... ⊗ gki = δn=k hfi, giiH . 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, fi g2 ⊗ ... ⊗ gn, `∗(f)(g) = hg, fi 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 ⊗ ... ⊗ hni = hf ⊗ g1 ⊗ ... ⊗ gn, h0 ⊗ h1 ⊗ ... ⊗ hni n Y = hf, h0i hgi, hii i=1 D E = g1 ⊗ ... ⊗ gn, hf, h0ih1 ⊗ ... ⊗ hn ∗ = 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, fji, 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, f2i Ω See Figure 6.

Note: since all fi ∈ HR, hfi, fji = hfj, fii.

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).

k X X Y ϕ [X(f1)X(f2) ...X(fn)] = w(path) = fu(i), fv(i)

Dyck paths(n) π∈NC2(n) i=1 π={(u(1),v(1))...(u(k),v(k))} u(i)

(again, order in the inner products irrelevant).

b. ϕ R [X(f1),X(f2),...,X(fn)] = δn=2 hf1, f2i .

c. Each X(f) has distribution S(0, kfk2).

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.

t Mean of a = vector (ϕ [a1] , . . . , ϕ [ad]) .

d h ϕ i Covariance of a = matrix ϕ [aiaj] − ϕ [ai] ϕ [aj] = R [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 ! 1 X X √ 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, fi W (f2, . . . , fn), For example, W (f) = X(f)W (∅) = X(f),

W (f1, f2) = X(f1)X(f2) − hf2, f1i .

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.

2 Fix f ∈ HR with kfk = 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 ( ⊗k ⊗n 0, k 6= n, = hUk(X, t)Ω,Un(X, t)Ωi = f , f = tn, n = k.

Corollary 6.14.

n X #matched steps X = t U#unmatched steps(X, t) incomplete Dyck paths(n) n n = U (X, t) + c tU (X, t) + c t2U (X, t) + .... n 2 1 n−2 4 2 n−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 r Y 2 W (fu(1), fu(2), . . . fu(n)) = 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 2 ⊗k Uk(X(f), kfk )g1 ⊗ ... ⊗ gn = f ⊗ g1 ⊗ ... ⊗ gn. Indeed, by induction, for k ≥ 1 2 2 2 2 Uk+1(X, kfk )g1 ⊗ ... ⊗ gn = XUk X(f), kfk − kfk Uk−1 X(f), kfk g1 ⊗ ... ⊗ gn

⊗k 2 ⊗(k−1) = X(f)f ⊗ g1 ⊗ ... ⊗ gn − kfk f ⊗ g1 ⊗ ... ⊗ gn ⊗(k+1) = f ⊗ g1 ⊗ ... ⊗ gn. n n Exercise 6.16. Let HR = R be a real Hilbert space, H = C 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 ⊕ ... ⊕ H. Show that | {z } N the second operator is a sum of N freely independent operators, and apply the free central limit theorem to conclude that X(f) is a semicircular element. Hint: what is f⊕f√⊕...⊕f ? 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

∞ X ∗ k−1 ∗ ∗ 2 Xµ = ` + rk(` ) = ` + r1 + r2` + r3(` ) + ... k=1 be a formal power series. Then n Xµ 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.

* ∞ !n + n X ∗ k−1 X X Y Xµ Ω, Ω = ` + rk(` ) Ω, Ω = w(path) = r|V |. k=1 Łukasiewicz paths(n) 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 ∞ X ∗ k−1 Xµ = `1 + rk[µ](`1) , k=1 ∞ X ∗ k−1 Xν = `2 + rk[ν](`2) . k=1

62 Then Xµ and Xν are freely independent, and Xµ + Xν has the same distribution as ∞ X ∗ k−1 Y = l1 + (rk[µ] + rk[ν])(l1) . 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 ∞ ∞ X ∗ k−1 X ∗ k−1 Z = Xµ + Xν = (`1 + `2) + rk[µ](`1) + rk[ν](`2) k=1 k=1 and note that ∞ ∞ X ∗ k−1 X ∗ k−1 Y = `1 + rk[µ](`1) + rk[ν](`1) . k=1 k=1 We verify that for all n, hZnΩ, Ωi = hY nΩ, 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 Schurmann’s¨ operator model. Freely in- ﬁnitely 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), non- crossing 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) = (Tif1) ⊗ 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.

ϕ R [Xi] = λi, ϕ R [Xi,Xj] = hξi, ξji , ϕ R [Xu(1),Xu(2),...,Xu(n)] = ξu(1),Tu(2) ...Tu(n−1)ξu(n) .

Proof. X ϕ Xu(1)Xu(2) ...,Xu(n) = Xu(1)Xu(2) ...,Xu(n)Ω, Ω = hw(path)Ω, Ωi , Motzkin paths(n)

64 ∗ where the rising step in position i has weight ` (ξu(i)), the falling step `(ξu(i)), a ﬂat “m” step weight Λ(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,T10ξ4i λ11 hξ9, ξ5i hξ8,T7ξ6i λ3 hξ2, ξ1i .

Which joint distributions arise in this way?

(1/n) Deﬁnition 6.30. A distribution µ ∈ D(d) is freely (or )-inﬁnitely divisible if for all n, µ exists, n that is, there exists νn ∈ D(d) such that νn = µ. Note: positivity is crucial. Proposition 6.31. The following are equivalent.

a. µ is -inﬁnitely divisible.

µ µt b. For all t ≥ 0, all k ∈ N, and |~u| = k, t R~u = R~u for some µt ∈ D(d). c. There exists a free convolution semigroup

{µt : t ≥ 0} µt µs = µt+s, µ1 = µ.

1 µ νn Proof. µ is -inﬁnitely divisible if and only if there exists νn with n R~u = R~u . Denote µ1/n = νn, and p for p/q ∈ Q+, µp/q = νq ∈ D(d) satisﬁes

µ p R p/q = Rµ. ~u q ~u 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

C0hx1, . . . , xdi = {P ∈ Chx1, . . . , xdi with no constant term.}

A self-adjoint linear functional ρ on Chxi is conditionally positive if it is positive on C0hxi. n Example 6.34. On C[x], let ρ [x ] = δn=2. Then ρ (x2 − 1)2 = −2, so ρ is not positive.

65 Lemma 6.35. Every P ∈ C0hx1, . . . , xdi is of the form

d d X X P = λixi + xiPij(x)xj,Pij ∈ Chx1, . . . , xdi. i=1 i,j=1

If ψ ∈ D(d), then ρ deﬁned arbitrarily on 1 and xi, and

" d # d X X ρ xiPijxj = 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 ∈ C0hx1, . . . , xdi,

" d !∗ d # " d # d ∗ X X X ∗ X ∗ ρ [P P ] = ρ Pixi Pjxi = ρ xiPi Pjxj = ψ [Pi Pi] ≥ 0. 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 (Schurmann’s¨ construction). Let ρ on C0hx1, . . . , xdi be conditionally positive. GNS construction for (C0hx1, . . . , xdi, ρ) produces a Hilbert space

H = (C0hx1, . . . , xdi/N ) with the inner product D E P,ˆ Qˆ = ρ [Q∗P ] .

Note that C0hx1, . . . , xdi 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 =x ˆi ∈ H,Ti = λ(xi) ∈ L(H). Then

ρ [xi] = λi,

ρ [xixj] = hξi, ξji , 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

µ ϕ R [xu(1), xu(2), . . . , xu(n)] = R [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 |π| µ µt xu(1)xu(2) . . . xu(n) = t Rπ[xu(1), xu(2), . . . , xu(n)]. π∈NC(n)

Then

µ µ 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 On the other hand, if P ∗P has no constant term

∗ ∗ d ∗ µt [P P ] − µ0 [P P ] µt [P P ] = lim ≥ 0 + dt t=0 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 µ µt positive, and so by the Fock space constructions above, tR = R 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) = (fg1) ⊗ 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 fifj = 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

n X Y hf1 ⊗ ... ⊗ fn, g1 ⊗ ... ⊗ gkis = δn=k fi, gα(i) . α∈Sym(n) i=1

For example

hf1 ⊗ f2, g1 ⊗ g2is = hf1, g1i hf2, g2i + hf1, g2i hf2, g1i .

67 This is a degenerate inner product, since kf ⊗ g − g ⊗ fks = 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

n − X a (f)(g1 ⊗ ... ⊗ gn) = hgi, fi g1 ⊗ ... ⊗ gˆj ⊗ ... ⊗ gn, j=1 and preservation operators

n X Λ(e T )(g1 ⊗ ... ⊗ gn) = g1 ⊗ ... ⊗ (T gj) ⊗ ... ⊗ gn j=1

Again, + − Xi = a (ξi) + a (ξi) + Λ(e Ti) + λi 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

µ k [xu(1), . . . , xu(n)] = ξu(1),Tu(2) ...Tu(n−1)ξu(n)

d Proposition 6.40. µ ∈ D(d) is ∗-inﬁnitely divisible if and only if µ comes from (λi, ξi,Ti)i=1 on a symmetric Fock space, if and only if kµ is a conditionally positive functional.

Corollary 6.41. The Bercovici-Pata bijection is the bijection

BP : {∗ − inﬁnitely divisible distributions} ↔ { − inﬁnitely divisible distributions} , where kµ = RBP[µ]. µ µ This is a bijection since both sets {k : µ ∗ −ID} and {R : 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.

d Exercise 6.42 (Boolean inﬁnite divisibility). Let H and (λi, ξi,Ti)i=1 be as before. The Boolean Fock space is simply CΩ ⊕ H, with Boolean creation operators a+(f)Ω = f, a+(f)g = 0,

68 and annihilation operators a−(f)Ω = 0, a−(f)g = hg, fi Ω.

Extend T ∈ L(H) to the Boolean Fock space by T Ω = 0. Finally, denote PΩ the projection onto Ω, PΩΩ = Ω, PΩg = 0. Deﬁne + − Xi = a (ξi) + a (ξi) + Ti + λiPΩ.

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 d in turn is equivalent to a representation on a free Fock space coming from the data (λi, ξi,Ti)i=1: one direction is Schurmann’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, . . . , xdi, µ) on a Hilbert space K with µ = h·Ω, Ωi. Let H = K CΩ,

λi = hλ(xi)Ω, Ωi ∈ C,

ξi = (λ(xi) − λiPΩ)Ω ∈ K, + − Ti = λ(xi) − a (ξi) − a (ξi) − λiPΩ ∈ L(K). Show that these satisfy all the conditions in Remark 6.28 and in part (a) (on H, not on K). In particular, show that ξi ∈ H and Ti maps H to itself. Show that on the Boolean Fock space of d H, CΩ ⊕ H ' K, the joint distribution of X1,...,Xd arising from this family (λi, ξi,Ti)i=1, with respect to the vacuum state, is µ. 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

1 ⊗ 1 ⊗ ... ⊗ 1 = en, | {z } n e0 = Ω, we identify F(C) = Span (en : n ≥ 0).

69 2 Choose the inner product so that {en : n ≥ 0} form an orthogonal basis, with kenk = γnγn−1 . . . γ1, ke0k = 1. Deﬁne

+ a en = en+1, − − a en = γnen−1, a e0 = 0, 0 a en = βnen.

Then (a+)∗ = a−, a0 is symmetric, and

X = a+ + a− + a0 is also symmetric, with some distribution µ, Z n n hX e0, e0i = x dµ(x). R Theorem 6.43 (Darboux, Chebyshev, Stieltjes, Viennot, Flajolet, etc.).

a. See Figure 6.

2 3 m1[µ] = β0, m2[µ] = β0 + γ1, m3[µ] = β0 + 2β0γ1 + β1γ1, and in general X mn[µ] = w(path). Motzkin paths(n)

b. The moment generating function of µ has a continued fraction expansion

∞ X 1 M(z) = m [µ]zn = n γ z2 n=0 1 1 − β0z − 2 γ2z 1 − β1z − 2 γ3z 1 − β z − 2 1 − ...

c. Any measure µ all of whose moments are ﬁnite arises in this way. The corresponding pair of sequences β β β ... J(µ) = 0 1 2 γ1 γ2 γ3 ... are the Jacobi parameters of µ.

n d. Let {Pn(x): n ≥ 0} be the monic orthogonal polynomials with respect to µ, that is Pn(x) = x +... 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) + βnPn(x) + γnPn−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,

γ1γ2 . . . γn = m2n − Poly(γ1, . . . , γn−1, β0, . . . , βn−1),

γ1γ2 . . . γnβn = m2n+1 − Poly(γ1, . . . , γn, β0, . . . , βn−1),

d. Note that by assumption, hPn(X)e0,Pk(X)e0i = 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 + βnen + γnen−1 = Pn+1(X)e0 + βnPn(X)e0 + γnPn−1(X)e0.

Example 6.44. Let βn = 0 and γn = t for all n. Thus

0 0 0 ... J(µ ) = t t t t . . .

Then X n/2 mn[µt] = t , 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(µ ) = t γt γt γt . . .

Then X #falling steps #ﬂat steps X mn[µt] = (γt) (βt) = Rπ[µt],

Motzkin paths(n) π∈NC1,2(n)

t where r1[µt] = βt, r2[µt] = γt, rn[µt] = 0 for n > 2. Thus µt = S(βt, γt) = S(β, γ) .

Example 6.46. Let β0 = βt, βn = βt + b, γn = γt, so that

βt βt + b βt + b . . . J(µ ) = t γt γt γt . . .

Then

X #falling steps #singleton ﬂat steps #middle ﬂat steps X mn[µt] = (γt) (βt) b = Rπ[µt], labeled Motzkin paths π∈NC(n)

n−2 t n−2 where r1[µt] = βt, rn[µt] = γb t. Thus µt = µ , where r1[µ] = β, rn[µ] = γb . µ is a shifted, scaled free Poisson distribution: ( a + r [ν], n = 1, r [D ν δ ] = 1 n α a n α 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(µ ) = t 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 t -inﬁnitely divisible. If c < 0, µ = µ−c/γ is not -inﬁnitely divisible, but µ 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.

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(µ ) = t t t − 1 t − 1 ... For t = 1, 0 0 0 ... J(µ ) = t 1 0 0 ... so that 1 M µ(z) = 1 − z2 and 1 1 z 1 1 1 G (z) = = = − . µ z 1 − z−2 z2 − 1 2 z − 1 z + 1 So 1 µ = (δ + δ ) 2 −1 1 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 β β β ... J(µ) = 0 1 2 , γ1 γ2 γ3 ...

show that its Boolean cumulants are B1[µ] = β0, Bn+2[µ] = γ1mn[ν], where β β ... J(ν) = 1 2 γ2 γ3 ... The map µ 7→ ν is the coefﬁcient stripping. b. Conclude that β t β β ... J(µ]t) = 0 1 2 , γ1t γ2 γ3 ...

73 6.5 q-deformed free probability.

See [BS91], [Ans01]. Deﬁnition 6.51. Let q be a number or a symbol. The q-integers are 1 − qn [n] = 1 + q + q2 + ... + qn−1 = q 1 − q The notion goes back to Euler (≈ 1750), Rogers (1894), Ramanujan (1913).

Example 6.52. [n]1 = n, ( 0, n = 0, [n]0 = 1, n ≥ 1.

n−1 Consider Dyck paths with weights 1 on rising steps, and weight [n]q = 1 + q + ... + q on a falling step at height n. From general theory, if X mn[µ(x|q)] = 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]qPn−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, X mn[µ(x|q)] = w(path), labeled Dyck paths n−1 where each falling step at height n may have one of the labels 1, q, . . . , q . Using this idea, on Falg(H), define a+(f) = `(f) but n − X i−1 a (f)(g1 ⊗ ... ⊗ gn) = q hfi, gi g1 ⊗ ... ⊗ gî ⊗ ... ⊗ gn. i=1 Note that on the full Fock space F(H), these are not adjoints. Definition 6.53. On the algebraic full Fock space, define the q-inner product n X inv(α) Y hf1 ⊗ ... ⊗ fn, g1 ⊗ ... ⊗ gkiq = δn=k q fi, gα(i) , α∈Sym(n) 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)}|

74 Theorem 6.54 (Bozejko,˙ 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 X(f)2nΩ, Ω = w(path) = qcr(π),

labeled Dyck paths π∈P2(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 P2(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 qj−1. Corollary 6.58. X inv(α) q = [n]q! = [n]q[n − 1]q ... [1]q. α∈Sym(n)

Proof. For kfk = 1,

X inv(α) ⊗n ⊗n − n ⊗n q = f , f q = (a (f)) f , Ω q . α∈Sym(n) Deﬁnition 6.59. For a measure µ, its q-cumulants are deﬁned via

X cr(π) Y (q) mn[µ] = q r|V |. π∈P(n) V ∈π

(0) (1) Note: rn = rn, rn = kn. Theorem 6.60. If ρ is conditionally positive, there exists a positive measure µ such that

(q) n rn+2[µ] = ρ [x ] .

Denote all measures arising in this way by IDq.

75 Corollary 6.61. If deﬁne µ q ν by

(q) (q) (q) rn [µ q ν] = rn [µ] + r [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 Levy´ 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 Levy´ 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 µ µt a -semigroup {µt : t ≥ 0}. Let ρ = R , R = 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 Levy´ 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 [b1ab2|B] = b1ϕ [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 Levy´ process is a Markov process.

b. For t ≥ s, 1 ∗ 1 ϕ C (X(r): r ≤ s) = . 1 − X(t)z + Rµt (z) 1 − X(s)z + Rµs (z)

1 That is, t 7→ 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, . . . , xdi → Chx1, . . . , xdi ⊗ Chx1, . . . , xdi,

78 deﬁned by X ∂xi (xu(1)xu(2) . . . xu(n)) = 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−1 n−1 X X xn − yn ∂ xn = xi ⊗ xn−i−1 ↔ xiyn−i−1 = , x x − y i=0 i=0 and so p(x) − p(y) ∂ p(x) = . x 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. P For A = C[x], these are ordinary polynomials An(x), such that ∂xAn(x) = Ai(x) ⊗ An−i−1(x) and R An(x) dµ(x) = 0 for n ≥ 1. Theorem 7.8.

a. ∞ X 1 A (x)zn = , n 1 − xz + Rµ(z) n=0 and a similar result holds in the multivariate case. b. n X X0A(X1,X2,...,Xn) = A(X0,X1,...,Xn) + 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 P n Proof. For (a), if F (x, z) = n An(x)z , then µ[F ] = 1 and

X n ∂xF (x, z) = Ai ⊗ An−1−iz = zF ⊗ F. n,i

1 These conditions uniquely determine F (x, z) = 1−xz+Rµ(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 [a1b1a2b2 . . . anbn] = Rπ[a1, b1, a2, b2, . . . , an, bn] π∈NC(2n) X X = Rσ[a1, a2, . . . , an] Rτ [b1, b2, . . . , bn] σ∈NC(n) τ∈NC(n) σ∪τ∈NC(2n) X X = Rσ[a1, a2, . . . , an] Rτ [b1, b2, . . . , bn] σ∈NC(n) τ∈NC(n) τ≤K[σ] X = Rσ[a1, a2, . . . , an]MK[σ][b1, b2, . . . , bn]. σ∈NC(n) P In particular, mn[µab] = 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 aiaj = 0 for i 6= j. Let s be a semicircular variable free from {ai : 1 ≤ i ≤ k}. Then {sa1s, sa2s, . . . , saks} 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. 2 2 2 M[sau(1)s, sau(2)s, . . . , sau(n)s] = τ sau(1)ssau(2)s . . . sau(n)s = τ au(1)s au(2)s . . . au(n)s X 2 2 2 = Mπ[au(1), au(2), . . . , au(n)]RK[π][s , s , . . . , s ] 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 {sa1s, sa2s, . . . , saks} are free. Moreover,

rn[sais] = R[sais, sais, . . . , sais] = M[ai, ai, . . . , ai] = mn[ai], so sais has a free compound Poisson distribution. ∗ ∞ R 1 Example 8.3. Let A = C (s) ∗ L [0, 1], with the free product state τ = ϕ ∗ 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 1 (Ae, τ) = (pAp, τ|pAp). e α Then 1 Rτe[pa p, pa p, . . . , pa p] = Rτ [α a , α a , . . . , α a ]. u(1) u(2) u(n) α u(1) u(2) u(n)

82 See Theorem 14.10 in [NS06] for the proof in the non-tracial case.

Proof in the tracial case. 1 M τe[pa p, pa p, . . . , pa p] = M τ [a p, a p, . . . , a p] u(1) u(2) u(n) α u(1) u(2) u(n) 1 X = Rτ [a , a , . . . , a ]M [p, p, . . . , p] α π u(1) u(2) u(n) K[π] π∈NC(n) 1 X = Rτ [a , a , . . . , a ]αn+1−|π| α π u(1) u(2) u(n) π∈NC(n) |π| X 1 = Rτ [α a , α a , . . . , α a ]. α π u(1) u(2) u(n) 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 µt µ sense that there exists µt ∈ D(d) with R = tR .

1 Proof. Choose a1, . . . , ad and p as in the proposition so that τa1,...,ad = Dtµ and τ [p] = t . Then in t (pAp, τe), (pa1p, pa2p, . . . , padp) has distribution µ . 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 left- hand-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

(1/α) τe τ µpap = (µαa) .

On the other hand, τ τe µpap = µpap = (1 − α)δ0 + αµpap. Thus (1/α) µpap = (1 − α)δ0 + αµαa

83 and 1 Gpap(z) = (1 − α) + αG (1/α) (z). z µαa Finally, 1 R (1/α) (z) = Rαa(z) = Ra(αz). µαa α

Using these results, we compute Gq, then Rq, then R (1/α) , then G (1/α) , and ﬁnally Gpqp. µαq µαq

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 µ -infinitely divisible ⇔ R = ρ is conditionally positive, while µ ] -infinitely divisible ⇔ Bµ = ρ is conditionally positive, and any µ is ]-infinitely divisible. So we can define a map

B : D(d) → ID via RB[µ] = Bµ. This is the Boolean Bercovici-Pata bijection.

Relation between free and Boolean cumulants.

Recall X M[a1, a2, . . . , an] = Rπ[a1, a2, . . . , an] π∈NC(n) while X M[a1, a2, . . . , an] = Bπ[a1, a2, . . . , an] π∈Int(n) Corollary 9.2. Denote n π o NC(f n) = π ∈ NC(n) : 1 ∼ n . Then X B[a1, a2, . . . , an] = Rπ[a1, a2, . . . , an]

π∈NC(g n)

85 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[µ] X B[µ] X µ B [xu(1), xu(2), . . . , xu(n)] = Rπ [xu(1), xu(2), . . . , xu(n)] = Bπ [xu(1), xu(2), . . . , xu(n)] π∈NC(g n) π∈NC(g n) Deﬁnition 9.5. On NC(n), deﬁne a new partial order: σ π ⇔ σ ≤ π and ∀ V ∈ π, min(V ) ∼σ max(V ). Example 9.6. NC(f n) = σ ∈ NC(n): σ 1ˆn .

Note also that if σ π and V ∈ π, then σ|V ∈ NC(f V ). For another example, see Figure 9. Deﬁnition 9.7. Denote by Out(σ) the outer blocks of σ. If σ ∈ NC(f 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 )} . In particular, for σ ∈ NC(f 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 In particular, if σ ∈ NC(f n), then X t|π|−1s|σ|−|π| = (t + s)|σ|−1. π∈NC(n) σπ

Proof. X X t|π|−|Out(σ)|s|σ|−|π| = t|S|−|Out(σ)|s|σ|−|S| π∈NC(n) Out(σ)⊂S⊂σ σπ |Inn(σ)| X |Inn(σ)| = tis|σ|−i = (t + s)|Inn(σ)|. i i=0

Bt transformation.

Deﬁnition 9.10. For t ∈ R, deﬁne the linear map (Belinschi-Nica remarkable transformation) ∗ Bt : D(d) → Chx1, . . . , xdi by Bt[µ] X |π|−1 µ B [xu(1), xu(2), . . . , xu(n)] = t Bπ [xu(1), xu(2), . . . , xu(n)]. π∈NC(g n)

Remark 9.11. B1 = B, while B0 is the identity transformation. Proposition 9.12. For t ≥ 0, ] 1 (1+t) 1+t Bt[µ] = µ . 1 2] 2 In particular, the Boolean Bercovici-Pata bijection is given by the formula B[µ] = µ .

Corollary 9.13. For t ≥ 0, Bt maps D(d) to itself, and so preserves positivity.

Proof of the Proposition.

](1+t) BBt[µ] = (1 + t)BBt[µ] (by deﬁnition of ]) X |π|−1 µ = (1 + t) t Bπ (by deﬁnition of Bt) π∈NC(g n) X Y X = (1 + t) t|π|−1 Rµ ( ). σi by Corollary 9.2 V ∈π π∈NC(g n) σi∈NC(g Vi)

88 Let σ = σ1 ∪ σ2 ∪ ... ∪ σ|π|. Then σ ≤ π, σ π, and σ ∈ NC(f n). So we can continue the expression above as

X X |π|−1 µ = (1 + t) t Rσ σπ σ∈NC(g n) X |σ|−1 µ = (1 + t) (1 + t) Rσ (by Corollary 9.9) σ∈NC(g n) X |σ| µ = (1 + t) Rσ σ∈NC(g n) X µ(1+t) = Rσ (by deﬁnition of ) σ∈NC(g n) (1+t) = Bµ (by Corollary 9.2). We conclude that ](1+t) (1+t) Bt[µ] = µ .

Proposition 9.14. {Bt : t ∈ R} form a group,

Bt ◦ Bs = Bt+s, −1 with the identity element B0 and Bt = B−t. In particular, {Bt : t ≥ 0} form a semigroup. Remark 9.15. It follows that

] 1 ] 1 (1+t) 1+t ] 1 µ(1+s) 1+s = µ(1+t+s) 1+t+s , so there is a commutation relation between the Boolean and free convolution powers.

Proof of the proposition.

(Bt◦Bs)[µ] X |π|−1 Bs[µ] B = t Bπ (by definition of Bt) π∈NC(g n) X Y X = t|π|−1 s|σi|−1Bµ ( ) σi by definition of Bs V ∈π π∈NC(g n) i σi∈NC(g Vi) X X |π|−1 |σ|−|π| µ = t s Bσ (by definition of ) σπ σ∈NC(g n) X |σ|−1 µ = (t + s) Bσ (by Corollary 9.9) σ∈NC(g n) Bt+s[µ] = B (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

Bt[µ] X |π|−1 µ R [xu(1), xu(2), . . . , xu(n)] = t Rπ[xu(1), xu(2), . . . , xu(n)]. π∈NC(g 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, . . . , xdi by d Φ[µ] X µ B (z1, z2, . . . , zd) = ziM (z1, z2, . . . , zd)zi. i=1 Φ[µ] Equivalently, B [xi] = 0, and

Φ[µ] µ B [xu(1), xu(2), . . . , xu(n)] = δu(1)=u(n)M [xu(2), . . . , xu(n−1)]. Remark 9.18. Recall that in one variable,

µ 1 M (z) = 2 γ0z 1 − β0z − 2 γ1z 1 − β1z − 2 γ2z 1 − β z − 2 1 − ... Thus 2 Φ[µ] z B (z) = 2 γ0z 1 − β0z − 2 γ1z 1 − β1z − 2 γ2z 1 − β z − 2 1 − ... and so 1 1 M Φ[µ](z) = = 1 − BΦ[µ](z) z2 1 − 2 γ0z 1 − β0z − 2 γ1z 1 − β1z − 2 γ2z 1 − β z − 2 1 − ...

90 In other words, 0, β β β ... J(Φ[µ]) = 0 1 2 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 BΦ[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].

Bt[Φ[ρ]] X |π|−1 Φ[ρ] B = t Bπ [xu(1), . . . , xu(n)](by deﬁnition of Bt) π∈NC(g n) X Y = t|π|−1 δ M ρ (by deﬁnition of Φ) u(min(Vi))=u(max(Vi)) Vi\{min(Vi),max(Vi)} π∈NC(g n) Vi∈π X Y = t|π|−1 Rγ1 M ρ {min(Vi),max(Vi)} Vi\{min(Vi),max(Vi)} π∈NC(g n) Vi∈π X Y X Y = t|π|−1 Rγ1 Rρ o(σi) U V ∈π U∈Inn(σ ) π∈NC(g n) i σi∈NC(g Vi) i ( δ , |o(σ )| = 2, (by the moment-cumulant formula and because Rγ1 = u(min(Vi)=u(max(Vi)) i ) o(σi) 0, otherwise

91 X Y X Y = t−1 Rγt Rρ (because Rγt = tRγ1 ) o(σi) U V ∈π U∈Inn(σ ) π∈NC(g n) i σi∈NC(g Vi) i

X X −1 Y γt Y ρ = t R RU (by deﬁnition of ) o(σ|Vi ) σπ V ∈π U∈Inn(σ| ) σ∈NC(g n) i Vi

X X −1 Y γt Y ρ = t RU RU (by Proposition 9.8) σ∈NC(g n) o(σ)∈S⊂σ U∈S U6∈S

X −1 γt Y γt ρ = t Ro(σ) RU + RU σ∈NC(g n) U∈Inn(σ) X −1 γt Y ργt = t Ro(σ) RU (by deﬁnition of ) σ∈NC(g n) U∈Inn(σ)

X γ1 Y ργt = Ro(σ) RU σ∈NC(g n) U∈Inn(σ) ργt Φ[ργt] = δu(1)=u(n)M [xu(2), . . . , xu(n−1)] = B .

Remark 9.21 (A 2012). Using

] 1 t t Bt[ρ] = (Bt−1 ◦ B)[ρ] = B[ρ] , the identity Bt[Φ[ρ]] = Φ[ρ γt] implies

t ]t B[Φ[ρ]] = Φ[ρ γt] .

Denote µ = B[Φ[ρ]]. Then µ is -inﬁnitely divisible, and

t ]t µt = µ = Φ[ρ γt] . What is ρ? In the single-variable case,

B[Φ[ρ]] Φ[ρ] ρ Rn = Bn = 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 = (ε(1), . . . , ε(n)), ε(i) ∈ {∗, ·} Denote P2(~ε) = {π ∈ P2(n): each block of π connects a ∗ and a ·} and NC2(~ε) = P2(~ε) ∩ NC2(n). See Figure 10. Example 10.3. Let u be a Haar unitary, so that

n ϕ[u ] = 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 ( 1, |{i : ε(i) = ·}| = |{i : ε(i) = ∗}| , µ x~ε = ϕ u~ε = ϕ u|{i:ε(i)=·}|−|{i:ε(i)=∗}| = 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, ( 1, path ends at height 0, µ x~ε = 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 ~ε 1, can reduce the word ` 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 = √1 (s + is ), where s , s are free 2 1 2 1 2 standard semicircular.

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 [(s + is )(s − is )] = 1, 2 1 2 1 2 with the rest all equal to zero. What are the ∗-moments? Using Notation 10.2,

~ε µc x = |NC2(~ε)| =?

Proposition 10.7. c∗c has a free Poisson distribution.

Note: this is not obvious, as 1 i c∗c = (s2 + s2) + (s s − s s ). 2 1 2 2 1 2 2 1

Proof. To show: ϕ [(c∗c)n] = s2n , where s is standard semicircular. Indeed, using the non-crossing property (see Figure 10),

∗ n 2n ϕ [(c c) ] = |NC2(∗, ·, ∗, ·,..., ∗, ·)| = |NC2(2n)| = ϕ s .

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 cp . . . c2c1c1c2 . . . 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 = qcr(π).

π∈P2(~ε)

Know very little about them.

Question 10.11. Compute

 1/n 2 ∗ ∗ n 1/n X cr(π) kY k = kY Y k = lim ϕ [(Y Y ) ] = lim  q  . n→∞ n→∞ π∈P2(∗,·,∗,·,...,∗,·)

96 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 )).

∞− 2 2 We can identify A ∈ MN (L (Ω,P )) with a N -tuple of random variables, so it has a N -dimensional distribution. But also: for N 1 1 X tr[A] = Tr[A] = A N N ii 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 d (X1,...,Xd) are jointly Gaussian with mean 0 and (positive deﬁnite) covariance matrix Q = (Qij)i,j=1 if

d ! 1 X dµ (x , . . . , x ) = exp − x (Q−1) x /2 dx . . . dx . X1,X2,...,Xd 1 d Z i ij j 1 d i,j=1

Remark 11.3. Z is the normalization constant, in this case

Z d ! X −1 Z = exp − xi(Q )ijxj/2 dx1 . . . dxd. d R 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 1 X `(z , . . . , z ) = z Q z . 1 d 2 i ij j i,j=1 t Hint: denote x = (x1, . . . , xd) , and the same for y, z. Re-write the jointly Gaussian density as

1 t −1 dµ(x) = e−x Q x/2 dx. Z

Recall from Remark 5.7 that Z `µ(z) = log extz 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∈P2(n) (i,j)∈π where P2(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

Moreover, if Y1,...,Yd are independent complex Gaussian (i.e. their real and imaginary parts are inde- ∗ pendent), 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,

h ε(1) ε(n)i X Y h ε(i) ε(j)i E Yu(1) ...Yu(n) = E Yu(i) Yu(j) . π∈P2(~ε) (i,j)∈π

Remark 11.8. The distribution of Y as a C ' R2-valued random variable is

1 2 2 e−(x +y )/2a 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 = Xji, each Xii real.

b. {Xii,

1 1 c. Xii ∼ N (0, N ),

Note that N 1 X ◦ tr[X2] = [X X ] = 1. E N E ij ji i,j=1 Remark 11.10. We can identify the space of self-adjoint N × N matrices with

N 2 {(xii : 1 ≤ 1 ≤ N,

101 2 the RN distribution of a GUE matrix is N 1 2 Y Y 1 2 = e−N Tr[X ]/2 dx d

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 UXU ∗. (You can use the deﬁnition, or Remark 11.10). See Remark 11.48. Example 11.12. We would like to compute 1 hX i ◦ tr[X4] = X X X X . E N E i(1)i(2) i(2)i(3) i(3)i(4) i(4)i(1) Apply the Wick formula:

( 1 N , if i(j) = i(k + 1), i(j + 1) = i(k), E Xi(j)i(j+1)Xi(k)i(k+1) = 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 1 X ◦ tr[X4] = X X X X δ E N E i(1)i(2) i(2)i(3) E i(3)i(4) i(4)i(1) i(1)=i(3) 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 + . N N 2 N 2

What about a general moment E ◦ tr[X2n]? 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 ∈ P2(2n), get a term 1 1 N # faces = 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, 1 N #F −#E−1 = N χ−2 = . N 2g Proposition 11.14. For X GUE,

n X 1 X 1 ◦ tr[X2n] = = |{π ∈ P (2n) of genus g}| . E N 2g(π) N 2g 2 π∈P2(2n) 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 ↔ P2(2n).

Thus we proved: for X GUE, n X 1 ◦ tr[X2n] = |M (2n)| . E N 2g g g=0 In particular, for large N

2n 2n E ◦ tr[X ] ≈ |M0(2n)| = #(planar maps) = |NC2(2n)| = cn = ϕ s , 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

2n 2n E ◦ tr[X ] → cn = ϕ s , 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[X2n] → 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 E [Xij] = 0, lim N E |Xij| − 1 = 0 N→∞ N 2 i,j=1 and for each n, h √ ni sup sup E NXij < ∞. N 1≤i,j≤N 1 PN Third, recall from Exercise 2.23 that for a self-adjoint matrix, its distribution is N i=1 δλi , where {λi} are its (real) eigenvalues. Similarly, we may identify the distribution of a random matrix as 1 µˆ = (δ + ... + δ ) , N N λ1 λN the random spectral distribution of X, which is a random measure. Then in fact, for a Wigner matrix, as N → ∞, 1 √ µˆ → 4 − x2 dx, N 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 ∗ Xii = √ si,Xij = √ cij,Xji = √ c for i < j. N N N ij

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 ◦ tr[X2n] = |{maps with 1 vertex, n edges, of genus g}| . E N 2g g=0 What about " k # n(1) n(2) n(k) Y n(i) E N Tr(X )N Tr(X ) ...N Tr(X ) = E N Tr(X ) i=1 k X = N 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(X4)NT r(X3)NT r(X3)]. 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 N #F = N #V −#E+#F . N #E In general, if a map has c components, c X 2c − gi = 2c − 2g = #V − #E + #F. i=1 Thus " k # ∞ ∞ Y X X 1 N Tr(Xn(i)) = |G (n(i), 1 ≤ i ≤ k)| , E N 2g−2c g,c i=1 g=0 c=1 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, k ! ∞ k !n X X 1 X exp t N Tr[Xi] = t N Tr[Xi] E i E n! i i=1 n=0 i=1 ∞ k X 1 X n Y n(i) = t (N Tr[Xi])n(i) E n! n(1), n(2), . . . , n(k) i n=0 n(1)+...+n(k)=n i=1 ∞ k n(i) X X Y t = i (N Tr[Xi])n(i) E n(i)! n=0 n(1)+...+n(k)=n i=1 ∞ ∞ ∞ k n(i) X X 1 X Y t = i |G ((i, n(i)), 1 ≤ i ≤ k)| . N 2g−2c n(i)! g,c g=0 c=1 n(1),...,n(k)=0 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 = Fπ. diagrams 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 1 X 1 m zn = exp k zn . n! n n! n n=0 n=1

It follows that the generating function for connected maps is

k ! ∞ k n(i) X X 1 X Y t log exp t N Tr[Xi] = i |M ((i, n(i)), 1 ≤ i ≤ k)| . E i N 2g−2 n(i)! g i=1 g=0 n(1),...,n(k)≥0 i=1 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) 1 X X 1 X Y t log exp t N Tr[Xi] = i |M ((i, n(i)), 1 ≤ i ≤ k)| . N 2 E i N 2g n(i)! g i=1 g=0 n(1),...,n(k)≥0 i=1 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 1 apply to N 2 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, n ! Z " n #! X i 2 X i E exp tiN Tr[X ] = exp N Tr −X /2 + tiX dX = Z, i=1 i=1

106 where " n #! 1 X exp N Tr −X2/2 + t Xi dX Z i i=1 is a probability measure. Z is called the partition function. It can be thought of as a normalization 1 constant, but N 2 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) 2 = V 0(y), y ∈ supp(µ). (11.2) y − x 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 equation 11.2. Hint: use the real part of the Cauchy transform. Note that care is required in even R dµ(x) deﬁning y−x for y real. Example 11.27. Let X be GUE, and D(1),...,D(n) (for each N) ﬁxed (non-random) matrices.

(1) (n) E ◦ tr XD ...XD =? Consider (1) (2) (3) (4) E ◦ tr XD XD XD XD . See Figure 11. According to the Wick formula, this expression is again a sum of three terms. The ﬁrst one is 1 X D X D X D X D N E i(1)i(2) i(2)i(2) i(2)i(1) i(1)i(3) i(3)i(4) i(4)i(4) i(4)i(3) i(3)i(1) 1 1 X X X = D D D D = tr[D(1)] tr[D(3)] tr[D(2)D(4)]. N N 2 i(2)i(2) i(4)i(4) i(1)i(3) i(3)i(1) 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 X D X D X D X D = tr D(1)D(4)D(3)D(2) . N E i(1)i(2) i(2)i(3) i(3)i(4) i(4)i(2) i(2)i(1) i(1)i(4) i(4)i(3) i(3)i(1) N 2 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 Y trα [A1,A2,...,An] = 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 = σ1σ2 . . . 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−1β| 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−1ν = (α−1β)(β−1ν),

−1 −1 −1 α ν ≤ α β + β ν , and −1 −1 α β = σ1σ2 . . . σk ⇔ β α = σk . . . σ2σ1. For part (b), by Remark 11.29, X n − #cyc(α) = (|V | − 1) ≥ |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−1βα| = |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 −1 |α| + α 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

−1 −1 |α| + α β + β 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−1γ| = n − 1, so that

−1 α = σ1σ2 . . . σk, α γ = σk+1 . . . σn−1.

Then γ = σ1σ2 . . . 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),

−1 K[π] ↔ Pπ γ.

Sketch of proof. See Figure 11. The proof is by induction. For α = σ1 = (i, j), i < j, α−1γ = (1 ... (i − 1)j . . . n)(i . . . (j − 1)) = K[α].

We want to build up to the case of

−1 α = σ1 . . . σk, α γ = σk+1 . . . σn−1.

−1 On the induction step, multiplying ασj connects two cycles of α, while σjα 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 Y trα [X1,...,Xn] = tr  Xi . cycles of α i∈ cycle

For example, tr(152)(34) [X1,X2,X3,X4,X5] = tr [X1X5X2] tr [X3X4] . 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. X 1 ◦ tr[X2n] = . E N n−#cyc(πγ)+1 π∈P2(2n) b. X 1 ◦ tr XD(1)XD(2) ...XD(2n) = tr D(1),D(2),...,D(2n) . E N n−#cyc(πγ)+1 πγ π∈P2(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

(1) (2) (2n) X E ◦ tr XD XD ...D = 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 1 X (1) (2) (2n) D D ...D N n+1 u(1)i(2) u(2)i(3) u(2n)i(1) π∈P2(2n) i(j)=u(π(j)) X 1 X (1) (2) (2n) = D D ...D N n+1 u(1)u(π(2)) u(2)u(π(3)) u(2n)u(π(1)) π∈P2(2n) u(j) X 1 X (1) (πγ(1)) (πγ)2(1) = D D D ... N n+1 u(1)u(πγ(1)) u(πγ(1))u(πγπγ(1)) u((πγ)2(1))u((πγ)3(1)) π∈P2(2n) u(j) X 1 = Tr D(1),D(2),...,D(2n) N n+1 πγ π∈P2(2n) X 1 = tr D(1),D(2),...,D(2n) . N n−#cyc(πγ)+1 πγ π∈P2(2n)

Remark 11.38. Note that the leading term in the expansion corresponds to g = 0, but also to π ∈ NC2(2n). Equivalently,

−1 n − #cyc(πγ) + 1 = 2n − #cyc(πγ) − n + 1 = |πγ| − n + 1 = π 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 (N) (N) (N) i ϕN au(1)au(2) . . . au(n) → ϕ au(1)au(2) . . . au(n)

as N → ∞. Equivalently, h i ϕN (N) (N) (N) ϕ R au(1), au(2), . . . , au(n) → R 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 ϕN (N) (N) (N) N→∞ R 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)), 1 ≤ u(i) ≤ k if i ∼π j ⇒ u(i) = u(j).

See Figure11. Denote such π by P~u(n). Then as in Theorem 11.37,

X 1 ◦ tr X X ...X = . E u(1) u(2) u(2n) N n−#cyc(πγ)+1 π∈P2,~u(2n) and X 1 ◦ tr X D(1)X D(2) ...X D(2n) = tr D(1),D(2),...,D(2n) . E u(1) u(2) u(2n) N n−#cyc(πγ)+1 πγ π∈P2,~u(2n)

Theorem 11.41. Let X1,...,Xp be entry-wise independent GUE matrices, and D1,...,Dq be non- random 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.

(i) Similarly, taking D = fi(D1,...,Dq) in the preceding remark, lim ◦ tr Xu(1)f1(D1,...,Dq)Xu(2)f2(D1,...,Dq) ...Xu(n)fn(D1,...,Dq) N→∞ E X = lim trπγ [f1(D1,...,Dq), f2(D1,...,Dq), . . . , fn(D1,...,Dq)] 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 ϕ = Rπ 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 δ → circular law 1 on . N λi π |z|≤1 C i=1 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 t appear in numerous applications, from cell phone communications to medicine. Let (y1, . . . yK ) be jointly K Gaussian, with the K × K covariance matrix Q. For j = 1, 2,...,N (with N ≥ K) let (Yij)i=1 be independent samples from this Gaussian distribution, and Y = (Yij) a K × N random matrix, whose 1 t columns are independent and identically distributed. Then the N × N matrix N Y Y is called a Wishart matrix. Note that the K × K matrix N 1 1 X (YY t) = Y Y N ij N ik jk k=1 1 t is the sample covariance matrix for this distribution, and in particular E N (YY ) = Q, and that the distributions of Y tY and YY 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 Y tY = XtPX, N 1 where X is a N ×N random matrix all of whose entries are independent N (0, N ) 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,Xt) → (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(VUV ∗) 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 U n dU = (λU)n d(λU) = λn U n dU, UN UN UN

115 and so Z n U 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) non- random 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, u1d1u1, . . . , updpup 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(VXV )dPX .

b. For example, if X is a GUE matrix,

1 2 dP = e−N Tr[X ]/2 dX X Z and Tr[(VXV ∗)2] = Tr[X2], so this is a unitarily invariant ensemble (compare with GOE, GSE). 1 −T r[V (X)] More generally, matrix models Z e 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 N 1 X 1 X µ = δ , µ = δ , A N λi B N ρi i=1 i=1

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 (1) ∗ (1) (2) ∗ (2) E ◦ tr UA U B UA U B ... , where A(i) = Au(i), B(i) = Bv(i). What is the analog of the Wick formula?

N Remark 11.51. Let U = (Uij)i,j=1 be an N × N Haar unitary matrix. For N ≥ n,

¯ ¯ E Ui(1)j(1) ...Ui(n)j(n)Uu(1)v(1) ... Uu(n)v(n) X −1 = δi(β(1))=u(1) . . . δi(β(n))=u(n)δj(α(1))=v(1) . . . δj(α(n))=v(n) Wg(N, βα ). α,β∈Sym(n)

117 Here for β ∈ Sym(n) and N ≥ n ¯ ¯ Wg(N, β) = E U11 ...UnnU1β(1) ... Unβ(n) is the Weingarten function.

Example 11.52. ¯ ¯ E U11U12U22U12 = 0, ¯ ¯ E U11U12U12U11 = 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) + O . N 2n−#cyc(α) N 2n−#cyc(α)+2

Moreover, for α = Pσ, β = Pπ ∈ SNC(n), µ(β) = µ(0ˆn, π) and more generally

−1 −1 µ(α β) = µ(Pσ Pπ) = µ(σ, π), the Mobius¨ function on NC(n).

118 Remark 11.54. Thus we can now compute (1) ∗ (1) (2) ∗ (2) (n) ∗ (n) E ◦ tr UA U B UA U B ...UA U B 1 X h (1) (1) (n) (n) i = U A U¯ B ...U A U¯ B N E i(1)j(1) j(1)v(1) u(1)v(1) u(1)i(2) i(n)j(n) j(n)v(n) u(n)v(n) u(n)i(1) ~i,~j ~u,~v

1 X (1) (1) (2) (2) (n) (n) = A B A B ...A B N j(1)v(1) u(1)i(2) j(2)v(2) u(2)i(3) j(n)v(n) u(n)i(1) ~i,~j ~u,~v ¯ ¯ ¯ E Ui(1)j(1)Uu(1)v(1)Ui(2)j(2)Uu(2)v(2) ...Ui(n)j(n)Uu(n)v(n) 1 X (1) (1) (2) (2) (n) (n) = A B A B ...A B N j(1)v(1) u(1)i(2) j(2)v(2) u(2)i(3) j(n)v(n) u(n)i(1) ~i,~j ~u,~v X −1 δi(β(1))=u(1) . . . δi(β(n))=u(n)δj(α(1))=v(1) . . . δj(α(n))=v(n) Wg(N, βα ) α,β∈Sym(n)

1 X X (1) (n) X (1) (n) = Wg(N, βα−1) A ...A B B N j(1)j(α(1)) j(n)j(α(n)) u(1)u(β−1γ(1)) u(n)u(β−1γ(n)) α,β∈Sym(n) ~j ~u

1 X −1 (1) (n) (1) (n) = Wg(N, βα ) Tr A ,...,A Tr −1 B ,...,B N α β γ α,β∈Sym(n)

1 X −1 #cyc(α)+#cyc(β−1γ) (1) (n) (1) (n) = Wg(N, βα )N tr A ,...,A tr −1 B ,...,B . N α β γ α,β∈Sym(n) 1 Using the Weingarten function asymptotics, the degree of N in the leading term is (2n − #cyc(α−1β)) + 1 − #cyc(α) − #cyc(β−1γ) −1 −1 = 2n + 1 − (n − α β ) − (n − |α|) − (n − β γ ) −1 −1 = |α| + α β + β γ − (n − 1). −1 −1 Recall that |α| + |α β| + |β 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 µ(P −1P ) tr A(1),...,A(n) tr B(1),...,B(n) + O . σ π σ K[π] N 2 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 (1) (n) (1) (n) (A ,...,A ) → (a1, . . . , an) ∈ (A, ϕ), (B ,...,B ) → (b1, . . . , bn) ∈ (A, ϕ), with tracial state ϕ, and UN be an N × N Haar unitary matrix. Then as N → ∞, (1) ∗ (1) (n) ∗ (n) X −1 E ◦ tr UA U B ...UA U B → µ(Pσ Pπ)ϕσ [a1, . . . , an] ϕK[π] [b1, . . . , bn] . σ,π∈NC(n) σ≤π

119 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 −1 µ(Pσ 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[gj(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 ∈ NC(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 [b1ab2] = 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

⊗ 2 ⊗ 3 A = Bhxi = C − Span (b0xb1x . . . , xbn : n ≥ 0) /(linearity relations) ' B ⊕ B C ⊕ B C ⊕ .... Example 12.4. (C, ϕ) an n.c. probability space. Then

(A = Mn(C), tr ◦ϕ) is also an n.c. probability space. Let

n B = Mn(C), E : A → B,A 7→ (ϕ[Aij])i,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 [b0xb1 . . . xbn] = E [b0Xb1 . . . Xbn] = b0 E [Xb1 ...X] bn. Same for joint distributions. Remark 12.7 (Generating functions). Moment-generating function: for b ∈ B, X −1 −1 M (b) = E (1 − bX) = µX (1 − bx) = 1 + b E[X] + b E[XbX] + b E[XbXbX] + .... Note: involves only the symmetric moments E[XbXb . . . bX], not general moments

E[Xb1Xb2 . . . bn−1X]. 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, a2i = E[a2a1]. Note that ∗ ∗ ∗ ∗ ∗ ∗ ∗ hb1a1b2, a2b3i = E [b3a2b1a1b2] = b3 E [(b1a2) a1] b2 = b3 ha1, b1a2i b2.

Remark 12.9 (Tensor products with amalgamation). V1,...,Vn B-bimodules, i.e. have commuting ac- tions B × Vi → Vi,Vi × B → Vi. Then ( k ) X (i) (i) (i) V1 ⊗B V2 ⊗B ... ⊗B Vn = ξ1 ⊗ ... ⊗ ξn : ξj ∈ Vj /(B − linearity), i=1 i.e.

ξ1 ⊗B ... ⊗B (b1ξb2 + b3ηb4) ⊗B ... ⊗B ξn

= ξ1 ⊗B ... ⊗B ξk−1b1 ⊗B ξ ⊗B b2ξk+1 ⊗B ... ⊗B ξn + ξ1 ⊗B ... ⊗B ξk−1b3 ⊗B η ⊗B b4ξk+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. n Remark 12.11 (Free product with amalgamation). (A, B, E) = ∗i=1(Ai, B, Ei). Denote ◦ Ai = {a ∈ Ai : Ei[a] = 0} .

Let W∅ = B, ◦ ◦ ◦ W~u = Au(1) ⊗B Au(2) ⊗B ... ⊗B Au(k) Then ∞ M M A = W~u, k=0 |~u|=k u(1)6=u(2)6=...6=u(k) and E[1] = 1, E[a1a2 . . . ak] = 0.

Complete positivity of the free product.

∗ n Lemma 12.12. Let B be a unital C -algebra. A matrix (Bij)i,j=1 ∈ Mn(B) is positive (as an element in the C∗-algebra) if and only if n X ∗ ∀ b1, b2, . . . , bn ∈ B, bi Bijbj ≥ 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. n X ∗ Aij = bkibkj, buv ∈ A. k=1 Then using notation from Deﬁnition 3.21, n n X ∗ n (En[A])i,j=1 = (E[bkibkj])i,j=1 , k=1 so ∀ b1, . . . , bn ∈ B, n " n n # X ∗ ∗ X X ∗ X bi E[bkibkj]bj = E (bkibi) (bkjbj) ≥ 0 in B. k=1 i=1 j=1

123 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[b0X1b1,X2b2,...,Xnbn] = b0M[X1, b1X2, . . . , bn−1Xn]bn = E[b0X1b1X2b2 ...Xnbn].

Deﬁne the free cumulant functionals

R[b0X1b1,X2b2,...,Xnbn] = b0R[X1, b1X2, . . . , bn−1Xn]bn by X M[b0X1b1,X2b2,...,Xnbn] = Rπ[b0X1b1,X2b2,...,Xnbn]. 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[b0X1b1,X2b2,X3b3,X4b4,X5b5,X6b6,X7b7,X8b8]

= b0R[X1b1R[X2b2,X3b3R[X4]b4X5]b5X6]b6R[X7b7,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,

X R2 [b] = E[XbX] = variance

X is a completely positive map R2 : B → B. Thus: let η : B → B be a completely positive map. The functional µ : Bhxi → B deﬁned by ( µ η(b1), n = 2, R [x, b1x, b2x, . . . , bn−1x] = 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[b0Xb1] = 0,

E[b0Xb1Xb2] = b0η(b1)b2,

E[b0Xb1Xb2Xb3Xb4] = b0η(b1)b2η(b3)b4 + b0η(b1η(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 ∗ ∗ hb1fb2, gb3i = b3 hf, b1gi b2 (compare with Remark 12.8), hf, gi∗ = hg, fi , hf, fi ≥ 0 in B. The full Fock module F(M) is the B-bimodule

∞ M ⊗Bk F(M) = M = B ⊕ M ⊕ (M ⊗B M) ⊕ (M ⊗B M ⊗B M) ⊕ ..., k=0 with the inner product

hf1 ⊗ ... ⊗ fn, g1 ⊗ ... ⊗ gki = δn=k hgk hgk−1 h... hg1, f1i , f2i . . . , fk−1i , fki

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:

∗ ∗ hb1ξb2, b3ξb4i = b4η(b3b1)b2.

On F(M), get the creation operator

∗ ` (b1ξb2ξ . . . bn−1ξbn) = ξb1ξb2ξ . . . bn−1ξbn and the annihilation operator

`(b1ξb2ξ . . . bn−1ξbn) = η(b1)b2ξ . . . bn−1ξbn,

∗ `(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.

µ µt Deﬁnition 12.21. µ : Bhxi → B is -inﬁnitely divisible if for all t ≥ 0, tR = R for some positive µt. t µt In that case write µt = µ . But remember: R ∈ 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.

t η b. If µ is -inﬁnitely divisible (i.e. µ ≥ 0), then automatically µ ≥ 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

(1+η) ](1+η) µ = (Bη[µ]) .

Method II (Shlyakhtenko 2012) is to use an operator-valued Voiculescu’s operator model from Proposi- tion 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,

B R au(1)d1, au(2)d2, . . . , au(n) ∈ D.

Then B D R au(1)d1, au(2)d2, . . . , au(n) = R 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[b1b2] = 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,

B D R [n1b1, n2b2, . . . , nk−1bk−1, nk] = R [n1E[b1], . . . , nk−1E[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

B µX+Y : Bhxi → B

B B B by µX+Y = µX µY . Then can also compute

n B n µX+Y [x ] = ϕ µX+Y [x ] ∈ 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 Feray,´ Lassalle, Sniady´ 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 Z −=z =Gµ(z) = = dµ(x) = 2 dµ(x). R z − x R |z − x|

Remark 13.2. Gµ is invertible in a Stolz angle + z ∈ C : |z| > α, =z > β | β |

128 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:

∂tG(z, t) + G(z, t) ∂zG(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 = µX1/2YX1/2 = µY 1/2XY 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 xz ψµ(z) = dµ(x) = Mµ(z) − 1, R+ or T 1 − xz

−1 χµ(z) = ψµ (z) (on a domain), and the S-transform 1 + z S (z) = χ (z). µ z µ 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: z z 1 + z ψ (z) = α , χ (z) = ,S (z) = . µ 1 − z µ α + z µ α + z Thus (1 + z)2 S (z) = , µν (α + z)(β + z) z(1 + z) χ (z) = , µν (α + z)(β + z) 1 − z(α + β) + p(az − 1)(bz − 1) ψ (z) = , µν 2(z − 1) where a, b = α + β − 2αβ ± p4αβ(1 − α)(1 − β). 1 1 Then G(z) = z 1 + ψ z gives 1 z − (α + β) + p(z − a)(z − b) G (z) = + , µν 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] = tλ, rn[µt] = tmn−2[ρ], in other words " ∞ #! X Z z R (z) = t λ˜ + ρ xn−2zn−1 = t λ˜ + dρ(x) , µt 1 − xz n=2 R where ρ is a finite positive measure. This formula still holds for general µ with finite variance (Kol- mogorov representation). In general, have a free Levy-Khinchin´ representation Z z + x Rµt (z) = t λ + dρ(x) . R 1 − xz 130 It follows that n o ID = νλ,ρ : λ ∈ , ρ finite Borel measure . R λ,ρ Can also define ν∗ via its Fourier transform Z 2 ixθ x + 1 F λ,ρ (θ) = exp iλθ + (e − 1 − ixθ) dρ(x) . ν∗ 2 R x 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:

λ,ρ a. µn ∗ µn ∗ ... ∗ µn → ν∗ weakly. | {z } kn

λ,ρ b. µn µn ... µn → ν 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 2µ. 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 = D2µ, so µ is stable. 1 1 G−1(w) = + α, G (z) = . µ w µ z − α

If α is real, then µ = δα. But if α is purely imaginary, α = ia, then 1 a − =z = = z − ia (

131 and 1 a dµ(x) = dx. π x2 + a2 These are the Cauchy distributions. They have a very special property that

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 −1 interval [a, b]. Fix n, and for 0 ≤ i ≤ n let ai = F (i/n). Now look at all the atomic measures 1 (δ + δ + ... + δ ) n x1 x2 xn with ai−1 ≤ xi ≤ ai. On the one hand, both the k-th moment of this measure and of µ lie

" n−1 n # 1 X 1 X ak, ak . n i n i i=0 i=1

k k k k Therefore the difference between them is at most (1/n)(an − a0) = (1/n)(b − a ). 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 i − 1 V = (a − a ) = F −1 − F −1 n i i−1 n n i=1 i=1

132 and so its logarithm is n X i i − 1 log F −1 − F −1 n n i=1 Now i i − 1 1 i F −1 − F −1 ≈ (F −1)0 , n n n n so the above expression is approximately

n X i log V ≈ log((F −1)0) − n log n. n n i=1 Therefore as n goes to inﬁnity, 1 Z Z Z log(V ) + log(n) → log((F −1)0(x))dx = − log(f(F −1(x)))dx = − log(f(x))dF (x). n n See (Voiculescu 1993) for the corresponding heuristics in the single-variable free entropy case. Remark 13.10 (Gaussian maximization). Now look at all the atomic measures with the second moment Pn 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 n/2 −1 (n−1)/2 n in R , An = 2π Γ(n/2) n . The corresponding normalized logarithmic volume converges 1 1 to 2 log(2π) + 2 , 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) sa = A1,...,An ∈ MN×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 χ(X1,...Xn) = inf inf lim sup 2 log vol Γ(X1,...,Xn; m, N, ε) + log N . m∈N ε>0 N→∞ N 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.

2 e. Normalize τ[Xj ] = 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

χ(X1 + εS1,...,Xn + εSn) δ(X1,...,Xn) = n + lim sup . ε↓0 |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.

1 b. For any set of generators, δ(matrix algebra) = 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 [ξiP (X1,...,Xn)] 2 (Schwinger-Dyson equations). Typically ξi is not a polynomial. Sometimes, ξi ∈ L (A, τ) exists. Then deﬁne the free Fisher information n n ∗ X 2 X ∗ Φ (X1,...,Xn) = kξik2 = τ[ξi ξi] i=1 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 ∗ 1/2 1/2 n 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 p−1 Theorem 13.25. More generally if p, q ≥ 2 and λ = q−1 , then

(L(Fp))λ ' L(Fq).

Using this idea, Dykema and Radulescu˘ 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), 1 1 where (1 − n2 ) + (1 − k2 ) = s. More generally, formulas for free products of ﬁnite dimensional and even hyperﬁnite algebras. Even with amalgamation.

Much, much more to be said.

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