The bijection

[math] \newcommand{\mathds}{\mathbb}[/math]

This article was automatically generated from a tex file and may contain conversion errors. If permitted, you may login and edit this article to improve the conversion.

12a. Cumulants

In this chapter we discuss the precise abstract relation between classical and free probability. This is something quite tricky, and as a starting point, we have the following statement, that we know from the above, and which is something very concrete:

Theorem

The moments of the main limiting measures in classical and free probability, real and complex, and discrete and continuous,

[[math]] \xymatrix@R=20pt@C=23pt{ &\mathfrak B_t\ar@{-}[rr]\ar@{-}[dd]&&\Gamma_t\ar@{-}[dd]\\ \beta_t\ar@{-}[rr]\ar@{-}[dd]\ar@{-}[ur]&&\gamma_t\ar@{-}[dd]\ar@{-}[ur]\\ &B_t\ar@{-}[rr]\ar@{-}[uu]&&G_t\ar@{-}[uu]\\ b_t\ar@{-}[uu]\ar@{-}[ur]\ar@{-}[rr]&&g_t\ar@{-}[uu]\ar@{-}[ur] } [[/math]]

are always given by the same formula, [math]M_k=\sum_{\pi\in D(k)}t^{|\pi|}[/math], where [math]D\subset P[/math] is a certain set of partitions associated to the measure, and where [math]|.|[/math] is the number of blocks.

Show Proof

This is something that we know well, the sets of partitions being:

[[math]] \xymatrix@R=20pt@C=3pt{ &\mathcal{NC}_{even}\ar[dl]\ar[dd]&&\ \ \ \mathcal{NC}_2\ \ \ \ar[ll]\ar[dd]\ar[dl]\\ NC_{even}\ar[dd]&&NC_2\ar[ll]\ar[dd]\\ &\mathcal P_{even}\ar[dl]&&\mathcal P_2\ar[ll]\ar[dl]\\ P_{even}&&P_2\ar[ll] } [[/math]]

For full details on all this, we refer to the previous chapters.

■

What is interesting with the above cube is that it provides us with some 3D orientation in noncommutative probability, taken at large. To be more precise, the 3 “coordinate axes” that we have there, corresponding to the 3 pairs of opposing faces, are:

(1) Real vs. complex.

(2) Discrete vs. continuous.

(3) Classical vs. free.

All this is very nice, and potentially fruitful. In what follows we will be mainly interested in what happens on the vertical, classical vs. free. And here, just by looking at the upper and lower faces of the cube, and how they are connected, we conclude that there should be a bijection between classical and free probability, having something to do with crossing and noncrossing partitions. Thus, we are led to: \begin{question} What is the exact bijection between classical and free limiting laws, which connects the upper and lower faces of the standard cube? \end{question} This is certainly a very interesting and fundamental question, and fortunately, there is a simple answer to it, known since the paper of Bercovici-Pata , who first axiomatized this bijection. Explaining all this, Bercovici-Pata bijection, will be our next task.

Getting to work now, what we have in Theorem 12.1 is of rather advanced nature, regarding some special measures. In order to explain the Bercovici-Pata bijection, which basically deals with arbitrary probability measures, it is better to forget Theorem 12.1, and go back to the basics. And talking basics now, probability and combinatorics at large, of quite general type, we have here the following key definition, due to Rota:

Definition

Associated to any real probability measure [math]\mu=\mu_f[/math] is the following modification of the logarithm of the Fourier transform [math]F_\mu(\xi)=E(e^{i\xi f})[/math],

[[math]] K_\mu(\xi)=\log E(e^{\xi f}) [[/math]]

called cumulant-generating function. The Taylor coefficients [math]k_n(\mu)[/math] of this series, given by

[[math]] K_\mu(\xi)=\sum_{n=1}^\infty k_n(\mu)\,\frac{\xi^n}{n!} [[/math]]

are called cumulants of the measure [math]\mu[/math]. We also use the notations [math]k_f,K_f[/math] for these cumulants and their generating series, where [math]f[/math] is a variable following the law [math]\mu[/math].

In other words, the cumulants are more or less the coefficients of the logarithm of the Fourier transform [math]\log F_\mu[/math], up to some normalizations. To be more precise, we have [math]K_\mu(\xi)=\log F_\mu(-i\xi)[/math], so the formula relating [math]\log F_\mu[/math] to the cumulants [math]k_n(\mu)[/math] is:

[[math]] \log F_\mu(-i\xi)=\sum_{n=1}^\infty k_n(\mu)\,\frac{\xi^n}{n!} [[/math]]

Equivalently, the formula relating [math]\log F_\mu[/math] to the cumulants [math]k_n(\mu)[/math] is:

[[math]] \log F_\mu(\xi)=\sum_{n=1}^\infty k_n(\mu)\,\frac{(i\xi)^n}{n!} [[/math]]

We will see in a moment the reasons for the above normalizations, namely change of variables [math]\xi\to -i\xi[/math], and Taylor coefficients instead of plain coefficients, the idea being that for simple laws like [math]g_t,p_t[/math], we will obtain in this way very simple quantities. Let us also mention that there is a reason for indexing the cumulants by [math]n=1,2,3,\ldots[/math] instead of [math]n=0,1,2,\ldots\,[/math], and more on this later, once we will have some theory and examples.

As a first observation, the sequence of cumulants [math]k_1,k_2,k_3,\ldots[/math] appears as a modification of the sequence of moments [math]M_1,M_2,M_3,\ldots\,[/math], the numerics being as follows:

Proposition

The sequence of cumulants [math]k_1,k_2,k_3,\ldots[/math] appears as a modification of the sequence of moments [math]M_1,M_2,M_3,\ldots\,[/math], and uniquely determines [math]\mu[/math]. We have

[[math]] k_1=M_1 [[/math]]

[[math]] k_2=-M_1^2+M_2 [[/math]]

[[math]] k_3=2M_1^3-3M_1M_2+M_3 [[/math]]

[[math]] k_4=-6M_1^4+12M_1^2M_2-3M_2^2-4M_1M_3+M_4 [[/math]]

[[math]] \vdots [[/math]]

in one sense, and in the other sense we have

[[math]] M_1=k_1 [[/math]]

[[math]] M_2=k_1^2+k_2 [[/math]]

[[math]] M_3=k_1^3+3k_1k_2+k_3 [[/math]]

[[math]] M_4=k_1^4+6k_1^2k_2+3k_2^2+4k_1k_3+k_4 [[/math]]

[[math]] \vdots [[/math]]

with in both cases the correspondence being polynomial, with integer coefficients.

Show Proof

Here all the theoretical assertions regarding moments and cumulants are clear from definitions, and the numerics are clear from definitions too. To be more precise, we know from Definition 12.3 that the cumulants are defined by the following formula:

[[math]] \log E(e^{\xi f})=\sum_{s=1}^\infty k_s(f)\,\frac{\xi^s}{s!} [[/math]]

By exponentiating, we obtain from this the following formula:

[[math]] E(e^{\xi f})=\exp\left(\sum_{s=1}^\infty k_s(f)\,\frac{\xi^s}{s!}\right) [[/math]]

Now by looking at the terms of order [math]1,2,3,4[/math], this gives the above formulae.

■

Obviously, there should be some explicit formulae for the correspondences in Proposition 12.4. This is indeed the case, but things here are quite tricky, and we will discuss this later, once we will have enough motivations for the study of the cumulants.

The interest in cumulants comes from the fact that [math]\log F_\mu[/math], and so the cumulants [math]k_n(\mu)[/math] too, linearize the convolution. To be more precise, we have the following result:

Theorem

The cumulants have the following properties:

• [math]k_n(cf)=c^nk_n(f)[/math].
• [math]k_1(f+d)=k_1(f)+d[/math], and [math]k_n(f+d)=k_n(f)[/math] for [math]n \gt 1[/math].
• [math]k_n(f+g)=k_n(f)+k_n(g)[/math], if [math]f,g[/math] are independent.

Show Proof

Here (1) and (2) are both clear from definitions, because we have the following computation, valid for any [math]c,d\in\mathbb R[/math], which gives the results:

[[math]] \begin{eqnarray*} K_{cf+d}(\xi) &=&\log E(e^{\xi(cf+d)})\\ &=&\log[e^{\xi d}\cdot E(e^{\xi cf})]\\ &=&\xi d+K_f(c\xi) \end{eqnarray*} [[/math]]

As for (3), this follows from the fact that the Fourier transform [math]F_f(\xi)=E(e^{i\xi f})[/math] satisfies the following formula, whenever [math]f,g[/math] are independent random variables:

[[math]] F_{f+g}(\xi)=F_f(\xi)F_g(\xi) [[/math]]

Indeed, by applying the logarithm, we obtain the following formula:

[[math]] \log F_{f+g}(\xi)=\log F_f(\xi)+\log F_g(\xi) [[/math]]

With the change of variables [math]\xi\to-i\xi[/math], we obtain the following formula:

[[math]] K_{f+g}(\xi)=K_f(\xi)+K_g(\xi) [[/math]]

Thus, at the level of coefficients, we obtain [math]k_n(f+g)=k_n(f)+k_n(g)[/math], as claimed.

■

At the level of examples now, we have the following result:

Proposition

The sequence of cumulants [math]k_1,k_2,k_3,\ldots[/math] is as follows:

• For [math]\mu=\delta_c[/math] the cumulants are [math]c,0,0,\ldots[/math]
• For [math]\mu=g_t[/math] the cumulants are [math]0,t,0,0,\ldots[/math]
• For [math]\mu=p_t[/math] the cumulants are [math]t,t,t,\ldots[/math]
• For [math]\mu=b_t[/math] the cumulants are [math]0,t,0,t,\ldots[/math]

Show Proof

We have 4 computations to be done, the idea being as follows:

(1) For [math]\mu=\delta_c[/math] we have the following computation:

[[math]] \begin{eqnarray*} K_\mu(\xi) &=&\log E(e^{c\xi})\\ &=&\log(e^{c\xi})\\ &=&c\xi \end{eqnarray*} [[/math]]

But the plain coefficients of this series are the numbers [math]c,0,0,\ldots\,[/math], and so the Taylor coefficients of this series are these same numbers [math]c,0,0,\ldots\,[/math], as claimed.

(2) For [math]\mu=g_t[/math] we have the following computation:

[[math]] \begin{eqnarray*} K_\mu(\xi) &=&\log F_\mu(-i\xi)\\ &=&\log\exp\left[-t(-i\xi)^2/2\right]\\ &=&t\xi^2/2 \end{eqnarray*} [[/math]]

But the plain coefficients of this series are the numbers [math]0,t/2,0,0,\ldots\,[/math], and so the Taylor coefficients of this series are the numbers [math]0,t,0,0,\ldots\,[/math], as claimed.

(3) For [math]\mu=p_t[/math] we have the following computation:

[[math]] \begin{eqnarray*} K_\mu(\xi) &=&\log F_\mu(-i\xi)\\ &=&\log\exp\left[(e^{i(-i\xi)}-1)t\right]\\ &=&(e^\xi-1)t \end{eqnarray*} [[/math]]

But the plain coefficients of this series are the numbers [math]t/n![/math], and so the Taylor coefficients of this series are the numbers [math]t,t,t,\ldots\,[/math], as claimed.

(4) For [math]\mu=b_t[/math] we have the following computation:

[[math]] \begin{eqnarray*} K_\mu(\xi) &=&\log F_\mu(-i\xi)\\ &=&\log\exp\left[\left(\frac{e^\xi+e^{-\xi}}{2}-1\right)t\right]\\ &=&\left(\frac{e^\xi+e^{-\xi}}{2}-1\right)t \end{eqnarray*} [[/math]]

But the plain coefficients of this series are the numbers [math](1+(-1)^n)t/n![/math], so the Taylor coefficients of this series are the numbers [math]0,t,0,t,\ldots\,[/math], as claimed.

■

At a more theoretical level, we have the following result, generalizing (3,4) above, and which is something very useful, when dealing with the compound Poisson laws:

Theorem

For a compound Poisson law [math]p_\nu[/math] we have

[[math]] k_n(p_\nu)=M_n(\nu) [[/math]]

valid for any integer [math]n\geq1[/math].

Show Proof

We can assume, by using a continuity argument, that our measure [math]\nu[/math] is discrete, as follows, with [math]t_i \gt 0[/math] and [math]z_i\in\mathbb R[/math], and with the sum being finite:

[[math]] \nu=\sum_i t_i\delta_{z_i} [[/math]]

By using now the Fourier transform formula for [math]p_\nu[/math] from chapter 11, we obtain:

[[math]] \begin{eqnarray*} K_{p_\nu}(\xi) &=&\log F_{p_\nu}(-i\xi)\\ &=&\log\exp\left[\sum_it_i(e^{\xi z_i}-1)\right]\\ &=&\sum_it_i\sum_{n\geq1}\frac{(\xi z_i)^n}{n!}\\ &=&\sum_{n\geq1}\frac{\xi^n}{n!}\sum_it_iz_i^n\\ &=&\sum_{n\geq1}\frac{\xi^n}{n!}\,M_n(\nu) \end{eqnarray*} [[/math]]

Thus, we are led to the conclusion in the statement.

■

12b. Inversion formula

Getting back to theory now, the sequence of cumulants [math]k_1,k_2,k_3,\ldots[/math] appears as a modification of the sequence of moments [math]M_1,M_2,M_3,\ldots\,[/math], and understanding the relation between moments and cumulants will be our next task. We recall from Proposition 12.4 that we have the following formulae, for the cumulants in terms of moments:

[[math]] k_1=M_1 [[/math]]

[[math]] k_2=-M_1^2+M_2 [[/math]]

[[math]] k_3=2M_1^3-3M_1M_2+M_3 [[/math]]

[[math]] k_4=-6M_1^4+12M_1^2M_2-3M_2^2-4M_1M_3+M_4 [[/math]]

[[math]] \vdots [[/math]]

Also, we have the following formulae, for the moments in terms of cumulants:

[[math]] M_1=k_1 [[/math]]

[[math]] M_2=k_1^2+k_2 [[/math]]

[[math]] M_3=k_1^3+3k_1k_2+k_3 [[/math]]

[[math]] M_4=k_1^4+6k_1^2k_2+3k_2^2+4k_1k_3+k_4 [[/math]]

[[math]] \vdots [[/math]]

In order to understand what exactly is going on, with moments and cumulants, which reminds a bit the Möbius inversion formula, we need to do some combinatorics, in relation with partitions. So, let us go back to the material from chapter 4, where some theory for the partitions was developed. We recall that we have the following definition:

Definition

The Möbius function of any lattice, and so of [math]P[/math], is given by

[[math]] \mu(\pi,\nu)=\begin{cases} 1&{\rm if}\ \pi=\nu\\ -\sum_{\pi\leq\tau \lt \nu}\mu(\pi,\tau)&{\rm if}\ \pi \lt \nu\\ 0&{\rm if}\ \pi\not\leq\nu \end{cases} [[/math]]

with the construction being performed by recurrence.

This is something that we already discussed in chapter 4, and as a first example here, the Möbius matrix [math]M_{\pi\nu}=\mu(\pi,\nu)[/math] of the lattice [math]P(2)=\{||,\sqcap\}[/math] is as follows:

[[math]] M=\begin{pmatrix}1&-1\\ 0&1\end{pmatrix} [[/math]]

At [math]k=3[/math] now, we have the following formula for the Möbius matrix [math]M_{\pi\nu}=\mu(\pi,\nu)[/math], once again written with the indices picked increasing in [math]P(3)=\{|||,\sqcap|,\sqcap\hskip-3.2mm{\ }_|\,,|\sqcap,\sqcap\hskip-0.7mm\sqcap\}[/math]:

[[math]] M=\begin{pmatrix} 1&-1&-1&-1&2\\ 0&1&0&0&-1\\ 0&0&1&0&-1\\ 0&0&0&1&-1\\ 0&0&0&0&1 \end{pmatrix} [[/math]]

In general, as explained in chapter 4, the Möbius matrix of [math]P(k)[/math] looks a bit like the above matrices at [math]k=2,3[/math], being upper triangular, with 1 on the diagonal, and so on.

Back to the general case now, the main interest in the Möbius function comes from the Möbius inversion formula, which can be formulated as follows:

Theorem

We have the following implication,

[[math]] f(\pi)=\sum_{\nu\leq\pi}g(\nu) \quad\implies\quad g(\pi)=\sum_{\nu\leq\pi}\mu(\nu,\pi)f(\nu) [[/math]]

valid for any two functions [math]f,g:P(n)\to\mathbb C[/math].

Show Proof

The above formula is in fact a linear algebra result, so let us start with some linear algebra. Consider the adjacency matrix of [math]P[/math], given by the following formula:

[[math]] A_{\pi\nu}=\begin{cases} 1&{\rm if}\ \pi\leq\nu\\ 0&{\rm if}\ \pi\not\leq\nu \end{cases} [[/math]]

Our claim is that the inverse of this matrix is the Möbius matrix of [math]P[/math], given by:

[[math]] M_{\pi\nu}=\mu(\pi,\nu) [[/math]]

Indeed, the above matrix [math]A[/math] is upper triangular, and when trying to invert it, we are led to the recurrence in Definition 12.8, so to the Möbius matrix [math]M[/math]. Thus we have:

[[math]] M=A^{-1} [[/math]]

Now by applying this equality of matrices to vectors, regarded as complex functions on [math]P(n)[/math], we are led to the inversion formula in the statement.

■

As a first illustration, for [math]P(2)[/math] the formula [math]M=A^{-1}[/math] appears as follows:

[[math]] \begin{pmatrix}1&-1\\ 0&1\end{pmatrix}= \begin{pmatrix}1&1\\ 0&1\end{pmatrix}^{-1} [[/math]]

At [math]k=3[/math] now, the formula [math]M=A^{-1}[/math] for [math]P(3)[/math] reads:

[[math]] \begin{pmatrix} 1&-1&-1&-1&2\\ 0&1&0&0&-1\\ 0&0&1&0&-1\\ 0&0&0&1&-1\\ 0&0&0&0&1 \end{pmatrix}= \begin{pmatrix} 1&1&1&1&1\\ 0&1&0&0&1\\ 0&0&1&0&1\\ 0&0&0&1&1\\ 0&0&0&0&1 \end{pmatrix}^{-1} [[/math]]

In general, the formula [math]M=A^{-1}[/math] looks quite similar, and we refer here to chapter 4.

With these ingredients in hand, let us go back to probability. We first have:

Definition

We define quantities [math]M_\pi(f),k_\pi(f)[/math], depending on partitions

[[math]] \pi\in P(k) [[/math]]

by starting with [math]M_n(f),k_n(f)[/math], and using multiplicativity over the blocks.

To be more precise, the convention here is that for the one-block partition [math]1_n\in P(n)[/math], the corresponding moment and cumulant are the usual ones, namely:

[[math]] M_{1_n}(f)=M_n(f)\quad,\quad k_{1_n}(f)=k_n(f) [[/math]]

Then, for an arbitrary partition [math]\pi\in P(k)[/math], we decompose this partition into blocks, having sizes [math]b_1,\ldots,b_s[/math], and we set, by multiplicativity over blocks:

[[math]] M_\pi(f)=M_{b_1}(f)\ldots M_{b_s}(f)\quad,\quad k_\pi(f)=k_{b_1}(f)\ldots k_{b_s}(f) [[/math]]

With this convention, following Rota and others, we can now formulate a key result, fully clarifying the relation between moments and cumulants, as follows:

Theorem

We have the moment-cumulant formulae

[[math]] M_n(f)=\sum_{\nu\in P(n)}k_\nu(f)\quad,\quad k_n(f)=\sum_{\nu\in P(n)}\mu(\nu,1_n)M_\nu(f) [[/math]]

or, equivalently, we have the moment-cumulant formulae

[[math]] M_\pi(f)=\sum_{\nu\leq\pi}k_\nu(f)\quad,\quad k_\pi(f)=\sum_{\nu\leq\pi}\mu(\nu,\pi)M_\nu(f) [[/math]]

where [math]\mu[/math] is the Möbius function of [math]P(n)[/math].

Show Proof

There are several things going on here, the idea being as follows:

(1) First, it is clear from our conventions, from Definition 12.10, that the first set of formulae is equivalent to the second set of formulae, by multiplicativity over blocks.

(2) The other observation is that, due to the Möbius inversion formula, from Theorem 12.9, in the second set of formulae, the two formulae there are in fact equivalent.

(3) Summarizing, the 4 formulae in the statement are all equivalent. In what follows we will focus on the first 2 formulae, which are the most useful, in practice.

(4) Let us first work out some examples. At [math]n=1,2,3[/math] the moment formula gives the following equalities, which are in tune with the findings from Proposition 12.4:

[[math]] M_1=k_|=k_1 [[/math]]

[[math]] M_2=k_{|\,|}+k_\sqcap=k_1^2+k_2 [[/math]]

[[math]] M_3=k_{|\,|\,|}+k_{\sqcap|}+k_{\sqcap\hskip-2.8mm{\ }_|}+k_{|\sqcap}+k_{\sqcap\hskip-0.5mm\sqcap}=k_1^3+3k_1k_2+k_3 [[/math]]

At [math]n=4[/math] now, which is a case which is of particular interest for certain considerations to follow, the computation is as follows, again in tune with Proposition 12.4:

[[math]] \begin{eqnarray*} M_4 &=&k_{|\,|\,|}+(\underbrace{k_{\sqcap\,|\,|}+\ldots}_{6\ terms})+(\underbrace{k_{\sqcap\,\sqcap}+\ldots}_{3\ terms})+(\underbrace{k_{\sqcap\hskip-0.5mm\sqcap\,|}+\ldots}_{4\ terms})+k_{\sqcap\hskip-0.5mm\sqcap\hskip-0.5mm\sqcap}\\ &=&k_1^4+6k_1^2k_2+3k_2^2+4k_1k_3+k_4 \end{eqnarray*} [[/math]]

As for the cumulant formula, at [math]n=1,2,3[/math] this gives the following formulae for the cumulants, again in tune with the findings from Proposition 12.4:

[[math]] k_1=M_|=M_1 [[/math]]

[[math]] k_2=(-1)M_{|\,|}+M_\sqcap=-M_1^2+M_2 [[/math]]

[[math]] k_3=2M_{|\,|\,|}+(-1)M_{\sqcap|}+(-1)M_{\sqcap\hskip-2.8mm{\ }_|}+(-1)M_{|\sqcap}+M_{\sqcap\hskip-0.5mm\sqcap}=2M_1^3-3M_1M_2+M_3 [[/math]]

Finally, at [math]n=4[/math], after computing the Möbius function of [math]P(4)[/math], we obtain the following formula for the fourth cumulant, again in tune with Proposition 12.4:

[[math]] \begin{eqnarray*} k_4 &=&(-6)M_{|\,|\,|}+2(\underbrace{M_{\sqcap\,|\,|}+\ldots}_{6\ terms})+(-1)(\underbrace{M_{\sqcap\,\sqcap}+\ldots}_{3\ terms})+(-1)(\underbrace{M_{\sqcap\hskip-0.5mm\sqcap\,|}+\ldots}_{4\ terms})+M_{\sqcap\hskip-0.5mm\sqcap\hskip-0.5mm\sqcap}\\ &=&-6M_1^4+12M_1^2M_2-3M_2^2-4M_1M_3+M_4 \end{eqnarray*} [[/math]]

(5) After all these preliminaries, time now to get to work, and prove the result. As mentioned above, our formulae are all equivalent, and it is enough to prove just one of them. We will prove in what follows the first formula, namely:

[[math]] M_n(f)=\sum_{\nu\in P(n)}k_\nu(f) [[/math]]

(6) In order to do this, we use the very definition of the cumulants, namely:

[[math]] \log E(e^{\xi f})=\sum_{s=1}^\infty k_s(f)\,\frac{\xi^s}{s!} [[/math]]

By exponentiating, we obtain from this the following formula:

[[math]] E(e^{\xi f})=\exp\left(\sum_{s=1}^\infty k_s(f)\,\frac{\xi^s}{s!}\right) [[/math]]

(7) Let us first compute the function on the left. This is easily done, as follows:

[[math]] \begin{eqnarray*} E(e^{\xi f}) &=&E\left(\sum_{n=0}^\infty\frac{(\xi f)^n}{n!}\right)\\ &=&\sum_{n=0}^\infty M_n(f)\,\frac{\xi^n}{n!} \end{eqnarray*} [[/math]]

(8) Regarding now the function on the right, this is given by:

[[math]] \begin{eqnarray*} \exp\left(\sum_{s=1}^\infty k_s(f)\,\frac{\xi^s}{s!}\right) &=&\sum_{p=0}^\infty\frac{\left(\sum_{s=1}^\infty k_s(f)\,\frac{\xi^s}{s!}\right)^p}{p!}\\ &=&\sum_{p=0}^\infty\frac{1}{p!}\sum_{s_1=1}^\infty k_{s_1}(f)\,\frac{\xi^{s_1}}{s_1!}\ldots\ldots\sum_{s_p=1}^\infty k_{s_p}(f)\,\frac{\xi^{s_p}}{s_p!}\\ &=&\sum_{p=0}^\infty\frac{1}{p!}\sum_{s_1=1}^\infty\ldots\sum_{s_p=1}^\infty k_{s_1}(f)\ldots k_{s_p}(f)\,\frac{\xi^{s_1+\ldots+s_p}}{s_1!\ldots s_p!} \end{eqnarray*} [[/math]]

(9) The point now is that all this leads us into partitions. Indeed, we are summing over indices [math]s_1,\ldots,s_p\in\mathbb N[/math], which can be thought of as corresponding to a partition of [math]n=s_1+\ldots+s_p[/math]. So, let us rewrite our sum, as a sum over partitions. For this purpose, recall that the number of partitions [math]\nu\in P(n)[/math] having blocks of sizes [math]s_1,\ldots,s_p[/math] is:

[[math]] \binom{n}{s_1,\ldots,s_p}=\frac{n!}{p_1!\ldots p_s!} [[/math]]

Also, when resumming over partitions, there will be a [math]p![/math] factor as well, coming from the permutations of [math]s_1,\ldots,s_p[/math]. Thus, our sum can be rewritten as follows:

[[math]] \begin{eqnarray*} \exp\left(\sum_{s=1}^\infty k_s(f)\,\frac{\xi^s}{s!}\right) &=&\sum_{n=0}^\infty\sum_{p=0}^\infty\frac{1}{p!}\sum_{s_1+\ldots+s_p=n}k_{s_1}(f)\ldots k_{s_p}(f)\,\frac{\xi^n}{s_1!\ldots s_p!}\\ &=&\sum_{n=0}^\infty\frac{\xi^n}{n!}\sum_{p=0}^\infty\frac{1}{p!}\sum_{s_1+\ldots+s_p=n}\binom{n}{s_1,\ldots,s_p}k_{s_1}(f)\ldots k_{s_p}(f)\\ &=&\sum_{n=0}^\infty\frac{\xi^n}{n!}\sum_{\nu\in P(n)}k_\nu(f) \end{eqnarray*} [[/math]]

(10) We are now in position to conclude. According to (6,7,9), we have:

[[math]] \sum_{n=0}^\infty M_n(f)\,\frac{\xi^n}{n!}=\sum_{n=0}^\infty\frac{\xi^n}{n!}\sum_{\nu\in P(n)}k_\nu(f) [[/math]]

Thus, we have the following formula, valid for any [math]n\in\mathbb N[/math]:

[[math]] M_n(f)=\sum_{\nu\in P(n)}k_\nu(f) [[/math]]

We are therefore led to the conclusions in the statement.

■

Summarizing, we have now a nice theory of cumulants, or rather a beginning of such a theory, and with this in hand, we can go back to the diagram in Theorem 12.1, see if we can now better understand what is going on there. However, this is a bit tricky:

(1) Our theory of cumulants as developed so far only applies properly to the “real classical” case, that is, to the measures [math]g_t,b_t[/math] there. In order to deal with the full classical case, comprising as well the measures [math]G_t,B_t[/math], we would have to upgrade everything into a theory of [math]*[/math]-cumulants, and this is something quite technical.

(2) Regarding the “free real” measures [math]\gamma_t,\beta_t[/math] and their complex analogues [math]\Gamma_t,\mathfrak B_t[/math], here the cumulant theory developed above gives nothing interesting. We will see in the next section, at least in the real case, that of [math]\gamma_t,\beta_t[/math], that the revelant theory which applies to them is a substantial modification of what we have, called free cumulant theory.

In short, technical problems in all directions, and we are not ready yet for better understanding Theorem 12.1. As a more modest objective, however, we have the quite reasonable question of understanding the moment formula [math]M_k=\sum_{\pi\in D(k)}t^{|\pi|}[/math] there for the measures [math]g_t,b_t[/math], by using the cumulant theory developed above. Which is in fact a non-trivial question too, with the answer involving the following result from :

Theorem

The uniform orthogonal easy groups [math]G\subset O_N[/math], and their associated categories of partitions [math]D\subset P[/math], all coming from subsets [math]L\subset\mathbb N[/math], are as follows,

[[math]] \xymatrix@R=50pt@C=50pt{ B_N\ar[r]&O_N\\ S_N\ar[u]\ar[r]&H_N\ar[u]}\quad \item[a]ymatrix@R=25pt@C=20pt{\\ :} \quad \item[a]ymatrix@R=50pt@C50pt{ P_{12}\ar[d]&P_2\ar[d]\ar[l]\\ P&P_{even}\ar[l]} \quad \item[a]ymatrix@R=25pt@C=20pt{\\ :} \quad \item[a]ymatrix@R=50pt@C50pt{ \{1,2\}\ar[d]&\{2\}\ar[d]\ar[l]\\ \mathbb N&2\mathbb N\ar[l]} [[/math]]

with [math]D[/math] consisting of the partitions [math]\pi\in P[/math] whose blocks have lengths belonging to [math]L\subset\mathbb N[/math].

Show Proof

Consider an arbitrary easy group, [math]S_N\subset G_N\subset O_N[/math]. This group must then come from a category of partitions, as follows:

[[math]] P_2\subset D\subset P [[/math]]

Now if we assume [math]G=(G_N)[/math] to be uniform, this category [math]D[/math] is uniquely determined by the subset [math]L\subset\mathbb N[/math] consisting of the sizes of the blocks of the partitions in [math]D[/math]. And as explained in , one can prove that the admissible sets are those in the statement, corresponding to the categories and the groups in the statement.

■

In relation now with cumulants, we have the following result, also from :

Theorem

The cumulants of the asymptotic truncated characters for the uniform easy groups [math]G=(G_N)[/math] are given by the formula

[[math]] k_n(\chi_t)=t\delta_{n\in L} [[/math]]

with [math]L\subset\mathbb N[/math] being the associated subset, and at the level of asymptotic moments this gives

[[math]] M_k(\chi_t)=\sum_{\pi\in D(k)}t^{|\pi|} [[/math]]

with [math]D\subset P[/math] being the associated category of partitions.

Show Proof

This is clear indeed from Theorem 12.12, by performing a case-by-case analysis, with the cases [math]G=O,S,H[/math] corresponding to the computations for [math]g_t,p_t,b_t[/math] from Proposition 12.5, and with the remaining case, that of the bistochastic groups, [math]G=B[/math], being similar. Again, for details on all this, we refer to .

■

Summarizing, we have now a good understanding of the formula [math]M_k=\sum_{\pi\in D(k)}t^{|\pi|}[/math] for the real classical limiting measures, based on cumulants, but with this involving however some more advanced mathematics. It is possible of course to reformulate all the above in terms of categories of partitions only, but this won't lead to any simplifications in the proofs, which are based on categories of partitions anyway, and would rather obscure the final results themselves, which are best thought of in terms of easy groups.

Finally, in order to extend the above results to the general the complex case, the cumulant theory must be upgraded into a [math]*[/math]-cumulant theory, which is something quite technical. We will discuss however such questions in chapter 15 below, directly in a more general setting, that of operator-valued noncommutative probability theory, following Speicher and others , , . In what regards the easy groups, and more generally easy quantum groups, in the general unitary setting, this is again a quite technical subject, and we will be back to this on several occasions, in the remainder of this book.

12c. Free cumulants

In what follows we discuss the free analogues of the above, following Speicher , and subsequent work. We first have the following definition:

Definition

The free cumulants [math]\kappa_n(a)[/math] of a variable [math]a\in A[/math] are defined by

[[math]] R_a(\xi)=\sum_{n=1}^\infty\kappa_n(a)\xi^{n-1} [[/math]]

with the [math]R[/math]-transform being defined as usual by the formula

[[math]] G_a\left(R_a(\xi)+\frac{1}{\xi}\right)=\xi [[/math]]

where [math]G_a(\xi)=\int_\mathbb R\frac{d\mu(t)}{\xi-t}[/math] with [math]\mu=\mu_a[/math] is the corresponding Cauchy transform.

As before with classical cumulants, we have a number of basic examples and illustrations, and a number of basic general results. Let us start with some numerics:

Proposition

The free cumulants [math]\kappa_1,\kappa_2,\kappa_3,\ldots[/math] appear as a modification of the moments [math]M_1,M_2,M_3,\ldots\,[/math], and uniquely determine [math]\mu[/math]. We have

[[math]] \kappa_1=M_1 [[/math]]

[[math]] \kappa_2=-M_1^2+M_2 [[/math]]

[[math]] \kappa_3=2M_1^3-3M_1M_2+M_3 [[/math]]

[[math]] \kappa_4=-5M_1^4+10M_1^2M_2-2M_2^2-4M_1M_3+M_4 [[/math]]

[[math]] \vdots [[/math]]

in one sense, and in the other sense we have

[[math]] M_1=\kappa_1 [[/math]]

[[math]] M_2=\kappa_1^2+\kappa_2 [[/math]]

[[math]] M_3=\kappa_1^3+3\kappa_1\kappa_2+\kappa_3 [[/math]]

[[math]] M_4=\kappa_1^4+6\kappa_1^2\kappa_2+2\kappa_2^2+4\kappa_1\kappa_3+\kappa_4 [[/math]]

[[math]] \vdots [[/math]]

with in both cases the correspondence being polynomial, with integer coefficients.

Show Proof

Here all theoretical assertions regarding moments and cumulants are clear from definitions, and the numerics are clear from definitions too, after some computations based on Definition 12.14. Let us actually present these computations, which are quite instructive, more complicated than the classical ones, and that we will need, later on:

(1) We know that the Cauchy transform is the following function:

[[math]] G(\xi)=\sum_{n=0}^\infty\frac{M_n}{\xi^{n+1}} [[/math]]

Consider the inverse of this Cauchy transform [math]G[/math], with respect to composition:

[[math]] G(K(\xi))=K(G(\xi))=\xi [[/math]]

According to Definition 12.14, the free cumulants [math]\kappa_n[/math] appear then as follows:

[[math]] K(\xi)=\frac{1}{\xi}+\sum_{n=1}^\infty\kappa_n\xi^{n-1} [[/math]]

Thus, we can compute moments in terms of free cumulants, and vice versa, by using either of the inversion formulae [math]G(K(\xi))=\xi[/math] and [math]K(G(\xi))=\xi[/math].

(2) This was for the theory. In practice now, playing with the original inversion formula from Definition 12.14, namely [math]G(K(\xi))=\xi[/math], proves to be something quite complicated, so we will choose to use instead the other inversion formula, namely:

[[math]] K(G(\xi))=\xi [[/math]]

Thus, the equation that we want to use is as follows, with [math]G=G(\xi)[/math]:

[[math]] \frac{1}{G}+\sum_{n=1}^\infty\kappa_nG^{n-1}=\xi [[/math]]

(3) With [math]\xi=z^{-1}[/math] our equation takes the following form, with [math]G=G(z^{-1})[/math]:

[[math]] \frac{1}{G}+\sum_{n=1}^\infty\kappa_nG^{n-1}=z^{-1} [[/math]]

Now by multiplying by [math]z[/math], our equation takes the following form:

[[math]] \frac{z}{G}+z\sum_{n=1}^\infty\kappa_nG^{n-1}=1 [[/math]]

Equivalently, our equation is as follows, with [math]G=G(z^{-1})[/math] as before:

[[math]] \frac{z}{G}+\sum_{n=1}^\infty\kappa_nz^n\left(\frac{G}{z}\right)^{n-1}=1 [[/math]]

(4) Observe now that we have the following formula:

[[math]] \frac{G}{z}=\frac{G(z^{-1})}{z}=\frac{\sum_{n=0}^\infty M_nz^{n+1}}{z}=\sum_{n=0}^\infty M_nz^n [[/math]]

This suggests introducing the following quantity:

[[math]] F=\sum_{n=1}^\infty M_nz^n [[/math]]

Indeed, we have then [math]G/z=1+F[/math], and our equation becomes:

[[math]] \frac{1}{1+F}+\sum_{n=1}^\infty\kappa_nz_n(1+F)^{n-1}=1 [[/math]]

(5) By expanding the fraction on the left, our equation becomes:

[[math]] \sum_{n=0}^\infty(-F)^n+\sum_{n=1}^\infty\kappa_nz_n(1+F)^{n-1}=1 [[/math]]

Moreover, we can cancel the 1 term on both sides, and our equation becomes:

[[math]] \sum_{n=1}^\infty(-F)^n+\sum_{n=1}^\infty\kappa_nz_n(1+F)^{n-1}=0 [[/math]]

Alternatively, we can write our equation as follows:

[[math]] \sum_{n=1}^\infty\kappa_nz_n(1+F)^{n-1}=-\sum_{n=1}^\infty(-F)^n [[/math]]

(6) Good news, this latter equation is something that we are eventually happy with. By remembering that we have [math]F=\sum_{n=1}^\infty M_nz^n[/math], our equation looks as follows:

[[math]] \begin{eqnarray*} &&\kappa_1z+\kappa_2z^2(1+M_1z+M_2z^2+\ldots)+\kappa_3z^3(1+M_1z+M_2z^2+\ldots)^2+\ldots\\ &=&(M_1z+M_2z^2+\ldots)-(M_1z+M_2z^2+\ldots)^2+(M_1z+M_2z^2+\ldots)^3-\ldots \end{eqnarray*} [[/math]]

(7) This was for the hard part, carefully fine-tuning our equation, as to have it as simple as possible, before getting to numeric work. The rest is routine. Indeed, by looking at the terms of order [math]1,2,3,4[/math] we obtain, instantly or almost, the formulae of [math]\kappa_1,\kappa_2,\kappa_3,\kappa_4[/math] in the statement. As for the formulae for [math]M_1,M_2,M_3,M_4[/math], these follow from these.

(8) To be more precise, the equations that we get at order [math]1,2,3,4[/math] are as follows:

[[math]] \kappa_1=M_1 [[/math]]

[[math]] \kappa_2=M_2-M_1^2 [[/math]]

[[math]] \kappa_2M_1+\kappa_3=M_3-2M_1M_2+M_1^3 [[/math]]

[[math]] \kappa_4+2\kappa_3M_1+\kappa_2M_2=M_4-2M_1M_3-M_2^2+3M_1^2M_2-M_1^4 [[/math]]

Thus, we are led to the formulae of [math]\kappa_1,\kappa_2,\kappa_3,\kappa_4[/math] in the statement, and then to the formulae of [math]M_1,M_2,M_3,M_4[/math] in the statement, as desired.

■

Observe the similarity with the formulae in Proposition 12.4. In fact, a careful comparison with Proposition 12.4 is worth the effort, leading to the following conclusion: \begin{conclusion} The first three classical and free cumulants coincide,

[[math]] k_1=\kappa_1\quad,\quad k_2=\kappa_2\quad,\quad k_3=\kappa_3 [[/math]]

but the formulae for the fourth classical and free cumulants are different,

[[math]] k_4=-6M_1^4+12M_1^2M_2-3M_2^2-4M_1M_3+M_4 [[/math]]

[[math]] \kappa_4=-5M_1^4+10M_1^2M_2-2M_2^2-4M_1M_3+M_4 [[/math]]

and the same happens at higher order as well. \end{conclusion} This is something quite interesting, and we will back later with a conceptual explanation for this, via partitions, the idea being that all this comes from:

[[math]] P(n)=NC(n)\iff n\leq 3 [[/math]]

But more on this later. At the level of basic general results, we first have:

Theorem

The free cumulants have the following properties:

• [math]\kappa_n(\lambda a)=\lambda^n\kappa_n(a)[/math].
• [math]\kappa_n(a+b)=\kappa_n(a)+\kappa_n(b)[/math], if [math]a,b[/math] are free.

Show Proof

This is something very standard, the idea being as follows:

(1) We have the following Cauchy transform computation:

[[math]] \begin{eqnarray*} G_{\lambda a}(\xi) &=&\int_\mathbb R\frac{d\mu_{\lambda a}(t)}{\xi-t}\\ &=&\int_\mathbb R\frac{d\mu_a(s)}{\xi-\lambda s}\\ &=&\frac{1}{\lambda}\int_\mathbb R\frac{d\mu_a(s)}{\xi/\lambda-s}\\ &=&\frac{1}{\lambda}\,G_a\left(\frac{\xi}{\lambda}\right) \end{eqnarray*} [[/math]]

But this gives the following formula, by using the definition of the [math]R[/math]-transform:

[[math]] \begin{eqnarray*} G_{\lambda a}\left(\lambda R_a(\lambda\xi)+\frac{1}{\xi}\right) &=&\frac{1}{\lambda}\,G_a\left(R_a(\lambda\xi)+\frac{1}{\lambda\xi}\right)\\ &=&\frac{1}{\lambda}\cdot\lambda\xi\\ &=&\xi \end{eqnarray*} [[/math]]

Thus we have the formula [math]R_{\lambda a}(\xi)=\lambda R_a(\lambda\xi)[/math], which gives (1).

(2) This follows from the standard fact, that we know well from chapter 9, that the [math]R[/math]-transform linearizes the free convolution operation.

■

Again in analogy with the classical case, at the level of examples, we have:

Theorem

The sequence of free cumulants [math]\kappa_1,\kappa_2,\kappa_3,\ldots[/math] is as follows:

• For [math]\mu=\delta_c[/math] the free cumulants are [math]c,0,0,\ldots[/math]
• For [math]\mu=\gamma_t[/math] the free cumulants are [math]0,t,0,0,\ldots[/math]
• For [math]\mu=\pi_t[/math] the free cumulants are [math]t,t,t,\ldots[/math]
• For [math]\mu=\beta_t[/math] the free cumulants are [math]0,t,0,t,\ldots[/math]

Also, for compound free Poisson laws the free cumulants are [math]k_n(\pi_\nu)=M_n(\nu)[/math].

Show Proof

The proofs are analogous to those from the classical case, as follows:

(1) For [math]\mu=\delta_c[/math] we have [math]G_\mu(\xi)=1/(\xi-c)[/math], and so [math]R_\mu(\xi)=c[/math], as desired.

(2) For [math]\mu=\gamma_t[/math] we have, as computed in chapter 9, [math]R_\mu(\xi)=t\xi[/math], as desired.

(3) For [math]\mu=\pi_t[/math] we have, also from chapter 11, [math]R_\mu(\xi)=t/(1-\xi)[/math], as desired.

(4) For [math]\mu=\beta_t[/math] this follows from the formulae in chapter 11, but the best is to prove directly the last assertion, which generalizes (3,4). With [math]\nu=\sum_ic_i\delta_{z_i}[/math] we have:

[[math]] \begin{eqnarray*} R_{\pi_\nu}(\xi) &=&\sum_i\frac{c_iz_i}{1-\xi z_i}\\ &=&\sum_ic_iz_i\sum_{n\geq0}(\xi z_i)^n\\ &=&\sum_{n\geq0}\xi^n\sum_ic_iz_i^{n+1}\\ &=&\sum_{n\geq1}\xi^{n-1}\sum_ic_iz_i^n\\ &=&\sum_{n\geq 1}\xi^{n-1}\,M_n(\nu) \end{eqnarray*} [[/math]]

Thus, we are led to the conclusion in the statement.

■

Observe in particular that the last formula in the above statement, [math]k_n(\pi_\nu)=M_n(\nu)[/math], which is something quite powerful, clarifies a discussion started in chapter 8, and then continued in chapter 11, in relation with the block-modified Wishart matrices.

As before in the classical case, we can define now generalized free cumulants, [math]\kappa_\pi(a)[/math] with [math]\pi\in P(k)[/math], by starting with the numeric free cumulants [math]\kappa_n(a)[/math], as follows:

Definition

We define free cumulants [math]\kappa_\pi(a)[/math], depending on partitions

[[math]] \pi\in P(k) [[/math]]

by starting with [math]\kappa_n(a)[/math], and using multiplicativity over the blocks.

To be more precise, the convention here is that for the one-block partition [math]1_n\in P(n)[/math], the corresponding free cumulant is the usual one, namely:

[[math]] \kappa_{1_n}(a)=\kappa_n(a) [[/math]]

Then, for an arbitrary partition [math]\pi\in P(k)[/math], we decompose this partition into blocks, having sizes [math]b_1,\ldots,b_s[/math], and we set, by multiplicativity over blocks:

[[math]] \kappa_\pi(a)=\kappa_{b_1}(a)\ldots\kappa_{b_s}(a) [[/math]]

With this convention, we have the following result, due to Speicher :

Theorem

We have the moment-cumulant formulae

[[math]] M_n(a)=\sum_{\nu\in NC(n)}\kappa_\nu(a)\quad,\quad \kappa_n(a)=\sum_{\nu\in NC(n)}\mu(\nu,1_n)M_\nu(a) [[/math]]

or, equivalently, we have the moment-cumulant formulae

[[math]] M_\pi(a)=\sum_{\nu\leq\pi}\kappa_\nu(a)\quad,\quad \kappa_\pi(a)=\sum_{\nu\leq\pi}\mu(\nu,\pi)M_\nu(a) [[/math]]

where [math]\mu[/math] is the Möbius function of [math]NC(n)[/math].

Show Proof

As before in the classical case, the 4 formulae in the statement are equivalent, via Möbius inversion. Thus, it is enough to prove one of them, and we will prove the first formula, which in practice is the most useful one. Thus, we must prove that:

[[math]] M_n(a)=\sum_{\nu\in NC(n)}\kappa_\nu(a) [[/math]]

(1) In order to prove this formula, let us get back to the construction of the free cumulants, from Definition 12.14. The Cauchy transform of [math]a[/math] is the following function:

[[math]] G_a(\xi)=\sum_{n=0}^\infty\frac{M_n(a)}{\xi^{n+1}} [[/math]]

Consider the inverse of this Cauchy transform [math]G_a[/math], with respect to composition:

[[math]] G_a(K_a(\xi))=K_a(G_a(\xi))=\xi [[/math]]

According to Definition 12.14, the free cumulants [math]\kappa_n(a)[/math] appear then as follows:

[[math]] K_a(\xi)=\frac{1}{\xi}+\sum_{n=1}^\infty\kappa_n(a)\xi^{n-1} [[/math]]

Thus, we can compute moments in terms of free cumulants by using either of the inversion formulae [math]G_a(K_a(\xi))=\xi[/math] and [math]K_a(G_a(\xi))=\xi[/math].

(2) In practice, as explained in the proof of Proposition 12.15, the best is to use the second inversion formula, [math]K_a(G_a(\xi))=\xi[/math], which after some manipulations reads:

[[math]] \begin{eqnarray*} &&\kappa_1z+\kappa_2z^2(1+M_1z+M_2z^2+\ldots)+\kappa_3z^3(1+M_1z+M_2z^2+\ldots)^2+\ldots\\ &=&(M_1z+M_2z^2+\ldots)-(M_1z+M_2z^2+\ldots)^2+(M_1z+M_2z^2+\ldots)^3-\ldots \end{eqnarray*} [[/math]]

We have already seen, in the proof of Proposition 12.15, how to exploit this formula at order [math]n=1,2,3,4[/math]. The same method works in general, and after some computations, this leads to the formula that we want to establish, namely:

[[math]] M_n(a)=\sum_{\nu\in NC(n)}\kappa_\nu(a) [[/math]]

(3) We are therefore led to the conclusions in the statement. All this was of course quite brief, and for details here, we refer for instance to Nica-Speicher .

■

Observe that the above result leads among others to a more conceptual explanation for Conclusion 12.16, with the equalities and non-equalities there simply coming from:

[[math]] P(n)=NC(n)\iff n\leq3 [[/math]]

Finally, in what regards more advanced aspects, in relation with the moment formula [math]M_k=\sum_{\pi\in D(k)}t^{|\pi|}[/math], this ideally requires quantum groups, and more specifically easy quantum groups, and we will talk about this in chapter 13 below. As an advertisement for that material, however, let us record in advance the following statement:

Theorem

The free uniform orthogonal easy quantum groups [math]G\subset O_N^+[/math], and their associated categories of partitions [math]D\subset P[/math], all coming from subsets [math]L\subset\mathbb N[/math], are

[[math]] \xymatrix@R=50pt@C=50pt{ B_N^+\ar[r]&O_N^+\\ S_N^+\ar[u]\ar[r]&H_N^+\ar[u]}\quad \item[a]ymatrix@R=25pt@C=20pt{\\ :} \quad \item[a]ymatrix@R=53pt@C43pt{ NC_{12}\ar[d]&NC_2\ar[d]\ar[l]\\ NC&NC_{even}\ar[l]} \quad \item[a]ymatrix@R=25pt@C=20pt{\\ :} \quad \item[a]ymatrix@R=50pt@C50pt{ \{1,2\}\ar[d]&\{2\}\ar[d]\ar[l]\\ \mathbb N&2\mathbb N\ar[l]} [[/math]]

with [math]D[/math] consisting of the partitions [math]\pi\in NC[/math] whose blocks have lengths belonging to [math]L\subset\mathbb N[/math]. The free cumulants of the corresponding measures are given by the formula

[[math]] \kappa_n=t\delta_{n\in L} [[/math]]

and at the level of moments this gives the formula [math]M_k=\sum_{\pi\in D(k)}t^{|\pi|}[/math].

Show Proof

Obviously, this is something informal, and we will be back to it, with details. However, with the plea of just believing us, the idea is that the easy quantum groups are abstract beasts of type [math]S_N^+\subset G\subset O_N^+[/math], coming from categories [math]NC_2\subset D\subset NC[/math], and so we are left with an algebraic and probabilistic study of these latter categories, which can be done exactly as in the classical case, and which leads to the above conclusions. More on this in a moment, and in the meantime, we refer to for all this.

■

There are many other things that can be said about free cumulants, and we will come back to this later on, in chapter 15 below, directly in a more general setting, that of the operator-valued free probability theory, following , when discussing free de Finetti theorems, which crucially use the free cumulant technology.

Importantly, everything that has been said above about free cumulants, be it a bit technical, is a mirror image of what can be said about classical cumulants. But at a more advanced level, things are far more interesting than this, for instance because of the key isomorphism [math]NC(k)\simeq NC_2(2k)[/math], that we already met in this book in some other contexts, having no classical counterpart. We will be back to this.

12d. The bijection

With the above classical and free cumulant theory in hand, we can now formulate the following simple definition, making the connection between classical and free:

Definition

We say that a real probability measure

[[math]] m\in\mathcal P(\mathbb R) [[/math]]

is the classical version of another measure, called its free version, or liberation

[[math]] \mu\in\mathcal P(\mathbb R) [[/math]]

when the classical cumulants of [math]m[/math] coincide with the free cumulants of [math]\mu[/math].

As a first observation, this definition fits with all the classical and free probability theory developed in the above, in this whole book so far, and notably with the measures from the standard cube, and to start with, we have the following result:

Theorem

In the standard cube of basic probability measures,

[[math]] \xymatrix@R=20pt@C=22pt{ &\mathfrak B_t\ar@{-}[rr]\ar@{-}[dd]&&\Gamma_t\ar@{-}[dd]\\ \beta_t\ar@{-}[rr]\ar@{-}[dd]\ar@{-}[ur]&&\gamma_t\ar@{-}[dd]\ar@{-}[ur]\\ &B_t\ar@{-}[rr]\ar@{-}[uu]&&G_t\ar@{.}[uu]\\ b_t\ar@{-}[uu]\ar@{-}[ur]\ar@{-}[rr]&&g_t\ar@{-}[uu]\ar@{-}[ur] } [[/math]]

the upper measures appear as the free versions of the lower measures.

Show Proof

This follows indeed from our various cumulant formulae found above.

■

In order to reach now to a more advanced theory, depending this time on a parameter [math]t \gt 0[/math], which is something essential, and whose importance will become clear later on, let us formulate, following Bercovici-Pata , and the subsequent work in :

Definition

A convolution semigroup of measures

[[math]] \{m_t\}_{t \gt 0}\quad:\quad m_s*m_t=m_{s+t} [[/math]]

is in Bercovici-Pata bijection with a free convolution semigroup of measures

[[math]] \{\mu_t\}_{t \gt 0}\quad:\quad \mu_s\boxplus\mu_t=\mu_{s+t} [[/math]]

when the classical cumulants of [math]m_t[/math] coincide with the free cumulants of [math]\mu_t[/math].

As before, this fits with all the theory developed so far in this book, and notably with the measures from the standard cube, and we have the following result:

Theorem

In the standard cube of basic semigroups of measures,

[[math]] \xymatrix@R=20pt@C=22pt{ &\mathfrak B_t\ar@{-}[rr]\ar@{-}[dd]&&\Gamma_t\ar@{-}[dd]\\ \beta_t\ar@{-}[rr]\ar@{-}[dd]\ar@{-}[ur]&&\gamma_t\ar@{-}[dd]\ar@{-}[ur]\\ &B_t\ar@{-}[rr]\ar@{-}[uu]&&G_t\ar@{.}[uu]\\ b_t\ar@{-}[uu]\ar@{-}[ur]\ar@{-}[rr]&&g_t\ar@{-}[uu]\ar@{-}[ur] } [[/math]]

the upper semigroups are in Bercovici-Pata bijection with the lower semigroups.

Show Proof

This is a technical improvement of Theorem 12.23, based on the fact that the upper measures in the above diagram form indeed free convolution semigroups, and that the lower measures form indeed classical convolution semigroups, which itself is something that we know well, from the various semigroup results established in above.

■

Back to the examples now, there are many other, and we will be back to this. But, before anything, let us formulate the following surprising result, from :

Theorem

The normal law [math]g_1[/math] is freely infinitely divisible.

Show Proof

This is something tricky, involving all sorts of not very intuitive computations, and for full details here, we refer here to the original paper .

■

The above result shows that the normal law [math]g_1[/math] should have a “classical analogue” in the sense of the Bercovici-Pata bijection. And isn't that puzzling. The problem, however, is that this latter law is difficult to compute, and interpret. See .

Still in relation with the Bercovici-Pata bijection, let us also mention that there are many interesting analytic aspects, coming from the combinatorics of the infinitely divisible laws, classical or free. For this, and other analytic aspects, we refer to .

Finally, as previously promised, let us briefly discuss the axiomatization of the standard cube, using quantum groups. Skipping some details, or rather leaving them for chapter 13 below, the idea is that we have a result as follows:

Theorem (Ground Zero)

Under a collection of suitable extra assumptions

[[math]] \xymatrix@R=16pt@C=16pt{ &K_N^+\ar[rr]&&U_N^+\\ H_N^+\ar[rr]\ar[ur]&&O_N^+\ar[ur]\\ &K_N\ar[rr]\ar[uu]&&U_N\ar[uu]\\ H_N\ar[uu]\ar[ur]\ar[rr]&&O_N\ar[uu]\ar[ur] } [[/math]]

are the unique easy quantum groups. Equivalently, under suitable extra assumptions

[[math]] \xymatrix@R=17pt@C4pt{ &\mathcal{NC}_{even}\ar[dl]\ar[dd]&&\mathcal {NC}_2\ar[dl]\ar[ll]\ar[dd]\\ NC_{even}\ar[dd]&&NC_2\ar[dd]\ar[ll]\\ &\mathcal P_{even}\ar[dl]&&\mathcal P_2\ar[dl]\ar[ll]\\ P_{even}&&P_2\ar[ll] } [[/math]]

are the unique categories of partitions. Also equivalently, under suitable assumptions

[[math]] \xymatrix@R=18pt@C=20pt{ &\mathfrak B_t\ar@{-}[rr]\ar@{-}[dd]&&\Gamma_t\ar@{-}[dd]\\ \beta_t\ar@{-}[rr]\ar@{-}[dd]\ar@{-}[ur]&&\gamma_t\ar@{-}[dd]\ar@{-}[ur]\\ &B_t\ar@{-}[rr]\ar@{-}[uu]&&G_t\ar@{.}[uu]\\ b_t\ar@{-}[uu]\ar@{-}[ur]\ar@{-}[rr]&&g_t\ar@{-}[uu]\ar@{-}[ur] } [[/math]]

are the unique main probability measures.

Show Proof

There is a long story here, first for formulating the precise statement, which is something non-trivial, and then of course for proving it, and for the whole story here, we refer to . We will be back with more details on all this in chapter 13 below.

■

As a conclusion to all this, with some ideas from combinatorics and quantum groups, we have managed to axiomatize the main laws in classical and free probability. Which is certainly something interesting, because we have now some clear ground, free of traps and abstractions, that we can build upon. We will discuss this a bit, in what follows.

General references

Banica, Teo (2024). "Calculus and applications". arXiv:2401.00911 [math.CO].