## Classical convolution VS Free Convolution

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We denote $\varphi:\mathbb R^2\rightarrow\mathbb R$ the addition of real numbers, and $\varphi_*:M_1(\mathbb R^2)\rightarrow M_1(\mathbb R)$ the induced push-forward map (where $M_1(\Delta)$ stands for the set of probability measures on $\Delta$).

Note that given $\mu$, $\nu\in M_1(\mathbb R)$, we have by definition of the (classical) convolution of measures $\varphi_*(\mu\otimes\nu)=\mu *\nu$.

Now, consider two random variables $X$ and $Y$ taking values in $\mathbb R$, and denote $\mu_X$, $\mu_Y\in M_1(\mathbb R)$ their laws. If $X$ and $Y$ independent, then the random variables $(X,Y)$ taking values in $\mathbb R^2$ has for law $\mu_{(X,Y)}=\mu_X\otimes\mu_Y$ and $\varphi_*(\mu_{(X,Y)})=\mu_X*\mu_Y$.

I was wondering if it is possible to describe the free (additive) convolution in the same setting :

Consider two auto-adjoint non-commutative random variables $a$ and $b$ which are free, and denote $\mu_a$, $\mu_b\in M_1(\mathbb R)$ their laws. Is there a (universal) bilinear map $\star:M_1(\mathbb R)\times M_1(\mathbb R)\rightarrow M_1(\mathbb R^2)$ such that $\mu_{(a,b)}=\mu_a\star\mu_b$, and moreover $\varphi_*(\mu_{a}\star\mu_b)=\mu_a\boxplus\mu_b$ ?

Somehow the vague question is "what is the analogue of the free product when one describe the elements of an operator algebra through their spectral measures ?"

And what about the free multiplicative convolution ? and the rectangular one ?

EDIT (after the comments). As Mikael de la Salle explained, there is no hope to obtain such operation $\star:M_1(\mathbb R)\times M_1(\mathbb R)\rightarrow M_1(\mathbb R^2)$ because of lack of bilinearity of $\boxplus$. Terry Tao also emphasis that $M_1(\mathbb R^2)$ is certainly not the good space to consider (there is no "space below" once we deal with non-commutativity !).

This motivates the following question :

Is there exists some "space" (let us stay vague) $\mathcal E$ which may represent the joint laws of two non-commutative random variables, equipped with $\star : \mathcal E\rightarrow M_1(\mathbb R)$, and a map $\varphi_*:\mathcal E\rightarrow M_1(\mathbb R)$ which "looks like the $\varphi_*$", such that we have the spliting $$\varphi_{*} \circ \star = \boxplus \qquad ?$$

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The spectral measure $\mu_{(a,b)}$ is not classically defined when $a$ and $b$ don't commute. About the best one can do in the general non-commutative situation is compute all the mixed *-moments of $a$ and $b$. At that level of abstraction, one could define $\ast$ by abstract nonsense (because the mixed moments of free variables $a,b$ are all polynomials in the individual moments of $a,b$) but this is a rather trivial way to answer the question. – Terry Tao Apr 14 2012 at 16:08

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@Terry: I think one can use that $a$ and $b$ are "free" to form $\mu_{(a,b)}$; as $\mu_a$ is only the spectral measure "at a vector state" (sorry, I'm not sure what the correct terminology should be), see page 8 of the survey arxiv.org/abs/0911.0087 However, I don't know precisely what the definition of $\mu_{(a,b)}$ is... – Matthew Daws Apr 14 2012 at 18:04

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Another reason why the answer to your question is no is that the map $(\mu,\nu)\mapsto \mu \boxplus \nu$ is not "bilinear". – Mikael de la Salle Apr 15 2012 at 20:03

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Somewhat embarrassingly, I realise that another good introduction to such notions is Terry's own notes: terrytao.wordpress.com/2010/02/10/… But I still don't see what $\mu_{(a,b)}$ is; could the OP clarify? – Matthew Daws Apr 16 2012 at 19:16

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@Matthew: I think the question was precisely whether there exists a natural measure $\mu_{(a,b)}$ on $\mathbf R^2$ such that $\varphi_*(\mu_{(a,b)})=\mu_a \boxplus \mu_b$. Terry's comment was that the natural analogue of $\mu_a \otimes \mu_b$ was not a measure on $\mathbf R^2$ (which is a linear form on $C(Sp a)\otimes C(Sp b)$), but rather a more complicated object which is a linear form on the free product of $C(Sp a)$ and $C(Sp b)$. My comment was that there is no (even "not natural") such measure $\mu_{(a,b)}$ if, as OP,one requires that the dependance in $\mu_a$ and $\mu_b$ be bilinear. – Mikael de la Salle Apr 17 2012 at 11:44
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## 1 Answer

The free analogue of the tensor product of measures is the free product.

Instead of the space $\mathbb{R}$, we consider the algebra $C_0(\mathbb{R})$ of continuous functions on $\mathbb{R}$, tending to 0 at infinity. In this framework, measures are modelled by states: continuous (for supremum norm) linear functionals $\varphi:C_0(\mathbb{R})\rightarrow \mathbb{C}$ with the properties that $\varphi(f)\geq 0$ whenever $f\geq 0$, and $\sup\{\varphi(f)\mid 0\leq f\leq 1\}=1$.

It is clear that integration wrt any probability measure on $\mathbb{R}$ satisfies this properties, and if i am not mistaken, every such state defines a measure on $\mathbb{R}$. So $S(C_0(\mathbb{R}))=M_1(\mathbb{R})$.

In fact, we can replace $C_0(\mathbb{R})$ by an arbitrary (possibly non-commutative) C$^\ast$-algebra $C$ and consider its state space as a "set of measures on some non-commutative space".

Now the free product of two C$^\ast$-algebras $C_1$ and $C_2$ is the universal C$^\ast$-algebra $C=C_1\star C_2$ that is generated by $C_1$ and $C_2$ and such that every pair of $\ast$-homomorphisms $\pi_i:C_i\rightarrow D$ extends to $\pi:C\rightarrow D$. It is classical that then also every pair of states $\varphi_i:C_i\rightarrow \mathbb{C}$ extends uniquely to a state $\varphi_1\star\varphi_2$ on $C$. (look up reduced free products) So the space $\mathcal{E}$ asked for is $S(C_0(\mathbb{R})\star C_0(\mathbb{R}))$, and the map $\star$ is the one given above. To be explicit, denote the canonical embeddings by $\psi_i:C_0(\mathbb{R})\rightarrow C_0(\mathbb{R})\star C_0(\mathbb{R})=C$.

Consider that map $i:\mathbb{R}\rightarrow\mathbb{C}:x\mapsto x$. Then $\psi_1(i)+\psi_2(i)\in C$ is an element with spectrum $\mathbb{R}$, so this defines an embedding $\varphi:C_0(\mathbb{R})\rightarrow C$ by the spectral theorem. The map $\varphi_\ast$ from the question is just the corresponding restriction map $\varphi_{\ast}:S(C)\rightarrow M_1(\mathbb{R})$.

This argument is not entirely correct because $i\not\in C_0(\mathbb{R})$, but this can be corrected using approximations. Someone with more expertise in C$^\ast$-algebras can explain this better than I can.

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Hi Steven ! Hope everything is fine in Sweden. – Adrien Hardy Apr 26 2012 at 19:06
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