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 $$ \varphi_{*} \circ \star = \boxplus \qquad ? $$

`$(\mu,\nu)\mapsto \mu \boxplus \nu$`

is not "bilinear". $\endgroup$ – Mikael de la Salle Apr 15 '12 at 20:03`$\varphi_*(\mu_{(a,b)})=\mu_a \boxplus \mu_b$`

. 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. $\endgroup$ – Mikael de la Salle Apr 17 '12 at 11:44