Free probability

Remarks

Example: let $F_2$ be the free group on generators $a$ and $b$, let $A = \C[F_2]$ be its group algebra, and let $\varphi(\sum_g c_g g) = c_e$, the coefficient of the identity. Then the subalgebras $A_1 = \C[\langle a \rangle]$ and $A_2 = \C[\langle b \rangle]$ are free. An element of $A_1$ with $\varphi = 0$ is a combination of powers $a^k$ with $k \ne 0$, and likewise for $A_2$. A product of such elements with alternating indices is a combination of alternating words $a^{k_1} b^{l_1} a^{k_2} \cdots$ with nonzero exponents. These are reduced words, hence not the identity, so $\varphi$ of the product is $0$. The same argument works for any free product of groups.

Non-example: classically independent variables are not free. Let $A = L^\infty(\Omega, P)$ for a probability space $(\Omega, P)$, $\varphi$ the expectation, and $x, y$ independent with mean $0$ and nonzero variance. Then $\varphi(xyxy) = \varphi(x^2)\varphi(y^2) \ne 0$, although the indices alternate, so the defining condition fails.

Often $A$ is a C*-algebra or a von Neumann algebra and $\varphi$ is a state, frequently a faithful trace. Freeness determines all mixed moments of free variables from their separate moments, as independence does classically.