Let $G$ be a countable discrete group and consider $l^2(G)$ and let $H$ be any Hilbert space. What is the explicit isomorphism between the Hilbert space $l^2(G,H)$ and $l^2(G)\otimes H$.
Is it always true that if $X$ is any measure space, then $L^2(X,H)$ isomorphic to $L^2(X)\otimes H$ as Hilbert spaces?
Let $X$ be a measure space with measure $\mu$ and let $S$ denote the simple $H$-valued functions on sets of finite measure.
Define $\varphi$ from $S$ to the algebraic tensor product by $v1_A\mapsto 1_A\otimes v$, where $v\in H,$ $A\subseteq X$ is measurable with finite measure and $v 1_A$ is constantly equal to $v$ on $A$ and $0$ else.
Note now that for $v,w\in H$ and $A,B\subseteq X$ measurable with finite measure, $$ \int_X \langle v1_A,w1_B\rangle_H\textrm{d}\mu=\mu(A\cap B)\langle v,w\rangle_H=\langle 1_A 1_B\rangle_{L^2(X)} \langle v,w\rangle_H= \langle 1_A\otimes v,1_B\otimes w\rangle_{L^2(X)\otimes H}, $$ which implies that $\varphi$ is an isomorphism of inner product spaces. Taking completions implies $L^2(X)\otimes H$ is isomorphic to $L^2(X,H)$.