Can the tensor product of two function spaces be regarded as a function space? Let $K,T$ be fields and $V:=\{g:K\to T\}$ be a vector space over T. Then take $W:=V\otimes V$, is this $W$ isomorphic to some function space?
Little background: In quantum mechanics the the state of a one-electron (half spin fermion) system at a given time is a function $\mathbb{R}^{3}\to \mathbb{C}^4$ (with possible constraints I can't recall). Then the state of an $N$-electron system is an element of $\{\mathbb{R}^{3}\to \mathbb{C}^4\}^{\otimes N}$, then mysteriously the state is regarded as a function $\mathbb{R}^{3N}\to (\mathbb{C}^4)^{\otimes N}$.
 A: Turns out it is generally not true that there is a one-to-one correspondence between $\{f:V\to W\}^{\otimes N}$ and $\{g:V^N\to W^{\otimes N}\}$ where $V,W$ are vector spaces and $N\in \mathbb N$ because (let $K:= \{f:V\to W\}$, $L:=\{g:V^N\to W^{\otimes N}\}$) $$\mathrm{dim } \,K=(\mathrm{dim }\,W)^{\left|{V}\right|}$$ moreover $$\mathrm{dim }\,K^{\otimes N}=(\mathrm{dim }\,K)^N=(\mathrm{dim } \,W)^{N\left|V\right|}$$ whereas $$\mathrm{dim }\,L=(\mathrm{dim }\,(W^{\otimes N}))^{\left| V^N\right|}=(\mathrm{dim }\, W)^{N\left|{V^N}\right|}$$ but since in my example $\left|\mathbb R^3\right| =\left|\mathbb R^{3N} \right|$ they have the same dimension ($2^\mathfrak c$) thus are isomorphic.
So for it to be true either $\left|V\right|$ or $\mathrm{dim}\, W$ has to be infinite.
A: In short, yes. You provide two logically equivalent descriptions with the only difference being whether you regard the system as a collection of states or as having one global complicated state. The difference is semantic, only.
