Symmetry of Tensor Space? Assume we have a tensor space $W=U\otimes V$, we ask what are "basic components" of $W$ be. One answer would be $U, V$ because they tensor into $W$. But is it possible that there are actually multiple choices of subspaces $U'$ and $V'$ in $W$ composed into $W$ and $(U,V)$ is just one of them?
My formulation goes as following.
Define a multiplication $*:W\times W\rightarrow W$ with linear embedding $\iota:U\rightarrow W$ and $\tau:V\rightarrow W$ that respect the tensor product,
$$
\forall u\in U,v\in V:\iota(u)*\tau(v)=u\otimes v.
$$
For now we pick $\iota:u\mapsto u\otimes 1_V$, $\tau: v\mapsto 1_U\otimes v$ and
$$
(u_1\otimes v_1) * (u_2\otimes v_2) := u_1u_2\otimes v_1v_2.
$$
We have the following property that mimic the original tensor product,


*

*$\iota (U)*\tau(V):=\{u*v: u\in U, v\in V\}=W$,

*$\iota(U)*\iota(U)\subseteq\iota(U)$.


My question is to ask for another "interesting" decomposition under this multiplication $*$. Is it possible that there exists some non-trivial proper sub-algebras $U', V'\subseteq W$ that
$$
U'* V' = W,
$$
and $\{\iota(U),\tau(V)\}\neq \{U', V'\}$? If so, could we have some characterization among relationship between these splitting subspaces?
Remark 1.
One candidate would be further decompose $U$ or $V$ respectively. But then we could ask the same question toward the smaller spaces recursively. Therefore WLOG one might assume $U, V$ are somehow "irreducible" so the choice of $U', V'$ must lie accross $U$ and $V$, that is to say,
$$
U'\not\subseteq\iota(U) \text{ and } V'\not\subseteq\tau(V).
$$
Remark 2. With different subspaces chosen, perhaps the tensor rank could vary. If so, then we directly have this algebraic version tensor rank defined as the smallest one,
$$
\mathrm{rank}(A):=\min_{U'*V'=W}\mathrm{rank}_{U',V'}(A).
$$
Simlarly, many other object related to tensor space such as locality of operators could be modifed to obtain its algebraic version.
 A: (The following tensor operations are over $\mathbb C$, ie, $\otimes :=\otimes_\mathbb C$.)
Consider $\mathbb C[x]\otimes\mathbb C[y]\simeq \mathbb C[x,y]$ with the natural inclusions $\mathbb C[x], \mathbb C[y]\hookrightarrow\mathbb C[x,y]$. Then $$\mathbb C[x+y]\otimes \mathbb C[x-y]\simeq \mathbb C[x+y, x-y]\overset \phi\simeq \mathbb C[x,y]$$ may be what you want, where $$\phi:\sum_{i,j\geq 0}a_{i,j}(x+y)^i(x-y)^j\mapsto \sum_{m,n\geq 0}b_{m,n}x^my^n$$ is defined by the usual polynomial expansion. Here $\phi$ is giving the inclusions $\mathbb C[x+y], \mathbb C[x-y]\hookrightarrow\mathbb C[x,y]$ by expanding polynomials.
A: You're asking if $U\otimes V \cong U' \otimes V'$ implies that $\begin{cases} U \cong U' \\ V \cong V'\end{cases}$.
The answer is no. Take $U=V=\mathbb R^4$. Then $$U\otimes V \cong \displaystyle\bigoplus_{i, j} \mathbb R (e_i\times e_j)$$ where $(e_i)_i$ is a basis for $\mathbb R^4$. But this is just $4$ copies of $\mathbb R^4$, so
$$U\otimes V \cong \displaystyle\bigoplus_{i, j} \mathbb R (e_i\otimes e_j) \cong \bigoplus_1^{4} \mathbb R^4 \cong \mathbb R^{16}.$$
But then consider the case $U'=\mathbb R$, $V'=\mathbb R^{16}$.
In this case $$U'\otimes V' \cong \displaystyle\bigoplus_{i, j} \mathbb R (e_i\otimes e_j) \cong \bigoplus_{j=1}^{16} \mathbb R(e_1 \otimes e_j) \cong \mathbb R^{16}.$$
Since $U\not\cong U'$ and $V\not\cong V'$, we see that we can get the same tensor product even if our factors are different.
More generally, if $U$ and $V$ are real vector spaces with $\dim(U)=n$ and $\dim(V)=m$, then $U\otimes V$ is a real $nm$-dimensional vector space, so $$U\otimes V \cong \mathbb R^{nm}.$$
