If I take a tensor product of vector spaces (for simplicity - this could be more general) $V\otimes W$ then of course it is a vector space, but it has additional structure. One way to think about this is that there are two distinguished sets $P$ and $E$ where $P=\{\,\mbox{elements of form}\, v\otimes w\,|\, v\in V,w\in W\}$ and E = the complement of P (so in quantum mechanics these are the pure and entangled states). These sets have certain properties, e.g. they are closed under multiplication by non-zero scalars; each of them generates $V\otimes W$ (I think, at least for char $\neq2$, and assuming $E\neq\emptyset$); if dim($V\otimes W$) is prime then $E=\emptyset$; etc.
My question is whether there is a result along the lines of "Given a vector space $X$ with two subsets $P$, $E$ with properties [to be specified], there exist vector spaces $V_1,...,V_n$ and an isomorphism $X\cong V_1\otimes...\otimes V_n$ of vector spaces, such that under that isomorphism the subsets $P$ and $E$ of $X$ correspond respectively to the pure and entangled states of $V_1\otimes...\otimes V_n$"?
(ps this is my first post so apologies if the question is inappropriate/ mis-placed)
Given two vector spaces $V,W$ the space of decomposable tensors in $V\otimes W$ is the Segre cone.
It is often convenient to projectivize the situation to the Segre embedding $i:\mathbb P(V)\times \mathbb P(W)\hookrightarrow \mathbb P(V\otimes W)$, whose image then corresponds to the pure states.
The question you ask is whether you can recognize abstractly in a projective space $\mathbb P(X)$ the image of $i$.
If the dimension of $X$ is 4, the answer is yes: any smooth quadric surface in $\mathbb P^3$ can be seen as the space of pure states , but I don't know the general answer.