What kind of a set do decomposable tensors form in the tensor product $V\otimes W$? Are decomposable tensors dense in finite dimensions?

Question

Q

If $V,W$ are finite dimensional over $\mathbb R$ or $\mathbb C$, how does the set of decomposable tensors $$S=\{v\otimes w:v\in V,w\in W\}\subset V\otimes W$$ "fit" in its ambient space $V\otimes W$? What kind of a subset is it, topologically? Is it dense? And if yes, then how much? Is it $G_\delta$? Or something in between?

Is there a measure-theoretic explanation on how big it is (if we focus on some unit ball to make measures finite) - is it zero-measure, or is almost every tensor a decomposable tensor in the measure-theoretic sense?

My thoughts

Algebraically, we can immediately observe that it's a union of 1-dimensional subspaces - that is, a set of lines through the origin, and it does not fit into any strict subspace - because it contains a basis for the whole $V\otimes W$.

Moreover, since every pair of vectors $(v,w)$ gives birth to a line in this set, I start to feel¹ the union of all such lines should cover the entire $V\otimes W$ - but it surely doesn't (Tensors that are not pure tensors), so I begin to expect that it's at least dense.

As a finite-dimensional space $V\otimes W$ can always be endowed with the unique (Hausdorff) topology, so it makes sense to ask for topological properties likes denseness.

Context

I came up with this question while following an online course on quantum computing. The state of two qubits lives in $\mathbb C^2 \otimes \mathbb C^2$, entangled qubits being described by indecomposable tensors therein. The course authors argue that there are "a lot more" entangled states (than pure product states) because the number of independent variables describing decomposable tensors grow linearly in the number of multipliers in $V\otimes\cdots\otimes V$, while a general tensor requires exponentially more data. I want to get a better grasp on this.

---

¹ gripping on 2- and 3-dimensional intuition, where $\mathbb R^2=\bigcup_{v,w\in\mathbb R^2} \textrm{span}\{v+w\}$ is a union of all lines generated by pairs of vectors.

Answer 1

The pure tensors in $V\otimes W$ are mapped to from $V\times W$, and the fibers are the equivalence classes with respect to the "anti-diagonal" scalar action $\lambda\cdot(v,w)=(\lambda v,\lambda^{-1}w)$. Therefore, we can say the pure tensors have dimension $\dim V+\dim W-1$, whereas all of $V\otimes W$ has dimension $(\dim V)(\dim W)$. If our field of scalars is the reals, then dimension can be defined topologically (note having positive codimension implies measure zero, so this is certainly measure zero), but more generally we can define dimensions of algebraic varieties.

In general, if $V^\ast=\hom(V,\mathbb{F})$ denotes the dual of a vector space, there is a natural isomorphism between $\hom(V,W)$ (the vector space of linear maps $V\to W$) and $V^\ast\otimes W$. Given $\varphi\in V^\ast$, $w\in W$, the tensor $\varphi\otimes w$ becomes the linear transformation $T(x)=\varphi(x)w$. While $V$ and $V^\ast$ are always isomorphic for finite-dimensional vector spaces, they are not naturally so $-$ usually one picks coordinates. But if we assume $V,W$ are more than just vector spaces $-$ real inner product spaces $-$ then $V\cong V^\ast$ is natural, and we can identify $V\otimes W$ with $\hom(V,W)$; $v\otimes w$ will correspond to $T(x)=\langle v,x\rangle w$.

Indeed, $V\otimes W$ becomes an inner product space with $\langle v_1\otimes w_1,v_2\otimes w_2\rangle=\langle v_1,v_2\rangle\langle w_1,w_2\rangle$ and $\hom(V,W)$ becomes an inner product space with Frobenius norm $\langle A,B\rangle=\mathrm{tr}(A^TB)$. The vector space isomorphism $V\otimes W\cong\hom(V,W)$ will also be an isometry! In coordinates, if $V\cong\mathbb{R}^n$ and $W=\mathbb{R}^m$ then $\hom(V,W)\cong\mathbb{R}^{m\times n}$ is represented by $m\times n$ matrices, and the isomorphism $V\otimes W\cong\hom(V,W)$ is given by taking the formal symbol $v\otimes w$ to the Kronecker product $v^T\otimes w$ (a matrix transformation). Thus, the nonzero pure tensors correspond to matrices with column rank $1$.

Every nonzero pure tensor is a unique positive real times a pure tensor with magnitude $1$, equivalently $v\otimes w$ with unit vectors $v,w$. Then the unit pure tensors are mapped to from the pair of unit spheres in $V$ and $W$. That is, the unit pure tensors are topologically $(S^{n-1}\times S^{m-1})/\mathbb{Z}_2$ (the "mod $\mathbb{Z}_2$" comes from $v\otimes w=(-v)\otimes(-w)$). This means that all pure tensors (including zero) are topologically a cone $C((S^{n-1}\times S^{m-1})/\mathbb{Z}_2)$.