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The Question is simple, yet I have encountered it multiple times in my mathematical life without finding an obvious answer, so I've decided to post it here.

Say $U, V$ are real vector spaces of dimension $m,n$. What is the largest possible dimension of a subspace $W\subset U \otimes V$ so that $W$ contains no nonzero element of the form $u \otimes v$?

For complex vector spaces, the question was asked and answered here on MSE. However, see the comments on the answer there, the correct number over $\mathbb{C}$, which is $(m-1)(n-1)$, is not the correct number over $\mathbb{R}$, as an explicit counterexample with $n=m=2$ shows.

https://math.stackexchange.com/questions/4080198/how-big-can-a-subspace-of-u-otimes-v-be-if-it-contains-no-non-zero-pure-tens

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    $\begingroup$ It might be better to use the phrase "pure tensors" instead of "elementary tensors" $\endgroup$ Commented Nov 25 at 22:48
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    $\begingroup$ Oh!!!! I misread and thought your "elementary vectors" were of the form $u\otimes v$ where $u$ and $v$ were elements of fixed bases, not arbitrary vectors. Ignore me!! $\endgroup$ Commented Nov 25 at 23:05
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    $\begingroup$ This can be rephrased as a question about subspaces $W\subseteq\mathbb R^{m\times n}$ satisfying $\operatorname{rank} (A)\ge 2$ for all $A\in W$, $A\not= 0$ (since rank $1$ matrices $uv^t$ correspond to pure tensors $u\otimes v$, and some very smart folks even write them this way). I found a lot of work on the analogous question with an upper bound on the rank with a perfunctory search, but not much that looked relevant to the actual question. $\endgroup$ Commented Nov 26 at 1:58
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    $\begingroup$ I think Meshulam has solved the "rank bounded below" problem with only mild conditions on the underlying field.: sciencedirect.com/science/article/pii/0024379589904655 $\endgroup$ Commented Nov 26 at 2:07
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    $\begingroup$ @PseudoNeo The paper is interesting, but looks at a different problem. He considers affine subspaces, not linear ones, and only the case m=n. For rank $\ge 2$, the bound he gets is $n^2-3$, which I'd be extremely surprised to be the correct number for the non-affine problem. $\endgroup$ Commented Nov 26 at 2:19

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Here's a fairly obvious comment about an easy case of this problem. I post it as answer because it's too long for the comment box.

In projective terms, you are looking for the largest dimension of a linear space $\Lambda$ in $\mathbb P^{mn-1}$ whose intersection with the Segre variety $\Sigma_{m-1,n-1}$ does not have any real points.

As the answer by Sasha to the linked question on M.SE explains, for any $\Lambda$ of codimension $\operatorname{dim}\left(\Sigma_{m-1,n-1}\right)=m+n-2$, the intersection $\Lambda \cap \Sigma_{m-1,n-1}$ has nonempty set of complex points.

Now suppose there is a $\Lambda$ of codimension $\leq m+n-2$ such that $\Lambda \cap \Sigma_{m-1,n-1}$ has no real points. Then clearly there is such a $\Lambda$ of codimension equal to $m+n-2$. I claim* that by perturbing $\Lambda$, we can arrange that the intersection $\Lambda \cap \Sigma_{m-1,n-1}$ still has no real points, and each complex point of intersection has multiplicity 1.

Under these hypotheses, the number of complex intersection points equals the degree of $\Sigma_{m-1,n-1}$, which equals ${m+n-2 \choose m-1}$. Now complex conjugation acts on this set of intersection points with no fixed point (because a fixed point would be real). So the degree of the Segre variety must be even.

Contrapositively, if the number ${m+n-2 \choose m-1}$ is odd, then the maximal dimension of a disjoint real subspace is the same as the maximal dimension of a disjoint complex subspace, i.e. $(m-1)(n-1)$.

Unfortunately, the set of pairs $(m,n)$ satisfying this condition seems to have has asymptotic density zero.

*: The set of linear subspaces intersecting $\Sigma$ transversely is a dense open set of the real Grassmannian, while the set of linear subspaces whose intersection has no real points is a Euclidean open set. So their intersection should be nonempty.

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    $\begingroup$ The same intersection-theoretic argument seems to give the following bound: if $i\leq m-1$ and $j\leq n-1$ are such that $\binom{i+j}{i}$ is odd, then any subspace of codimension $i+j$ must have a pure tensor. The largest possible value of $i+j$ (for fixed values of $m$ and $n$) one can achieve this way is the bitwise OR of $m-1$ and $n-1$ (i.e., the number whose binary expansion has a $1$ in any given binary place iff either $m-1$ or $n-1$ (or both) do.) $\endgroup$ Commented Nov 27 at 4:00
  • $\begingroup$ @dhy: you could edit this answer to add details of your extended argument, if you are so inclined. (I for one would like to see them.) $\endgroup$ Commented Nov 27 at 9:48
  • $\begingroup$ This does settle the case $m=2$, $n\ge 2$. Your answer shows that the maximal dimension is $n-1$ when $n$ is odd. On the other hand, for even $n$, the answer is $n$. This is an almost immediate consequence of the point of view suggested in my last comment above. (The expression $2\lfloor n/2\rfloor$ works in both cases.) $\endgroup$ Commented 2 days ago
  • $\begingroup$ The fact that there is already something going on in the simple looking case $m=2$ perhaps suggests that it is too ambitious to hope for a formula for general $(m,n)$. $\endgroup$ Commented 2 days ago

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