TITLE: A question about subspace in ${\bigwedge}^2({\mathbb R}^n)$
QUESTION [33 upvotes]: Let $E$ be a linear subspace of ${\bigwedge}^2({\mathbb R}^n)$. What is the minimal dimension of $E$ that guarantees $E$ contains a nonzero element of the form $X\wedge Y$, with $X, Y\in{\mathbb R}^n$?
When $n=3$, dimension $1$ is enough. When $n=4$ we would need dimension $4$. For general $n$, it is easy to see $E$ having dimension $\frac{(n-1)(n-2)}{2}+1$ is sufficient, but I don't know if that is optimal.

REPLY [13 votes]: One more exact answer: for $n=8$ the minimal dimension is $22$,
again attaining the "easy" bound ${n-1 \choose 2} + 1$
and the same as the value for the next dimension $n=9$.
First to explain why ${n-1 \choose 2} + 1$ is enough,
not just for real vector spaces but for an $n$-dimensional vector space $V$
over any field.
Fix nonzero $X_0 \in V$.  Then $X_0 \wedge V = \{ X_0 \wedge Y : Y \in V \}$
is a linear subspace of dimension $n-1$ in $\bigwedge^2 V$.
Thus if $E \subset \bigwedge^2 V$ is a linear subspace of codimension $n-2$
then it must have nonzero intersection with $X_0 \wedge V$.
Now take $n=8$ and identify $V$ with the Cayley octonions.
Let $V_0 \subset V$ consist of the "purely imaginary" octonions,
so $V$ is the orthogonal direct sum of $\bf R$ with $V_0$.
Write $\bigwedge^2 V = ({\bf R} \wedge V_0) \oplus \bigwedge^2 V_0$,
and let $E$ be the kernel of the homomorphism $h: \bigwedge^2 V \to V_0$
that takes $1 \wedge X$ to $X$ and $X \wedge Y$ to the imaginary part of
the octonion $XY$, for any $X,Y \in V_0$.
[This is well-defined because $XY + Y\!X \in \bf R$ for all $X,Y \in V_0$,
so $h(Y\wedge X) = - h(X \wedge Y)$.]  Then $E$ has dimension $21$.
I claim that $E$ contains no nonzero pure tensors.  Indeed a pure tensor in
$\bigwedge^2 V$ has the form $1 \wedge X$, $X \wedge Y$, or $(1+X) \wedge Y$
for some $X,Y \in V_0$ which are linearly independent (so in particular nonzero).
Certainly $h(1\wedge X) = X$ is nonzero.  So is $h(X \wedge Y)$,
because if $XY \in \bf R$ for $X,Y \in V_0$ then $X$ and $Y$ are proportional.
Finally $h\bigl((1+X) \wedge Y\bigr) = Y + h(X \wedge Y)$ cannot be zero because
the imaginary part of $XY$ is orthogonal to $Y$.
It also follows that for $n=6,7$ the minimal $\dim E$ is at least
${n \choose 2} - 6 = 9, 15$, and thus exceeds the lower bound
${n-2 \choose 2} + 1 = 7, 11$ coming from the dimension of the Grassmannian.
Replacing the Cayley octonions by the Hamilton quaternions
recovers the answer of $4$ for $n=4$.