Maximal common isotropic subspace for a finite family of skewforms Let $V$ be  a vector space of dimension $n$  over a field $F$. An alternating bilinear form  $\alpha\colon V \times V \rightarrow F $ will be called a skewform. A subspace $W$ is isotropic for $\alpha$ if $\alpha(x,y) = 0$ for all $x,y \in W$.
More generally, if $\alpha = \{\alpha_{1}, \alpha_{2},\dotsc,\alpha_{k}\}$ is a collection of $k$ skew forms on $V$  then a subspace $ W $ is isotropic for $ \alpha $ if it is isotropic for each $ \alpha_{i} $.
Given a collection  $ \alpha $ of $k$ skewforms, let $ m(\alpha) $ be the dimension of the largest isotropic subspace for $\alpha$. Then we define  $d(F,n,k) = \min_{\alpha} m(\alpha) $.
We set  $ s_{n}(F) $ to be that positive integer $r$ satisfying  $d(F,n,r)= 1$ and $d(F,n,r-1)  \geq 2$.
Now I am interested when $ F $ is $\mathbb{Q}$. We know that $s_{2n}(\mathbb{Q}) = 2n$ or $2n-1$ . My question is:

*

*For which values of $2n$, $ s_{2n}(\mathbb{Q} )= 2n-1 $? That is for which values of $2n$ there exists $2n-1$ skewforms which have no common isotropic subspaces of dimension $> 1$. Already I know that for $2n=4$ or $ 8 $, $ s_{2n}(\mathbb{Q} )= 2n-1$.

*What is the value of $s_{6}(\mathbb{Q})$?

 A: $s_{2n}(\mathbb R) \geq  2n$ for $n \neq 1,2,4$.
In particular, one can't show that $s_{2n}(\mathbb Q) = 2n-1$ for these $n$ using real methods.
Proof: Suppose for contradiction that there were $\alpha_1,\dotsc, \alpha_{2n-1}$ skew-symmetric forms on $\mathbb R^{2n}$ such that for any $u$, $v$ linearly independent, we have $\alpha_i (u, v)\neq 0$ for some $i$.
Then we would obtain a parallization of $S^{2n-1}$, as follows: For $u \in S^{2n-1} \subset \mathbb R^{2n}$ (embedded as the unit sphere, say), we can map the tangent space of $S^{2n-1}$ at $u$ to $\mathbb R^{2n-1}$ by mapping a vector $v$ orthogonal to $u$ to $\alpha_1(u,v),\dots, \alpha_{2n-1}(u,v)$. This is injective by assumption, hence an isomorphism, and it varies continuously with $u$, thus gives an isomorphism of the vector bundle with the trivial bundle.
By a classical theorem of algebraic topology, $n=1$, $2$, or $4$. QED.
I explained in the comments (1 2) how to check that the $u$ such that there exists $v$ linearly independent from $u$ with $\alpha_i (u,v)=0$ for all $i$ form a hypersurface of degree $2n-2$ in $\mathbb P^{2n-1}$. If this hypersurface is smooth, then it is Fano and the Brauer–Manin obstruction vanishes (as long as $n>2$). The above argument shows it has a real point. If it has a $p$-adic point for each prime $p$ (which perhaps can be checked separately), then a conjecture of Colliot-Thélène suggests it should have a rational point.
However, this conjecture is very hard to prove in most special cases. Thus I suspect $s_{2n}(\mathbb Q) = 2n$ to be very hard to prove for these $n$ even if it is true, unless the hypersurface in question has some special geometric structure I am missing that simplifies the problem.
EDIT: Here is what is probably a better way to obtain a hypersurface.
Given skew-symmetric forms $\alpha_1,\dots, \alpha_{2n-1}$ on a $2n$-dimensional vector space $V$ over a field $F$, a vector $u$ lies in an isotropic subspace of rank $\geq 2$ if and only if the map $F^{2n-1} \to V^\vee$ that sends the unit vector $e_i$ to the linear form $v \mapsto \alpha_i (u, v)$ has image of codimension at least $2$ (since its image is contained in the perpendicular to the isotropic), in other words if and only if it has kernel of dimension at least $1$.
Thus, there is a two-dimensional maximal isotropic if and only if some nontrivial linear combination of the $\alpha_i$s is degenerate.
A skew-symmetric form is degenerate if and only if the Pfaffian vanishes. The Pfaffian of a linear combination of the $\alpha_i$s is a polynomial of degree $n$ in $2n-1$ variables, and thus defines a hypersurface of degree $n$ in $\mathbb P^{2n-2}$. We have $s_{2n}(F) = 2n-1$ if and only if, for some collection of $\alpha_i$s, this hypersurface does not have an $F$-rational point.
One advantage of this perspective is that it seems geometrically simpler, being a lower-degree hypersurface in a lower-degree projective space.
Another is that we can describe the singular locus more clearly. Inside the space of all skew-symmetric matrices, the singular locus of the Pfaffian hypersurface is the locus of matrices of rank $\leq 2n-4$, which has codimension $6$.
Thus for a generic $\alpha_1,\dots,\alpha_{2n-1}$, the singular locus of this hypersurface in $\mathbb P^{2n-1}$ has codimension $6$. In particular, for $n=3$, generically we get a smooth cubic threefold. I don't know any approach to the existence of rational points on such varieties.
