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Let $V = \mathbb{C}^6$ and let $\alpha_1, \alpha_2, \alpha_3 : V \times V \to \mathbb{C}$ be three alternating bilinear forms. Is there always a 3-dimensional subspace $U \le V$ such that $\alpha_i|U = 0$ for $i = 1, 2, 3$?

The answer is yes if we believe the Main Theorem of Buhler-Gupta-Harris (1987) [1], which asserts that if you have $k > 1$ skewforms defined on an $n$-dimensional vector space over an algebraically closed field $F$ then there is always a common isotropic subspace of dimension $\lfloor (2n+k) / (k+2)\rfloor$. Really I am interested in this general result, not just the illustrative special case above.

The proof is some straightforward-looking Schubert calculus, which I'm not experienced with, but there are several steps in the proof that puzzle me:

  1. On p. 272 the authors use some structure results about the even-codim cohomology groups, but here the classes have codim $\binom{m}{2}$, which is not even in general. In the example above, $m=3$, so $\binom{m}{2} = 3$.
  2. They refer to "standard ring isomorphisms" $H^{\text{even}}(G(m,n)) \cong \Lambda / I \cong \Lambda_m / (h_{n-m+1})$, where $I$ is the ideal generated by all Schur functions whose diagram does not fit inside the $m \times (n-m)$ rectangle. Is this really right? The last ring doesn't even look finite-dim to me.
  3. This is what worries me most. At the top of p. 274 they claim that $k$ copies of the staircase partition of size $\binom{m}{2}$ fit inside an $m \times (n-m)$ rectangle provided only that $k \binom{m}{2} \le m (n-m)$, and they have a specious picture as demonstration, but surely this only works if $k$ is even! For the example above, the claim is that $3$ copies of the diagram of the partition $(2,1)$ fit together inside a $3\times3$ square.
  4. They have some cool version of the Littlewood--Richardson rule (Littlewood--Roe?) that somehow allows you to rotate things around at will. I guess this is probably standard but it's unfamiliar to me, and it's unclear to me how it's working on p. 274, even when $k$ is even. The displayed equation certainly has some typos.

I did at least check (or Urban Jezernik checked anyway) that $s_{(2,1)}^3$ has a nonzero coefficient attached to $s_{(3,3,3)}$. Does this somehow prove the answer to the titular question is yes?

[1] Buhler, Joe; Gupta, Ranee; Harris, Joe, Isotropic subspaces for skewforms and maximal abelian subgroups of (p)-groups, J. Algebra 108, 269-279 (1987). ZBL0612.20009.

I would write to the authors but the paper is so old that it feels unfair to interrogate them about it, and it seems likely that there are many people here who understand these things much better than I do and can help.

This question is related but much more specialized: Maximal common isotropic subspace for a finite family of skewforms

Edit:

The comment section has persuaded me that points 1--2 are not important, as I suspected. In the special case in which either $k$ is even or $m = \lfloor (2n+k) / (k+2) \rfloor$ is odd, the theorem is then equivalent to the following concrete question about Schur polynomials.

Let $\lambda$ be the partition $(m-1, \dots, 2, 1)$ of $m(m-1)/2$. Let $\kappa$ be the partition whose diagram is a $k(m-1)/2 \times m$ rectangle. Is it true that $\langle s_\lambda^k, s_\kappa\rangle > 0$?

The stated inner product is not generally equal to $1$, as claimed by BGH. For example $\langle s_{(1)}^4, s_{(2,2)}\rangle = 2$.

In particular, the answer to the titular question is yes, because $\langle s_{(2,1)}^3, s_{(3,3,3)}\rangle = 2$. This implies (assuming I understand everything right) that there are generically exactly $2$ common isotropic $3$-dim subspaces to the three given skewforms.

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    $\begingroup$ The cycle class map from the group of rational equivalence classes of cycles of codimension $c$ takes values in cohomology $H^{2c}$ (for every Weil cohomology theory). So the relevant part of the cohomology ring is the subring of even-degree summands. $\endgroup$ Commented Jan 28 at 23:42
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    $\begingroup$ For 2. a good reference is [W. Fulton, Intersection theory]. $\endgroup$ Commented Jan 29 at 14:25
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    $\begingroup$ For the last part of your question, the non-vanishing of $s_{(2,1)}^3$ does indeed imply a positive answer, at least if all the bilinear forms are non-degenerate. This seems to be assumed implicitly in the paper (and might be harmless). (And I agree that the proof in the paper, even for $k$ odd, is not clear.) $\endgroup$ Commented Jan 30 at 2:10
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    $\begingroup$ @naf Right, it's the equality case. But note that one can reduce $n$ at liberty, just be restricting to an arbitrary smaller-dim subspace. Therefore without loss of generality $n$ is the smallest integer satisfying $km(m-1)/2 \le m (n-m)$, and if $k$ is even or $m$ is odd this means that equality holds. $\endgroup$ Commented Jan 30 at 10:56
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    $\begingroup$ @SeanEberhard "This implies (assuming I understand everything right) that there are generically exactly 2 common isotropic 3-dim subspaces to the three given skewforms." -- this is true, and can be checked in another way. The locus you are looking for is the zero loci of three copies of $\wedge^2 U^*$ on $Gr(3,6)$. You can directly check (Koszul, Riemann-Roch, etc) that this finite set has indeed degree 2. $\endgroup$ Commented Jan 30 at 14:40

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I have persuaded myself that the theorem is correct, and I will record my notes here. If any of the people who gave helpful pointers in the comments section would like to summarize their comments in an answer I would be happy to accept it.

Suppose we have $k > 1$ skewforms on an $n$-dimensional vector space over $F$ (an algebraically closed field). Then there is a common $m$-dimensional isotropic subspace provided only that $n \ge m + k (m-1)/2$ (and this is best possible -- this part of the paper is not in question). By restricting to a subspace there is no loss in assuming that $n = \lceil m + k(m-1)/2\rceil$. Write $\def\eps{\epsilon} n = m + k(m-1)/2 + \eps$, where $\epsilon = 0$ if $k$ is even or $m$ is odd and otherwise $\epsilon = 1/2$.

Let us assume $\alpha_1, \dots, \alpha_k$ are all nondegenerate. Maybe there is a way of reducing to this case, but in any case it's intuitively the worst case. We can also assume $\alpha_1, \dots, \alpha_k$ are in general position, because this really is the worst case (see comments section).

In the paper below, it is proved that if $\alpha$ is a nondegenerate skewform on $F^n$ and $X \subset \mathrm{Gr}(n, m)$ is the variety of $\alpha$-isotropic subspaces, then the the cohomology class $[X]$ is just the Schubert cycle $\sigma_\tau$ where $\tau$ is the triangle partition $(m-1, m-2, \dots, 1)$ of total weight $m(m-1)/2$.

Harris, Joe; Tu, Loring W., On symmetric and skew-symmetric determinantal varieties, Topology 23, 71-84 (1984). ZBL0534.55010.> Blockquote

Moreover, the cohomology ring of $\mathrm{Gr}(n, m)$ is isomorphic to $\Lambda / I$, where $\Lambda$ is the ring of symmetric functions with coefficients in $\mathbf Z$ and $I$ is the ideal generated by all Schur functions whose associated diagram does not fit inside the rectangle $\kappa$ of dimensions $m \times (n-m)$. The Schubert cycle $\sigma_\tau$ corresponds to the Schur function $s_\tau$. It follows that we just need to check that $s_\tau^k$ is nonzero mod $I$, or equivalently that $s_\tau^k$ has a nonzero coefficient attached to some Schur function $s_\lambda$ such that the diagram of $\lambda$ fits inside that of the rectangle $\kappa$. Note that if $\epsilon = 0$ the only possibility is $\lambda = \kappa$, while if $\epsilon = 1/2$ then there are several possibilities. We claim in both cases that $$\langle s_\tau^k s_{(\eps m)}, s_\kappa\rangle > 0,$$ which is sufficient.

Observation: $s_\tau^2$ has a nonzero coefficient attached to $s_\mu$, where $\mu$ is an $m \times (m-1)$ rectangle. Indeed, $\langle s_\tau^2,s_\mu\rangle = \langle s_\tau, s_{\mu / \tau}\rangle = \langle s_\tau, s_\tau\rangle = 1$ (this follows from the fact that one can rotate skew Schur diagrams by 180 degrees). Moreover, products of Schur polynomials are Schur-positive by the Littlewood--Richardson rule, so if $\kappa'$ denotes a rectangle of dimensions $m \times (n-m - (m-1))$ then $$\langle s_\tau^k s_{(\eps m)}, s_\kappa\rangle \ge \langle s_\mu s_\tau^{k-2} s_{(\eps m)}, s_\kappa \rangle = \langle s_\tau^{k-2} s_{(\eps m)}, s_{\kappa / \mu} \rangle = \langle s_{\tau}^{k-2} s_{(\eps m)}, s_{\kappa'}\rangle.$$ Therefore the theorem is true for $k$ whenever it is true for $k-2$. By induction we are done if $k$ is even, and if $k$ is odd we are reduced to the case $k = 3$.

Now assume $k = 3$. Then $\kappa$ is an $m \times (3(m-1)/2 + \eps)$ rectangle. Let $\lambda$ be the partition obtained by rotating the skew diagram $\kappa / \tau$ by $180$ degrees. Then $$\langle s_\tau^3 s_{(\eps m)}, s_{\kappa}\rangle = \langle s_{\tau}^2 s_{(\eps m)}, s_\lambda\rangle.$$ First assume $m$ is odd, so that $\eps = 0$. Then by the Littlewood--Richardson rule we just need to check that there is at least one Littlewood--Richardson tableau of skew shape $\lambda / \tau$ (a parallelogram) and weight $\tau$. For example, the positive answer to the question in the title follows from the existence of the following Littlewood--Richardson diagram: $$\begin{aligned} &**1\\ &*1\\ &2 \end{aligned}$$

It turns out that if you just do this by hand for the first several cases (i.e., small $m$), always making a lexically minimal choice, then a pattern emerges, and it becomes clear that a solution exists for all $m$. Rather than try to describe the pattern, let me just illustrate the case $m = 9$ with suggestive colouring:

$$\begin{aligned} &********\color{blue}{1111}\\ &*******\color{red}{1}\color{blue}{222}\\ &******\color{red}{12}\color{blue}{33}\\ &*****\color{red}{123}\color{blue}{4}\\ &****\color{green}{1234}\\ &***\color{green}{2345}\\ &**\color{green}{3456}\\ &*\color{green}{4567}\\ &\color{green}{5678}\\ \end{aligned}$$

(If somebody is good with mathjax alignment, please feel free to edit!)

The case in which $m$ is even and $\epsilon = 1/2$ is similar, but with the height of the parallelogram reduced by $1$. Here is the case $m = 8$:

$$\begin{aligned} &*******\color{blue}{1111}\\ &******\color{red}{1}\color{blue}{222}\\ &*****\color{red}{12}\color{blue}{33}\\ &****\color{red}{123}\color{blue}{4}\\ &***\color{green}{2345}\\ &**\color{green}{3456}\\ &*\color{green}{4567}\\ &\color{green}{8888}\\ \end{aligned}$$

Here the 8's come from the slab given by $s_{(\eps m)}$.

I imagine/hope there is some cleaner version of this argument using well-known rules for skew Schur functions and Littlewood--Richardson diagrams, but this will suffice for now.

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