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Let $n\geqslant 5$ and let $E_4(n)$ be a linear subspace of $(n\times n)$- real skew-symmetric matrices such that $$ rank(A)=4,\text{ for all }A\in E_4(n),A\neq 0. $$ I'm curious about the following question:

QUESTION:

What can be said about the dimension of $E_4(n)$? Of course, it is easy to check that $\operatorname*{dim}E_4(n)\leqslant \binom{n-1}{2}$. But what is the best possible value? I'm specially interested in the case $n=5,6,7$.

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As you probably already know, there's a lot of literature on questions like this. One starting point might be the paper "On symmetric degeneracy loci, spaces of symmetric matrices of constant rank and dual varieties" by B. Ilic and J.M. Landsberg. (In spite of the title, they do say something about skew-symmetric matrices, albeit over the complex numbers.) But the bibliography referes to a number of papers treating related questions. –  Artie Prendergast-Smith Feb 12 '13 at 17:24
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3 Answers

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Partial progress: It's easy to achieve $n-3$. Consider matrices of the form

$$\begin{pmatrix} 0 & 0 & r_1 & r_2 & \cdots & r_{n-3} & 0 \\ 0 & 0 & 0 & r_1 & \cdots & r_{n-4} & r_{n-3} \\ r_1 & 0 & & & & & \\ r_2 & r_1 & & & & & \\ \vdots & \vdots & & & {\LARGE 0}& & \\ r_{n-3} & r_{n-4} & & & & & \\ 0 & r_{n-3} & & & & & \\ \end{pmatrix}$$ where the bottom right square is entirely $0$. If this has rank $<4$, then the upper-left $4 \times 4$-submatrix implies $r_1^4=0$, so $r_1=0$. Then inductively $r_2^4=0$, and so forth.

For $n=5$, this only gives a $2$ dimensional subspace, and I argued in the comments on my other answer that a generic $3$ dimensional subspace of the $5 \times 5$ matrices should work. Right now, though, I can't see how to do better for $n \geq 5$.

Ah, slight improvement. For $n$ even, and taking advantage of the fact that I'm working over the reals, I can do $n-2$: $$\begin{pmatrix} 0 & 0 & a_1 & b_1 & \cdots & a_{(n-2)/2} & b_{(n-2)/2} \\ 0 & 0 & -b_1 & a_1 & \cdots & -b_{(n-2)/2} & a_{(n-2)/2} \\ a_1 & -b_1 & & & & & \\ b_1 & a_1 & & & & & \\ \vdots & \vdots & & & {\LARGE 0}& & \\ a_{(n-2)/2} & -b_{(n-2)/2} & & & & & \\ b_{(n-2)/2} & a_{(n-2)/2} & & & & & \\ \end{pmatrix}$$ If this has rank $<4$, then $(a_i^2+b_i^2)^2=0$ for all $i$, so over the reals this can only happen when it is $0$.


A potential strategy: What are (up to conjugation) the maximal subspaces of $n \times n$ skew-symmetric matrices on which the rank is always $\leq 4$? Then we can focus our efforts on finding large subspaces on each of these which miss the rank $2$ locus.

So far, I have only been able to find four maximal subspaces. I'll describe them all as block matrices with the size and nature of the blocks indicated: $$\begin{pmatrix} 5 \times 5 & 0 \\ 0 & 0 \end{pmatrix}$$ $$\begin{pmatrix} 0 & 2 \times (n-2) \\ (n-2) \times 2 & 0 \end{pmatrix}$$ $$\begin{pmatrix} 0 & 3 \times 3,\ \mbox{skew symmetric} & 0 \\ 3 \times 3,\ \mbox{skew symmetric} & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}$$ $$\begin{pmatrix} 3 \times 3 & 0 & 0 \\ 0 & 3 \times 3 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}$$

If these are the only options, then the only one which grows with $n$ is the second case so that's the one we need to concentrate on.

In that case, the question is equivalent to "what is the largest linear subspace of the $2 \times (n-2)$ matrices which includes no rank $1$ submatrices?" I can show that the answer to that question is $n-3$, over $\mathbb{C}$, and is $2 \lfloor (n-2)/2 \rfloor$ over $\mathbb{R}$; I'll post the argument if anyone cares.

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Let $J_4(n)$ be the $n \times n$ matrix $$\begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 & \cdots \\ -1 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 1 & 0 & 0 & \cdots \\ 0 & 0 & -1 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{pmatrix}$$ where all hidden entries are $0$. Then $E_4(n)$ is the orbit of $J_4(n)$ under $GL_n$ acting by $g : X \mapsto g X g^T$. (See any linear algebra text that works out the classification of skew symmetric forms.)

The stabilizer of $J_4(n)$ is $$\begin{pmatrix} S & T \\ 0 & U \end{pmatrix}$$ where $S$ is in $Sp(4)$, $T$ is an arbitrary $4 \times (n-4)$ matrix and $U$ is an invertible $(n-4) \times (n-4)$ matrix.

The dimension of $GL_n$ is $n^2$. The dimension of $Sp(4)$ is $10$. So the dimension of the orbit is $$n^2 - \left( 10 + 4(n-4) + (n-4)^2 \right) = 4n-10 .$$

Sanity check: A generic $4 \times 4$ skew symmetric matrix is invertible, so $E_4(4)$ is $\binom{4}{2} = 6$ dimensional. A $5 \times 5$ skew symmetric matrix is not invertible, since odd size skew symmetric matrices are always singular; generically there is no reason for it not to have rank $4$. So $E_4(4)$ is $\binom{5}{2} = 10$ dimensional. A $6 \times 6$ skew symmetric matrix has rank $6$ if and only if its Pfaffian does not vanish; if the Pfaffian does vanish than it has rank $\leq 4$ and generically has rank $4$. So $E_4(6)$ is an open dense locus inside a hypersurface in $\mathbb{R}^{15}$, and we conclude that $\dim E_4(6) = 14$.

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I do not understand one point. You mentioned that $E_4(5)=10$. Let us take $S:=e_1\wedge\mathbb{R}^5$. Then, S is a subspace of $5\times 5$ skew-symmetric matrices that has trivial intersection with $E_4(5)$. This implies that dimension of $E_4(5)$ has to be less than or equal to 10-4=6. –  tatin Feb 12 '13 at 16:32
    
If I understand correctly, the OP is asking about linear subspaces of matrices of fixed rank. Your answer gives the dimension of the locus of such matrices, but that doesn't seem to answer the question. Am I missing something? –  Artie Prendergast-Smith Feb 12 '13 at 16:32
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For $n=5$, the answer is $3$. I claim that a generic $3$ dimensional subspace of the $5 \times 5$ symmetric matrices misses the rank $2$ locus, and no $4$ dimensional subspace does. Proof: Let's work projectively. We want to find a linear subspace of $\mathbb{P}^{10-1}$ which misses the projectivization of the rank $2$ locus. That projectivization of the rank $2$ locus is the Grassmannian $G(2,5)$, with dimension $6$, and forms a degree $5$ subvariety of $\mathbb{P}^{10-1}$. So any $\mathbb{P}^3$ hits this subvariety, generically doing so in $5$ points, but a generic $\mathbb{P}^2$ misses it. –  David Speyer Feb 12 '13 at 16:58
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David: that argument definitely works for complex matrices (and can be generalised), but could the real case be more complicated? (One could imagine the P^3 intersecting only in non-real points.) Maybe there is a way to get around this, though. –  Artie Prendergast-Smith Feb 12 '13 at 17:08
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Conveniently, $5$ is odd, so at least one of the intersection points is real. (More specifically, this immediately gives it for a transverse intersection, and the condition of having a real intersection point is closed.) Definitely a good thing to point out, though. –  David Speyer Feb 12 '13 at 17:22
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Denote by $\DeclareMathOperator{\uso}{\underline{so}}$ $\uso(n)$ the space of symmetric $n\times n$ matrices. $\DeclareMathOperator{\rank}{rank}$ Note that for $A\in \uso(n)$

$$ \rank A=n-\dim\ker A, $$

$$\dim \ker A \equiv n\bmod 2. $$

Denote by $ \uso(n)_k $ the space consisting of matrices $X\in\uso(n)$ such that $\dim \ker X= k$, where $k\equiv n\bmod 2$.

You are interested in the space $\uso(n)_{n-4}$.

We have a diffeo $\uso(n)_k\to\uso(n-k)_0$ which associates to a matrix $X\in\uso(n)_k$ its restriction to the orthogonal complement of the kernel. We deduce

$$\dim \uso(n)_k=\dim \uso(n-k)_0=\dim \uso(n-k)=\binom{n-k}{2}. $$

Thus, in your case

$$\dim \uso(n)_{n-4}=\binom{4}{2}= 6. $$

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