8
$\begingroup$

Let $A_{n}(\mathbb{Q}) $ denote the $n$ times $n$ skew symmetric matrices over the rational number field. Let $N$ be a subspace of $A_{n}(\mathbb{Q}) $.

  1. If all the non-zero matrices in $N$ are invertible, what is the maximum the dimension of $N$ can be? You can assume $ n $ is even.

  2. Does there exist a subspace $M$ of $A_{n}(\mathbb{Q}) $ of dim $ n-1 $ with all the non-zero matrices in $M$ are invertible? You can assume $n$ is even.

Note that if $A_{n}(\mathbb{R}) $ denotes the $n$-times-$n$ skew-symmetric matrices over the real number field, then, for $ n= 4 $ and $ n = 8 $, the answer of the second question is 'yes', but, for $ n= 6 $, there is no such subspace.

$\endgroup$

1 Answer 1

7
$\begingroup$

For a field $\mathbb{F}$, let $\mu_\mathbb{F}(n)$ denote the maximal dimension of a subspace $N\subset A_n(\mathbb{F})$ such that all the nonzero elements of $N$ are invertible. For simplicity, I will assume that the characteristic of $\mathbb{F}$ is not $2$.

Then, because $\det(a) = (-1)^n\det(-a) = (-1)^n\det(a^\mathsf{T})= (-1)^n\det(a)$, we have $\mu_\mathbb{F}(2m{+}1) = 0$.

Meanwhile, clearly, $\mu_\mathbb{F}(2m)\ge 1$, and, as the OP points out, $\mu_\mathbb{R}(4m)\ge 3$ and $\mu_\mathbb{R}(8m)\ge7$ due to the existence of normed division algebras $\mathbb{H}$, of dimension $4$ over $\mathbb{R}$, and $\mathbb{O}$, of dimension $8$ over $\mathbb{R}$.

When $n=2m$, the polynomial function $\det:A_n(\mathbb{F})\to\mathbb{F}$ is the square of a polyomial $\mathrm{Pf}:A_n(\mathbb{F})\to\mathbb{F}$ homogeneous of degree $m=n/2$, unique up to a choice of sign. In fact, $\mathrm{Pf}$ is defined over the integers, $\mathrm{Pf}:A_n(\mathbb{Z})\to\mathbb{Z}$, as a polynomial with integer coefficients with the property that $\mathrm{Pf}(mam^\mathsf{T}) = \det(m)\,\mathrm{Pf}(a)$ for $a\in A_{2m}(\mathbb{Z})$ and $m\in M_{2m}(Z)$. Consequently, this property holds with $\mathbb{Z}$ replaced by $\mathbb{F}$ for any field $\mathbb{F}$.

It follows that $\mu_\mathbb{R}(4m+2)=1$, since, in this case, $\mathrm{Pf}$ is a polynomial of odd degree, implying that, for any pair $a,b\in A_{4m+2}(\mathbb{R})$, the homogeneous polynomial $p(s,t)=\mathrm{Pf}(sa+tb)$ of odd degree $2m{+}1$ will vanish for some real ratio $[s:t]$.

If $\mathbb{F}$ is an ordered field (more generally, if every nontrivial sum of squares in $\mathbb{F}$ is nonzero), then the standard Clifford algebra construction (using a definite quadratic form) shows that $\mu_\mathbb{F}(n)\ge\rho(n){-}1$, where $\rho(n)$ is the Radon-Hurwitz number. In particular, $\mu_\mathbb{Q}(n)\ge\rho(n){-}1$. Meanwhile, J. F. Adams has shown that $\mu_\mathbb{R}(n)=\rho(n){-}1$. Thus, $\mu_\mathbb{Q}(n)\ge\mu_\mathbb{R}(n)$, but, in general, equality does not hold.

Claim: $\quad 2m{-}1\ge\mu_\mathbb{Q}(2m)\ge m$.

In particular, $\mu_\mathbb{Q}(6)\ge 3 > \mu_\mathbb{R}(6) = 1$, thus verifying that $\mu_\mathbb{Q}(2m)$ can be strictly greater than $\mu_\mathbb{R}(2m)$.

The claim follows from the fact that the characteristic polynomial of a generic element $a\in A_{2m}(\mathbb{Q})$ is irreducible over $\mathbb{Q}$. For, when the characteristic polynomial of $a$ is irreducible over $\mathbb{Q}$, then $I, a, a^2,\ldots, a^{2m-1}$ spans a field $\mathbb{Q}(a)\subset M_{2m}(\mathbb{Q})$, and hence every nonzero linear combination of these matrices is invertible. Meanwhile, $N(a) = \mathbb{Q}(a)\cap A_{2m}(\mathbb{Q})$ is a vector space with basis $a, a^3, \ldots a^{2m-1}$ and hence has dimension $m$. Thus, $\mu_\mathbb{Q}(2m)\ge m$.

The upper bound follows from the fact that any subspace $N\subset A_{2m}(\mathbb{Q})$ of dimension greater than $2m{-}1$ must intersect nontrivially with the subspace of matrices with the first column equal to zero, since that subspace has codimension $2m{-}1$.

Remark 1: It seems likely that the 'generic' $m$-dimensional subspace of $A_{2m}(Q)$ has all of its nonzero elements invertible, but, perhaps this depends on some carefully defined notion of 'generic'.

Remark 2: Since $\mu_\mathbb{Q}(n)\ge \mu_\mathbb{R}(n)$, the lower bound in the Claim cannot always be strengthened to equality. For example, $\mu_\mathbb{Q}(4)\ge \mu_\mathbb{R}(4) = 3 > 2$. Thus, $\mu_\mathbb{Q}(4)=3$. Similarly, since $\mu_\mathbb{R}(8)=7$, we have $\mu_\mathbb{Q}(8)=7$. (This answers Question 1 for $n=4$ and $n=8$.)

Note that the OP's Question 2 asks whether $\mu_\mathbb{Q}(2m)\ge 2m{-}1$, presumably provoked by the observation that $\mu_\mathbb{R}(2m) = 2m{-}1$, when $m=2$ and $m=4$. However, these low dimensions can be very misleading. For all other values of $m$, we have $\mu_\mathbb{R}(2m) < 2m{-}1$, and, in fact, for all but a finite set of values of $m$, we have $\mu_\mathbb{R}(2m) < m$, and in general, as $m$ grows, the lim inf of $\mu_\mathbb{R}(2m)/m$ equals $0$. On the other hand, $\mu_\mathbb{Q}(2m)/m\ge 1$ for all $m$.

Remark 3: I'm including this last remark at the request of the OP, but, not being a number theorist, I do not have any realy confidence that this can be turned into a rigorous argument.

I do not know whether $\mu_\mathbb{Q}(6)>3$, however, a very heuristic speculation leads me to suspect that this is true and that it might even be true that $\mu_\mathbb{Q}(6)=5$.

The Grassmannian $G_4(15)$ of $4$-dimensional subspaces of $A_\mathbb{Q}(6)$ is a rational variety of dimension $4\cdot (15-4) = 44$. Meanwhile, the group $\mathrm{SL}(6,\mathbb{Q})$ has dimension $35$ and it acts on $A_\mathbb{Q}(6)$ via $m\cdot a = mam^\mathsf{T}$ preserving $\mathrm{Pf}:A_\mathbb{Q}(6)\to\mathbb{Q}$. The induced action of $\mathrm{SL}(6,\mathbb{Q})$ on $G_4(15)$ has generic orbits of dimension $35$, so the 'moduli space' $\mathscr{M}$ of orbits has formal dimension $44-35 = 9$. Meanwhile, the restriction of $\mathrm{Pf}$ to a subspace $N\subset A_\mathbb{Q}(6)$ is a rational cubic form on $N$, generically nondegenerate. The moduli of cubic forms of rank $4$ under $\mathrm{GL}(4,\mathbb{Q})$ equivalence has formal dimension $20 - 16 = 4 < 9$, and it is known that there are rational cubic forms of rank 4 that do not represent $0$ rationally. It seems that the map assigning to a generic $4$-plane $N\subset A_6(\mathbb{Q})$ the rational cubic form $\mathrm{Pf}:N\to\mathbb{Q}$ is 'dominant'. For this reason, it seems likely to me that a 'generic' $4$-plane $N\subset A_6(\mathbb{Q})$ will have the property that $\mathrm{Pf}:N\to\mathbb{Q}$ will not represent $0$ rationally (and hence the nonzero elements of $N$ would all be invertible).

However, it's not that easy to determine whether a given rational cubic forms of rank $4$ represents $0$ rationally, so just choosing a $4$-plane $N$ 'at random' and testing whether its Pfaffian represents $0$ seems to be a very labor intensive way to try to find an example.

All of the above is very speculative, but one could go on to make a similar argument for $5$-planes in $A_6(\mathbb{Q})$. There, it seems even harder to test for when a given rational cubic form of rank $5$ represents $0$ rationally, though.

$\endgroup$
14
  • $\begingroup$ dear @Robert can you explain why $\mu_{\mathbb{Q} }(n) \geq \mu_{\mathbb{R}} (n) $? can you give some example that $\mu_{\mathbb{Q} }(6)= 5 $ or prove that $\mu_{\mathbb{Q} }(6)= 4 $. $\endgroup$
    – Sky
    May 19, 2021 at 20:52
  • 1
    $\begingroup$ @SugataMandal: We know that $\mu_\mathbb{R}(n)=\rho(n)$ is achieved in every case by a Clifford algebra construction, and the coefficients of the Clifford algebra are rational (in fact, integral) in a suitable basis. Thus, $\mu_\mathbb{Q}(n)\ge \mu_\mathbb{R}(n)$. As I wrote above, I do not know that $\mu_\mathbb{Q}(6)>3$, so I certainly don't have an example of a $4$- or $5$-dimensional subspace $N\subset A_6(\mathbb{Q})$ such that all of its non-zero elements are invertible. $\endgroup$ May 19, 2021 at 23:35
  • $\begingroup$ dear @Robert I think $\mu_{\mathbb{R}} (n)= \rho(n) - 1 $ if your notation $ \rho(n) $ denotes Hurwitzs number . As $ \rho(4) = 4 $ ,$ \rho(8) = 8$ but $\mu_{\mathbb{R}} (4) =3 $, $\mu_{\mathbb{R}} (n) =7 $ . I thnik the quaternions and octonions valied for $ \mathbb{Q} $ also, I cant contruct this type of exmples for other values of $n$ . $\endgroup$
    – Sky
    May 20, 2021 at 20:29
  • $\begingroup$ @SugataMandal: Oh, yes, you are right about the definition of $\rho(n)$. I should have used $\rho(n)-1$ where I wrote $\rho(n)$ above. It's not surprising that you can't find other values of $n$ for which $\rho(n) = n-1$ because this only happens for $n=1,2,4,8$, by a famous theorem of Adams. $\endgroup$ May 21, 2021 at 1:01
  • $\begingroup$ Yes @Robert , But I think Adams proved that the result for $ \mathbb{R} $ , I think he used K theory , but it may not be true for the $ \mathbb{Q} $ but we can say that the result is true for $ n=4 $ , $ n= 8 $ , quaternions and octonions valied for ℚ. For the other case I cant find any solid argument.. $\endgroup$
    – Sky
    May 21, 2021 at 6:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.