8
$\begingroup$

Edit:
In all likelihood, the original question does not have a positive answer (see comment by abx).

Modified question: Let $\rho_H(n)$ be the maximal dimension of a space of symmetric real matrices contained in $\mathrm{GL}_n(\mathbb{R})\cup\{0\}$. Are there any upper bounds on $\rho_H(n)$, in terms of $\rho_H(m)$ and $\rho(m)$, that can be derived by elementary methods?

For example, in On Matrices Whose Real Linear Combinations are Nonsingular (Proc. Amer. Math. Soc., 16(2), 1965), Adams, Lax and Phillips prove that $\rho_H(n) \le \rho(8n)-7$.
(This, together with $\rho(16n)\le \rho(n)+8$ implies $\rho_H(n)\le \rho(n/2)+1$, which was the original motivation for my question.)

Original post:
For the purpose of this question, let us define the Radon-Hurwitz number $\rho(n)$ to be the maximal dimension of a subspace $W$ of the the real vector space $\mathbb{R}^{n\times n}$ of $n\times n$ matrices, such that $W\subset\mathrm{GL}_n(\mathbb{R})\cup \{0\}$.

Question: Is there an elementary proof of the inequality $\rho(16n) \le \rho(n) + 8$?

Comments:

  1. By "elementary" proof I mean one that does not rely, directly or indirectly, on $K$-theory.
  2. From the general formula for $\rho(n)$ it is plain that $\rho(16n) = \rho(n) + 8$. My question is if the inequality can be derived without using the formula, and ideally only using methods from linear algebra and/or elementary topology.
  3. Other, similar inequalities could be of interest too, if derived by elementary methods.
$\endgroup$
2
  • $\begingroup$ This inequality implies that there is no division algebra over $\mathbb{R}$ of dimension $2^n$ with $n\geq 4$. I don't think that there exists an ''elementary" (in your sense) proof of that fact. $\endgroup$
    – abx
    Jan 15, 2021 at 9:05
  • $\begingroup$ @abx Indeed, you are probably right about that. $\endgroup$
    – Erik D
    Jan 15, 2021 at 10:25

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy