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$\newcommand{\GLp}{\operatorname{GL}_n^+}$ $\newcommand{\SO}{\operatorname{SO}_n}$

Let $n>2$, and Let $A \in \GLp$ be an invertible real $n \times n$ matrix, which commutes with $\SO$.

Is it true that $A= \lambda Id$ for some $\lambda \in \mathbb{R}$ ?

An equivalent requirement is that $A$ commutes with every skew-symmetric matrix.

One direction is obtained by differentiating a path of orthogonal matrices starting at the identity. The converse implication comes from the fact that every element of $\SO$ equals to $\exp(M)$ for some skew-symmetric $M$.

Note that if we assume that $A \in \SO$, then the answer is positive: we must have $A=\pm Id$ .

Asaf Shachar
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  • If you write it instead as $Q = A^{-1}B Q B^{-1} A = A^{-1} B Q ( A^{-1} B)^{-1}$. Then denote $C = A^{-1} B$ and then we have the system $C Q C^{-1} = Q$. So it seems to be true that if the matrix $A^{-1}B$ is commutative things work out. So perhaps instead of saying $B = \pm Id$, we must have $A^{-1}B = \pm id \implies B = \pm A$. Oh I think this is the answer – TrostAft Dec 20 '18 at 20:39
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    You could have $B=kA$ for any nonzero $k$. – Angina Seng Dec 20 '18 at 20:42
  • @LordSharktheUnknown Oh yes, I was too hasty in concluding $A^{-1} B$ commutative implies $A^{-1}B = \pm Id$. – TrostAft Dec 20 '18 at 20:44
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    When $n=2$, $SO_n$ consists of the $\cos t I+\sin t J$ for $J=\pmatrix{0&1\-1&0}$. Then the centraliser of all these matrices consists of $\Bbb RI +\Bbb RJ$. – Angina Seng Dec 20 '18 at 20:46
  • Simultaneously diagonalizable matrices are also commutative, must it be true that if $A, B$ simultaneously diagonalizable, then $A = \lambda B, \lambda \in \mathbb{R}$? I don't think so right? – TrostAft Dec 20 '18 at 20:47
  • @LordSharktheUnknown You are right; I forgot that for $n=2$ $SO \cong S^1$ is commutative. I will add $n>2$. Thank you for your useful comments. – Asaf Shachar Dec 20 '18 at 20:50
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    As every element of $SO_n$ is $\exp(M)$ for a skew-symmetric $M$, then $C$ commutes with all of $SO_n$ iff $C$ commutes with all skew-symmetric matrices. – Angina Seng Dec 20 '18 at 20:55
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    The proof sketch in the linked question applies to this case as well – Dap Dec 21 '18 at 14:30

1 Answers1

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This is a representation theory question: slightly generalized (there's no need to restrict our attention to $GL_n^{+}$), you're asking what the endomorphisms of $\mathbb{R}^n$ as a representation of the Lie group $SO(n)$ (or, equivalently, the Lie algebra $\mathfrak{so}(n)$) are.

This representation is always irreducible, so by Schur's lemma the endomorphisms form a division algebra over $\mathbb{R}$, which by the Frobenius theorem must be $\mathbb{R}, \mathbb{C}$, or $\mathbb{H}$. The latter two cases can't happen if $n$ is odd (because $\mathbb{C}$ and $\mathbb{H}$ only act on $\mathbb{R}^n$ when $n$ is divisible by $2$ or $4$ respectively).

If $n = 2k \ge 4$ is even we can argue as follows: if the endomorphism ring contains $\mathbb{C}$, then $SO(2k)$ must embed into $GL_k(\mathbb{C})$ and hence into the unitary group $U(k)$, by compactness, and similarly on the level of Lie algebras. But this is impossible by a dimension count: $SO(2k)$ has dimension $k(2k-1)$, but $U(k)$ has dimension $k^2$, and for $k \ge 2$ we have $2k-1 > k$. (For $k = 1$ they are equal, reflecting the coincidence $SO(2) = U(1)$.) So the endomorphism ring must be $\mathbb{R}$. Probably a simpler argument is possible here.

Qiaochu Yuan
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  • Hi, I know it has been some time ago, but coming to this question again, I see that I don't understand two things here: (1) Why the fact the endomorphism ring contains $\mathbb C$ implies that $\text{SO}(2k)$ embeds in $\text{GL}_k(\mathbb C)$? Can you describe the embedding more explicitly (I guess in terms of an endomorphism $J:\mathbb R^n \to \mathbb R^n$ whose square is $-1$)? (2) Why can $\mathbb H$ only act on $\mathbb R^n$ when $n$ is divisible by $4$? (If I understand correctly, the reason why $\mathbb C$ can only act on $\mathbb R^n$ for even $n$ is that... – Asaf Shachar Jan 30 '19 at 08:42
  • if you have a $J \in \text{GL}(\mathbb R^n)$ , $J^2=-1$ you can take determinants and see what happens. What is the argument for the quaternionic case? I appreciate your help. – Asaf Shachar Jan 30 '19 at 08:43
  • @AsafShachar Concerning your second question: both $\mathbb C$ and $\mathbb H$ are division algebras; if a division algebra $A$ over $\mathbb R$ acts on an $\mathbb R$-vector space $V$, $V$ in turn is an $A$-vector space. If $V$ has dimension $n$ over $A$, it has dimension $n \cdot \dim A$ over $\mathbb R$. – lisyarus Feb 12 '19 at 12:34