# Which matrices commute with $\operatorname{SO}_n$?


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 $$A=\pm Id$$ .

• 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
• You could have $B=kA$ for any nonzero $k$. – Lord Shark the Unknown Dec 20 '18 at 20:42
• 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 $\Bbb RI +\Bbb RJ$. – Lord Shark the Unknown Dec 20 '18 at 20:46
• 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. – Lord Shark the Unknown Dec 20 '18 at 20:55
• The proof sketch in the linked question applies to this case as well – Dap Dec 21 '18 at 14:30

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.
• 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 at 8:42
• 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 at 8:43
• @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 at 12:34