Consider a skew-symmetric $(4n+2) \times (4n+2)$ block-diagonal real matrix $A$ in normal form: $$A = \begin{bmatrix} \Lambda_1 & 0 & \cdots & 0\\ 0& \Lambda_2 & \cdots & 0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\Lambda_{2n+1} \end{bmatrix},$$ where the diagonal $\Lambda_j$ are $2\times 2$ blocks $$\Lambda_j = \begin{bmatrix} 0 & -\lambda_j\\ \lambda_j & 0 \end{bmatrix}.$$ We assume that the $\lambda_j$ are nonzero integers, not necessarily the same (or different). I would like to prove that the elements of the orthogonal group $\mathrm{O}(4n+2)$ that commute with such a matrix in fact all lie within the special orthogonal component $\mathrm{SO}(4n+2)$. If that's not the case, I'd like a counterexample.

I feel there should be a relatively simple proof that a matrix commuting with $A$ can't have negative determinant, but I am persistently missing it.

P.S.: In case this problem seems unnatural, the real motivation is to understand the number of components of the normalizer in $\mathrm{SO}(4n+2)$ of a "generic" circle subgroup that does not lie in any $\mathrm{SO}(4n)$ subgroup. This number of components will be one if the elements of $\mathrm{O}(4n+2)$ centralizing my matrix $A$ are all special orthogonal, and two otherwise.


1 Answer 1


I've found two proofs of this result now. First, the geometric one.

Conjugate $A$ by permutation matrices to arrange the the $\Lambda_j$ in order of increasing absolute value of $\lambda_j$, and within each absolute value, by regular value. (So, e.g., $-1 < 1 \leq 1 < 2 < -5 \leq -5 < 5 \leq 5$, for example.)

Then if a matrix $B$ commutes with $A$, it is block-diagonal in such a way that one block corresponds to each absolute value. For example, the potentially nonzero blocks in a $B$ corresponding to the example order of $\lambda_j$ above are of side lengths $6$ (for $\lambda = 1$), $2$ (for $\lambda = 2$) and $8$ (for $\lambda = 5$).

Thus we only need to find those matrices commuting with a block-diagonal

$$A_j = \Lambda_{-j} \oplus \cdots \oplus \Lambda_{-j} \oplus \Lambda_{j} \oplus \cdots \oplus \Lambda_{j}.$$

Say a $2 \times 2$ matrix is of type $C$ if it is an $\mathbb R$-linear combination of $I$ and $\Lambda_1$. Matrices of type $C$ commute with the $\Lambda_j$. Say a $2 \times 2$ matrix is of type $J$ if it is $\begin{bmatrix}0 & 1\\1&0\end{bmatrix}$ times a matrix of type $C$; these anticommute with the $\Lambda_j$.

A matrix $B$ commuting with $A_j$ has a decomposition into $2 \times 2$ blocks of types $C$ and $J$. For example, the matrices commuting with $A_7 = \Lambda_{-7} \oplus \Lambda_7 \oplus \Lambda_7$ decompose as

$$B_7 = \begin{bmatrix} C&J&J\\ J&C&C\\ J&C&C \end{bmatrix}; $$

here $C$ and $J$ represent types of blocks; they aren't all the same two matrices.

Now if we demand $B_7 \in \mathrm{O}(6)$, then it follows that all rows are in $S^5 \subset \mathbb{R}^6$ and mutually orthogonal. Note that the first row uniquely determines the second, the third the fourth, and the fifth the sixth, because of the symmetry in matrices of types $C$ and $J$. Given a first row, the third rows mutually orthogonal to the first and second rows form a space of homeomorphism type $S^3 \subset \mathbb{R}^4$, and given a first and a third row, the space of fifth rows orthogonal to the first through fourth rows form a space of homeomorphism type $S^1 \subset \mathbb R^2$.

It follows that the centralizer $Z = Z_{\mathrm{O}(6)}(A_7)$ has an iterated fiber bundle structure $\require{AMScd}$ \begin{CD} S^1 @>>>Y @>>> Z\\ @. @VVV @VVV\\ @. S^3 @. S^5. \end{CD}

Since a bundle of connected fibers over a connected base has connected total space, $Z$ is connected. But there was nothing special about $\lambda = 7$ or the number (three) of $\pm 7$'s in $A_7$ in this argument.

As each space of blocks centralizing a block $A_j$ is connected, so is their product, which is the centralizer of $A$ in $\mathrm{O}(4n+2)$. As the identity matrix centralizes $A$, it follows the entire centralizer lies in the identity component $\mathrm{SO}(4n+2)$.


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