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Suppose that $S$ is the set of all real $n\times n$ matrices $A$ such that $A^2=I_n$. Since these matrices are diagonalizable with $\pm 1$ as eigenvalues, we get the partition $S=\cup_{0\leq k \leq n} S_k$ where $S_k$ is the set of all matrices of the form $PJ_k P^{-1}$ where $P$ is a real invertible matrix and $J_k$ the diagonal $(1,\cdots,1,-1,\cdots,-1)$ (the first $k$ components are equal to $1$, and the other $n-k$ components are equal to $-1$).

My question is : are the $S_k$ the connected components of $S$ ? (and if yes, why ?) If not, what are the connected components of $S$?

whpowell96
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Phil-W
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1 Answers1

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I think they are.

Firstly, the trace function is a continuous map from $S$ to $\Bbb Z$, and $S_k$ is the preimage of $\{2k - n\}$, which is clopen in $\Bbb Z$. So every connected component is certainly contained in $S_k$ for some $k$.

Your remark shows that the function $f: \mathrm{GL}_n(\Bbb R) \to S_k$ defined by $f(P) = P J_k P^{-1}$ is surjective. $f$ is continuous, so this almost shows that $S_k$ is connected, but not quite - the problem is that $\mathrm{GL}_n(\Bbb R)$ has two connected components.

We can fix this by restricting $f$ to the set of matrices of positive determinant. This map remains surjective, for example because $J_k$ always has a matrix of negative determinant in its centraliser in $\mathrm{GL}_n(\Bbb R)$. Explicitly: let $Q$ be the diagonal matrix with diagonal entries $(-1, 1, 1, \dotsc, 1)$. Then for all $P$, we have $PQ J_k (PQ)^{-1} = P (Q J_k Q) P^{-1} = P J_k P^{-1}$. So if $P$ had negative determinant, just replace it by $PQ$, which has positive determinant.

So $f$ is still surjective. But the set of matrices of positive determinant is connected, and hence $S_k$ is connected. So we are done.

Izaak van Dongen
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