# A conjugacy class under O(n), orthogonal projection

This is not really a homework question per se but I wasn't sure where else to put it:

In a script I'm reading the following set is defined:

$P(n)_k := \{p \in S(n) | p^2 = p, \text{trace } p = k\}$

(i.e. the set of all real orthogonal projection matrices with trace k).

Now the following statement is made:

"$P(n)_k$ is a submanifold of the affine space $S(n)_k$ since it is the conjugacy class of the matrix $p_0 = \begin{pmatrix}I_k && 0 \\ 0 && 0\\ \end{pmatrix}$, i.e. the orbit of $p_0$ under the action of the group O(n) on S(n) by conjugation."

($S(n)_k$ is the set of all real symmetric matrices with trace k.)

I don't understand the second part of this statement. The conjugacy class should be:

$\{Ap_0A^{-1} | A \in O(n)\} = \{Ap_0A^{t} | A \in O(n)\} =$

$\{\begin{pmatrix}a_{11}^2 && 0 && ... && 0 && ... && 0 \\ 0 && a_{22}^2 && ... && 0 && ... \\ 0 && 0 && ... && a_{kk}^2 && ... && 0 \\ 0 && 0 && 0 && 0 && ... && 0\end{pmatrix} | (a_{ij}) \in O(n)\}$

and I don't see why this equals $P(n)_k$. (Or maybe my calculation is wrong.)

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I'm lost at how you found

$Ap_0A^{t}=\begin{pmatrix}a_{11}^2 && 0 && ... && 0 && ... && 0 \\ 0 && a_{22}^2 && ... && 0 && ... \\ 0 && 0 && ... && a_{kk}^2 && ... && 0 \\ 0 && 0 && 0 && 0 && ... && 0\end{pmatrix}$

That doesn't seem correct at all.

You're right. I don't know what happened there. I'll take a look at this again.

Some keywords which might help: "diagonalization of symmetric matrices"

Hey! First of all, thank you for your help! I didn't get back to this in the last couple of days but I will take a closer look tomorrow.

Ok, I figured this out. Thank you.

Just one more thing: I can't really follow why this makes the conjugacy class a submanifold. I know that certain kinds of orbit spaces are manifolds but this is not true for every group action, and I couldn't find anything on conjugacy classes always being submanifolds.