Is there a 'natural' projection from $O(n)$ into $S_n$?

Question

Is there an easily definable projection $F$ from the orthogonal group $O(n)$ into the permutation group $S_n$, which has the following properties?

1. $F(P_\sigma) = \sigma$ for all $\sigma \in S_n$

2. $F^{-1}(\sigma)$ contains an open subset of $O(n)$ around $P_\sigma$

3. $\mu(F^{-1}(\sigma)) = \mu(F^{-1}(\varphi))$ for all $\sigma, \varphi \in S_n$, where $\mu$ is the Haar measure on $O(n)$

4. Optional: For any diagonal matrix $A$ the matrices $HAH^T$ and $P_{F(H)} \, A \, P_{F(H)}^T$ are in some sense close to each other. I realize this doesn't make much sense, since one is diagonal and the other is not, but maybe there is a projection where the diagonal entries of $HAH^T$ and $P_{F(H)} \, A \, P_{F(H)}^T$ are somewhat close to each other.

Any help is much appreciated.

PS. I would also be interested in the same setting with $U(n)$ instead of $O(n)$.

Answer 1

Here is a natural map $F\colon O(n)\to S_n$ which does not have the specified properties, because I did not read them correctly at first. Nonetheless, it may be of interest, if only as an example of something to avoid.

The idea is to use the Bruhat decomposition $GL_n(\mathbb{R})=\coprod_{\sigma\in S_n}B\sigma B$.

In more detail, consider two flags $0=U_0<U_1<\dotsb<U_n=\mathbb{R}^n$ and $0=V_0<V_1<\dotsb<V_n=\mathbb{R}^n$. For $0\leq i,j<n$ put $$ Q_{ij} = \frac{U_i\cap V_j}{(U_i\cap V_{j-1})+(U_{i-1}\cap V_j)} \simeq \frac{U_{i-1}+(U_i\cap V_j)}{U_{i-1}+(U_i\cap V_{j-1})} \simeq \frac{(U_i\cap V_j)+V_{j-1}}{(U_{i-1}\cap V_j)+V_{j-1}} $$ One can check that there is a unique permutation $\sigma$ such that $Q_{ij}=0$ unless $i=\sigma(j)$ in which case $Q_{ij}\simeq\mathbb{R}$. (Indeed, the third description of $Q_{ij}$ makes it clear that there is a unique function $\sigma$ with this property, the second description makes it clear that there is a unique function $\tau$ such that $Q_{ij}=0$ unless $j=\tau(i)$, and then we conclude that $\sigma$ and $\tau$ are inverse to each other and so must be permutations.)

Now consider an element $g\in O(n)$. Let $U_i$ be the span of $e_1,\dotsc,e_i$, and let $V_i$ be the span of $ge_1,\dotsc,ge_i$. Let $F(g)$ be the permutation corresponding to this pair of flags. This is a fairly natural map $O(n)\to S_n$ that is the identity on $S_n$. Everything works in essentially the same way over $\mathbb{C}$.

Answer 2

I give a positive answer when $n=3$. I work with $O(\mathbb{R}^3)$ (orthogonal endomorphisms) rather than $O(n)$ (orthogonal matrices).

I define $\mu$-almost surely $F$ as follows.

Given a reflexion $s$ with regard to $D^\perp$, set $F(s)$ equal to $(1~2)$ or $(1~3)$ or $(2~3)$ according the closer line to $D$ (in terms of angles between two lines) is $\mathbb{R}(e_1-e_2)$ or $\mathbb{R}(e_1-e_3)$ or $\mathbb{R}(e_2-e_3)$.

Given a rotation $r$ with axis $D$, call $u$ the unit vector of $D$ which closer to $e_1+e_2+e_3$, and $\theta \in ]-\pi,\pi[$ the angle (oriented by the vector $u$). Set $F(s)$ equal to $(3~2~1)$ or Id or $(1~2~3)$ according $-\pi < \theta < -\alpha$ or $-\alpha < \theta < \alpha$ or $\alpha < \theta < \pi$. The real number $\alpha$ is chosen in such a way that each possibility for a randomly chosen rotation occurs with probability $1/3$.

Since the distribution of $\theta$ admits the density $\theta \mapsto (1-\cos\theta)/(2\pi)$ over $]-\pi,\pi[$, the real number $\alpha$ is given by $(\alpha - \sin\alpha)/(2\pi)= 1/6$. Since $0 < \alpha <2\pi/3$, condition 1 holds.