# canonical action of symmetric groups on orthogonal groups

There is a canonical faithful orthogonal representation of the symmetric group $S_{n+1}$, for $n\geq 1$: $$S_{n+1}\to O(n)$$ given as follows.

(1). I regard $O(n)$ as the isometry group of the unit sphere $S^{n-1}$ in $\mathbb{R}^{n}$.

(2). Let $\Delta^n$ be a regular $n$-simplex embedded in $\mathbb{R}^{n}$ such that

(i). all its edges are of the same length;
(ii). all its $(n+1)$-vertices are in $S^n$;
(iii). its center is the 0.


(3). I observe that any permutation on the $(n+1)$-vertices of $\Delta^n$ can be uniquely extended to an isometry of $S^n$. Hence I get an embedding of $S_{n+1}$ into $O(n)$.

Regarding $O(n)$ as a manifold, we have a canonical action of $S_{n+1}$ on $O(n)$. Hence we have a covering map $$O(n)\to O(n)/S_{n+1}.$$ Let the vector bundle (the action of $S_{n+1}$ on $\mathbb{R}^{n+1}$ is given by permutation of coordinates of $\mathbb{R}^{n+1}$) $$\eta: \mathbb{R}^{n+1}\to O(n)\times _{S_{n+1}}\mathbb{R}^{n+1}\to O(n)/S_{n+1}.$$

Question: I want to know the Stiefel-Whitney class of $w(\eta)$. How to compute it?

• I can see a canonical homomorphism from $S_n$ to $O(n)$, namely by imbedding as a subgroup of the hyper-octahedral group; how do you define one from $S_{n+1}$ to $O(n)$? – John Jiang Sep 21 '15 at 4:19
• Hi Prof., could I embed $S_{n+1}$ in such a way? – Shi Q. Sep 21 '15 at 6:26
• @JohnJiang The map is in the original post. Another way to see this is the canonical homomorphism you mention, which is usually called a \it{unreduced regular representation}, fixes a hyperplane $x_1+x_2+⋯x_n=0$, so by restricting, you get a map from Sn to $O(n−1)$, which is called {\it reduced regular representation}. – user43326 Sep 21 '15 at 8:43
• @JohnJiang: I think the embedding exists, and this is my way to see it. If $x$ and $y$ are two vertices of the $n$-simplex $\Delta$, I can define the `axis' $\pi(x,y)$ to be the hyperplane joining the middle point between $x$ and $y$, with the remaining vertices of $\Delta$. Then you can map the transposition $(i,j)$ from $S_n$ into the reflection with respect to the hyperplane $\pi(x_i,x_j)$ ($x_i$ vert. of $\Delta$). Since the transpositions generates $S_n$ (mathoverflow.net/questions/24101/…) this is (almost) enough to define the embedding. – Giovanni Moreno Sep 21 '15 at 14:25
• What do you mean by "compute"? As I said elsewhere, there is a spectral sequence for $H^*(O(n)/S_{n+1})$. The SW classes in question are the images of the SW classes in $H^*BS_{n+1}$ of the standard representation of $S_{n+1}$, so they lie on the bottom line of the spectral sequence. You need to understand something about the behaviour of the spectral sequence before you can formulate any further questions. – Neil Strickland Sep 22 '15 at 7:04

I think the bundle you consider is trivial, because the representation of $S_{n+1}$ on $\mathbb R^{n+1}$ by permutation of coordinates extends to a representation of $O(n)$. I am not completely sure about the first part of the argument, however.
I think the map $S_{n+1}\to O(n)$ you describe actually comes from the $n$-dimensional standard representation of $S_{n+1}$. Indeed, permutation of coordinates gives a homomorphism $S_{n+1}\to O(n+1)$, but this has actually values in the stabilizer of the vector $v:=(1,\dots,1)$ and that stabilizer is isomorphic to $O(n)$, via the action on $v^\perp$. Now I think that if you orthogonally project the unit vectors to this hyperplane, you get (up to a multiplication by a constant factor depending on $n$) the $n+1$ points in the sphere that you describe.
If this is correct, then the representation of $S_{n+1}$ on $\mathbb R^{n+1}$ indeed extends to $O(n)$ as the direct sum of the defining representation of $O(n)$ on $v^\perp$ and the trivial representation on the line spanned by $v$. But then it follows from general principles that the bundle $O(n)\times_{S_{n+1}}\mathbb R^{n+1}$ can be trivialized. To do this, consider the map $O(n)\times\mathbb R^{n+1}\to (O(n)/S_{n+1})\times\mathbb R^{n+1}$ defined by $(g,v)\mapsto (gS_{n+1},g\cdot v)$. This evidently factorizes to an isomorphism on the induced bundle.