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Let $M$ be a $m$-dimensional manifold whose cohomology ring and cell structure are well-understood, such that there is a free action of the symmetric group $S_n$ on $M$. Then we have a $n!$-sheeted covering space $$ \pi:M\to M/S_n. $$ Let $S_n$ act on the Euclidean spaces $\mathbb{R}^n$ and $\mathbb{C}^n$ by permuting the order of coordinates. We obtain vector bundles by attaching Euclidean spaces as fibres to the covering $\pi$: $$ \xi:\mathbb{R}^n\to M\times_{S_n}\mathbb{R}^n\to M/S_n, $$ $$ \xi^\mathbb{C}: \mathbb{C}^n\to M\times_{S_n}\mathbb{C}^n\to M/S_n. $$ Question: (1). Any methods to know the Stiefel-Whitney classes $$ w(\xi)? $$ Question: (2). Any methods to know the Chern classes $$ c(\xi^\mathbb{C})? $$

Note: if we take coefficients to be $\mathbb{Q}$, then all the Chern class vanish. Hence we take the coefficients to be $\mathbb{Z}/p$ for $p$ prime.

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    $\begingroup$ A vector bundle that arises this way is flat, and the Pontryagin or Chern classes (as appropriate) of a flat vector bundle on a smooth manifold vanish rationally by Chern-Weil theory. Alternatively, by hypothesis these vector bundles trivialize when pulled back to $M$, so these characteristic classes all vanish when pulled back to $M$. The map $H^{\bullet}(M/S_n, \mathbb{Q}) \to H^{\bullet}(M, \mathbb{Q})$ is injective, so again we conclude that the Pontryagin and Chern classes vanish rationally. So the only possible interesting classes are torsion classes that vanish when pulled back to $M$. $\endgroup$ Oct 8, 2015 at 3:08
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    $\begingroup$ Also, $\xi^{\mathbb{C}}$ is the complexification of $\xi$, so understanding its Chern classes reduces to understanding the Pontryagin classes of $\xi$. $\endgroup$ Oct 8, 2015 at 3:19

1 Answer 1

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When you have an Euclidean vector space of type: $$\mathbb{R}^n\rightarrow M\times_{S_n}\mathbb{R}^n\rightarrow M/S_n$$ Then its classifying map $\phi:M/S_n\rightarrow BO(n)$ factors as: $$M/S_n\rightarrow BS_n\stackrel{\rho_n}{\rightarrow} BO(n)$$ where $\rho_n$ is induced by the regular reprsentation $S_n\rightarrow O(n)$.

Thus what you want to compute is the morphism $$\rho_n^*:H^*(BO(n),\mathbb{F}_2)\rightarrow H^*(BS_n,\mathbb{F}_2)$$ A nice way to formulate such a result is to put all $n$ together. Then you have a map of Hopf rings $$\rho:H^*(BO(\bullet),\mathbb{F}_2)\rightarrow H^*(BS_{\bullet},\mathbb{F}_2).$$ Then set $w_{i,n}=\rho^*_n(w_i)\in H^i(BS_n,\mathbb{F}_2)$ where $w_i$ is the i-th Stiefel-whitney class and $w(k,l)=w_{2^k(2^l-1),2^{k+l}}$. Then Giusti, Salvatore and Sinha proved in their paper "Mod-2 cohomology of symmetric groups as a Hopf ring" (theorem $10.7$):

"The classes $w(k,l)$ generate $H^*(BS_{\bullet},\mathbb{F}_2)$ as a Hopf ring."

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  • $\begingroup$ Well, it still remains to compute $H^*(BS_n)\rightarrow H^*(M)$ , right? $\endgroup$
    – user43326
    Oct 8, 2015 at 13:18
  • $\begingroup$ You're right and that's another story. $\endgroup$
    – David C
    Oct 8, 2015 at 15:05

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