Let $M$ be a manifold.

Let $F(M,3)=\{(m_1,m_2,m_3)\mid m_1, m_2, m_3\in M, m_i\neq m_j, \text{ for any } i\neq j\}$.

Let $S_3$ be the symmetric group of order $3$.

Let $S_3$ act on $F(M,3)$ by $\sigma(m_1,m_2,m_3)=(m_{\sigma(1)},m_{\sigma(2)},m_{\sigma(3)})$, for $\sigma\in S_3$.

Let $F(M,3)/S_3$ be the quotient space of $F(M,3)$ under the action of $S_3$.

Let $S_3$ act on $\mathbb{C}^3$ by $\sigma(z_1,z_2,z_3)=(z_{\sigma^{-1}(1)},z_{\sigma^{-1}(2)},z_{\sigma^{-1}(3)})$.

Then we have a $3$-dimensional complex vector bundle $\xi$:

$$F(M,3)\times_{S_3}\mathbb{C}^3\to F(M,3)/S_3.$$

Let $c_1,c_2,\cdots$ be the Chern classes of $\xi$ with rational coefficients $\mathbb{Q}$.

Does $c_1=0$?

Does $c_2=0$?

Or depend on $M$?


1 Answer 1


Consider the covering $\pi :F(M,3)\rightarrow F(M,3)/S_3$. By construction $\pi ^*\xi $ is the trivial bundle, so $\pi ^*c_i(\xi )=0$ for $i>0$. But $H^*(F(M,3)/S_3,\mathbb{Q})$ is just the $S_3$-invariant subspace of $H^*(F(M,3),\mathbb{Q})$, so $c_i(\xi )=0$ for $i>0$.

  • 1
    $\begingroup$ So your point is that $\pi^*$ is injective, which follows from existence of an averaging homomorphism on rational cochains, left-inverse to $\pi^*$. $\endgroup$
    – ThiKu
    Apr 10, 2015 at 8:25

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