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 $M$? 



Chern classes of three (two) dimensional complex vector bundles
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 $i>0$.


