Let $\Sigma_k$ be the symmetric group on $k$-letters. Let $M$ be a manifold with a free $\Sigma_k$-action. Then we can form a $k$-dimensional vector bundle $$ \xi:\mathbb{R}^k\longrightarrow M\times_{\Sigma_k}\mathbb{R}^k\longrightarrow M/\Sigma_k. $$ We notice that the transition functions of $\xi$ are given in the form $$ (a_{i,j}=\delta_{j,\sigma(i)})_{k\times k} $$ where $\sigma\in \Sigma_k$ and $\delta_{j,\sigma(i)}=1$ if $j=\sigma(i)$ and $0$ otherwise. These matrices have determinants both $1$ and $-1$.

Question. For any $k\geq 2$, can we conclude that $\xi$ is always non-orientable?

Thanks for the answer given by Will Sawin! In his solution, I do not understand the following part:

Suppose $M$ is connected and the covering map from $M$ to $M/\Sigma_k$ induces a surjective homomorphism \begin{eqnarray*} h: \pi_1(M/\Sigma_k)\longrightarrow \Sigma_k. \end{eqnarray*} (I obtained this surjective homomorphism by Prop. 1.40 (c), Algebraic Topology, A. Hatcher).

Let $r: \Sigma_k\longrightarrow O(k)$ be the regular representation of $\Sigma_k$ given by permuting the coordinates of $\mathbb{R}^k$.

Question: Does a $k$-dimensional vector bundle over $M/\Sigma_k$ with structure group $\Sigma_k$ is uniquely determined by a map from $\pi_1(M/\Sigma_k)$ to $\Sigma_k$? Why the bundle $\xi $ comes from the map $r\circ h$?


1 Answer 1


Yes if $M$ is connected, no otherwise.

If $M$ is connected, the covering gives a surjective homomorphism $\pi_1(M/\Sigma_k)\to \Sigma_k$. The vector bundle comes from a representation of $\pi_1$ defined by composing this with the permutation representation of $\Sigma_k$. Taking determinants, the determinant line bundle is the same homomorphism composed with the sign representation of $\Sigma_k$. Because the homomorphism is surjective and the sign representation is nontrivial, the determinant line bundle is nontrivial. (Here I am using the fact that the map from homomorphisms $\pi_1(X) \to \pm 1$ to isomorphism classes of line bundles is a bijection.)

If $M$ is disconnected then the vector bundle could be trivial. For instance, $M$ might itself equal the symmetric group.

  • $\begingroup$ Thanks, Sawin! What means "the map from homomorphisms $\pi_1(X) \to \pm 1$ to isomorphism classes of line bundles is a bijection"? What is this map? $\endgroup$ Mar 9, 2016 at 9:34

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