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? 




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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 
 $r\circ h$?



non-orientability of vector bundles induced from a symmetric group action
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.


