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Let $S_n$ be the $n$-th symmetric group. Suppose we have a $n!$-sheeted covering space $$ S_n\to M\to M/S_n $$ where $M$ is a manifold. Let $\mathbb{K}$ be the real numbers $\mathbb{R}$, complex numbers $\mathbb{C}$ or quaternions $\mathbb{H}$. We construct an associated vector bundle $$ \xi:\mathbb{K}^n\to M\times _{S_n} \mathbb{K}^n\to M/S_n $$ where $S_n$ acts on $\mathbb{K}^n$ by permuting the order of coordinates. Then the structure group of $\xi$ is $S_n\subset O(\mathbb{K}^n)$. Let $SO(\mathbb{K}^n)$ denote the sub-Lie group of $O(\mathbb{K}^n)$ consisting of elements with determinant $1$.

Question: can we conclude that:

(1). since the determinant of elements in the structure group can be either $1$ or $-1$, the structure group of $\xi$ cannot be reduced to $SO(\mathbb{K}^n)$?

(2). Moreover, it follows from (1) and by http: //mathoverflow.net/questions/220117 first Chern class of complex vector bundles and first Pontrjagin class of quaternionic vector bundles, $$ w_1(\xi), c_1(\xi),q_1(\xi)\neq 0 $$ for $\mathbb{K}=\mathbb{R},\mathbb{C},\mathbb{H}$ respectively?

(Note: I have a counterexample for the statement (2). But I do not know why it is false.)

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    $\begingroup$ Statements (1) and (2) can not be made generally, they depend on the nature of the covering. The covering is classified by a map $M/S_n\to BS_n$, the associated vector bundle (with metric) is classified by the composition $M/S_n\to BS_n\to BO(\mathbb{K}^n)$ (where I disagree with the choice of notation). The determinant fact in (1) only means that the associated $O(\mathbb{K}^n)$-bundle over $BS_n$ does not have a reduction of structure group. However, the composition $M/S_n\to BO(\mathbb{K}^n)$ could factor through $BSO(\mathbb{K}^n)$, as happens e.g. for the trivial $S_n$-covering. $\endgroup$
    – Matthias Wendt
    Oct 6, 2015 at 9:22

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