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I am reading now Tyler Lawson's $E_n$ ring spectra and Dyer-Lashof operations form the Handbook of Homotopy Theory and I've got a question on the Remark 1.4.19.

We have an operad $\mathcal{O}$ and $\Sigma_k$ acts freely and properly discontinuosly on $\mathcal{O}(k)$. Let $V\subset\mathbb{R}^k$ consisting of elements summing up to $0$ and let $\rho\to B\Sigma_k$ be associated vector bundle of dimension $k-1$ (1).

Now we can form for any $m$ an associated vector bundle $\mathbb{R}^m\otimes\rho$. If we define $P(k)$ as $\mathcal{O}(k)/\Sigma_k$, then there is a virtual bundle $m\rho$ on $P(k)$. Then the Thom spectrum $P(k)^{m\rho}$ is canonically equivalent to the spectrum $\Sigma^{-m}\Sigma^{\infty}_+\mathcal{O}(k)\otimes_{\Sigma_k}(S^m)^k$, where the latter appears in the definition of power operations (2).

So my question(s) are:

  1. What is the associated vector bundle $\rho$ appearing in (1)? Is it just subbundle of the trivial bundle?

  2. How do we get the equivalence in (2)?

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    $\begingroup$ Have you looked at the "associated vector bundle" construction? That might answer both your questions. By the way, $\rho$ is not a subbundle of a trivial bundle. For example when $k$=2 then $\rho$ is the universal line bundle over $\mathbb{RP}^\infty$. $\endgroup$ – John Greenwood Apr 10 at 13:10
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If a group $G$ acts on a vector space $V$, and $X$ is a space where $G$ acts properly discontinuously, then the map $$ (V \times X) / G \to X/G $$ can be given the structure of a vector bundle: it inherits this structure from the vector space $V$. The pullback of this bundle to $X$ is the trivial bundle $X \times V$, but it is probably not trivial on $X/G$.

(If you prefer to think about vector bundles in terms of classifying spaces, the action of $G$ on $V$ is a group homomorphism $G \to GL(V)$, and there's an induced map $BG \to BGL(V)$. The space $X/G$ has a map to $BG$ classifying the principal $G$-bundle $X \to X/G$, and the composite $X/G \to BG \to BGL(V)$ classifies this associated bundle.)

This recipe for the associated bundle also gives you a recipe for the Thom space: the Thom space is $$(S^V \wedge X_+)/G$$ where $S^V$ is the one-point compactification. We sometimes write this as $S^V \wedge_G X_+$.

In the case of this vector bundle $\bar \rho$ that you've written down, there a $\Sigma_k$-equivariant isomorphism between $\Bbb R^k$ and $\Bbb R \oplus V$, where $\Sigma_k$ acts on the factor of $\Bbb R$ trivially. The associated sphere is $S^k$, and this decomposition determines a $\Sigma_k$-equivariant isomorphism of one-point compactifications $S^k \simeq S^1 \wedge S^V$. The Thom space of the bundle $\rho$ associated to $\Bbb R^k$ then satisfies an identity: $$Th(\rho) \cong S^k \wedge_{\Sigma_k} X_+ \cong S^1 \wedge (S^V \wedge_{\Sigma_k} X_+) \cong S^1 \wedge Th(\bar \rho)$$ On the level of Thom spectra, we can desuspend both sides and turn this into an identity $$ X^{\bar \rho} = \Sigma^\infty Th(\bar \rho) \simeq S^{-1} \wedge \Sigma^\infty Th(\rho) \simeq \Sigma^{-1} X^\rho. $$

The version where we multiply the vector bundle by an integer $m$ is roughly the same, except that we get more trivial factors and we have to take care that we get everything with virtual bundles correct when when $m < 0$.

Sorry for the confusion!

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  • $\begingroup$ Thank you for the clarification! $\endgroup$ – Igor Sikora Apr 11 at 14:42

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