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)?



Thom spectrum in the definition of power operations
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!


