# Thom spectrum in the definition of power operations

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)?

• 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 $\mathbb{RP}^\infty$. – John Greenwood Apr 10 at 13:10

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!

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