# Why the Thom spectrum of $-\xi$ (or more generally of a virtual bundle) is defined as it is?

As the title suggests, I'm trying to find motivation on the definition of the Thom spectrum of $$-\xi$$, or more generally on the definition of the Thom spectrum of a virtual bundle.

In this paper by S. Bauer (middle of page 7) he defines the Thom spectrum for a virtual bundle of the form $$\xi - \Bbb R^m\times X$$ as the desuspension: $$Th(\xi-\Bbb R^m\times X) = \sum^{-m}\Bbb S\wedge X^{\xi}$$

with this definition, let $$\eta$$ be a bundle over $$X$$ such that $$\xi\oplus \eta\cong X\times \Bbb R^N$$, then we have we have that $$Th(-\xi) = \sum^{-N}\Bbb S\wedge X^{\eta}$$

I'm wondering what is the reason of this definition and if it was somehow related with the concept of duality morphism (Rudyak, page 47, def 2.3)

My intuition is that with this definition we should have that the two spectra are dual in the sense above. This would explain for example why Bauer in the paper mentioned above, or Crabb & Knapp in this paper (page 90, Lemma 1.1) claim that there is a pairing given by cap product: $$\widetilde{h}^0(X^{-\xi})\times \widetilde{h}^r(X^{\xi})\to \widetilde{h}^r(X)$$ since we could interpret the first one as an homology group and then cap product is well defined.

So I tried proving that we have this aforementioned duality. In fact we have something resembling a duality between $$Th(-\xi)$$ and $$Th(\xi)$$: $$Th(-\xi)\wedge Th(\xi)=\sum^{-N}\Bbb S\wedge X^{\eta}\wedge \Bbb S\wedge X^{\xi}$$ $$= \sum^{-N}\Bbb S\wedge X^{\eta}\wedge X^{\xi}$$ $$= \sum^{-N}\Bbb S\wedge (X\times X)^{\eta \times \xi}$$

but my problem is that $$(X\times X)^{\eta \times \xi}$$ is not quite $$X_+\wedge \sum^N\Bbb S$$, instead we have $$X_+\wedge \sum^N\Bbb S = \Delta^*(X\times X)^{\eta \times \xi}\wedge \Bbb S$$ where $$\Delta \colon X \to X\times X$$ is the diagonal map, and this prevents me for proving that we have an actual duality.

I'm slightly confused here so I apologise in advance if something is not super precise.

• Unfortunately these are not dual. For example, Milnor and Spanier showed that, if $M$ is a compact manifold, then $M^{-TM}$ is dual to $\Sigma^{\infty}_+M = M^{\mathbb{R}^d}$ where $d=\mathrm{dim}(M)$. This is certainly not $M^{TM}$ for most $M$. As you say, it is true that we have interesting maps like $X_+ \to X^{\xi} \wedge X^{-\xi}$ coming from the diagonal map- it's just not part of a duality datum in general. (Also it looked like you were trying to prove the even stronger statement that $X^{\xi} \wedge X^{-\xi} = X_+$ which is almost never gonna be true). – Dylan Wilson Oct 1 '18 at 19:59
• Mmh I see your point Dylan. Thanks. Is it still possible to regard $h^0(X^{-\xi})$ as the $0$-th homology of some reasonable space? Since according to the paper of Crabb & Knapp the pairing should be given by cap product but all can I see there are cohomology groups – Riccardo Oct 1 '18 at 20:11
• @Dylan I suspect you mean $\Sigma^∞_+M=M^{\mathbb{R}^0}$ (sorry for the nitpicking :)) – Denis Nardin Oct 1 '18 at 20:36
• @DenisNardin whoops, of course you're right! – Dylan Wilson Oct 1 '18 at 21:24
• my own nitpick: the theorem you attribute to Atiyah was proved a year earlier by Milnor and Spanier (as a lemma, no less). Atiyah's contribution was to extend to the case when the manifold has boundary (though the proof is the same... maybe Atiyah gets the credit because he went on to say some interesting things about vector fields and James periodicity in the same paper) – Dylan Wilson Oct 1 '18 at 21:28

Just a quick answer to explain the original reason behind the definition and why our modern understanding of Thom spectra vindicates it.

Let $$X$$ be a space and $$\xi$$ a virtual vector bundle over $$X$$. The definition you give is the only possible definition of $$X^\xi$$ such that

• $$X^\xi$$ coincide with the classical notion of Thom spectrum (the suspension spectrum of the one-point compactification of $$\xi$$) when $$\xi$$ is a vector bundle
• The formula $$X^{\xi\oplus \mathbb{R}}\cong \Sigma X^\xi$$ (which is known to be true for the case when $$\xi$$ is a vector bundle) holds.

This definition is in fact very useful, because it allows us to state Atiyah duality in a very simple and elegant way:

Theorem (Atiyah duality): Let $$M$$ be a $$d$$-dimensional closed smooth manifold. Then the Spanier-Whitehead dual of $$\Sigma^\infty_+M$$ is the Thom spectrum $$M^{-TM}$$ where $$TM$$ is the tangent bundle over $$M$$.

This theorem can be seen as a generalization of Poincaré duality for nonorientable manifolds (and homology theories): it's saying that for every spectrum $$E$$ there are natural isomorphisms $$\tilde{E}^{-*}(M^{-TM})\cong E_*M$$ When $$TM$$ is $$E$$-orientable, the Thom isomorphism gives us the classical statement of Poincaré duality. There is also a version for manifolds with boundary recovering Lefschetz duality.

I think that the map that's puzzling you to comes from the map of spectra $$\Sigma^\infty_+X\to X^\xi\wedge X^{-\xi}\cong (X\times X)^{\xi \boxplus -\xi}$$ induced by the map of spaces equipped with a virtual vector bundle $$(X,0)\to (X\times X,\xi\boxplus -\xi)$$ given by the diagonal $$X\to X\times X$$ (since the pullback of $$\xi\boxplus -\xi$$ along the diagonal is $$\xi\oplus -\xi\cong 0$$). Note that this map does not come from a duality map.

Finally let me mention that it turns out that the Thom spectrum can also be defined as $$X^\xi=\mathrm{hocolim}_{x\in X} \mathbb{S}^{\xi_x}$$ where $$\mathbb{S}^{\xi_x}$$ is the suspension spectrum of the 1-point compactification of the fiber $$\xi_x$$ over $$x\in X$$. This definition is very convenient for proving a lot of properties. For example see this paper using it to study the multiplicative structures on Thom spectra.