A well known theorem in algebraic topology relates the (co)homology of the Thom space $X^\mu$ of a orientable vector bundle $\mu$ of dimension $n$ over a space $X$ to the (co)homology of $X$ itself: $H_\ast(X^\mu) \cong H_{\ast-n}(X)$ and $H^\ast(X^\mu) \cong H^{\ast-n}(X)$.

This isomorphism can be proven in many ways: Bott & Tu has an inductive proof using good covers for manifolds and I learned on MathOverflow that one can use a relative Serre spectral sequence. However, I believe that there should also be a proof using stable homotopy theory, in the case of homology by directly constructing a isomorphism of spectra $X^\mu \wedge H\mathbb{Z} \to X_+ \wedge \Sigma^{-n} H\mathbb{Z}$, where $X^\mu$ denotes the Thom spectrum, $H\mathbb{Z}$ the Eilenberg-Mac Lane spectrum for $\mathbb{Z}$ and $X_+$ the suspension spectrum of $X$ with a disjoint basepoint added.

Is there an explicit construction of such a map implementing the Thom isomorphism on the level of spectra? I am interested in such a construction for both homology and cohomology. If so, is there a similar construction for generalized (co)homology theories? I would also be interested in references.

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    $\begingroup$ You'd probably be interested in (at least the introduction of) arxiv.org/abs/0810.4535 , along with its lengthy bibliography. $\endgroup$ Nov 21, 2010 at 0:44
  • $\begingroup$ If we have the same standard of what is meant by "explicit", then there is not an explicit construction of such an equivalence of spectra. This is because in particular it would give an explicit construction of the Thom class $X^\mu \to K(\mathbb{Z},n)$, for which there is none. $\endgroup$ Nov 21, 2010 at 10:52
  • $\begingroup$ @Oscar. Is it a theorem that there is no such construction or is it simply too hard to give one when our models of $K(\mathbb{Z},n)$ get more complicated as $n$ increases? $\endgroup$
    – skupers
    Nov 21, 2010 at 19:33
  • $\begingroup$ @skupers: No, it is not a theorem. $\endgroup$ Nov 21, 2010 at 21:19

3 Answers 3


There is a construction for both Thom isomorphisms, homological and cohomological, via classical stable homotopy theory. You find the details in Rudyaks book "On Thom spectra, orientability, and cobordism", chapter V, §1. The Thom class is a map $X^{\mu} \to\Sigma^{n} H \mathbb{Z}$. Moreover, there is a map of spectra $X^{\mu} \to X_+ \wedge X^{\mu}$ which is induced from the map of vector bundles $\mu \to \mathbb{R}^0 \times \mu$ over the diagonal map $X \to X \times X$. Here is the definition of the homological Thom isomorphism; the cohomological one is in the same spirit. Consider the composition

$X^{\mu} \wedge H \mathbb{Z} \to X_+ \wedge X^{\mu} \wedge H\mathbb{Z} \to X_+ \wedge \Sigma^n H \mathbb{Z} \wedge H \mathbb{Z} \to X_+ \wedge \Sigma^n H \mathbb{Z} $. On homotopy groups, it induces a map lowering the degree by $n$ (there is a sign mistake in your question that confused me for some minutes).

It is clear that this works for orientations with respect to other ring spectra as well.

  • $\begingroup$ This is interesting, Johannes. May I ask what $H\sigma^n\mathbb{Z}$ is and what the map $\Sigma^n H\mathbb{Z}\wedge H\mathbb{Z}\to H\sigma^n\mathbb{Z}$ does? One likes to have a map $f:X^\mu\wedge H\mathbb{Z}\to X_+\wedge \Sigma^n H\mathbb{Z}$ in the end to get the homology isomorphism, right? Do you say that $X^\mu\to \Sigma^n H\mathbb{Z}$ can be defined as a Thom class iff the map $f$ is a weak homotopy equivalence? Thank you. $\endgroup$
    – roger123
    Nov 23, 2010 at 12:27
  • $\begingroup$ That was a typo, $H \sigma^n\bZ=\Sigma^n H\bZ$. The composition is a weak equivalence iff $X^{\mu} \to \Sigma^n H \bZ$ is a Thom class. $\endgroup$ Nov 23, 2010 at 13:10

Johannes points out that it works quite generally, and perhaps it is useful to see that generality. Pardon me for going back to basics here. There are three ingredients:

  1. Given an n-plane bundle $\xi \to B$ we have the Thom diagonal $T(\xi) \to T(\xi) \wedge B_+ $ induced by the diagonal map $B_+ \to B_+ \wedge B_+$, expressing $T(\xi)$ as a comodule over $B_+$ and the natural map $B_+ \to T(\xi)$ as a comodule map.

  2. A ring spectrum $E$.

  3. An orientation $T(\xi) \to \Sigma^n E$. This is a cohomology class in $E^n T(\xi)$ which restricts to a generator of $\pi_*E$ when composed with the natural map $S^n \to T(\xi)$ induced by the inclusion of a point into $B$. (See pp. 253-255 (following Prop. 10.5) of Adams' "Stable Homotopy and Generalised Homology" for a nice discussion of why this is the right way to define an orientation.)

The `geometric' Thom isomorphism is then the homotopy equivalence obtained by composing these:

$E \wedge T(\xi) \to E \wedge T(\xi) \wedge B_+ \to E \wedge \Sigma^n E \wedge B_+ \to E \wedge \Sigma^n B_+$

Applying $\pi_* $ we obtain the Thom isomorphism $E_* T(\xi) \cong E_* \Sigma^n B_+$.

Since it is an $E$-module map, we can apply $F_E(-,E)$ and the equivalence $F_E(E\wedge X,E) \simeq F(X,E)$ to get an equivalence $F(\Sigma^n B_+,E) \to F(T(\xi),E)$ inducing the Thom isomorphism in $E$-cohomology, $E^*\Sigma^n B_+ \cong E^* T(\xi) $.

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    $\begingroup$ I should point out that this was published by Mark Mahowald and Nige Ray in their Proc. AMS paper "A Note on the Thom Isomorphism", in 1982: Proc. AMS V. 82 (1981), no. 2, pp 307-308. $\endgroup$ Dec 9, 2011 at 16:33

I just noticed this question, so my apologies for a very belated answer. Proposition 20.5.5 of http://www.math.uchicago.edu/~may/EXTHEORY/MaySig.pdf states that a $k$-orientation of a spherical fibration $X$ over a base space $B$ specifies a $k$-trivialization of $X$. Here $k$ is any (homotopy) commutative ring spectrum, and the notion of a $k$-orientation is the standard one. However, a $k$-trivialization is an equivalence $k\wedge X\simeq k\wedge S^n_B$ of parametrized spectra over $B$. Intuitively, after smashing with $k$ over $B$, $X$ thinks that it is the trivial fibration $B \times S^n \to B$.The Thom isomorphism follows directly.


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