Let $f \colon X \rightarrow Y$ be a map of topological spaces. Let's say that they are (closed) manifolds (not necessarily orientable), though to be honest I'm really interested in the more general setup of $\mathbb{Z}/2$-Poincare duality spaces.

(From now on all cohomology coefficients are $\mathbb{Z}/2$.) Poincare duality induces, from the usual pullback, a pushforward map

$$ f_* \colon H^*(X) \rightarrow H^*(Y) .$$

While pullback commutes with Steenrod square, the pushforward unfortunately does not.

Question: can one "quantify" the failure of this commutativity? (i.e. some nice expression for the difference $f_* Sq - Sq f_*$?)

I have an answer to my question in terms of Wu classes, but I want something that doesn't involve them. Here's an example of the kind of thing that I'm looking for.

Suppose now that $X \hookrightarrow Y$ is a closed embedding of smooth manifolds. A reliable friend told me the following nice formula measuring the failure of this commutativity. If $N_{X/Y}$ is the normal bundle of $X$ in $Y$, then by the Gysin map, tubular neighborhood, excision etc. we get a composition

$$H^*(X) \xrightarrow{\text{Gysin}} H^{*+r}(N_{X/Y}, N_{X/Y}-X) \cong H^*(Y, Y-X) \xrightarrow{g^*} H^*(Y)$$

and the claim is that this composition is $f_*$. In particular, this can be used to write down a formula for the "error term" in commuting Sq with pushforward: if $u$ is the Thom class of $N_{Y/X}$ then

$$ f_*(Sq \alpha) = g^* (Sq \alpha \cup u) = Sq (g^* \alpha \cup Sq^{-1} u) $$

Now by definition $Sq^{-1} u = w^{-1} u$ where $w$ is the (total) Stiefel-Whitney class of $N_{Y/X}$, so

$$ f_* (Sq \alpha) = Sq ( g^* \alpha \cup w^{-1} u ) = Sq (f_*(\alpha \cup w^{-1})) $$

Unfortunately this doesn't make literally make sense if $X,Y$ are not smooth (since there's no $N_{X/Y}$). However, some things make sense; for instance, we can say that $g^* u = f_*(1)$. In my situation $f_*(1)$ is very simple; for instance I know that $Sq (f_*(1)) = f_*(1)$. (This is basically why I suspect that ``$w"=1$ in my case, so that $f_*$ and Sq should commute.)

Question: is there a "salvage" of the above formula to the case where $X,Y$ are not smooth?

I would prefer an answer that is formulated at the level of Poincare duality spaces, since that's more like my situation (and the question makes sense at that level).

  • 3
    $\begingroup$ Is there a particular reason why you don't want an answer involving the Wu classes? I think they are well defined for a Poincaré duality space, aren't they? $\endgroup$ May 28, 2016 at 13:53
  • 1
    $\begingroup$ Oh yes, the reason is that I actually want to prove a Wu formula. Not the usual one; I'm actually interested in smooth algebraic varieties, embedded in PD spaces via etale homotopy type, so it makes some sense to talk about "tangent bundle", etc. $\endgroup$
    – user84144
    May 28, 2016 at 16:20

1 Answer 1


If $f: X \longrightarrow Y$ is a map whose homotopy fibers are all Poincaré duality spaces then one gets a wrong-way map in the category of spectra: $\Sigma_+^{\infty}Y \longrightarrow X^{\nu}$ where $\nu$ is the stable spherical fibration given by the Spivak normal bundle of this map. On cohomology we get $g: H\mathbb{F}_2^*(X^{\nu}) \longrightarrow H\mathbb{F}_2^*(Y)$ which commutes with all Steenrod operations since it comes from a map of spectra. Composing with the Thom isomorphism $H\mathbb{F}_2^*(X) \cong H\mathbb{F}_2^*(X^{\nu})$ (where there's a degree shift) gives your umkehr map coming from Poincaré duality. The Thom isomorphism comes from cupping with the Thom class $U_{\nu}$, so you could define some notion of Stiefel-Whitney class by $Sq^{-1}U_{\nu}$ and get a formula similar to the one you've got for an embedding of smooth manifolds.

Here's a sketch of a construction of that pushforward map. If $Y$ is a point we'd like a stable spherical fibration $\nu$ on $X$ together with a map $S^0 \longrightarrow X^{\nu}$ of spectra sending the generator in homology to the fundamental class under the Thom isomorphism (which we can arrange if $X$ is simply connected so that $\nu$ is orientable). Such a thing exists and is unique up to fiberwise stable homotopy and it's called the Spivak normal fibration. You can also get something when $X$ isn't simply connected, and you can arrange for these to vary in a family over a base $Y$ and get the map I've indicated.

In the case of a map of manifolds this goes back at least to Atiyah: factor $f$ as $X \hookrightarrow \mathbb{R}^N \times Y \longrightarrow Y$ and let $\nu_N$ be the normal bundle to the first inclusion. Then Pontryagin-Thom collapse gives a map $S^N \wedge Y_+ \longrightarrow X^{\nu_N}$, and then let $N \rightarrow \infty$ to get the map of spectra above.

  • $\begingroup$ Intriguing! Apologies for my ignorance in topology, but after staring at this a few times I've finally identified what part my brain isn't digesting. Where can I found out about stable spherical fibrations, Spivak normal bundles, and what this wrong-way map of spectra is? $\endgroup$
    – user84144
    May 30, 2016 at 15:55
  • $\begingroup$ Tilman gives some references on his answer here: mathoverflow.net/a/64757/6936 $\endgroup$ May 30, 2016 at 16:01
  • $\begingroup$ (Also: in case it isn't clear, $X^{\nu}$ denotes either the Thom space or Thom spectrum of some spherical fibration, depending on the context.) $\endgroup$ May 30, 2016 at 17:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.