## Are the stiefel-Whitney classes of the tangent bundle determined by the mod 2 cohomology?

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Let $G=\mathbb{Z}/2\mathbb{Z}$. Let $f\colon L \to N$ be a smooth map of connected smooth closed $n$-dimensional manifolds such that the induced map

$f^* \colon H^*(N,G) \to H^*(L,G)$

is an isomorphism.

Question: Are the pull back of the Stiefel-Whitney classes of the tangent bundle of $N$ the Stiefel-Whitney classes of the tangent bundle of $L$?.

This is in fact true for the first Stiefel-Whitney class by considering coverings and degrees, but what about the higher degree classes?

Motivation: This came up because (relative) spin is important in defining Floer homology with $\mathbb{Z}$ coefficients. So I am in fact mostly interested in the following sub-question.

Question: In particular what about the second Stiefel-Whitney class in the case where both $N$ and $L$ are also assumed to be oriented? and if the answer is negative: what extra conditions do I need to make it positive?

The idea is that I apriori have to use $G$ coefficients, but can prove that it is a $G$-cohomology equivalence, and want to use that to start the argument over again with other coefficients, but for that I need this property of the second Stiefel-Whitney class.

This sub-question and the relation to Floer homology is related to orientations in real $K$-theory and delooping in the following sense: take a map $h\colon X \to U/O$ by delooping we get a map $\Omega h \colon \Omega X \to \Omega U/O \simeq \mathbb{Z}\times BO$ which classifies a virtual bundle over the loop space of $X$. This bundle is oriented iff the original map composed with the canonical map $U/O \to BO$ classified a virtual $0$-dimensional bundle with vanishing second Stiefel-Whitney class. This is the main point of why orientations in Floer homology is initimitely linked with spin! In the case of a Lagrangian sub-manifold $L\subset T^*N$ the difference of the tangent bundles precisely defines such a map $L \to U/O$ ($U(n)/O(n)$ classifies Lagrangians in $\mathbb{R}^{2n}$) such that the composition to $BO$ classifies the virtual bundle $TN-TL$. So in fact you may add this lifting property as an extra condition to the sub-question if you like, and then I would lose no generality. I believe that this condition implies that all the relative Prontryagin classes vanishes, which may be helpfull.

ADDED: in light of the answer, all this motivation and these extra possible assumptions are not important nor relevant for the actual question.

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"Floer homology is oriented iff you have spin!" Um... Relative pin structures are sufficient to determine coherent orientations, but I'm not sure I'd care to formulate a converse. Two disjoint Lagrangians have a natural (but trivial!) Floer complex over the integers. – Tim Perutz May 30 2010 at 18:48
By the way, the (excellent!) idea that one should prove that nearby Lagrangians are mod 2 cohomology equivalent and hence relatively spin is I believe one that Fukaya-Seidel-Smith were aware of when they wrote their papers on nearby Lagrangians. They couldn't use it because they invoke a theorem that requires char $\neq 2$. But it comes up at the very end of Abouzaid's preprint arxiv.org/pdf/1005.0358. – Tim Perutz May 30 2010 at 19:04
@Tim: The Floer homology I am refering to is the one for the action of a given Hamiltonian on a symplectic manifold $M$, not the intersection Floer homology for two Lagrangians, which I admit is a not obvious from the above since I am explicitly considering a Lagrangian in $T^*N$ (this is also what Viterbo did in "Exact Lagrange submanifolds, periodic orbits and the cohomology of the free loop space", and I am working in the same spirit), in this case the Floer homology can be given coherent orientations without employing any tricks iff $M$ is spin, – Thomas Kragh May 31 2010 at 7:51
and in the case Viterbo considers the generating functions are related by an oriented bundle iff $L\to N$ is relative spin. Thanks for the reference I had not seen that one, I am looking forward to seing if the passage from homology equivalence to homotopy equivalence can work in my frame work as well. – Thomas Kragh May 31 2010 at 7:56
I realize now that something is not quite right about this statement because Floer homology can be oriented in the case of $T^*N$ even if $N$ is not spin. I am not sure anymore what precisely the general statement is, but at least the orientation is intimitely linked to spin, because of what I write, and the relative spin statement about generating functions is valid in the Viterbo case. (I have changed the statement) – Thomas Kragh May 31 2010 at 8:05
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## 3 Answers

The answer to the question is positive, due to Wu's formula. See e.g. Milnor-Stasheff, Characteristic classes, lemma 11.13 and theorem 11.14. In fact, all one needs to compute the Stiefel-Whitney classes of a smooth compact manifold (orientable or not) is the cohomology mod 2 (as an algebra) and the action of the Steenrod algebra on it. Both structures are preserved under cohomology isomorphisms induced by continuous maps.

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How does this relate to exotic differentiable structures on the sphere? It seems that none of the things you mention as sufficient to calculate S.-W. classes depend on the differentiable structure at all. Are S.-W. classes of the tangent bundle not enough to detect weird smoothness structures? How is the situation different for Pontryagin classes (which is I think what Milnor used to distinguish them)? – Ilya Grigoriev May 30 2010 at 18:29
Ilya, homotopy spheres are stably parallelizable. Milnor distinguished exotic 7-spheres via Pontryagin numbers of bounding 8-manifolds. – Tim Perutz May 30 2010 at 18:51

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Ilya -- yes, the (integral) Pontrjagin classes depend on the smooth structure, but the Stiefel-Whitney classes don't and neither do the rational Pontrjagin classes. – algori May 30 2010 at 18:57
The Stiefel-Whitney classes don't even depend on the topological structure: They are preserved by homotopy equivalences between closed manifolds. This follows from the positive answer to the question asked. Another perspective on this is that these characteristic classes can be defined for spherical fibrations; you don't need a vector bundle (or even a topological disk bundle or microbundle). They can be defined as the classes that correspond by the Thom isomorphism to the Steenrod squares of the Thom class (and you need only a spherical fibration to get a Thom class and Thom isomorphism). – Tom Goodwillie Jul 20 2010 at 20:25

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But my last comment does not apply at all to the rational Pontrjagin classes. You can give examples of homotopy-equivalent smooth closed manifolds where one is stably parallelizable and the other has nontrivial Pontrjagin classes: the total spaces of the sphere bundles over two vector bundles over a sphere -- one bundle trivial, the other corresponding to a nontrivial element of the kernel of $\pi_{4k}(BO_j)\to pi_{4k}(BG_j)$. – Tom Goodwillie Jul 20 2010 at 20:32
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The most conceptual way of understanding the relation between the mod 2 Wu and Stiefel-Whitney classes of a manifold and the action of the Steenrod algebra on the mod 2 cohomology is to use the homotopy theory of Poincare duality spaces and the Spivak normal fibration, and also the chain homotopy theory of chain complexes with symmetric Poincare complexes and the normal chain bundle expounded in my 1980 paper The algebraic theory of surgery (Part I, Part II). A map $f:L\to N$ of $n$-dimensional manifolds which induces isomorphisms in $Z_2$-coefficient cohomology also induces a chain equivalence of $n$-dimensional symmetric Poincare complexes over $Z_2$. Such a chain equivalence automatically preserves the Spivak normal chain bundles. The mod 2 Wu and Stiefel-Whitney classes of the manifolds are preserved by $f$ because they only depend only on the underlying chain homotopy structure. It is also worth reminding ourselves that Atiyah's 1960 paper Thom complexes established the $S$-duality between the Thom space of the normal bundle of a manifold $X$ and $X_+=X \cup {*}$, and so proved a conjecture of Milnor and Spanier: the stable fibre homotopy type of the tangent sphere bundle of a differentiable manifold $X$ depends only on the homotopy type of $X$.

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There was a fundamental error in my answer.

The error was in misunderstanding the naturality of the $w_i$. $f: L \to N$ inducing an isomorphism in cohomology does not imply anything about the induced map of the tangent bundles. This is where i made my fundamental error. Please see the comments for details or look at earlier versions of this answer.

I had hoped that there would be a more "axiomatic" proof in the sense of Milnor and Stasheff. If anyone comes up with one please feel free to put one here.

Thanks to Tom and Dan for the comments!

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This doesn't answer the question, because a smooth map $f:N\to L$ need not give a bundle isomorphism, even if it induces isomorphisms on cohomology.  – Tom Goodwillie Jul 20 2010 at 22:24
Are you saying that f being an isomorphism in mod 2 cohomology implies that `$f^*(\tau_N) = \tau_L$`? – Dan Ramras Jul 20 2010 at 22:27
(My comment is the same as Tom's) – Dan Ramras Jul 20 2010 at 22:28
I did not mean to imply that an isomorphism in cohomology implied that the bundles were isomorphic. My mistake. Do the equations i have above actually require that the map of bundles be an isomorphism of bundles? I believe i only need the bundle map.  If this is wrong, is there a way it can be fixed? – Sean Tilson Jul 20 2010 at 23:56
Any map $X\to Y$ of spaces is covered by a bundle map, for any vector bundles on $X$ and $Y$. (Zero map.) Maybe some books say bundle map when they mean fiberwise isomorphism? – Tom Goodwillie Jul 21 2010 at 1:59
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