Are the stiefel-Whitney classes of the tangent bundle determined by the mod 2 cohomology? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T22:06:36Z http://mathoverflow.net/feeds/question/26461 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26461/are-the-stiefel-whitney-classes-of-the-tangent-bundle-determined-by-the-mod-2-coh Are the stiefel-Whitney classes of the tangent bundle determined by the mod 2 cohomology? Thomas Kragh 2010-05-30T15:32:47Z 2013-05-05T08:03:50Z <p>Let $G=\mathbb{Z}/2\mathbb{Z}$. Let $f\colon L \to N$ be a smooth map of connected smooth <strong>closed</strong> $n$-dimensional manifolds such that the induced map</p> <p>$f^* \colon H^*(N,G) \to H^*(L,G)$</p> <p>is an isomorphism.</p> <p><strong>Question</strong>: Are the pull back of the Stiefel-Whitney classes of the tangent bundle of $N$ the Stiefel-Whitney classes of the tangent bundle of $L$?.</p> <p>This is in fact true for the first Stiefel-Whitney class by considering coverings and degrees, but what about the higher degree classes?</p> <p><strong>Motivation</strong>: 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.</p> <p><strong>Question</strong>: 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?</p> <p>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. </p> <p>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.</p> <p>ADDED: in light of the answer, all this motivation and these extra possible assumptions are not important nor relevant for the actual question.</p> http://mathoverflow.net/questions/26461/are-the-stiefel-whitney-classes-of-the-tangent-bundle-determined-by-the-mod-2-coh/26469#26469 Answer by algori for Are the stiefel-Whitney classes of the tangent bundle determined by the mod 2 cohomology? algori 2010-05-30T16:33:45Z 2010-05-30T17:40:24Z <p>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.</p> http://mathoverflow.net/questions/26461/are-the-stiefel-whitney-classes-of-the-tangent-bundle-determined-by-the-mod-2-coh/32682#32682 Answer by Sean Tilson for Are the stiefel-Whitney classes of the tangent bundle determined by the mod 2 cohomology? Sean Tilson 2010-07-20T19:41:16Z 2010-07-21T16:40:51Z <p>There was a fundamental error in my answer.</p> <p>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.</p> <p>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.</p> <p>Thanks to Tom and Dan for the comments!</p> http://mathoverflow.net/questions/26461/are-the-stiefel-whitney-classes-of-the-tangent-bundle-determined-by-the-mod-2-coh/129694#129694 Answer by Andrew Ranicki for Are the stiefel-Whitney classes of the tangent bundle determined by the mod 2 cohomology? Andrew Ranicki 2013-05-05T05:19:46Z 2013-05-05T08:03:50Z <p>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 <em>The algebraic theory of surgery</em> (<a href="http://www.maths.ed.ac.uk/~aar/papers/ats1.pdf" rel="nofollow">Part I</a>, <a href="http://www.maths.ed.ac.uk/~aar/papers/ats2.pdf" rel="nofollow">Part II</a>). 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 <em><a href="http://www.maths.ed.ac.uk/~aar/papers/atiyahthom.pdf" rel="nofollow">Thom complexes</a></em> 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: <em>the stable fibre homotopy type of the tangent sphere bundle of a differentiable manifold $X$ depends only on the homotopy type of $X$</em>. </p>