Which sets of Stiefel-Whitney characteristic numbers can be realized as coming from a manifold? - MathOverflowmost recent 30 from mathoverflow.net2022-01-27T02:39:28Zhttps://mathoverflow.net/feeds/question/40539https://creativecommons.org/licenses/by-sa/4.0/rdfhttps://mathoverflow.net/q/4053910Which sets of Stiefel-Whitney characteristic numbers can be realized as coming from a manifold?Dylan Wilsonhttps://mathoverflow.net/users/69362010-09-29T20:56:28Z2013-07-16T22:11:03Z
<p>EDIT: The original question was answered very quickly (and very nicely!) but the answer leads to a pretty obvious subsequent question, which I will now ask. The original question is maintained for motivational purposes below.</p>
<p>I now know that not every sequence of zeros and ones can be realized as the Stiefel-Whitney numbers for some manifold- as I'm sure many of you all already knew. What I don't know, and what I suspect is a more delicate question, is: Which ones are? Is there a relatively easy necessary condition? Any sufficient conditions? </p>
<p>Along similar lines: are there estimates as to the number of cobordism classes in any dimension that are tighter than the number of "possible" Stiefel-Whitney numbers? A tighter bound, as it were.</p>
<p>Thanks!</p>
<hr>
<p>(original question)</p>
<p>Well I just learned a very cool fact over tea: apparently there are finitely many (unoriented) cobordism classes of compact manifolds in any given dimension! The cobordism class is completely determined by the Stiefel-Whitney characteristic numbers (which were explained to me as "the various numbers one gets by cupping characteristic classes of the tangent bundle together and applying them to the fundamental class, all mod 2")... so that's pretty awesome.</p>
<p>While I get over this initial shock, I was wondering if anyone knew the answer to the following: we have an upper bound on the number of cobordism classes by looking at the number of possible Stiefel-Whitney numbers. But is this upper bound realized?</p>
<p>In other words, given a sequence of zeros and ones (the right number of them), can I always construct a manifold that has precisely that sequence of zeros and ones as its Stiefel-Whitney numbers?</p>
https://mathoverflow.net/questions/40539/-/40540#405405Answer by Oscar Randal-Williams for Which sets of Stiefel-Whitney characteristic numbers can be realized as coming from a manifold?Oscar Randal-Williamshttps://mathoverflow.net/users/3182010-09-29T21:00:11Z2010-09-29T21:00:11Z<p>No: for example, there is no 1-manifold with Stiefel--Whitney number for $w_1$ equal to 1.</p>
https://mathoverflow.net/questions/40539/-/40571#405713Answer by Dev Sinha for Which sets of Stiefel-Whitney characteristic numbers can be realized as coming from a manifold?Dev Sinhahttps://mathoverflow.net/users/49912010-09-30T03:51:21Z2010-09-30T03:51:21Z<p>Recall Thom's result that cobordism groups are homotopy groups of Thom spaces/ spectra. The characteristic numbers of $[M]$ are encoding the image of the corresponding element of $\pi_* MO$ under the Hurewicz(/Boardman) map to $H_* (MO)$ (which you may recall is isomorphic to $H_*(BO)$). So your question is essentially "what is the image of this Hurewicz map?" The complex analogue of this question was answered by Adams in his "Stable Homotopy" book, using the framework of formal group laws; the $MO$ question has a similar answer.</p>
https://mathoverflow.net/questions/40539/-/40578#405787Answer by Christian Nassau for Which sets of Stiefel-Whitney characteristic numbers can be realized as coming from a manifold?Christian Nassauhttps://mathoverflow.net/users/88242010-09-30T06:20:08Z2010-09-30T06:56:58Z<p>The Steenrod operations on mod 2 cohomology imply the vanishing of some characteristic numbers. Specifically, if $p(w_1,w_2,\ldots)\in H^k(M^n;Z/2)$ for $k\lt n$ then $0=\langle \sum_{i+j=n-k}u_i{\rm Sq}^{j} p, [M^n]\rangle$ where $u_i$ is the Wu class of $M$ (but take this with a grain of salt, since I'm quoting from memory here). Thom showed that <em>all</em> relations between characteristic numbers arise in this way. This allowed him to compute the bordism ring of unoriented manifolds exactly: it is $Z/2[m_k\vert k$ not of the form $2^j-1]$. This gives you the tight bound you were asking for.</p>
<p>But, as others have already pointed out, there are plenty of good books & papers on this, so you should probably just dive into the library.</p>
https://mathoverflow.net/questions/40539/-/136910#1369102Answer by András Szűcs for Which sets of Stiefel-Whitney characteristic numbers can be realized as coming from a manifold?András Szűcshttps://mathoverflow.net/users/369502013-07-16T21:02:52Z2013-07-16T21:02:52Z<p>Here are the data from MathScinet of Dold's paper proving that precisely the so called Wu relations give a complete set of restrictions on the Stiefel Whitney numbers of manifolds:</p>
<p>MR0079765 (18,143a) Reviewed
Dold, Albrecht
Vollständigkeit der Wuschen Relationen zwischen den Stiefel-Whitneyschen Zahlen differenzierbarer Mannigfaltigkeiten. (German)
Math. Z. 65 (1956), 200–206. </p>