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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.

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?

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.

Thanks!

(original question)

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.

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?

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?

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  • $\begingroup$ Whoops, I put "classes" when I meant "numbers." I don't know if your comment would give a necessary condition for when some sequence of zeros and ones can be realized as the Stiefel-Whitney numbers of some manifold. Would it? $\endgroup$
    – Dylan Wilson
    Sep 29, 2010 at 23:04
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    $\begingroup$ I'll assume you already know that Thom's original paper both proves that the Stiefel Whitney numbers determine the unoriented bordism class and also determines which numbers occur. Take a look at WIkipedia's "cobordism" page. $\endgroup$
    – Paul
    Sep 29, 2010 at 23:11
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    $\begingroup$ I mean Thom determines, not the numbers! For any sequence $(n_1, n_2, n_3,..n_k)$ where $n_1 +2n_2+3n_3+...+kn_k=d$, you get a Stiefel whitney number of a d dimensional manifold by evaluating w_1^n_1 U ... U w_k^n_k on the fundamental class. The collection of {0,1} you get from all such "partions" of d give a monomorphism $Omega_d\to Z/2^p$, i.e. they determine the bordism class for a given manifold. But this need not be onto; eg for d=1, p=1. Thom determined the image. $\endgroup$
    – Paul
    Sep 30, 2010 at 3:27
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    $\begingroup$ THere are some pretty great notes by Haynes Miller called "Notes on Cobordism" that explain all of this and more. $\endgroup$
    – Daniel Pomerleano
    Sep 30, 2010 at 3:48
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    $\begingroup$ Dylan, if you have a new question, you should submit it as a new question, rather than writing over the old one. $\endgroup$
    – S. Carnahan ♦
    Sep 30, 2010 at 8:24

4 Answers 4

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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 all 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.

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.

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  • $\begingroup$ Excellent! This is all crazy cool... A different flavor of algebraic topology than the "apply a functor to show your desired construction doesn't exist" type. Thanks! $\endgroup$
    – Dylan Wilson
    Sep 30, 2010 at 15:11
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No: for example, there is no 1-manifold with Stiefel--Whitney number for $w_1$ equal to 1.

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  • $\begingroup$ Thanks for the quick answer! I guess I was too busy trying to think of weird manifolds that I forgot about trivial ones... Please see the new version of the question... perhaps that will have a more interesting answer? $\endgroup$
    – Dylan Wilson
    Sep 29, 2010 at 21:21
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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.

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  • $\begingroup$ Thank you very much! Any chance you could give a brief description of what this image looks like, or is it ugly enough that I should just check out Adams' book? $\endgroup$
    – Dylan Wilson
    Sep 30, 2010 at 5:24
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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:

MR0079765 (18,143a) Reviewed Dold, Albrecht Vollständigkeit der Wuschen Relationen zwischen den Stiefel-Whitneyschen Zahlen differenzierbarer Mannigfaltigkeiten. (German) Math. Z. 65 (1956), 200–206.

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