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Consider an oriented manifold $X$. To calculate its Euler characteristic, one might integrate the Euler class. Now if $X$ were a complex manifold, and given as a section of some complex bundle $E$ over $Y$, we would have

$\chi(X) = \int\limits_X \mbox{ } c_* (TY) / c_* (E)$

where $c_*(\cdot)$ is the total Chern class.

Can anything of the sort be said if $X$ is a real manifold?

Presumably, if one wants only the Euler characteristic modulo 2, one can use the Stiefel-Whitney classes instead of the Chern classes. On the other hand, it seems to me that the topology of $TY$ and $E$ as bundles over $Y$ cannot suffice to carry the information of the Euler characteristic of the zero locus of a section of $E$. So I guess what I'm really asking is:

What should I know about a section $\sigma:Y\to E$ in order to know the Euler number of its intersection with the zero section, assuming this is transverse?

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Perhaps you meant "... a section $\sigma:Y\to E$ ..."? –  Somnath Basu May 21 '11 at 2:42
    
Your question isn't clear to me. What does "real manifold" mean? And "Can anything of the sort be said..." of course, but what sort of thing are you interested in? –  Ryan Budney May 21 '11 at 4:04
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It's noteworthy that for a rank $r$ oriented bundle $E$ on an $n$-dimensional compact oriented manifold $Y$ the Euler characteristic of the zero locus $X$ of a transverse section is given by integrating the Euler class of $TY$ over $X=Y$ in the extreme case $r=0$, and by integrating the Euler class of $E$ over the zero-dimensional oriented manifold $X$ in the other extreme case $r=n$. When $0<r<n$ I can't imagine what an answer could look like. If $n=4$ and $r=2$ then the Euler characteristic of $X$ can be any even integer, for any choice of $Y$ and $E$. –  Tom Goodwillie May 21 '11 at 4:05
    
@Sommath, yes, thanks. @Ryan, I just meant to emphasize it wasnt a complex manifold. –  Vivek Shende May 21 '11 at 6:17
    
Vivek -- a small remark: the Euler characteristic of the mod 2 cohomology coincides with the Euler characteristic of the cohomology with any field coefficients and with the Euler characteristic of the integral cohomology (defined as the alternating sum of the ranks). –  algori May 21 '11 at 13:39

1 Answer 1

up vote 1 down vote accepted

As far as I understand the question, it is asking: given a(n oriented) vector bundle $E$ on a(n oriented) manifold, what information on the Euler characteristic of the zero locus of a transversal section of $E$ can we deduce from the characteristic classes of $E$?

I'm afraid the answer to that is "in general, not much, apart from the parity". For instance, every orientable surface $S$ is the zero locus of a function (i.e., a section of the trivial 1-bundle) on the 3-sphere: embed the surface in the sphere and take the "oriented distance" function. In a similar way one can realize $S$ as the zero locus of two functions on $S^4$: realize $S^3$ as the equator in $S^4$, take one of the functions to be the oriented distance function extended to $S^4$ and the other a height function whose zero locus is $S^3$.

The reason one can compute the Chern classes of the zero locus in the holomorphic case is that the total Chern class is invertible, so the total Chern class of tangent bundle of the zero locus of a transversal section comes from the ambient manifold. The Euler class however is not invertible.

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