2
$\begingroup$

I am wondering if there is a link between euler characteristics of orientable manifolds X and Y if there is a map of degree k from X to Y. I know about the result for a n-sheet covering map. Does the result work for maps with degrees aswell?

Many thanks

$\endgroup$
  • $\begingroup$ There are maps $f : S^2 \to S^2$ having every possible integer as their degree, but $\chi S^2 = 2$. $\endgroup$ – Ryan Budney May 30 '11 at 3:15
  • $\begingroup$ @Ryan what about the orientable genus? as a hint to this question: "For each k show there is some natural number g(k) and a degree k map from the genus_g(k) to genus_2". they set eulercharac(genus_2)*k=eulercharac(genus_g(k)). Why? thanks $\endgroup$ – El Moro May 30 '11 at 3:30
  • 2
    $\begingroup$ You've totally lost me. Earlier you said you were wondering about something, now you're giving a hint to a question? If you're doing homework, please let people know beforehand. In particular you might as well give a precise reference for your homework problem -- is it out of a textbook, for a particular course, etc. $\endgroup$ – Ryan Budney May 30 '11 at 3:46
2
$\begingroup$

As Ryan commented, there are maps of every degree from $S^2$ to itself, and so we cannot work with arbitrary maps. However, the Riemann-Hurwitz formula allows you to generalize to ramified covering maps (i.e. maps which are coverings after removing a finite number of points) of surfaces. Where the cover is ramified, you need correction terms, and these terms can be given explicitly in terms of how the sheets of the covering come together.

These two examples show that degree is only one aspect of how maps affect Euler characteristic. The Riemann-Hurwitz theorem is closely related to the Riemann-Roch theorem, and so the Grothendieck-Riemann-Roch theorem might be in some sense the best "generalization" of the Riemann-Hurtiwz formula, although I wouldn't not consider it a direct generalization in the usual sense.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.