16
$\begingroup$

Suppose that $M$ and $N$ are closed connected oriented surfaces. It is well-known that if $f \colon M \to N$ has degree $d > 0$, then $\chi(M) \le d \cdot \chi(N)$.

What is an elementary proof of this?

I found references to some H.Kneser's articles (1928, 1929, 1930). However, I am not sure that these give a complete proof. Also they are written in German with obsolete notation. Is there a survey on maps of surface where I can find the proof in English?

Michael Albanese suggests that we use the Gromov norm. This path seems complicated (to me). Is it possible that the original proof (before Gromov) is more elementary?

For example, if $\chi(M) > \chi(N)$, then one can easily prove that $d$ is zero, using the quadratic form on $H^1(N ; \mathbb{Z})$. Can we use this approach, or something similar, to get a precise estimate on $d$ in general?

$\endgroup$
2
  • 2
    $\begingroup$ An improved version of Kneser’s result was proved by Allan Edmonds, and implies this result: mathscinet.ams.org/mathscinet/article?mr=541331 $\endgroup$
    – Ian Agol
    Jul 26, 2023 at 5:00
  • 2
    $\begingroup$ Just to highlight this: the original poster (Andrey Ryabichev) has now written a nice article - arxiv.org/abs/2308.07813 - answering the question. It also has pictures and an overview of the literature. +1 $\endgroup$
    – Sam Nead
    Aug 16, 2023 at 10:43

3 Answers 3

12
$\begingroup$

The best, elementary, self-contained, and clarifying proof of Kneser's result, including the desired inequality, is due to Richard Skora. See

Skora, Richard, The degree of a map between surfaces, Math. Ann. 276(1987), no.3, 415–423.

He careful handles the cases when the surfaces are not necessarily orientable, using absolute degree. He also includes a self-contained, general version of my "pinch followed by branched covering" theorem mentioned above, correcting my misstatement in the non-orientable case.

$\endgroup$
1
  • 1
    $\begingroup$ I thought up more simple proof of the factorisation theorem which goes in one step without induction. Now I wrote it down arxiv.org/abs/2308.07813 Also here I prove some properties of the geometric degree, the proofs of which in old articles that I found are based on the absolute degree and ferefore too complicated. I will be grateful for any comments on this text! $\endgroup$ Aug 16, 2023 at 5:30
17
$\begingroup$

I like the Gromov norm approach. Another approach uses the Milnor-Wood inequality (here I am assuming $\chi(M), \chi(N) < 0$ and $M, N$ are connected WOLG). A discrete faithful representation of $\pi_1(N)$ to $PSL_2(\mathbb{R})$ gives a flat circle bundle of euler number $-\chi(N)$. Pulling back to $M$ via $f$ multiplies the Euler number of the pulled back bundle by $d$. By the Milnor-Wood inequality, $-d\chi(N) \leq -\chi(M)$.

As I indicated in a comment above, Edmonds proved that any map between surfaces is homotopic to a composition of a pinch map and a branched cover, improving the results of Kneser that you refer to. One can see that a pinch map increases Euler characteristic (for negative Euler characteristic surfaces) and branched covers satisfy the inequality by the Riemann-Hurwitz formula.

One could probably also give a proof using the interpretation of $\chi(M)$ as the self-intersection number of $\Delta(M)$ in $M\times M$. This can be interpreted in terms of cup product of the Poincaré dual, and one should be able to give a fairly elementary proof.

$\endgroup$
1
  • $\begingroup$ Ian, sorry, I just noticed that you mention Edmonds as well. $\endgroup$ Jul 26, 2023 at 7:46
12
$\begingroup$

Allan Edmonds shows that such a map decomposes as the composition of a pinch (i.e., pinching off handles) and a branched cover, which immediately gives what you need.

Edmonds, Allan L. Deformation of maps to branched coverings in dimension two. Ann. of Math. (2) 110 (1979), no. 1, 113–125.

His work certainly relies on Kneser. If there is any modern presentation of Kneser's work, it would be in a survey by Edmonds.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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