2
$\begingroup$

Let $M$ be an orientable surface of genus $g>1$, I can assume compact. Let $f$ be a continuous map from $M$ to $M$. I want to prove that the degree of $f$ is $1$, $0$ or $-1$.

For a surface of genus $1$, a torus $S^1\times S^1$, it is easy to construct a map of any degree. So I have to use topological properties of the genus.

Any help is appreciated

$\endgroup$
1
  • $\begingroup$ Mathematical formulae look better in $\LaTeX$. Here is a quick tutorial. $\endgroup$ – Τίμων Apr 9 '16 at 9:41
4
$\begingroup$

If $M$ is a closed, connected, oriented manifold, let $\|M\|$ denote its Gromov norm (also known as simplicial volume). One of the key properties of the Gromov norm is that if $N$ is another closed, connected, oriented manifold of the same dimension, then for any continuous map $f : M \to N$, the degree of $f$ satisfies

$$|\deg f|\|N\| \leq \|M\|.$$

In particular, if $f : M \to M$, then either $\|M\| = 0$ in which case we get no information about the degree of $f$, or $\|M\| \neq 0$ and so $|\deg f| \leq 1$, i.e. $\deg f \in \{-1, 0, 1\}$.

Gromov proved that the Gromov norm of a closed oriented hyperbolic manifold $M$ of dimension $n$ satisfies

$$\|M\| = \frac{\operatorname{vol}(M)}{\nu_n}$$

where $\nu_n$ is the supremal volume of all geodesic $n$-simplices in hyperbolic $n$-space (which is finite). In particular, $\|M\| > 0$.

For $g \geq 2$, the surface $\Sigma_g$ is hyperbolic and hence $\|\Sigma_g\| \neq 0$. So by the above, any map $f : \Sigma_g \to \Sigma_g$ has degree $-1$, $0$, or $1$.

In fact, as $\Sigma_g$ is hyperbolic for $g \geq 2$, it admits a metric of constant Gauss curvature $-1$, so by Gauss-Bonnet

$$\operatorname{vol}(\Sigma_g) = \int_{\Sigma_g}1 = -\int_{\Sigma_g}-1 = -2\pi\chi(\Sigma_g) = -2\pi(2 - 2g) = (4g - 4)\pi.$$

As $\nu_2 = \pi$, we see that $\|\Sigma_g\| = 4g - 4$.

In contrast, $S^2$ and $S^1\times S^1$ admit self-maps of any degree, so their Gromov norm is zero.

$\endgroup$
0
$\begingroup$

There is a nice formula for surfaces which is called Riemann-Hurwitz formula. Let $X$ and $Y$ surfaces (now the genus can be less than $2$) and let $f:X\longrightarrow Y$ a continuous map which is surjective and not a branched covering, then:

$$\chi(X)=d\chi(Y),$$

where $d$ is the degree of the map. Suppose now that $M$ is a surfaces of hyperbolic type: $i.e.$ $g(M)\ge2$. Then, by the formula above $\chi(M)=d\chi(M)$; so $d=1$.

The case of $d=-1$ appear when you choose different orientation on $M$. Infact the map $f$ induce a map $f_\sharp$ in cohomology: $$f_\sharp: H^2(M,\mathbb{Z})\longrightarrow H^2(M,\mathbb{Z})$$ such that $f_\sharp([M])=d[M]$, where $[M]$ is a fundamental class. Choosing an orientation means choosing a generator of $H^2$. So with different orientations the degree of the map is $-1$.

Finally the degree $0$ appear when you consider continuous map which are homotopic to constant map.

$\endgroup$
2
  • 1
    $\begingroup$ Thank you for your answer. This formula is true for covering maps. How do you use the genus >1 to assume that f is a covering map ? And if f is not surjective, can we show that the degree of f is always 0 ? $\endgroup$ – P.L.D Apr 9 '16 at 15:09
  • $\begingroup$ Since the surfaces are compact the map f is proper. Moreover if f is locally is one-to-one then f is a local homeomorphism, so is a covering map, otherwise is a branched covering and the Riemann-Hurwitz formula is similar. $\endgroup$ – InsideOut Apr 10 '16 at 19:46

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.