# Degree of map between surfaces of genus $g>1$ is $1$, $0$ or $-1$

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

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

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.

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.

• 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 ? – P.L.D Apr 9 '16 at 15:09
• 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. – InsideOut Apr 10 '16 at 19:46