# Degrees of maps between surfaces

Let $\Sigma$ and $\Sigma'$ be closed oriented surfaces of genus $g$ and $g'$ respectively. I know that if $g < g'$ then every map $\Sigma \to \Sigma'$ has degree 0. What are the possible degrees of maps $\Sigma \to \Sigma'$ when $g \geq g'$?

• You can find a degree one map, and thus a map of any degree. – user98602 Jun 6 '17 at 18:24
• @MikeMiller: I don't think that's true. You can get a degree one map, but not every manifold admits self-maps of every degree. – Michael Albanese Jun 6 '17 at 21:23
• @MichaelAlbanese oops, how embarrassing. Thanks for the correction! – user98602 Jun 6 '17 at 21:30

First note that if there is a map $$M \to N$$ of non-zero degree, then $$b_i(M) \geq b_i(N)$$; in particular, if $$g < g'$$, then every map $$\Sigma_g \to \Sigma_{g'}$$ has degree zero.

If $$g \geq g'$$, then $$\Sigma_g = \Sigma_{g'}\#\Sigma_{g-g'}$$. There is a degree one map $$\Sigma_g \to \Sigma_{g'}$$ given by crushing $$\Sigma_{g-g'}$$ to a point.

If $$g' = 0$$, then for every $$k \in \mathbb{Z}$$, there is a degree $$k$$ map $$\Sigma_g \to S^2$$ because the sphere admits self-maps of every degree. To see this, note that $$S^2$$ can be identified with the quotient $$S^1\times[0, 1]/\sim$$ where $$(z, 0) \sim (z', 0)$$ and $$(z, 1) \sim (z', 1)$$; this is the suspension of $$S^1$$. The map $$S^1\times[0, 1] \to S^1\times [0, 1]$$ given by $$(z, t) \mapsto (z^k, t)$$ descends to a map $$S^2 \to S^2$$ and has degree $$k$$; this is the suspension of the map $$S^1 \to S^1$$ given by $$z \mapsto z^k$$. Thinking of $$t$$ as a height parameter, this map restricts to $$z \mapsto z^k$$ on every circle of latitude.

If $$g' = 1$$, then for every $$k \in \mathbb{Z}$$, there is a degree $$k$$ map $$\Sigma_g \to S^1\times S^1$$ because the torus admits self-maps of any degree, namely the maps $$S^1\times S^1 \to S^1\times S^1$$ given by $$(z, w) \mapsto (z^k, w)$$.

This leaves us with the case $$g \geq g' \geq 2$$. Here we can use the Gromov norm (also known as simplicial volume). If $$\Sigma_g$$ has genus $$g \geq 2$$, then it has Gromov norm $$\|\Sigma_g\| = 4g - 4$$.

In general, if $$f : M \to N$$ is a map of degree $$k$$, then $$|k|\|N\| \leq \|M\|$$. In particular, if $$f : \Sigma_g \to \Sigma_{g'}$$ with $$g, g' \geq 2$$, then $$|k|(4g' - 4) \leq (4g - 4)$$ so

$$|k| \leq \left\lfloor\frac{g - 1}{g' - 1}\right\rfloor.$$

This inequality gives a bound on the possible degrees of maps $$\Sigma_g \to \Sigma_{g'}$$; note, this also recovers the result that if $$g < g'$$, every map has degree zero.

In fact, for any $$k$$ satisfying this necessary condition, there is a map $$\Sigma_g \to \Sigma_{g'}$$ with degree $$k$$. To see this, first note that if $$k > 0$$, then $$g \geq k(g'-1) + 1$$, so there is a degree one map $$\Sigma_g \to \Sigma_{k(g'-1)+1}$$. As explained in this answer, there is a degree $$k$$ covering map $$\Sigma_{k(g'-1)+1} \to \Sigma_{g'}$$; the composition of these two maps is a degree $$k$$ map $$\Sigma_g \to \Sigma_{g'}$$. If $$k < 0$$, compose this map with a degree $$-1$$ map $$\Sigma_{g'} \to \Sigma_{g'}$$; an example of such a map is the restriction of the map $$(x, y, z) \to (x, y, -z)$$ to the image of a symmetric embedding $$\Sigma_{g'} \to \mathbb{R}^3$$.

Example: If $$g = 6$$ and $$g' = 3$$, then $$|k| \leq \left\lfloor\frac{5}{2}\right\rfloor = 2$$, so every map $$\Sigma_6 \to \Sigma_3$$ has degree $$-2, -1, 0, 1,$$ or $$2$$, and each of these possibilities arise.

• My comment is true when $g' <2$, so this is a full answer. – user98602 Jun 6 '17 at 21:30
• @MikeMiller: Nice. I've included this observation in my answer. Thanks. – Michael Albanese Jun 6 '17 at 22:14