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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 $g \geq g'$?

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  • $\begingroup$ You can find a degree one map, and thus a map of any degree. $\endgroup$ – user98602 Jun 6 '17 at 18:24
  • $\begingroup$ @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. $\endgroup$ – Michael Albanese Jun 6 '17 at 21:23
  • $\begingroup$ @MichaelAlbanese oops, how embarrassing. Thanks for the correction! $\endgroup$ – user98602 Jun 6 '17 at 21:30
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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.

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

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