# The biggest degree of a map between fixed surfaces

Let $S_1$ and $S_2$ be compact surfaces (real manifolds of dimension 2). None of them is a sphere.

Question: What is the biggest degree of a smooth map from $S_1$ to $S_2$?

Comment 1. I have a conjecture. Integer part of $\frac{ \chi (S_1) } { \chi(S_2) }$. Denote this number as k. At least I can construct a map of degree $k$. Each surface (which is not a sphere) has an $n$-fold covering for any $n$. So we can consider $\tilde{S}_2 -$ $k$-fold covering of $S_2$. It is easy to see that $S_1$ has not less handles than $\tilde{S}_2$ does. So there is a map of degree 1 from $S_1$ to $\tilde{S}_2$ which contracts some handles.

Comment 2. If we require $S_1$ and $S_2$ to be Riemann surfaces and would consider holomorphic maps, than this result is trivial from Riemann-Hurwitz formula.

At least for mappings into a torus we can produce all integers as degrees. Let $T=S^1\times S^1$ be the torus. We see $S^1\subset\mathbb C$ and then we have $f_k:T\to T$ given by $f_k(z,w)=(z^k,w)$, which has degree $k$, for any $k\in \mathbb Z$. On the other hand, for any orientable surface $S_g$ with $g>1$ holes, one can collapse all the surface but a nbhd of the first hole onto a point, which gives a torus $T$, so getting a degree $1$ map $h:S_g\to T$. Now compose to get a map $h_k=f_k\circ h:S_g\to T$ of degree $k$. The latter is not smooth, but a good enough approximation will have the same degree.

On the other hand, suppose $h:T\to S$ is a map of degree $d\ne0$ into some other surface $S$. Then the composite $h\circ f_k:T\to S$ has degree $kd$. We see that the bound exists only if all mappings $h:T\to S$ have degree $0$ (and the bound is $0$).

• However, it follows from math.stackexchange.com/questions/1230760/… that, actually, all maps $T\rightarrow S$ have degree $0$. According to that link, any such map $T\rightarrow S$ is NOT injective on the $H_1$ level, which persists even if we use $\mathbb{Q}$ coefficients. By dualizing, this implies the map on the $H^1$ level is not surjective, and since $H^1(T;\mathbb{Q}) = \mathbb{Q}^2$, this implies that the image of $H^1(S;\mathbb{Q})$ in $H^1(T;\mathbb{Q})$ is a line. In particular, – Jason DeVito Apr 15 '15 at 13:31
• the cup product $f^\ast(v)\cup f^\ast(w) = 0$ for any $v,w\in H^1(S;\mathbb{Q})$. Since $H^2$ is generated by $H^1$ as an algebra for both $T$ and $S$, it follows that the map on the $H^2$ level is he $0$ map. Dualizing again, it's the $0$ map on the $H_2$ level, so is degree $0$. – Jason DeVito Apr 15 '15 at 13:32
• You're right, of course. I was just trying to suggest the fact (which in that case fits the bound in the conjecture). You should post the comment as an answer, I think. – Jesus RS Apr 15 '15 at 13:37
• I think I agree.. writing it up now... – Jason DeVito Apr 15 '15 at 13:47
• Furthermore, this gives interesting "counterexamples" to the Hopf Theorems for mappings $T\to S^2$, that is essential mappings $T\to S$ of degree $0$. – Jesus RS Apr 15 '15 at 13:54

Let me confirm your conjecture when $S_1 = T$ is a torus. Then, the conjecture asserts that every map $f:T\rightarrow S$ is degree $0$ if $S$ is not a torus or a sphere.

(As Jesus RS pointed out, if $S$ is a torus, you can get maps of any degree. It is also true that if $S = S^2$, then you get can get any degree.)

This MSE question proves that for every map $f:T\rightarrow S$, $f_\ast:H_1(T)\rightarrow H_1(S)$ is not injective. It follows that the same is true if we change coefficient from $\mathbb{Z}$ to $\mathbb{Q}$.

Dualizing, we see that no $f$ induces a surjection from $f^\ast:H^1(S;\mathbb{Q})\rightarrow H^1(T;\mathbb{Q} = \mathbb{Q}^2$. But, since $f^\ast$ is linear and not surjective, the image must lie on a line in $\mathbb{Q}^2$.

Since the cup product is graded anti-commutative, it follows that for any $v,w\in H^1(S;\mathbb{Q})$, that $f^\ast(v)\cup f^\ast(w) = 0 \in H^2(T)$.

Now, picking $v$ and $w$ so that $v\cup w$ generates $H^2(S;\mathbb{Q})$, we see that $f^\ast:H^2(S;\mathbb{Q})\rightarrow H^2(T;\mathbb{Q})$ must be the $0$ map on $H^2$. Thus, $f$ has degree $0$.

• Doesn't something of the type work for mappings $S_1\to S_2$ with genus of $S_1$ strictly smaller than genus of $S_2$? – Jesus RS Apr 15 '15 at 13:59
• @JesusRS: I suspect so, but don't immediately see how to make Mike's answer in the MSE link work in this case. – Jason DeVito Apr 15 '15 at 14:55
• It is worth remarking (surely the owner of the question was aware) that for the standard $m$-sheeted coverings $h_d:S_k\to S_g$ with $k=m(g-1)+1$ for $g\ge2$, we have $\frac{2-2k}{2-2g}=d=$ degree$(h)$. Exactly the bound. – Jesus RS Apr 15 '15 at 17:43

Jason DeVito's answer dealt with the case where $S_1 = S^1\times S^1$ and Jesus RS's answer dealt with the case where $S_2 = S^1\times S^1$. You said that neither surface is $S^2$, so this answer addresses the remaining case: $S_1 = \Sigma_g$ and $S_2 = \Sigma_h$ with $g, h \geq 2$.

The Gromov norm $\|\cdot\|$, also known as simplicial volume, is a useful tool for tackling questions about degree due to the following fact:

Let $M$ and $N$ be closed, oriented manifolds of the same dimension and $f : M \to N$ a continuous map. Then $|\deg f|\|N\| \leq \|M\|$.

The Gromov norm of $\Sigma_g$, for $g \geq 2$, is $\|\Sigma_g\| = 4g - 4$ (see this answer). So for any $f : \Sigma_g \to \Sigma_h$ with $g, h \geq 2$, we have $|\deg f|(4h - 4) \leq (4g - 4)$ and therefore

$$|\deg f| \leq \left\lfloor\frac{g-1}{h-1}\right\rfloor = \left\lfloor\frac{2 - 2g}{2 - 2h}\right\rfloor = \left\lfloor\frac{\chi(\Sigma_g)}{\chi(\Sigma_h)}\right\rfloor =: L.$$

In particular, if $g < h$, then $f$ has degree zero which agrees with Mike Miller's answer.

This upper bound is actually achieved, i.e. there is a map $f : \Sigma_g \to \Sigma_h$ of degree $L$. This can be seen by combining the following two facts:

• If $a \geq b$, then there is a degree one map $\Sigma_a \to \Sigma_b$ (write $\Sigma_a = \Sigma_b\#\Sigma_{a-b}$ then crush $\Sigma_{a-b}$ to a point).
• There is a $k$-sheeted covering map $\Sigma_a \to \Sigma_b$ if and only if $\chi(\Sigma_a) = k\chi(\Sigma_b)$, i.e. $a = k(b - 1) + 1$ (see Example $1.41$ of Hatcher's Algebraic Topology).

As $L(h-1) + 1 \leq g$, there is a degree one map $\Sigma_g \to \Sigma_{L(h-1)+1}$ by the first point. By the second point, there is an $L$-sheeted covering map $\Sigma_{L(h-1)+1} \to \Sigma_h$. The composition of these two gives a map $\Sigma_g \to \Sigma_h$ of degree $L$.

In conclusion, you are correct.

If $g < h$, maps $M_g \to M_h$ necessarily have degree zero.

Suppose we had a map $f: M_g \to M_h$ with nonzero degree $d$. Let $K = \text{Im}(f_*: \pi_1(M_g) \to \pi_1(M_h))$, and lift $f$ to $\tilde f: M_g \to M_K$, the surface corresponding to the covering $p: M_K \to M_h$ with $p_*(\pi_1(M_K)) = K$. $M_K$ must be compact, lest the map on top homology factor through $H_2(M_K) = 0$ and force $d=0$; and so we have a surjection $\pi_1(M_g) \to \pi_1(M_K)$. By an Euler characteristic argument the genus of $M_K$ is greater than the genus of $M_h$; call it $g'$. $\pi_1(M_g)$ is generated by $2g$ elements, so such a surjection would generate $H_1(M_K) = \Bbb Z^{2g'}$ with $2g$ elements. But this is impossible. So $f$ must have had degree zero all along.

Jason DeVito sent me an alternative cohomological argument: a map $f: M_g \to M_h$ induces a map $f^*: H^1(M_h) \to H^1(M_g)$; because of the ranks involved, this has some $x$ in the kernel. Poincare duality (or knowing the cohomology ring) gives you some $y$ such that $x \smile y$ is nonzero in $H^2(M_h)$; then $f^*(x \smile y) = f^*(x) \smile f^*(y) = 0$, so the map must have degree zero.