Is a positive degree self map on a Riemann surface homotopic to a holomorphic self map?

Question

Let $S$ be a compact Riemann surface and $f:S\to S$ be a continuous self map of positive degree. Is $f$ homotopic to a holomorphic map on $S$?

Motivation: I had intention to consider this question for every map $f:S\to S^2$ where $S$ is an arbitrary complex manifold. The homotopy class of a map $f:S\to S^2\sim \mathbb{C}P^1$ determines a unique complex line bundle on $S$ up to isomorphism. So an arbitrary line bundle over $S$ is potentially a holomorphic bundle if the corresponding map $f$ metioned above is homotopic to a holomorphic map. (So after I received the comment by Nicolast I am thinking to associate a natural number, the order of holomorphic map or some thing similar, to a given line bundle on $S$). With this motivation I initialy presented the question for self maps. This motivation can be generalized for vector arbitrary rank vector bundle with replacement of complex Grassmanian with complex projective space.

Answer 1

The answer is "yes" if $S$ is the Riemann sphere. This is because a map $f$ of degree $d$ from the sphere to itself is homotopic to $z \mapsto z^d$.

The answer is "basically no" if $S$ has genus two or higher. There are no self-maps on $S$ with degree two or higher. See here. Maps of degree one are homotopic to homeomorphisms (by a result of Edmonds, mentioned by Agol here). Self-homeomorphisms of surfaces give mapping classes; most of these have infinite order, so cannot be homotopic to a holomorphic map.

The answer is "interesting" if $S$ has genus one. (See Speyer's answer for a useful condition for this case.) Again applying the result of Edmonds, any self-map of positive degree is homotopic to a covering map. Sometimes these are homotopic to holomorphic coverings, and sometimes they are not. For example, every elliptic curve has a self-covering of degree $d^2$ (for any $d$). The square torus has a self covering of degree two... but most elliptic curves do not. In degree one the situation is the same as in higher genus - there are only finitely many self-homeomorphisms that are realised by holomorphic maps (and these all give symmetries of the elliptic curve - see Nicolast's comment).

Answer 2

The answer is no. This is a corrected version of Nicolast's comment.

Let $E$ be an elliptic curve, let $f: E \to E$ be an endomorphism and let $H_1(f) : H_1(E) \to H_1(E)$ be the induced map on $H_1$. Let the characteristic polynomial of $H_1(f)$ be $x^2 - b x + c$. Clearly, the coefficients $b$ and $c$ are homotopy invariants of $f$.

Lemma If $f$ is holomorphic, then $b^2 - 4c \leq 0$.

Proof The map $f$ induces a map $\tilde{f} : \mathbb{C} \to \mathbb{C}$ from the universal cover of $E$ to itself. If $f$ is holomorphic, then this map is holomorphic, and thus must be of the form $z \mapsto \lambda z + a$. Then the roots of $x^2 - b x + c$ are $\lambda$ and $\overline{\lambda}$. A quadratic with complex conjugate roots has non-positive discriminant. $\square$

(Nicolast's comment claimed that $H_1(f)$ has to be finite order, which is not true; the eigenvalues of $H_1(f)$ must be complex conjugate but they don't need to be roots of unity.)

So, for example, the endomorphism of $\mathbb{R}^2/\mathbb{Z}^2$ induced by $\left[ \begin{smallmatrix} 2&0 \\ 0&1 \end{smallmatrix} \right]$ has a homotopy class which can't be induced by any holomorphic self map of an elliptic curve.