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I have already asked this question on stack exchange, but I didn’t get any answer.

Let $X$ be a compact connected complex manifold.

Let $f:X \to X$ be a surjective holomorphic map. Is it true that $f$ is a finite map (i.e., every point has finitely many preimages)?

If $X$ is one dimensional then the answer to the above question is yes. If $X$ is a complex projective space then the answer is again yes.

If we don’t assume surjectivity then we can easily construct non-constant, non-finite self-maps (just consider the product manifold and $f$ to be projection on one of the coordinates).

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    $\begingroup$ In this paper ihes.fr/~gromov/wp-content/uploads/2018/08/1024.pdf Gromov claims that "More generally, if $V$ and $V'$ are complex (not necessarily compact or Kaehler) manifolds of equal dimensions and their even Betti numbers are flnite and equal (i.e. $b_{2 i} = b'_{2i}$ ) then every proper surjective holomorphic map $f \colon V \to V'$ is finite-to-one ", see p. 223 $\endgroup$
    – Francesco Polizzi
    Commented Nov 24, 2020 at 19:05
  • $\begingroup$ In particular, this applies to your case. $\endgroup$
    – Francesco Polizzi
    Commented Nov 24, 2020 at 19:06
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    $\begingroup$ Well, take the blow-up of $\mathbb{P}^2(\mathbb{C})$ at one point. $\endgroup$
    – Francesco Polizzi
    Commented Nov 24, 2020 at 19:19
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    $\begingroup$ Well, Gromov just states the result mentioned by Francesco Polizzi, he doesn't prove it. It would be nice if one can actually prove Gromov's claim. $\endgroup$
    – user48958
    Commented Nov 24, 2020 at 21:09
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    $\begingroup$ I added a sketch of proof $\endgroup$
    – Francesco Polizzi
    Commented Nov 25, 2020 at 7:16

1 Answer 1

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Let me give a sketch of proof for Gromov's claim in the case where $X$ is Kähler. More precisely, let me prove the following ${}$

Proposition [G03, p.223]. Let $X$, $Y$ be two complex manifolds (not necessarily compact) of the same dimension and having the same even Betti numbers. If $X$ is Kähler, then every proper surjective holomorphic map $f \colon X \to Y$ is finite-to-one.

Proof. By a result of Wells [W74, Theorem 3.1], the map $f \colon X \to Y$ induces an injection $$f^* \colon H^{r}(Y, \, \mathbb{C}) \to H^{r}(X, \, \mathbb{C})$$ for all $r$. Using the duality $$H_r(X, \, \mathbb{C})=\mathrm{Hom}(H^r(X, \, \mathbb{C}), \, \mathbb{C})$$ we infer that the induced map in homology $$f_* \colon H_{r}(X, \, \mathbb{C}) \to H_{r}(Y, \, \mathbb{C})$$ is surjective for all $r$.

If $f \colon X \to Y$ contracts a subvariety $Z$ of positive complex dimension $i$ to a point, then its fundamental class would give an element $[Z]$ in the kernel of $$f_* \colon H_{2i}(X, \, \mathbb{C}) \to H_{2i}(Y, \, \mathbb{C})$$

Now $Z$ is compact by the properness assumption and so, pairing with the Kähler
form on $X$, we can deduce that $[Z]$ is non-zero. Hence $b_{2i}(X) > b_{2i}(Y)$, contradiction. $\square$

Remark. The first version of this answer did not assume that $X$ was Kähler (in fact, Gromov does not make this assumption). However, if $X$ is not Kähler it may happen that the fundamental class of a compact subvariety of $X$ is homologically trivial, see the example of Hopf's surface given in Michael Albanese answer to [MSE]: in this case, $X$ contains a torus $C$ whose fundamental class $[C]$ is zero in $H_2(X, \, \mathbb{C})$, simply because the last group is trivial. So, if $X$ is not Kähler, the last part of the proof breaks down.

In fact, I do not know how to make the argument work without the Kähler assumption for $X$, so I asked a new MO question about this problem.

References.

[MSE] https://math.stackexchange.com/q/1556561/456212

[G03] Gromov, M: On the entropy of holomorphic maps, Enseign. Math. II. Sér. 49, No. 3-4, 217-235 (2003). ZBL1080.37051.

[W74] Wells, R. O. jun: Comparison of de Rham and Dolbeault cohomology for proper surjective mappings, Pac. J. Math. 53, 281-300 (1974). ZBL0261.32005.

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  • $\begingroup$ You should remove the word "proper" in the proposition: since the domain $X$ is compact, properness is automatic. $\endgroup$
    – Moishe Kohan
    Commented Nov 25, 2020 at 7:36
  • $\begingroup$ I removed "compact", that is a too restrictive assumption. Thanks. $\endgroup$
    – Francesco Polizzi
    Commented Nov 25, 2020 at 8:57
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    $\begingroup$ Yeh, I meant to ask how do you know that the class is nontrivial, but, occasionally, I have to sleep. $\endgroup$
    – Moishe Kohan
    Commented Nov 25, 2020 at 19:25
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    $\begingroup$ @user48958 I disagree. The restriction of the Kaehler form $\omega$ to $Z$ is Kaehler, so $Z$ is a cpt. Kaehler manifold. Thus, if $\dim(Z)=k>0$, integrating $\omega^k$ over $[Z]$ we get $k! \mathrm{vol}(Z)$. This is Wirtinger's theorem, we are in the Kaehler world, that is much better than the ordinary manifold world. If $[Z]$ were $0$ in homology, we would get that the Riemannian volume of $Z$ is zero, which is a contradiction since $Z$ is positive-dimensional. See for instance Ballmann's book, Theorem 4.23 (2), p. 50. people.mpim-bonn.mpg.de/hwbllmnn/archiv/kaehler0609.pdf $\endgroup$
    – Francesco Polizzi
    Commented Nov 25, 2020 at 22:37
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    $\begingroup$ @user48958 Let me answer to the second point you raised. One of the assumptions of part 2 of Ballman's theorem is the existence of a compact complex submanifold $N$ of dimension $k$ on $M$. If $M=\mathbb{C}^n$, there exists no such a submanifold unless $k=0$. $\endgroup$
    – Francesco Polizzi
    Commented Nov 27, 2020 at 7:24

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