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Good evening,

I would like to ask the following questions.

Let $X$ be a compact Kahler manifold. Denote by Aut(X) the group of all the biholomorphisms of $X.$

1) What can we say about this group? E.g, Is it a Lie group?

2) Does there exist a manifold $X$ with $Aut(X)$ trivial?

3) Let $A$ be an analytic subset of $X$ and $x_0\in A$ some point. Does there exist an automorphism $\gamma\in Aut(X)$ with $\gamma(x_0) \not\in A$?

4) Finally, is there a way to produce automorphisms of $X$?

Any help is appreciated. Thanks in advance.

Duc Anh

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2 Answers 2

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The answer to $(1)$ is 'yes'. It is a Lie group. This is true for any compact complex manifold. Basically, this is because the equations for a holomorphic vector field on a compact manifold always have a finite dimensional space of solutions.

Related to this is the question $(4)$: Is there any way to 'produce' automorphisms of $X$? If by this you mean some 'effective' way, you'll need to tell me how you are effectively describing the complex manifold $X$. Whether you can 'effectively' construct the biholomorphism group of $X$ depend on how much you know about $X$. For example, is it described by algebraic equations in some projective space?

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  • $\begingroup$ Thank you very much. At this moment, I'm only interested in a general compact Kahler manifold. My motivation comes from the following: I want to move a litte bit (in a reasonable way, e.g. in 3) ) the image of an entire curve $f\colon \mathbb{C}\to X,$ but I don't know any tool. So my questions are very vague. $\endgroup$
    – Đức Anh
    Commented Mar 11, 2013 at 15:54
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    $\begingroup$ Well, there is a sense in which 'most' compact Kähler manifolds have a trivial biholomorphim group. In such a case, there is no hope to use the automorphism group of the target to 'move' any curve, much less an entire curve. You probably will need to know more about your target manifold $X$ before you can hope to get anywhere with this. $\endgroup$
    – Robert Bryant
    Commented Mar 11, 2013 at 16:07
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    $\begingroup$ There is a complete proof that the automorphism group of a compact complex manifold is a complex Lie group acting holomorphically in the book Kobayashi, Shoshichi(1-CA), Transformation groups in differential geometry. Reprint of the 1972 edition. Classics in Mathematics. Springer-Verlag, Berlin, 1995. viii+182 pp. ISBN: 3-540-58659-8 $\endgroup$
    – Ben McKay
    Commented Mar 11, 2013 at 17:24
  • $\begingroup$ Thank you all very much for useful informations. I should try another approach to my problems. $\endgroup$
    – Đức Anh
    Commented Mar 11, 2013 at 18:17
  • $\begingroup$ @RobertBryant Would you mind commenting on in what sense 'most' compact Kähler manifolds have a trivial biholomorphism group? $\endgroup$
    – Josh Lackman
    Commented Oct 24 at 9:59
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You can get compact Riemann surfaces of genus $g\geq 3$ (thanks, bryant) whose automorphism group is trivial (in fact, this happens generically). The fundamental group of a generic R.s of genus $g\geq 3$ is in fact a maximal discrete subgroup of $PSL(2,{\mathbb R})$ (I think this is a theorem of Wolpert). That answers 2 and 3.

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    $\begingroup$ Ah, actually, a compact Riemann surface of genus $g=2$ is hyperelliptic, so it always has a nontrivial automorphism group. You should have written '$g\ge3$' in your answer. $\endgroup$
    – Robert Bryant
    Commented Mar 11, 2013 at 15:45
  • $\begingroup$ Thank you very much. So the automorphism group may be bad (?) Btw, I would like to explain my motivation. Consider an entire curve $f\colon \mathbb{C}\to X.$ There is some place in $X$ where we can control the image of the curve. So, is there any way to perturbate the curve, or to move the curve in a reasonable way? $\endgroup$
    – Đức Anh
    Commented Mar 11, 2013 at 15:57
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    $\begingroup$ I can't recall a precise reference, but I recall that there are holomorphic maps $X \to Y$ of compact Kaehler manifolds, with generic fiber a genus 2 compact Riemann surface but with finitely many fibers being reducible curves consisting of elliptic curves and Riemann spheres meeting at various points, and with $Y$ Kobayashi hyperbolic. If I am right about that, then every entire curve has image in those elliptic curves and Riemann spheres, and doesn't move. $\endgroup$
    – Ben McKay
    Commented Mar 11, 2013 at 17:30
  • $\begingroup$ interesting informations! Thank you very much. $\endgroup$
    – Đức Anh
    Commented Mar 11, 2013 at 18:19

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