Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question

2 Answers 2

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?

share|improve this answer
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. –  Đức Anh Mar 11 '13 at 15:54
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. –  Robert Bryant Mar 11 '13 at 16:07
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 –  Ben McKay Mar 11 '13 at 17:24
Thank you all very much for useful informations. I should try another approach to my problems. –  Đức Anh Mar 11 '13 at 18:17

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.

share|improve this answer
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. –  Robert Bryant Mar 11 '13 at 15:45
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? –  Đức Anh Mar 11 '13 at 15:57
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. –  Ben McKay Mar 11 '13 at 17:30
interesting informations! Thank you very much. –  Đức Anh Mar 11 '13 at 18:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.