# Why the moduli space of complex structure in a compact complex manifold is of finite dimension

I admit the statement in the title might be much too unclear. I just heard from my teacher that we can form a finite dimension moduli space of all the complex structure in a compact complex manifold. He said it is a standart fact in complex geometry but I failed to find out both the correct statement and the proof. So I'm asking for both a theorem you think is closest to this kind of meaning and a reference of its proof. Thank everyone in advance. (Should I add some more tags for such kind of vague question?)

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By Kodaira-Spencer theory, you can compute the space of first-order deformations of the complex structure as a cohomology group. For a fixed choice of complex structure, this space is (more or less) the tangent space of the moduli space of complex structures at the point that describes the chosen complex. structure. – Scott Carnahan Aug 1 '11 at 15:06
Any references about this? And I still have no idea how to put all the complex structure together.(Its topology? Manifold structure?) – Honglu Aug 2 '11 at 3:55
I do not know the answer, but i can add that in general such a moduli space need not be a manifold. In the case of a torus (which is the edge of my knowledge) it is an "orbifold", which is roughly supposed to be a manifold with corners. To see a picture, see en.wikipedia.org/wiki/…, but glueing around the edges in some way is required.. – Joachim Sep 23 '12 at 17:21
the space of all complex structure on a compact complex manifold is an infinite dimensional complex manifold, with the $L^2$ norm defined by using the Beltrami differenctial, it can be shown that this infinite dimensional space is actually kahler, but i can not see how it could be finite dimensional. – Vicky Cheung Sep 24 '13 at 2:06

I just asked this question to Siu at Harvard. He responded immediately that Brieskorn gave an infinite diml example in math annalen 1965. Deformation theory implies the space is locally finite dimensional, but the local dimension goes to infinity in Brieskorn's example.Dennis Sullivan

Here is Yum Tong Siu's precise response:

If a complex structure is required for the space of complex structures, one can only use Kuranishi's semi-universal moduli space. For such a setup, there is the following situation.

One starts out with a compact complex algebraic manifold X_1 and forms its semi-universal moduli space M_1 whose dimension at the point corresponding to X_1 is d_1.

Along M_1 the manifold X_1 is deformed to X_2 whose semi-universal moduli space M_2 has dimension d_2 at the point corresponding to X_2.

Then one continues the process and gets d_{j+1}>d_j to end up with a topologically connected component whose analytic branches have dimension going to infinity.

A concrete example for this situation is Briskorn's generalization of Hirzebruch surfaces in Math. Annalen 157 (1965), 343-357.

The relevant statements in Brieskorn's 1965 paper are:

Satz 2.6 on p.349

Satz 6.1 i) on p.354

Satz 6.2 on p.355.

Yum Tang Siu.

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Wow... I asked this question while I was an undergrad... Great example and great timing, as I can understand quite better than 5 years ago. Thanks! – Honglu Jun 15 at 1:59