I've had difficulty finding sources which treat the classification of holomorphic disc bundles over (compact and noncompact) Riemann surfaces. Note that by "bundle", I mean a holomorphic fiber bundle, which means it is locally holomorphically trivial.

I'm really just looking for information about holomorphic disc bundles, but if you want a specific question:

What is the classification of holomorphic disc bundles over a Riemann surface?

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    $\begingroup$ Isn't $PSL_2(\mathbb{R})$ the same as the (oriented) isometry group of the hyperbolic disc? $\endgroup$ – Johannes Ebert Aug 31 '11 at 18:41
  • $\begingroup$ Yes - let me look again at the references and I'll post what I find. $\endgroup$ – Marco Gualtieri Aug 31 '11 at 18:46

There's a paper of H. L. Royden

"Holomorphic fiber bundles with hyperbolic fiber", Proc. AMS, Volume 43, Number 2, April 1974

which proves

"Theorem: The holomorphic fiber bundles with hyperbolic fiber M and base B are in natural one-to-one correspondence with the homomor- phisms of the fundamental group of B into the group of biholomorphic automorphisms of M onto itself. "

In other words, it appears that holomorphic disc bundles are naturally flat.

Note: the first projection $p_1: \mathbb{B}^2\to \mathbb{D}$ of the real 4-ball to the disc is not a holomorphic bundle of discs, because it is not locally trivial. To see why, suppose we have an isomorphism $f:p_1^{-1}(D)\to D\times \mathbb{D}$ of bundles over a sufficently small disc $D$ of radius $r<1$ around zero. Write $f(x,y) = (x, f_2(x,y))$.

Now choose $y_0\in p_1^{-1}(0)$ with $\sqrt{1-r^2} <|y_0|< 1$, so that it is sent via the second projection $p_2:\mathbb{B}^2\to\mathbb{D}$ outside the disc of radius $\sqrt{1-r^2}$.

By assumption, $p_1$ has a section $s$ over $D$ passing through $y_0$ -- this is just a level set of $f_2$. Composing, we obtain $ h=p_2\circ s : D\to \mathbb{D}$. I claim $h$ fails the maximum modulus principle for holomorphic functions. Why? Any circle of sufficiently large radius in $D$ must be sent by $h$ inside a disc of radius strictly less than $|y_0|$ around $0$. Meanwhile, the center of the disc $D$ is sent to $p_2(y_0)$, which has norm equal to $|y_0|$. This goes against the maximum principle or the Gauss mean value theorem, if you like.

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    $\begingroup$ Something is not clear to me: is the projection of the 2-dimensional ball $\mathbb B^2\subset \mathbb C^2$ onto the disc $\mathbb B^2 \to \mathbb D:(z,w)\mapsto z$ a disc bundle? $\endgroup$ – Georges Elencwajg Aug 31 '11 at 20:21
  • $\begingroup$ Wonderful proof! This is reassuring, because if the ball were a disc bundle, Royden's theorem would imply that it is trivial and the ball would be holomorphically isomorphic to a bidisc, which is well known to be false $\endgroup$ – Georges Elencwajg Sep 1 '11 at 17:18

If I don't misunderstand the question, I don't believe that any kind of such classification can exists in the case when the normal bundle of the surface has positive degree. This moduli is infinite-dimensional.

Even in the case of negative degree there are problems. If we take a disk bundle over $\mathbb CP^1$ of degree -1, and contract $\mathbb CP^1$ we will get in particular plenty of complex balls in $\mathbb C^2$. Can one classify them up to biholomorphism?

  • $\begingroup$ Perhaps the issues you raise do not arise with disc bundles, which are equipped with a holomorphic map to the base which is locally trivial? $\endgroup$ – Marco Gualtieri Aug 31 '11 at 21:57
  • $\begingroup$ It may seem crazy but I don't think the Hermitian disc bundle on $\mathcal{O}(-1)$ is actually a locally trivial holomorphic fiber bundle (see my response to Georges above.) $\endgroup$ – Marco Gualtieri Aug 31 '11 at 22:05
  • $\begingroup$ Marco it seems to me that I understand what you say, but it would be rally good if you write down in your question explicitly what kind of bundles you consider. Should they be holomorphically locally trivial? My examples are of course not of this type. And it is not surprising that hololomorphically locally trivial guys over surfaces are flat. $\endgroup$ – Dmitri Panov Sep 1 '11 at 7:10

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