30
$\begingroup$

I hope these are not to vague questions for MO.

Is there an analog of the concept of a Riemannian metric, in algebraic geometry?

Of course, transporting things literally from the differential geometric context, we have to forget about the notion of positive definiteness, cause a bare field has no ordering. So perhaps we're looking to an algebro geometric analog of semi- Riemannian geometry.

Suppose to consider a pair $(X,g)$, where $X$ is a (perhaps smooth) variety and $g$ is a nondegenerate section of the second symmetric power of the tangent bundle (or sheaf) of $X$.

What can be said about this structure? Can some results of DG be reproduced in this context? Is there a literature about this things?

Improve this question
$\endgroup$
5
  • $\begingroup$ It is my impression that a lot of results in Riemannian geometry rely on partitions of unity, which don't exist in the algebraic or even holomorphic cases. $\endgroup$
    – Harry Gindi
    Mar 25, 2010 at 18:22
  • 1
    $\begingroup$ @fpqc: A lot of results of differential topology rely on partitions of unity. - You do have a well developped theory of holomorphic-symplectic (and algebraic-symplectic) manifolds, though. $\endgroup$
    – Qfwfq
    Mar 25, 2010 at 18:33
  • 1
    $\begingroup$ (continued) What I wanted to say, actually, is that you may have a sensible notion of "algebraic semiRiemannian manifold" (after all, you have algebraic-symplectic manifolds) even though global existence of such a structure is not a priory granted (whereas in the differentiable category you can always have global existence of tensors with "convex" properties just by partitions of unity). So even the global existence of such a structure would impose -I guess- severe restrictions on the variety, as it happens in the alg.-symplectic case. $\endgroup$
    – Qfwfq
    Mar 25, 2010 at 18:40
  • $\begingroup$ I think this is a very very good question. I hope that it gets good answers. $\endgroup$
    – Kevin H. Lin
    Mar 26, 2010 at 8:33
  • $\begingroup$ From a lecture of Yom-Tung Siu, differential geometers tend to prove theorems by taking smooth approximations or resolutions of singular things; when seeking analogous results in algebraic geometry, the tendency is to try to concentrate curvature in a subvariety of lower dimension. Not being an algebraic geometer myself, I can't (alas) produce a clear example of this practice off the top of my head. $\endgroup$
    – some guy on the street
    Apr 15, 2010 at 16:02

4 Answers 4

Reset to default
9
$\begingroup$

Joel Kamnitzer had a very similar question a couple years ago, that prompted a nice discussion at the Secret Blogging Seminar. I'm afraid no one ended up citing any literature, and I have been unable to find anything with a quick Google search, but that doesn't rule out the possibility of existence.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Thanks! I think it's very spontaneous question: given that there are algebraic analogues of symplectic forms, why not to consider the algebraic analogue of the (perhaps) more intuitive structure of a "metric"? $\endgroup$
    – Qfwfq
    Apr 5, 2010 at 10:12
5
$\begingroup$

If a holomorphic Riemannian metric $g=g_{ij}(z) dz^i dz^j$ on a compact Kaehler manifold $X$ is everywhere nondegenerate, then the metric has a holomorphic Levi-Civita connection, so the Atiyah class of the tangent bundle of $X$ is zero. Therefore a finite etale cover of $X$ is a complex torus, and the metric pulls back to be translation invariant. Hence $X$ is a quotient of such a torus by a finite group acting as affine isometries without fixed points. On the other hand, if you allow degeneracies of the holomorphic Riemannian metric, I suppose anything could happen. If you allow $X$ to be a compact complex manifold, perhaps not Kaehler, then you might look at the papers of Sorin Dumitrescu where you find a low dimensional classification.

Improve this answer
$\endgroup$
2
  • $\begingroup$ So, in a sense, a holomorphic Riemannian Kaehler manifold has to be "flat" right? That may be a reason why people in integrable systems -if I'm not mistaken- only consider Frobenius manifolds (of which a piece of data is a metric, holomorphic in the above sense in the holomorphic setting) that are flat? - just wondering $\endgroup$
    – Qfwfq
    Jun 18, 2016 at 15:12
  • 1
    $\begingroup$ Yes, it has to be flat. Frobenius manifolds are not required to be compact or Kaehler, so I don't think there is any relationship. In low dimensional (and perhaps in any dimension), any holomorphic Riemannian metric on any compact complex manifold has constant curvature, even if the manifold is not Kaehler. $\endgroup$
    – Ben McKay
    Jun 19, 2016 at 11:32
4
$\begingroup$

This topic in the affine case is extensively studied in Ernst Kunz unpublished book "Algebraic Differential Calculus". You can get it as a collection of several PS files at his webpage (scroll to the bottom):

Kunz' webpage

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Chapter 4.3 seems to be the relevant starting point $\endgroup$
    – Konrad Voelkel
    Nov 24, 2010 at 19:48
4
$\begingroup$

It is also rather natural to look at holomorphic conformal structure given locally by holomorphic Riemannian metrics up to multiplication by invertible functions. More precisely, a holomorphic conformal structure is given by a nowhere degenerate section $\omega \in H^0(X,Sym^2\Omega^1_X \otimes \mathcal L)$ where $\mathcal L$ is a line-bundle.

The classification of holomorphic conformal structures on compact complex surfaces is carried out by Kobayashi and Ochiai in Holomorphic structures modeled after hyperquadratics, Tohoku Math. J. 34, 587-629 (1982).

There is also a classification in the case of projective $3$-folds in this paper.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ The flat holomorphic conformal structures on smooth complex projective varieties are classified by Jahnke and Radloff here: arxiv.org/abs/1502.07843 $\endgroup$
    – Ben McKay
    Jul 25, 2017 at 8:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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