# Algebraic (semi-) Riemannian geometry ?

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 $X$.

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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. –  Harry Gindi Mar 25 '10 at 18:22
@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. –  Qfwfq Mar 25 '10 at 18:33
(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. –  Qfwfq Mar 25 '10 at 18:40
I think this is a very very good question. I hope that it gets good answers. –  Kevin H. Lin Mar 26 '10 at 8:33
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. –  some guy on the street Apr 15 '10 at 16:02

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.

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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"? –  Qfwfq Apr 5 '10 at 10:12

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

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Chapter 4.3 seems to be the relevant starting point –  Konrad Voelkel Nov 24 '10 at 19:48

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.

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