# Algebraic Riemannian manifolds?

Some of us, especially those of us who have been hanging out with Nick Proudfoot and Roman Bezrukavnikov, are familiar with the notion of an algebraic symplectic manifold. Let me review this concept.

Recall that a symplectic manifold is a manifold along with a non-degenerate closed 2-form. An algebraic symplectic manifold is smooth algebraic variety over C (or any other field), again with a non-degenerate closed 2-form. In particular algebraic symplectic manifolds are always even complex dimensional and hence their real dimension is a multiple of 4. They are closely related to the differential geometry notion of hyperKahler manifold. There are many nice exampls of algebraic symplectic manifolds. The simplest is the cotangent bundle of a smooth variety.

The other day I was wondering whether anyone has ever considered “algebraic Riemannian manifolds”. By this I mean a smooth algebraic variety with a non-degenerate symmetric bilinear form on each tangent space (note that we drop positive definite, since this makes no sense over C).

I don’t really have much to say about such things, I just wanted to throw it out here to see if anyone can tell me whether such a concept exists or if not, why it is a bad idea to consider such a notion. So any answers?

## 19 thoughts on “Algebraic Riemannian manifolds?”

1. Allen Knutson says:

Just as in the symplectic case, these guys are automatically Calabi-Yau, right? Or maybe having a nonvanishing volume form isn’t considered enough to be C-Y, you need be Kahler as well?

2. I don’t see why the canonical bundle is trivial, but perhaps I am not thinking right.

One similarity to the symplectic case is that there will be Lagrangians (at least when the overall dimension is even). Moreover those Lagrangians will have their normal bundles isomorphic to their contangent bundles (just as in the symplectic case). This is actually why I was thinking about these guys.

3. The canonical bundle is trivial for the same reason it’s trivial on an algebraic symplectic manifold: the top wedge power of the Riemannian form is a holomorphic volume form.

4. But the Riemannian form is not a 2-form. It’s a section of $Sym^2(T^\star M)$, not $\Lambda^2$. So you can produce sections of $\Lambda^k Sym^2$ but that is not what you want.

5. Chris Brav says:

A nowhere vanishing section of $Sym^{2}(T^{\ast}M)$ is part of the data of a Frobenius manifold, so I’d look at Manin’s book on quantum cohomology or Dubrovin’s notes on TQFT for examples of where such a thing comes up in algebraic geometry.

6. Good point.

Now that I look again, it seems you might not be guaranteed to have trivial canonical bundle, just that the canonical and anti-canonical bundles are isomorphic. Thus, the canonical bundle just has to be order 2 in the Picard group.

7. Chris Brav says:

I like Ben’s last comment. This means that our variety already has trivial canonical bundle or is double-covered by a variety with trivial canonical bundle. In other words, it is a so-called canonical quotient of order two.
Canonical quotients are described nicely in Huybrechts’s book on Fourier-Mukai transforms.

I don’t see immediately whether the converse is true. If a variety is double covered by a Calabi-Yau, must it admit the desired kind of metric?

8. I don’t see immediately whether the converse is true. If a variety is double covered by a Calabi-Yau, must it admit the desired kind of metric?

That seems pretty optimistic, though I’ll admit, I can’t think of an obstruction right this second.

9. Such a metric would imply that the tangent bundle was isomorphic to its dual, right? We should be able to come up with a CY example where this doesn’t hold. Being isomorphic to your dual means that all the odd Chern classes are 2-torsion, while the condition of being 2-covered by a CY just means that the first chern class is 2-torsion. So there might be a good counter-example in CY-three folds.

10. I’m sure it is not true that all CY or all varieties with canonical and anticanonical bundle isomorphic are of this form. I’m just curious whether any interesting examples of algebraic Riemannian manifolds exists or if there is some reason why not.

11. Is every positive genus curve Riemannian? If I’m not mistaken, Sym^2 Omega is the tensor square of Omega, and it has a 2g dimensional space of sections.

12. Scott —

As I understood the rules of the game, we are requiring the “metric” to be everywhere nondegenerate. For genus two or greater, all of those sections vanish somewhere.

Here is a question that strikes me as somewhat interesting: Is there an example of an algebraic variety where Wedge^2 T^* admits an everywhere nondegenerate section but Sym^2 T^* doesn’t? All of the topological obstructions I can think of really obstructions to (T^*)^{\otimes 2} having a nondegerate section.

(It is easy to construct examples in the reverse direction, where Sym^2 T^* has a nondegen section and Wedge^2 T^* does not. Take something odd dimensional, or take a two-fold quotient of a CY which is not itself CY.)

13. David-

Since the canonical bundle is of finite order in Picard group, it must be degree 0, no? That alone rules all non-elliptic curves.

14. I think David was pointing out a specific gap in my reasoning rather than the more holistic crash-and-burn that the Picard group provides.

Do all of these manifolds admit a Ricci-flat metric (as real manifolds)?

15. Given stuff that John Nash proved, I bet any smooth Riemannian metric on a compact manifold can be approximated arbitrarily well in the C-infinity topology by a smooth real affine algebraic variety with its induced metric (coming from the usual Euclidean metric on R^n).

Does anyone know if this is true?

(I know you guys are talking about complex algebraic Riemannian manifolds, but we can also talk about real ones.)

16. I thought that admitting a Ricci-flat metric as a real manifold (at least in the Kahler case) was just equivalent to being CY. But we already agreed that (probably) CY is not required to have such a section.

17. Oops, I was thinking if you have a Ricci-flat complex connection then that means CY. Sorry.

18. Given stuff that John Nash proved, I bet any smooth Riemannian metric on a compact manifold can be approximated arbitrarily well in the C-infinity topology by a smooth real affine algebraic variety with its induced metric (coming from the usual Euclidean metric on R^n).

Nash proved that every compact, smooth manifold is a component of a smooth real algebraic variety. You can prove this result using the methods of Whitney to improve a C^1 manifold M to a smooth manifold or even real analytic. Namely, embed M in Euclidean space, then realize it as a non-singular solution set to smooth equations, then approximate the smooth equations by polynomials. Nash then merely conjectured that every compact, smooth manifold is all of a smooth real algebraic variety, i.e., that the variety is connected. This conjecture was proved by Tognoli.

There is in this theory something called Nash structure, which is a weird cross between an algebraic variety and a more usual kind of manifold. Namely, the manifold has an analytic atlas of charts, rather than a Zariski atlas, but the gluing maps are required to be real algebraic. Every compact, smooth manifold M has a ton of Nash structures; but again by the Whitney-type arguments, there is a unique Nash structure relative to which M embeds in Euclidean space.

Every Nash structure induces a real analytic structure. It is a fantastic theorem of Morrey and Grauert that all paracompact real analytic manifolds embed in Euclidean space, with the consequence that if they are diffeomorphic, they are real-analytic-omorphic.

Exercise to understand these ideas: R/Z inherits a Nash structure from R, but not the same Nash structure as the unit circle in R^2. But they are the same real analytic structure.

19. Wayne says:

Would this become interesting if we consider a variety over a field with nonzero characteristic? I think at least some Riemannian geometry concepts in is known for some time, but for symplectic geometry I am not sure.