Even and odd structures on supermanifolds

[ajb]

We are all aware of the importance of geometry in the formulation of modern physics. Prime examples include general relativity and Yang-mills theory.

 

Supermanifolds owe their conception in quantum physics where a semi-classical understanding of fermions is required. Specifically, the notion of how to integrate fermions in needed in the path integral formulation.

 

Loosely, one should think of supermanifolds as "manifolds" with both commuting and anticommuting coordinates. In the physical language that is both bosonic and fermonic degrees of freedom. Via the inclusion of minus signs due to the anticommuting nature of some of the coordinates most of the theory of classical manifolds carries over to supermanifolds.

 

A little more correctly, we are working in the [math]\mathbb{Z}_{2}[/math]-graded category. That is everything comes in one of two "flavours": bosonic or fermionic, aka even and odd.

 

Now, the classical structures of symplectic forms and Riemannian metrics carries over to supermanifolds very directly. The only difference now is that we can consider both even and odd structures. On a manifold only even structures exist.

 

Here is the interesting part. A kind of maths-physics "dictionary",

 

1) Even symplectic geometry <-> Classical mechanics.

2) Odd symplectic geometry <-> Classical field theory and perturbative quantum field theory.

3) Even (pesudo)Riemannian geometry <-> metric theories of gravity, general relativity and supergravity.

4) Odd Riemannian geometry <-> ???

 

We see something is missing. What does odd Riemannian geometry correspond to? As far as I know, no application in physics has been found. This seems strange to me. Their is a piece of the "dictionary" missing.

 

It turns out that even and odd Reimannian geometry are for the most part identical. The Levi-Civita theorem holds, the Reimannian curvature tensors are both even and have the same symmetries. However at the level of the Ricci tensor things start to diverge. The Ricci tensor is even/odd for an even/odd metric and has different symmetries. Then their is a remarkable theorem. The Ricci scalar for an odd metric is necessarily constant. (see Asorey and Lavrovb arXiv:0805.2297 for example).

 

This makes me wonder about the notion of odd Einstein field equations and odd Einstein manifolds, all things for later.

 

Anyway, given the success of the analogues of classical symplectic and even Riemannian structures I "conjecture" that odd Riemannian structures will find an application in physics.

 

As a final remark, I have only listed homogeneous structures of "degree two": sympelctic structures are 2-forms, metric are also 2-tensors. On supermanifolds there appears structures that are just either even or odd and otherwise inhomogeneous. Some of these have also found applications in physics.

---

Merged post follows:

Consecutive posts merged

Another important departure is the Laplace--Beltrami operator.

 

With a slight abuse of language, we will think of "degenerate" Riemannian metrics. In particular, we will not worry about inverting such things and we will assume an independent volume [math]\underline{\rho}[/math] is supplied. In local coordinates we have [math]\underline{\rho} = \rho(x)D(x)[/math], where [math]D(x)[/math] is the "coordinate volume".

 

So, you can think of an even Riemannian metric as [math]g^{AB}[/math] such that [math]g^{AB} = (-1)^{\widetilde{A}\widetilde{B}}g^{BA}[/math]. (see my thread here for some notation). You can think of it as a function on [math]T^{*}M[/math] that is quadratic in fibre coordinates.

 

Similarly, an odd Riemannian metric can be thought of as [math]\chi^{AB}[/math] such that [math]\chi^{AB} = (-1)^{(\widetilde{A}+1)(\widetilde{B}+1)}\chi^{BA}[/math]. Then it can be thought of as a quadratic on [math]\Pi T^{*}M[/math].

 

Let us look at the even case first. Up to signs we will define the even Laplace--Beltrami operator as

 

[math]\triangle_{\underline{\rho}}^{0} f = \pm \frac{1}{\rho} \frac{\partial}{\partial x^{A}}\left(\rho g^{AB}\frac{\partial f}{\partial x^{B}} \right)[/math],

 

where [math]f[/math] is a function on our supermanifold. Expanding this out we get

 

[math]\triangle_{\underline{\rho}}^{0} f = \pm g^{AB}\frac{\partial^{2}f}{\partial x^{B}\partial x^{A}} \; +[/math] lower order terms.

 

The even Laplace--Beltrami operator is second order. This agrees completely with classical geometry.

 

Now for the odd Laplace--Beltrami operator.

 

[math]\triangle_{\underline{\rho}}^{1} f = \pm \frac{1}{\rho} \frac{\partial}{\partial x^{A}}\left(\rho \chi^{AB}\frac{\partial f}{\partial x^{B}} \right)[/math],

 

which looks the same so far. Now expand as before

 

[math]\triangle_{\underline{\rho}}^{1} f = \pm \chi^{AB}\frac{\partial^{2}f}{\partial x^{B} \partial x^{A}} \pm \frac{1}{\rho}\frac{\partial}{\partial x^{A}} \left( \rho \chi^{AB} \right)\frac{\partial f}{\partial x^{B}}[/math].

 

Now, due to the symmetric properties of the derivatives and the antisymmetric properties of the odd metric the second order term vanishes. (You are contracting over something symmetric and antisymmetric thus it vanishes). Then

 

[math]\triangle_{\underline{\rho}}^{1} f = \pm \frac{1}{\rho}\frac{\partial}{\partial x^{A}} \left( \rho \chi^{AB} \right)\frac{\partial f}{\partial x^{B}}[/math]

 

which shows that the odd Laplace--Beltrami operator is a first order differential operator and not second order as in the even case.

 

So, the odd analogue of the Poisson and Laplace equation on a supermanifold is going to be first order and not second!

[michel123456]

It seems you are flying too high AJB.

I like the "Their is a piece of the "dictionary" missing." but i can't help you on this, except agreeing with you: there is a piece missing.

[ajb]

It seems you are flying too high AJB.

I like the "Their is a piece of the "dictionary" missing." but i can't help you on this, except agreeing with you: there is a piece missing.

 

Didn't expect much help, just wondered if anyone else was interested .

 

I may post more on odd Riemannian geometry as I work more out for myself.

 

It does seem that from the point of view of connection theory both even and odd Riemannian geometry work as well as each other. Only when we get to the Ricci tensor and scalar curvature does the basic geometry diverge.

 

The odd Laplace--Beltrami operators are like modular vector fields on Poisson manifolds. I don't know if we have theory of modular classes. I wonder if something goes wrong because of the parity.

 

Geometrically we can think of an odd Riemannian structure as defining a fibrewise map

 

[math]TM \rightarrow \Pi T^{*}M[/math]

[ajb]

Let us see how far we can push the analogy between odd Riemannian geometry and Poisson geometry.

 

Mimicking Poisson geometry I define the odd metric bracket as

 

[math]<f|g> = (-1)^{\widetilde{A}(\widetilde{f}+1)}\frac{\partial f}{\partial x^{A}} \chi^{AB}\frac{\partial g}{\partial x^{B}}[/math],

 

with [math]f,g \in C^{\infty}(M)[/math]. This bracket is odd, and antisymmetric

 

[math]<f|g> = (-1)^{(\widetilde{f}+1)(\widetilde{g}+1)}<g|f>[/math].

 

Warning This bracket does not satisfy any Jacobi-like identities. In particular, it does not define a Lie algebra.

 

We can then define an odd metric vector field viz

 

[math]<f|g> = (-1)^{\widetilde{f}}X_{f}[g][/math].

 

In local coordinates we have

 

[math]X_{f}^{A} = (-1)^{\widetilde{f} + \widetilde{B}(\widetilde{f}+1)}\frac{\partial f}{\partial x^{B}} \chi^{BA}[/math].

 

Warning This does not provide a map as Lie algebras between the metric brackets and the Lie bracket on vector fields. Simply as the metric bracket is not a Lie bracket. Thus we have to be more careful than simply apply some of the well-known identities in Poisson geometry.

 

Given a density we can proceed to define the odd Poisson--Beltrami operator viz

 

[math]L_{X_{f}} \underline{\rho} = (\triangle_{\rho}f) \underline{\rho}[/math].

 

Statement [math]\triangle_{\rho}[/math] is an odd vector field.

 

I will skip details about Lie derivatives of densities on supermanifolds, but the theory is essentially the same as for manifolds. Just to quote the result (same as earlier posts)

 

[math]\triangle_{\rho} = (-1)^{\widetilde{A}+1}\left ( \frac{\partial \chi^{AB}}{\partial x^{A}} + \frac{\log(\rho)}{\partial x^{A}}\chi^{AB} \right)\frac{\partial }{\partial x^{B}}[/math].

 

Proposition

Under changes of density of the form

 

[math] \rho \rightarrow \rho' = e^{- \kappa(x)}\rho[/math]

 

the respective odd Poisson--Beltrami operators differ by a odd metric vector field.

 

I won't prove the above in detail, just give you the result.

 

[math]\triangle_{\rho}- \triangle_{\rho'} = X_{\kappa} = (-1)^{\widetilde{A}}\frac{\partial \kappa}{\partial x^{A}}\chi^{AB}\frac{\partial}{\partial x^{B}}[/math].

 

So, at this stage it looks like the ideas of Weinstein's modular classes seems to apply to odd Riemannian geometry. However, I do not think there is a good notion of "odd Riemannian cohomology". Thus, we do not really have a theory of modular classes.

 

Proposition

 

[math]L_{\triangle_{\rho}}\underline{\rho} =0[/math]

 

I won't give a proof as it is just a boring calculation in local coordinates. The statement is that the odd Poisson--Beltrami operator (with respect to a given density) generates infinitesimal diffeomorpisms that preserve the density. They are "volume preserving".

 

What I have said so far is not surprising or unexpected. All we have done is lost some of the features of Poisson geometry, but it seems plenty of them remain.

 

There are a few other things I will be thinking about. I expect there is some further important roles of the odd Poisson--Beltrami operator in odd Riemannian geometry.

 

To recap. Odd Riemannian geometry looks like Poisson geometry, but the brackets are not Lie algebras and we do not have a good theory of modular classes.

 

(Thanks for reading this!)

Edited January 21, 2010 by ajb

[rozoord]

We analyze from a general perspective all possible supersymmetric generalizations of symplectic and metric structures on smooth manifolds. There are two different types of structures according to the even/odd character of the corresponding quadratic tensors. In general we can have even/odd symplectic supermanifolds, Fedosov supermanifolds and Riemannian supermanifolds.

 

The geometry of even Fedosov supermanifolds is strongly constrained and has to be flat. In the odd case, the scalar curvature is only constrained by Bianchi identities. However, we show that odd Riemannian supermanifolds can only have constant scalar curvature. We also point out that the supersymmetric generalizations of AdS space do not exist in the odd case.

 

need tent structures

[ajb]

rozoord, thanks for posting the abstract of a paper by Asorey and collaborators. I am aware of the papers they wrote. I have not really thought much about Fedosov manifolds and geometric quantisation.

 

Anyway, let us continue.

 

Another question that naturally arises is if there is an odd metric version of the Koszul--Schouten bracket on differential forms. See my preprint arXiv:0910.1992v1 [math-ph] for a discussion of such brackets for the higher Poisson case. The point is that this construction does pass over to an odd Riemannian metric.

 

However, not unexpectedly the resulting brackets on differential forms do not form a Lie algebra.

 

So, does any of what I have said explain why odd Riemannian manifolds have yet to find a place in physics? I say partly.

 

From the point of view of classical Riemannian geometry, even and odd metrics look very similar, only when you get to the Ricci tensor and Ricci scalar do things start to look different.

 

From the point of view of classical Poisson geometry any brackets one can naturally form are not Lie brackets. The Laplace--Beltrami operator is first order and looks akin to modular vector fields, but there is no simple notion of a relevant cohomology here. (The direct analogue of the Poisson coboundary operator does not work ).

 

These gives us a clue as to why mathematicians have not really explored odd Riemannian geometry, also not many know it exists. Largely the theory looks boring.

 

Can this really be enough to explain why no physics?