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  Extending General Relativity with Kahler Manifolds?

+ 6 like - 0 dislike
109 views

Standard general relativity is based on Riemannian manifolds.

However, the simplest extension of Riemannian manifolds seems to be Kahler manifolds, which have a complex (hermitian) structure, a symplectic structure, a riemannian structure, all these structures being compatible. So my question is :

What is the "simplest" theory, extending general relativity, based on Kahler manifolds ?


This post imported from StackExchange Physics at 2016-01-07 09:54 (UTC), posted by SE-user Trimok

asked Nov 5, 2012 in Theoretical Physics by Trimok (950 points) [ revision history ]
edited Jan 7, 2016 by Dilaton
Kähler manifolds are no "extension" of Riemannian ones! Every Kähler manifold is Riemannian, but not every Riemannian manifold is Kähler. So we are speaking about a (small) subset of Riemannian manifolds, for which the default general relativity-formalism naturally exists. I think I didn't understand your question :)

This post imported from StackExchange Physics at 2016-01-07 09:54 (UTC), posted by SE-user Tobias Diez
@altertoby: What I mean is that the compatible complex hermitian, symplectic and riemannian structures of Kahler manifolds, are an extension of the simple riemannian structure of Riemannian manifoldS. So I am interesting with theories which make use of all these compatible structures of Kahler manifolds, and not only the Riemannian structure part.

This post imported from StackExchange Physics at 2016-01-07 09:54 (UTC), posted by SE-user Trimok
Giampiero Esposito's "Complex Geometry of Nature and General Relativity" arXiv:gr-qc/9911051 reviews some recent approaches to complex differential geometric treatments of GR, which may be interesting to the OP...although by no means are they the "simplest" approach!

This post imported from StackExchange Physics at 2016-01-07 09:54 (UTC), posted by SE-user Alex Nelson

3 Answers

+ 4 like - 0 dislike

A Kahler manifold is essentially a complex manifold with a metric that may be written as,

$$g_{ij}=\frac{\partial^2}{\partial z^i \partial z^j}\mathcal{K}$$

where $\mathcal{K}$ is denoted the Kahler potential. The form of $g_{ij}$ ensures the Kahler form (which is a natural choice for a volume form) is closed. Kahler manifolds are a specific type of Riemannian manifold and hence are not extensions of the concept of a Riemannian manifold. As such, if we wanted to perform general relativity on a Kahler manifold, it would require no modification of the field equations,$^{\dagger}$

$$R_{ij}-\frac{1}{2}g_{ij}R = 8\pi T_{ij}$$

I would think it is not sensible to take a spacetime manifold to be a complex manifold, as in reality we are 'living' on real pseudo-Riemannian manifolds. However, sometimes it is convenient to use complex coordinates. For example, in string theory, we define new complex wordsheet coordinates,

$$z=\tau+i\sigma \quad \bar{z}=\tau-i\sigma$$

which are Euclidean analogues of lightcone coordinates. If we consider $z$ and $\bar{z}$ as separate independent variables, then we are going from $\mathbb{R}^2$ to $\mathbb{C}^2$. However, at the end, we need to keep in mind we are on the real slice which is a subset, i.e. $\mathbb{R^2}\subset \mathbb{C}^2$. I guess one could apply a similar procedure to a solution to the Einstein field equations, and in that sense we are, in a manner, temporarily extending ourselves to a complex manifold.


$\dagger$ In units where $c=G=1$.

This post imported from StackExchange Physics at 2016-01-07 09:54 (UTC), posted by SE-user JamalS
answered May 13, 2014 by JamalS (885 points) [ no revision ]
+ 4 like - 0 dislike

Kähler manifolds are not an extension of Riemannian manifolds. On the contrary, they are a restriction : every Kähler is Riemannian, but the opposite certainly isn't the case.

Actually, being Kähler even imposes some very strict topological restrictions on the manifold (such as all even rank de Rham cohomology groups being nonzero), whereas any differentiable manifold can be given a Riemannian structure.

This post imported from StackExchange Physics at 2016-01-07 09:54 (UTC), posted by SE-user Youloush
answered May 13, 2014 by Youloush (40 points) [ no revision ]
In the comment to his own question the original poster clarifies what he means to ask. Btw, the Kähler condition certainly imposes topological restrictions, but not that odd cohomology groups vanish. E.g. for a Riemann surface the rank of the first cohomology group, equal to twice the genus, is the topological invariant (I am sure you know that, maybe you meant something else).

This post imported from StackExchange Physics at 2016-01-07 09:54 (UTC), posted by SE-user doetoe
Ouch that's corrected.

This post imported from StackExchange Physics at 2016-01-07 09:54 (UTC), posted by SE-user Youloush
+ 1 like - 0 dislike

Years ago I came across a second hand book (Springer lecture notes in physics 46) called "Hermitian and Kählerian Geometry in Relativity" by E.J. Flaherty. I bought it, and although I never read it I think it addresses your question to a certain extent. Skimming the text, it looks like the following is established: spacetime (having a metric of Lorentz signature) doesn't admit an almost Hermitian structure, in particular it doesn't admit a Kählerian structure, but the author proposes a modification by means of a complex Lorentz transformation which gives rise to a complex valued Hermitian structure that formally behaves just like an ordinary Hermitian structure and is compatible with the metric. This allows for an adaptation of some results in Kähler geometry to what he calls Kähler spacetimes.

This post imported from StackExchange Physics at 2016-01-07 09:54 (UTC), posted by SE-user doetoe
answered May 13, 2014 by doetoe (125 points) [ no revision ]
+1 for the informations.

This post imported from StackExchange Physics at 2016-01-07 09:54 (UTC), posted by SE-user Trimok

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