Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,352 answers , 22,786 comments
1,470 users with positive rep
820 active unimported users
More ...

  Compact manifold taken as an Einstein Manifold

+ 2 like - 0 dislike
880 views

In Kaluza-Klein theories I often see that the compact space is assumed to be an Einstein manifold, that is, its Ricci tensor is proportional to its metric. 

So, why is this done?

asked Apr 25, 2014 in Theoretical Physics by Dmitry hand me the Kalashnikov (735 points) [ no revision ]

2 Answers

+ 2 like - 0 dislike

Let K denotes the compact space in a Kaluza-Klein theory. In its simplest form, Kaluza-Klein theory is a pure gravity theory on MxK where M is the Minkowski space (or dS or AdS depending on the cosmological inclination). At low energy, pure gravity is general relativity. A vacuum of such a theory should be solution of Einstein equations. In a Kaluza-Klein theory, one generally assumes that the metric on MxK is a product of the metric on M by a metric on K, so that the metric on K should be a solution of Einstein equations. But a solution of Einstein equations (possibly with a cosomological constant) is by definition the same as an Einstein metric. To take for the compact space a Einstein manifold is simply to take a solution of Einstein equations.

Another remark: on of the goals of a Kaluza-Klein theory is to produce a gauge theory  of group G in the non-compact space M. It is possible by taking for K a Riemannian manifold of isometry group G. The simplest way to construct such a K is to take for a G-invariant metric on a quotient K = G/H (such that G acts transitively on G/H and H is the isotropy subgroupr of this action). In many cases (for example if the action of H on the tangent space at a given point of K is irreducible), such a metric is necessarely Einstein (the metric and the Ricci curvature are two quadratic forms on the tangent space at a point of K, both invariant under the action of H. If the action is irreducible, they necessarely are proportional). This explains why Einstein metrics naturally happen when one searches for metric with many symmetries, as it is the case in Kaluza-Klein theory.

One can show that a compact Einstein manifold of negative cosmological constant has no continous isometry group. This implies that the most relevant Einstein manifolds for a Kaluza-Klein theory are those of non-negative cosmological constant. This is the case for the Einstein metrics G-invariant on K=G/H.

Many interesting Einstein metrics are not of constant curvature, for example the projective spaces. 

answered Apr 25, 2014 by 40227 (5,140 points) [ revision history ]
+ 0 like - 0 dislike

If I recall correctly, it is assumed that the manifold has a constant Riemann curvature, which automatically implies that the manifold is an Einstein manifold. The assumption that the manifold has a constant Riemann curvature has to do with creating new symmetries.  

answered Apr 25, 2014 by dimension10 (1,985 points) [ no revision ]

Can the constant have any value ...? I remember having heard about some issues with curved manyfilds, leading to a nonvanishing Ricci flow etc ...

@Dilaton I don't understand what you mean. The ricci flow expands negatively curved regions and shrinks positively curved regions; the normalised ricci flow surely vanishes for constantly curved manifolds, doesn't it?  

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ys$\varnothing$csOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...