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Compact manifold taken as an Einstein Manifold

+ 2 like - 0 dislike

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 (720 points) [ no revision ]

2 Answers

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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 (4,660 points) [ revision history ]
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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 (1,950 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?  

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