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  Is the distinction between the Poincaré group and other internal symmetry groups artificial?

+ 2 like - 0 dislike

For instance, given a theory and a formulation thereof in terms of a principal bundle with a Lie group $G$ as its fiber and spacetime as its base manifold, would a principle bundle with the Poincaré group as its fiber and $\mathcal{M}$ as its base manifold, where $\mathcal{M}$ is a manifold the group of whose isometries is $G$, lead to an equivalent formulation? Why? Why not?

On a related note, can any Lie group be realized as the group of isometries of some manifold?

This post has been migrated from (A51.SE)
asked Nov 8, 2011 in Theoretical Physics by Arpan Saha (50 points) [ no revision ]
On the related note, the answer is trivially true and not very interesting. Any finite-dimensional Lie group is a group of isometries of any left-invariant metric on its underlying manifold.

This post has been migrated from (A51.SE)
Thank you. Yes, that is trivial indeed. Sorry, didn't see that.

This post has been migrated from (A51.SE)
I don't understand why you expect this sort of duality to exist. Normally, switching spacetime and the target space in a non-linear sigma model leads to something completely differently. Mappings from M to N are one thing, mappings from N to M completely other.

This post has been migrated from (A51.SE)
I am just wondering - isn't trying to write a theory with the Poincare group as the fibre on the space-time the "same" as doing Einstein's gravity? Of course the later part of the question doesn't make sense to me - I mean in the usual theory of connections on some G-bundle (i.e Yang-Mill's theory!) I don't see how the gauge group is acting as isometry on the space-time!? That doesn't look right at all. .

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I would think that the vielbein language of gravity is sort of the correct formulation in which these two pictures are manifest that in some sense $G\times Poincare\text{ }Group$ is the local gauge group of a Yang-Mill's theory with the gauge group $G$ on a space-time. I am not sure. I would love to be corrected!

This post has been migrated from (A51.SE)

1 Answer

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I think no. In a local trivialization of your G-bundle $\mathcal{M}\times G$, we have a right action $g(x,h)=(x,gh)$ - in other words, $G$ does not act on $\mathcal{M}$ in that sense. So moving to the second case must change the base space to some manifold $\mathcal{M}$ whose Poincare group is isomorphic to $G$ (see what Squark said). They have equivalent fibers but inequivalent base spaces.

This post has been migrated from (A51.SE)
answered Jan 22, 2012 by cduston (160 points) [ no revision ]

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