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Kac-Moody algebras in 5 dimensional Kaluza Klein

+ 3 like - 0 dislike
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I am trying to make sense to the issue of how does the Kac-Moody algebra encode the symmetries of the non-truncated theory.

Let's contextualize a little bit. Ok, so in the 5 dimensional Kaluza-Klein we start with pure (5 dimensional) gravity. Then we assume a $M^4\times{}S^1$ ground state. The topology of this ground state breaks the 5D diffeomorphism invariance because now only periodic transformations are allowed. So considering a coordinate change

$x\to{}x'=x+\xi$

we can Fourier expand

$\xi(x,\theta)=\sum{}\xi(x)e^{in\theta}$

It is known that if we truncate at $n=0$ we get a theory of gravitation in 4 dimensional space-time, electromagnetism and a scalar field (sometimes called the Dilaton).

So, the remnants of the initial 5 dimensional diffeo invariance are 4 dimensinal diffeo invariance,  U(1) gauge and a scale invariance for the dilaton.

In the non truncated theory we will keep all the modes in the previous Fourier expansion and we will have more symmetries. The 4d diffeo and U(1) gauge do survive in the non truncated version but not the scale invariance of the dilaton.

So far so good.

But, how to know the symmetries of the full theory? well,  with Kac-Moody generalization of Poincaré  algebras. For example in http://arxiv.org/pdf/hep-th/9410046v1.pdf says this. Now the question, to get the symetries of the full theory I just have to gauge what this algebra gives me? And if this is so, in particular, in n=0 we get a Poincaré algebra (with U(1)) , so is 4 dimensional diffeo a gauged global Poincaré? And if this is so, could you prove me that gauging Poincaré we get 4 dimensional diffeo?

asked Apr 5, 2014 in Theoretical Physics by Dmitry hand me the Kalashnikov (720 points) [ no revision ]

1 Answer

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As the very formulation of your question makes clear, we know what the actual algebra of local symmetries is. It is the five-dimensional diffeomorphism invariance assuming the $M^4\times S^1$ topology of the five-dimensional spacetime.

The term "Kač-Moody generalization of an algebra" is nothing else than an alternative name for this algebra, especially for the form of the algebra that is obtained by considering not generators as continuous functions of the fifth coordinate $\theta$, but using the discrete Fourier modes in this direction labeled by $n$.

The original, Kač-Moody-non-generalized algebra has commutators of the generators like $$ [G_i,G_j] = f_{ijk} G_k $$ with the appropriately raised indices. In your case, all $G_i$ are linear combinations of generators of diffeomorphisms, i.e. integrals of the stress-energy tensor with some tensor-valued coefficients as functions of the 4D spacetime.

The Kač-Moody generalization arises when all the generators are allowed to depend on the extra coordinate $\theta$ or, equivalently, to depend on the Fourier mode integer index $n$. Then the generalization replaces $G_i$ by $G^n_i$ and the commutator becomes something like $$ [G_i^m,G_j^n] = (\pm m \pm n)^{\text{0 or 1}} f_{ijk} G_k^{m+n} $$ There may also be $n^3$ terms weighted by $\delta_{m,-n}$ etc. but I don't want to present all possible subtleties and generalizations of Kač-Moody algebras here.

The appearance of $m+n$ as the superscript on the right hand side is guaranteed by the $\theta\to \theta+c$ translational symmetry, the $U(1)$ gauge subalgebra you mentioned. Otherwise it's not suprising that we must get a similar algebra to the original 4D one, just with some extra indices $m,n,m+n$ moderately inserted and with some moderate coefficients.

To check the equivalence of the two descriptions, one only needs to know some basics of integrals and derivatives or the Fourier transform etc.

This post imported from StackExchange Physics at 2014-06-27 15:38 (UCT), posted by SE-user Luboš Motl
answered Jun 26, 2014 by Luboš Motl (10,248 points) [ no revision ]
Hi Lubos, thanks for the answer! I would like to clarify one thing so please bear with me. Consider the non massive case. What we get is $Poincaré\otimes{}U(1)$'s algebra (I have read somewhere that it is actually $Poincaré\otimes{}SO(1,2)$ but nevermind that). The thing I wanna know is how do we get diffeo4 out of this. Maybe it is stupid but a seeing explicitly how diffeo 4 comes out of the non massive case would be great. Also I have edited the question and inserted the explicit form of the infinite parameter algebra I am considering.

This post imported from StackExchange Physics at 2014-06-27 15:38 (UCT), posted by SE-user silvrfück
Hi, diffeo4 is just the subgroup of diffeo5 of transformations independent of the fifth coordinate theta or, equivalently, the transformations generated by the generators with the index $n=0$. I don't know what you mean by "we get Poincare times U(1)". Poincare is a finite-dimensional group only. You may restrict yourself to the isometries like Poincare but the original theory is a theory of gravity, so it allows one to apply any diffeomorphism.

This post imported from StackExchange Physics at 2014-06-27 15:38 (UCT), posted by SE-user Luboš Motl

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