Betti multiplets in Kaluza Klein compactifications

Question

It is well known that if the compactification manifold of a supergravity theory has non-zero Betti numbers, this may lead to the so called Betti multiplets in the spectrum of the low dimensional theory. A famous example is compactification of IIB supergravity on $T^{1,1}$, where a Betti multiplet shows up because of the nonzero second Betti number of $T^{1,1}$.

My question is this: is it the $L^2$-Betti numbers that necessitate Betti multiplets in the low dimensional theory, or just normal Betti numbers? In particular, do Betti numbers generated by discrete identification (orbifolding) of trivial manifolds lead to Betti multiplets? (I am actually not even sure if orbifolding trivial topologies can yield non-zero Betti numbers.) Is there a good reference I can look into for that?

This post imported from StackExchange Physics at 2014-03-07 13:10 (UCT), posted by SE-user Arash Arabi

Answer 1

No physicist is using $L^2$ Betti numbers, and unless he is a (semi)professional mathematician at the same time, he doesn't even know what these $L^2$ Betti numbers are. So it's surely ordinary Betti numbers that matter in physics.

Otherwise compact (compactification) manifolds always have some nonzero Betti numbers. It is not clear why you think that the Betti numbers should be zero for "orbifolds of trivial topologies". Compact manifolds never have "quite" trivial topologies. The sphere $S^k$ could perhaps be viewed as one with the "nearly trivial" topology similar to the infinite space and it has the maximum number of vanishing Betti numbers, indeed. But aside from the sphere, pretty much all compact manifolds have some nonzero Betti numbers even if we don't count $b_0$ and $b_d$, the zero- and highest-dimensional ones.

The Euler characteristic tends to be divided by the order of the group for orbifolds but the behavior of the general Betti numbers for an orbifold may be very general and complex.

This post imported from StackExchange Physics at 2014-03-07 13:10 (UCT), posted by SE-user Luboš Motl

Comments

Thanks for the response Lubos. Let me explain why I think $L^2$ Betti numbers should matter, and not normal Betti numbers. Start from round $S^k$ ($k>1$) and it has a trivial fundamental group. Now a smooth quotient of the round sphere by a discrete identification leads to a non-trivial fundamental group, hence a non-zero first Betti number. However, smooth quotients do not lead to Betti multiplets in the massless sector; $S^5/Z_3$ compactification of IIB supergravity is an example. Therefore a non-zero Betti number doesn't necessitate a Betti multiplet, but a non-zero $L^2$ Betti number might

This post imported from StackExchange Physics at 2014-03-07 13:10 (UCT), posted by SE-user Arash Arabi

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I think $L^2$ Betti numbers are roughly Betti numbers on the universal cover (I'm not sure how precise or even correct this sentence is actually). So the complexities with orbifolding procedures are avoided once one speaks of $L^2$ Betti numbers.

This post imported from StackExchange Physics at 2014-03-07 13:10 (UCT), posted by SE-user Arash Arabi

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Sorry, Arash, we are not studying orbifolds in order to "avoid their complexities". The complexities - the orbifolds' being new solutions that are interesting and shouldn't be overlooked - is the very purpose why people study them. The orbifolds, especially those of flat manifolds, still lead to simpler analyses than their generic resolutions. So whether some other manifolds have simpler Betti numbers of one kind or another is completely irrelevant.

This post imported from StackExchange Physics at 2014-03-07 13:10 (UCT), posted by SE-user Luboš Motl

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Otherwise your paper arxiv.org/abs/arXiv:1304.1540 about the Z3 orbifold is "solving" a completely trivial problem. The spectrum of SUGRA in an orbifold is just the subspace of the spectrum on the covering space that is invariant under the action of the orbifolding group $\Gamma$. For a $Z_3$, it just means to divide the space to 3 possible eigenvalues and pick one. Nontrivial physics with orbifolds starts in string theory's twisted sectors or, which is what this reduces to in the SUGRA limit, in physics of possible resolutions of the fixed points (orbifold singularities).

This post imported from StackExchange Physics at 2014-03-07 13:10 (UCT), posted by SE-user Luboš Motl

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These fixed points play an important role in the existence of new states and modified Betti numbers. For example, the Betti numbers of a four-torus T4 are (1,4,6,4,1) but the orbifolds T4/Z2 and T4/Z3 when properly done are K3 manifolds whose Betti numbers are (1,0,22,0,1), very different. Some of them have dropped, some of them have increased, and the exact nature of the orbifolding group and the physics localized near the fixed points were needed to "adjust" the original Betti numbers to the orbifolds' Betti numbers. It's the genuine Betti numbers (1,0,22,0,1) that dictate physics on a K3.

This post imported from StackExchange Physics at 2014-03-07 13:10 (UCT), posted by SE-user Luboš Motl

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Your comment on orbifolds of T4 and their relation to K3 was very interesting.

This post imported from StackExchange Physics at 2014-03-07 13:10 (UCT), posted by SE-user Arash Arabi

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I think I should correct my second comment. As you correctly said, we don't want to avoid complexities with singular orbifolds, they are quite interesting in their own right. What I meant was that for smooth orbifolds, there is no added massless Betti multiplets while there are (sometimes) added Betti numbers, hence putting the Betti multiplets in correspondence with Betti numbers doesn't seem correct; but maybe putting Betti multiplets with $L^2$ Betti numbers works.

This post imported from StackExchange Physics at 2014-03-07 13:10 (UCT), posted by SE-user Arash Arabi

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I also see now, thanks to your comments, that the proposal that $L^2$ Betti numbers are the right ones to use, requires that when the orbifold action is singular, $L^2$ Betti numbers are `not' simply the Betti numbers of the universal cover (otherwise if the cover is smooth, we would loose the Betti multiplets in the twisted sector). I don't know the relation to the universal cover for singular orbifolds though.

This post imported from StackExchange Physics at 2014-03-07 13:10 (UCT), posted by SE-user Arash Arabi