# Betti multiplets in Kaluza Klein compactifications

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 smooth (fixed point free) discrete identification (orbifolding) of trivial manifolds lead to Betti multiplets? (I am actually not even sure if smooth orbifolding of trivial topologies can yield non-zero Betti numbers.) Is there a good reference I can look into for that?

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.
• 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 – user24155 Jan 25 '14 at 21:37
• 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. – user24155 Jan 25 '14 at 21:39
• 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). – Luboš Motl Jan 26 '14 at 7:29