9

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?

1 Answers1

2

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.

Luboš Motl
  • 182,599