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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?

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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.

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  • $\begingroup$ 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 $\endgroup$ – user24155 Jan 25 '14 at 21:37
  • $\begingroup$ 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. $\endgroup$ – user24155 Jan 25 '14 at 21:39
  • $\begingroup$ 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. $\endgroup$ – Luboš Motl Jan 26 '14 at 7:01
  • $\begingroup$ 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). $\endgroup$ – Luboš Motl Jan 26 '14 at 7:29
  • $\begingroup$ 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. $\endgroup$ – Luboš Motl Jan 26 '14 at 7:33

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