Thursday, May 20, 2021

The seemingly infinite volume of \({\mathrm{PSL}}(2,\RR)\) is \(-2 \pi^2\)

And you need just a few short steps to calculate the D-brane tensions from that fact

The today's most joyful hep-th abstract was written by two names that didn't click with me (Eberhardt, Pal; their publication records initiated around 2017 seem impressive) but their affiliation did, the IAS at Princeton (that's a place where Einstein has worked for 1/2 of his career). It's an appropriate place for the authors to work because I am really happy about the paper which is titled simply
The Disk Partition Function in String Theory
and it discusses a theme that I found cool since the high school: many seemingly infinite expressions must actually be assigned finite values. The most famous example is the sum of positive integers\[ 1+2+3+4+5+ \dots = -\frac{1}{12}. \] This identity is important enough in string theory, including the (natural, bureaucracy-free) calculation of the critical spacetime dimension. I've played with some generalizations before I started to submit some serious papers. You may also check a recursive evaluation of zeta of negative integers.

Other seemingly singular sums, products, and integrals have finite natural (regularized) values, too. For example, the product of all positive integers is \(\sqrt{2\pi}\), yes, it is the same factor that you see in the Stirling approximation for the factorial (including the factorial of infinity) and you may even determine the product of all primes which is \(4\pi^2\). Because the number PI appears everywhere in this blog post, you should notice that we are currently using the newest MathJax 3.1.4 ;-). The current version of \(\rm\TeX\) is 3.141592653, I clearly share some of Donald Knuth's intellectual humor.

This new paper discusses an identity with a very similar result (as these regularized sums and products),\[ {\rm Vol} ({\mathrm{PSL}}(2,\RR)) = -2 \pi^2. \] Cool. For the result to be this unambiguous, they had to pick a (usual) convention for the metric on the group manifold of this "projective special linear" group. The convention is basically equivalent to "the unit radius of all the spheres inside". The group is noncompact which means that its volume is infinite. However, this infinity is just an informal way of describing "that the group is noncompact" while the right value of the "volume" that you need to substitute to formulae is a finite number. Like in the case of the products of integers, it is a product of powers of rational numbers and \(\pi\); and like in the sum of integers, the right value turns out to be negative. Yes, the volume is \(-2\pi^2\).

Well, they dedicate the Appendix B to this constant. I can give you a motley, three-line derivation. The group is really the same as \({\mathrm{PSU}}(1,1)\). That differs from \({\mathrm{PSU}}(2)\) by a simple flip of the signature. But \({\mathrm{SU(2)}}\) is just a three-sphere whose surface is \(+2\pi^2\). I may get the volume of the noncompact group simply by Wick-rotating 2 of the 4 dimensions where \(SU(2)\) is embedded, and each Wick rotation produces a factor of \(i\). And \(i^2=-1\) gives me the minus sign. (Later on Thursday, the authors have written to me about some subtleties, e.g. that I should really divide the volume by 2 because of the P for projective, and I surely agree with that, and they seem to be infidels when it comes to the very possibility of fixing such a derivation.)

Who you trying to get crazy with ése? Don't you know I'm loco? Insane in the membrane, insane in the brane.

Note that in this post about the Euler characteristic, I argued that the Euler characteristic may be interpreted as a regularized number of points of a manifold (which is also seemingly infinite for almost all manifolds).

This volume enters calculations in perturbative string theory, too – and that is why they are doing it. Note that the oldest nontrivial result of string theory (in the Prague Spring year of 1968) was the Veneziano amplitude, the formula for the scattering amplitude of 4 open string tachyons in bosonic string theory. The result is \(B(u,v)\) where \(u,v\) are linear functions of the Mandelstam variables. It is the Euler Beta function, a ratio of the Gamma functions (generalized factorials).

You may obtain the Veneziano amplitude from a disk-shaped world sheet via a path integral. You need to insert those 4 vertex operators for the external four tachyons somewhere at the boundary of the disk. Three of them may be placed to arbitrary positions, by using the \({\mathrm{PSL}}(2,\RR)\) residual symmetry (which has 3 free real parameters). The location of the fourth vertex operator cannot be fixed and must be integrated over. So the Beta function may be expressed as a one-dimensional (real) integral.

Cool. If you considered the scattering amplitude of 3 tachyons, there would be no residual integral – but the (spacetime) angles between the tachyons would have to be very specific for them to be on-shell and for the energy-momentum conservation to hold. But can't you go below 3 tachyons? Like 0 tachyons? If you do, the path integral has the unfixed residual symmetry group (more generally, a part of it which is 1,2,3-dimensional if you only place 2,1,0 tachyonic vertex operators on the disk, respectively). Because the configurations related by the \({\mathrm{PSL}}(2,\RR)\) are physically eequivalent (gauge symmetry), you need to avoid multiple counting and that is why you divide the (finite) integrand by the volume of the group. Because that denominator seems infinite, you naively get a zero.

But just like the sum of integers isn't really "infinite" in the deep physical sense, the volume of this group manifold isn't infinite, either. It is finite, it is \(-2\pi^2\) in the usual conventions, and you may substitute this correct value to the path integral and calculate the D-brane tensions from the simplest possible disk-shaped path integral in the appropriate string theory. They actually do so. The fact that the regularized volume of \({\mathrm{PSL}}(2,\RR)\) is negative is essential for the fact that the known D-brane tensions end up being positive.

They verify their result for the D-brane tension against Joe's Big Book of String which I have previously verified as well – well, I sent 128 valid corrections to Joe (almost beating the rest of the Earth combined), including numerous corrections of formulae dealing with D-brane tensions, indeed. ;-)

But yes, the simplest forms of the path integrals should be dealt in this way. The infinite values such as the volumes of similar noncompact groups are spurious and the right, finite volumes should be widely known (instead of computing the simple disk diagram without insertions, the usual method to determine the D-brane tensions involve much more complex and "reducible" diagrams). Note that \(-2\pi^2\) is a number of a similar form as the volumes of \(d\)-dimensional spheres. And indeed, the group manifold is basically a hyperboloid which is a Wick rotation of a sphere which is really why you had to get a similar form of the result. However, you need to be careful about the world sheets of the spherical topology. In superstring theory, you clearly get a vanishing result from those, after all (supersymmetry is a spacetime argument why it needs to be so).

So, an update: Yes, I think that the purely bosonic regulated volume of \(\mathrm{SL}(2,\CC)\) should better be related to the volume of \(\mathrm{SO}(3,1)\) which is a different signature of \({\mathrm{SO}}(4)\) which is \(\mathrm{SU}(2)\times {\mathrm{SU}}(2)\) which is two 3-spheres, so up to the powers of two (and, less likely, other rational factors), I do think that the volume of \(\mathrm{SL}(2,\CC)\) should be \(4\pi^4\) or so, as a product of two disks! I would be surprised if the regulated volumes were legitimate but this tempting configuration of "closed strings equal open squared" in the case of "sphere is a disk squared" were totally violated for volumes. (I won't write the group names with the mathrm command again because it is a pain in the aß.)

No comments:

Post a Comment