31
$\begingroup$

Bosonic string theory lives in 26 dimensions, and it gives a conformal field theory where the field is a map from a Riemann surface to $\mathbb{R}^{24}$. The Leech lattice $L$ is an even unimodular lattice in $\mathbb{R}^{24}$. We can form a conformal field theory where the field is a map from a Riemann surface to the torus $T = \mathbb{R}^{24}/L$, and this theory almost has the Monster group as its symmetry group. In fact we need to go one step further and replace $T$ by the orbifold where we mod out by the involution of $T$ coming from the transformation $x \mapsto -x$ of $\mathbb{R}^{24}$. In this case Frenkel, Lepowsky and Meurman showed we get a conformal field theory, or more technically a vertex operator algebra, whose symmmetry group includes the Monster group.

There could be a supersymmetric analogue of this, and it's probably been studied. What group does that give?

More precisely:

Superstring theory lives in 10 dimensions, and it should give a conformal field theory where the field is a map from a Riemann surface to $\mathbb{R}^{8}$, or actually a super-vector space $V$ with $\mathbb{R}^8$ as its even part. The $\mathrm{E}_8$ lattice is an even unimodular lattice in $\mathbb{R}^8$. I suspect we should be able to form a form a conformal field theory where the field is a map from a Riemann surface to the 'supertorus' $T_\mathrm{super} = V/\mathbb{E}_8$. Is the symmetry group of the corresponding vertex operator algebra known? We may have to replace $T_\mathrm{super}$ by a super-orbifold, e.g. mod out by an involution, to get a really interesting group.

$\endgroup$
4
  • 1
    $\begingroup$ Some remark: in the monster story, the even unimodularity condition guarantees the modularity of the holomorphic conformal field theory obtained by compactifying holomorphic bosons over the torus. To have a modular holomorphic conformal field theory, the central charge has to be divisible by 8 (for exemple holomorphic bosons over the E_8 torus define an holomorphic conformal field theory: the affine E_8 current algebra at level 1). Conformal field theory on a supertorus of dimension 8 has central charge: 8+4=12 and so cannot give rise to a modular holomorphic conformal field theory. $\endgroup$
    – user25309
    May 20, 2016 at 20:27
  • 5
    $\begingroup$ The theorem you cite was not Borcherds, but Frenkel-Lepowsky-Meurman. Borcherds's role was (a) defining vertex algebras and (b) showing that the moonshine module constructed by Frenkel-Lepowsky-Meurman satisfied the Conway-Norton Monstrous Moonshine conjectures. $\endgroup$
    – S. Carnahan
    May 21, 2016 at 6:25
  • $\begingroup$ @user25309 You'll never get full $SL_2(\mathbb{Z})$-invariance from a supersymmetric theory, since odd states restrict translation symmetry to $\tau \mapsto \tau + 2$. However, even the free fermions of central charge $1/2$ satisfy a property analogous to holomorphicity, namely that the representation category is equivalent to vector spaces. $\endgroup$
    – S. Carnahan
    May 21, 2016 at 15:38
  • $\begingroup$ I fixed the reference to Borcherds - thanks! $\endgroup$
    – John Baez
    May 23, 2016 at 5:07

3 Answers 3

18
$\begingroup$

There is a super analog constructed just as you describe with the Conway group $Co_0$ replacing the Monster and commuting with the superconformal algebra. The construction is described in detail in:

and in John Duncan's paper:

$\endgroup$
1
  • 1
    $\begingroup$ Nora Ganter's work on categorifying Co_0 is surely relevant to this somehow. $\endgroup$
    – David Roberts
    May 21, 2016 at 1:53
3
$\begingroup$

The paper Beauty and the Beast (open access) shows that the Moonshine module contains a copy of the super-Virasoro algebra, and so in some sense is already supersymmetric. I don't know how to interpret it in terms of a susy string, however.

$\endgroup$
2
  • 5
    $\begingroup$ The super-Virasoro algebra is not contained in the Moonshine module, but in a larger structure whose $\mathbb{Z}/2$-fixed points form the Moonshine module. That object only has symmetries given by an extraspecial extension of Conway, rather than the (substantially larger) monster. $\endgroup$
    – S. Carnahan
    May 21, 2016 at 6:21
  • $\begingroup$ @S.Carnahan Thank you for the correction! I only briefly skimmed that paper. $\endgroup$ May 23, 2016 at 14:52
3
$\begingroup$

This may be of interest:

Monstrous BPS-Algebras and the Superstring Origin of Moonshine
by Natalie M. Paquette, Daniel Persson, Roberto Volpato

http://arxiv.org/abs/1601.05412

$\endgroup$
5
  • $\begingroup$ Also the followup: "3D String Theory and Umbral Moonshine" arxiv.org/abs/1603.07330 $\endgroup$ May 23, 2016 at 15:52
  • $\begingroup$ I see the Paquette et al paper mentions heterotic string theory. Indeed my next question was going to be about "heterotic Moonshine" where we treat the bosonic left-movers using the Leech lattice and the supersymmetric right-movers using the E8 lattice, and then (I guess) do a Z/2 orbifold. What symmetry group does this theory have? I guess the answer should at least include the symmetries of the two theories separately, which seem to be the Monster group and Conway's $\mathrm{Co}_0$ group. $\endgroup$
    – John Baez
    May 23, 2016 at 21:04
  • 2
    $\begingroup$ This is correct except you should do a $Z/2 \times Z/2$ orbifold with the first factor acting on the left to give the Monster VOA and the second factor acting on the right to give the Conway Super VOA. $\endgroup$ May 24, 2016 at 1:25
  • $\begingroup$ Jeff Harvey - is there a place to read about that construction? $\endgroup$
    – John Baez
    May 24, 2016 at 5:13
  • $\begingroup$ JohnBaez-it is discussed in the above paper, but the construction is pretty obvious. The Monster VOA is constructed as a $Z/2$ asymmetric orbifold of the bosonic string on the Leech lattice and the Conway super VOA as an asymmetric $Z/2$ orbifold of the $E_8$ theory and given the known cancellation of modular anomalies in the heterotic string it is obvious you can take the tensor product to construct a consistent heterotic background with Monster x Conway symmetry. $\endgroup$ May 24, 2016 at 11:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.