The largest prime in the order of the Monster group is $71$. This number $71$ shows up at various places:

  • The minimal faithful representation has dimension $196883 = 47.59.71$

  • The Monster group can be realised as a Galois group $Gal$ $L(71)/{\mathbb{Q}(\sqrt{-71})}$ where $L(71)$ is a suitable field.

The appearance of $71$ in the above two cases (and possibly others) is not very surprising and might be reasoned out.

But the appearance of $71$ in a different area seems very intriguing:

The Monster is intimately connected to a special class of conformal field theories. These are the meromorphic $c = 24$ CFTs. The Monster here arises as the discrete automorphism group of the vertex operator algebra of one of the c =24 CFTs.

Schellekens in 1992 enumerated such CFTs and he found $71$ such CFTs! All these CFTs have a partition function of the form $$ Z(\tau) = j(\tau) + \mathcal{N} $$ where $j$ is modular invariant and $\mathcal{N} \geq -744$ is an integer. But any value of $\mathcal{N}$ won't work. Schellekens found $71$ values of $\mathcal{N}$ which will work.

Unfortunately, it is still not clear if the enumeration Schellekens made is exhaustive, i.e. if there are only exactly $71$ such theories.

Is the appearance of $71$ here just a coincidence? Or is it again connected to the Monster? It is hard to believe that this is just a coincidence.

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    $\begingroup$ As far as I know the 71 in Schelleken's classification is not related to the other occurrences you mention. Some recent work on the construction of Schelleken's examples can be found here arxiv.org/pdf/1708.05990.pdf $\endgroup$ Sep 15, 2017 at 19:44
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    $\begingroup$ Schellekens remarks on this coincidence towards the end of his paper. But that was 25 years ago, and I wanted to know if there was any progress. @ Jeff Harvey The article is dedicated to Griess on the occasion of his 71st birthday, cool! It looks like a very useful article, thank you! $\endgroup$ Sep 15, 2017 at 19:52
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    $\begingroup$ One might also ask why, at magic number 24, there are n=24 self-dual even lattices of dimension 24. I know how most numbers 24 are related, but I don't yet know if |{lattices}| is the "same" 24 as all the others. $\endgroup$ Sep 16, 2017 at 3:29
  • $\begingroup$ @TheoJohnson-Freyd For that matter, is it also related to the $576=24^2$-periodicity of TMF? $\endgroup$ Sep 16, 2017 at 4:29
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    $\begingroup$ @მამუკაჯიბლაძე Here's how I want to explain the number 576. I will say it in condensed matter. As warm-up, Bott periodicity. If you allow interacting fermions, then there are 2 = \pi_2(stable sphere) invertible phases of (1+1)-dimensional matter. But there are 8 phases "protected by time-reversal symmetry" (which is the condensed matter way of saying "real"). Similarly, there should be 24 = \pi_3(stable sphere) invertible phases of (2+1)-dimensional matter. I expect this to expand to 576 when you protect by the appropriate symmetry. The symmetry group should be related to SL(2,Z)... $\endgroup$ Sep 16, 2017 at 4:39

1 Answer 1


Schellekens' enumeration is exhaustive in the following sense: the degree 1 subspace of the meromorphic CFT/vertex algebra is naturally a Lie algebra, and it is known that this Lie algebra must be one of the 71 that Schellekens wrote down.

Each of these 71 Lie algebras is realised as the weight 1 piece of some holomorphic c=24 vertex algebra, but it is still an open conjecture that the vertex algebra is unique. (Though it is verified in the large majority of cases.)

Regarding possible doubts about machine precision arithmetic and such in the linear programming part of Schellekens' calculations: In https://arxiv.org/abs/1507.08142 one of the things my collaborators and I do is to independently confirm Schellekens' result, this time with exact arithmetic, and perhaps you might say a streamlined proof. But it's still a computer proof. We go on to use the result in our construction of some new holomorphic vertex algebras as orbifolds.

I find the appearance of 71 as tantalising as you do. I'd certainly like for it not be a coincidence, but for the moment at least I just don't know.

  • $\begingroup$ Perhaps this answer should be updated with the uniqueness claim. $\endgroup$ Jan 9, 2020 at 22:12

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