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This is connected to my MO post "Monstrous Moonshine for $M_{24}$ and K3?". In page 44 of this paper, eqn(7.16) and (7.19) yield,

$$\begin{aligned}h^{(2)}(\tau)&=\frac{\vartheta_2(0,p)^4-\vartheta_4(0,p)^4}{\eta(\tau)^3}-\frac{24}{\vartheta_3(0,p)}\sum_{n=-\infty}^\infty\frac{q^{n^2/2-1/8}}{1+q^{n-1/2}}\\ &=\color{red}{m}\,q^{-1/8}(-1+45q+231q^2+770q^3+2277q^4+\dots)\end{aligned}$$

It was observed by Eguchi, Ooguri, and Tachikawa that the first five coefficients of the RHS are equal to the dimensions of irreducible representations of $M_{24}$.

I assume that $q = p^2$, nome $p = e^{\pi i \tau}$, Jacobi theta functions $\vartheta_n(0,p)$, Dedekind eta function $\eta(\tau)$, and 30 coefficients $a_i$ of the RHS are given by OEIS A212301 as $2a_i$.

Question:

The paper implies that $m=1$. However, if I test it with $\tau=\sqrt{-n}$ for various positive integer $n$, then it seems m varies as well. In particular, if $\tau=\sqrt{-1}$, then apparently $m=2$. Which of my assumptions are wrong, and how do we fix it? (Or is it a bug in Mathematica again?)

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    $\begingroup$ How can $m$ vary in a $q$-series expansion? Anyway, I get $q^{-1/8}(-1-28p+\cdots)$. This confirms the $-1$ (with $m=1$), but not the rest. Note: I rearranged $\sum_{n=-\infty}^\infty\frac{p^{n^2}}{1+p^{2n-1}} = \sum_{n=1}^\infty\frac{p^{n^2}}{1+p^{2n-1}} + \sum_{n=0}^\infty\frac{p^{n^2+2n+1}}{p^{2n+1}+1} = 2\sum_{n=1}^\infty\frac{p^{n^2}}{1+p^{2n-1}}$. $\endgroup$
    – ccorn
    Commented Nov 23, 2013 at 5:18
  • $\begingroup$ I also do not get what you mean by varying $m$. $\endgroup$
    – anon
    Commented Nov 23, 2013 at 5:29
  • $\begingroup$ Wondering why Will Orrick's answer has not been accepted yet. Just in case, I have checked with Pari/GP that the series do match with the correction in Will's answer. The authors apparently mean summation over $\mathbb{Z}_{>0}$. Ironically, my first (manual) checks used a wrong $\vartheta_4$ expansion which was also due to such confusion of summation ranges... $\endgroup$
    – ccorn
    Commented Nov 23, 2013 at 8:52
  • $\begingroup$ @ccorn: When Orrick answered, it was past bedtime where I live. :) $\endgroup$
    – Tito Piezas III
    Commented Nov 23, 2013 at 14:06

1 Answer 1

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Neither you nor Mathematica are to blame.

If you change the coefficient $24$ in front of the sum to $12,$ the expression produces the desired series. Alternatively, following ccorn's comment, you can change the sum so that it runs from $1$ to $\infty$ rather than $-\infty$ to $\infty$. In the paper you cite, the authors write $\displaystyle\sum_{n\in\mathbb Z}.$ Perhaps they meant $\displaystyle\sum_{n\in\mathbb Z^+}.$

(I observed this by expanding both terms in the expression and noting that, after removing the overall factor $p^{-1/4},$ both contain undesired odd powers of $p.$ These undesired terms, however, differ between the two terms only by a factor of $2,$ and will cancel if the second term is halved.)

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  • $\begingroup$ After factoring out the $q^{1/8}$ from the $\eta^3$ and the explicit sum, I find $\vartheta_2^4(0,p)-\vartheta_4^4(0,p)=-1+20p +\operatorname{O}(p^2)$. The rest of $\eta^3$ is $1 + \operatorname{O}(q)$. Furthermore, $\vartheta_3(0,p)=1+\operatorname{O}(p)$, and the sum seems to be $2p + \operatorname{O}(p^2)$. Thus the $24$ would have to be changed to $10$ for cancellation, according to my calculations. $\endgroup$
    – ccorn
    Commented Nov 23, 2013 at 6:28
  • $\begingroup$ I think that $\vartheta_2^4(0,p)-\vartheta_4^4(0,p)=-1+24p+O(p^2).$ We have $\vartheta_2(0,p)=2p^{1/4}+2p^{9/4}+2p^{25/4}+\ldots$ and $\vartheta_4(0,p)=1-2p+2p^4-2p^9+\ldots$. Therefore $\vartheta_2^4(0,p)=16p+O(p^2)$ and $\vartheta_4^4(0,p)=1-8p+O(p^2).$ $\endgroup$
    – Will Orrick
    Commented Nov 23, 2013 at 6:41
  • $\begingroup$ Right, I missed the $2$ in $\vartheta_4$. Sorry for the confusion. +1 $\endgroup$
    – ccorn
    Commented Nov 23, 2013 at 6:42
  • $\begingroup$ Thanks, Orrick! I believe the authors missed the tiny positive sign of $\mathbb{Z^+}$. (Murphy's Law strikes again.) They could have just used the more mistake-proof $\displaystyle\sum_{n=1}.$ P.S. The motivation for this post (as I mentioned in MO), was that I wanted to know if there was an evaluation of $h^{(2)}(\tau)$ for some $\tau$ like $\tau = \frac{1+\sqrt{-163}}{2}$ in terms of known constants. There doesn't seem to be any known. $\endgroup$
    – Tito Piezas III
    Commented Nov 23, 2013 at 14:21
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    $\begingroup$ @Tito, I don't know of any results along the lines you mention in your comment involving the values of mock modular forms at singular moduli, but if there were such results they might appear for the non-holomorphic modular completion ${\hat h}^{(2)}(\tau)$ rather than for the holomorphic, mock modular $h^{(2)}(\tau)$. $\endgroup$
    – Jeff Harvey
    Commented Nov 23, 2013 at 16:16

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