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Let $X(2)$ denote the compact Riemann surface obtained by compactifying $Y(2) = \Gamma(2)\mathfrak{h}$ by adding cusps.

The modular $\lambda$-function on the complex upper half plane $\mathfrak{h}$ induces an isomorphism $\lambda:X(2)\longrightarrow \mathbf{P}^1$. It has a $q$-expansion $$\lambda = \sum_{j=0}^\infty a_j q^j$$ on the complex upper half plane. Its first couple of terms are given on the corresponding wikipedia page

http://en.wikipedia.org/wiki/Modular_lambda_function

Aim. I would like to know a lower bound for the convergence radius of $\sum_{j=0}^\infty a_j q^j$. More precisely, I would like to have a real number $R$ such that, for all $\tau\in \mathfrak{h}$ with $\vert q(\tau)\vert \leq R$, we have that $\lambda(\tau) = \sum_{j=0}^\infty a_j q^j(\tau)$.

I need this number for computing the Arakelov degree of a certain line bundle on a cover of $X(2)$.

I think there are programs (such as SAGE) that can compute this number, but I don't know how to do this.

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    $\begingroup$ Sloane gives the coefficients at oeis.org/A115977 including a definition in terms of $\eta$ as 16 times the coefficient of $\eta(q)^8\eta(q^4)^{16}/\eta(q^2)^{24}$. This can yield their growth, and I suspect it is an exercise. I don't know a reference. If you want exact bounds, rather than asymptotics then it is harder, but for convergence radius, this is more than abides. $\endgroup$
    – Junkie
    Commented Jun 22, 2011 at 10:41
  • $\begingroup$ Thank you for your comment. It didn't occur to me to search Sloane's database. $\endgroup$
    – Ariyan Javanpeykar
    Commented Jun 22, 2011 at 10:48
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    $\begingroup$ As with many modular computations, it appears you can bound $a_j$ by a measure like $j^{\sqrt j}$, so that indeed for any $R<1$ the series will converge, by comparison $\sum_j j^{\sqrt j}e^{2\pi i\tau j}$. $\endgroup$
    – Junkie
    Commented Jun 22, 2011 at 11:11
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    $\begingroup$ The classical reference for growth of coefficients is Rademacher with the $j$-invariant via the method of Farey arcs jstor.org/stable/2371313 , and Brisebarre and Philibert perso.ens-lyon.fr/nicolas.brisebarre/Publi/fujijrms.pdf don't even bother to replicate the proof when noting that it holds for all modular functions of weight 0 for the full modular group (Theorem 5.1). I suspect varying it to your case should not yield tumultuous difficulties. $\endgroup$
    – Junkie
    Commented Jun 22, 2011 at 11:32
  • $\begingroup$ Thank you very much! I'll accept S. Carnahan's answer below and thank you for your answer again (which was posted before his). Thnx! $\endgroup$
    – Ariyan Javanpeykar
    Commented Jun 22, 2011 at 12:27

1 Answer 1

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Sorry about the one-sentence answer. Here is an expanded version that uses the following facts: The modular function $\lambda$ is holomorphic on the upper half-plane, invariant under the action of $\Gamma(2)$, and non-singular at infinity.

From the invariance under the subgroup of $\Gamma(2)$ generated by $\binom{12}{01} = (z \mapsto z + 2)$, we can conclude that $\lambda$ has a $q$-expansion in powers of $q^{1/2}$, and from regularity at infinity, these powers are nonnegative. In fact, the periodicity combined with holomorphicity implies the map $\lambda: \mathcal{H} \to \mathbb{C}$ factors as a composite of $(\tau \mapsto e^{i \pi \tau}): \mathcal{H} \to D_{q^{1/2}}$ followed by $\bar{\lambda}: D_{q^{1/2}} \to \mathbb{C}$, where $D_{q^{1/2}}$ is the open unit disc of radius one with coordinate $q^{1/2}$, and $\bar{\lambda}$ is a holomorphic function uniquely defined by this factorization. It's power series expansion at $q^{1/2}=0$ is given by the $q$-expansion of $\lambda$, and because the function is holomorphic on the open unit disc, the radius of convergence of the series is at least one.

To show that the radius of convergence is exactly one, we note that $\lambda$ is singular on a $\Gamma(2)$-orbit of cusps, and these cusps are taken to a dense subset of the unit circle under the $\tau \mapsto e^{i \pi \tau}$ map. The fact that $\lambda$ is singular at a nonempty set of cusps follows from the fact that $\lambda$ factors through the $\Gamma(2)$-quotient map $\mathcal{H} \twoheadrightarrow Y(2)$, the compactification $Y(2) \hookrightarrow X(2)$, and an analytic isomorphism $X(2)\overset{\simeq}{\longrightarrow} \mathbb{P}^1$. Since $Y(2)$ is dense in $X(2)$, the composite has dense image but the fact that $\lambda$ is holomorphic and nonconstant on $\mathcal{H}$ implies the image approaches but does not include infinity.

Regarding more precise asymptotic estimates for coefficients, if you want to apply the Rademacher circle method to get asymptotics of a form of level greater than one (such as $\lambda$), you need to work with the singularities at all cusps. There is an extremely brief description of this method in section 5 of Borcherds's paper Automorphic forms on $O_{s+2,2}(\mathbf{R})$ and infinite products. Alternatively, you may turn $\lambda$ into a vector-valued form of level one with respect to a representation of $\Gamma(1)$.

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  • $\begingroup$ So, for all tau in the complex upper half plane, we have that lambda(tau) = sum a_j q(tau)^j? (Here I use that |q(tau)| < R if and only if Im(tau) > log(1/R)/(2pi) .) $\endgroup$
    – Ariyan Javanpeykar
    Commented Jun 22, 2011 at 12:05
  • $\begingroup$ I see Junkie made the same comment. $\endgroup$
    – Ariyan Javanpeykar
    Commented Jun 22, 2011 at 12:25
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    $\begingroup$ Alternatively: clearly the series converges for $|q|<1$, because $\lambda$ is a function on the upper half plane. However the coefficients are integers and $\lambda$ is surely known not to be a polynomial (I believe this is a general fact about modular functions), so clearly the function will never converge absolutely for $|q|=1$ and you're home. $\endgroup$
    – Kevin Buzzard
    Commented Jun 22, 2011 at 18:57

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