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I am trying to examine the behavior of the theta function $\theta(z)=\sum_{n\in\mathbb{Z}} e^{2\pi i n^2 z}$, which is modular for $\Gamma_0(4)$ of weight 1/2, at the cusps 0 and 1/2. My calculations seem to show that it vanishes at least at one of these cusps. I would like to calculate the order of vanishing.

Apostol discusses this theta function in some detail in his second number theory book. He gives a transformation formula as $\theta(-1/z)=(-iz)^{1/2}\theta(z)$. This transformational formula doesn't make clear what the function's behavior is at the cusp at $\infty$.

Koblitz talks some about the order's of vanishing at the 0 cusp, but doesn't mention the order of vanishing at other cusps.

Does anyone have a good reference for this problem?

I've also tried looking for the valence formula for $\Gamma_0(4)$ to help with the calculation, but couldn't find it written down anywhere. A source for this would be helpful as well.

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This sounds like a finite but nasty computation. I would suggest to consider instead $\theta^2$ or even a higher power $\theta^{2r}$ and then divide the result by $2r$. Also note that $\theta$ is nonzero throughout the upper half plane. Lots of things concerning theta functions are computed in Farkas and Kra's book on Theta constants. –  Robert Kucharczyk Mar 9 '12 at 22:36
    
Also, the very definition of $\theta$ allows you to estimate its behaviour around $\infty$ quite easily: in the local parameter $q=\mathrm{exp}πiz$ you have $\theta (z)=1+2q+2q^4+\ldots$ (btw, your formula in the first line is wrong, surely a typo) –  Robert Kucharczyk Mar 9 '12 at 22:45
    
Here's a reference, wfu.edu/~rouseja/cv/sturmfinal.pdf –  i707107 Mar 9 '12 at 22:54
    
@Nathan : There seems to be a problem in your definition of $\theta(z)$. Are you considering $\theta(z) = \sum_{n \in \mathbf{Z}} e^{2i\pi n^2 z}$ ? –  François Brunault Mar 10 '12 at 11:56
    
Edited only to fix the typo in the title and to change the formula for $\theta$ to the closest thing that matches "modular theta function". –  Noam D. Elkies Mar 11 '12 at 7:31
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4 Answers 4

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The usual way to investigate the order of vanishing of a modular form at a cusp other than $\infty$ is to find an element of $\mathrm{SL}_2(\mathbb{Z})$ that maps $\infty$ to your cusp and "recenter" your form at $\infty$ using this element. If your element is $\gamma$, then look at $j(\gamma,z)^{-1}f(\gamma z)$ (or something like this...) where $j$ is the cocyle by which the form $f$ transforms. With any luck you can work out a formula for this from which the behavior at $\infty$ (and hence the behavior of $f$ at your chosen cusp) is evident.

For $\theta$ the easiest way to do this is to look at a table of "theta functions with characteristics" (like the one in Mumford's Tata Lectures on Theta, Volume I if I recall correctly). The result is that the $q$-expansions of $\theta$ at $\infty$, $0$, and $1/2$ are, respectively, $\sum q^{n^2}$, $\sum (q^{1/4})^{n^2}$, and $q^{1/4}\sum q^{n^2+n}$, where the sums are over $\mathbb{Z}$ and the latter two are really only defined up to some constant (perhaps just a fourth root of unity).

Now each cusp on a modular curve comes with a "width" $h$ - its ramification index over $X(1)$, and usually the $q$-expansion of a modular form at a cusp is an expansion in $q^{1/h}$. In this case the cusp $1/2$ has width $1$, and clearly something else is going on here. It appears that $\theta$ vanishes to order $1/4$ at this cusp!

While this turns out to be an oddly useful perspective (at least it has to me), it's nonsense on its face. To remedy this you need to be careful about setting up $\theta$ as a section of some complex-analytic line bundle on $X_0(4)$. As usual, this thing is just cooked up using the cocyle by which $\theta$ transforms and when you work out the local picture around the cusp $1/2$ you'll see that the section $\theta$ vanishes to order $1$ there as a section of this bundle.

To read more about the goofy phenomenon at $1/2$ search around for "irregular cusp." The underlying reason for the problem is that, from a moduli point of view, the object classified by this cusp has non-trivial automorphisms (so complex-analytically $X_0(4)$ is better thought of as an orbifold). On the other hand, one can see this disparity on the power of $q$ needed to expand a modular form very explicitly in the integral weight case by messing with local calculations for the sheaf $\omega^{\otimes k}$ around the cusp. This was done very nicely in some early draft of Brian Conrad's book on the Ramanujan-Petersson conjectures. I don't know what the status of that book is these days though...

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[Edited for consistency with the normalization $\theta(z) = \sum_{n=-\infty}^\infty e^{\pi i n^2 z}$ of the proposer (and of my own article!); expansions in powers of $e^{2\pi i z}$ are more common, but for this modular form $q = e^{\pi i z}$ does seem to be the better choice.]

The function $$ \theta(z) = \sum_{n=-\infty}^\infty q^{n^2} $$ (where $q = e^{\pi i z}$) has a product formula converging in $|q|<1$, and thus for all $z$ in the upper half-plane: $$ \theta(z) = \frac{\eta(z)^5}{\bigl(\eta(z/2)\eta(2z)\bigr)^2} = \frac{1+q}{1-q} \cdot \frac{1-q^2}{1+q^2} \cdot \frac{1+q^3}{1-q^3} \cdot \frac{1-q^4}{1+q^4} \cdot \frac{1+q^5}{1-q^5} \cdot \frac{1-q^6}{1+q^6} \cdots $$ (where $\eta(z) = e^{\pi i z/12} \prod_{m=1}^\infty (1-e^{2\pi i m z})$ as usual). It follows from this formula that $\theta$ cannot vanish except at a cusp: as long as $|q|<1$, the factors converge quickly to $1$ and none of them has a zero or pole. But $\theta$ is a modular form of positive weight, so it must vanish somewhere. It is clear from the sum formula that $\theta \neq 0$ as $q \rightarrow 1$ and $q \rightarrow 0$, so $\theta$ does not vanish at the cusps $z=0$ and $z=\infty$. Hence it must vanish at the remaining cusp of $\Gamma(2)$. Indeed each factor $(1 \pm q^n) / (1 \mp q^n)$ vanishes at $q = -1$ (and stays in $(0,1)$ for $-1 < q < 0$), so $\theta$ must vanish at the corresponding cusp $z=1$.

The author of the question also asked for a source for "a good reference for this problem". It's not clear what level of exposition is most appropriate here. Perhaps this paper might be of use, since I needed to use the vanishing order of $\theta$ in a context where I could not assume the reader had any background in modular forms:

Elkies N.D.: A characterization of the ${\bf Z}^n$ lattice, Math. Research Letters 2 (1995), 321-326 (arxiv:math.NT/9906019).

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Thanks Noam, this is a very clear exposition. Very helpful as well. –  Nathan Green Mar 12 '12 at 15:41
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Let $\theta(z) := \sum_{n \in \mathbf{Z}} e^{2i\pi n^2 z}$. It is known that $\theta^4$ is a holomorphic modular form of weight $2$ for $\Gamma_0(4)$. In other words $\omega :=\theta^4(z) dz$ defines a meromorphic differential form on the modular curve $X_0(4)$. This modular curve has $3$ cusps : $0$, $1/2$ and $\infty$. The vanishing of $\theta$ at a cusp $x$ is equivalent to saying that $\omega$ is regular at $x$. By definition $\theta$ doesn't vanish at $\infty$ so that $\omega$ has a simple pole there. The functional equation implies that $\theta$ doesn't vanish at $0$, so that $\omega$ also has a simple pole there. Since the modular curve $X_0(4)$ has genus $0$ and it has been noted that $\theta$ doesn't vanish on the upper half plane, it follows that $\omega$ is regular at the cusp $1/2$, so that $\theta$ vanishes at $1/2$.

A more direct way is the following. Put $\theta_{1/2}(z) = \theta(z+1/2)$. It is easy to check that $\theta(z)+\theta_{1/2}(z)=2\theta(4z)$. Now substitute $z$ by $-1/z$ to get

$$\theta_{1/2}(-1/z) = 2\theta(-4/z) - \theta(-1/z).$$

Using the functional equation of $\theta$, we see that the right hand side tends to $0$ when $z \to i\infty$, so that $\theta$ vanishes at $1/2$.

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The usual definition of $\theta$ is $$\theta(z):=\sum_{n\in\mathbb{Z}}e^{i\pi n^2 z},\qquad \Im z>0.$$ Sometimes $\theta(z)$ is defined to be $\theta(2z)$ above, which is modular for $\Gamma_0(4)$. At any rate, the definition above shows that $\theta(z)=1+O(e^{i\pi z})$ for $\Im z>1$, hence it equals $1$ at the cusp $\infty$. Coupled with the transformation rule you mentioned, it follows that $\theta(it)$ is asymptotically $t^{-1/2}$ as $t\to 0$. That is, $\theta(z)$ blows up at the cusp $0$ with the given rate. (The fact that it blows up at this rate is also immediate from the definition, just use $z=it$ and positivity of the terms there.)

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Isn't there a third cusp to be dealt with, the cusp $1/2$? –  David Loeffler Mar 10 '12 at 11:25
    
The cusp $1/2$ can be dealt with similarly, but I was lazy to do it. –  GH from MO Mar 10 '12 at 12:36
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