Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The Eisenstein series $$ G_{2k} = \sum_{(m,n) \neq (0,0)} \frac{1}{(m + n\tau)^{2k}} $$ are modular forms (if $k>1$) of weight $2k$ and quasi-modular if $k=1$. It is clear that given modular forms $f,g$ of weight $2k$ and $2\ell$ that $f\cdot g$ is a modular form of weight $2(k + \ell)$.

We can also define modular forms of half-integral weight if we are a little more careful. However, the functional equation $$ f\Big(\frac{az + b}{cz+d}\Big) = (cz+d)^{2k}f(z) $$ must be replaced with something more subtle.

In particular, the Dedekind $\eta$-function is a modular form of weight 1/2; it satisfies $$ \eta(z + 1) = e^{\frac{\pi i}{12}}\eta(z) \qquad \eta\Big(-\frac{1}{z}\Big) = \sqrt{-iz}\ \eta(z) $$ Now, it would be nice if $\eta^4$ were a modular form of weight 2; however, an easy check using the above relations shows that this is only the case up to roots of unity, and so $\eta^4$ is not a multiple of $G_2$.

My question is then the following: What is the relation between modular forms of half-integral weight and (quasi-)modular forms of even integer weight? I know that $\eta^{24}$ is an honest modular form of weight 12, so I'm more curious about the general setting, or even what can be said about things like $\eta^4$.

Edit: As was pointed out in the comments, $\eta^4$ is a modular form for a congruence subgroup of $SL_2(\mathbb{Z})$, but not for the full modular group. The space of modular forms for the full modular group is generated by $G_4$ and $G_6$ (with $G_2$ thrown in if we are looking at the space of quasi-modular forms); is there a corresponding statement for modular forms on congruence subgroups?

share|improve this question
1  
I'm not sure what you are getting at, but $\eta^4$ is an honest modular form of weight $2$, but not for the whole modular group $\mathrm{SL}_2(\mathbb{Z})$ but for a congurence subgroup thereof. –  Robin Chapman Sep 8 '10 at 18:47
    
Ah, that might be my misunderstanding. –  Simon Rose Sep 8 '10 at 19:22

3 Answers 3

One of the best-known relations between modular forms of half integral weight and even weight is the Shimura correspondence taking certain weight k+1/2 forms to weight 2k forms. See Koblitz's book or Shimura's paper for details.

By the way, η4 is indeed a modular form of weight 2 for the whole modular group SL2(Z), for a nontrivial character of this group. For half-integral weight forms one needs to use a character of the metaplectic double cover in the functional equation; this is what the funny roots of unity in the functional equation of η are doing.

share|improve this answer

To answer Scott; yes, every modular form of integer weight $k$ for a congruence group can be expressed rationally in terms of $\eta(r\tau)$ for rational $r$. For a start, $g=f/\eta^{2k}$ is a modular function for some $\Gamma(N)$, so it suffices to consider these. Replacing $g$ by $g(N\tau)$ we can reduce to the case where $g$ is a modular function for some $\Gamma_0(M)$. The field of modular functions for $\Gamma_0(M)$ is generated by $j(\tau)$ and $j(M\tau)$. So all we need is that $j(\tau)$ expressible in terms of the $\eta(r\tau)$. This is well-known; one can express $j$ in terms of the Weber functions, and them in terms of $\eta(\tau)$, $\eta(2\tau)$ and $\eta(\tau/2)$.

Note that this argument is essentially the same as the proof for a very similar theorem in this paper by Kilford.

share|improve this answer

Regarding the question you added later, the ring of modular forms for a congruence group is finitely generated, but for large level it can be computationally difficult to find explicit generators. I believe you can always write the generators as rational functions in $\eta(r\tau)$ for $r$ ranging over rational numbers, but it is not clear to me that this is a useful method of presentation.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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