Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

To be specific, do the weights have some geometric meanings? A modular form $f$ satisfies $f(\frac{az+b}{cz+d})(cz+d)^{-2k}=f(z)$, where $z\in \mathbb{C}$. $k$ or $2k$ is called the weight of $f$.

share|improve this question

1 Answer 1

Well, it depends on what you mean by geometric. As was recently discussed in another question, modular forms of weight $k$ can be identified with sections of the $k^{th}$ tensor power of a certain line bundle, but I don't know much about this so I won't say more.

Here's another answer. Recall that $G = \text{SL}_2(\mathbb{R})$ acts transitively on the upper half plane $\mathbb{H}$ with stabilizer $K = \text{SO}(2)$, so that we may identify $\mathbb{H}$ with the homogeneous space $G/K$ and modular forms with certain functions $f : G/K \to \mathbb{C}$ which transform in a nice way under the action of $H = \text{SL}_2(\mathbb{Z})$. In other words, modular forms can be identified with certain functions on $G$ which are invariant under $K$ and almost invariant under $H$.

It turns out we can trade "invariant" and "almost invariant" above. Define

$$\phi_f \left( \left[ \begin{array}{cc} a & b \\\ c & d \end{array} \right] \right) = f \left( \frac{ai + b}{ci + d} \right) (ci + d)^{-k}.$$

This defines a new function on $G$ which is now invariant under $H$ and almost invariant under $K$: now $K$ acts according to the representation corresponding to $k$. This is a special case of a general construction that I don't fully understand. I think has something to do with induced representations; I don't understand that connection either.

share|improve this answer
    
You mentioned that $G=SL_2(\mathbb{R})$ acts transitively on the upper half plane with stabilizer $K$.Can you say a little more on this? And on the $f:G/K\longrightarrow \mathbb{C}$,any referrence? Thank you very much for your kind help. –  user14242 Aug 6 '11 at 10:46
    
A matrix in $\text{SL}_2(\mathbb{R})$ acts by $z \mapsto \frac{az+b}{cz+d}$. It's a good exercise to show that this action is transitive, and you can compute that the stabilizer of $i$ is isomorphic to the subgroup of rotation matrices. The only reference I know for this stuff only mentions it very, very briefly, so I don't think it would help much. –  Qiaochu Yuan Aug 6 '11 at 14:30

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.