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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 $f$.

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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.

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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

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