We have a reductive group $G/\mathbb{Q}$ and a representation space $V$ of this group.

Let $K$ be an open subgroup of $G(\mathbb{A}_{\mathbb{Q}})$ (where $\mathbb{A}_{\mathbb{Q}}$ are adeles of $\mathbb{Q}$) with some nice properties that wont concern us.

Define the space of "algebraic modular forms":

$\{f: G(\mathbb{A}_{\mathbb{Q}})\rightarrow V\,|\,f(gk) = f(g) \text{ for all } k\in K, g\in G(\mathbb{A}_{\mathbb{Q}}) \text{ and } f(\gamma g) = \gamma f(g) \text{ for all } \gamma\in G(\mathbb{Q})\}$

Now assume that $G(\mathbb{Q}) \backslash G(\mathbb{A}_{\mathbb{Q}})/K$ is finite, with reps $z_1, z_2, ..., z_h\in G(\mathbb{A}_{\mathbb{Q}})$.

The paper I am reading claims that you determine an algebraic modular form $f$ as soon as you specify the values $f(z_1), f(z_2), ..., f(z_m)\in V$.

I can't see why though.

Suppose $g\in G(\mathbb{A}_{\mathbb{Q}})$. Then $g = \gamma z_i k$ tells us that $f(g) = \gamma f(z_i)$.

Surely we have a dependence on $\gamma$ too? I hope I haven't missed anything simple.

  • $\begingroup$ Can you give a link to said paper? $\endgroup$ – Eugene Jun 9 '12 at 19:04
  • $\begingroup$ neil-dummigan.staff.shef.ac.uk/simpletrace3.pdf $\endgroup$ – fretty Jun 9 '12 at 19:05
  • $\begingroup$ I don't think the claim is that you can freely choose the $f(z_i)$, only that if you happen to have some algebraic modular form $f$ and you happen to learn the values $f(z_i)$ then you already know all the other values (which you've already shown). $\endgroup$ – Qiaochu Yuan Jun 9 '12 at 19:06
  • $\begingroup$ Maybe you might want to read this? $\endgroup$ – Eugene Jun 9 '12 at 19:08
  • $\begingroup$ I know you can't freely choose them, this is what the next part of the paper says (you have to choose the $f(z_i)$'s to be certain eigenvectors of things). I think I see your point, for any $g$ you can work out the corresponding $\gamma$ and then you will know $f(g)$. For some reason I just saw a dependence on $\gamma$. $\endgroup$ – fretty Jun 9 '12 at 19:09

Maybe a toy analogue will help here:

Proposition: Let $f$ be a function $\mathbf{R} \to \mathbf{R}$ such that $f(cx) = c f(x)$ for all $c \in \mathbf{R}$. Then $f$ is uniquely determined by $f(1)$.

I won't insult your intelligence by proving this! The point is that the statement you're trying to prove is essentially a more complicated version of the same thing. In our toy example it's not true that $f(x) = f(1)$ for all $x$, or anything like that, and you certainly need to know $x$ in order to know $f(x)$; but the function f is uniquely determined by its value at 1. And it's the same with algebraic modular forms.

  • $\begingroup$ Thanks for this, it just turns out I was just being dumb! I am actually Neil's current (first year) student and getting to grips with all of this stuff...most of it is completely new to me. I recognise your name from the paper (and from NRICH a long time ago). I would be very interested to hear about the calculations you did here... $\endgroup$ – fretty Jun 10 '12 at 11:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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