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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.

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Can you give a link to said paper? –  Eugene Jun 9 '12 at 19:04
    
    
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). –  Qiaochu Yuan Jun 9 '12 at 19:06
    
Maybe you might want to read this? –  Eugene Jun 9 '12 at 19:08
    
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 $\gamma$. –  fretty Jun 9 '12 at 19:09

1 Answer 1

up vote 3 down vote accepted

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.

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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... –  fretty Jun 10 '12 at 11:31

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