Modular forms, Hecke Operator, translation property

[binbagsss]

Homework Statement

I am trying to follow the attached solution to show that $T_{p}f(\tau+1)=T_pf(\tau)$

Where $T_p f(\tau) p^{k-1} f(p\tau) + \frac{1}{p} \sum\limits^{p-1}_{j=0}f(\frac{\tau+j}{p})$

Where $M_k(\Gamma)$ denotes the space of modular forms of weight $k$
(So we know that $f(\tau+1)=f(\tau)$ )

Homework Equations

look above, look below,

The Attempt at a Solution

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QUESTION

1) Using $f(\tau+1)=f(\tau)$, similarly to what has been done for the first term in going from the second line to the first line , see solution below, I don't understand why we can't don't have :

$\frac{1}{p}\sum\limits^{p-1}_{j=0} f(\frac{\tau+j+1}{p}) =\frac{1}{p}\sum\limits^{p-1}_{j=0} f(\frac{\tau+j}{p})$ via $\tau'=\frac{\tau+j}{p}$ so then we have:

$\frac{1}{p}\sum\limits^{p-1}_{j=0} f(\frac{\tau'+1}{p}) =\frac{1}{p}\sum\limits^{p-1}_{j=0} f(\frac{\tau'}{p})$Solution attached:

Many thanks

 

[Dick]

QUESTION

1) Using $f(\tau+1)=f(\tau)$, similarly to what has been done for the first term in going from the second line to the first line , see solution below, I don't understand why we can't don't have :

$\frac{1}{p}\sum\limits^{p-1}_{j=0} f(\frac{\tau+j+1}{p}) =\frac{1}{p}\sum\limits^{p-1}_{j=0} f(\frac{\tau+j}{p})$ via $\tau'=\frac{\tau+j}{p}$ so then we have:

$\frac{1}{p}\sum\limits^{p-1}_{j=0} f(\frac{\tau'+1}{p}) =\frac{1}{p}\sum\limits^{p-1}_{j=0} f(\frac{\tau'}{p})$

I don't know a lot about modular forms, but I do follow their proof. As for your question, I could see why $f(\frac{\tau+j}{p}+1)=f(\frac{\tau+j}{p})$ but I don't see why you think $f(\frac{\tau+j+1}{p})=f(\frac{\tau+j}{p})$.

 

[binbagsss]

I don't know a lot about modular forms, but I do follow their proof. As for your question, I could see why $f(\frac{\tau+j}{p}+1)=f(\frac{\tau+j}{p})$ but I don't see why you think $f(\frac{\tau+j+1}{p})=f(\frac{\tau+j}{p})$.

Thank you for your reply.
I see $f(\frac{\tau+j}{p}+1)=f(\frac{\tau+j}{p})$

But I thought $f(\frac{\tau+j+1}{p})=f(\frac{\tau+j}{p})$ is true too for the same reason that $f(p(t+1))=f(p(t))$- factoring out the $p$.. so i thought it would also be true that $f(\frac{1}{p}(t+1))=f(\frac{1}{p}t)$ and then using this but were we shift $t$ by $j$?

 

[Dick]

Thank you for your reply.
I see $f(\frac{\tau+j}{p}+1)=f(\frac{\tau+j}{p})$

But I thought $f(\frac{\tau+j+1}{p})=f(\frac{\tau+j}{p})$ is true too for the same reason that $f(p(t+1))=f(p(t))$- factoring out the $p$.. so i thought it would also be true that $f(\frac{1}{p}(t+1))=f(\frac{1}{p}t)$ and then using this but were we shift $t$ by $j$?

$f(p(t+1))=f(p(t))$ isn't true because of any 'factoring out of $p$'. It's true because $f(p(t+1))=f(pt+p)$ and p is an integer. If $f(t+1)=f(t)$ then also $f(t+2)=f(t+1)=f(t)$. So $f(t+n)=f(t)$ for $n$ any integer. $f$ is periodic with period 1.