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.