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Modular forms, Hecke Operator, translation property

  1. Jul 6, 2017 #1
    1. The problem statement, all variables and given/known data

    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)## )



    2. Relevant equations

    look above, look below,


    3. The attempt at a solution

    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:

    heksol.png

    Many thanks
     
  2. jcsd
  3. Jul 6, 2017 #2

    Dick

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    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})##.
     
  4. Jul 6, 2017 #3
    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##?
     
  5. Jul 6, 2017 #4

    Dick

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    ## 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.
     
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