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Theta function squared is a weight $1$ modular form


eisenstein part of theta functionModular forms on the theta groupOn the square coeffecients of a modular formModular forms on $Gamma_0(N)$ with character in SageReverse the twisting of modular formCusp condition of modular forms of half-integral weightWriting an integer as a sum of two squares: Why is this a modular form of weight 1?*WHY* are half-integral weight modular forms defined on congruence subgroups of level 4N?Hecke operator on half integral weight modular formGenerating function for sum of two squares is a modular form













2












$begingroup$


Let
$$vartheta(tau) = sum_{ninmathbb{Z}}e^{pi in^2tau}.$$
I know that $vartheta$ satisfies the transfromation properties
$$vartheta(tau + 2) = vartheta(tau), quad varthetaleft(-frac{1}{tau}right) = sqrt{frac{tau}{i}}vartheta(tau).$$
What I am interested in is the transformation properties of
$$f(tau) = vartheta(2tau)^2.$$
I have seen it mentioned in some notes by Kevin Buzzard and elsewhere that $f(tau)$ is a weight $1$ modular form on the congruence subgroup $Gamma_0(4)$ with Nebentypus $chi$, the Dirichlet character coming from the Legendre symbol mod $4$. In other words,
$$fleft(frac{atau + b}{ctau + d}right) = chi(d)(ctau + d)f(tau), quad begin{pmatrix} a & b \ c & d end{pmatrix} in Gamma_0(4).$$
I cannot find a reference for this fact nor does it seem obvious from the transformation properties of $vartheta(tau)$. I know that given a modular form $g(tau)$ of weight $k$ on $Gamma_0(N)$ for some $N$, the function $g(2tau)^2$ is modular of weight $2k$ on $Gamma_0(2N)$. But the problem is that the subgroup on which $vartheta(tau)$ is modular (of weight $1/2$) is not one of the congruence subgroups $Gamma_0(N)$. I would greatly appreciate any help tracking down a reference for this fact.










share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Let
    $$vartheta(tau) = sum_{ninmathbb{Z}}e^{pi in^2tau}.$$
    I know that $vartheta$ satisfies the transfromation properties
    $$vartheta(tau + 2) = vartheta(tau), quad varthetaleft(-frac{1}{tau}right) = sqrt{frac{tau}{i}}vartheta(tau).$$
    What I am interested in is the transformation properties of
    $$f(tau) = vartheta(2tau)^2.$$
    I have seen it mentioned in some notes by Kevin Buzzard and elsewhere that $f(tau)$ is a weight $1$ modular form on the congruence subgroup $Gamma_0(4)$ with Nebentypus $chi$, the Dirichlet character coming from the Legendre symbol mod $4$. In other words,
    $$fleft(frac{atau + b}{ctau + d}right) = chi(d)(ctau + d)f(tau), quad begin{pmatrix} a & b \ c & d end{pmatrix} in Gamma_0(4).$$
    I cannot find a reference for this fact nor does it seem obvious from the transformation properties of $vartheta(tau)$. I know that given a modular form $g(tau)$ of weight $k$ on $Gamma_0(N)$ for some $N$, the function $g(2tau)^2$ is modular of weight $2k$ on $Gamma_0(2N)$. But the problem is that the subgroup on which $vartheta(tau)$ is modular (of weight $1/2$) is not one of the congruence subgroups $Gamma_0(N)$. I would greatly appreciate any help tracking down a reference for this fact.










    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Let
      $$vartheta(tau) = sum_{ninmathbb{Z}}e^{pi in^2tau}.$$
      I know that $vartheta$ satisfies the transfromation properties
      $$vartheta(tau + 2) = vartheta(tau), quad varthetaleft(-frac{1}{tau}right) = sqrt{frac{tau}{i}}vartheta(tau).$$
      What I am interested in is the transformation properties of
      $$f(tau) = vartheta(2tau)^2.$$
      I have seen it mentioned in some notes by Kevin Buzzard and elsewhere that $f(tau)$ is a weight $1$ modular form on the congruence subgroup $Gamma_0(4)$ with Nebentypus $chi$, the Dirichlet character coming from the Legendre symbol mod $4$. In other words,
      $$fleft(frac{atau + b}{ctau + d}right) = chi(d)(ctau + d)f(tau), quad begin{pmatrix} a & b \ c & d end{pmatrix} in Gamma_0(4).$$
      I cannot find a reference for this fact nor does it seem obvious from the transformation properties of $vartheta(tau)$. I know that given a modular form $g(tau)$ of weight $k$ on $Gamma_0(N)$ for some $N$, the function $g(2tau)^2$ is modular of weight $2k$ on $Gamma_0(2N)$. But the problem is that the subgroup on which $vartheta(tau)$ is modular (of weight $1/2$) is not one of the congruence subgroups $Gamma_0(N)$. I would greatly appreciate any help tracking down a reference for this fact.










      share|cite|improve this question









      $endgroup$




      Let
      $$vartheta(tau) = sum_{ninmathbb{Z}}e^{pi in^2tau}.$$
      I know that $vartheta$ satisfies the transfromation properties
      $$vartheta(tau + 2) = vartheta(tau), quad varthetaleft(-frac{1}{tau}right) = sqrt{frac{tau}{i}}vartheta(tau).$$
      What I am interested in is the transformation properties of
      $$f(tau) = vartheta(2tau)^2.$$
      I have seen it mentioned in some notes by Kevin Buzzard and elsewhere that $f(tau)$ is a weight $1$ modular form on the congruence subgroup $Gamma_0(4)$ with Nebentypus $chi$, the Dirichlet character coming from the Legendre symbol mod $4$. In other words,
      $$fleft(frac{atau + b}{ctau + d}right) = chi(d)(ctau + d)f(tau), quad begin{pmatrix} a & b \ c & d end{pmatrix} in Gamma_0(4).$$
      I cannot find a reference for this fact nor does it seem obvious from the transformation properties of $vartheta(tau)$. I know that given a modular form $g(tau)$ of weight $k$ on $Gamma_0(N)$ for some $N$, the function $g(2tau)^2$ is modular of weight $2k$ on $Gamma_0(2N)$. But the problem is that the subgroup on which $vartheta(tau)$ is modular (of weight $1/2$) is not one of the congruence subgroups $Gamma_0(N)$. I would greatly appreciate any help tracking down a reference for this fact.







      number-theory modular-forms theta-functions






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 7 at 4:58









      Ethan AlwaiseEthan Alwaise

      6,446717




      6,446717






















          1 Answer
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          active

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          1












          $begingroup$

          $g = vartheta^2(2tau)$ is invariant under $Gamma_1(4)$ is easily checked: since $Gamma_1(4)$ is generated by
          $$begin{pmatrix}1 & 1 \ 0 & 1 end{pmatrix}qquad begin{pmatrix}1 & 0 \ 4 & 1 end{pmatrix}$$
          Obviously $g$ is weight-$1$ invariant under the first one. For second one,
          $$begin{aligned}gleft[begin{pmatrix}1 & 0 \ 4 & 1 end{pmatrix}right]_1 (tau)&= (4tau +1)^{-1}vartheta^2(frac{2tau}{4tau+1}) \&= (4tau +1)^{-1} frac{i(4tau+1)}{2tau}vartheta^2(-frac{4tau+1}{2tau})
          \& = frac{i}{2tau}vartheta^2(frac{-1}{2tau}) = vartheta^2(2tau)
          end{aligned}$$



          To check $g$ has nebentypus $chi$, it suffices to show $$gleft[begin{pmatrix}-1 & 0 \ 4 & -1 end{pmatrix}right]_1(tau)=-g(tau)$$
          I shall leave this checking to you.






          share|cite|improve this answer











          $endgroup$













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            1












            $begingroup$

            $g = vartheta^2(2tau)$ is invariant under $Gamma_1(4)$ is easily checked: since $Gamma_1(4)$ is generated by
            $$begin{pmatrix}1 & 1 \ 0 & 1 end{pmatrix}qquad begin{pmatrix}1 & 0 \ 4 & 1 end{pmatrix}$$
            Obviously $g$ is weight-$1$ invariant under the first one. For second one,
            $$begin{aligned}gleft[begin{pmatrix}1 & 0 \ 4 & 1 end{pmatrix}right]_1 (tau)&= (4tau +1)^{-1}vartheta^2(frac{2tau}{4tau+1}) \&= (4tau +1)^{-1} frac{i(4tau+1)}{2tau}vartheta^2(-frac{4tau+1}{2tau})
            \& = frac{i}{2tau}vartheta^2(frac{-1}{2tau}) = vartheta^2(2tau)
            end{aligned}$$



            To check $g$ has nebentypus $chi$, it suffices to show $$gleft[begin{pmatrix}-1 & 0 \ 4 & -1 end{pmatrix}right]_1(tau)=-g(tau)$$
            I shall leave this checking to you.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              $g = vartheta^2(2tau)$ is invariant under $Gamma_1(4)$ is easily checked: since $Gamma_1(4)$ is generated by
              $$begin{pmatrix}1 & 1 \ 0 & 1 end{pmatrix}qquad begin{pmatrix}1 & 0 \ 4 & 1 end{pmatrix}$$
              Obviously $g$ is weight-$1$ invariant under the first one. For second one,
              $$begin{aligned}gleft[begin{pmatrix}1 & 0 \ 4 & 1 end{pmatrix}right]_1 (tau)&= (4tau +1)^{-1}vartheta^2(frac{2tau}{4tau+1}) \&= (4tau +1)^{-1} frac{i(4tau+1)}{2tau}vartheta^2(-frac{4tau+1}{2tau})
              \& = frac{i}{2tau}vartheta^2(frac{-1}{2tau}) = vartheta^2(2tau)
              end{aligned}$$



              To check $g$ has nebentypus $chi$, it suffices to show $$gleft[begin{pmatrix}-1 & 0 \ 4 & -1 end{pmatrix}right]_1(tau)=-g(tau)$$
              I shall leave this checking to you.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                $g = vartheta^2(2tau)$ is invariant under $Gamma_1(4)$ is easily checked: since $Gamma_1(4)$ is generated by
                $$begin{pmatrix}1 & 1 \ 0 & 1 end{pmatrix}qquad begin{pmatrix}1 & 0 \ 4 & 1 end{pmatrix}$$
                Obviously $g$ is weight-$1$ invariant under the first one. For second one,
                $$begin{aligned}gleft[begin{pmatrix}1 & 0 \ 4 & 1 end{pmatrix}right]_1 (tau)&= (4tau +1)^{-1}vartheta^2(frac{2tau}{4tau+1}) \&= (4tau +1)^{-1} frac{i(4tau+1)}{2tau}vartheta^2(-frac{4tau+1}{2tau})
                \& = frac{i}{2tau}vartheta^2(frac{-1}{2tau}) = vartheta^2(2tau)
                end{aligned}$$



                To check $g$ has nebentypus $chi$, it suffices to show $$gleft[begin{pmatrix}-1 & 0 \ 4 & -1 end{pmatrix}right]_1(tau)=-g(tau)$$
                I shall leave this checking to you.






                share|cite|improve this answer











                $endgroup$



                $g = vartheta^2(2tau)$ is invariant under $Gamma_1(4)$ is easily checked: since $Gamma_1(4)$ is generated by
                $$begin{pmatrix}1 & 1 \ 0 & 1 end{pmatrix}qquad begin{pmatrix}1 & 0 \ 4 & 1 end{pmatrix}$$
                Obviously $g$ is weight-$1$ invariant under the first one. For second one,
                $$begin{aligned}gleft[begin{pmatrix}1 & 0 \ 4 & 1 end{pmatrix}right]_1 (tau)&= (4tau +1)^{-1}vartheta^2(frac{2tau}{4tau+1}) \&= (4tau +1)^{-1} frac{i(4tau+1)}{2tau}vartheta^2(-frac{4tau+1}{2tau})
                \& = frac{i}{2tau}vartheta^2(frac{-1}{2tau}) = vartheta^2(2tau)
                end{aligned}$$



                To check $g$ has nebentypus $chi$, it suffices to show $$gleft[begin{pmatrix}-1 & 0 \ 4 & -1 end{pmatrix}right]_1(tau)=-g(tau)$$
                I shall leave this checking to you.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Mar 12 at 5:28

























                answered Mar 12 at 5:19









                piscopisco

                11.9k21743




                11.9k21743






























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