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I hope that this question is appropriate here. I am Interested in the Proof for the named theorem only using modular forms. I read on Wikipedia that the Proof actually breaks down to an Identitiy for theta series on the congruence sub Group $\Gamma_1(4)$ and weight $k=2$. I have to say that I only know some Basic stuff About modular forms and only very little to nothing About theta series at this point. Also I want to ask if one needs Lambert series for this Proof? I would appreciate any help. And maybe you know a good Source for a Proof using this method.

Davood
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deavor
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2 Answers2

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This is detailed p.11 and p.95 of Diamond & Shurman book.

Let $\vartheta(z) = \sum_{n=-\infty}^\infty e^{2i \pi n^2 z}$ and $r_4(n) = \# \{v \in \mathbb{Z},n = \sum_{j=1}^4 v_j^2 \}$ and $f(z) = \sum_{n=1}^\infty r_4(n)e^{2i \pi n z}$. Then $$\vartheta(z)^4 = f(z)$$

  • Using the Poisson formula show that $\vartheta(z)$ is modular of weight $1/2$ for $\Gamma_0(4)$, and together with the fact that $\vartheta(z)^4$ is holomorphic at the cusp, you get that $f(z) \in M_2(\Gamma_0(4))$ the vector space of modular forms of weight $2$ for $\Gamma_0(4)$.

  • Furthermore $G_{2,2}(z) = G_2(z) - 2 G_2(2z)$ and $G_{2,4}(z) = G_2(z) - 4 G_2(4z)$ are linearly independent and are in $M_2(\Gamma_0(4))$

  • The Riemann-Roch theorem applied to the compact Riemann surface $X(\Gamma_0(4))$ which is $Y(\Gamma_0(4))=\mathcal{H}\setminus \Gamma_0(4)$ compactified with the cusps added, give $\text{dim}(M_2(\Gamma_0(4))=2$.

  • And hence $f(z) = \alpha G_{2,2}(z)+\beta G_{2,4}(z)$ for some $\alpha,\beta \in \mathbb{C}$. Computing the first few coefficients give $\alpha = 0,\beta=\frac{-1}{\pi^2}$, i.e. with the Fourier expansion of $G_k(z) = 1+C_k\sum_{n=1}^\infty e^{2i \pi n z} \sum_{d |n} d^{k-1}$ : $$r_4(n) = 8 \sum_{d | n, 4 \nmid d} d$$

Community
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reuns
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  • Thank you for your work. I have one question. Is showing that $\vartheta(z) $is modular of weight $1/2$ for $\Gamma_0(4)$ equivalent to showing that $\vartheta(z)^4$ is modular of weight 2 for the Thetagroup $\Gamma_\vartheta$? – deavor Nov 23 '16 at 01:36
  • @Deavor What do you mean exactly with "showing that $\vartheta(z)^4$ is modular of weight 2 for the Thetagroup $Γ_\vartheta$" ? And there are good chances for showing $\vartheta(z)^4$ is modular for something, you'll use that $\vartheta(\frac{-1}{4z}) = (-2i z)^{1/2} \vartheta(z)$ using the Poisson summation formula (from which you have the functional equation for $\zeta(s)$) – reuns Nov 23 '16 at 01:41
  • I mean showing these conditions: $(M1)$ $f$ is holomorphic on $\mathbb{H}$. $(M2)$ $f|{k}L=\chi(L) \cdot f$ for all $L\in \Lambda.$ $(M3)$ $f|{k}M$ is holomorphic for all $M \in \Gamma$ in $\infty$. Where $\Lambda=\Gamma_\vartheta$ the group generated by $$S=\begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix} \ \text{and} \ T^2=\begin{pmatrix} 1 & 2 \ 0 & 1 \end{pmatrix}.$$ – deavor Nov 23 '16 at 01:43
  • @Deavor Yes, more or less. $\vartheta(z+1) = \vartheta(z)$ (note here it is $2i \pi n^2 z$) i.e. $f|_2 \alpha(z) = f(z)$ with $\alpha = \begin{pmatrix} 1 & 1 \ 0 &1 \end{pmatrix}$, and $\vartheta(-1/4z) = (-2i \pi z)\vartheta(z)$, combining those two you have $f|_2 \gamma(z) = f(z)$ for $\gamma = \begin{pmatrix} 1 & 0 \ -4 &1 \end{pmatrix}$, and those two matrices generate $\Gamma_0(4)$ (download 'diamond shurman a first course in modular forms pdf' you'll have the details) – reuns Nov 23 '16 at 01:49
  • Ah ok. Just one last question if you don't mind. Whats is the difference between defining $\vartheta(z)=\sum_{n\in \mathbb{z}}q^{n^2}$, with $q=e^{\pi \mathrm{i}z}$ or with $q=e^{2\pi\mathrm{i}z}$ except for $1\neq2$ – deavor Nov 23 '16 at 01:50
  • @Deavor Any modular form (for any congruence subgroup) is $1$ periodic ($ \begin{pmatrix} 1 & 1 \ 0 &1 \end{pmatrix}$ invariant) , otherwise you have to consider $\vartheta(2z)$. And I corrected an important typo for $\gamma$ – reuns Nov 23 '16 at 01:52
  • Hmm another definition I got for $\Gamma_\vartheta$ is $$\Gamma_\vartheta:=\left{M \in \Gamma: M \equiv \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\ \ (\operatorname{mod}2 ) \ \text{or} \ M\equiv \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \ (\operatorname{mod}2 )\right}.$$ So $\begin{pmatrix} 1 & 1 \ 0 &1 \end{pmatrix} \notin \Gamma_\vartheta$. Furthermore, I took $q=e^{\pi \mathrm{i}z}$ and got $$\vartheta^4|2T^2(z)=\vartheta^4(T^2z)=\vartheta^4(z+2)=\chi\vartheta(T^2)\vartheta^4(z)$$ – deavor Nov 23 '16 at 01:55
  • @Deavor but $ \begin{pmatrix} 1 & 2 \ 0 &1 \end{pmatrix}$ is in your group, so with $z \to z/2$ or something it should be the same. And you can consider a smaller group $\Gamma$, but you'll need an other dimension formula for $M_k(\Gamma)$ and a known basis of modular forms, for proving that $f(z) = \beta G_{2,4}$ – reuns Nov 23 '16 at 01:56
  • I think I got it. So Its actually nearly the same. Thank you for your patience – deavor Nov 23 '16 at 02:01
  • I used a difference theorem: If $f$ is analytic and $(1)$ $f(z+2)=f(z), \ f\left(-\frac{1}{z}\right)=\sqrt{\frac{z}{\mathrm{i}}}^rf(z),$ $(2)$$\lim_{y\to \infty}f(z)$ exists, $(3)$ $\lim_{y\to \infty}\sqrt{\frac{z}{\mathrm{i}}}^{-r} f\left(1-\frac{1}{z}\right)e^{-\frac{\pi\mathrm{i}rz}{4}}$ exists.

    Then $$f(z)=c\cdot \vartheta(z)^r, \ \text{with} \ c=\lim_{y\to \infty}f(z).$$ I then took the solution and just prove that it satisfy the conditions. But I see your way is the one to approach a actual solution. However, until I am at this level, I have to learn more.

     – deavor Nov 23 '16 at 02:02
  • @Deavor Do you have some references for this theorem ? Yes what I wrote is a way (the modular way) to prove it for $r = 4$, but for a larger $r$ I would have some troubles. – reuns Nov 23 '16 at 02:06
  • @Deavor next time you should give a little more context, tell exactly what you want to do and learn (because as you see the field is really large) – reuns Nov 23 '16 at 02:15
  • Yes you are right. However, I learned much from our conversation and your answer and all in all there are no questions left. – deavor Nov 23 '16 at 02:17
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Don Zagier's chapter 'Elliptic Modular Forms and Their Applications' from the book '1-2-3 of modular forms' has a section on theta series and how it can be used to prove the sum of two and four squares theorem.

winawer
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