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In p. 476 of volume 3 of Gauss's collected works, appear several interesting identities on Jacobi theta functions which were used by the czech mathematician Karel Petr in an 1904 article "Bemerkung zur einer Gausschen Formel über die Thetafunktionen" to derive relations giving the number of representations of a number $N$ by three quaternary quadratic forms: $x^2 + y^2 + 9z^2 + 9u^2$, $x^2+y^2+z^2+9u^2$, $x^2+9y^2+9z^2+9u^2$. The identities by Gauss are:

$$(\frac{3P^2-P^0\cdot P^0}{2})^2= p^4-4(\frac{pqr}{2})^{\frac{4}{3}}$$ $$(\frac{3Q^2-Q^0\cdot Q^0}{2})^2=q^4+4(\frac {pqr}{2})^{\frac{4}{3}}$$

where the relevant quantities are defined to be:

$$P(x^3,1)=P , P(x,1)=p, P^0=P(x^{\frac{1}{3}},1)$$ $$Q(x^3,1)=Q , Q(x,1)=q, Q^0 = Q(x^{\frac{1}{3}},1)$$ $$R(x^3,1)=R , R(x,1)= r$$

and the three functions $P(x,y),Q(x,y),R(x,y)$ are equivalent to Jacobi's theta functions $\vartheta_3,\vartheta_4,\vartheta_2$ and defined to be:

$$P(x,y)=1+x(y+\frac{1}{y})+x^4(y^2+\frac{1}{y^2})+x^9(y^3+\frac{1}{y^3})+...$$ $$Q(x,y)= 1-x(y+\frac{1}{y})+x^4(y^2+\frac{1}{y^2})-x^9(y^3+\frac{1}{y^3})+...$$ $$R(x,y)=x^{\frac{1}{4}}(y^{\frac{1}{2}}+y^{-\frac{1}{2}})+x^{\frac{9}{4}}(y^{\frac{3}{2}}+y^{-\frac{3}{2}})+x^{\frac{25}{4}}(y^{\frac{5}{2}}+y^{-\frac{5}{2}})+...$$

To see the equivalency between Gauss's notation and Jacobi's theta functions, look at the post Interpretation of a certain general theorem used by Gauss in his work on theta functions.. User Paramanand Singh helped me understand the meaning of Gauss's identities and remarked that they are essentialy a "mixed modular equation" which connects theta functions of $x^{\frac{1}{3}},x,x^3$ (or, equivalently, $\tau,\tau^3,\tau^9$), and said that Ramanujan also gave this modular equation.

According to Paramanand Singh's comments, the following identity (from p. 142 of the book "pi and the AGM") of Ramanujan is equivalent to Gauss's:

$$\frac{\theta_3(q)}{\theta_3(q^9)} - 1 = (\frac{\theta_3^4(q^3)}{\theta_3^4(q^9)}-1)^{\frac{1}{3}}$$

However, he didn't provide proof of equivalency. Therefore, it's not certain this is the desired identity of Ramanujan, and it's not even certain that Ramanujan stated an equivalent identity at all.

Therefore, my questions are:

  • Can anyone familiar with Ramanujan's results on mixed modular equations say which of Ramanujan's theorems is equivalent to Gauss's identities? and where can i find the relevant fragment of Ramanujan's writings on the internet?
  • Can someone also give a proof of equivalency between Ramanujan's theorem and Gauss's identities? this doesn't have to be a proof of correctness of Gauss's identities (which can be quite complicated), just a proof of equivalency to Ramanujan's theorem.
user2554
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  • Do you know how to use the Sturm bound to prove such algebraic equations between modular forms ? So all you need is to prove the modularity of a few theta series then take a power of your identities and multiply with $\theta_3^n$ to obtain identifies in $M_k(\Gamma)$ with integer $k$ then compare the first few coefficients. – reuns Dec 16 '20 at 14:57
  • @reuns - no, i don't know how to use the Sturm bound, and my knowledge on such advanced mathematical topics like modular forms tends to zero. Anyway, feel free to give additional points of view, as other perspectives can also shed light on Gauss's motivations and not just on a proof using modern tools. – user2554 Dec 16 '20 at 21:50

3 Answers3

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The Broweins' pi and the AGM identity

$$ \frac{\theta_3(q)}{\theta_3(q^9)} - 1 = \left(\frac{\theta_3^4(q^3)}{\theta_3^4(q^9)}-1\right)^{\frac{1}{3}} \tag{1} $$

appears in a slightly modified form in Bruce Berndt, Ramanujan's Notebooks, Part III, page 218, equation (24.29)

$$ \frac{\phi(q^\frac13)}{\phi(q^3)} = 1 + \left(\frac{\phi^4(q)}{\phi^4(q^3)} - 1\right)^\frac13. \tag{2}$$

The Gauss identity (with a typo fixed)

$$ \left(\frac{3P^2-P^0\cdot P^0}{2}\right)^2= p^4-4\left(\frac{p\,q\,r}2\right)^{\frac{4}{3}} \tag{3} $$

after translating the theta functions into the Ramanujan notation

$$ q = x^{\frac13}, \quad \phi(q) := f(q,q), \quad f(-q) := f(-q,-q^2) \tag{4} $$

where the general Ramanujan theta function is

$$ f(a,b) := 1 +(a+b) +(a^2+b^2)(ab) +(a^3+b^3)(ab)^3 +\dots \tag{5} $$

and using the identity $$ \frac{p\,q\,r}2 = \left(q^\frac14 f(-q^6)\right)^3 \tag{6} $$

becomes

$$ \left(\frac{3\phi(q^9)^2 - \phi(q)^2}2\right)^2 = \phi(q^3)^4 - 4\,q\,f(-q^6)^4. \tag{7} $$

However, the identity in equation $(2)$ is very different than this equation $(7)$. In the notation of my eta product identity database, equation $(2)$ is $\texttt{q36_44_420}$ while equation $(7)$ is not in the database but would be $\texttt{x36_52_520}$. However, it is possible that several Ramanujan results combined would imply the Gauss identity.

Somos
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  • thanks for your confirmation that Gauss's identity is not equivalent to Ramanujan's identity. Can you please make some comments about interpretation of Gauss's identities? because these identities belong the deepest part of his work on theta functions so i'm very curius to understand more about their meaning. – user2554 Dec 15 '20 at 20:09
  • Thanks for translating in Ramanujan notation. I checked both Borwein's book as well as Berndt's books which give a host of such mixed modular equations. @user2554 : It appears that the one by Gauss is not listed or particularly well known and it needs to be derived using techniques given by Berndt. – Paramanand Singh Dec 16 '20 at 04:54
  • I had made an edit and later rolled back. Your identity $(7)$ is correct and verified by me in another answer here. However I am not sure how $4(pqr)^{4/3}$ equals $4qf(-q^6)^{4}$. The equation $(6)$ is also correct and based on this the term should be $2^{10/3}qf(-q^6)^4$. – Paramanand Singh Dec 16 '20 at 06:19
  • The proof in my answer is quite tedious and based on mostly complicated algebra (borrowed from Berndt). Do you think translating the identity in terms of Dedekind eta functions will help find a simpler proof? There is the approach based on modular forms which requires the evaluation of coefficients upto a certain order (mostly using symbolic software), but I am not even a novice in modular forms. – Paramanand Singh Dec 16 '20 at 13:31
  • @user2554 I have just written an answer to your other question linked about interpretation. – Somos Dec 16 '20 at 19:38
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The identity of Gauss has a typo and the correct form is $$\left(\frac{3P^2-P^0P^0}{2}\right)^2=p^4-4\left(\frac{pqr}{2}\right)^{4/3}\tag{1}$$ As in Somos's answer this translates to $$(3\phi^2(q^9)-\phi^2(q))^2=4\phi^4(q^3)-16qf^{4}(-q^6)\tag{2}$$ where $$\phi(q) =\vartheta_3(q)=\sum_{n\in\mathbb {Z}} q^{n^2},f(-q)=\prod_{n=1}^{\infty} (1-q^n)\tag{3}$$ We will use Ramanujan's notation and the proof borrows heavily from work of Dr. Bruce C. Berndt and his collaborators.

Given the nome $q$, the corresponding elliptic modulus $k$ and elliptic integral $K$ (or Ramanujan's $z$) are given in terms of theta functions $$\vartheta_2(q)=\sum_{n\in\mathbb {Z}} q^{(n+(1/2))^2},k=\frac{\vartheta_2^2(q)} {\vartheta_3^2(q)}, z=\frac{2K}{\pi}=\phi^2(q)\tag{4}$$ Let the parameter $k^2$ for nomes $q, q^3,q^9$ be denoted by $\alpha, \beta, \gamma $ respectively (so that for example $\beta=\vartheta_2^4(q^3)/\vartheta_3^4(q^3)$). Further let the parameter $z$ corresponding to these nomes be denoted by $z_1,z_3,z_9$ (so that $z_3=\phi^2(q^3)$).

The modular equations of degree 3 allow us to express $\alpha, \beta, \gamma $ in terms of multiplier $m$. Let us define $$m=\frac{z_1}{z_3},m'=\frac{z_3}{z_9}\tag{5}$$ Berndt gives us $$\alpha = \frac{(3 + m)^{3}(m - 1)}{16m^{3}}, \beta =\frac{(m - 1)^{3}(3 + m)}{16m}\tag{6}$$ and $$1-\alpha=\frac{(3 - m)^{3}(m + 1)}{16m^{3}}, 1-\beta =\frac{(m + 1)^{3}(3 - m)}{16m}\tag{7}$$ If we replace $m$ by $m'$ in above equations then $\alpha$ changes to $\beta$ and $\beta$ changes to $\gamma$. It should also be clear that if $0<q<1$ then we have $1<m,m'<3$.

Next Berndt uses another set of parameters $$u=\frac{m-1}{2}+\left(\frac{m-1}{2}\right)^2, v=\frac{m'-1}{2}+\left(\frac{m'-1}{2}\right)^2\tag{8}$$ (Berndt actually uses intermediate parameters $p, p'$ with $m=1+2p,m'=1+2p'$ and $q=p+p^2, q'=p'+p'^2$ instead of $u, v $, but this will conflict with the usage of $q$ for nome and hence we avoid it).

We have the obvious relations $$m=\sqrt {1+4u},m'=\sqrt{1+4v}\tag{9}$$ and the more difficult relations $$\alpha(1-\alpha)=u\left(\frac {2-u}{1+4u}\right)^3,\beta(1-\beta)=u^3\left(\frac{2-u}{1+4u}\right)\tag{10}$$ and $$\beta(1-\beta)=v\left(\frac{2-v}{1+4v}\right)^3,\gamma(1-\gamma)=v^3\left(\frac{2-v}{1+4v}\right)\tag{11}$$ Using the two expressions for $\beta(1-\beta)$ one gets a polynomial equation between $u, v$ of degree $4$ which Berndt successfully (!) solves using another parameter $t$ given by $v=2t^3$.

The algebra involved is tedious and I reproduce the necessary results directly from Berndt's Ramanujan's Notebooks Vol 3, page 354: $$m=\frac{(1+2t)^2}{\sqrt{1+8t^3}},m'=\sqrt{1+8t^3}\tag{12}$$ and $$\beta(1-\beta)=16t^3\left(\frac{1-t^3}{1+8t^3}\right)^3\tag{13}$$ The relation $(2)$ can be transcribed easily as $$\left(\frac{3}{m'}-m\right)^2=4-2^{8/3}(\beta(1-\beta))^{1/3}\tag{14}$$ which is easily verified using $(12),(13)$.


I am sure that Gauss would have provided a simpler proof using his relations between theta functions.

  • Thanks! i don't know what happened to me but i had a copy mistake and the identity of Gauss really has the term $4(\frac{pqr}{2})^{\frac{4}{3}}$ instead of what i wrote. – user2554 Dec 16 '20 at 07:53
  • @user2554: I have given a proof in my revised answer. Please do have a look. Also you need to have Berndt's books to have a full understanding of the proof. – Paramanand Singh Dec 16 '20 at 10:44
  • where can i find Berndt's books? because in order to make a truly deep reading of your proof i need to read some sources as a background. Anyway, obviously i'll try to understand your proof (i already had a look at it), although without background it might take some time. – user2554 Dec 16 '20 at 18:18
  • @user2554: these books should be available in some library. Otherwise they are a bit costly to obtain. I have presented a theory of Ramanujan theta functions on my blog (based on Berndt books and some other sources). Search for Ramanujan in this blog archive page. You may also search for theta functions and elliptic functions on blog. – Paramanand Singh Dec 17 '20 at 00:55
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I just want to describe the arithmetic information yielded by Gauss's identities. As in Somos's and Paramanand Singh's answers, Gauss's identity translates into the equation:

$$(3\phi^2(q^9)-\phi^2(q))^2 = 4\phi^4(q^3)-16qf^4(-q^6)$$

After opening the parentheses, this equation becomes:

$$9\phi^4(q^9)- 6\phi^2(q^9)\phi^2(q) +\phi^4(q) = 4\phi^4(q^3)-16qf^4(-q^6)\implies 6\phi^2(q^9)\phi^2(q) = 9\phi^4(q^9)+\phi^4(q)-4\phi^4(q^3)+16qf^4(-q^6)$$

Now, the coefficient of the $N$ power in the series expansion of the left side of the second equation, $6\phi^2(q^9)\phi^2(q)$, is none other than $6$ times the number of representations of $N$ by the quaternary quadratic form $x^2+y^2+9z^2+9u^2$. Denote this number as $\varphi(N)$. In addition the coefficient of the $N$ power in the series expansion of the term $\phi^4(q)$ is none other than the usual sum of four squares function, $r_4(N)$ (number of representations of $N$ by $x^2+y^2+z^2+u^2$), so let us denote this as $\chi(N)$.

For $N$ not divisible by $3$, the term $9\phi^4(q^9)-4\phi^4(q^3)$ contribute nothing to $\varphi(N)$ (these theta powers generate only powers of $q$ with exponents divisible by $3$). In addition, the coefficient of the $N$ power in the series expansion of the term $16qf^4(-q^6)$ can be interpreted as $16$ times the number of partitions of $\frac{N-1}{6}$ with at most $4$ repetitions of the same summand, with each partition counted as $(-1)$ or $(+1)$ if the partition is into an odd or even number of parts, respectively. Denote this as $16\gamma(\frac{N-1}{6})$.

Than we get:

$$\varphi(N)=\frac{1}{6}(\chi(N)+16\gamma(\frac{N-1}{6})), N\ne 0 \pmod 3$$

In p.300 of volume 2 of L. Dickson "history of the theory of numbers" you can find Karel Petr's exhaustive result for all the cases of $N$, including a different form for the case $N\ne 0 \pmod 3$ (which I was not able to prove from my interpretation of $qf^4(-q^6)$). Unfortunately, I was not able to find Karel Petr's original article about his interpretation of Gauss's equation, nor I am able to understand what led Gauss to this particular equation.

user2554
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