Computing an order complete basis for $M^{\infty}(N)$ and Applications

TL;DR

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Contribution

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Abstract

This paper gives a quick way to construct all modular functions for the group $\Gamma_0(N)$ having only a pole at $\tau = i \infty$. We assume that we are given two modular functions $f,g$ for $\Gamma_0(N)$ with poles only at $i \infty$ and coprime pole orders. As an application we obtain two new identities from which one can derive that $p(11n+6)\equiv 0\pmod{11}$, here $p(n)$ is the usual partition function.

Equations

$$t:=q^{-5}\prod _{n=1}^{\infty }{\left(\frac{1-q^n}{1-q^{11n}}\right)}^{12}.$$

$$h:=qt\prod _{n=1}^{\infty }(1-q^{11k})\sum _{n=0}^{\infty }p(11n+6)q^n$$

$$f:=(dt/dq)\prod _{n=1}^{\infty }(1-q^n{)}^{-2}(1-q^{11n}{)}^{-2}.$$

$$b(\tau ):=q\prod _{n=1}^{\infty }(1-q^n{)}^2(1-q^{11n}{)}^2,q=e^{2\pi i\tau }$$

$$b\Bigl{(}\frac{a\tau+b}{c\tau+d}\Bigr{)}=(c\tau+d)^{2}b(\tau)$$

$$t\Bigl{(}\frac{a\tau+b}{c\tau+d}\Bigr{)}=t(\tau).$$

$$(c\tau+d)^{-2}t^{\prime}\Bigl{(}\frac{a\tau+b}{c\tau+d}\Bigr{)}=t^{\prime}(\tau)$$

$$(t^{\prime}/b)\Bigl{(}\frac{a\tau+b}{c\tau+d}\Bigr{)}=(t^{\prime}/b)(\tau).$$

$$f\Bigl{(}\frac{a\tau+b}{c\tau+d}\Bigr{)}=f(\tau)$$

$$f\Bigl{(}\frac{a\tau+b}{c\tau+d}\Bigr{)}=\sum_{n=m(\gamma)}^{\infty}a_{\gamma}(n)q^{\frac{\gcd(c^{2},n)n}{N}}.$$

$$\eta (-1/\tau )=(-i\tau {)}^{1/2}\eta (\tau ).$$

$$t(-1/\tau)=t^{-1}\Bigl{(}\frac{\tau}{11}\Bigr{)}$$

$$b(-1/\tau )=-\frac{1}{11}b(\frac{\tau }{11}){\tau }^2.$$

$${\tau }^{-2}{t}^{\prime }(-1/\tau )=\frac{1}{11}{t}^{-2}(\frac{\tau }{11})t^{\prime }(\frac{\tau }{11})$$

$$t^{\prime }(-1/\tau )=\frac{1}{11}{\tau }^2{t}^{-2}(\frac{\tau }{11})t^{\prime }(\frac{\tau }{11}).$$

$$(t^{\prime }/b)(-1/\tau )=-(t^{\prime }/b)(\frac{\tau }{11})t^{-2}(\frac{\tau }{11}).$$

$$f(-1/\tau )=-f(\tau /11)t^{-2}(\tau /11)=5q^{4/11}+O(q^{5/11}).$$

$$p(x,y)=y^5+170x{y}^4+9345x^2{y}^3+167320x^3{y}^2+(5^5{x}^2-7903458x+5^{5}11^6)x^4.$$

$$b_2=$$

$$b_3=$$

$$b_4=$$

$$h=11q^{-4}+165q^{-3}+748q^{-2}+1639q^{-1}+3553+O(q)$$

$$h-11b_4-165b_3-748b_2-3553b_0=O(q).$$

$$h=11b_4+165b_3+748b_2+3553b_0.$$

$$h=qt\prod _{n=1}^{\infty }(1-q^{11k})\sum _{n=0}^{\infty }p(11n+6)q^n=11^4+55t\left(5\frac{t-11^3}{f+47t}-\frac{2(71t+3f)(t+11^3)}{f^2+89ft+1424t^2}\right).$$

$$E_4:=1+240\sum _{n=1}^{\infty }\frac{n^3{q}^n}{1-q^n}$$

$$\Delta :=q\prod _{n=1}^{\infty }(1-q^n{)}^{24}.$$

$$f:=J{t}^3.$$

$$E_{4}\Bigl{(}\frac{a\tau+b}{c\tau+d}\Bigr{)}=(c\tau+d)^{4}E_{4}(\tau)$$

$$\Delta\Bigl{(}\frac{a\tau+b}{c\tau+d}\Bigr{)}=(c\tau+d)^{12}\Delta(\tau)$$

$$J\Bigl{(}\frac{a\tau+b}{c\tau+d}\Bigr{)}=J(\tau)$$

Full text

$\mbox{}^{\dagger}$ Supported by NSF 1618657.

$\mbox{}^{\ddagger}$ Supported by grant SFB F50-06 of the Austrian Science Fund (FWF)

1. Description of the Problem

For basic notions about modular functions used in this paper we refer to [14]. In this paper we show how to obtain an order complete basis for $M^{\infty}(N)$ with an application to the case $N=11$. We use this basis to obtain two new Ramanujan type identities for $\sum_{n=0}^{\infty}p(11n+6)q^{n}$. Such bases have also been constructed by other authors [1, 2, 4, 7, 8, 9, 11, 12] by using various tricks to produce sufficiently many new modular functions $f_{1},f_{2},\ldots\in M^{\infty}(N)$ until $\mathbb{C}[f_{1},f_{2},\ldots]$ becomes equal to $M^{\infty}(N)$. The advantage of our approach is that we need only two functions in $t,f\in M^{\infty}(N)$. Then $\mathbb{C}[t,f]$ will generally be a proper subset of $M^{\infty}(N)$, but instead of searching for more modular functions, we fill this gap with a normalized integral basis.

Let $t$ and $f$ be modular functions for the group $\Gamma_{0}(N)$ with poles only at $\tau=i\infty$, in other words, let $t,f\in M^{\infty}(N)$. Suppose that the pole orders are $n$ and $m$ respectively, and that $\gcd(n,m)=1$, such functions always exist [13, Example 2.3]. Then there exists an irreducible polynomial $p=p(x,y)\in\mathbb{C}[x,y]$ with $p(t,f)=0$, ${\rm deg}_{x}(p)=m$, and ${\rm deg}_{y}(p)=n$ by [21, Lemma 1]. One can compute $p$ from the $q$-expansions of $t$ and $f$ by making an Ansatz for the unknown coefficients of $p$ and solving a system of equations where each equation is a coefficient in the $q$-expansion of $p(t,f)$. We use $p$ to compute in the function field $\mathbb{C}(t,f)\cong\mathbb{C}(x)[y]/(p)$.

The function field $\mathbb{C}(t,f)$ contains $M^{\infty}(N)$ see [13, Prop 4.3], here $M^{\infty}(N)$ is the set of all modular functions for the group $\Gamma_{0}(N)$ with a pole only at $i\infty$. Obtaining all modular functions for the group $\Gamma_{0}(N)$ having a pole only at $i\infty$ is equivalent to finding all modular functions $h\in\mathbb{C}(t,f)$ that are integral over $\mathbb{C}[t]$ (which means there is a monic polynomial $g(X)\in\mathbb{C}[t][X]$ for which $g(h)=0$). Thus, one starts by computing an integral basis, which is a basis $b_{1},\dots,b_{n}\in\mathbb{C}(t,f)$ of the $\mathbb{C}[t]$-module of all $h\in\mathbb{C}(t,f)$ that are integral over $\mathbb{C}[t]$. There are several algorithms to compute an integral basis [5, 20] and implementations in several computer algebra systems. Then every $h$ that is integral over $\mathbb{C}[t]$ can be written as $h=p_{1}(t)b_{1}+\dots+p_{n}(t)b_{n}$ for some polynomials $p_{1},\ldots,p_{n}$. Given the $q$-expansions of $h$ and $b_{1},\ldots,b_{n}$ the algorithm described in [16, Alg. MW] can find $p_{1},\dots,p_{n}$ provided that $\operatorname{ord}_{i\infty}(b_{1})<\operatorname{ord}_{i\infty}(b_{2})<\dots<\operatorname{ord}_{i\infty}(b_{n})$. We call such an integral basis order complete.

After computing an integral basis, we can find an order complete basis by using normalization at infinity from Trager’s PhD thesis [19, Chapter 2, Section 3], see Section 1.2 for details.

1.1. Notations

$K=\mathbb{C}(x)[y]/(p)$ where $p\in\mathbb{C}[x,y]$ is irreducible.

${O}_{K}$ is ring of all elements of $K$ that are integral over $\mathbb{C}[x]$.

$R_{\infty}$ is the ring of all $h\in\mathbb{C}(x)$ that have no pole at $x=\infty$.

${O}_{\infty}$ is ring of all elements of $K$ that are integral over $R_{\infty}$.

To compute a basis of ${O}_{\infty}$ as $R_{\infty}$-module, first substitute $x\mapsto 1/\tilde{x}$, then compute a local integral basis at $\tilde{x}=0$ (most integral basis implementations allow the option of computing a local integral basis). After that, replace $\tilde{x}$ by $1/x$.

1.2. Normalize an integral basis at infinity

The process of normalizing an integral basis at infinity was introduced in [19] in order to compute a Riemann-Roch space that was needed for integrating algebraic functions. For completeness we will describe this process:

Algorithm: Normalize an integral basis at infinity.

1. (1)

Let $b_{1},\ldots,b_{n}$ be a basis of ${O}_{K}$ as $\mathbb{C}[x]$-module. 2. (2)

Let $b_{1}^{\prime},\ldots,b_{n}^{\prime}$ be a basis of ${O}_{\infty}$ as $R_{\infty}$-module. 3. (3)

Write $b_{i}=\sum_{j=1}^{n}r_{ij}b_{i}^{\prime}$ with $r_{ij}\in\mathbb{C}(x)$. 4. (4)

Let $D\in\mathbb{C}[x]$ be a non-zero polynomial for which $a_{ij}:=Dr_{ij}\in\mathbb{C}[x]$ for all $i,j$. Now $Db_{i}=\sum_{j=1}^{n}a_{ij}b_{i}^{\prime}$. 5. (5)

For each $i\in\{1,\ldots,n\}$, let $m_{i}$ be the maximum of the degrees of $a_{i1},\ldots,a_{in}$. Now let $V_{i}\in\mathbb{C}^{n}$ be the vector whose $j$’th entry is the $x^{m_{i}}$-coefficient of $a_{ij}$. Let $d_{i}:=m_{i}-{\rm deg}_{x}(D)$. 6. (6)

If $V_{1},\ldots,V_{n}$ are linearly independent, then return $b_{1},\ldots,b_{n}$ and $d_{1},\ldots,d_{n}$ and stop.

Otherwise, take $c_{1},\ldots,c_{n}\in\mathbb{C}$, not all 0, for which $c_{1}V_{1}+\cdots c_{n}V_{n}=0$. 7. (7)

Among those $i\in\{1,\ldots,n\}$ for which $c_{i}\neq 0$, choose one for which $d_{i}$ is maximal. For this $i$, do the following

1. (a)

Replace $b_{i}$ by $\sum_{k=1}^{n}c_{k}x^{d_{i}-d_{k}}b_{k}$. 2. (b)

Replace $a_{ij}$ by $\sum_{k=1}^{n}c_{k}x^{d_{i}-d_{k}}a_{kj}$ for all $j\in\{1,\ldots,n\}$. 8. (8)

Go back to step 5.

The $b_{1},\ldots,b_{n}$ remain a basis of ${O}_{K}$ throughout the algorithm because the new $b_{i}$ in step 7a can be written as a nonzero constant times the old $b_{i}$ plus a $\mathbb{C}[x]$-linear combination of the $b_{j}$, $j\neq i$. When we go back to step 5 the non-negative integer $d_{i}$ decreases while the $d_{j}$, $j\neq i$ stay the same. Hence the algorithm must terminate.

Let $b_{1},\ldots,b_{n}$ and $d_{1},\ldots,d_{n}$ be the output of the algorithm. By construction, the number $d_{i}$ in the algorithm is the smallest integer for which $b_{i}\in x^{d_{i}}{O}_{\infty}$. If $\beta\in{O}_{K}$ with $\beta\neq 0$ then we can write $\beta=c_{1}b_{1}+\cdots+c_{n}b_{n}$ for some $c_{1},\ldots,c_{n}\in\mathbb{C}[x]$. Denote $d_{\beta}$ as the maximum of ${\rm deg}_{x}(c_{j})+d_{j}$ taken over all $j$ for which $c_{j}\neq 0$. Then $\beta\in x^{d_{\beta}}{O}_{\infty}$ by construction. Since the vectors $V_{1},\ldots,V_{n}$ in the algorithm are linearly independent when the algorithm terminates, there can not be any cancellation, which means that ${d_{\beta}}$ is the smallest integer for which $\beta\in x^{d_{\beta}}{O}_{\infty}$. Because of this, we get the following:

If $d$ is a positive integer, then the set $B_{d}:=\{x^{j}b_{i}\ |\ 0\leq j\leq d-d_{i},\ 1\leq i\leq n\}$ is a basis of ${O}_{K}\bigcap x^{d}{O}_{\infty}$ as $\mathbb{C}$-vector space.

Note that $B_{d}$ is a basis of the Riemann-Roch space of the pole-divisor of $x^{d}$. So computing $B_{d}$ can be interpreted as (i): a direct application of a normalized integral basis, or (ii): a special case of algorithms [3, 6] for Riemann-Roch spaces. The two interpretations are equivalent because the first step in computing Riemann-Roch spaces is to compute a normalized integral basis.

We can take $q$-expansions for each of the elements of $B_{d}$, and then make a change of basis so that the new basis $B_{d}^{\rm REF}$ will have $q$-expansions in Reduced Echelon Form. This means that if $b\in B_{d}^{\rm REF}$ and $b=a_{r}q^{r}+a_{r+1}q^{r+1}+\cdots$ with $a_{r}\neq 0$ then $a_{r}=1$ and all other basis elements have a zero coefficient at $q^{r}$. Then $B_{d}^{\rm REF}$, for suitable $d$, is an order complete basis. For an implementation and two examples see:

www.math.fsu.edu/$\sim$hoeij/files/OrderComplete

2. New Identities

We will give two identities of Ramanujan type found using our algorithm (the second one is only on our website). Let $p(n)$ be the partition function. Define

[TABLE]

and

[TABLE]

and

[TABLE]

Both $h$ and $t$ are modular functions in $M^{\infty}(11)$, see [14, Lemma 3.1].

To prove that $f$ is in $M^{\infty}(11)$ as well, first note that by [10, Prop. 3.1.1]

[TABLE]

satisfies

[TABLE]

for all \left(\begin{array}[]{cc}a&b\\ c&d\end{array}\right)\in\Gamma_{0}(11). Since $t\in M^{\infty}(11)$, we have

[TABLE]

The derivative with respect to $\tau$ is:

[TABLE]

Multiplying (2) by $(c\tau+d)^{2}$ and dividing by (1) gives

[TABLE]

Since $\frac{d}{d\tau}=2\pi iq\frac{d}{dq}$, it follows that $t^{\prime}/b=2\pi if$. Therefore

[TABLE]

for all \left(\begin{array}[]{cc}a&b\\ c&d\end{array}\right)\in\Gamma_{0}(11). Furthermore, since $b(\tau)$ has no zeros in the upper half plane and $t(\tau)$ is holomorphic in the upper half plane it follows that $f$ is holomorphic in the upper half plane. Hence the first condition of being a modular function for $\Gamma_{0}(11)$ according to the definition in [14] is satisfied. The second condition is equivalent to showing that for \gamma=\left(\begin{array}[]{cc}a&b\\ c&d\end{array}\right)\in\mathrm{SL}_{2}(\mathbb{Z}) we have an expansion of the form

[TABLE]

As seen in [14], if this property hold for \left(\begin{array}[]{cc}a&b\\ c&d\end{array}\right), then it also holds for \left(\begin{array}[]{cc}a^{\prime}&b^{\prime}\\ c^{\prime}&d^{\prime}\end{array}\right), if there exists \left(\begin{array}[]{cc}A&B\\ C&D\end{array}\right)\in\Gamma_{0}(11) such that $\frac{A\frac{a}{c}+B}{C\frac{a}{c}+D}=\frac{a^{\prime}}{c^{\prime}}$. So we need to find representatives of the orbits of the action of $\Gamma_{0}(11)$ on $\mathbb{Q}\cup\{i\infty\}$, that is, the cusps of $\Gamma_{0}(N)$. From [17] we find that these representatives are [math] and $i\infty$. Then it suffices to show (3) for two cases: \left(\begin{array}[]{cc}1&0\\ 0&1\end{array}\right) and \left(\begin{array}[]{cc}0&-1\\ 1&0\end{array}\right). The first case holds because $f$ is a $q$-series. For the second case we need to show that $f(-1/\tau)$ is a Laurent series in $q^{1/11}$ with finite principal part. By [15] we have

[TABLE]

This implies

[TABLE]

and

[TABLE]

The derivative of (4) is

[TABLE]

which is equivalent to

[TABLE]

This implies

[TABLE]

Hence

[TABLE]

So the last condition for $f$ being a modular function for $\Gamma_{0}(11)$ is verified. In order for $f$ to be in $M^{\infty}(11)$ we need the order of $f$ to be nonnegative at all cusps except $i\infty$. That only leaves the cusp [math] where the order is $4$. This shows $f\in M^{\infty}(11)$.

We want to express $h$ as an element of $\mathbb{C}(t,f)$. The pole orders of $t$ and $f$ are $5$ and $6$ so $p(x,y)=\sum_{i=0}^{6}\sum_{j=0}^{5}a_{ij}x^{i}y^{j}$ is an Ansatz for the algebraic relation $p(t,f)=0$. Solving linear equations coming from $q$-expansions gives

[TABLE]

We use $p(x,y)$ to compute in $\mathbb{C}(t,f)\cong\mathbb{C}(x)[y]/(p)$. We compute $B^{\rm REF}_{d}$ from the previous section with $d=1$ and obtain $b_{0},b_{2},b_{3},b_{4},b_{5}$ where $b_{0}=1$ and $b_{i}=q^{-i}+c_{i}q^{-1}+O(q^{1})$ for $i=2,\ldots,5$ for some constants $c_{i}$. Since $h$ has a pole of order 4, we can write it as a linear combination of $b_{0},b_{2},b_{3},b_{4}$. We have $b_{0}=1$ and

[TABLE]

Like in [16, Alg. MW], we use

[TABLE]

to find

[TABLE]

This expression in $M^{\infty}(11)$ has no poles and a root at $\tau=i\infty$ (at $q=0$) hence it is the zero function. Therefore

[TABLE]

Replacing $b_{0},b_{2},b_{3},b_{4}$ with their corresponding expressions in terms of $t$ and $f$ gives

[TABLE]

This implies $p(11n+6)\equiv 0\pmod{11}$. Other expressions for $h$ that prove this congruence were already in [1, 9], however, our expression in terms of $t,f$ is novel.

For our second example, take $t$ and $h$ be as before and let

[TABLE]

be the usual Eisenstein series. Let

[TABLE]

Let $J:=E_{4}^{3}/\Delta=q^{-1}+\cdots$ and

[TABLE]

Next we show that $f\in M^{\infty}(11)$. From the last chapter of [18] we find

[TABLE]

and

[TABLE]

for all \left(\begin{array}[]{cc}a&b\\ c&d\end{array}\right)\in\mathrm{SL}_{2}(\mathbb{Z}). These two identities imply

[TABLE]

for all \left(\begin{array}[]{cc}a&b\\ c&d\end{array}\right)\in\mathrm{SL}_{2}(\mathbb{Z}). Since $\mathrm{SL}_{2}(\mathbb{Z})$ has only one cusp, $i\infty$, and since $J$ is a $q$-series it follows that $J$ is a modular function on $\mathrm{SL}_{2}(\mathbb{Z})$ and thus on $\Gamma_{0}(11)$.

Since $t(\tau)$ is already a modular function on $\Gamma_{0}(11)$, it follows that $f$ is a modular function on $\Gamma_{0}(11)$. To show that $f$ is in $M^{\infty}(11)$ it suffices to show that the order of $f$ at the cusp [math] is nonnegative. Since $J(-1/\tau)=(q^{-1/11})^{11}+O(1)$ the order of $J$ at [math] is $-11$. The order of $t$ at [math] is $5$, so the order of $f$ at the cusp [math] is $-11+3\cdot 5=4\geq 0$. This shows $f\in M^{\infty}(11)$.

The only pole of $f$ is at $i\infty$, it has order 16. We compute the algebraic relation $p(t,f)=0$ with the Ansatz method, and use $p$ to compute $B_{d}^{\rm REF}$. Then we express $h$ in terms of the $t$ and the new $f$. This relation, and the Maple file that computes it, are given at www.math.fsu.edu/$\sim$hoeij/files/OrderComplete.