Jacobi Theta函数

格点上的Poisson求和

对充分好的函数, 我们有Fourier变换

$$ \widehat f(x)=\int_{-\infty}^{\infty} f(y)e^{-2\pi ixy}dy. $$

命题 1 (Parseval恒等式). $\sum_{n=-\infty}^{\infty} f(n)=\sum_{n=-\infty}^{\infty}\widehat f(n).$

考虑周期函数

$$ F(x)=\sum_{n=-\infty}^{\infty} f(x+n). $$

$F$可以视作$\mathbb{R}/\mathbb{Z}$上的函数, 有Fourier展开

$$ F(x)=\sum_{m=-\infty}^{\infty} a_m e^{2\pi imx}, $$

$$ a_m=\int_0^1 F(x) e^{-2\pi imx}dx=\int_0^1 \sum_{n=-\infty}^{\infty} f(x+n)e^{-2\pi im(x+n)}dx=\int_{-\infty}^\infty f(x)e^{-2\pi imx}dx=\widehat f(m). $$

从而

$$ \sum_{n=-\infty}^{\infty}f(n)=F(0)=\sum_{m=-\infty}^{\infty}a_m=\sum_{n=-\infty}^\infty \widehat f(n). $$

命题得证.

一般的, 对于任意$n$维实向量空间, 配备平移不变测度$\mu,$ 有Fourier变换

$$ \widehat f(x)=\int_V e^{-2\pi i\left<{}y,x\right>}f(y)\mu(y). $$

此时$\widehat f$是在$V^\ast $上充分好的函数.

称$\Gamma$为$V$中的格点, 若$\Gamma$以$V$中一组线性无关向量$\{v_1,\cdots,v_n\}$为$\mathbb{Z}$-基. $V^\ast $中的对偶格点$\Gamma'$定义为

$$ \Gamma':=\{x\in V^\ast \mid \left<{}y,x\right>\in \mathbb{Z}, \,\forall\,y\in \Gamma\}. $$

此时记$A=[v_1,\cdots,v_n],$ 那么$(A^{\ast })^{-1}$的列向量就是$\Gamma'$的一组$\mathbb{Z}$-基, 其中${}^\ast $为对偶阵. 当$V=\mathbb{R}^n,$ $\left<{}-,-\right>$取标准内积时, ${}^\ast $即为转置.

命题 2 (格点上的Poisson求和). 格点$\Gamma$在$V$中的体积记为$\mu(V/\Gamma).$ 若函数$f$有Fourier变换, 我们有

$$ \sum_{y\in\Gamma} f(y)=\frac{1}{\mu(V/\Gamma)}\sum_{x\in \Gamma'}\widehat f(x). $$

做尺度变换, 等式两侧保持(Fourier变换中有测度项), 因此总可设$\mu(V/\Gamma)=1.$ 记$\Gamma$的基拼为矩阵$A,$ 那么

$$ \sum_{y\in \Gamma}f(y)=\sum_{y\in \mathbb{Z}^n}f(Ay)=\sum_{y\in\mathbb{Z}^n}\widetilde f(y),\quad \widetilde f(y)=:f(Ay). $$

$$ \widehat f(x)=\int_V e^{-2\pi i\left<{}y,x\right>}f(y)\mu(y)=\int_V e^{-2\pi i\left<{}Ay,x\right>}\widetilde f(y)\mu(y)=\int_V e^{-2\pi i\left<{}y,A^\ast x\right>}\widetilde f(y)\mu(y)=\mathcal{F}[\widetilde f] (A^\ast x). $$

$$ \sum_{x\in\Gamma'}\widehat f(x)=\sum_{x\in\mathbb{Z}^n}\widehat f((A^{\ast })^{-1}x)=\sum_{x\in \mathbb{Z}^n}\mathcal{F}[\widetilde f] (x)=\sum_{y\in\mathbb{Z}^n}\widetilde f(y)=\sum_{y\in\Gamma}f(y). $$

其中利用到了格点$\mathbb{Z}^n\subset \mathbb{R}^n$上的标准Poisson求和, 证明和前一命题证明类似.

Theta函数

设$V$上有一个对称正定非退化二次型$B(x,y)=x\cdot y$ (与测度相容), 那么此时$V$可以和$V^\ast $等同起来, $\Gamma'$也是$V$中的格点. 对任意格点$\Gamma,$ 定义

$$ \Theta_\Gamma(t)=\sum_{x\in \Gamma} e^{-\pi tx\cdot x},\quad t>0,\quad t\in \mathbb{R}. $$

定理 3. $\Theta_\Gamma(t)=t^{-\frac{n}{2} }\frac{1}{\mu(V/\Gamma)}\Theta_{\Gamma'}(t^{-1}).$ 特别地, 若$\Gamma=\Gamma'=\mathbb{Z}\subset \mathbb{R},$ 那么$\Theta(t)=\frac{1}{\sqrt{t} }\Theta(\frac{1}{t}).$

对函数$f(x)=e^{-\pi x\cdot x}$使用Poisson求和, 那么

$$ \Theta_\Gamma(t)=\sum_{x\in \sqrt{t}\Gamma} f(x)=\frac{1}{\mu(V/\sqrt{t}\Gamma)}\sum_{x\in(\sqrt{t}\Gamma)'}\widehat f(x)=t^{-\frac{n}{2} }\frac{1}{\mu(V/\Gamma)} \sum_{x\in\Gamma'}\widehat f\left(\frac{1}{\sqrt{t} }x\right)=t^{-\frac{n}{2} }\frac{1}{\mu(V/\Gamma)}\Theta_{\Gamma'}(t^{-1}). $$

最后的等式是由于$f=\widehat f,$

$$ \Theta_{\Gamma'(t^{-1})}=\sum_{x\in\Gamma'} e^{-\pi \frac{1}{t}x\cdot x}=\sum_{x\in\Gamma'}\widehat f\left(\frac{1}{\sqrt{t} }x\right). $$

记二次型的gram矩阵为$A=(a_{ij}),$ $a_{ij}=e_i\cdot e_j,$ 其中$\{e_i\}$为$\Gamma$的基底. 那么$\Gamma$的体积为$\sqrt{\det(A)}.$ 若$B$为$A$的逆阵, 那么$B=(b_{ij})$是对偶基对应的gram阵, $\Gamma'$由对偶基$e_i'=b_{ij}e_j$拼成. $\Gamma'$的体积为$\det(A)^{-\frac{1}{2} }.$

记$r_\Gamma(m)$为使得$x\cdot x=2m$的$x\in\Gamma$的个数, $m\ge 0$. $r_\Gamma(m)$由关于$m$的多项式控制, Serre给出了$O(m^\frac{n}{2})$的界. 于是级数$\sum_{m=0}^\infty r_\Gamma(m)q^m$在$|q|<1$时收敛. 取$q=e^{2\pi iz},$ 那么对于$z\in\mathbb{H},$ 我们定义了全纯函数

$$ \theta_\Gamma(z)=\sum_{m=0}^\infty r_\Gamma(m)q^m=\sum_{x\in\Gamma} e^{\pi iz(x\cdot x)}=\sum_{x\in\Gamma} q^{(x\cdot x)/2}. $$

这里我们需要要求$x\cdot x\equiv 0\mod 2,$ $\,\forall\,x\in \Gamma.$ 这即是要求$A=(e_i\cdot e_j)$对角线上均为偶数.

之前的Theta函数是实参数$t,$ 此时的小theta函数满足

$$ \Theta_\Gamma(t)=\theta_\Gamma(it),\quad \Theta_\Gamma\left(\frac{1}{t}\right)=\theta_\Gamma\left(\frac{i}{t}\right)=\theta_\Gamma\left(-\frac{1}{it}\right). $$

进一步假设$\Gamma$是自对偶的, 即$\Gamma'=\Gamma.$ 这要求$A=(e_i\cdot e_j)$是整系数的, 且行列式为$1.$ 在这样的假设下, 由前面的Poisson和公式, 有

$$ \theta_\Gamma\left(-\frac{1}{it}\right)=t^{\frac{n}{2} }\theta_\Gamma(it). $$

由于$\theta_\Gamma(-\frac{1}{z}),$ $(-iz)^{\frac{n}{2} }\theta_\Gamma(z)$都是解析函数, 现证明了它们在纯虚轴(正半轴)上相等, 由解析连续性, 它们在上半平面上都相等.

命题 4. 对满足上述条件的格点$\Gamma,$ $\theta_\Gamma\left(-\frac{1}{z}\right)=(-iz)^{\frac{n}{2} }\theta_\Gamma(z),$ $\,\forall\,z\in\mathbb{H}.$

事实上, 可以证明满足自对偶且$x\cdot x\equiv 0\mod 2,$ $\,\forall\,x\in \Gamma$的格点$\Gamma\subset V$必须是$8k$维的, $k\in\mathbb{Z}_+.$

文章最后更新于 2023-09-05 15:45:49

Poisson Summation on Lattices

For sufficiently good functions, we have the Fourier transform

$$ \widehat f(x)=\int_{-\infty}^{\infty} f(y)e^{-2\pi ixy}dy. $$

Proposition 1 (Parseval identity). $\sum_{n=-\infty}^{\infty} f(n)=\sum_{n=-\infty}^{\infty}\widehat f(n).$

Consider a periodic function

$$ F(x)=\sum_{n=-\infty}^{\infty} f(x+n). $$

$F$ can be regarded as a function on $\mathbb{R}/\mathbb{Z}$, with Fourier expansion

$$ F(x)=\sum_{m=-\infty}^{\infty} a_m e^{2\pi imx}, $$

$$ a_m=\int_0^1 F(x) e^{-2\pi imx}dx=\int_0^1 \sum_{n=-\infty}^{\infty} f(x+n)e^{-2\pi im(x+n)}dx=\int_{-\infty}^\infty f(x)e^{-2\pi imx}dx=\widehat f(m). $$

thereby

$$ \sum_{n=-\infty}^{\infty}f(n)=F(0)=\sum_{m=-\infty}^{\infty}a_m=\sum_{n=-\infty}^\infty \widehat f(n). $$

The proposition is proved.

Generally, for any $n$-dimensional real vector space, equipped with a translation-invariant measure $\mu,$, there is a Fourier transform

$$ \widehat f(x)=\int_V e^{-2\pi i\left<{}y,x\right>}f(y)\mu(y). $$

At this time $\widehat f$ is a sufficiently good function on $V^\ast $.

Call $\Gamma$ the grid point in $V$, If $\Gamma$ takes a set of linearly independent vectors $\{v_1,\cdots,v_n\}$ in $V$ as $\mathbb{Z}$-basis. The dual grid point $\Gamma'$ in $V^\ast $ is defined as

$$ \Gamma':=\{x\in V^\ast \mid \left<{}y,x\right>\in \mathbb{Z}, \,\forall\,y\in \Gamma\}. $$

Record $A=[v_1,\cdots,v_n],$ at this time Then the column vector of $(A^{\ast })^{-1}$ is a set of $\mathbb{Z}$-basis of $\Gamma'$, Where ${}^\ast $ is the dual matrix. When $V=\mathbb{R}^n,$ When $\left<{}-,-\right>$ takes the standard inner product, ${}^\ast $ is the transpose.

Proposition 2 (Poisson summation on grid points). The volume of grid point $\Gamma$ in $V$ is recorded as $\mu(V/\Gamma).$. If the function $f$ has Fourier transformation, we have

$$ \sum_{y\in\Gamma} f(y)=\frac{1}{\mu(V/\Gamma)}\sum_{x\in \Gamma'}\widehat f(x). $$

Perform scale transformation and maintain both sides of the equation (there are measurement terms in the Fourier transformation), Therefore, we can always set $\mu(V/\Gamma)=1.$ and record the basis of $\Gamma$ as matrix $A,$. Then

$$ \sum_{y\in \Gamma}f(y)=\sum_{y\in \mathbb{Z}^n}f(Ay)=\sum_{y\in\mathbb{Z}^n}\widetilde f(y),\quad \widetilde f(y)=:f(Ay). $$

The standard Poisson summation on the grid point $\mathbb{Z}^n\subset \mathbb{R}^n$ is used, The proof is similar to the proof of the previous proposition.

Theta function

Suppose there is a symmetric positive definite non-degenerate quadratic form $B(x,y)=x\cdot y$ on $V$ (compatible with the measure), Then $V$ can be equated with $V^\ast $ at this time, and $\Gamma'$ is also a grid point in $V$. Definition for any grid point $\Gamma,$

$$ \Theta_\Gamma(t)=\sum_{x\in \Gamma} e^{-\pi tx\cdot x},\quad t>0,\quad t\in \mathbb{R}. $$

Theorem 3. $\Theta_\Gamma(t)=t^{-\frac{n}{2} }\frac{1}{\mu(V/\Gamma)}\Theta_{\Gamma'}(t^{-1}).$ In particular, if $\Gamma=\Gamma'=\mathbb{Z}\subset \mathbb{R},$ then $\Theta(t)=\frac{1}{\sqrt{t} }\Theta(\frac{1}{t}).$

Use Poisson summation on function $f(x)=e^{-\pi x\cdot x}$, then

The final equation is due to $f=\widehat f,$

$$ \Theta_{\Gamma'(t^{-1})}=\sum_{x\in\Gamma'} e^{-\pi \frac{1}{t}x\cdot x}=\sum_{x\in\Gamma'}\widehat f\left(\frac{1}{\sqrt{t} }x\right). $$

Remember the gram matrix of the quadratic type as $A=(a_{ij}),$ $a_{ij}=e_i\cdot e_j,$ Where $\{e_i\}$ is the base of $\Gamma$. Then the volume of $\Gamma$ is $\sqrt{\det(A)}.$ If $B$ is the inverse matrix of $A$, then $B=(b_{ij})$ is the gram matrix corresponding to the dual basis, $\Gamma'$ is composed of the dual basis $e_i'=b_{ij}e_j$. The volume of $\Gamma'$ is $\det(A)^{-\frac{1}{2} }.$

Let $r_\Gamma(m)$ be the number of $x\in\Gamma$ that makes $x\cdot x=2m$, $m\ge 0$. $r_\Gamma(m)$ is controlled by a polynomial with respect to $m$, and Serre gave a bound for $O(m^\frac{n}{2})$. So the series $\sum_{m=0}^\infty r_\Gamma(m)q^m$ converges at $|q|<1$. Taking $q=e^{2\pi iz},$ then for $z\in\mathbb{H},$ we define the holomorphic function

$$ \theta_\Gamma(z)=\sum_{m=0}^\infty r_\Gamma(m)q^m=\sum_{x\in\Gamma} e^{\pi iz(x\cdot x)}=\sum_{x\in\Gamma} q^{(x\cdot x)/2}. $$

Here we need to require $x\cdot x\equiv 0\mod 2,$ $\,\forall\,x\in \Gamma.$ This requires $A=(e_i\cdot e_j)$ to be even numbers on the diagonal.

The previous Theta function was a real parameter $t,$. The small theta function at this time satisfies

$$ \Theta_\Gamma(t)=\theta_\Gamma(it),\quad \Theta_\Gamma\left(\frac{1}{t}\right)=\theta_\Gamma\left(\frac{i}{t}\right)=\theta_\Gamma\left(-\frac{1}{it}\right). $$

Further assume that $\Gamma$ is self-dual, that is, $\Gamma'=\Gamma.$ This requires $A=(e_i\cdot e_j)$ to be an integral coefficient, and the determinant is $1.$. Under this assumption, From the previous Poisson sum formula, we have

$$ \theta_\Gamma\left(-\frac{1}{it}\right)=t^{\frac{n}{2} }\theta_\Gamma(it). $$

Due to $\theta_\Gamma(-\frac{1}{z}),$ $(-iz)^{\frac{n}{2} }\theta_\Gamma(z)$ are all analytical functions, It is now proved that they are equal on the pure imaginary axis (positive semi-axis). By analytical continuity, They are all equal in the upper half plane.

Proposition 4. For grid points that meet the above conditions $\Gamma,$ $\theta_\Gamma\left(-\frac{1}{z}\right)=(-iz)^{\frac{n}{2} }\theta_\Gamma(z),$ $\,\forall\,z\in\mathbb{H}.$

In fact, it can be shown that it satisfies self-duality and $x\cdot x\equiv 0\mod 2,$ Grid point $\Gamma\subset V$ of $\,\forall\,x\in \Gamma$ must be $8k$ dimensional, $k\in\mathbb{Z}_+.$

The article was last updated on 2023-09-05 15:45:49