$∑\cos\left(\pi\frac{k+k'}nm\right)+\cos\left(\pi\frac{k-k'}nm\right)$

hbghlyj

$n,k,k'\inN,2n\nmid k+k',2n\nmid k-k'$
\[
\sum_{m=0}^n\left[\cos\left(\pi\frac{k+k'}nm\right)+\cos\left(\pi\frac{k-k'}nm\right)\right]\\
=1+\cos((k+k')\pi)
\]如何证明

kuing

漏了 `k'\inN` 的说明。证明很简单，易知
\[\sum_{m=a}^b\cos(mx)=\frac{-\sin\bigl((2a-1)\frac x2\bigr)+\sin\bigl((2b+1)\frac x2\bigr)}{2\sin\frac x2},\]
所以
\[\sum_{m=0}^n\cos\left(\pi\frac{k+k'}nm\right)=\frac12+\frac{\sin\bigl((2n+1)\pi\frac{k+k'}{2n}\bigr)}{2\sin\bigl(\pi\frac{k+k'}{2n}\bigr)}=\frac12+\frac{\sin\bigl((k+k')\pi+\pi\frac{k+k'}{2n}\bigr)}{2\sin\bigl(\pi\frac{k+k'}{2n}\bigr)},\]
若 `k+k'` 为偶数，则显然上式 `= 1`，若 `k+k'` 为奇数，则上式 `= 0`，而 `k-k'` 与 `k+k'` 同奇同偶，所以另一个和式也是一样，因此当 `k+k'` 为偶数时原式 `= 2`，奇数时原式 `=0`，这就与 `1+\cos\bigl((k+k')\pi\bigr)` 一致。

hbghlyj

两边乘2，用$2\cos x=e^x+e^{-x}$化为$$\sum_{m=0}^{n} \left[\left(e^{i\pi\frac{k+k'}{n}}\right)^m+\left(e^{i\pi\frac{k-k'}{n}}\right)^m+\left(e^{i\pi\frac{-k+k'}{n}}\right)^m+\left(e^{i\pi\frac{-k-k'}{n}}\right)^m\right]=2\left[1+\cos (k+k')\pi\right]$$
左边可以分解$$\label1\tag1\sum_{m=0}^{n} \left[\left(e^{i\pi\frac{k}{n}}\right)^m+\left(e^{i\pi\frac{-k}{n}}\right)^m\right]\left[\left(e^{i\pi\frac{k'}{n}}\right)^m+\left(e^{i\pi\frac{-k'}{n}}\right)^m\right]=2\left[1+\cos (k+k')\pi\right]$$
是否能写成某个群的character orthogonality relation？

Let ${\displaystyle G}$ be a finite group and ${\displaystyle k}$ a field whose characteristic does not divide the order of ${\displaystyle G}$. Let ${\displaystyle \varphi _{1}}$ and ${\displaystyle \varphi _{2}}$ be two inequivalent irreducible linear representations of ${\displaystyle G}$ over ${\displaystyle k}$ and let ${\displaystyle \chi _{1}}$ and ${\displaystyle \chi _{2}}$ denote their characters.
$${\displaystyle \sum _{g\in G}\chi _{1}(g){\overline {\chi _{2}(g)}}=0}$$
$${\displaystyle \sum _{g\in G}\chi _{1}(g){\overline {\chi _{1}(g)}}=|G|}$$

en.wikipedia.org/wiki/Schur_orthogonality_relations
ncatlab.org/nlab/show/Schur+orthogonality+relation
见Characters.pdf的Theorem 1.5.

hbghlyj

特征标表的性质之一为其在行与列上都会有着正交关系。

二重循环群(dicyclic group)${\displaystyle \operatorname {Dic} _{n}=\left\langle a,x\mid a^{2n}=1,\ x^{2}=a^{n},\ x^{-1}ax=a^{-1}\right\rangle \,\!}$的不同的二维表示：groupprops.subwiki.org/wiki/Linear_representation_theory_of_dicyclic_groups
对于每个${\displaystyle k=1,2,\dots ,n-1},$ 群$\mathrm{Dic}_{n}$的生成元$x,a$有二维表示$\phi:\operatorname {Dic} _{n}\to\mathrm{GL}(2,\mathbb{C})$：
$\phi(a)=\begin{pmatrix}e^{i\pi k/n}&0\\0&e^{-i\pi k/n}\\\end{pmatrix},\qquad \phi(x)=\begin{pmatrix}0&-1\\1&0\\\end{pmatrix},$
群$\mathrm{Dic}_{n}$有$4n$个元素：$a^m$和$xa^m$，对于$m=1,\dots,2n$，它们的表示为：
$\phi(a^m)=\begin{pmatrix}e^{i\pi km/n}&0\\0&e^{-i\pi km/n}\\\end{pmatrix},\qquad\phi(xa^m)=\begin{pmatrix}0&-e^{-i\pi km/n}\\e^{i\pi km/n}&0\\\end{pmatrix},$
于是
$\chi(a^m)=\operatorname{tr}(\phi(a^m))=e^{ikm\pi/n}+e^{-ikm\pi/n},$
$\chi(xa^m)=\operatorname{tr}(\phi(xa^m))=0,$
取$k,k'\in\{1,2,\dots,n-1\},k\ne k'$，用 特征标表中的行 的正交关系${\displaystyle \sum _{g\in G}\chi _{1}(g){\overline {\chi _{2}(g)}}=0}$得$$\tag2\label2\sum_{m=1}^{2n} \left[\left(e^{i\pi\frac{k}{n}}\right)^m+\left(e^{i\pi\frac{-k}{n}}\right)^m\right]\left[\left(e^{i\pi\frac{k'}{n}}\right)^m+\left(e^{i\pi\frac{-k'}{n}}\right)^m\right]=0$$
对比$m$和$m+n$时的式子：
$$\left[\left(e^{i\pi\frac{k}{n}}\right)^{m+n}+\left(e^{i\pi\frac{-k}{n}}\right)^{m+n}\right]=e^{ik\pi}\left[\left(e^{i\pi\frac{k}{n}}\right)^{m}+\left(e^{i\pi\frac{-k}{n}}\right)^{m}\right]$$
$$\left[\left(e^{i\pi\frac{k'}{n}}\right)^{m+n}+\left(e^{i\pi\frac{-k'}{n}}\right)^{m+n}\right]=e^{ik'\pi}\left[\left(e^{i\pi\frac{k'}{n}}\right)^{m}+\left(e^{i\pi\frac{-k'}{n}}\right)^{m}\right]$$
因此$$\sum_{m=n+1}^{2n} \left[\left(e^{i\pi\frac{k}{n}}\right)^m+\left(e^{i\pi\frac{-k}{n}}\right)^m\right]\left[\left(e^{i\pi\frac{k'}{n}}\right)^m+\left(e^{i\pi\frac{-k'}{n}}\right)^m\right]=e^{i(k+k')\pi}\sum_{m=1}^{n} \left[\left(e^{i\pi\frac{k}{n}}\right)^m+\left(e^{i\pi\frac{-k}{n}}\right)^m\right]\left[\left(e^{i\pi\frac{k'}{n}}\right)^m+\left(e^{i\pi\frac{-k'}{n}}\right)^m\right]$$
将\eqref{2}分为两部分，$m=1,\dots,n$和$m=n+1,\dots,2n$，并使用上式：
$$(1+e^{i(k+k')\pi})\sum_{m=1}^{n} \left[\left(e^{i\pi\frac{k}{n}}\right)^m+\left(e^{i\pi\frac{-k}{n}}\right)^m\right]\left[\left(e^{i\pi\frac{k'}{n}}\right)^m+\left(e^{i\pi\frac{-k'}{n}}\right)^m\right]=0$$
当$k+k'$为奇数时，$1+e^{i(k+k')\pi}=0$，就得到了$0=0$.
当$k+k'$为偶数时，约去$1+e^{i(k+k')\pi}=2$得到
$$\label3\tag{3}\sum_{m=1}^{n} \left[\left(e^{i\pi\frac{k}{n}}\right)^m+\left(e^{i\pi\frac{-k}{n}}\right)^m\right]\left[\left(e^{i\pi\frac{k'}{n}}\right)^m+\left(e^{i\pi\frac{-k'}{n}}\right)^m\right]=0$$
上式不含$m=0$，而在$m=0$时为$2\times2$，就得到了公式\eqref{1}.

---

上面假定$k\ne k'$，在$k= k'$的情况则Schur relation变为${\displaystyle \sum _{g\in G}\chi _{1}(g){\overline {\chi _{1}(g)}}=|G|}$，这里$\abs{\operatorname {Dic} _n}=4n$，\eqref{2}变为$$\tag{2'}\label{2'}\sum_{m=1}^{2n} \left[\left(e^{i\pi\frac{k}{n}}\right)^m+\left(e^{i\pi\frac{-k}{n}}\right)^m\right]^2=4n$$
式\eqref{3}变为$$\tag{3'}\label{3'}2\sum_{m=1}^{n} \left[\left(e^{i\pi\frac{k}{n}}\right)^m+\left(e^{i\pi\frac{-k}{n}}\right)^m\right]^2=4n$$
即
\[
\sum_{m=1}^n\left[\cos\left(\pi\frac{2k}nm\right)+\cos0\right]\\
=n
\]
即
\[
\sum_{m=1}^n\cos\left(\pi\frac{2k}nm\right)\\
=0
\]
显然成立。表明上面的推导应该是对的

hbghlyj

用其它的群，写出特征标表中的正交关系，能否得到更多类似的公式？
例如多重循环群(metacyclic group)

hbghlyj

hbghlyj 发表于 2025-1-16 21:04
能否得到更多类似的公式？
例如多重循环群(metacyclic group)  ...

首先需要弄清楚，metacyclic group有哪些irreducible representation呢