weight 3以下の交代多重ゼータ値を全て求める

ここでは, weight 3以下の交代多重ゼータの値を全て求めたいと思います. 仮定する知識は, 調和積とシャッフル積です. ここにおいて求める, というのは, $\pi ,\ln 2,\zeta (3)$の多項式で表す, という意味です. weight 1の場合は, $\zeta (\bar{1})$しかありません. weight 2の場合は,
$$\begin{array}{r}\zeta (2),\zeta (\bar{2}),\zeta (1,\bar{1}),\zeta (\bar{1},\bar{1})\end{array}$$の4つがあります. weight 3の場合は,
$$\begin{array}{r}\zeta (3),\zeta (\bar{3}),\zeta (1,2),\zeta (1,\bar{2}),\zeta (\bar{1},2),\zeta (\bar{1},\bar{2}),\zeta (2,\bar{1}),\zeta (\bar{2},\bar{1}),\zeta (1,1,\bar{1}),\zeta (1,\bar{1},\bar{1}),\zeta (\bar{1},1,\bar{1}),\zeta (\bar{1},\bar{1},\bar{1})\end{array}$$があります. まず, weightの小さい方から求めていきます.

weight 1

$\ln x$のMaclaurin展開,
$$\begin{array}{r}-\ln (1-x)=\sum _{0<n}\frac{x^n}{n}\end{array}$$より,
$$\begin{array}{}\text{(1)}& \zeta (\bar{1})=\sum _{0<n}\frac{(-1{)}^n}{n}=-\ln 2\end{array}$$
が分かります.

weight 2

$$\begin{array}{}\text{(2)}& \zeta (2)=\frac{{\pi }^2}{6}\end{array}$$となることはよく知られています.
$$\begin{array}{rcl}\zeta (2)+\zeta (\bar{2})& =& \sum _{0<n}\frac{2}{(2n{)}^2}\\ \text{(3)}& & =& \frac{1}{2}\zeta (2)\end{array}$$より, $\zeta (\bar{2})=-{\pi }^2/12$が分かります. シャッフル積により,
$$\begin{array}{}\text{(4)}& \zeta (\bar{1}{)}^2=2\zeta (1,\bar{1})\end{array}$$だから, $\zeta (1,\bar{1})=(\ln 2{)}^2/2$となります. また調和積により,
$$\begin{array}{}\text{(5)}& \zeta (\bar{1}{)}^2=\zeta (2)+2\zeta (\bar{1},\bar{1})\end{array}$$より, $\zeta (\bar{1},\bar{1})=(\ln 2{)}^2/2-{\pi }^2/12$となります.

weight 3

まず, 多重ゼータ値の双対性より,
$$\begin{array}{}\text{(6)}& \zeta (1,2)=\zeta (3)\end{array}$$となることはよく知られています. また,
$$\begin{array}{rcl}\zeta (3)+\zeta (\bar{3})& =& \sum _{0<n}\frac{2}{(2n{)}^3}\\ \text{(7)}& & =& \frac{1}{4}\zeta (3)\end{array}$$より, $\zeta (\bar{3})=-3\zeta (3)/4$が分かります. そして,
$$\begin{array}{rcl}\zeta (1,2)+\zeta (1,\bar{2})+\zeta (\bar{1},2)+\zeta (1,\bar{2})& =& \sum _{0<n<m}\frac{2}{2n}\frac{2}{(2m{)}^2}\\ & =& \frac{1}{2}\zeta (1,2)\end{array}$$つまり,
$$\begin{array}{}\text{(8)}& \frac{1}{2}\zeta (3)+\zeta (1,\bar{2})+\zeta (\bar{1},2)+\zeta (1,\bar{2})& =& 0\end{array}$$が分かります. さて, ここから調和積とシャッフル積をもちいていきます. 調和積により,
$$\begin{array}{}\text{(9)}& \zeta (\bar{1})\zeta (2)& =& \zeta (\bar{1},2)+\zeta (2,\bar{1})+\zeta (\bar{3})\text{(10)}& \zeta (\bar{1})\zeta (\bar{2})& =& \zeta (\bar{1},\bar{2})+\zeta (\bar{2},\bar{1})+\zeta (3)\text{(11)}& \zeta (\bar{1})\zeta (1,\bar{1})& =& 2\zeta (1,\bar{1},\bar{1})+\zeta (\bar{1},1,\bar{1})+\zeta (1,2)+\zeta (\bar{2},\bar{1})\text{(12)}& \zeta (\bar{1})\zeta (\bar{1},\bar{1})& =& 3\zeta (\bar{1},\bar{1},\bar{1})+\zeta (\bar{1},2)+\zeta (2,\bar{1})\end{array}$$シャッフル積により,
$$\begin{array}{}\text{(13)}& \zeta (\bar{1})\zeta (2)& =& \zeta (\bar{1},2)+\zeta (\bar{1},\bar{2})+\zeta (\bar{2},\bar{1})\text{(14)}& \zeta (\bar{1})\zeta (\bar{2})& =& 2\zeta (1,\bar{2})+\zeta (2,\bar{1})\text{(15)}& \zeta (\bar{1})\zeta (1,\bar{1})& =& 3\zeta (1,1,\bar{1})\text{(16)}& \zeta (\bar{1})\zeta (\bar{1},\bar{1})& =& 2\zeta (\bar{1},1,\bar{1})+\zeta (\bar{1},\bar{1},\bar{1})\end{array}$$$(9),(10),(13)$により,
$$\begin{array}{rcl}\zeta (2,\bar{1})& =& \frac{{\pi }^2}{12}\ln 2-\frac{1}{4}\zeta (3)\\ \zeta (\bar{1},2)& =& \zeta (3)-\frac{{\pi }^2}{4}\ln 2\end{array}$$$(14)$により,
$$\begin{array}{r}\zeta (1,\bar{2})=\frac{1}{8}\zeta (3)\end{array}$$$(8)$により,
$$\begin{array}{r}\zeta (\bar{1},\bar{2})=\frac{{\pi }^2}{4}\ln 2-\frac{13}{8}\zeta (3)\end{array}$$$(10)$により,
$$\begin{array}{r}\zeta (\bar{2},\bar{1})=\frac{5}{8}\zeta (3)-\frac{{\pi }^2}{6}\ln 2\end{array}$$$(15)$により,
$$\begin{array}{r}\zeta (1,1,\bar{1})=-\frac{1}{6}{\ln }^{3}2\end{array}$$$(12)$により,
$$\begin{array}{r}\zeta (\bar{1},\bar{1},\bar{1})=\frac{{\pi }^2}{12}\ln 2-\frac{1}{6}{\ln }^{3}2-\frac{1}{4}\zeta (3)\end{array}$$$(16)$により,
$$\begin{array}{r}\zeta (\bar{1},1,\bar{1})=\frac{1}{8}\zeta (3)-\frac{1}{6}{\ln }^{3}2\end{array}$$$(11)$により,
$$\begin{array}{r}\zeta (1,\bar{1},\bar{1})=\frac{{\pi }^2}{12}\ln 2-\frac{1}{6}{\ln }^{3}2-\frac{7}{8}\zeta (3)\end{array}$$
これで全ての値を求めることができました. 以下, まとめていきたいと思います.

weight 1,2の値まとめ

$$\begin{array}{rcl}\zeta (\bar{1})& =& -\ln 2\\ \zeta (\bar{2})& =& -\frac{{\pi }^2}{12}\\ \zeta (1,\bar{1})& =& \frac{1}{2}{\ln }^{2}2\\ \zeta (\bar{1},\bar{1})& =& \frac{1}{2}{\ln }^{2}2-\frac{{\pi }^2}{12}\end{array}$$

weight 3の値まとめ

$$\begin{array}{rcl}\zeta (1,2)& =& \zeta (3)\\ \zeta (1,\bar{2})& =& \frac{1}{8}\zeta (3)\\ \zeta (\bar{1},2)& =& \zeta (3)-\frac{{\pi }^2}{4}\ln 2\\ \zeta (\bar{1},2)& =& \zeta (3)-\frac{{\pi }^2}{4}\ln 2\\ \zeta (\bar{1},\bar{2})& =& \frac{{\pi }^2}{4}\ln 2-\frac{13}{8}\zeta (3)\\ \zeta (2,\bar{1})& =& \frac{{\pi }^2}{12}\ln 2-\frac{1}{4}\zeta (3)\\ \zeta (\bar{2},\bar{1})& =& \frac{5}{8}\zeta (3)-\frac{{\pi }^2}{6}\ln 2\\ \zeta (1,1,\bar{1})& =& -\frac{1}{6}{\ln }^{3}2\\ \zeta (1,\bar{1},\bar{1})& =& \frac{{\pi }^2}{12}\ln 2-\frac{1}{6}{\ln }^{3}2-\frac{7}{8}\zeta (3)\\ \zeta (\bar{1},1,\bar{1})& =& \frac{1}{8}\zeta (3)-\frac{1}{6}{\ln }^{3}2\\ \zeta (\bar{1},\bar{1},\bar{1})& =& \frac{{\pi }^2}{12}\ln 2-\frac{{\ln }^{3}2}{6}-\frac{1}{4}\zeta (3)\end{array}$$
調和積とシャッフル積とちょっとで, 意外と簡単にできるんですね. みんなもいろいろやってみてね！