# Is $\zeta(6,2)$ the lowest multiple zeta function that cannot be expressed in terms of single zeta values?

Consider a following Euler sum: $$\sum\limits_{m=1}^\infty \frac{H_m^{(2)}}{m^6} = \zeta(6,2) + \zeta(8)$$ where $\zeta(,)$ is the multiple zeta function. As a result of calculations in Calculating alternating Euler sums of odd powers I was not able to reduce that quantity to single zeta functions. The question is to do that or otherwise to prove that it is impossible.

• Just a reformulation: is the integral $$I = \int_{0}^{1}\frac{\text{Li}_2(x)\log(x)^5}{1-x}\,dx$$ expressible in terms of $\zeta$ values only? – Jack D'Aurizio May 24 '17 at 17:59
• One can also write:\begin{eqnarray} I &=& \frac{1}{12} \int\limits_0^1 [\log(\frac{1}{\xi} )]^3 \cdot \frac{Li_2(\xi)^2}{\xi} d\xi - \frac{1}{5!} \int\limits_0^1 [\log(\frac{1}{\xi} )]^5 \cdot \frac{\log(1-\xi)^2}{\xi} d\xi +\zeta(8)\\ &=& \frac{1}{12} \int\limits_0^1 [\log(\frac{1}{\xi} )]^3 \cdot \frac{Li_2(\xi)^2}{\xi} d\xi - \frac{\pi^8}{6300} + 2 \zeta(3) \zeta(5) \end{eqnarray} where I used math.stackexchange.com/questions/1902096/… .Unfortunately I do not know how to compute the remaining integral in the rhs. – Przemo May 25 '17 at 10:17

The weight of a multiple zeta value $\zeta(s_1,\ldots,s_k)$ is $s_1+\ldots +s_k$. It is known that every multiple zeta value of weight at most $7$ can be written in terms of single zeta values, but conjecturally $\zeta(6,2)$ cannot be written in terms of single zeta values. In the MZV literature one can find this statement for $\zeta(5,3)$, but this implies the statement for $\zeta(6,2)$ because $$\zeta(6,2)=-\frac{42}{125}\zeta(2)^2+2\zeta(3)\zeta(5)-\frac{2}{5}\zeta(5,3).$$ I read the identity on the previous line off from the MZV data mine.
To prove $\zeta(6,2)$ is not in the algebra generated by single zeta values is an open problem, and problems like this are expected to be quite hard. It is not even known that $\zeta(6,2)$ is irrational. However, this statement would follow from a form of the Grothendieck period conjecture.
• This is interesting. I have analyzed the quantities in question up to weight twelve in math.stackexchange.com/questions/2169507/… and what I found is that the "new " quantities are $\zeta(6,2)$, $\zeta(8,2)$ and $\zeta(10,2)$. Maybe there is some pattern in that? Maybe , in general, the quantities $\zeta(2n, 2)$ for $n \ge 3$ are new. I do not know if there is any significance in that? – Przemo May 25 '17 at 11:44
Let us fix the weight at $P$. Here we assume that $P$ is even since the odd case has been solved. From the considerations in Calculating alternating Euler sums of odd powers we know that following recursion relations hold: \begin{eqnarray} &&{\bf H}^{(2 q)}_{P-2 q}(+1) +\sum _{l=1}^{P-2 q} {\bf H}^{(l)}_{P-l}(+1) \cdot\text{A2}(P,q,l)=\\ &&\sum_{l=P-2 q}^{\frac{P}{2}-1} (-1)^l \zeta (l) \zeta (P-l) \binom{l-1}{P-2 q-1}+\frac{1}{2} (-1)^{P/2} \zeta \left(\frac{P}{2}\right)^2 \binom{\frac{P}{2}-1}{P-2 q-1}+\zeta (P)+\\ &&\sum _{l=2}^{\frac{P}{2}-1} (-1)^l \zeta (l) \zeta (P-l) \text{A1}(P,q,l)+\sum _{l=2}^{P-2 q} (-1)^0 \zeta (l) \zeta (P-l) \text{A2}(P,q,l) \end{eqnarray} where $q=1,\cdots,P/2-1$. Likewise \begin{eqnarray} &&{\bf H}^{(2q+1)}_{P-2q-1}(+1)+\sum _{l=1}^{P-2 q-2} {\bf H}^{(P-l)}_l(+1) \text{B2}(P,q,l) = \\ &&\sum _{l=(P-2 q-1)\vee 2}^{\frac{P}{2}-1} (-1)^{l+1} \zeta(l)\zeta(P-l) \binom{l-1}{P-2 q-2} +\frac{1}{2} (-1)^{\frac{P}{2}+1} \zeta(\frac{P}{2})^2\binom{\frac{P}{2}-1}{P-2 q-2} +\zeta(P)+\\ &&\sum _{l=0}^{\frac{P}{2}-1} (-1)^{l+1} \text{B1}(P,q,l) \zeta(P-l) (\zeta(l) 1_{l\ge 2}-\delta_{l,0}) \end{eqnarray} where $q=0,\cdots,P/2-1$. The coefficients read: \begin{eqnarray} A1(P,q,l) &:=& \sum _{j=\max (l-q,0)}^{\frac{P}{2}-q-1} \left(\frac{1}{2} \binom{j+q-1}{2 j-1}+\binom{j+q-1}{2 j+0}\right) \binom{-j-l+P-q-1}{j-l+q}\\ A2(P,q,l)&:=&\sum _{j=0}^{-\left\lfloor \frac{l}{2}\right\rfloor +\frac{P}{2}-q} \left(\frac{1}{2} \binom{j+q-1}{2 j-1}+\binom{j+q-1}{2 j+0}\right) \binom{-j-l+P-q-1}{j+q-1}\\ B1(P,q,l) &:=& \sum _{j=\max (0,\text{l}-q-1)}^{\frac{P}{2}-q-2} \left(\frac{1}{2} \binom{j+q+0}{2 j+0}+\binom{j+q+0}{2 j+1}\right) \binom{-j-\text{l}+P-q-2}{j-\text{l}+q+1}\\ B2(P,q,l) &:=& \sum _{j=0}^{\left\lfloor -\frac{\text{l}}{2}+\frac{P}{2}-q-1\right\rfloor } \left(\frac{1}{2} \binom{j+q+0}{2 j+0}+\binom{j+q+0}{2 j+1}\right) \binom{-j-\text{l}+P-q-2}{j+q+0} \end{eqnarray} Altogether we have $(P-1)$ equations for $(P-1)$ unknowns. Now it is very easy to solve those equations using any symbolic manipulation package like Mathematica for example. What appears is that the number of linearly dependent equations $\nu(P)$ depends on the weight $P$. In fact we have: \begin{eqnarray} \nu(P) = \left\{ \begin{array}{rr} 0 & \mbox{if $P=2,4,6$}\\ 1 & \mbox{if $P=8,10,12$}\\ 2 & \mbox{if $P=14,16,18$}\\ \vdots \end{array} \right. \end{eqnarray} This explains why the lowest 2D zeta value that is not reducible to single zeta values occurs at weight eight. At this weight the rank of the corresponding matrix jumps down by one.
Now if we, for example, take $P=12$ the following identities hold: \begin{eqnarray} \begin{array} -\zeta (11,1)-\zeta (5) \zeta (7)-\zeta (3) \zeta (9)+\frac{9 \zeta (12)}{4}=0\\ 0=0\\ -\zeta (9,3)-\frac{9}{2} (\zeta (10,2)+\zeta (12))+12 \zeta (5) \zeta (7)+9 \zeta (3) \zeta (9)-\frac{151 \zeta (12)}{8}=0\\ -\zeta (8,4)+8 (\zeta (10,2)+\zeta (12))-28 \zeta (5) \zeta (7)-16 \zeta (3) \zeta (9)+\frac{84023 \zeta (12)}{2073}=0\\ -\zeta (7,5)-7 (\zeta (10,2)+\zeta (12))+28 \zeta (5) \zeta (7)+14 \zeta (3) \zeta (9)-\frac{324319 \zeta (12)}{8292}=0\\ \frac{12 \zeta (12)}{691}-\zeta (6,6)=0\\ -\zeta (5,7)+7 (\zeta (10,2)+\zeta (12))-27 \zeta (5) \zeta (7)-14 \zeta (3) \zeta (9)+\frac{316027 \zeta (12)}{8292}=0\\ -\zeta (4,8)-8 (\zeta (10,2)+\zeta (12))+28 \zeta (5) \zeta (7)+16 \zeta (3) \zeta (9)-\frac{335375 \zeta (12)}{8292}=0\\ -\zeta (3,9)+\frac{9}{2} (\zeta (10,2)+\zeta (12))-12 \zeta (5) \zeta (7)-8 \zeta (3) \zeta (9)+\frac{143 \zeta (12)}{8}=0\\ -\zeta (2,10)-\zeta (10,2)+\frac{893 \zeta (12)}{1382}=0\\ -\zeta (1,11)+\zeta (5) \zeta (7)+\zeta (3) \zeta (9)-\frac{13 \zeta (12)}{4} =0 \end{array} \end{eqnarray} where in the last line $\zeta(1,11) := -\zeta(12)+\lim_{t \rightarrow 1} \left({\bf H}^{(11)}_1(t) + \log(1-t)\cdot Li_{11}(t)\right)$. This is the highest weight where only one unknown parameter enters the solutions, namely $\zeta(10,2)$. For weights $P=14,16,18$ there will be two unknown parameters, i.e. $\zeta(P-2,2)$ and $\zeta(P-4,4)$ that will enter the solution. Then again for weights $P=20,22,24$ there will be three unknown parameters,i.e. $\zeta(P-2,2)$, $\zeta(P-4,4)$ and $\zeta(P-6,6)$, in the solutions. It is not hard to see the generic pattern in that.