A conjecture relating Multiple Zeta Values and the Polya Enumeration Theorem

Question

Let me state my motivation. I believe that the Polya Enumeration Theorem and Multiple Zeta Values (the classic being the Basel problem and the values of the Riemann zeta function at the even integers) are among the most intriguing objects in mathematics.

With this message I present a conjecture that connects these two (PET and MZVs). Most likely there is a proof somewhere in the literature, which I invite readers to submit in text form if it admits a compact presentation or as a reference if it does not.

To understand this conjecture you need to learn about multiset cycle indices, which I presented in a different context at this MSE link.

Multisets and their cycle indices form a combinatorial species like any other, e.g. cycles, sequences and sets. They represent multisets and are identified by the partition that corresponds to the multiplicities of the elements of the multiset, e.g. $\mathfrak{M}_{1,2,3}$ is a multiset that contains three elements, one of which in two copies and another in three. Substitution of an OGF into a multiset cycle index yields the generating function of the multiset.

There is a special multiset cycle index which is written $\mathfrak{M}_{1,1,1,\ldots,1}$ which gives the multiset where all elements are unique, i.e. the species of sets. This cycle index has been known for quite some time and is used to compute the set operator $\mathfrak{P}.$ The cycle index $Z(P_n)$ of the set operator $\mathfrak{P}_{=n}$ is the difference between the cycle index $Z(A_n)$ of the alternating group and the cycle index $Z(S_n)$ of the symmetric group on $n$ elements. It admits the following simple recursive definition:

$$Z(P_0) = 1 \quad\text{and}\quad Z(P_n) = \frac{1}{n} \sum_{l=1}^n (-1)^{l+1} a_l Z(P_{n-l}).$$

Here are the cycle indices $Z(P_3), Z(P_4)$ and $Z(P_5):$ $$\begin{array}{|l|l|} \hline Z(P_3) & \frac{1}{6}\,{a_{{1}}}^{3}-1/2\,a_{{2}}a_{{1}}+1/3\,a_{{3}}\\ \hline Z(P_4) & \frac{1}{24}\,{a_{{1}}}^{4}-1/4\,a_{{2}}{a_{{1}}}^{2} +1/3\,a_{{3}}a_{{1}}+1/8\,{a_{{2}}}^{2}-1/4\,a_{{ 4}}\\ \hline Z(P_5) & {\frac {1}{120}}\,{a_{{1}}}^{5}-\frac{1}{12}\,a_{{2}}{a_{{1}}}^{3} +1/6\,a_{{3}}{a_{{1}}}^{2}+1/8\,a_{{ 1}}{a_{{2}}}^{2}-1/4\,a_{{4}}a_{{1}} -1/6\,a_{{2}}a_{{3}}+1/5\,a_{{5}}\\ \hline\end{array}$$

With these definitions we are now ready to state the conjecture which is quite simply that $$\large \color{blue}{\zeta(s, s, s, \ldots, s) = Z(P_n)(\zeta(s), \zeta(2s), \ldots, \zeta(ns))}$$ i.e. the MZV of a unique argument $s$ repeated $n$ times is equal to the substituted cycle index $Z(P_n)$ of the set operator $\mathfrak{P}_{=n}$ according to the rule $a_l = \zeta(ls).$

Note that the case $n=3$ is found on the Wikipedia page for MZVs that I linked to in the introduction.

Example. The conjecture says e.g. that $$\zeta(5,5,5,5) = 1/24\, \zeta \left( 5 \right) ^{4}-1/4\,\zeta \left( 10 \right) \zeta \left( 5 \right) ^{2}+1/3\,\zeta \left( 15 \right) \zeta \left( 5 \right) +1/8\, \zeta \left( 10 \right) ^{2}-1/4\,\zeta \left( 20 \right).$$

Important observation. The reader may well ponder the statement of the conjecture and say that there is nothing to prove here, which I will accept as proof, if accompanied by a brief explanation why this should be so.

Addendum. An interesting application of the above result is found at this MSE link.

Answer 1

This actually has nothing to do with zeta functions, it is pure symmetry - a special case of

$$e_n=Z(P_n)(p_1,p_2,\cdots,p_n) \tag{1}$$

where $e_k$ and $p_k$ are the elementary and power-sum symmetric polynomials, in infinitely many variables $x_1,x_2,\cdots$, respectively. For zeta values we simply evaluate at $x_k:=k^{-s}$.

The recursive identity you wrote is the content of Newton's identities, which yields explicitly

$$e_n=\frac{1}{n}\sum_{l=1}^n(-1)^{l+1}p_le_{n-l}. \tag{2}$$

A more explicit formula is possible using generating functions. Observe

$$\begin{align} \sum_{n=0}^\infty T^ne_n & = \prod_{i=1}^\infty(1+Tx_i) \tag{3} \\ & = \exp\left[\sum_{i=1}^\infty \log(1+Tx_i)\right] \tag{4} \\ & = \exp\left[\sum_{i=1}^\infty \sum_{\ell=1}^\infty (-1)^{\ell+1}\frac{(Tx_i)^\ell}{\ell}\right] \tag{5} \\ & = \exp\left[\sum_{\ell=1}^\infty(-1)^{\ell+1}\frac{T^\ell}{\ell}\sum_{i=1}^\infty x_i^\ell\right] \tag{6} \\ & = \exp\left[\sum_{\ell=1}^\infty(-1)^{\ell+1}\frac{T^\ell}{\ell}p_\ell\right] \tag{7} \\ & = \prod_{\ell=1}^\infty\exp\left[(-1)^{\ell+1}\frac{T^\ell}{\ell}p_\ell\right] \tag{8} \\ & = \prod_{\ell=1}^\infty \left[\sum_{c_\ell=0}^\infty \frac{1}{c_\ell!}\left((-1)^{\ell+1}\frac{T^\ell}{\ell}p_\ell\right)^{c_\ell}\right] \tag{9} \\ & = \sum_{n=0}^\infty T^n \sum_{1c_1+2c_2+\cdots=n} \prod_{\ell=1}^\infty \frac{(-1)^{(\ell+1)c_\ell}}{c_\ell!}\left(\frac{p_\ell}{\ell}\right)^{c_\ell} \tag{10} \\ & = \sum_{n=0}^\infty \frac{T^n}{n!} \sum_{1c_1+\cdots+nc_n=n} (-1)^{c_2+c_4+\cdots}\frac{n!}{1^{c_1}2^{c_2}\cdots n^{c_n}c_1!c_2!\cdots c_n!}p_1^{c_1}\cdots p_n^{c_n} ~~ \tag{11} \\ & = \sum_{n=0}^\infty \frac{T^n}{n!}\sum_{\sigma\in S_n} {\rm sgn}(\sigma)p_1^{c_1(\sigma)}\cdots p_n^{c_n(\sigma)} \tag{12} \end{align}$$

where $c_k(\sigma)$ is the number of length $k$ cycles in $\sigma$'s disjoint cycle decomposition, the sign of $\sigma$ is equal to $(-1)^{c_2+c_4+\cdots}=(-1)^{2c_1+3c_2+4c_3+\cdots}$, and the number of permutations in $S_n$ with the cycle type $(\underbrace{n,\cdots,n}_{c_n},\cdots,\underbrace{1,\cdots,1}_{c_1})$ is given $n!/(1^{c_1}2^{c_2}\cdots n^{c_n}c_1!\cdots c_n!)$ (see here).

Therefore we may conclude

$$e_n=\frac{1}{n!}\sum_{\sigma\in S_n}{\rm sgn}(\sigma)p_1^{c_1(\sigma)}\cdots p_n^{c_n(\sigma)}. \tag{13}$$

The more general polynomials in your additional MZV identities are called the monomial symmetric functions $m_\mu$ for integer partitions $\mu$. These form a linear basis for the space of symmetric functions. I am not aware of a direct formula for monomial symmetric functions in terms of power sums, but in terms of standard constants and functions there is an indirect way: writing monomials in terms of Schur polynomials $s_\lambda$, and writing those with power-sums.

The Kostka numbers are defined by the relation $s_\lambda=\sum_\mu K_{\lambda\mu}m_\mu$, where the sum is over integer partitions. If one inverts the Kostka matrix we can write $m_\mu=\sum_\lambda J_{\mu\lambda}s_\lambda$ for some values $J_{\mu\lambda}$.

Schur polynomials in turn can be decomposed as

$$s_\lambda=\frac{1}{n!}\sum_{\sigma\in S_n}\chi_\lambda(\sigma)p_1^{c_1(\sigma)}\cdots p_n^{c_n(\sigma)} \tag{14}$$

where $\chi_\lambda$ is the irreducible character of $S_n$ corresponding to $\lambda$.

Another route for envelope calculation is to decompose $m_\mu$ into the variables $e_k$ (which is possible thanks to the fundamental theorem of symmetric polynomials) using this recursive algorithm, and then writing the $e_k$s in terms of $p_k$s as in $(13)$. This should be faster, although likely there are even faster, more direct ways of writing monomials in terms of power-sums. For more literature related to this you'll want to read deeper into the combinatorics of symmetric polynomials as well as the representation theory of symmetric groups.

Out of theoretical interest, representation theory illustrates $(14)$ is a trace formula:

$$\chi_{\Bbb S_\lambda(V)}(g)=\frac{1}{n!}\sum_{\sigma\in S_n} \chi_\lambda(\sigma)\chi_V(g^1)^{c_1(\sigma)}\cdots\chi_V(g^n)^{c_n(\sigma)}$$ where $\Bbb S_\lambda(V)=V^{\otimes n}\otimes_{\Bbb C[S_n]}M_\lambda$ is the Schur functor applied to $V$ considered as a representation of the group ${\rm GL}(V)$, and $M_\lambda$ is the irreducible representation of $S_n$ associated to $\lambda$.

Answer 2

By way of enrichment of this discussion I would like to present some additional MZV identities, the purpose being to verify if these are as easy to prove as what I had in my first post. They were computed with multiset cycle indices as described in that post and the link provided there. Please indicate possible proofs. Code available on request.

$$\begin{array}{|l|l|} \hline \mathfrak{M}_{3,3,3} & 1/6\, \left( \zeta \left( 3\,s \right) \right) ^{3}+1/3\,\zeta \left( 9\,s \right) -1/2\,\zeta \left( 3\,s \right) \zeta \left( 6\,s \right) =\zeta \left( 3\,s,3\,s,3\,s \right)\\ \hline \mathfrak{M}_{2,2,5} & 1/2\, \left( \zeta \left( 2\,s \right) \right) ^{2}\zeta \left( 5\,s \right) -1/2\,\zeta \left( 4\,s \right) \zeta \left( 5\,s \right) -\zeta \left( 2\,s \right) \zeta \left( 7 \,s \right) +\zeta \left( 9\,s \right) \\ &=\zeta \left( 2\,s,2\,s, 5\,s \right) +\zeta \left( 2\,s,5\,s,2\,s \right) +\zeta \left( 5\,s,2\,s,2\,s \right)\\ \hline \mathfrak{M}_{2,3,4} & \zeta \left( 2\,s \right) \zeta \left( 3\,s \right) \zeta \left( 4\,s \right) -\zeta \left( 2\,s \right) \zeta \left( 7 \,s \right) -\zeta \left( 3\,s \right) \zeta \left( 6\,s \right) -\zeta \left( 4\,s \right) \zeta \left( 5\,s \right) + 2\,\zeta \left( 9\,s \right)\\& =\zeta \left( 2\,s,3\,s,4\,s \right) +\zeta \left( 2\,s,4\,s,3\,s \right) +\zeta \left( 3\, s,2\,s,4\,s \right) \\&+\zeta \left( 3\,s,4\,s,2\,s \right) +\zeta \left( 4\,s,2\,s,3\,s \right) +\zeta \left( 4\,s,3\,s,2\,s \right)\\ \hline \mathfrak{M}_{1,1,1,2,2} & 1/12\, \left( \zeta \left( s \right) \right) ^{3} \left( \zeta \left( 2\,s \right) \right) ^{2}-1/4\,\zeta \left( s \right) \left( \zeta \left( 2\,s \right) \right) ^{3}-1/2\, \left( \zeta \left( s \right) \right) ^{2}\zeta \left( 2\,s \right) \zeta \left( 3\,s \right) \\&+2/3\, \left( \zeta \left( 2\,s \right) \right) ^{2}\zeta \left( 3\,s \right) +1/2\,\zeta \left( s \right) \left( \zeta \left( 3\,s \right) \right) ^{ 2}-1/12\, \left( \zeta \left( s \right) \right) ^{3}\zeta \left( 4\,s \right) \\&+5/4\,\zeta \left( s \right) \zeta \left( 2\,s \right) \zeta \left( 4\,s \right) -7/6\,\zeta \left( 3\,s \right) \zeta \left( 4\,s \right) +1/2\, \left( \zeta \left( s \right) \right) ^{2}\zeta \left( 5\,s \right) -3/2\,\zeta \left( 2\,s \right) \zeta \left( 5\,s \right)\\& -3/2\,\zeta \left( s \right) \zeta \left( 6\,s \right) +2\,\zeta \left( 7 \,s \right) =\zeta \left( s,s,s,2\,s,2\,s \right) +\zeta \left( s,s,2\,s,s,2\,s \right) \\&+\zeta \left( s,s,2\,s,2\,s,s \right) +\zeta \left( s,2\,s,s,s,2\,s \right) +\zeta \left( s, 2\,s,s,2\,s,s \right) \\&+\zeta \left( s,2\,s,2\,s,s,s \right) + \zeta \left( 2\,s,s,s,s,2\,s \right) +\zeta \left( 2\,s,s,s,2\, s,s \right) \\&+\zeta \left( 2\,s,s,2\,s,s,s \right) +\zeta \left( 2\,s,2\,s,s,s,s \right)\\ \hline \mathfrak{M}_{1,1,3,3} & 1/4\, \left( \zeta \left( s \right) \right) ^{2} \left( \zeta \left( 3\,s \right) \right) ^{2}-1/4\,\zeta \left( 2\,s \right) \left( \zeta \left( 3\,s \right) \right) ^{2}-\zeta \left( s \right) \zeta \left( 3\,s \right) \zeta \left( 4\,s \right)\\& +1/2\, \left( \zeta \left( 4\,s \right) \right) ^{2}+ \zeta \left( 3\,s \right) \zeta \left( 5\,s \right) -1/4\, \left( \zeta \left( s \right) \right) ^{2}\zeta \left( 6\,s \right) +1/4\,\zeta \left( 2\,s \right) \zeta \left( 6\,s \right) \\&+\zeta \left( s \right) \zeta \left( 7\,s \right) -3/2 \,\zeta \left( 8\,s \right) =\zeta \left( s,s,3\,s,3\,s \right) +\zeta \left( s,3\,s,s,3\,s \right) +\zeta \left( s,3 \,s,3\,s,s \right) \\&+\zeta \left( 3\,s,s,s,3\,s \right) +\zeta \left( 3\,s,s,3\,s,s \right) +\zeta \left( 3\,s,3\,s,s,s \right)\\ \hline \mathfrak{M}_{1,1,2,2} & 1/4\, \left( \zeta \left( s \right) \right) ^{2} \left( \zeta \left( 2\,s \right) \right) ^{2}-1/4\, \left( \zeta \left( 2 \,s \right) \right) ^{3}-\zeta \left( s \right) \zeta \left( 2 \,s \right) \zeta \left( 3\,s \right) +1/2\, \left( \zeta \left( 3\,s \right) \right) ^{2}\\&-1/4\, \left( \zeta \left( s \right) \right) ^{2}\zeta \left( 4\,s \right) +5/4\,\zeta \left( 2\,s \right) \zeta \left( 4\,s \right) +\zeta \left( s \right) \zeta \left( 5\,s \right) -3/2\,\zeta \left( 6\,s \right)\\& =\zeta \left( s,s,2\,s,2\,s \right) +\zeta \left( s,2 \,s,s,2\,s \right) +\zeta \left( s,2\,s,2\,s,s \right)\\& +\zeta \left( 2\,s,s,s,2\,s \right) +\zeta \left( 2\,s,s,2\,s,s \right) +\zeta \left( 2\,s,2\,s,s,s \right)\\ \hline \mathfrak{M}_{2,2,2,2} & 1/24\, \left( \zeta \left( 2\,s \right) \right) ^{4}-1/4\, \left( \zeta \left( 2\,s \right) \right) ^{2}\zeta \left( 4\, s \right) +1/8\, \left( \zeta \left( 4\,s \right) \right) ^{2} \\&+1/3\,\zeta \left( 2\,s \right) \zeta \left( 6\,s \right) -1/4 \,\zeta \left( 8\,s \right) =\zeta \left( 2\,s,2\,s,2\,s,2\,s \right)\\ \hline \end{array}$$ Another one which corresponds to $\mathfrak{M}_{1,2,2,5}$ is $$ \left( \zeta \left( 5\,s \right) \right) ^{2}+1/2 \,\zeta \left( 4\,s \right) \zeta \left( 6\,s \right) +\zeta \left( 3\,s \right) \zeta \left( 7 \,s \right) +2\,\zeta \left( 2\,s \right) \zeta \left( 8\,s \right) \\+\zeta \left( s \right) \zeta \left( 9\,s \right) -\zeta \left( 2\,s \right) \zeta \left( 3\,s \right) \zeta \left( 5\,s \right) -1/2\,\zeta \left( s \right) \zeta \left( 4\,s \right) \zeta \left( 5\,s \right) -1/2\, \left( \zeta \left( 2\,s \right) \right) ^{2} \zeta \left( 6\,s \right)\\ -\zeta \left( s \right) \zeta \left( 2\,s \right) \zeta \left( 7\,s \right) -3\,\zeta \left( 10\,s \right) +1/2\,\zeta \left( s \right) \left( \zeta \left( 2\,s \right) \right) ^{2}\zeta \left( 5\,s \right) \\= \zeta \left( s,2\,s,2\,s,5\,s \right) +\zeta \left( s,2\,s,5\,s,2\,s \right) +\zeta \left( s,5 \,s,2\,s,2\,s \right) +\zeta \left( 2\,s,s,2\,s,5\, s \right) \\+\zeta \left( 2\,s,s,5\,s,2\,s \right) + \zeta \left( 2\,s,2\,s,s,5\,s \right) +\zeta \left( 2\,s,2\,s,5\,s,s \right) +\zeta \left( 2\,s ,5\,s,s,2\,s \right) \\+\zeta \left( 2\,s,5\,s,2\,s,s \right) +\zeta \left( 5\,s,s,2\,s,2\,s \right) + \zeta \left( 5\,s,2\,s,s,2\,s \right) +\zeta \left( 5\,s,2\,s,2\,s,s \right).$$