Repeated-Index Multiple Zeta Values and Dirichlet Lambda Analogues

Repeated-Index Multiple Zeta Values and Dirichlet Lambda Analogues Gyeongmin Yang Abstract Our first contribution is a unified derivation of generating functions for ζ({s}n), ζ({¯s}n), and λ({s}n), obtained through Newton-type recursion formulas that refine classical symmetric-function identities. We further establish a closed combinatorial expression for the expansion coefficients of λ({2p}n), which highlights the structural role of root-of-unity symmetries. Finally, for λ({3}n)we prove that no such representation exists in the commutative setting, thereby motivating a non-commutative framework. Within this framework, the commutation condition [m(x), m(−x)] = 0 leads naturally to the notion of pairwise centrality, while the sum-zero and full 2-step nilpotent constraints subsequently emerge as higher-order structural requirements ensuring global commutativity. MSC(2020) : Primary 11M32, Secondary 33E20, 16T10 Keywords : multiple zeta values(MZVs), Dirichlet lambda function, noncommutative exponential 1 Introduction The Riemann zeta function, defined for <(s)>1by ζ(s) = ∞ X k=1 k−s,(1) admits analytic continuation to the complex plane except for a simple pole at s= 1. For positive integers s1, s2, . . . , snwith sn≥2, multiple zeta values (MZVs) and multiple zetastar value (MZSV) are defined as follows [1]: ζ(s1, s2, . . . , sn) := X 0<m1<m2<···<mn 1 ms1 1ms2 2···msn n ,(2) ζ∗(s1, s2, . . . , sn) := X 1≤m1≤m2≤···≤mn 1 ms1 1ms2 2···msn n .(3) 1 2 Derivations of λ({s}n)from ζ({s}n)and ζ({¯s}n) 2.1 Multiple T-values (odd-denominator MZVs) Let r≥1and k1, . . . , kr∈Z≥1. Following the odd-denominator restriction (often called tor T-values), set [2] T(k1, . . . , kr−1;s) := X 1≤m1<···<mr miodd 1 mk1 1···mkr−1 r−1ms r ,(<(s)>1).(4) For s=kr∈Zand repeated indices (cf. [3, §1]), we write T({s}n) := T(s, . . . , s | {z } ntimes ) := λ({s}n). 2.2 Case ζ({s}n) In [4], the following identity furnishes the ordinary generating function (as a formal power series) for the multiple zeta values with repeated index s, expressed equivalently as an infinite product and as the exponential of a Dirichlet series: X n≥0 xsnζ({s}n) = Y j≥11 + xs js= exp X k≥1 (−1)k−1xskζ(sk) k(5) with <(s)>1and ζ({. . . }0) = 1. Remark 2.1 (Conventions).Unless otherwise noted, we impose <(s)>1and |x|<1to guarantee absolute convergence. In particular, all generating functions are understood as formal power series under these conditions. Substituting xby x/2in equation (5): X n≥0x 2sn ζ({s}n) = Y j≥11 + (x/2)s js = exp X k≥1 (−1)k−1(x/2)skζ(sk) k. (6) Taking the logarithms of (5) and (6), subtracting (6)) from (5), and subsequently applying the exponential function yield the following expression: X n≥0 xsnλ({s}n) = Y j≥11 + xs (2j−1)s= exp X k≥1 (−1)k−1xskλ(sk) k.(7) For even integers s=m= 2p, let w2p= exp(iπ/2p), where i=√−1. We then let (7) denote the trigonometric product: [5] X n≥0 x2pnλ({2p}n) = p Y j=1 cos(w2p)2j−1π 2x.(8) 2 2.3 Case ζ({¯s}n) In [4], ζ({¯s}n)is defined as follows: X n≥0 xsnζ({¯s}n) = Y j≥11+(−1)jxs js = exp X k≥12(x/2)2sk−sζ(2sk −s) 2k−1−xskζ(sk) k (9) with <(s)>0and ζ({. . . }0) = 1. Taking the logarithms of (6) and (9), subtracting (9) from (6), and then applying the exponential function yield the following expression: X n≥0 xsnζ∗ odd({s}n) = Y j≥11−xs (2j−1)s−1= exp X k≥1 xskλ(sk) k(10) where ζ∗ odd denotes the multiple zeta-star value restricted to odd denominators mi∈ {1,3,5, . . . }. The coefficients in (10) follow directly from the generating-series identity for the complete homogeneous symmetric functions. 2.4 Unified Newton recursions Write u:= xs. Let H(t) = Pn≥0hntn=Qi≥1(1 −xit)−1and E(t) = Pn≥0entn= Qi≥1(1+xit)be the complete and elementary symmetric generating series, and let pr=Pixr i denote the power sums. The H,Eidentities [6, p. 19, p. 21, Eq. (2.11)] give: H0(t) H(t)=X r≥1 prtr−1⇐⇒ n hn= n X r=1 prhn−r, E0(t) E(t)=X r≥1 (−1)r−1prtr−1⇐⇒ n en= n X r=1 (−1)r−1pren−r. (11) Theorem 2.2. For any sequence φ={φr}r≥1, define Fφ(u) := exp nX r≥1 φr ruro=X n≥0 c(φ) nun, c(φ) 0= 1. Then the coefficients satisfy the Newton-type recursion n c(φ) n= n X r=1 φrc(φ) n−r(n≥1).(12) Sketch. From d du log Fφ(u) = Pr≥1φrur−1and logPn≥0c(φ) nun, multiply by Pm≥0c(φ) mum and compare coefficients. Equivalently, (12) is (11) with the specialization pr7→ φrand hn7→ c(φ) n. 3 We now record three specializations of (12) relevant to zeta-type inputs. (I) ζon the E(t).Set φ(ζ) r:= (−1)r−1ζ(rs)and Fζ(u) = exp nX r≥1 (−1)r−1 rζ(rs)uro=X n≥0 c(ζ) nun. Then by (12) (see also [3, Eq. (2.7)]), n c(ζ) n= n X r=1 (−1)r−1ζ(rs)c(ζ) n−r(n≥1) (13) (II) λon the H(t).With λ(t) = (1 −2−t)ζ(t), set φ(λ,+) r:= λ(rs)and Fλ,+(u) = exp nX r≥1 λ(rs) ruro=X n≥0 c(λ,+) nun. Then (12) yields n c(λ,+) n= n X r=1 λ(rs)c(λ,+) n−r(n≥1).(14) (III) λon the E(t)(alternating). Set φ(λ,−) r:= (−1)r−1λ(rs)and Fλ,−(u) = exp nX r≥1 (−1)r−1 rλ(rs)uro=X n≥0 c(λ,−) nun. Then (12) gives n c(λ,−) n= n X r=1 (−1)r−1λ(rs)c(λ,−) n−r(n≥1).(15) Remark 2.3 (Interpretation and relations).(i) Cases (I) and (III) arise from the E(t)of (11), hence the alternating sign (−1)r−1. Case (II) comes from the H(t)side and carries no alternation. (ii) The inputs are linked by λ(rs) = (1 −2−rs)ζ(rs), but the resulting recursions govern distinct coefficient families associated with the three generating functions above. (iii) In the region <(s)>1and |u|<1, the series in all three cases converge absolutely, legitimizing the coefficient extraction and the use of (12). 2.5 Derivation of the λ({2p}n)coefficients cp,α In the following discussion, we shall, for convenience, substituting πx/2in equation (8) by x. We consider the product Λp(x) = p Y j=1 cos(ω2p)2j−1xω2p=eπi/(2p).(16) 4 Our goal is to determine the Maclaurin coefficients cp,α of the expansion Λp(x) = ∞ X k=0 cp,α x2pα.(17) Using the identity cos x= (eix +e−ix)/2, we can rewrite the product as Λp(x) = 2−p p Y j=1 exp(i(ω2j−1 2p)x) + exp(−i(ω2j−1 2p)x). Expanding the product corresponds to choosing a sign εj∈ {±1}for each factor, which yields Λp(x) = 2−pX ε1,...,εp∈{±1} exp ix p X j=1 εj(ω2p)2j−1!. In this formulation, setting εj= +1 amounts to choosing exp(i(ω2j−1 2p)x), while εj=−1 corresponds to choosing exp(−i(ω2j−1 2p)x). Applying the Maclaurin series for the exponential, we obtain Λp(x) = 2−pX ε ∞ X m=0 (ixΨ(ε))m m!, where Ψ(ε) = p X j=1 εj(ω2p)2j−1. Hence, the coefficient of xmis cp,m =1 m! 2pX ε (iΨ(ε))m. Due to the symmetry of the 2p-th roots of unity, the above sum vanishes unless mis divisible by 2p. Consequently, the expansion of Λp(x)only contains terms of the form x2pα. We therefore arrive at the combinatorial expression cp,α =1 (2pα)! 2pX ε1,...,εp=±1 i p X j=1 εj(ω2p)2j−1!2pα .(18) This shows that only terms of order 2pα survive in the expansion, reflecting the inherent symmetry of the construction. 3 Towards a λ({3}n) 3.1 Absence of a classical solution in the commutative setting Substituting p= 3 into equation (16) yields Λ3(x) = 1 + 3 4 26 6! x6+3 4 212 12!x12 +3 4 218 18!x18 . . . (19) 5 and the expansion shows that every term of degree divisible by six carries the common factor 3/4. Assuming the abelian ansatz X n≥0 x3nλ({3}n) = j(x) = 1 4(eαx +eβx +eγx +eδx),(20) where α, β, γ, δ are abelian parameters in a unital ring. The product j(x)j(−x)enforces the algebraic constraints (α−β)6= (α−γ)6= (α−δ)6= (β−γ)6= (β−δ)6= (γ−δ)6=c, α3n−2+β3n−2+γ3n−2+δ3n−2= 0, α3n−1+β3n−1+γ3n−1+δ3n−1= 0 (21) for some constant c. This set of relations defines an infinite hierarchy of algebraic conditions, which leads to an infinite system of polynomial constraints on the parameters α, β, γ, δ. However, such a system admits no simultaneous solution for distinct values of α, β, γ, δ, which indicates that the commutative framework is too restrictive to accommodate a nontrivial realization. Lemma 3.1. Let z, w ∈Cwith |z|=|w|= 1 and z6=w. Then |z−w|= 1 ⇐⇒ z w∈ {eiπ/3, e−iπ/3}. In particular, if z1, z2, z3lie on the unit circle and are pairwise distinct, it is impossible to have |z1−z2|=|z1−z3|=|z2−z3|= 1. Proof. Compute |z−w|2= (z−w)(z−w)=2−zw −zw = 2 −2<(zw). Hence |z−w|= 1 iff <(zw) = 1 2, i.e. arg(zw)∈ {±π/3}, which is the stated equivalence. For the “in particular” part, assume |z1|=|z2|=|z3|= 1 and |z1−z2|=|z1−z3|= |z2−z3|= 1. Then by the first part z1 z2 ,z1 z3 ,z2 z3∈ {e±iπ/3}. But (z1/z2)·(z2/z3) = z1/z3forces a product of two factors in {e±iπ/3}to again lie in that same set, which is impossible since e±iπ/3·e±iπ/3=e±i2π/3/∈ {e±iπ/3}. Contradiction. Proposition 3.2 (No abelian solution).Fix n≥1. There do not exist pairwise distinct α, β, γ, δ ∈Cand a nonzero constant csuch that (α−β)2n= (α−γ)2n= (α−δ)2n= (β−γ)2n= (β−δ)2n= (γ−δ)2n=c6= 0. Proof. By translation and scaling, set α= 0 and c= 1; then β, γ, δ ∈µ2nand all pairwise differences among {0, β, γ, δ}lie in µ2n. In particular, |β|=|γ|=|δ|= 1 and also |β− γ|=|β−δ|=|γ−δ|= 1, since every root of unity has modulus 1. This contradicts Lemma 3.1. Hence such four distinct points cannot exist. (If c= 0, the only possibility is α=β=γ=δ.) 6 Structural origin of the obstruction. The failure of the abelian ansatz in Proposition 3.2 is not a numerical accident, but a manifestation of an intrinsic geometric obstruction. The equal-difference conditions of order six enforce a rigid overdetermined symmetry that cannot be simultaneously realized by distinct commuting parameters α, β, γ, δ in any commutative ring. In other words, the abelian framework lacks the algebraic degrees of freedom necessary to balance these sixth-order constraints. This suggests that the obstruction is structural rather than algebraic: it arises from the requirement that all pairwise differences commute, which effectively collapses the parameter space to a single degenerate configuration. Recognizing this structural limitation motivates the passage to the noncommutative setting, where nested commutators provide precisely the additional degrees of freedom needed to restore consistency. Remark 3.3 (Motivation for the noncommutative extension).The impossibility established in Proposition 3.2 reveals that even the simplest abelian realization cannot support a nontrivial system of parameters satisfying the sixth-power equal-difference relations. This obstruction is structural rather than numerical: it originates from the requirement that all pairwise differences commute, which forces every admissible configuration to collapse into trivial degeneracy. From this perspective, the passage to a noncommutative setting is not merely optional but necessary: allowing non-commuting generators aiintroduces additional algebraic degrees of freedom specifically, non-central commutators that can compensate for the symmetry constraints that make the abelian system overdetermined. Hence the noncommutative formulation arises as the minimal relaxation under which a λ({3}n)-type hierarchy may consistently exist. 3.2 Commutation condition [m(x), m(−x)] = 0 According to the Baker–Campbell–Hausdorff (BCH) expansion, for two elements X, Y one has [7, Eq. (5.3)] exp(X) exp(Y) = exp X+Y+1 2[X, Y ] + 1 12[X, [X, Y ]] −1 12[Y, [X, Y ]] + ···. This formula makes it evident that in the noncommutative setting higher-order commutators inevitably appear. In what follows we investigate the averaged exponential m(x) = 1 N N X i=1 eaix, with noncommuting elements a1, . . . , aN, and analyze its commutator [m(x), m(−x)] in relation to structural constraints such as 2-step nilpotency. Fix distinct indices i6=j. Consider the homomorphism (cf. [8, Ch. 1]) Φij :\ Cha1...,aNi −→ \ Chu, vi defined by Φij(ai) = u,Φij(aj) = v, and Φij(ak) = 0 for all k /∈ {i, j}. Such homomorphisms exist by the universal property of the free associative algebra. Then Φij(m(x)) = eux +evx/2 =: gu,v(x). 7 Hence if [m(x), m(−x)] ≡0, it follows that [gu,v(x), gu,v(−x)] ≡0 in the two-generator setting. 3.3 Coefficient extraction in the two-generator case Let gu,v(x) = (eux +evx)/2. Expanding [gu,v(x), gu,v(−x)] = euxe−vx +evxe−ux −e−uxevx −e−vxeux/4 from the series definition we obtain [xn]euxe−vx = n X k=0 uk(−v)n−k k! (n−k)! , and analogous formulas for the other three terms. Collecting them yields 1 4 n X k=0 1 k! (n−k)!uk(−v)n−k+vk(−u)n−k−(−u)kvn−k−(−v)kun−k. Using (−u)m= (−1)mumand (−v)m= (−1)mvm, the expression can be reorganized into commutators: [xn] [gu,v(x), gu,v(−x)] = 1 4 n X k=0 (−1)n−k k! (n−k)![uk, v n−k]+[vk, un−k]. This identity shows that every coefficient of the commutator [gu,v(x), gu,v(−x)] is itself a linear combination of higher commutators of monomials in uand v. In particular, the thirdorder coefficient reduces to [x3] [gu,v(x), gu,v(−x)] = [u, v2]+[v, u2]/4, and higher-order cases (e.g. n= 5,7) follow analogously. Since monomials in the free algebra are linearly independent, the vanishing of this coefficient forces [u, [u, v]] = 0,[v, [u, v]] = 0. Proposition 3.4 (Pairwise centrality).If [m(x), m(−x)] ≡0, then for every i6=jone necessarily has [ai,[ai, aj]] = [aj,[ai, aj]] = 0. That is, pairwise centrality is a necessary (but not sufficient) condition for the global commutation of m(x)and m(−x). 8 3.4 Global commutation constraints beyond pairwise centrality The pairwise centrality condition, together with the lowest-order balancing relation Pi,j [ai, aj](ai−aj)=0derived above, ensures the vanishing of all third-order coefficients in the expansion of [m(x), m(−x)]. However, as higher-order terms in the Baker–Campbell– Hausdorff (BCH) expansion involve quadratic and higher powers of the commutators [ai, aj], pairwise centrality alone does not guarantee global commutativity of m(x)and m(−x). In this subsection we extract the next-order obstructions and identify additional algebraic constraints that must hold if the entire commutator is to vanish. 3.4.1 Full 2-step nilpotency and sufficiency Higher-order BCH terms may survive unless all commutators [ai, aj]are central in the algebra generated by {a1, . . . , aN}. Assume henceforth that the system is fully 2-step nilpotent, i.e [ap,[aq, ar]] = 0 for all p, q, r. (22) Under this hypothesis, every exponential factor satisfies [7, Thm. 5.1] eaixe−ajx=e−ajxeaixe−1 2x2[ai,aj], and all factors e−1 2x2[ai,aj]commute with the remaining terms because [ai, aj]are central. Hence, in the averaged product m(x)m(−x), all nontrivial dependence on the noncommutativity of the ai’s enters only through the central elements {[ai, aj]}. A short calculation then yields [m(x), m(−x)] = 1 N2 N X i,j=1 e−x2[ai,aj]−1e−ajxeaix.(23) Since the factors e−x2[ai,aj]are central, the noncommutative contribution reduces to a central combination of commutators. Therefore, the expression vanishes iff the central elements satisfy the moment hierarchy (24). Proposition 3.5 (Exact equivalence under full two-step nilpotency).Assume that the generators {a1, . . . , aN}satisfy the full two-step nilpotent relations [ap,[aq, ar]] = 0 for all p, q, r, so that each commutator Cij := [ai, aj]is central. Define m(x) = 1 N N X i=1 eaix. Then the following statements are equivalent: (A) The averaged exponentials commute globally: [m(x), m(−x)] ≡0. 9 plays exactly the same role as the double-shuffle compatibility hierarchy in DMR theory. 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