[proofplan] We construct the cumulants by Möbius inversion on the finite lattice $NC(n)$. First, for each noncrossing partition, we define a blockwise moment functional by multiplying the moments of the ordered subwords determined by its blocks. We then invert the zeta relation on $NC(n)$ to obtain blockwise cumulant functionals, and the desired formula is the special case corresponding to the one-block partition. Uniqueness follows because the contribution of the one-block partition is the only term involving $\kappa_n$, while all other terms involve cumulants of lower order. [/proofplan] [step:Define blockwise moment functionals on noncrossing partitions] Fix $n\geq 1$. Let $P_n:=NC(n)$, ordered by refinement: for $\sigma,\pi\in P_n$, write $\sigma\leq\pi$ when every block of $\sigma$ is contained in a block of $\pi$. Let $0_n\in P_n$ denote the partition into singleton blocks and let $1_n\in P_n$ denote the partition with the single block $\{1,\dots,n\}$. For each $\pi\in P_n$, define the blockwise moment functional \begin{align*} \varphi_\pi:\mathcal A^n\to\mathbb C \end{align*} as follows. If $V=\{i_1<\cdots<i_r\}$ is a block of $\pi$, define \begin{align*} \varphi_V[a_1,\dots,a_n]=\varphi(a_{i_1}\cdots a_{i_r}). \end{align*} Then set \begin{align*} \varphi_\pi[a_1,\dots,a_n]=\prod_{V\in\pi}\varphi_V[a_1,\dots,a_n]. \end{align*} Since $\varphi$ is linear in one variable and multiplication in $\mathcal A$ is multilinear in each factor, each map $\varphi_\pi:\mathcal A^n\to\mathbb C$ is multilinear. [/step] [step:Invert the zeta relation on the finite lattice $NC(n)$] We recall the finite-poset inversion needed here and apply it to $P_n$. Let $\zeta_n$ denote the zeta function of $P_n$, defined by \begin{align*} \zeta_n(\sigma,\pi)=1 \end{align*} when $\sigma\leq\pi$, and $\zeta_n(\sigma,\pi)=0$ otherwise. Since $P_n$ is finite, the zeta function has a unique two-sided convolution inverse \begin{align*} \mu_n:P_n\times P_n\to\mathbb Z \end{align*} in the incidence algebra of $P_n$. Thus $\mu_n(\sigma,\pi)=0$ unless $\sigma\leq\pi$, and for all $\sigma,\pi\in P_n$ with $\sigma\leq\pi$ it satisfies both identities \begin{align*} \sum_{\sigma\leq\rho\leq\pi}\mu_n(\sigma,\rho)=\mathbb{1}_{\{\sigma=\pi\}} \end{align*} and \begin{align*} \sum_{\sigma\leq\rho\leq\pi}\mu_n(\rho,\pi)=\mathbb{1}_{\{\sigma=\pi\}}. \end{align*} This function is the Möbius function of $P_n$. [claim:Finite-poset Möbius inversion gives the inverse zeta relation] Let $P$ be a finite poset. Suppose $F:P\to E$ is a function into a complex [vector space](/page/Vector%20Space) $E$, and define $G:P\to E$ by \begin{align*} G(x)=\sum_{y\leq x}F(y). \end{align*} If $\mu:P\times P\to\mathbb Z$ is the two-sided Möbius function of $P$, meaning the convolution inverse of the zeta function of $P$, then \begin{align*} F(x)=\sum_{y\leq x}\mu(y,x)G(y) \end{align*} for every $x\in P$. [/claim] [proof] Fix $x\in P$. Expanding the definition of $G(y)$ inside the proposed inverse formula gives \begin{align*} \sum_{y\leq x}\mu(y,x)G(y)=\sum_{y\leq x}\mu(y,x)\sum_{z\leq y}F(z). \end{align*} Because $P$ is finite, the double sum may be rearranged by grouping terms according to $z\leq x$. Thus \begin{align*} \sum_{y\leq x}\mu(y,x)G(y)=\sum_{z\leq x}F(z)\sum_{z\leq y\leq x}\mu(y,x). \end{align*} The defining identity for the Möbius function gives \begin{align*} \sum_{z\leq y\leq x}\mu(y,x)=\mathbb{1}_{\{z=x\}}. \end{align*} Therefore all terms with $z\neq x$ vanish and the remaining term is $F(x)$. [/proof] [guided] The role of the Möbius function is to undo summation over all refinements. In our situation, the poset is $P_n=NC(n)$, and a relation of the form \begin{align*} \varphi_\pi=\sum_{\sigma\leq\pi}K_\sigma \end{align*} will express each blockwise moment functional as a sum of finer blockwise cumulant functionals. Since $P_n$ is finite, the inverse relation is purely algebraic and does not require any limiting argument. For a general finite poset $P$, define the zeta function by $\zeta(x,y)=1$ if $x\leq y$ and $\zeta(x,y)=0$ otherwise. The Möbius function $\mu$ is the two-sided inverse of $\zeta$ under convolution in the incidence algebra. In particular, it satisfies \begin{align*} \sum_{x\leq z\leq y}\mu(x,z)=\mathbb{1}_{\{x=y\}} \end{align*} and also \begin{align*} \sum_{x\leq z\leq y}\mu(z,y)=\mathbb{1}_{\{x=y\}}. \end{align*} The second identity is the one that cancels all lower-order summation terms in the displayed computation below. To see the cancellation directly, suppose $G(x)=\sum_{y\leq x}F(y)$. Then \begin{align*} \sum_{y\leq x}\mu(y,x)G(y)=\sum_{y\leq x}\mu(y,x)\sum_{z\leq y}F(z). \end{align*} Because the poset is finite, we may regroup the finite sum according to the lower index $z$. This gives \begin{align*} \sum_{y\leq x}\mu(y,x)G(y)=\sum_{z\leq x}F(z)\sum_{z\leq y\leq x}\mu(y,x). \end{align*} The inner sum is $1$ when $z=x$ and $0$ otherwise. Therefore only the term $F(x)$ survives. This is the exact algebraic mechanism that will isolate the cumulant functional from the moment functionals. [/guided] [/step] [step:Define cumulant functionals by Möbius inversion] For each $\pi\in P_n$, define the blockwise cumulant functional \begin{align*} K_\pi:\mathcal A^n\to\mathbb C \end{align*} by \begin{align*} K_\pi[a_1,\dots,a_n]=\sum_{\sigma\leq\pi}\varphi_\sigma[a_1,\dots,a_n]\mu_n(\sigma,\pi). \end{align*} Each $K_\pi$ is multilinear because it is a finite complex-linear combination of the multilinear functionals $\varphi_\sigma$. Applying the finite-poset inversion claim to the finite poset $P_n$, with $E$ equal to the vector space of multilinear maps $\mathcal A^n\to\mathbb C$, gives \begin{align*} \varphi_\pi[a_1,\dots,a_n]=\sum_{\sigma\leq\pi}K_\sigma[a_1,\dots,a_n] \end{align*} for every $\pi\in P_n$ and every $a_1,\dots,a_n\in\mathcal A$. [/step] [step:Recover the scalar cumulants from the one-block functionals] Define \begin{align*} \kappa_n:\mathcal A^n\to\mathbb C \end{align*} by \begin{align*} \kappa_n(a_1,\dots,a_n)=K_{1_n}[a_1,\dots,a_n]. \end{align*} This map is multilinear because $K_{1_n}$ is multilinear. We now identify $K_\pi$ with the multiplicative extension of the maps $\kappa_r$. Let $\pi\in P_n$. For a block $V=\{i_1<\cdots<i_r\}$ of $\pi$, let $\kappa_V[a_1,\dots,a_n]$ denote $\kappa_r(a_{i_1},\dots,a_{i_r})$. Define \begin{align*} \kappa_\pi[a_1,\dots,a_n]=\prod_{V\in\pi}\kappa_V[a_1,\dots,a_n]. \end{align*} The defining Möbius inversion is compatible with restriction to the blocks of $\pi$: the interval $[0_n,\pi]$ in $NC(n)$ decomposes as the product of the lattices $NC(V)$ over the blocks $V$ of $\pi$, where $NC(V)$ denotes the finite lattice of noncrossing partitions of the finite linearly ordered set $V$ with the order inherited from $\{1,\dots,n\}$. This decomposition holds because a noncrossing partition refined by $\pi$ is exactly a choice of a noncrossing partition inside each block. The zeta function and its inverse therefore factor over these blocks. Consequently, \begin{align*} K_\pi[a_1,\dots,a_n]=\kappa_\pi[a_1,\dots,a_n]. \end{align*} [/step] [step:Evaluate the inversion formula at the one-block partition] Take $\pi=1_n$ in the inversion relation \begin{align*} \varphi_\pi[a_1,\dots,a_n]=\sum_{\sigma\leq\pi}K_\sigma[a_1,\dots,a_n]. \end{align*} Since every $\sigma\in NC(n)$ satisfies $\sigma\leq 1_n$, and since $\varphi_{1_n}[a_1,\dots,a_n]=\varphi(a_1\cdots a_n)$, we obtain \begin{align*} \varphi(a_1\cdots a_n)=\sum_{\sigma\in NC(n)}K_\sigma[a_1,\dots,a_n]. \end{align*} Using the identification $K_\sigma=\kappa_\sigma$ from the previous step gives \begin{align*} \varphi(a_1\cdots a_n)=\sum_{\sigma\in NC(n)}\kappa_\sigma[a_1,\dots,a_n]. \end{align*} This is the desired moment-cumulant formula. [/step] [step:Prove uniqueness by isolating the one-block contribution] Suppose $(\lambda_n)_{n\geq 1}$ is another family of multilinear maps $\lambda_n:\mathcal A^n\to\mathbb C$ satisfying the same moment-cumulant formula, with multiplicative extensions $\lambda_\pi$. We prove $\lambda_n=\kappa_n$ for all $n\geq 1$ by induction on $n$. For $n=1$, the only noncrossing partition of $\{1\}$ is $1_1$, so the formula gives \begin{align*} \varphi(a_1)=\lambda_1(a_1)=\kappa_1(a_1) \end{align*} for every $a_1\in\mathcal A$. Assume now that $n\geq 2$ and that $\lambda_r=\kappa_r$ for all $1\leq r<n$. Fix $a_1,\dots,a_n\in\mathcal A$. In the moment-cumulant formula for order $n$, the term indexed by $1_n$ is exactly the $n$-linear functional $\lambda_n(a_1,\dots,a_n)$, while every partition $\pi\neq 1_n$ has all blocks of size strictly less than $n$. Hence every term $\lambda_\pi[a_1,\dots,a_n]$ with $\pi\neq 1_n$ agrees with $\kappa_\pi[a_1,\dots,a_n]$ by the induction hypothesis. Subtracting the common lower-block terms from the two moment-cumulant formulas yields \begin{align*} \lambda_n(a_1,\dots,a_n)=\kappa_n(a_1,\dots,a_n). \end{align*} Since $a_1,\dots,a_n$ were arbitrary, $\lambda_n=\kappa_n$. By induction, the family $(\kappa_n)_{n\geq 1}$ is unique. [/step]