Let $V$ be a finite dimensional vector space over a field of characteristic $0$, and let $T(V)$ be the tensor algebra (also called the free associative algebra) on $V$. This is actually a Hopf algebra, where the coproduct is defined on words by splitting them into two pieces in all possible ways: $$\Delta(v\_{I})=\sum\_{I=I'\amalg I''} v\_{I'}\otimes v\_{I''}.$$ Iterating the coproduct yields maps $\Delta^k:T(V)\to T(V)^{\otimes(k+1)}$ which break a word apart into $k+1$ pieces in all possible ways. Define the operator $\hat{\Delta}^k\colon T(V)\to T(V)^{\otimes (k+1)}$ which breaks a word apart into $k+1$ *nontrivial* pieces in all possible ways. Let $m^{k}\colon T(V)^{\otimes {k+1}}\to T(V)$ be the multiplication operator, and define $\zeta\colon T(V)\to T(V)$ by $$\zeta=\sum\_{i=0}^\infty \frac{(-1)^i}{i+1}m^i\hat{\Delta}^i.$$ Because the operator $\hat\Delta^{i}$ is locally nilpotent, this will be a finite sum when applied to any given element of $T(V)$ so there is no need to worry about convergence issues. Also note that $m^0\hat{\Delta}^0$ is by convention the number of ways to split a word into one nontrivial piece, which means that it is the identity on $T^+(V)$ and trivial on the base field sitting in degree $0$. Writing everything out in terms of basic definitions, we have $$\zeta(v\_I)=v\_I-1/2\sum v\_{I\_1}v\_{I\_2}+1/3\sum v\_{I\_1}v\_{I\_2}v\_{I\_3}+\cdots, $$
where each sum is over all ways to split $I$ into nontrivial pieces $I\_1,\ldots, I\_k$. Low degree calculations show that 


* $\zeta(v)=v$
* $\zeta(v\_1v\_2)=\frac{1}{2}[v\_1,v\_2]$
* $\zeta(v\_1v\_2v\_3)=\frac{1}{3}[v\_1,[v\_2,v\_3]]-\frac{1}{6}[v\_2,[v\_1,v\_3]]$


In particular, it looks like $\zeta$ lands in the space of iterated commutators, which in this case is the same as the primitive elements in the Hopf algebra. So this is my question: 


*How can one show that the image of $\zeta$ lies in the subspace of primitive elements?*


Conceivably one could either show that $\Delta(\zeta(v\_I))=1\otimes \zeta(v\_I)+\zeta(v\_I)\otimes 1$, or directly show that $\zeta(v\_I)$ is an iterated commutator. 



An identity in the free associative algebra
$\newcommand{\id}{\operatorname{id}}$ Just adding in a couple steps missing in the answer by "an eulerian idempotent":


Your map $\zeta$ can be rewritten as the logarithm of the map $\id : T(V) \to T(V)$ in the *convolution* algebra $\left(\operatorname{Hom}\left(T(V), T(V)\right), \star\right)$. (In fact, if we let $u$ denote the unit map and $\epsilon$ the counit map of $T(V)$, then every $i$ satisfies $\widehat{\Delta}^i = \left(\id - u\epsilon\right)^{\otimes \left(i+1\right)} \circ \Delta^i$, so that $m^i \circ \widehat{\Delta}^i = m^i \circ \left(\id - u\epsilon\right)^{\otimes \left(i+1\right)} \circ \Delta^i = \left(\id - u \epsilon\right)^{\star \left(i+1\right)}$, and thus your formula defining $\zeta$ reveals itself as the Mercator series for the logarithm of $\id$ in the convolution algebra.) So yes, this is the (first) Eulerian idempotent. To add two more references to the fact that it projects $T(V)$ onto its primitives:


* [Gérard Henry Edmond Duchamp, Vincel Hoang Ngoc Minh, Christophe Tollu, Bùi Chiên, Nguyen Hoang Nghia, *Combinatorics of $\varphi$-deformed stuffle Hopf algebras*, arXiv:1302.5391v7](http://arxiv.org/abs/1302.5391), page 12 ("We first prove that $\pi\_{1, \mathcal A}$ is a projector $\mathcal A \to \operatorname{Prim}\left(\mathcal B\right)$"; in your case it is obvious that $\mathcal B = \mathcal A$). This gives a self-contained proof
* [Victor Reiner and your humble servant, *Hopf algebras in combinatorics*, June 5, 2014](http://web.mit.edu/~darij/www/algebra/HopfComb-sols.pdf), Exercise 5.35 (the numbering is subject to change, so search for "e is a projection from A").


