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.

  • $\begingroup$ Have you checked whether you obtain in this way an idempotent in the group ring of the symmetric group ? If yes, then the proper context may be "Lie idempotents in the Descent algebras". Do you know the Malvenuto-Reutenauer Hopf algebra on permutations ? $\endgroup$ – F. C. Jun 27 '14 at 19:46
  • $\begingroup$ @F.C.: I hadn't thought to check, but yes, now that you suggest it, this does seem to represent an idempotent in the group ring. (I checked it up to degree 3 anyway.) Thanks for the insight and the keywords. $\endgroup$ – Jim Conant Jun 27 '14 at 23:24

$\newcommand{\id}{\operatorname{id}}$ 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:

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$\zeta$ is the "first Eulerian idempotent" defined by Patras [1]. Schmitt [2, Theorem 9.4] shows that it projects a graded cocommutative algebra onto its subspace of primitive elements.

[1] Patras, Frédéric. "La décomposition en poids des algèbres de Hopf." Annales de l'institut Fourier 43.4 (1993): 1067-1087. http://eudml.org/doc/75026

[2] William R. Schmitt, Incidence Hopf algebras, Journal of Pure and Applied Algebra, Volume 96, Issue 3, 21 October 1994, Pages 299-330, ISSN 0022-4049, http://dx.doi.org/10.1016/0022-4049(94)90105-8

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  • $\begingroup$ Great! Would you mind pointing me to the definition in Patras's paper? I skimmed through it, but I don't read French very well and I didn't immediately see it. $\endgroup$ – Jim Conant Jun 28 '14 at 1:52
  • $\begingroup$ I wasn't aware of the reference to Schmitt; the proof he gives seems very nice! @JimConant: Lemma 4.1 in Patras says $\operatorname{Prim} H = H^{\left(1\right)}$, but the definitions of §3 easily yield $H^{\left(1\right)} = e^1\left(H\right)$, and finally $e^1 = \frac{\left(\log \operatorname{id}\right)^{\star 1}}{1!} = \log \operatorname{id}$. $\endgroup$ – darij grinberg Jun 28 '14 at 1:56

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