In the comments, Tom Copeland asked if, six years later, I had any further insights into the diagrammatics I proposed near the end of my question. So I figured I'd mention what I know, which is yet another (marginally) diagrammatic description of BCH multiplication and in particular the map mathcalUmathfrakgtomathcalSmathfrakg. What I'll describe is some part of http://dx.doi.org/10.1090/pspum/088 (also available at http://arxiv.org/abs/1307.5812). Some pictures are available in the last chapter of my thesis, but note that that chapter has a subtle error (the details of which I haven't yet sussed out) which I have corrected in the linked paper; the error does not affect the case of (duals of) Lie algebras, but does affect Poisson structures whose Taylor expansions include quadratic or higher terms.
I will describe some homological algebra, and then unpack into diagrams. Let mathfrakG=mathfrakgotimesOmegamathrmcpt(mathbbR)[1] denote the cochain complex of mathfrakg-valued compactly supported de Rham forms on mathbbR, shifted so that its cohomology is concentrated in degree 0. Then mathrmH0(mathfrakG)=mathfrakg, and the projection to homology is given by integrating de Rham forms — there is a cohomological-degree-0 map int:mathfrakGtomathfrakg. Note also that Omegamathrmcpt(mathbbR) has a non-unital graded-commutative multiplication (wedge), through which mathfrakG picks up a graded-symmetric (!) cohomological-degree-(+1) map delta:mathfrakGotimesmathfrakGtomathfrakG by delta(xotimesalpha,yotimesbeta)=[x,y]otimes(alphawedgebeta) (up to a sign that depends on a choice of conventions about how to handle elements of shifted complexes). Consider the (graded-commutative) symmetric algebra mathcalSmathfrakG. I will denote its usual differential (which is the extension of de Rham-form differentiation as a derivation) by partialmathrmdR.
Integration of de Rham forms extends to an algebra homomorphism int:(mathcalSmathfrakG,partialmathrmdR)tomathcalSmathfrakg.
The map delta has a canonical extension to a second-order differential operator on mathcalSmathfrakG which I will also call delta. (A second order differential operator on a symmetric algebra is unique determined by its values on constant, linear, and quadratic terms. We declare that delta vanishes on constants and linears, and set it to be the original delta on quadratics.) Because wedge plays well with partialmathrmdR, delta and partialmathrmdR graded-commute. The Jacobi identity implies delta2=0. So we have a new differential partialq=partialmathrmdR+delta on mathcalSmathfrakG.
Pick any alphainOmega1mathrmcpt(mathbbR) with intalpha=1. (Henceforth I will call such one-forms bumps.) The map xmapstoxotimesalpha from mathfrakgtomathfrakG extends to an algebra homomorphism mathcalSmathfrakgtomathcalSmathfrakG splitting int. There is a unique contracting homotopy etaalpha:Omegamathrmcpt(mathbbR)toOmegamathrmcpt(mathbbR) (of cohomological degree −1) such that the graded commutator [partialmathrmdR,etaalpha]=mathrmid−(alphaotimes)circint; it vanishes on Omega0mathrmcpt and on Omega1mathrmcpt satisfies etaalpha(beta)=partial−1mathrmdR(beta−(intbeta)alpha); note that int(beta−(intbeta)alpha)=0, so the one-form beta−(intbeta)alpha has a unique antiderivative among compactly-supported functions.
We can extend etaalpha to mathcalSmathfrakG in many ways, and the choice can be proven not to matter. To make a choice, declare that on the constants mathcalS0mathfrakG we have etaalpha=0, and on mathcalSnmathfrakG we have
etaalpha(beta1odotdotsodotbetan)=frac1nsumibeta1odotdotsodotetaalpha(betai)odotdotsodotbetan,
where
odot denotes the symmetric multiplication in
mathcalS. Note that this is not the extension of
etaalpha as a derivation.
Since etaalpha always attaches 1-forms and delta involves wedge multiplication, etaalpha:mathcalSmathfrakgto(mathcalSmathfrakG,partialq) is a chain map. The integration map int:mathcalSmathfrakGtomathcalSmathfrakg splitting this map is not a chain map from (mathcalSmathfrakG,partialq). But with the above choices we can choose a different splitting, namely intcirc(mathrmid−deltaetaalpha)−1. (I have a 50% chance of getting that minus sign wrong.) I will leave checking that this is a chain map splitting etaalpha to you. Note that (mathrmid−deltaetaalpha)−1=sumN(deltaetaalpha)N converges on mathcalSmathfrakG, since deltaetaalpha drops polynomial degree by 1.
Pick bumps alpha1,dots,alphan such that the support of alphai is in [i−1,i], and pick one final bump alpha arbitrarily. One can prove that the map mathcalUmathfrakgtomathcalSmathfrakg is given on monomials x1dotsxn (with multiplication in mathcalU) by
mathcalUmathfrakgnix1dotsxnmapstointcirc(mathrmid−deltaetaalpha)−1bigl((alpha1otimesx1)odotdotsodot(alphanotimesxn)bigr)inmathcalSmathfrakg
(or I might be off by a sign somewhere). In general, similar formulas describe the entire product on
mathcalSmathfrakg given by transporting the one from
mathcalUmathfrakg along the symmetrization isomorphism.
Let me now unpack this formula, or rather give the answer after some unpacking. (Proving that this is a valid unpacking is straightforward: you need to track the numerical factors coming from etaalpha, understand how to apply a second-order differential operator to a monomial, and also include a brief "degree reasons" argument to get etaalpha and delta to apply always to the same things at the same time.)
Define an n-leaf binary heighted forest, abbreviated forest, to be set of binary rooted trees whose leaves are put in bijection with the set 1,dots,n and whose nodes are totally ordered (I mean: totally order the set of all nodes) such that in a given tree, and path from root to leaf is increasing for the total ordering. Arbitrarily choose for each node which of its two branches is left and which right (the choice will cancel out).
Given a forest and the list x1,dots,xn of elements in mathfrakg, there is an obvious element of mathcalSmathfrakg given by putting the xis at the leaves and reading the forest as instructions of who to bracket with whom (then multiply the "root" outputs).
Now I will describe, for each forest, how to compute a number. Consider the map Omega1mathrmcpt(mathbbR)otimesOmega1mathrmcpt(mathbbR)toOmega1mathrmcpt(mathbbR) given by beta1otimesbeta2mapstobeta1wedgeetaalpha(beta2)−beta2wedgeetaalpha(beta1). Place this map at each vertex, and alphai at the ith leaf, and let the forest tell you how to apply this map to end up with a bunch of 1-forms at the roots. Then integrate all these 1-forms to get numbers, and multiply those numbers together.
Finally, suppose there are k roots (and hence (n−k) nodes). Then multiply by frac1nfrac1n−1dotsfrac1n−k.
Note that neither the number nor the element of mathcalSmathfrakg determined by a forest depends on the height ordering. But I now want you to sum over all forests with total node-ordering of the product of these two numbers. That sum computes the map mathcalUmathfrakgtomathcalSmathfrakg above. (If I had a good way to count the number of total orderings of the nodes for a given un-heighted forest, I would have used it.)
This is similar to, but not the same as, Kontsevich's star product. In particular, note that my forests have no wheels, whereas Kontsevich does not describe the symmetrization map mathcalSmathfrakgcongmathcalUmathfrakg, but rather this map twisted by some traces in the adjoint representation.