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Let $x$ be a formal (or small, since the function is analytic) variable, and consider the power series $$ A(x) = \frac{x}{1 - e^{-x}} = \sum_{m=0}^\infty \left( -\sum_{n=1}^\infty \frac{(-x)^n}{(n+1)!} \right)^m = 1 + \frac12 x + \frac1{12}x^2 + 0x^3 - \frac1{720}x^4 + \dots $$ where I might have made an arithmetic error in expanding it out.

  1. Are all the coefficients egyptian, in the sense that they are given by $A^{(n)}(0)/n! = 1/N$ for $N$ an integer? The answer is no, unless I made an error, e.g. the third coefficient. But maybe every non-zero coefficient is egyptian?

  2. If all the coefficients were positive eqyptian, then the sequence of denominators might count something — one hopes that the $n$th element of any sequence of nonnegative integers counts the number of ways of putting some type of structure on an $n$-element set.

Of course, generating functions really come in two types: ordinary and exponential. The difference is whether you think of the coefficients as $\sum a_n x^n$ or as $\sum A^{(n)} x^n/n!$. If it makes more sense as an exponential generating function, that's cool too.

So my question really is: is there a way of computing the $n$th coefficient of $A(x)$, or equivalently of computing $A^{(n)}(0)/n!$, without expanding products of power series the long way?

Where you might have seen this series

Let $\xi,\psi$ be non-commuting variables over a field of characteristic $0$, and let $B(\xi,\psi) = \log(\exp \xi \exp \psi)$ be the Baker-Campbell-Hausdorff series. Fixing $\xi$ and thinking of this as a power series in $\psi$, it is given by $$B(\xi,\psi) = \xi + A(\text{ad }\xi)(\psi) + O(\psi^2)$$ where $A$ is the series above, and $\text{ad }\xi$ is the linear operator given by the commutator: $(\text{ad }\xi)(\psi) = [\xi,\psi] = \xi\psi - \psi\xi$.

More generally, $B$ can be written entirely in terms of the commutator, and so makes sense as a $\mathfrak g$-valued power series on $\mathfrak g$ for any Lie algebra $\mathfrak g$. It converges in a neighborhood of $0$ when $\mathfrak g$ is finite-dimensional over $\mathbb R$, in which case $\mathfrak g$ is a (generally noncommutative) "partial group".

(More generally, you can consider the "formal group" of $\mathfrak g$. Namely, take the commutative ring $\mathcal P(\mathfrak g)$ of formal power series on $\mathfrak g$; then $B$ defines a non-cocommutative comultiplication, making $\mathcal P = \mathcal P(\mathfrak g)$ into a Hopf algebra. Or rather, $B(\mathcal P)$ does not land in the algebraic tensor product $\mathcal P \otimes \mathcal P$. Instead, $\mathcal P$ is cofiltered, in the sense that it is a limit $\dots \to \mathcal P_2 \to \mathcal P_1 \to \mathcal P_0 = 0$, where (over characteristic 0, anyway) $\mathcal P_n = \text{Poly}(\mathfrak g)/(\mathfrak g \text{Poly}(\mathfrak g))^n$, where $\text{Poly}(\mathfrak g)$ is the ring of polynomial functions on $\mathfrak g$, and $\mathfrak g \text{Poly}(\mathfrak g)$ is the ideal of functions vanishing at $0$. Then $B$ lands in the cofiltered tensor product, which is just what it sounds like. (In arbitrary characteristic, $\mathcal P$ is the cofiltered dual of the filtered Hopf algebra $\mathcal S \mathfrak g$, the symmetric algebra of $\mathfrak g$, filtered by degree.))

Why I care

When $\mathfrak g$ is finite-dimensional over $\mathbb R$, and $U$ is the open neighborhood of $0$ in which $B$ converges, then $\mathfrak g$ acts as left-invariant derivations on $U$, where by left-invariant I mean under the multiplication $B$. Hence there is a canonical identification of the universal enveloping algebra $\mathcal U\mathfrak g$ with the algebra of left-invariant differential operators on $U$. Since $\mathfrak g$ is in particular a vector space, the "symbol" map gives a canonical identification between the algebra of differential operators on $U$ and the algebra of functions on the cotangent bundle $T^{\ast} U$ that are polynomial (of uniformly bounded degree) in the cotangent directions. Left-invariance then means that the operators are uniquely determined by their restrictions to the fiber $T^{\ast}_0\mathfrak g = \mathfrak g^{\ast}$, and the space of polynomials on $\mathfrak g^{\ast}$ is canonically the symmetric algebra $\mathcal S \mathfrak g$. This gives a canonical PBW map $\mathcal U \mathfrak g \to \mathcal S \mathfrak g$, a fact I learned from J. Baez and J. Dolan.

(In the formal group language, the noncocommutative cofiltered Hopf algebra $\mathcal P(\mathfrak g)$ is precisely the cofiltered dual to the filtered algebra $\mathcal U\mathfrak g$, whereas with its cocommutative Hopf structure $\mathcal P(\mathfrak g)$ is dual to $\mathcal S \mathfrak g$. But as algebras these are the same, and unpacking the dualizations gives the PBW map $\mathcal U\mathfrak g \cong \mathcal S \mathfrak g$, and explains why it is actually an isomorphism of coalgebras.)

Anyway, in one direction, the isomorphism $\mathcal U\mathfrak g \cong \mathcal S \mathfrak g$ is easy. Namely, the map $\mathcal S \mathfrak g \to \mathcal U \mathfrak g$ is given on monomials by the "symmetrization map" $\xi_1\cdots \xi_n \mapsto \frac1{n!} \sum_{\sigma \in S_n} \prod_{k=1}^n \xi_{\sigma(k)}$, where $S_n$ is the symmetric group on $n$ letters, and the product is ordered. (In this direction, the isomorphism of coalgebras is obvious. In fact, the corresponding symmetrization map into the full tensor algebra is a coalgebra homomorphism.)

In the reverse direction, I can explain the map $\mathcal U \mathfrak g \to \mathcal S \mathfrak g$ as follows. On a monomial $\xi_1\cdots \xi_n$, it acts as follows. Draw $n$ dots on a line, and label them $\xi_1,\dots,\xi_n$. Draw arrows between the dots so that each arrow goes to the right (from a lower index to a higher index), and each dot has either 0 or 1 arrow out of it. At each dot, totally order the incoming arrows. Then for each such diagram, evaluate it as follows. What you want to do is collapse each arrow $\psi\to \phi$ into a dot labeled by $[\psi,\phi]$ at the spot that was $\phi$, but never collapse $\psi\to \phi$ unless $\psi$ has no incoming arrows, and if $\phi$ has multiple incoming arrows, collapse them following your chosen total ordering. So at the end of the day, you'll have some dots with no arrows left, each labeled by an element of $\mathfrak g$; multiply these elements together in $\mathcal S\mathfrak g$. Also, multiply each such element by a numerical coefficient as follows: for each dot in your original diagram, let $m$ be the number of incoming arrows, and multiply the final product by the $m$th coefficient of the power series $A(x)$. Sum over all diagrams.

Anyway, the previous paragraph is all well and cool, but it would be better if the numerical coefficient could be read more directly off the diagram somehow, without having to really think about the function $A(x)$.

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I am sort of astonished that you gave so much background without mentioning the name of this sequence: – Qiaochu Yuan Dec 18 '09 at 1:57
@Qiaochu: See, I'm neither a combinatorialist nor a number theorist, and although I guess I've seen the Bernoulli numbers before, I never really encoded them in memory. Anyway, I've accepted Pete's answer below, but I'm secretly hoping that someone will connect it with the diagrams I described. – Theo Johnson-Freyd Dec 18 '09 at 2:55
@Theo: I didn't actually remember these were the Bernoulli numbers until I did the expansion (by computer, of course) and saw the mysterious numerator 691. – Michael Lugo Dec 18 '09 at 3:27
Given your background you might be interested to know that this power series is used to define the Todd class: – Steve Huntsman Dec 18 '09 at 6:22
Another place to see this series, though shifted by two: the Planck black-body distribution.'s_law – Allen Knutson Feb 26 '10 at 4:51

15 Answers 15

up vote 26 down vote accepted

Two people have pointed it out already, but somehow I can't resist: your formal power series is precisely the defining power series of the Bernoulli numbers:

Accordingly, they are far from Egyptian: as came up recently in response to the question

When does the zeta function take on integer values?

the odd-numbered terms (except the first) are all zero, whereas the even-numbered terms alternate in sign and grow rapidly in absolute value, so only finitely many are reciprocals of integers.

I find it curious that you are looking at this sequence from such a sophisticated perspective and didn't know its classical roots. I feel like there should be a lesson here, but I don't know exactly what it is. Here's a possibility: every young mathematician should learn some elementary number theory regardless of their primary interests. Comments?

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You win. I know a lot about Lie algebras, and I've never studied any number theory. I think I've seen the Bernoulli numbers once or twice, but never really encoded them. – Theo Johnson-Freyd Dec 18 '09 at 2:50
Bernoulli numbers are fairly ubiquitous. They come up, for example, in very basic real analysis; namely in Euler-Maclaurin summation formula. So I am not sure if these numbers should be thought of as pertaining to number theory. – Idoneal Jan 3 '10 at 4:19
Well, certainly not only to number theory, anyway. – Pete L. Clark Jan 3 '10 at 4:29
I'd certainly agree with your last suggestion (and in particular wish I knew more about number theory than I do). Next, take a roomful of mathematicians, get all their suggestions for fields that should be added to "elementary number theory" here. What would you guess is the probability that any one of the mathematicians in the room has any real knowledge of all the fields that have been named? – Mark Meckes Jun 3 '10 at 17:02
I would add that it's nearly impossible to learn the BCH formula (with the proof, of course) and $\mathit{not}$ to see Bernoulli numbers mentioned. There is another lesson here, possibly that one needs to read textbooks systematically rather than just pick up bits and pieces. – Victor Protsak Jul 15 '10 at 18:11

Here's another way to get at the answer. You think you have a sequence of rationals that may be familiar:

1, 1/2,1/12,0,-1/720,...

The denominators seem more interesting than the numerator, so maybe the "right" sequence is:


You go to Sloane's Encyclopedia and enter the sequence, to no avail. You could now try superseeker, which looks at many transformations of the sequence, but for this few terms that will return too many hits. Let's try the one transformation you mentioned, and look at the exponential generating function, whose coefficients have denominators:

1, 2, 6, 1, 30, ...

Sloane's immediately identifies that sequence as the denominators of Bernoulli numbers, giving not only the generating function you started with but many other interesting factoids and references.

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I think Sloane's should always be consulted when faced with an unknown sequence of integers or of numbers from which integers may be reasonably extracted. – Omar Antolín-Camarena Mar 9 '10 at 12:40

I am adding the following remark because it may be of some interest to the number theorists who recognized the Bernoulli numbers to know that the relationship with Lie theory explained in the question has number-theoretic substance: namely, in his article on the thrice-punctured sphere, Deligne uses the Lie algebra point of view on Bernoulli numbers described in the question (together with other ingredients, of course, and applied to a specific Lie algebra) to derive Euler's formula for the values of $\zeta(2n)$.

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The Bernoulli numbers are closely related to alternating permutations. That is to say, permutations like $1524376$ where the numbers alternately go up and down. Specifically, if $A_n$ is the number of such permutations of an $n$ element set, then $$B_{2n} = (-1)^{n-1} \frac{2n}{4^{2n}-2^{2n}} A_{2n-1}.$$ It's possible you could somehow relate your sums over diagrams to alternating permutations.

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My favourite introduction to the Bernoulli numbers is section 3 of Pierre Cartier's paper Mathemagics. I quote:

I claim that they are defined by the equation $(B + 1)^n = B^n$ for $n \geq 2$, together with the initial condition $B^0 = 1$. The meaning is the following: expand $(B + 1)^n$ by the binomial theorem, then replace the power $B^k$ by $B_k$.

There's a whole lot of great stuff in this paper, besides Bernoulli numbers.

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Ah, yes, I have seen that definition. What I've never done is calculated out more than the first two or so terms, and 1, 1/2, 1/12 is meaningless, and when today I got 1, 1/2, 1/12, 0, -1/720, I still didn't have anything with which to recognize it. – Theo Johnson-Freyd Dec 18 '09 at 5:48
You might already know this, but that 1/12 is part of the "reason" for the appearances of 12 and 24 in mathematics, as described by John Baez here: – Qiaochu Yuan Dec 18 '09 at 8:43
John Baez and I had a discussion about giving a species interpretation of X/(1 - e^X) here… – David Corfield Dec 18 '09 at 8:59

This is an old question, but since no one seems to have given this answer, I thought I'd add it to the list, since it's where I first ran across Bernoulli numbers. Everyone (in mathematics) runs across the formulas $$ 1+2+\cdots+n=\frac{n(n+1)}{2}=\frac12n^2+\frac12n $$ and $$ 1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}=\frac13n^3+\frac12n^2+\frac16n. $$ One is naturally led to ask if $$ 1^k+2^k+\cdots+n^k $$ is also a polynomial in $n$. For a number theorist, there is no justification needed for looking at this question, but I'd suggest that this sort of sums of powers appears ubiquitously in mathematics, since a fundamental tool in studying just about any sequence is to look at its $k$'th moments. So now we're faced with the question of determining polynomials $P_k(T)$ satisfying $$ 1^k+2^k+\cdots+n^k = P_k(n). $$ Many areas of mathematics teach us that when confronted with a sequence (in this case, of polynomials), we should amalgamate them into a power series and study them simultaneously. In this case, a little experimentation shows that an exponential power series works nicely, so we look at $$ F(X,T) = \sum_{k=0}^\infty \frac{P_k(T)}{k!}X^k. $$ Then $$ F(X,n) = \sum_{k=0}^\infty \frac{P_k(n)}{k!}X^k = \sum_{k=0}^\infty \sum_{j=1}^{n} \frac{(jX)^k}{k!} = \sum_{j=1}^{n} e^{jX} = \frac{1-e^{(n+1)X}}{1-e^X}-1 = \frac{1-e^{(n+1)X}}{X}\cdot\frac{X}{1-e^X}-1. $$ The Taylor expansion of $\frac{1-e^{(n+1)X}}{X}$ is easy, so the interesting quantities that appear in the formula for $P_k(T)$ are the coefficients of your power series $\frac{X}{1-e^X}$. More generally, one defines Bernoulli polynomials whose coefficients are Bernoulli numbers and so that $P_k(T)$ is more-or-less a Bernoulli polynomial. Anyway, the moral is that the power series expansion of $\frac{X}{1-e^X}$ appears naturally even in the very elementary problem of finding a formula for a finite sum of $k$'th powers.

Just noticed that there's a closely related question Why do Bernoulli numbers arise everywhere? in which several people note that Bernoulli numbers occur automatically when looking at sums of powers. But I'll leave this answer, since it provides more details.

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To have a geometric interpretation of this generating function in Lie theory you do not need to work over reals, in fact any commutative ground ring containing rationals suffices. For a version of such interpretation utilizing functors representing a version of "formal schemes" see chapters 7-10 (and introduction) to our paper

N. Durov, S. Meljanac, A. Samsarov, Z. Škoda, A universal formula for representing Lie algebra generators as formal power series with coefficients in the Weyl algebra, Journal of Algebra 309, Issue 1, pp.318-359 (2007), math.RT/0604096

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This isn't an answer, but I saw that you said a couple of times that you were doing the expansion by hand. I'll just point out that you can get a free sage notebook account, and then do

f(x) = x/(1 - e^(-x))


in a new worksheet to get the expansion to the $x^{10}$ term. Wolfram Alpha probably has something similar. Of course, one sometimes learns something by doing things manually, but it is often useful to have an easy way to check your answer.

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You will find what you describe in the first reference. This describes how the Bernouilli numbers arise when studying the universal enveloping algebra. I have seen unpublished notes on this from a talk by Kostant in the '70s. This is a strong form of the PBW theorem and is closer to Poincare's result. This is discussed in the second reference. This is an early version of universal quantisation.

MR2301242 (2008d:17015) Durov, Nikolai ; Meljanac, Stjepan ; Samsarov, Andjelo ; Škoda, Zoran . A universal formula for representing Lie algebra generators as formal power series with coefficients in the Weyl algebra. J. Algebra 309 (2007), no. 1, 318--359.

MR1793103 (2001f:01039) Ton-That, Tuong ; Tran, Thai-Duong . Poincaré's proof of the so-called Birkhoff-Witt theorem. Rev. Histoire Math. 5 (1999), no. 2, 249--284 (2000).

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For those who don"t bother to read other answers and comments closely, (so this will probably itself be overlooked by the intended audience, lol), the first reference had been presented by one of the authors (Skoda) in his answer here. – Tom Copeland Sep 27 at 22:44
+1 for the second ref though. – Tom Copeland Sep 28 at 17:54

If you expand this out a bit further, you get

$ 1 + {1 \over 2} x + {1 \over 12} x^2 - {1 \over 720} x^4 + {1 \over 30240} x^6 - {1 \over 1209600} x^8 + {1 \over 47900160} x^{10} - {691 \over 1307674368000} x^{12} + \cdots $

Notice that the nonzero coefficients are alternating in sign.

In fact it turns out that the sequence you call $A^{(n)}$ are exactly the Bernoulli numbers.

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Nice. I was surprised when the first 0 showed up, since I've been expanding by hand. – Theo Johnson-Freyd Dec 18 '09 at 2:52

You can generate the Bernoullis through the combinatorics of permutahedra and graphical interpretations of surjections presented in OEIS-A133314 (cf. A049019, A019538, A008292) weighted by the reciprocal integers. (See also MOQ-61252.)

This is equivalent to determining the reciprocal of the exponential generating function

$$\frac{e^t-1}{t}=1+\frac{1}{2}t+ \frac{1}{3}\frac{t^2}{2!}+ \frac{1}{4}\frac{t^3}{3!}+\cdots\;\;.$$

Naturally, it's an involution, so you can go in the reverse direction from the Bernoullis to the reciprocal integers by the same surjections weighted by the Bernoulli numbers.

Alternatively, consider the e.g.f. $ \displaystyle \frac{2}{1+e^{2x}}$ for $ \displaystyle a(n-1)= \frac{2^n-4^n}{n} B_n$ , which without the zeros and signs are the tangent or zag numbers A000182 (a different normalization gives A002105). The reciprocal e.g.f. encodes $ \displaystyle b(n) = 1,2^0,2^1,2^2, ... $, which can be used as signed weights of surjection mappings to reconstitute the $a(n)$. These mappings are in turn encoded in the signed, refined face (partition) polynomials of the permutahedra of A133314.

But, with the o.g.f., using the normalized Bernoulli numbers, compositional inversion (A134264), rather than reciprocation enters the picture and, therefore, weighted noncrossing partitions and Dyck lattice paths (and myriad other related combinatoric structures).

These number arrays can be related to volumes of structures, as well as the Bernoullis (see Noam Elkies).

Also see these papers relating the Bernoullis to permutations, decomposition of hypercube volumes, and quantum algebras (related to A119467 and A119879):

Hodges and Sukumar, Sukumar and Hodges, Hetyei.

(Edit 10/7/2015)

What I find most interesting about Theo's question and follow-up answer are his attempts to connect the reciprocal integers and the Bernoulli numbers through topology/geometry since the two sequences are intimately paired through the formalism of Appell Sheffer sequences, two pairs with Lie and quantum algebras written all over them, and to the algebraic topology/geometry of exotic spheres through the Kervaire-Milnor formula related to the order of homotopy groups and multiplicative (elliptic) genera.

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Connections to Witten"s work on quantum guage theories in 2-D and the Kervaire-Milnor formula for homotopy groups of hyperspheres are noted at my website. Removed from a second newer anawer, – Tom Copeland Oct 8 at 3:34

Here you can find all sorts of information, different representations and connections concerning the Bernoulli numbers:

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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 $\mathcal U \mathfrak g \to \mathcal S \mathfrak g$. What I'll describe is some part of (also available at 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 $\mathfrak G = \mathfrak g \otimes \Omega_{\mathrm{cpt}}(\mathbb R)[1]$ denote the cochain complex of $\mathfrak g$-valued compactly supported de Rham forms on $\mathbb R$, shifted so that its cohomology is concentrated in degree $0$. Then $\mathrm H^0(\mathfrak G) = \mathfrak g$, and the projection to homology is given by integrating de Rham forms — there is a cohomological-degree-$0$ map $\int : \mathfrak G \to \mathfrak g$. Note also that $\Omega_{\mathrm{cpt}}(\mathbb R)$ has a non-unital graded-commutative multiplication ($\wedge$), through which $\mathfrak G$ picks up a graded-symmetric (!) cohomological-degree-$(+1)$ map $\delta : \mathfrak G \otimes \mathfrak G \to \mathfrak G$ by $\delta(x\otimes \alpha,y\otimes \beta) = [x,y] \otimes (\alpha \wedge \beta)$ (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 $\mathcal S \mathfrak G$. I will denote its usual differential (which is the extension of de Rham-form differentiation as a derivation) by $\partial_{\mathrm{dR}}$. Integration of de Rham forms extends to an algebra homomorphism $\int : (\mathcal S\mathfrak G, \partial_{\mathrm{dR}}) \to \mathcal S \mathfrak g$.

The map $\delta$ has a canonical extension to a second-order differential operator on $\mathcal S \mathfrak G$ 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 $\partial_{\mathrm{dR}}$, $\delta$ and $\partial_{\mathrm{dR}}$ graded-commute. The Jacobi identity implies $\delta^2 = 0$. So we have a new differential $\partial_q = \partial_{\mathrm{dR}} + \delta$ on $\mathcal S \mathfrak G$.

Pick any $\alpha \in \Omega_{\mathrm{cpt}}^1(\mathbb R)$ with $\int\alpha = 1$. (Henceforth I will call such one-forms bumps.) The map $x \mapsto x \otimes \alpha$ from $\mathfrak g \to \mathfrak G$ extends to an algebra homomorphism $\mathcal S \mathfrak g \to \mathcal S \mathfrak G$ splitting $\int$. There is a unique contracting homotopy $\eta_\alpha : \Omega_{\mathrm{cpt}}(\mathbb R) \to \Omega_{\mathrm{cpt}}(\mathbb R)$ (of cohomological degree $-1$) such that the graded commutator $[\partial_{\mathrm{dR}},\eta_{\alpha}] = \mathrm{id} - (\alpha \otimes) \circ \int$; it vanishes on $\Omega_{\mathrm{cpt}}^0$ and on $\Omega_{\mathrm{cpt}}^1$ satisfies $\eta_\alpha(\beta) = \partial_{\mathrm{dR}}^{-1}(\beta - (\int \beta)\alpha)$; note that $\int(\beta - (\int \beta)\alpha) = 0$, so the one-form $\beta - (\int \beta)\alpha$ has a unique antiderivative among compactly-supported functions.

We can extend $\eta_\alpha$ to $\mathcal S\mathfrak G$ in many ways, and the choice can be proven not to matter. To make a choice, declare that on the constants $\mathcal S^0\mathfrak G$ we have $\eta_\alpha = 0$, and on $\mathcal S^n\mathfrak G$ we have $$ \eta_\alpha(\beta_1 \odot \dots \odot \beta_n) = \frac1n \sum_i \beta_1 \odot \dots \odot \eta_\alpha(\beta_i) \odot \dots \odot \beta_n, $$ where $\odot$ denotes the symmetric multiplication in $\mathcal S$. Note that this is not the extension of $\eta_\alpha$ as a derivation.

Since $\eta_\alpha$ always attaches 1-forms and $\delta$ involves wedge multiplication, $\eta_\alpha : \mathcal S \mathfrak g \to (\mathcal S \mathfrak G,\partial_q)$ is a chain map. The integration map $\int : \mathcal S \mathfrak G \to \mathcal S \mathfrak g$ splitting this map is not a chain map from $(\mathcal S \mathfrak G,\partial_q)$. But with the above choices we can choose a different splitting, namely $\int \circ (\mathrm{id} - \delta \eta_\alpha)^{-1}$. (I have a 50% chance of getting that minus sign wrong.) I will leave checking that this is a chain map splitting $\eta_\alpha$ to you. Note that $(\mathrm{id} - \delta \eta_\alpha)^{-1} = \sum_N (\delta \eta_\alpha)^N$ converges on $\mathcal S \mathfrak G$, since $\delta \eta_\alpha$ drops polynomial degree by $1$.

Pick bumps $\alpha_1, \dots,\alpha_n$ such that the support of $\alpha_i$ is in $[i-1,i]$, and pick one final bump $\alpha$ arbitrarily. One can prove that the map $\mathcal U \mathfrak g \to \mathcal S \mathfrak g$ is given on monomials $x_1 \dots x_n$ (with multiplication in $\mathcal U$) by $$ \mathcal U \mathfrak g \ni x_1 \dots x_n \mapsto \int \circ (\mathrm{id} - \delta \eta_\alpha)^{-1} \bigl( (\alpha_1 \otimes x_1) \odot \dots \odot (\alpha_n \otimes x_n) \bigr) \in \mathcal S \mathfrak g$$ (or I might be off by a sign somewhere). In general, similar formulas describe the entire product on $\mathcal S\mathfrak g$ given by transporting the one from $\mathcal U \mathfrak g$ 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 $\eta_\alpha$, understand how to apply a second-order differential operator to a monomial, and also include a brief "degree reasons" argument to get $\eta_\alpha$ 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 $x_1,\dots,x_n$ of elements in $\mathfrak g$, there is an obvious element of $\mathcal S \mathfrak g$ given by putting the $x_i$s 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 $\Omega_{\mathrm{cpt}}^1(\mathbb R) \otimes \Omega_{\mathrm{cpt}}^1(\mathbb R) \to \Omega_{\mathrm{cpt}}^1(\mathbb R)$ given by $\beta_1 \otimes \beta_2 \mapsto \beta_1 \wedge \eta_\alpha (\beta_2) - \beta_2 \wedge \eta_\alpha(\beta_1) $. Place this map at each vertex, and $\alpha_i$ at the $i$th 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 $\frac1 n \frac1{n-1} \dots \frac1{n-k}$.

Note that neither the number nor the element of $\mathcal S \mathfrak g$ 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 $\mathcal U \mathfrak g \to \mathcal S \mathfrak g$ 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 $\mathcal S \mathfrak g \cong \mathcal U \mathfrak g$, but rather this map twisted by some traces in the adjoint representation.

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Must somehow be related to "Expansions that grow on trees" by Iserles ( and "The Magnus expansion, trees, and the Knuth correspondence" by Ebrahimi-Fard and Manchon ( ? Your trees must be enumerated on the OEIS. If not,please add an entry there for other researchers. – Tom Copeland Oct 3 at 0:54

Bernoulli numbers are especial case of Poly-Bernoulli numbers which are very important in combinatorics.

In definition of Poly-Bernoulli numbers

$${Li_{k}(1-e^{-x}) \over 1-e^{-x}}=\sum_{n=0}^{\infty}B_{n}^{(k)}{x^{n}\over n!}$$

he used of poly-logarithm which has some relationships with poly-gamma , where The $B_{n}^{(1)}$ are the usual Bernoulli numbers.

A combinatorial interpretation is that the poly-Bernoulli numbers of negative index enumerate the set of n by k (0,1)-matrices uniquely reconstructible from their row and column sums.

Here you can find combinatorics of Poly-Bernoulli numbers

For $\zeta(2n+1)$ Poly-Bernoulli numbers are more functional , see here

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I edited it , thanks – Hassan Jolany 海桑乔朗丽 Sep 28 at 21:43
Do you have an OEIS entry for the enumeration of the (0,1)-matrices? – Tom Copeland Sep 29 at 18:33
Tom, I don't OEIS, what is it? – Hassan Jolany 海桑乔朗丽 Sep 29 at 20:40
A necessity for conscientious mathematicians: . – Tom Copeland Sep 29 at 22:47

You can look for the insight here.

In short, it we introduce a non-Archimedean numerical system that interprets generalized summations of infinite series (in Ramanujan's sense or Zeta function regularization) as "standard part" of the sum of the series, then in this system the quantity of all natural numbers $\omega_-$ has the standard part $-1/2$ and the quantity of non-negative integers $\omega_+$ is greater by 1 (for zero), so has the standard part $1/2$.

Consequently, from Faulhaber's formula for Ramanujan's summation,



where Where $B_n$ are the first Bernoulli numbers and $B^∗_n$ are the second Bernoulli numbers.

So the Bernoulli numbers are the standard part of the powers of the quantity of naturals.


$$\operatorname{st} e^{z\omega_-}=\frac{z}{e^{z}-1}$$


$$\operatorname{st} e^{-z\omega_-}=\frac{z}{1-e^{-z}}$$

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