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In Prospects in Mathematics (AM-70), Hirzebruch gives a nice discussion of why the formal power series $f(x) = 1 + b_1 x + b_2 x^2 + \dots$ defining the Todd class must be what it is. In particular, the key relation $f(x)$ must satisfy is that

($\star$) the coefficient of $x^n$ in $(f(x))^{n+1}$ is 1 for all $n$.

As Hirzebruch observes, there is only one power series with constant term 1 satisfying that requirement, namely $$f(x) = \frac{x}{1-e^{-x}} = 1 + \frac{x}{2}+\sum_{k\geq 2}{B_{k}\frac{x^{k}}{k!}} = 1 + \frac{x}{2} + \frac{1}{6}\frac{x^2}{2} - \frac{1}{30}\frac{x^4}{24} + \dots,$$ where the $B_k$ are the Bernoulli numbers.

The only approach I see to reach this conclusion is:

  1. Use ($\star$) to find the first several terms: $b_1 = 1/2, b_2 = 1/12, b_3 = 0, b_4 = -1/720$.
  2. Notice that they look suspiciously like the coefficients in the exponential generating function for the Bernoulli numbers, so guess that $f(x) = \frac{x}{1-e^{-x}}$.
  3. Do a residue calculation to check that this guess does satisfy ($\star$).

My question is whether anyone knows of a less guess-and-check way to deduce from ($\star$) that $f(x) = \frac{x}{1-e^{-x}}$.

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The Todd power series arises as the solution to an ODE while studying the logarithm of a Lie group. It's sort of a mess, but you can untangle it yourself by looking at problems 8, 9 on pg 36 of math.berkeley.edu/~theojf/LieGroupsBook.pdf . A rational genus on complex manifolds is a map $MU_* \to \mathbb{Q}$, it's known that rationally $MU_*$ is the ring with a universal example of a logarithm over it, and your condition ($\star$) corresponds to selecting the standard logarithm for the formal multiplicative gp. That's my guess at finding a structured answer; no doubt these are connected. –  Eric Peterson Apr 3 '11 at 20:34
    
@Eric Interesting. Thanks for the link. –  Dan Kneezel Apr 3 '11 at 21:05
1  
Incidentally, there is an updated and expanded version of my Lie Groups notes available at math.berkeley.edu/~theojf/LieQuantumGroups.pdf . This summer I plan on completing a draft of Part II: Quantum Groups (unedited notes for Part II are available on my website). That said, the section referenced is largely unchanged between the two versions. –  Theo Johnson-Freyd Apr 3 '11 at 23:31
    
The following paper was recently posted to the arXiv: arxiv.org/abs/1204.6522 . Its tail might contain some relevant information for this question, and even if it turns out not to, it's quite neat and is a short read. –  Eric Peterson May 1 '12 at 9:16
    
Are the Todd generator and Planck's law related? –  John McKay Sep 27 '12 at 19:10

4 Answers 4

up vote 65 down vote accepted

Since you mention playing around with residues, I'm probably not telling you anything you don't already know. But there is a systematic way to extract the power series $f$ from the coefficients of $x^{n-1}$ in $f(x)^{n}$, which goes by the name of the Lagrange inversion formula.

Assume that the constant term of $f$ is invertible, and define $g(x) = \frac{x}{f(x)}$. Then $g(x)$ is a power series which has a compositional inverse. Denote this inverse by $h$, so that if $y = g(x)$ then $x = h(y)$. Write $h(y) = c_1 y + c_2 y^2 + c_3 y^3 + \cdots$. For every integer $n$, the product $n c_n$ is the residue of the differential $\frac{1}{y^n} h'(y) dy = \frac{1}{g(x)^{n}} dx = \frac{ f(x)^{n} }{x^n} dx$, which is the coefficient of $x^{n-1}$ in $f(x)^{n}$.

In your example, you get $c_n = \frac{1}{n}$, so that $h(y) = y + \frac{y^2}{2} + \frac{y^3}{3} + \cdots = - \log(1-y)$. Then $g(x) = 1 - e^{-x}$, so that $f(x) = \frac{x}{1-e^{-x}}$.

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That is very nice indeed. Thanks! –  Dan Kneezel Apr 3 '11 at 20:45
    
very nice explanation! –  SGP Apr 3 '11 at 21:00
    
For the general theory of Hirzebruch, the Lagrange inversion can be extended to formal power series, and, when expressed as such in the indeterminates, the fascinating connections to myriad combinatorial structures are revealed and the general interplay between number theory and topology, as Hirzebruch encourages us to explore as he reminisces. The analytic inverse function theorem then serves as a guide to even more interesting vistas. –  Tom Copeland 2 days ago

The following is basically taking a standard proof of Lagrange inversion and specializing it to your case, but it might amuse. You can rewrite $(\star)$ as $$\frac{1}{2 \pi i} \oint \left(\frac{f(x)}{x}\right)^{n+1} dx =1$$ for all $n$, where the contour surrounds $0$ and is small enough to avoid all other poles of $f$. Multiplying by $y^{n+1}$ and summing on $n$, $$\frac{1}{2 \pi i} \oint \sum_{n=0}^{\infty} \left(\frac{y f(x)}{x}\right)^{n+1} dx =\frac{y}{1-y}$$ or $$\frac{1}{2 \pi i} \oint \frac{dx}{1-y f(x)/x} = \frac{y}{1-y}.$$ Set $g(x)=x/f(x)$. By the holomorphic inverse function theorem, $g$ is invertible near zero, set $h=g^{-1}$.

The only pole of the integrand near $0$ is at $x=h(y)$. The residue at that pole is $$\frac{-1}{y \frac{d}{du} g(u)^{-1}} = \frac{1}{y g(x)^{-2} g'(x)} = \frac{h'(y)}{y y^{-2}} = y h'(y).$$

So $y h'(y) = \frac{y}{1-y}$, $h'(y) = \frac{1}{1-y}$, $h(y) = -\log(1-y)$ (no constant of integration since $h(0)=0$), $g(x) = 1-e^{-x}$ and $f(x) = x/(1-e^{-x})$.

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I think you missed the sign in the last displayed equation, namely the residue is $-1$ times what you wrote. Also, I would not say that $h=g^{-1}$ exists by the inverse function theorem. Instead, I would sat that $h=g^{-1}$ exists by the argument principle (or Rouché's theorem). –  GH from MO Nov 24 at 15:36
    
@GHfromMO Thanks for the sign correction. Changed "inverse function theorem" to "holomorphic inverse function theorem". Rouche or the argument principle shows that $h$ exists, but how do you know it is differentiable, let alone holomorphic, if you cite those theorems? –  David Speyer Nov 24 at 19:31
    
To be honest, I never heard of a holomorphic inverse function theorem. Is this standard? Anyways, the argument principle not only gives that $h=g^{-1}$ exists, but it also gives that it is continuous. Hence for $z\to z_0$ we have $h(z)\to h(z_0)$ and so $g(h(z))-g(h(z_0))=(c+o(1))(h(z)-h(z_0))$ with $c:=g'(h(z_0))\neq 0$. Hence $h(z)-h(z_0)=(c^{-1}+o(1))(z-z_0)$, so $h'(z_0)$ exists and equals $c^{-1}$. –  GH from MO Nov 24 at 19:52

Although I had favorited this entry some time ago because of its relation to the Bernoullis, I had completely forgotten about it until Zurab Silagdaze pointed out the Hirzebruch paper in a comment to me a couple of days ago. The excellent answer by Jacob Lurie using Lagrange's inversion formula prompted me to look at some old notes of mine relating the inversion to a formal Borel-Laplace transform. With that I'd like to make a note of how the Bernoullis and the integer reciprocals are really two sides of the same coin and how both play a role in the Todd class criterion, and then note the relation to some important combinatorics, again through a Lagrange inversion formula (LIF).

At the heart of this is an Appell sequence pair for the Bernoulli polynomials and the normalized reverse face polynomials of the simplices (see this MOQ for background) with their base number sequences being the Bernoullis $B_n$ and the reciprocals $R_n = 1/(n+1)$ with e.g.f.s

$$e^{B.t}=\frac{t}{e^t-1}, \;\;\;\; e^{R. t}=\frac{e^t-1}{t}.$$

Consider a function $f(x)$ with compositional inverse $f^{-1}(x)$ such that $f(0)=0$ and $f_x(0)=1$ (this is necessary for the associated Appell sequences to have nice properties). Let $z\;u=f^{-1}(\nu)$. Then $\nu=f(z\; u)$ and $z\; du = D_{\nu}f^{-1}(\nu)\;d\nu=D_{\nu}f^{-1}(\nu)\;z\; f^{'}(z\;u)\; du$, the inverse function theorem, and the formal Borel-Laplace transform under a change of variables is a weighting of the inverse function relation giving the standard LIF

$$\int_{0}^{\infty }\frac{1}{z} \exp\left ( -\nu \frac{1}{z} \right )D_{\nu}f^{-1}(\nu)\;d\nu=\int_{0}^{\infty }\exp\left [ - \frac{f\left (z\;u \right ) }{z}\right ]\;du$$

$$=\int_{0}^{\infty }\exp[z\;u\;D_{\omega=0}]\;\exp\left [ -u \; \frac{f(\omega)}{\omega} \right ]\;du$$

$$=\sum_{n\geq 0}z^n\;D^n_{\omega=0}\int_{0}^{\infty }\frac{u^n}{n!}\;\exp\left [ -u \; \frac{f(\omega)}{\omega} \right ]\;du$$

$$=\sum_{n\geq 0}z^n\;D^n_{\omega=0}\left [ \frac{\omega}{f(\omega)} \right ]^{n+1},$$

so we have the LIF with four key expressions, but here we focus on just three

$$\int_{0}^{\infty }\frac{1}{z} \exp\left ( -\nu \frac{1}{z} \right )D_{\nu}f^{-1}(\nu)\;d\nu=\sum_{n\geq 0}z^n\;D^n_{\omega=0}\int_{0}^{\infty }\frac{u^n}{n!}\;\exp\left [ -u \; \frac{f(\omega)}{\omega} \right ]\;du$$

$$=\sum_{n\geq 0}z^n\;D^n_{\omega=0}\left [ \frac{\omega}{f(\omega)} \right ]^{n+1}.$$

With $f(x)=e^x-1$ and $f^{-1}(x)=ln(1+x)$,

$$\int_{0}^{\infty }\frac{1}{z} \exp\left ( -\nu \frac{1}{z} \right )\frac{1}{1+\nu}\;d\nu=\sum_{n\geq 0}(-1)^{n} n! \; z^n $$

$$=\sum_{n\geq 0}z^n\;D^n_{\omega=0}\int_{0}^{\infty }\frac{u^n}{n!}\;\exp\left [ -u \; \frac{e^{\omega}-1}{\omega} \right ]\;du$$

$$=\sum_{n\geq 0}z^n\;D^n_{\omega=0}\left [ \frac{\omega}{e^{\omega}-1} \right ]^{n+1}.$$

Equivalently,

$$(-1)^{n} =\; \frac{D^n_{\omega=0}}{n!}\int_{0}^{\infty }\frac{u^n}{n!}\;\exp\left [ -u \; e^{R. \omega} \right ]\;du= \frac{D^n_{\omega=0}}{n!}\left [e^{B.\;\omega} \right ]^{n+1}.$$

Now we can see how the reciprocal derivatives in the inverse fct. theorem $f_u(u)=\frac{1}{f^{-1}_{\nu}{(\nu)}}$ are related to the reciprocal Appell sequences and the reciprocation performed by the Borel-Laplace transform. This links together the compositional inversion with a multiplicative inversion, both "performed" by the Borel-Laplace transform weighting of the inverse function relation, and shows how reciprocal expressions related to a reciprocal Appell sequence pair are related to Hirzebruch's criterion for the Chern and Todd class relation. The same two Appell sequences with the basic number sequences being the Bernoullis and the reciprocal integers are at the heart of the Euler-MaClaurin formalism.

On the other hand, performing the inversion by using the coefficients of the power series expansion of $h(x) = x/(e^x-1)$ leads to the formalism of the LIF of OEIS-A134264 (see Example 3) with all sorts of related combinatorial structures, including those of noncrossing partitions and Dyck lattice paths, related to the Eulerian, Narayana, and Fuss-Catalan numbers among others. The LIF is used by Ardila, Rincon, and Williams to show the relation between cardinalities of connected and disconnected positroids. The partition polynomials themselves are a general Appell sequence that can be used to construct a "trajectory" through various classic number arrays and polynomials, with far reaching implications, I believe, one enticing fact being that the Eulerians and Narayanas are related to volumes of polytopes.

Almost forgot to mention that this skewed view of the arguments with the Appell umbral approach reveals some interesting structure. The Bernoulli polynomials (any Appell sequence) have the interesting property $(B.(0)+x)^n= B_n(x)$, so with our notation $B_n=B_n(0)$ and

$$\left [e^{B.\;\omega} \right ]^{n+1}=exp\left [ (B.(0) + B.(0) +\; \cdots\; + B.(0) \right)\omega ],$$

giving the coefficient of $\omega^n$ as

$$[ B.(0) + B.(0) +\; \cdots \; + B.(0)) ]^n = B_n(B.(B.(B.(\cdots B.(0)))))=(-1)^n n!$$

with $n+1$ summands on the left and $n+1$ umbral substitutions on the right.

The L.H.S. can be expanded as a multinomial treating each umbra as independent and not evaluating them until all monomial summands of degree $n$ have been formed. When that is done, the superscripts can be dropped to subscripts. This is equal to the R.H.S. which is iterated umbral substitution. For example, for $n=2$,

$$(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)=3B_2+6(B_1)^2=2$$

and

$$(B.(B.(B.(0))))^2=B_2(B.(B.(0)))= 1/6 - (B.(B.(0)))^1 + (B.(B.(0)))^2=1/6-B_1(B.(0))+B_2(B.(0))=1/6- [B_1(0) -1/2] + 1/6 - B_1(0) + B_2(0)=2. $$

For $n=3$, the result is $4\;B_3+36\;B_1B_2+24\;B_1^3=−3!$. The explicit expressions for these umbral reductions of the homogeneous monomial symmetric polynomials, which hold for any Appell sequence, are those of the LIF of OEIS-A248120, with its various combinatorial interpretations.

This reveals associations among the symmetric polynomials, convolutions, umbral compositions, and umbra, and sheds light on the nature of the Hirzebruch criterion and why algebraically the Bernoulli sequence is the only sequence to satisfy it.

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see also mathoverflow.net/questions/10630/… –  Tom Copeland Nov 23 at 16:54
    
For notes on Blissard's characterization of $((e^x-1)/x)^{m}$, see "Umbral presentations of polynomial sequences" by B. D. Taylor arxiv.org/abs/math/9908131 . –  Tom Copeland Nov 24 at 1:21

Now let's bypass Hirzebruch's criterion above and connect the Bernoullis, using their basic operational definition rather than their e.g.f., to the Todd genus and more through formal group laws and associated Lie ops.

First, define the Bernoulli polynomials as the Appell sequence, $(B.(0)+x)^n=B_n(x)$, such that, $$f(B.(x+1))-f(B.(x))={f}'(x)$$ when convergent. Action on $f(x)=e^{xt}$ gives the e.g.f. since $$e^{B.(x+1)t}-e^{B(x)t}=t\;e^{xt}$$ implies $$e^{B.(x)t}(e^t-1)=t\;e^{xt}$$ and $$e^{B.(x)t}=\frac{t}{e^t-1}e^{xt}.$$

Action on $f(x)=-ln(1-xt)$ gives

$$-ln\left [1-B.(x+1)t \right ]+ln\left [1-B.(x)t \right ]= ln\left [ \frac{1-B.(x)t}{1-(B.(x)+1)t} \right ]=\frac{d\left [ -ln(1-xt) \right ]}{dx}=\frac{t}{1-xt}\;,$$

an iconic o.g.f., and using the special linear fractional (Mobius) transformation $L(t,x)=\frac{t}{1+xt}$, whose inverse in $t$ is $L(t,-x)$, this can be expressed succinctly as

$$ln\left [ 1+L[t,-(B.(x)+1)] \right ]=ln\left [ \frac{1-B.(x)t}{1-(B.(x)+1)t} \right ]=L(t,-x)=\frac{t}{1-xt},$$

with compositional inverse in $t$

$$L[e^t-1,B.(x)+1]=\frac{e^t-1}{1+(B.(x)+1)(e^t-1)}=L(t,x)=\frac{t}{1+xt}.$$

Together they comprise the formal group law

$$FGL(y,z;x)=L[L(y,-x)+L(z,-x),x]=\frac{y+z-2x\;yz}{1-x^2yz},$$

which (according to Lenart and Zainoulline) corresponds to the Euler characteristic.

For $x=0$, $(B.(0))^n=B_n$ are the Bernoulli numbers, $(B.(0)+1)^n=(B.(1))^n=(-B.(0))^n=(-1)^nB_n$, and the FGL specializes to

$$FGL(y,z;0)=y+z,$$

the fundamental additive FGL associated with the infinitesimal generator $d/dt=D_t$ and the iterated op $(D_t)^n$ with action of translation $exp(x\;D_t)f(t)=f(t+x)$.

For $x=-1$, this specializes to

$$FGL(y,z;-1)=\frac{y+z+2yz}{1-yz},$$

a fundamental FGL associated to the infinitesimal generator $(1+t)^2\frac{d}{dt}$, and, with a shift in coordinates, to the iterated op $(t^2D_t)^n=t^{2n}Lah(:tD_t:)$, related to the Lah polynomials, with the action $exp(x\;t^2D_t)f(t)=f(t/(1-xt))$, the special linear fractional transformation.

More generally for indeterminates $a_n$ with $a_0=1$, action on $f(a.,t)=-ln(1-a.\;t)$

gives (precisely when convergent and formally usefully otherwise)

$$ln\left [ \frac{1-B.(a.)t}{1-(B.(a.)+1)t} \right ]=\frac{t}{1-a.t}=\sum_{n \ge 0} a_n t^{n+1},$$

an o.g.f. for the series, which itself can be extended as an Appell sequence defined by the o.g.f.

$$\boldsymbol{O}_A(t,x)=\frac{t}{1-(a.+x)t}=\sum_{n \ge 0} (a.+x)^n t^{n+1}=\sum_{n \ge 0} A_n(x) t^{n+1}.$$

Letting $a_n=\bar{B}.(x)$, the umbral compositional inverse for the Bernoulli polynomials, i.e., $B_n(\bar{B}.(x))=x^n=\bar{B}_n(B.(x))$, we get

$$\boldsymbol{O}_\bar{B}(t,x)=ln\left [ \frac{1-B.(\bar{B}.(x))t}{1-(B.(\bar{B}.(x))+1)t} \right ]=ln\left [ \frac{1-xt}{1-(x+1)t} \right ]$$

$$=\sum_{n \ge 0} \frac{(x+1)^{n+1}-x^{n+1}}{n+1} t^{n+1},$$

and, consistently,

$$\bar{B}_n(B.(x))=\frac{(B.(x)+1)^{n+1}-(B.(x))^{n+1}}{n+1}=\frac{d}{dx}\;\frac{x^{n+1}}{n+1}=x^n.$$

The compositional inverse, through the same substitution above, is

$$\boldsymbol{O}_{\bar{B}}^{(-1)}(t,x)=\frac{e^t-1}{1+(x+1)(e^t-1)}.$$

Together they comprise the formal group law

$$FGL(y,z;x)=\boldsymbol{O}_{\bar{B}}^{(-1)}[\boldsymbol{O}_\bar{B}(y,x)+\boldsymbol{O}_\bar{B}(z,x),x]=\frac{y+z-(1+2x)\;y z}{1-x(1+x)\;yz}.$$

For $x=-1$, this specializes to

$$FGL(y,z;-1)=y+z+yz,$$ the multiplicative FGL associated with the Todd genus and the infinitesimal generator $(1+t)\frac{d}{dt}$ at the identity related by a coordinate shift to the iterated op. $(t\frac{d}{dt})^n=\phi_n(:tD_t:)$, the Bell or Stirling polynomials of the second kind, with action $exp(x\;tD_t)f(t)=f(e^xt)$, a dilation.

For $x=-1/2$,

$$FGL(y,z;-1/2)=\frac{y+z}{1+\;yz/4},$$

the Lorentz FGL, related to the Atiyah-Singer signature (Lenart and Zainoulline, "Towards generalized cohomology Schubert calculus via formal root polynomials").

So the Bernoullis are taking us on a tour through some basic formal groups, associated genera, and the conformal Lie algebra $sl_2(C)$ and associated group, the conformal global subgroup of the Virasoro algebra.

(There's a relation to elliptic functions, and more, as discussed by L $ Z. For infinite (and finite) matrix reps related to the Pascal matrix, see my mini-arxiv at my website--the notes on Infinigens.)

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