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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 . 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
Incidentally, there is an updated and expanded version of my Lie Groups notes available at . 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: . 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

3 Answers 3

up vote 77 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

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 '14 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 '14 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 '14 at 19:52
See pg. 15 of Manifolds and Modular Forms by Hirzebruch, Berger, and Jung for example, for the general argument and discussion. – Tom Copeland Oct 18 at 19:18

Hirzebruch was big on the use of number theory and special functions in topology (publishing even in his eighties on Eulerian polynomials and algebraic topology) and was familiar with Norlund's work (Vorlesungen uber Differenzenrechnung, 1924)--citing it in his book--containing a formula for the falling factorial in terms of the generalized Bernoulli polynomials (which appear to be a perennially popular topic in certain math communities):

(in umbral notation here)

$$ (x-1) \cdots (x-n) = (x+\hat{B}.)^n=\hat{B}_n^{(n)}(x)$$

where $$ e^{\hat{B}^{(n)}.(x)t}= ( \frac{t}{e^t-1})^{n+1} e^{xt} .$$

Clearly, $$D^n_{t=0}(\frac{t}{e^t-1})^{n+1}=\hat{B}^{(n)}_n(0)=(-1)^n n! \; ,$$ which is what he required.

(Given the relations between the elementary symmetric functions, the falling factorial, the symmetric power functions and the exterior algebra and characteristic classes, it would be a natural formula for him to note and retain in memory. In fact, the number of $m$-dimensional faces of the $n$-dimensional simplex (with $n+1$ vertices) is $\binom{n+1}{m+1} = \frac{\hat{B}_{m+1}^{(m+1)}(n+2)}{(m+1)!}$ and the number of k-letter words defined by permuting letters attached to the vertices (A068424) is $\frac{(n+1)!}{(n+1-k)!} = \hat{B}_k^{k}(n+2)$.)

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