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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
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@Eric Interesting. Thanks for the link. –  Dan Kneezel Apr 3 '11 at 21:05
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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

2 Answers 2

up vote 70 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 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

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