Let $T_k(x_1,\ldots,x_n)$ be the Todd polynomials, $e_k(x_1,\ldots,x_n)$ the elementary symmetric polynomials and $p_k(x_1,\ldots, x_n)$ the power sums of degree $k$.

We have the following generating formulas \begin{align*} \sum_{k\geq 0}T_k(x_1,\ldots,x_n)t^k = \prod_{i=1}^n\frac{tx_i}{1-e^{-tx^i}}\,,\\ \sum_{k\geq 0}e_k(x_1,\ldots,x_n)t^k = \prod_{i=1}^n(1+tx_i)\,,\\ \sum_{k\geq 0}\frac{1}{k!}p_k(x_1,\ldots,x_n)t^k = \sum_{i=1}^ne^{tx_i}\,. \end{align*} There is an explicit relation between $(1/k!)p_k$ and $e_k$ in terms of Newton's identities which can be expressed using generating series (see e.g. wikipedia). Is there some similar expression for $T_k$ in terms of $e_k$ or $p_k$?

For example if $X$ is a hyperkähler complex manifold. Replacing $x_i$ with $\alpha_i$ the roots of $c(TX)$, one can show that (see (3.13)): $$ td(X) = \text{exp}\Big[-2\sum_{n\geq0} b_{2n}\text{ch}_{2n}(TX)\Big]\,, $$ where $b_{2n}$ are the modified Bernoulli numbers. Is there an explicit formula without any assumptions on the geometry?

  • $\begingroup$ I don't see a "3.13" anywhere in your linked pdf, or a reference to any Bernoulli numbers...? $\endgroup$ Nov 22, 2020 at 3:35
  • $\begingroup$ Thank you for pointing it out. I corrected the reference now. $\endgroup$
    – Arkadij
    Nov 22, 2020 at 9:03

1 Answer 1


We have

$$\log \sum_{k \ge 0} T_k t^k = \sum_{i=1}^n \log \frac{x_i t}{1 - e^{-x_i t}}$$

so if we write

$$\log \frac{x_i t}{1 - e^{-x_i t}} = \log \sum_{k \ge 0} B_k^{+} x_i^k \frac{t^k}{k!} = \sum_{k \ge 1} b_k x_i^k \frac{t^k}{k!}$$

(using the sign conventions explained on Wikipedia) then we straightforwardly have

$$\sum_{k \ge 0} T_k t^k = \exp \left( \sum_{k \ge 1} b_k p_k \frac{t^k}{k!} \right).$$

WolframAlpha gives that the generating function of the series $b_k$ begins

$$b(t) = \log \frac{t}{1 - e^{-t}} = \frac{t}{2} - \frac{t^2}{24} + \frac{t^4}{2880} - \frac{t^6}{181440} \pm $$

which gives $b_2 = - \frac{1}{12}, b_4 = \frac{1}{120}, b_6 = - \frac{1}{252}$. Plugging the denominators into the OEIS gives A006953, the sequence of denominators of $\frac{B_{2k}}{2k}$, which suggests the following. We have

$$b'(t) = \frac{d}{dt} \left( \log t - \log (1 - e^{-t}) \right) = \frac{1}{t} - \frac{e^{-t}}{1 - e^{-t}} = \frac{1}{t} \left( 1 - \frac{t}{e^t - 1} \right) = - \sum_{k \ge 1} B_k^{-} \frac{t^{k-1}}{k!}$$

which gives $b_k = - \frac{B_k^{-}}{k}$ for $k \ge 1$. Altogether we have

$$\boxed{ \sum_{k \ge 0} T_k t^k = \exp \left( \sum_{k \ge 1} - \frac{B_k^{-}}{k} p_k \frac{t^k}{k!} \right) }.$$

As a sanity check, expanding the terms in WolframAlpha up to $t^4$ in terms of elementary symmetric polynomials / Chern classes gives

$$T_1 = \frac{c_1}{2}$$ $$T_2 = \frac{c_1^2 + c_2}{12}$$ $$T_3 = \frac{c_1 c_2}{24}$$ $$T_4 = \frac{-c_1^4 + 4c_2 c_1^2 + c_3 c_1 + 3c_2^2 - c_4}{720}$$

which agrees with the formulas for the first few terms of the Todd class on Wikipedia.

Edit: I believe this is the same as the result you quote except that the linear term in your result is zero (equivalently, $c_1$ vanishes for a hyperkahler manifold).

  • $\begingroup$ Wow, soon to enter the exclusive 100k club ... $\endgroup$
    – J.J. Green
    Nov 22, 2020 at 10:53

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