Hirzebruch's $L$-polynomial is the formal power series \begin{equation} L(x) = \frac{x}{\tanh x} = 1 + \frac{x^2}{3} + \cdots \end{equation}

This defines a multiplicative sequence and a genus $L(M)$ for an oriented manifold. See the Wikipedia page.

Let $n = 2k$ be an even positive integer. Hirzebruch's signature theorem, applied to the $4k$-manifold $\mathbb{C}P^n$, tells us that \begin{equation} \langle L(\mathbb{C}P^n), [\mathbb{C}P^n] \rangle = 1. \end{equation}

I'm wondering whether this can be shown directly without appealing to the signature theorem. For example, in the case of a 4-manifold $M$, direct computation with symmetric polynomials shows that \begin{equation*} L_2(M) = \frac{c_1^2 - 2c_2}{3} \end{equation*}

I also know that the Chern classes of $\mathbb{C}P^n$ are $c(\mathbb{C}P^n) = (1+y)^{n+1}$ for a generator $y \in H^2(\mathbb{C}P^n)$, so specializing to $n = 2$, $M = \mathbb{C}P^2$, \begin{equation*} L_2(\mathbb{C}P^2) = \frac{(3y)^2 - 2(3y^2)}{3} = y^2 \end{equation*} which when paired with the fundamental class gives $1$, as wanted.

When $n$ is larger, the computations get very messy. Is there some trick to painlessly compute $\langle L(\mathbb{C}P^n), [\mathbb{C}P^n] \rangle = 1$?

  • $\begingroup$ You can conjecture the signature theorem by conjecturing that the $L_i$ exist and determining what they must be using the fact that the signature of $\mathbb{CP}^n$ is always $1$. The computation involves using Lagrange inversion, and will also prove this fact. $\endgroup$ Aug 19, 2015 at 20:59

1 Answer 1


I found a method in Dan Freed's notes on bordism. It's very, very slick, and I'll sketch it below for completeness.

The main difficulty with the direct calculation is the use of "formal Chern roots" for the tangent bundle $T\mathbb{C}P^n$. If $T\mathbb{C}P^n$ were a sum of line bundles, then there would be no need to symmetrize the product and the calculation would be easier. The key insight is that while $T\mathbb{C}P^n$ is not a sum of line bundles, it is stably equivalent to $(L^*)^{\oplus n+1}$, where $L$ is the tautological line bundle over $\mathbb{C}P^n$. And since Chern classes are stable, the computation of the $L$-class can be done with $(L^*)^{\oplus n+1}$ instead.

In other words, since $c_1(L^*) = y$, the generator in $H^2 (\mathbb{C}P^n)$, we have \begin{equation*} L(\mathbb{C}P^n) = L((L^*)^{\oplus n+1}) = \left(\frac{y}{\tanh y}\right)^{n+1} \end{equation*}

To find the coefficient of $y^n$ in this product, we can use the method of residues (the integral is quite fun to do), and we conclude that if $n$ is even then $\langle L(\mathbb{C}P^n), [\mathbb{C}P^n] \rangle$ is 1.


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