# Hirzebruch's $L$-polynomial and $\mathbb{C}P^n$

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

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 $$\langle L(\mathbb{C}P^n), [\mathbb{C}P^n] \rangle = 1.$$

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 $\langle L(\mathbb{C}P^n), [\mathbb{C}P^n] \rangle = 1$?

• 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. Aug 19, 2015 at 20:59

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.