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Consider the $k$-th chern class $c_k:=c_k(\mathcal{T}_{\mathbb{P}^n})$ of the tangent sheaf of projective space $\mathbb{P}^n=\mathbb{P}^n_\Bbbk$ over some (algebraically closed, if you want) field $\Bbbk$. I am then wondering what the degree of $\prod_{k=1}^n c_k^{\nu_k}$ is, given that $\sum_{k=1}^n k\nu_k=n$. For instance, I would already be happy to see how to compute $c_2c_1^{n-2}$.

This is certainly well-known, but I cannot find a good reference for it and I am not quite comfortable yet computing it myself, I am looking for a good example to get some more "hands-on" experience with chern numbers.

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up vote 8 down vote accepted

Edit: as implicitly pointed out in the other answer, I messed up when I claimed that Milnor's class was a hyperplane class. It's the negative of a hyperplane class. I think I fixed that below.

Write $a$ for the negative of the cohomology class of a hyperplane. In Milnor's legendary book "Characteristic Classes", theorem 14.10 states that the $k$th Chern class of the tangent sheaf of complex projective space is $$c_k(\mathcal{T}_{\mathbb{P}^n})=\left( \begin{matrix} n+1 \\ k \end{matrix} \right) a^k.$$ Therefore the degree of the class corresponding to the partition of $n$ with multiplicity sequence $\nu$ is $$(-1)^n \prod \left( \begin{matrix} n+1 \\ k \end{matrix} \right)^{\nu_k}.$$ In particular, the degree of $c_2 c_1^{n-2}$ is $$\mathrm{deg}(c_2 c_1^{n-2})=(-1)^n (n+1)^{n-2} \left( \begin{matrix} n+1 \\ 2 \end{matrix} \right). $$

The proof of Milnor's theorem 14.10 is not hard (it was worth it for me to work it out a couple of times on my own): it uses the standard identification of the tangent bundle with the bundle of homomorphisms from the universal (sometimes, "tautological") bundle to its orthogonal complement (this is in fact precisely the same thing as the Euler sequence, so the two answers are basically the same).

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Perfect! Thanks in particular for the reference, I never heard of that book until now. – Jesko Hüttenhain Mar 4 '12 at 11:08

On $\mathbb{P}^n$, there is the so-called Euler sequence - a short exact sequence of vector bundles $$ 0\to \mathcal{O}_{\mathbb{P}^n}\to \mathcal{O}_{\mathbb{P}^n}(1)^{n+1} \to \mathcal{T}_{\mathbb{P}^n} \to 0. $$

For a short exact sequence of vector bundles $0\to A\to B\to C\to 0$, we have the equality $c(B) = c(A)\cdot c(C)$ where $c$ is the total Chern class: $$ c(E) = 1 + c_1(E) + c_2(E) + \ldots $$

Let $H\in H^2(\mathbb{P}^n)$ be the class of a hyperplane. Then $c(\mathcal{O}_{\mathbb{P}^n}(1)) = 1+H$, so $$ c(\mathcal{T}_{\mathbb{P}^n}) = c(\mathcal{O}_{\mathbb{P}^n})/c(\mathcal{O}_{\mathbb{P}^n}(1)) = 1/(1+H)^{n+1}. $$ This tells you that $c_k(\mathcal{T}_{\mathbb{P}^n}) = (-1)^k \binom{n+1}{k}H^k$. The cohomology groups $H^{2k}(\mathbb{P}^n)$ are free spanned by $H^k$ for $k\leq n$ and zero for $k>n$.

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urg. yes, minus a hyperplane class. – Stephen Mar 4 '12 at 0:55
BTW, there was not really a need for a separate answer, basically equivalent to the first. You could have just pointed out my mistake. – Stephen Mar 4 '12 at 1:06
Actually, I take it back. I think it's good to use the Euler sequence instead, since only over the complex numbers does my argument work. Sorry about that. – Stephen Mar 4 '12 at 1:19
I fixed a tiny mistake in the last sentence. – user64687 Dec 10 '13 at 21:35
As you wrote (almost), $c_t(\cdot)=1+c_1(\cdot)\,t+\dots$ is multiplicative wrt short exact sequences. Then the Euler sequence should give $c_t(\mathcal{O}_{\mathbb{P^n}}(1))^{n+1}=c_t(\mathcal{O}_{\mathbb{P^n}}) c_t(\mathcal{T}_{\mathbb{P^n}})$. Assuming $c_t(\mathcal{O}_{\mathbb{P^n}})=1$ and $c_t(\mathcal{O}_{\mathbb{P^n}}(1))=1+H$ gives then $c_k(\mathcal{T}_{\mathbb{P^n}})=\binom{n+1}{k}H^k$. – A.P. Jan 12 '14 at 16:39

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