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one can show that the relation between first Chern class and second Chern class of $CP^n$ is

$\frac{2(n+1)}{n} c_2 (M)=c_1 (M)^2$

here $c_1 (M)^2=c_1 (M)∧c_1 (M)$. So is there any recurrence formula for i-th Chern class of $CP^n$ ?

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up vote 5 down vote accepted

There is no need for a recurrence formula. If $h\in \mathrm H^2(\mathbb C\mathbb P^n, \mathbb Z)$ is the Poincaré dual of a hyperplane section, then $h^i$ generates $\mathrm H^{2i}(\mathbb C\mathbb P^n, \mathbb Z)$ for all $i = 0, \dots, n$. Then it follows from the Euler sequence that $$ \mathrm c_i(\mathbb C\mathbb P^n) = \binom{n+1}{i}h^i. $$

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nice, thanks dear Prof Angelo – Hassan Jolany 海桑乔朗丽 Jul 1 '12 at 13:14
Angelo -- I completely agree with what you say. However, perhaps the OP was also thinking about how to express powers of $c_1$ in terms of the other classes: a problem that leads directly to the Segre classes (in the relative setting). – Jason Starr Jul 1 '12 at 14:57
Dear Jason, would you please explain with more details – Hassan Jolany 海桑乔朗丽 Jul 1 '12 at 17:21
To Jason: well, in this particular case $c_1^i$ is a multiple of $c_i$, so you don't see the Segre classes. – Angelo Jul 1 '12 at 17:30
Dear Jason, for more information you can see the reason of Angelo in following explanatory paper – Hassan Jolany 海桑乔朗丽 Jul 1 '12 at 18:11

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