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Let $E$ be a rank $r$ complex vector bundle over the complex projective plane, $X=\mathbb{C}\mathbb{P}^2$, $c_1(E)$ and $c_2(E)$ its Chern classes. What is the effect on the Chern classes of tensoring with the twisting sheaf $\mathcal{O}_X(n)$ (i.e. what are $c_i(E(n))$)? Any answers or references would be appreciated.

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Are you familiar with the Chern character? Its definition and basic properties are outlined here https://stacks.math.columbia.edu/tag/02UM.

Since we are on $\mathbb P^2$, I will write $c_1(E) = c_1H$ and $c_2(E) = c_2H^2$, where $c_1,c_2 \in \mathbb Z$ and $H$ is the hyperplane class. Now the Chern character of a rank $r$ bundle on $\mathbb P^2$ is $$ r + c_1H + \frac{(c_1H)^2 - 2c_2H^2}{2} = r + c_1H + \frac{c_1^2 - 2c_2}{2}H^2. $$

The reason we use the Chern character is because of nice formal properties like $ch(E\otimes F) = ch(E)ch(F)$. We have $ch(\mathcal O(n)) = 1 + nH$, so multiplying out (and using $H^3 = 0$) we get $$ ch(E(n)) = r + (c_1 + rn)H + \frac{c_1^2 + 2nc_1 - 2c_2}{2}H^2. $$

Now, writing $d_iH^i = c_i(E(n))$, we can immediately read off $d_1 = c_1 + rn$ (by comparing to the formula for $ch$ of an arbitrary bundle). Plugging this into the relation obtained from setting $d_1^2 - 2d_2$ equal to the coefficient of $H^2$, we get a relation for $d_2$ which, if my scribbled algebra is correct, can be solved to yield $$ d_2 = c_2 + (r-1)nc_1 + \frac{r^2n^2}{2}. $$

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  • $\begingroup$ It is easier to use Chern roots : $(x + nH) + (y +nH) = (x + y) + 2nH = c_1 + 2nH$ and similarly $(x + nH)(y + nH) = xy + (x+y)nH + n^2H^2 = c_2 + nHc_1 + n^2H^2$. $\endgroup$
    – Sasha
    Commented Aug 4, 2019 at 7:39

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