# effect on Chern classes of tensoring with $\mathcal{O}(n)$

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.

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 $$d_2 = c_2 + (r-1)nc_1 + \frac{r^2n^2}{2}.$$
• It is easier to use Chern roots : $(x + nH) + (y +nH) = (x + y) + 2nH = c_1 + 2nH$ and similarly $(x + nH)(y + nH) = xy + (x+y)nH + n^2H^2 = c_2 + nHc_1 + n^2H^2$. – Sasha Aug 4 at 7:39