How to calculate the Euler class, Euler characteristic and top Chern class of $End(E)$? It's me again. Could someone please ilustrate the relationships between these concepts through the following example:
Let $E$ be a rank 2 holomorphic vector bundle on $\mathbb{CP}^2$. Find the Euler Characteristic $\chi (E \otimes E^*)$.
What I thought it was true was that the $k$-th Chern class $c_k$ where $k=\min \{ Rank(E \otimes E^*),dim(\mathbb{CP}^2) \}=2$ is equal to the Euler class of $E \otimes E^*$. Is this true? If yes, then I can show that: $c_2(E \otimes E^*) = e(E \otimes E^*) = 4 c_2(E) - c_1(E)^2$
And my other question assuming this is correct: how could I find $\chi (E \otimes E^*)$ knowing that $e(E \otimes E^*)=4 c_2(E) - c_1(E)^2$?  Thanks in advance!
 A: Question: "Sorry but I am confused. Could you be a little more explicit. I have some questions.. What is the subscript 2 in your first comment? How do I use (and what does it even mean) your second comment? And I guess you have a typo in your third comment, you have a0+a1t+a2t2. @hm2020"
Answer: You find this in Hartshorne, Appendix A. Since $S:=\mathbb{P}^2_{\mathbb{C}}$ is projective you may assume $E$ to be algebraic.
Let $k$ be the rational numbers. There is by the projective bundle formula an isomorphism $CH^*(S):=CH^*(\mathbb{P}^2)_k \cong k[t]/(t^3) \cong k\{1,t,t^2\}$. It follows $CH^i(\mathbb{P}^2) \cong kt^i$ for $i=0,1,2$ and
$$CH^*(S)_k\cong CH^0(S) \oplus CH^1(S) \oplus CH^2(S).$$
The Chern classes $c_i(E) \in CH^i(S)$ hence you may write
$$c(E):=c_0(E)+c_1(E)+c_2(E)=a_0+a_1t+a_2t^2 \in CH^*(S)_k.$$
Similarly
$$c(E^*)=b_0+b_1t+b_2t^2\text{ and }td(T_S)=c_0+c_1t+c_2t^2\in CH^*(S)_k.$$
As an example (this is in HH,page 432)
$$Ch(E)=2+c_1(E)+\frac{1}{2}(c_1(E)^2-2c_2(E))$$
and similar for $Ch(E^*)$. There is moreover a formula for the Todd class $Td(T_S)$ of the tangent bundle $T_S$. The Hirzebruch-Riemann-Roch formula says
$$\chi(E\otimes E^*)=deg(Ch(E)Ch(E^*)Td(T_S))_2$$
by the 2-subscript we mean the following: When you write
$$Ch(E)Ch(E^*)Td(T_S)=d_0+d_1t+d_2t^2$$
you let
$$\chi(E\otimes E^*)=deg(Ch(E)Ch(E^*)Td(T_S))_2:=d_2.$$
The element $d_2t^2$ lives in $CH^2(S)_k$ but the element $d_2$ is an integer: You get the formula
$$\chi(E\otimes E^*):= \sum_{i\geq 0}(-1)^i dim_{\mathbb{C}}(H^i(S,E\otimes E^*))=d_2.$$
If the "Chern roots" of $E$ is $L_1,L_2$ you should get an integer $d_2$ expressed in terms of $c_1(L_i)$ (and $c_1(L_i^*)=-c_1(L_i)$) and $c_i(T_S)$.
The formulas you get are quite involved. If everything is correct you shold get (using the above notation) something like
$$\chi(E\otimes E^*)=a_0b_1c_1+a_1b_0c_1+ a_1b_1c_0+ a_0b_0c_2 +a_0b_2c_0+ a_2b_0c_0.$$
Then you must express the numbers $a_i,b_i$in terms of $c_1(L_i)$.
There are further simplifications that can be made: There is on projective $n $-space $X$ (HH, pg 176) an exact sequence
$$0 \rightarrow \mathcal{O}_X \rightarrow \mathcal{O}_X(1)^{n+1} \rightarrow T_X \rightarrow 0$$
and this gives the following formula:
$$c_i(T_X)=(n+1)c_i(\mathcal{O}_X(1))-c_i(\mathcal{O}_X)$$
etc.
Hence
$$c_i(T_S)=(n+1)c_i(\mathcal{O}_X(1))-c_u(\mathcal{O}_X)=0,$$
for $i \geq 2$ and
$$c_1(T_S)=(n+1)c_1(\mathcal{O}_X(1)).$$
Hence
$$Td(T_S)=1+\frac{1}{2}c_1(T_S)+\frac{1}{12}c_1(T_S)^2.$$
