Show that $c_1(L\otimes E) = rc_1(L) + c_1(E)$ if $E$ is a vector bundle of rank $r$ and $L$ is a line bundle Let $L$ be a line bundle and $E$ a vector bundle of rank $r$, then how can we prove that $$c_1(L\otimes E)=rc_1(L)+c_1(E)?$$ Here $c_1$ means the first Chern class.
 A: Below I give two methods for establishing the identity you asked about, but these methods can also be used to establish the general identity $c_1(E_1\otimes E_2) = \operatorname{rank}(E_2)c_1(E_1) + \operatorname{rank}(E_1)c_1(E_2)$.
$1.$ Using the splitting principle.

Splitting Principle: Let $E \to Y$ be a rank $r$ complex vector bundle. There is a space, $X$ (namely the total space of the flag bundle of $E$), and a continuous map $f : X \to Y$ such that

*

*$f^*E = L_1\oplus\dots\oplus L_r$ where each $L_i$ is a complex line bundle, and


*$f^* : H^*(Y; \mathbb{Z}) \to H^*(X; \mathbb{Z})$ is injective.

As $c_1$ is additive with respect to direct sums and tensor products of line bundles, we see that
\begin{align*}
f^*(c_1(L\otimes E)) &= c_1(f^*(L\otimes E))\\
&= c_1(f^*L\otimes f^*E)\\
&= c_1(f^*L\otimes(L_1\oplus\dots\oplus L_r))\\
&= c_1(f^*L\otimes L_1\oplus\dots\oplus f^*L\otimes L_r)\\
&= c_1(f^*L\otimes L_1) + \dots + c_1(f^*L\otimes L_r)\\
&= c_1(f^*L) + c_1(L_1) + \dots + c_1(f^*L) + c_1(L_r)\\ 
&= rc_1(f^*L) + c_1(L_1) + \dots + c_1(L_r)\\
&= rc_1(f^*L) + c_1(L_1\oplus\dots\oplus L_r)\\
&= rc_1(f^*L) + c_1(f^*E)\\
&= rf^*c_1(L) + f^*c_1(E)\\
&= f^*(rc_1(L) + c_1(E)).
\end{align*}
As $f^*(c_1(L\otimes E)) = f^*(rc_1(L) + c_1(E))$ and $f^*$ is injective, we see that
$$c_1(L\otimes E) = rc_1(L) + c_1(E).$$
$2.$ Using the Chern character.
The Chern character is given by $\operatorname{ch}(E) = \operatorname{rank}(E) + c_1(E) + \dots$ where the dots indicate higher order terms, and has the property that $\operatorname{ch}(E_1\otimes E_2) = \operatorname{ch}(E_1)\operatorname{ch}(E_2)$.
So, by definition, we have
$$\operatorname{ch}(L\otimes E) = r + c_1(L\otimes E) + \dots$$
On the other hand, using the multiplicative property of the Chern character, we have
$$\operatorname{ch}(L\otimes E) = \operatorname{ch}(L)\operatorname{ch}(E) = (1 + c_1(L) + \dots)(r + c_1(E) + \dots) = r + (rc_1(L) + c_1(E)) + \dots$$
Comparing the degree two parts of both expressions, we see that $c_1(L\otimes E) = rc_1(L) + c_1(E)$.
A: In smooth case the first Chern class of bundle $B \rightarrow M$ is equal up to constant to the class in de-Rham cohomology of the trace of any curvature form $\Theta$:
$$c_1(B)=\left[ \frac{2 \pi}{i} \operatorname{tr} \Theta \right] \in H_{dR}^2 (M),$$
here $\Theta$ is a differential 2-form with values in endomorphisms of $B$, so you can consider it as a matrix of usual 2-forms and take its trace, which will be usual 2-form.
Because you can suppose
$$\Theta_{L \otimes E}=\Theta_E \otimes I_1+I_r \otimes\Theta_L,$$
your formula is very transparent.
You can find details in Griffiths, Harris "Principles of algebraic geometry-1", chapter 3.3, this formula is one page before Gauss-Bonnet formula.
