I'm trying to prove this seemingly innocent result that for line bundles $L $and $L'$ the Chern class of the tensor product bundle is given by $$c_1(L \otimes L') = c_1(L) + c_1(L').$$
I managed to find this proven in few places, but all of the proofs are done in a quite a different framework compared to the one I'm working in.
I've defined $c_k(E) = \left[f_k\left(\frac{i}{2\pi}\Omega\right)\right]$ where $\Omega$ is the curvature matrix and $f_k$ the k'th elementary symmetric polynomial.
Bott and Tu does this using the Euler class for which they have the expression $$e(E)=-\frac{1}{2\pi i} \sum_\gamma d(\rho_\gamma d\log g_{\gamma \alpha})$$ and use the cocycle on the product bundle $L \otimes L'$ to conclude, but I would like to have a more direct way instead of deriving this expression $e(E)=-\frac{1}{2\pi i} \sum_\gamma d(\rho_\gamma d\log g_{\gamma \alpha})$ for the Euler class. Is this doable? I didn't find at least any easy way to proceed.
Let $(E,\nabla)$ be a vector bundle with a flat connection, i.e., $\nabla\colon \mathfrak X(M)\times\Gamma(E)\to \Gamma(E)$. Now the tensor product of $(E_1,\nabla_1)$ and $(E_2,\nabla_2)$ is $(E_1\otimes E_2,\nabla_1\otimes1+1\otimes\nabla_2)$, i.e., with the connection (for $s_i\in\Gamma(E_i)$) $$\nabla_X(s_1\otimes s_2)=\nabla_Xs_1\otimes s_2+s_1\otimes\nabla_Xs_2.$$ Now given $(E,\nabla)$, the curvature $R_E$ is, for $X,Y\in\mathfrak X(M)$, $$R_E(X,Y):=[\nabla_X,\nabla_Y]-\nabla_{[X,Y]}\colon \Gamma(E)\to\Gamma(E).$$ Now, $$\begin{align*} R_{E_1\otimes E_2}(X,Y)&=[\nabla_X\otimes1+1\otimes\nabla_X,\nabla_Y\otimes1+1\otimes\nabla_Y]-(\nabla_{[X,Y]}\otimes1+1\otimes\nabla_{[X,Y]})\\ &=[\nabla_X,\nabla_Y]\otimes1+1\otimes[\nabla_X,\nabla_Y]-(\nabla_{[X,Y]}\otimes1+1\otimes\nabla_{[X,Y]})\\ &=R_{E_1}(X,Y)\otimes1+1\otimes R_{E_2}(X,Y). \end{align*}$$ Thus, upon taking traces (recall that $c_1(E)=\frac{i}{2\pi}\mathrm{tr}\ \Omega_E$ where $\Omega_E$ is the matrix representing $R_E$), we obtain $$\Omega_{E_1\otimes E_2}(X,Y)=r_2\Omega_{E_1}(X,Y)+r_1\Omega_{E_2}(X,Y),$$ where $r_i$ is the rank of $E_i$, so passing to cohomology, $$c_1(E_1\otimes E_2)=r_2c_1(E_1)+r_1c_1(E_2).$$