chern character of wedge product of bundles

Question

It is well known that the, for Chern character of vector bundles, $$ ch (E \otimes F)= ch(E) \wedge ch(F)$$ holds true. What about $E \wedge E$ when $E$ is holomorphic vector bundle?

Answer 1

It's easy to get it via the splitting principle. Supposing $E=L_1\oplus L_2\oplus\dots\oplus L_n$, where $L_k$'s are line bundles, $ch(L_k)=e^{x_k}$ ($x_k$ the 1st Chern class of $L_k$) we get $ch(E\wedge E)=\sum_{i<j}e^{x_i+x_j}=\bigl((\sum_{k}e^{x_k})^2-\sum_{k}e^{2x_k}\bigr)/2$, hence $$ch(E\wedge E)=\bigl((ch(E))^2-r_2\cdot ch(E)\bigr)/2$$ where $r_2$ acts on $H^{2k}$ by multiplying by $2^k$.