Why is $\bar\partial_E=\bar\partial+A^{0,1}$?

26 Views Asked by At

For a connection$\nabla$, we have $\nabla=d+A$, $\nabla=\nabla^{1,0}+\nabla^{0,1}$.

In particular, for a Chern connection, we have $\nabla=\nabla^{1,0}+\bar\partial_E$, which means $\bar\partial_E=\bar\partial+A^{0,1}$.

But by $\nabla\xi(f)=\sum \nabla\xi^i(f)e_i=\sum [d\xi^i(f)\otimes e_i+\xi^i(f)\sum A_{ij}(f)\otimes e_j]$, I can't tell the difference between $\bar\partial_E$ and the $\bar\partial$ in $\bar\partial_E=\bar\partial+A^{0,1}$. It seems they are the same when acting on $\xi$. Can you tell me the difference?