Isn't $∇^{0,1}=\bar\partial_E+A^{0,1}$?

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A connection ∇ on a holo bundle $E$ is called compatible with holo structure if $∇^{0,1}=\bar\partial_E$. And such a connection is called a Chern connection. (reference) p.17

And we know $\nabla=d+A$. So for a chern connection, we have $\bar\partial_E=∇^{0,1}=\bar\partial+A^{0,1}$.

However, 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 think the $\nabla^{0,1}$ is just $\bar\partial_E+A^{0,1}$.

And on Huybrechts' book about the proof of Nakano inequality, he does use the relation:$\bar\partial_E=\bar\partial+A^{0,1}$

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Now I feel confused. Where did I go wrong for $∇^{0,1}=\bar\partial_E+A^{0,1}$?

If $∇^{0,1}=\bar\partial_E+A^{0,1}$ is not true, can you tell me the difference between $\bar\partial_E(\xi(f))$ and $\bar\partial(\xi(f))$?

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For Chern connection on hol vector bundle,we have $\nabla^{0,1}=\bar\partial _E $,locally,$\nabla$ denoted by $d+A$,where $A$ is the corresponding connection matrix.For local smmoth section $\left\{ s_i \right\}$ of $E$,$f_i\in A^{p,q}(X)$,we have $$\nabla^{0,1}(f_is_i)=\bar\partial _E(f_is_i)=\bar\partial(f_i)s_i+f_i(A^{0,1})_{ij}s_j, $$ which means $$\nabla^{0,1}=\bar\partial _E=\bar\partial+A^{0,1}.$$

Here,in Huybrechts' proof,he takes the local smooth sections instead of local holomorphic sections, for local holomorphic section s of $E$,we have $\bar\partial s=0$,so for $\alpha=f_is_i\in A^{p,q}(E)$,where $f_i\in A^{p,q}(X)$,we can define global operator $\bar\partial _E$.

Also,we must watch out for Chern connection on $E$,the local hol section $s$ and the local smooth section $t$,$$\nabla s=\bar\partial s=0,$$which means that $A^{0,1}=0$.

$$\nabla t=\bar\partial t\not=0,$$which means that $A^{0,1}\not=0$.