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The conjugate bundle $\bar{E}$ of a complex vector bundle E is defined as a complex vector bundle which is the same smooth manifold as E and same projection map but on each fiber has the complex conjugate action.

Here are some question I have in mind for conjugate bundle:

  1. Suppose E is holomorphic vector bundle. Is its conjugate bundle still holomorphic?

  2. I know that $\bar{E}$ is isomorphic to $E^{*}$ as complex vector bundle when $E$ is equipped with a Hermitian structure. If $E$ is holomorphic , then there is a fact $E^{*}$ is also holomorphic. My question is can the isomorphism between $\bar{E}$ and $E^{*}$ be a holomorphic isomorphism?

  3. Suppose $E$ is the holomorphic cotangent bundle $(T_{\mathbb{C}}^{(1,0)}M)^{*}$ of some complex manifold $M$. What is $\bar{E}$ then? Is it $(T_{\mathbb{C}}^{(0,1)}M)^{*}$?

Thanks for any explaination!

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I thought I figured some out.

  1. The transition function for $\bar{E}$ should be ($\bar{g_{ij}}$) if the transition function for $E$ is ($g_{ij}$). So if $E$ is holomorphic, i.e. ($g_{ij}$) holomorphic, then ($\bar{g_{ij}}$) must be anti holomorphic. So for the first one, the answer is negative.

  2. As $E^{*}$ is holomorphic, this isomorphism is not a holomorphic one. The regularity of the isomorphic should correspond to the regularity of the Hermitian metric. This isomorphism takes ($\bar{g}$) to $(g^{-1})^{*}$. So it's eqivalent to say it provides a reduction for the structure group from GL(2n,R) to O(2n), where n= complex rank of $E$, when we view the hermitian structure as a bilinear map on $E \times \bar{E}$.

  3. For the third one, the answer is yes if checking the transition function of $(T_{\mathbb{C}}^{(0,1)}M)^{*}$

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