Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $M$ be a complex manifold of dimension $n$. Furthermore assume that we have a action of a Lie-Group $G$ on $M$ i.e. $G \times M \rightarrow M$, which is differential, meaning that for every $g \in G$ the map $g:M \rightarrow M$, $g:x \mapsto g \cdot x$ is differential (not holomorphic). Consider now a $(p,q)$-form $\alpha$ on $M$. My question is the following: Let $g \in G$ is it true that $g^{*}\overline{\partial} \alpha = \overline{\partial} g^{*}\alpha$ ? Or do we need that our group action is holomorphic? What is right?

bill

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

Let's try a simple example: $M=\mathbb C$, $\alpha=z$ (a $0$-form). Now take $g(z)=\bar z$ and observe that $g^* \bar \partial \alpha=0$ while $\bar \partial g^*\alpha=d\bar z$. In fact, $\bar \partial g^*\alpha=0$ exactly when $g$ is holomorphic.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.