MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a $G$-homogeneous space $M$, for $G$ a (Lie) group, we have a canonical $G$-action on the tangent bundle $T(M)$ of $M$. If $M$ is a complex manifold, then we have a decomposition of $T(M) \otimes_{R} C$ into $T^{(1,0)}(M)$ and $T^{(0,1))}(M)$. If this decomposition is equivariant with respect to the $G$-action, then we say that we have an equivariant complex structure. As far as I can see, the set of all equivariant decompositions of $T(M) \otimes_{R} C$ into two ismorphic parts, corresponds to set of all possible equivariant almost-complex structures for $M$.

More generally though, what can we say about the relationship between equivariant decompositions of $T(M)$ (ot $T(M) \otimes_{R} C$) and the structure of $M$. When do they exist? Does the number of components in an irreducible equivariant decomposition of $T(M)$ have any geometric meaning?

share|cite|improve this question
You probably mean to say "...decomposition of $T(M)\otimes_{\mathbb{R}} \mathbb{C}$ into..." – Johan Mar 23 '12 at 19:17
Yes, I do. I spend most of my time in the complex world and sometimes forget the uncomplexified tangent bundle even exists. – Jean Delinez Mar 23 '12 at 19:30

Your Answer


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

Browse other questions tagged or ask your own question.