Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

For a complex manifold $M$, the transition functions of the tangent bundle $T(M)$ come from the Jacobian of the change-of-coordinate maps. Does there exist a related description of the transition functions of $T^{(0,1)}$ and $T^{(1,0)}$? An example would also be nice, maybe $\mathbb{CP}^1$.

share|improve this question

1 Answer 1

up vote 4 down vote accepted

For a real manifold $M$ the transition functions of the tangent bundle $T(M)$ come from the Jacobian of the change-of-coordinate maps.

When $M$ is complex, it has a complex tangent bundle $T_{\mathbb{C}}M$, which can be identified with the holomorphic vector bundle $T^{(1,0)} \subset TM \otimes \mathbb{C}$. The transition functions on $T_{\mathbb{C}}M$ are given by the (complex) Jacobian of the change-of-coordinate maps, so the same is true for $T^{(1,0)}$.

Since $T^{(0,1)}$ is the complex conjugate bundle, its transition functions are the complex conjugate of the same Jacobian.

share|improve this answer
    
To see if I understand I am going to take the example of $\mathbb{CP}^1$. The change-of-coordinate map from $\phi_1(U_1) \subset \mathbb{C}$ to $\phi_2(U_2) \subset \mathbb{C}$ is $z \mapsto z^{-1}$. The complex Jacobian of this map is $z \mapsto -z^{-2}$. This gives the bundle $\mathcal{O}(2)$? The conjugate is $z \mapsto \overline{z}^{-1}$, which is not $\mathcal{O}(-2)$ as I thought it should be. –  Aston Smythe Feb 1 '10 at 23:05
1  
Indeed the vector bundle $T^{(0,1)}$ is a complex vector bundle, but is it not holomorphic, rather antiholomorphic. So, no chance for it to be $\mathcal{O}(-2)$. –  Andrea Ferretti Feb 1 '10 at 23:38
    
... and a similar situation holds for $\Omega^{(1,0)}(M)$ and $\Omega^{(0,1)}(M)$? –  Aston Smythe Feb 2 '10 at 0:31
    
Yes but by David Speyer's answer in mathoverflow.net/questions/8484/… the two bundles are isomorphic as smmoth bundles, if not as holomorphic ones. –  Jean Delinez Feb 2 '10 at 6:27

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.