# Transition Functions and Complex Structure

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 $\mathbb{CP}^1$.

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## 1 Answer

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.

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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
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