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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asked Feb 1, 2010 at 7:29
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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.
answered Feb 1, 2010 at 14:13
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$\begingroup$ 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. $\endgroup$ Commented Feb 1, 2010 at 23:05
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1$\begingroup$ 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 $\mathcal{O}(-2)$. $\endgroup$ Commented Feb 1, 2010 at 23:38
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$\begingroup$ ... and a similar situation holds for $\Omega^{(1,0)}(M)$ and $\Omega^{(0,1)}(M)$? $\endgroup$ Commented Feb 2, 2010 at 0:31
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$\begingroup$ 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. $\endgroup$ Commented Feb 2, 2010 at 6:27