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Given a complex manifold $X$,we have the holomorphic tangent bundle which is given by sets of cocycles $$\rho_{ij}:U_{ij} \to GL(n,\Bbb{C})$$

Which are holomorphic map with $\rho_{ij}(z) = d_z(\phi_i\circ \phi_{j}^{-1})$ ,where $\phi_i:U_i \to \Bbb{C}^n$ gives the holomorphic chart for $X$.


If we consider $X$ simply as a smooth manifold , we have the real tangent bundle $T_{\Bbb{R}}X$, and natural complex structure $I$ induced from the coordinate chart (where $I$ is real linear map), such that under this chart $I\frac{\partial}{\partial x_i} = \frac{\partial}{\partial y_i}$.


We can consider the complexified bundle $T_{\Bbb{C}}X$, such that the local frame are $\frac{\partial}{\partial z_i}, \frac{\partial }{\partial \bar{z}_i}$, which are defined to be $$\frac{\partial}{\partial z_i} =\frac{1}{2}\left(\frac{\partial}{\partial x_i} -i\frac{\partial}{\partial y_i}\right) \\\frac{\partial}{\partial \bar{z_i}} =\frac{1}{2}\left(\frac{\partial}{\partial x_i} +i\frac{\partial}{\partial y_i}\right) $$

With the subbundle $T^{1,0}X$ spaned by the local vector fields $\{\frac{\partial}{\partial z_i}\}$ (which is eigen bundle for the complexified $I_{\Bbb{C}}$ with eigenvalue $i$).


There are three different complex vector bundle

(1) the cocyle definition (2) the real tangent bundle with complex structure $I$, makes it a complex vector bundle $(T_{\Bbb{R}}X , I)$ . (3) $T^{(1,0)}M$

I want to show three of them are equivalent.

The idea is simple if the cocycle are equivalent then they define the same vector bundle, therefore only needs to compute the cocycle for $T^{1,0}X$, with a bit long computation I can show that the transition map really given by the first definition.

Is there some better approach to prove that they are equivalent?

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1 Answer 1

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First the isomorphism between the complex bundle $(T_{\Bbb{R}}X,I)$ and $T^{(1,0)}X$ is really the linear algebra, pick the frame $$(T_{\Bbb{R}}X,I) \to T^{(1,0)}X\\\frac{\partial}{\partial x_i}\mapsto \frac{1}{2}(\frac{\partial}{\partial x_i } - i \frac{\partial}{\partial y_i})$$

If can be checked this is a complex isomorphism, as the local frame is isomorphic the vector bundle is also isomorphic.


For the first one and the third one :

Since $$\frac{\partial}{\partial z_i} =\frac{1}{2}\left(\frac{\partial}{\partial x_i} -i\frac{\partial}{\partial y_i}\right)$$

We know in the real case $$\frac{\partial}{\partial x_j} = \frac{\partial \tilde{x}_j}{\partial x_i} \frac{\partial}{\partial \tilde{x_j}}$$ Similar for $y_i$, Substitute into the above definition gives that $$\frac{\partial }{\partial z_i} = \frac{\partial \tilde{z}_j}{\partial z_i} \frac{\partial}{\partial \tilde{z_j}}$$

Therefore the cocycle is really given in the first definition,

therefore we see these three definitions are equivalent.


Maybe these three different characterizations have their own strength

for example the cocycle characterization will be useful in building some abstract theory about vector bundles (for example when dealing with classification problems).

The third one is very useful in local computation since we are familiar with differential calculus on the local coordinate for the real case, the calculation will naturally extend to the complexified case in the third setting, that's why this is useful.

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