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I am currently reading a book which deals with complex manifolds. Since I am fairly new to the topic I don't know exactly the meaning of the followinig:

Suppose we have a holomorphic vector bundle $V$ over the manifold $M$ with frames $s_\alpha$ over each trivialization $U_\alpha \subset M$

We can construct a Hermitian metric $h$ on $V$, and the author says this is given locally as

$h_\alpha = (s_\alpha,s_\alpha) $.

Then a connection 1 - form is defined locally by

\begin{equation} \omega_\alpha = \partial h_\alpha h_\alpha^{-1} \end{equation} where \begin{equation} d = \partial + \bar{\partial} \end{equation} and \begin{equation} \partial(f) = \sum_j \frac{\partial f}{\partial z^j}dz^j \end{equation} (more generally $\partial \colon C^\infty(\Lambda^{p,q}) \to C^\infty(\Lambda^{p+1,q}) $. It is then shown in the book that these 1-forms patch together to form a connection $\triangledown_h$.

Now comes the bit where I am struggeling with, to the extend that I can't read on without a bad feeling:

From the definition, one should see that \begin{align} (\triangledown_h s_\alpha, s_\alpha) + (s_\alpha, \triangledown_h s_\alpha) &= \omega_\alpha h_\alpha + h_\alpha \omega^*_\alpha \\ &= \partial h _\alpha + \bar{\partial}h _\alpha = dh _\alpha \end{align}

I am afraind I don't know enough about connections yet, in particular I don't really understand how to get from the first expression to the second. If anyone could fill in a little more details into the lines above that would be very helpful!

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1 Answer 1

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Let $s_\alpha = (s_1,\ldots, s_k)$ be a local frame. The connection 1-form $\omega_\alpha$ is a matrix of one forms $\omega_i^j$. The connection is defined by the equation $\nabla s_i = \omega_i^j s_j$ (summation implied). Therefore $$ (\nabla s_i, s_j) = (\omega_i^k s_k, s_j) = \omega_i^k h_{kj}. $$ So the expression $(\nabla s_\alpha, s_\alpha) = \omega_\alpha h_\alpha$ is the same thing but in matrix notation.

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Ah ok, so then is it true that I have $(s_j,\triangledown s_i) = (s_j, \omega^k_i s_k) = (s_j, s_k) \bar{\omega}^k_i = h_{kj} \bar{\omega}^k_i $ ? –  harlekin Jan 16 '12 at 17:34
Yep, except $h_{kj}$ should be $h_{jk}$. –  Eric O. Korman Jan 16 '12 at 18:28
oh, ya - of course. Great that really helped, I can now go on with my reading, tks a lot! –  harlekin Jan 16 '12 at 18:45

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