# How to work with Connections

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

$$\omega_\alpha = \partial h_\alpha h_\alpha^{-1}$$ where $$d = \partial + \bar{\partial}$$ and $$\partial(f) = \sum_j \frac{\partial f}{\partial z^j}dz^j$$ (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!

-

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.
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