Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
up vote 1 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.