8
$\begingroup$

Does anyone know of some good references for computing Christoffel symbols for Hermitian metrics?

A quick Google search turns up this. The following formula appears on page 4:

$$\Gamma_{AB}^C = \frac{1}{2}h^{CE}\left(\frac{\partial h_{AE}}{\partial z^B} + \frac{\partial h_{BE}}{\partial z^A} - \frac{\partial h_{AB}}{\partial z^E}\right)$$

where $A, B, C, E \in$ {$1, \dots, n, \bar{1}, \dots, \bar{n}$} and $z^{\bar{i}} = \bar{z}^i$. From this they get

$$\Gamma_{\bar{i}j}^k = \frac{1}{2}h^{k\bar{l}}\left(\frac{\partial h_{j\bar{l}}}{\partial \bar{z}^i} - \frac{\partial h_{j\bar{i}}}{\partial \bar{z}^l}\right)$$

How do they obtain this? Are they regarding $h$ as a map $(T^{1,0}M\oplus T^{0,1}M) \times (T^{1,0}M\oplus T^{0,1}M) \to \mathbb{C}$ where $h_{ab} = 0$, $h_{\bar{a}\bar{b}} = 0$, and $h_{\bar{a}b} = \overline{h_{b\bar{a}}}$? Even if they do, I don't see how they get the second term.

Everything else I have found deals only with Kähler metrics, in which case $\Gamma_{ab}^c$ and $\Gamma_{\bar{a}\bar{b}}^{\bar{c}}$ are the only non-trivial symbols.

More generally, are there any treatments of Hermitian geometry which take this coordinate approach (as is common in Riemannian geometry texts)?

Improve this question
$\endgroup$
12
  • 3
    $\begingroup$ When I was learning this stuff, I found it extremely confusing, especially working with the complexified tangent bundle. I also found references useful only as overall guidance and not for details. I suggest two things: a) First, do all of the calculations in the original "real" tangent bundle and figure out all of the symmetries satisfied by the metric tensor and Christoffel symbol. b) Work out in painful detail what happens in complex dimensions 1 and 2. $\endgroup$
    – Deane Yang
    Commented Jun 29, 2012 at 16:40
  • $\begingroup$ Are you planning to use the Levi-Civita connection or the Chern connection? If you use the Chern connection, the only nonzero Christoffel symbols (in complex coordinates) are the usual $\Gamma^i_{jk}$, and there are no major difficulties in computing with them (only that there is torsion in general). If on the other hand you use the Levi-Civita connection, like the reference you quote, then it becomes quite messy, and it is usually very painful to perform nontrivial calculations. $\endgroup$
    – YangMills
    Commented Jul 2, 2012 at 20:03
  • $\begingroup$ Also, the equation you wrote $h_{\overline{a}b}=\overline{h_{b\overline{a}}}$ is not correct. The correct one is $h_{a\overline{b}}=h_{\overline{b}a}$, which then allows you to derive the second equation from the first one. $\endgroup$
    – YangMills
    Commented Jul 2, 2012 at 21:14
  • 1
    $\begingroup$ You have all the reason to be confused, since the authors of that paper are not being consistent with their notation! Indeed, with their convention (2.3), then their equation (2.4) is wrong. I would use the definition $h_{a\overline{b}}=g(\partial/\partial z^a, \partial/\partial \overline{z}^b)$ (where $g$ is the $\mathbb{C}$-bilinear extension of the Riemannian metric), and compute from there. You can add a factor of $2$ if you wish. $\endgroup$
    – YangMills
    Commented Jul 3, 2012 at 14:40
  • 1
    $\begingroup$ With my definition the terms $h_{ab}=\overline{h_{\overline{a}\overline{b}}}$ vanish. This is because $g$ is assumed to be Hermitian, so $g(JX,JY)=g(X,Y)$ for all $X,Y\in TM$ (and so also in $TM\otimes\mathbb{C}$ by bilinear extension). So if $X,Y$ are both of type $(1,0)$ (e.g. $X=\partial/\partial z^i$), which just means that $JX=iX$ and the same for $Y$, then $g(X,Y)=0$. $\endgroup$
    – YangMills
    Commented Jul 3, 2012 at 17:47

4 Answers 4

Reset to default
4
$\begingroup$

Try to look at the book "Foundation of differential geometry" by Kobayashi and Numizu. You will find what you need (at least in the case of Kähler metrics) in Volume II, Chapter IX, Section 5.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks diverietti, I will take a look. I was hoping for something that would deal with general Hermitian metrics. As I mentioned, the situation is much simpler for Kähler metrics. I'd like to figure out why $\Gamma_{\bar{i}j}^k$ is of the form above (as well as understanding how all the other Christoffel symbols fit in to the general form stated). Maybe Kobayashi and Nomizu will be able to help anyway. $\endgroup$
    – Michael Albanese
    Commented Jun 29, 2012 at 13:24
  • $\begingroup$ I'll try to work it out by myself and write you a more complete answer. $\endgroup$
    – diverietti
    Commented Jun 29, 2012 at 18:19
4
$\begingroup$

You might find something in "Lectures on Kahler Geometry" by Andrei Moroianu.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks Daniel. I've had a look and section 12.2 (The curvature tensor in local coordinates) has been helpful, but it only deals with Kähler metrics. $\endgroup$
    – Michael Albanese
    Commented Jul 3, 2012 at 5:06
1
$\begingroup$

You might want to try Complex Manifolds by Kodaira and Morrow. I seem to recall that a fair amount is done in local coordinates.

Improve this answer
$\endgroup$
0
$\begingroup$

Just another reference: the book of Bochner and Yano, curvature and Betti numbers. In the chapter where they deal with kählerian metrics, they do some calculations in coördinates which you might find helpful.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .