0
$\begingroup$

We always can define the connection coefficient using such a formula:

$$D_{ \mu} e_\nu(x)= \frac{\partial e_\nu(x)}{\partial x^\mu}-\Gamma^\rho_{\mu\nu}(x)e_\rho(x)=0$$

Here is a problem, the definition of derivative $\frac{\partial }{\partial x^\mu}$ depends on a coordinate system, and the basis $ e_\nu(x)$ does not always has corresponding coordinate system, in other words, it may be a non-coordinate basis. so:

(1) If $e_\nu(x)$ is a coordinate basis, the connection coefficient $\Gamma^\rho_{\mu\nu}(x)$ will be symmetric in indices $\mu,\nu$. And it can be change to zero under coordinate transformation.

(2) If $e_\nu(x)$ is a non-coordinate basis, we can use an orthonormal basis $e_a$, and $ e_\mu(x)= e_\mu^a e_a$, then we will have: $$e^a_\mu \left(\frac{\partial e_\nu(x)}{\partial x^a}- \Gamma^\rho_{a\nu}(x)e_\rho(x)\right)=0$$, so the connection coefficient $\Gamma’^\rho_{\mu\nu}(x)= e_\mu^a \Gamma^\rho_{a\nu}(x)$, which is not symmetric in $\mu,\nu$. It can be transformed to zero under a frame transformation $ e_\mu^a $.

(3) For some connection coefficients, we can’t find a global basis field to calculate the connection, we only can find a local basis to describe the connection. Then:

$$e^a_\mu \left(\frac{\partial e_\nu(x)}{\partial x^a}- \Gamma^\rho_{a\nu}(x)e_\rho(x)\right)=0$$

Because $e^a_\mu$ only defined locally, so we can't define a global connection coefficient $\Gamma^\rho_{\mu\nu}(x)$, but we can define a connection $\Gamma^\rho_{a\nu}(x)$. It can’t be change to zero under any coordinate or frame transformation.

The first two classes of connection can be change to zero under coordinate transformation or frame transformation, so does it means that their corresponding curvature is zero?

To define curvature we should parallel a vector along a closed path in the space. In a coordinate system, it is very easy to describe a closed path. For example we can define a parallelogram which contains four infinitesimal segments: $\epsilon e_a , \epsilon e_b , -\epsilon e_a , -\epsilon e_b $. But in a space only equipped with a non-coordinate basis, because $[ e_\mu ,e_\nu]\neq 0$, so the path $\epsilon e_\mu , \epsilon e_\nu , -\epsilon e_\mu , -\epsilon e_\nu $ is not a closed path, so how to define a closed path?

For the third kind of connection, it can’t be described with a global basis field, so we can assert that this kind of connection also can’t be described with a metric, so how to judge if a connection can be described with a metric(here we also take the torsion part into account)?

$\endgroup$
0
$\begingroup$

Typically, a tangent frame field is introduced which are orthogonal everywhere, namely, $$e_{a}(x)=e^{\mu}_b(x)\frac{\partial} {\partial x^\mu}$$ and its dual $$\theta^a(x)=\theta^a_\mu(x)dx^\mu.$$

The frame indices (Latin) are raised and lowered by the flat Minkowski metric and the space-time (Greek) indices by the space-time metric tensor.

I don't under any of the questions because you haven't defined your notation - and there are simply to many questions.

What are you trying to calculate - the spin connections?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ For a curvilinear coordinate system $\{x^u\}$, the corresponding basis is a coordinate basis, and $e_\mu=\frac{\partial}{\partial x^\mu}=e_\mu^a e_a$. $\{e_a\}$ is an orthonormal basis corresponding to a Cartesian coordinate system, To an arbitrary coordinate transformation $ e_\mu $ must be orthogonal? $\endgroup$ – Jianbingshao Jun 13 at 7:01
  • $\begingroup$ My question is: for some connection coefficients, they can be transform to zero under a coordinate transformation, but others can't, then what is the difference between them?some connection can be change to zero under the frame transformation but others can't. then what is the difference.? Some connection can be defined with a global basis field. but others can't, then what is the difference between them. It just like the vector field, some vector field are gradient vector field but others not. the difference between them is the curl of gradient vector field is zero. $\endgroup$ – Jianbingshao Jun 13 at 7:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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