# Classification of connection coefficients

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)?

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 $$\theta^a(x)=\theta^a_\mu(x)dx^\mu.$$
• 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? – Jianbingshao Jun 13 at 7:01