Meaning and application of the connection coefficients (Christoffel symbols)

I know that in polar coordinates, it is $$\frac{\partial \,{{\mathbf{e}}_{r}}}{\partial \theta }={{\mathbf{e}}_{\theta }}$$ and $$\frac{\partial \,{{\mathbf{e}}_{\theta }}}{\partial \theta }=-{{\mathbf{e}}_{r}}$$ where $${{\mathbf{e}}_{r}}$$ and $${{\mathbf{e}}_{\theta }}$$ are the basis unit vectors.

Anyway, using the definition of the connection coefficients (Christoffel symbols) it should also be

$$\frac{\partial \,{{\mathbf{e}}_{r}}}{\partial \theta }={{\Gamma }^{r}}_{r\theta }\,{{\mathbf{e}}_{r}}+{{\Gamma }^{\theta }}_{r\,\theta }\,{{\mathbf{e}}_{\theta }}$$ and $$\frac{\partial \,{{\mathbf{e}}_{\theta }}}{\partial \theta }={{\Gamma }^{r}}_{\theta \,\theta }\,{{\mathbf{e}}_{r}}+{{\Gamma }^{\theta }}_{\theta \,\theta }\,{{\mathbf{e}}_{\theta }}$$

And since it is $${{\Gamma }^{r}}_{\theta \,\theta }=-r$$ , $${{\Gamma }^{\theta }}_{r\,\theta }=\frac{1}{r}$$ , $${{\Gamma }^{r}}_{r\,\theta }=0$$, $${{\Gamma }^{\theta }}_{\theta \,\theta }=0$$ (calculated with the metric) it should be

$$\frac{\partial \,{{\mathbf{e}}_{r}}}{\partial \theta }=\frac{1}{r}{{\mathbf{e}}_{\theta }}$$ and $$\frac{\partial \,{{\mathbf{e}}_{\theta }}}{\partial \theta }=-r\,{{\mathbf{e}}_{r}}$$

Where am I wrong?

You are using two different sets of basis vectors. $$\frac{\partial\mathbf e_r}{\partial \theta} = \mathbf e_\theta$$ and $$\frac{\partial \mathbf e_\theta}{\partial \theta} = -\mathbf e_r$$ hold for the orthonormal polar basis, in which the metric takes the form

$$g_{ij} = \pmatrix{1 & 0 \\ 0 & 1}$$

The connection coefficients you quote arise from the polar coordinate basis $$\left\{\frac{\partial}{\partial r},\frac{\partial}{\partial \theta}\right\}$$ which is not orthonormal, and in which the metric takes the form

$$g_{ij} = \pmatrix{1 & 0 \\ 0 & r^2}$$

The two bases are related via $$\mathbf e_r = \frac{\partial}{\partial r}$$ and $$\mathbf e_\theta = r\frac{\partial }{\partial \theta}$$.

An important thing to recognize is that the basis $$\left\{\frac{\partial}{\partial r},\frac{\partial}{\partial \theta}\right\}$$ arises naturally as the basis induced by the polar coordinates $$(r,\theta)$$. On the other hand, the orthonormal basis $$\{e_r,e_\theta\}$$ is not induced by a coordinate system; there is no set of coordinates $$(u,v)$$ such that $$e_r = \frac{\partial}{\partial u}$$ and $$e_\theta =\frac{\partial}{\partial v}$$. This is an example of a non-holonomic basis.

The reason that this is important is that in your first pass through GR, you will likely start off by using holonomic bases exclusively. Accidentally using a non-holonomic basis can lead to some apparent contradictions.

• yes, and just to be clear: the vectors in the two sets have same directions but different sizes Sep 12, 2020 at 19:06
• @AndrewSteane Yes, thanks. I've added to my answer to make that more clear. Sep 12, 2020 at 20:29
• j-murray, @AndrewSteane Thanks. It's all more clear now. I didn't expect such subtleties between the polar coordinate basis vectors (not unitary) and the polar unit vectors (not a coordinate basis). Now I know I must be careful...Many thanks. It's all more clear now. I didn't expect such subtleties in the difference between the polar coordinate basis vectors (not unitary) and the polar unit vectors (not a coordinate basis). Now I know I must be careful... Sep 13, 2020 at 11:54
• This other related post can be useful: physics.stackexchange.com/questions/198280/… Sep 13, 2020 at 12:12