Implication of the Symmetry of the Christoffel Symbol

Question

The Christoffel symbols are defined by $\frac{\partial \mathbf{e}_\alpha}{\partial x^\mu} \equiv \Gamma_{\mu \alpha}^\nu \mathbf{e}_\nu$. We also know that the Christoffel symbols are symmetric in their lower two indices. This appears to imply that $\frac{\partial \mathbf{e}_\alpha}{\partial x^\mu} = \frac{\partial \mathbf{e}_\mu}{\partial x^\alpha}$ for all coordinate bases, which seems unlikely. Is this true?

Answer 1

The Christoffel symbols are a set of coefficent that describe a connection. To understand the latter, you can read the most general case (at least known to me) on Natural Operations in Differential Geormetry (chapters II and III).

However to grasp the concept there is no need to use such formality. One can simply define a map $\nabla: \mathfrak{X}(M) \times \mathfrak{X}(M) \rightarrow \mathfrak{X}(M) :(X,Y) \mapsto \nabla_X(Y)$ with the following properties: $$ \forall f \in C^{\infty}(M) \space\ \nabla_{fX}Y=f \nabla_{X}Y, \space\ \nabla_{X}fY=X(f)+ f\nabla_{X}Y, $$ $$ \nabla_{X}(Y+Z)=\nabla_{X}Y+\nabla_{X}Z, \space\ \nabla_{X_1+X_2}Y=\nabla_{X_1}Y+\nabla_{X_2}Y $$ If you express in coordinate $(x^{\mu},\partial_{\mu}:=\frac{\partial}{\partial x^{\mu}})$ this map by defining $\nabla_{\partial_{\mu}}\partial_{\nu}:= \Gamma_{\mu \nu}^{\alpha} \partial_{\alpha}$, you get that $$ \nabla_XY=X^{\mu}\frac{\partial Y^{\alpha}}{\partial x^{\mu}}\partial_{\alpha}+X^{\mu}Y^{\nu}\Gamma_{\mu \nu}^{\alpha} \partial_{\alpha} $$ given locally $X=X^{\mu}\partial_{\mu}$ and $X=X^{\nu}\partial_{\nu}$. In general there is no need for the coefficients $\Gamma_{\mu \nu}^{\alpha}$ to be symmetric, however when this happens we say that the connection is torsionless. It can be shown that this happens iff the torsion $T:\mathfrak{X}(M) \times \mathfrak{X}(M) \rightarrow \mathfrak{X}(M)$ defined by $T(X,Y)=\nabla_XY-\nabla_YX - [X,Y]$ is zero. Among torsionless connections there is one which is special, that is the Levi-Civita connection for which $$ \Gamma_{\mu \nu}^{\alpha}=\frac{1}{2} g^{\alpha \beta} (\partial_{\mu}g_{\nu \alpha}+\partial_{\nu}g_{\alpha \mu}-\partial_{\alpha}g_{\mu \nu} ) $$ where $g$ is a Riemannian metric, if that happens we call the connection metric. Such expressions take the name of Christoffel symbols. Back to your question, any torsionless connection, thus not only for metric connections, $\nabla$ will have $\nabla_{\partial_{\mu}}\partial_{\nu}= \Gamma_{\mu \nu}^{\alpha} \partial_{\alpha}=\Gamma_{\nu \mu}^{\alpha} \partial_{\alpha}=\nabla_{\partial_{\nu}}\partial_{\mu}$.