# Symmetry of a connection and Christoffel's symbols

To prove that the symmetry of a connection is equivalent to the symmetry of the Christofell's symbols in $$i$$ and $$j$$, i.e $$\nabla_XY-\nabla_YX-[X,Y]=0\iff \Gamma_{ij}^k=\Gamma_{ji}^k$$ I have thought the following. In coordinates we have $$X=\xi^i\partial_i$$ and $$Y=\eta^j\partial_j$$, so:

$$\color{red}{\nabla_XY}-\color{blue}{\nabla_Y X}-\color{green}{[X,Y]}=\color{red}{\xi^i(\eta^j\Gamma_{ij}^k\partial_k+\eta^{j}_{,i}\partial_j)}-\color{blue}{\eta^j(\xi^i\Gamma_{ji}^k\partial_k+\xi^{i}_{,j}\partial_i)}-\color{green}{(\xi^i\eta^j_{,i}-\eta^i\xi^j_{,i})\partial_j}=\xi^i\eta^j\Gamma_{ij}^k\partial_k-\eta^j\xi^i\Gamma_{ji}^k\partial_k=\xi^i\eta^j(\Gamma_{ij}^k-\Gamma_{ji}^k)\partial_k=0\iff \Gamma_{ij}^k=\Gamma_{ji}^k$$ Do you think it is all right? I am not sure that I can write $$\xi^i\eta^j\Gamma_{ij}^k\partial_k-\eta^j\xi^i\Gamma_{ji}^k\partial_k=\xi^i\eta^j(\Gamma_{ij}^k-\Gamma_{ji}^k)\partial_k$$ I think yes since $$\xi^i \eta^j$$ is a product of functions and so the commutative property holds. Can you confirm me this or disprove? Thanks in advance!

• Yes, the proof is correct. As long as you're working with scalars (not necessarily invariant scalars, but scalar components), you can commute, take common factors out... and the indices take care of the non-commutativities that you might've found if you didn't work in components. Commented Apr 1, 2021 at 15:02

Yes, That is correct. Just plug in carefully $$\nabla_XY=(X^i(\partial_iY^k)+X^iY^j\Gamma_{ij}^k)\partial_k$$ and note that the components are product of real functions so $$X^iY^j=Y^jX^i$$ (like $$f(x,y,z)\times g(x,y,z)=g(x,y,z)\times f(x,y,z)$$).
You can also verify it just for $$\partial_i$$ and $$\partial_j$$ for simplicity because $$T(X,Y):=\nabla_XY-\nabla_YX-[X,Y]$$ is a tensor (known as torsion tensor): $$T_{ij}=\nabla_{\partial_i}\partial_j-\nabla_{\partial_j}\partial_i-[\partial_i,\partial_j]=\nabla_{\partial_i}\partial_j-\nabla_{\partial_j}\partial_i=(\Gamma_{ij}^k-\Gamma_{ji}^k)\partial_k.$$
The RHS is a vector, so it is zero iff its components is. therefore $$T=0\stackrel{\text{it is tensor}}{\iff} T_{ij}=0\iff \Gamma_{ij}^k=\Gamma_{ji}^k$$.