Check Christoffel symbol defines Levi-Civita connection I am trying to prove the existence of Levi-Civita connection. The hint says given $(U_\beta,\phi_\beta)$ be altas of $M$, for $X=x^i\partial_i,V=v^j\partial_j$, we define $$D_VX=v^i(\partial_i x^k+\Gamma_{ij}^k x^j)\partial_k$$
where $\Gamma_{ij}^k$ is Christoffel symbol.
Then the notes said it's easy to check $D_VX$ doesn't not depend on the coordinate $(U_\beta,\phi_\beta)$, hence it's a connection $D$ on $M$.
To check this fact, I think this is what I should do: given another $(V_\alpha,\psi_\alpha)$, I need to write out the base $\partial_i$, $x^i, v^j, \Gamma_{ij}^k$ in new chart (which I think should be related with $\psi_\alpha,\phi_\beta$), and show it's the same as $D_VX$. Is it correct?
And I am not sure how to write them out. Could you give me a demonstration? Thanks.
 A: Too lengthy for a comment...
I'm not sure I understand the question in your comment. A vector field, in coordinates, is given by $V= V^i \frac{\partial}{\partial x^i}$, and if you change coordinates you get, by the chain rule, $V^i \frac{\partial y^j}{\partial x^i}\frac{\partial}{\partial y^j}$ (summation convention is used). From this calculation people use to say that this is how a vector field transforms, i.e.
$$ \tilde{ V}^k = V^i \frac{\partial y^k}{\partial x^i} $$ 
Now the point is, that the partial derivatives of the $V^i$ do not transform that way (they depend on the coordinate system or chart). The components of covariant derivative do. For this you need to know how the Christoffel symbols transform and then compute to see that the components of $D_V X$ transform correctly...this is, again, nothing else but plugging the definitions and the application of the chain and product rule. It looks terrifying and complicated but it is, in fact, just a straightforward though lengty calculation.
If you want to see written down explicitly Spivaks 'Comprehensive Intrdoction to Differential Geometry' (2nd volume and Chapter 9 of volume 1, if I recall correctly) is a good source for this.
A: Calculating the different symbol can work out as you wish. But we may be stuck in complex computation.
As your mention, choose a coordinate charts $\{(U,\phi)\}$ and we can define a connection on each local chart, denoted $\bigtriangledown^U$. Now we just need to prove

For any $(U_1,\phi_1)$ and $(U_2,\phi_2)$ in charts, we have $\bigtriangledown^{U_1}_{U_1\cap U_2}=\bigtriangledown^{U_2}_{U_1\cap U_2}$.

There are two points


*

*$\bigtriangledown^{U_1}_{U_1\cap U_2}$ means the induced connection on $U_1\cap U_2$ from $\bigtriangledown^{U_1}$.

*The proof is based on the uniqueness of Levi-Civita connection. If you have proved the uniqueness, we can easily draw the conclusion because there is a unique connection on $U_1\cap U_2$.
