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.

  • $\begingroup$ Basically, yes. In the coordinate formalism you need to verify that the coordinate expression for $D_V X$ 'transforms correctly' (i.e. like a vector field) when you change coordinates. $\endgroup$ – Thomas Jan 26 '15 at 9:43
  • $\begingroup$ @Thomas So may I know what the basis $\partial_i$ will be under the transformation? $\endgroup$ – John Jan 26 '15 at 10:08

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 $U_1\cap U_2$.

  • $\begingroup$ But I think the induced connection on a sub-manifold is only valid if we already have a connection. Currently I only defined a map based on specific chart, which may not be a connection ( I have to argue it's independent of chart and satisfies those axioms). So do you think I can argue like that? $\endgroup$ – John Feb 2 '15 at 9:01
  • $\begingroup$ @JohnZHANG Yes, very right. We should firstly prove the $\bigtriangledown^U$ is a Levi-Civita connection on $U$. By your mean, check the axioms. $\endgroup$ – gaoxinge Feb 2 '15 at 9:56
  • $\begingroup$ But I think I also need to check the map is well-defined for vector fields, which means independent of choice of chart (as I am requested to do in the question) before claiming it's a connection. So I think the way you suggest is like a loop. $\endgroup$ – John Feb 2 '15 at 9:59
  • $\begingroup$ @JohnZHANG $M$ with $\{(U,\phi)\}$ and $U$ are different, since $U$ has a single coordinate chart $(U,\phi)$. So by using the global chart $(U,\phi)$, we can define a connection on $U$. But we should check the "compatibility" of connection on $M$ between different charts because $(U,\phi)$ is local. $\endgroup$ – gaoxinge Feb 2 '15 at 11:04
  • $\begingroup$ I see. Thanks for your nice explanation! $\endgroup$ – John Feb 2 '15 at 11:12

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.