Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Christoffel Connection and Curvature concept q

  1. Jan 18, 2015 #1
    I'm looking at lecture notes on General Relativity by Sean M. Carroll, and after defining the Riemannanian tensor in the usual theorem, the extent to which the partial derivatives of a vector field fail to commute, it says ' having defined curvature tensor as something which characterizes the connection, in GR we are most concerned with the Christoffel connection. In this case the connection is derived from the metric, and the associated curvature may be thought of as that of the metric itself'.

    I'm having trouble understanding this. What does the curvature describe otherwise? I've only heard of it being the metric itself.

    Also, I think, 'in GR we are mot concerned with the Christoffel connection' comes from having the fundamental theorem of Riemannian geometry holding , so we get that simple formula relating the connection to the metric. The conditions of the metric in this theorem are that it be 'symmetric, differentiable and non-degenerate'.
    So, the unextended Schwarzschild metric has a singulaity at r=2GM, it's determinant is in indeterminate form, and is degenerate. So doesn't this mean that at a curavture singularity, because the conditions for the fundamental theorem of Riemannian geometry to hold are not satisfied, the connection can not be given by the metric, and if I have interpreted the above statement correctly, the curvature computed from the Riemannian tensor will not be that associated with the metric? What does it represent then?

  2. jcsd
  3. Jan 18, 2015 #2


    User Avatar
    Science Advisor
    Gold Member

    First, partial derivatives always commute (as far as physicists are concerned anyways, since we are almost always dealing with smooth functions), and so the Riemann curvature tensor has nothing to do with whether partial derivatives commute or not. It has to do with whether covariant derivatives commute. An equivalent way of thinking about it is, the Riemann curvature tensor deals with the difference between a vector parallel transported around an infinitely small loop and the original vector itself.

    For the purposes of GR, the connection will always arise from a metric, because in GR, a metric is always defined on a manifold. However, on a general manifold, there need not be a defined metric in order for a connection to be defined. In that case, the connection is not the "Christoffel connection" (I think the more generally used term would be the "Levi-Civita connection"). In a manifold without metric, a notion of curvature can still be defined the same way as with a metric, except the connection will no longer arise from a metric.

    The fundamental theorem of Riemannian geometry basically says that in any Riemannian manifold, the torsion free, metric compatible, connection (called the Levi-Civita connection) is unique. The fact that the metric must be symmetric and non-degenerate is in the definition of the metric, it has nothing to do with the fundamental theorem of Riemannian geometry. Didn't you create another thread where I went over this?

    I don't know what you're trying to ask with your Schwarzschild metric question. The "singularity at r=2GM" is not a real singularity but a coordinate singularity.
  4. Jan 18, 2015 #3


    User Avatar
    Science Advisor

    Curvature is better thought of as a property of the connection not of the metric. Any connection has a curvature tensor whether or not there is a metric. When one has a metric then one can derive the connection from it, which means that one can derive a connection that is both compatible with the metric and is torsion free. This is the Levi-Civita connection associated with the metric. From the Levi-Civita connection one gets the curvature in the same way as with any connection.

    In gauge theory curvature of a connection (gauge field) is the field strength - though I would ask someone who actually knows some physics to explain this.
  5. Jan 19, 2015 #4
    Ok, I'm getting confused because the theorem includes these conditions in the statement, rather than just saying the metric
  6. Jan 19, 2015 #5
    And the formula is given by the fundamental theorem of riemann geometry? So is this just in Riemann geometry? Or whenever you have a metric is it Riemann geometry? Or what other formulas exist to get the connection from the metric?
  7. Jan 19, 2015 #6


    User Avatar
    Science Advisor
    Gold Member

    Probably whatever source you are reading this stuff from included these statements as a simple reminder. For example http://en.wikipedia.org/wiki/Fundamental_theorem_of_Riemannian_geometry does not mention the non-degenerate and symmetrical nature of the metric, since it is assumed.
  8. Jan 19, 2015 #7


    User Avatar
    Science Advisor
    Gold Member

    The fundamental theorem of Riemannian geometry guarantees that if you specify a torsion free, metric compatible connection on a Riemannian or Pseudo-Riemannian manifold, this connection will be the Levi-Civita connection. That's all it says.

    Any manifold with a positive definite metric is called a Riemannian manifold. If the metric is not positive definite then the manifold is pseudo-Riemannian. If the manifold has no metric, then it is neither Riemannian nor pseudo-Riemannian.
  9. Jan 19, 2015 #8


    User Avatar
    Science Advisor

    You get Riemannian geometry when the connection is compatible with a metric and is torsion free. As you said this is the fundamental theorem. But if the connection is not compatible with the metric you do not have Riemannian geometry.

    When you have a connection then for each local frame there is a matrix of 1-forms - called the connection forms - that satisfies the formula

    ## ∇e_{i} = Σ_{j}ω_{ij}e_{j}##

    If the connection is compatible with a metric and if ## {e_{i}}## is an orthonormal frame then
    ## ω_{ij} ## is a skew symmetric matrix. This will not be true for a general connection.

    For a connection on the dual bundle, torsion free can be expressed with the additional formula

    ## de_{i} = Σ_{j} ω_{ij}##Λ##e_{j}##

    This again will not be true for a general connection.

    For any connection, the curvature matrix is defined by the equation

    ## Ω = dω + ω## Λ ##ω##

    ## Ω ## is a matrix of 2 forms and ##ω## is the connection matrix. ##ω## Λ ##ω## is the matrix product using the wedge product instead of ordinary multiplication.

    Note that the formula for the curvature matrix does not depend on a metric but only on the connection.
    Last edited: Jan 20, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Christoffel Connection and Curvature concept q