6
$\begingroup$

When aren't Christoffel symbols symmetrical with respect to its bottom indexes? Why isn't the symmetry of second derivatives true in this case?

  • 1
    When the connection has torsion (assuming you are in a coordinate basis). But the metric connection is always torsion-free.2011-09-15
  • 1
    The tensor $T^\mu_{\nu \lambda} = \Gamma^\mu_{\nu \lambda} - \Gamma^\mu_{\lambda \nu}$ is called [torsion](http://en.wikipedia.org/wiki/Torsion_tensor).2011-09-15
  • 0
    @Zhen Lin It is of course the matter of definition, but torsion free requirement imposes relation between metric and connection. One can choose metric and connections to not agree. I remember back in undergrad school, a professor teaching Lie groups, showed two examples of metric+connection on the same group, one having zero torsion, the other having zero curvature tensor.2011-09-15
  • 0
    By definition Christoffel symbols is $\Gamma^{i}_{kj}=-\frac{\partial x^p}{\partial z^k}\frac{\partial x^q}{\partial z^j}\frac{{\partial}^2 z^i}{\partial x^p \partial x^q}$, So $\Gamma^{i}_{[kj]}=0$ if (iff) $\frac{{\partial}^2 z^i}{\partial x^p \partial x^q}=\frac{{\partial}^2 z^i}{\partial x^q \partial x^p}$. Am I right?2011-09-15
  • 0
    @Alyushin: I can't say I've ever seen a formula like that. What are $x$ and $z$ supposed to be?2011-09-15
  • 0
    @Zhen Lin: $(x^1,\ldots, x^n)$ and $(z^1,\ldots, z^n)$ are coordinate systems (old and new respectively)2011-09-15
  • 1
    @Alyushin: I'm afraid that really makes no sense. Let $x$ and $z$ be the same coordinate system; then your Christoffel symbols become identically zero. Are you sure you have the right definition?2011-09-15
  • 0
    Hm.. its strange2011-09-15
  • 0
    @Zhen Lin: Are you defined Christoffel symbols in terms metric tensor?2011-09-15
  • 1
    I think of Christoffel symbols as the coefficients of some given affine connection, but usually the Levi–Civita connection. In that case $\Gamma^i_{\phantom{i}jk}$ is given in terms of partial derivatives of the metric, yes.2011-09-15
  • 0
    @Zhen Lin: i.e. $\Gamma^{i}_{kl}=\frac{1}{2}g^{im}(g_{mk,l}+g_{ml,k}-g_{kl,m})$. Thus $\Gamma^{i}_{[kl]}=0$ if metric tensor is symmetrical ($g_{[ij]}=0$). Can you give me an example of anti-symmetrical metric?2011-09-15
  • 0
    More exactly: example of metric such that $A^{i}_{[kl]}\neq 0$, where $A^{i}_{[kl]}=g^{im}g_{kl,m}$.2011-09-15
  • 1
    @Zhen: the formula that Alyushin wrote down looks somewhat similar to, but is not the same as, the change of variables formula for Christoffel symbols, assuming we start with a flat coordinate system $x^i$. In particular, if $(x^i)$ is the standard coordinate system on $\mathbb{R}^n$ and $(z^i)$ a separate coordinate system, you have that the Christoffel symbols of $x$ is zero, and that for $z$ is given by $$\Gamma_{ij}^k = \frac{\partial z^k}{\partial x^m} \frac{\partial^2 x^m}{\partial z^i\partial z^j} $$which is symmetric in $ij$ because the Levi-Civita connection is torsion free.2011-09-15
  • 1
    @Alyushin: the Levi-Civita connection is by definition torsion free, and symmetric in those indices. One can define connections on the tangent bundle which is not Levi-Civita. See ["Torsion"](http://en.wikipedia.org/wiki/Torsion_tensor) and [this discussion](http://www.physicsforums.com/showthread.php?t=492320).2011-09-15

1 Answers 1