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

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When the connection has torsion (assuming you are in a coordinate basis). But the metric connection is always torsion-free. –  Zhen Lin Sep 15 '11 at 14:22
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The tensor $T^\mu_{\nu \lambda} = \Gamma^\mu_{\nu \lambda} - \Gamma^\mu_{\lambda \nu}$ is called torsion. –  Sasha Sep 15 '11 at 14:23
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@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? –  Zhen Lin Sep 15 '11 at 15:22
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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. –  Zhen Lin Sep 15 '11 at 15:32
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@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. –  Willie Wong Sep 15 '11 at 16:03
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1 Answer 1

Let $\pi:E\rightarrow M$ be an arbitrary vector bundle, not necessarily the tangent bundle $TM$ of $M$. Locally, say over an open subset $U\subset M$, one can choose a family of point-wise linearly independent sections $\{E_i:U\rightarrow E\}$, which is called a local frame in $E$ (over $U$). Then each section $X:U\rightarrow E$ can be uniquely represented in terms of $\{E_i\}$ $$ X=X^i E_i \quad \mbox{(Einstien summation assumed)} $$

In presence of a connection $\nabla$ in $E$ the covariant derivatives can be computed in terms of the Christoffel symbols of $\nabla$ with respect to the local frame $\{E_i\}$ defined as $$ \nabla_{E_i}{E_j} = \Gamma^k_{ij} E_k $$

In general, the Christoffel symbols need not to be symmetric since torsion may not vanish, as noted by the commenters.

For the case of Riemannian manifolds, the Riemannian metric $g$ induces a preferred connection in the tangent bundle which is called the Levi-Civita connection. We prefer this one because it is torsion-free (the torsion tensor $T(X,Y)=\nabla_X Y - \nabla_Y X -[X,Y]$ vanishes) and the metric tensor is parallel w.r.t. to it: $\nabla g =0$.

Connections in the tangent bundle are sometimes termed linear.

In a coordinate patch $(U,x^i)$ we have the standard coordinate frame $\{ \partial_i \}$ (of the tangent bundle) where vector fields $\partial_i$ act as partial derivatives on functions. This implies that a linear connection is torsion-free if and only if the Christoffel symbols with respect to any coordinate frame (!) are symmetric w.r.t the bottom indices (see J.Lee, Riemannian manifolds. An introduction to curvature, Springer 1987, p.63) (Hint: $[\partial_i, \partial_j]=0)$

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