Lie derivative in terms of covariant derivative and the symmetry of Christoffel symbols

Question

I want to verify that if a manifold is torsion-free with a metric compatible derivative operator $\nabla_a$, the Lie derivative of a vector $W^a$ along $V^a$ can be written as

$$L_V W^a = V^\nu\nabla_\nu W^\alpha - W^\nu\nabla_\nu V^\alpha\tag{1}$$

and the $\nabla$ will reduce to $\partial$ when the Christoffel symbols cancel. However, I cannot see how the Christoffel symbols cancel. I get

$$V^\nu\partial_\nu W^\alpha - W^\nu\partial_\nu V^\alpha - V^\nu\Gamma_{\nu c}^\alpha W^c + W^\nu\Gamma_{\nu c}^\alpha V^c$$

$$= V^\nu\partial_\nu W^\alpha - W^\nu\partial_\nu V^\alpha - \frac{1}{2}(V^\nu\partial_\nu W^\alpha + V^\alpha \partial_\nu W^\nu - V^\nu\partial^\alpha W_\nu) + \frac{1}{2}(W^\nu\partial_\nu V^\alpha + W^\alpha \partial_\nu V^\nu - W^\nu\partial^\alpha V_\nu).\tag{2}$$

I checked and re-checked this calculation, but could not find an error.

Based on my reading, it sounds like the Christoffel symbols are supposed to cancel due to the symmetry of their lower indices. But I'm having trouble understanding how that leads to any cancellations here. How does this expression reduce to the Lie derivative?

Answer 1

The indices $\nu$ and $c$ in the items that contain Christoffel symbols are dummy indices, so we can make a switch $ \nu \leftrightarrow c$ in one of these items:

\begin{align*} V^{\nu}\nabla_{\nu}W^{\alpha} - W^{\nu}\nabla_{\nu}V^{\alpha} &= V^\nu(\partial_\nu W^\alpha + \Gamma_{\nu c}^\alpha W^c) - W^\nu(\partial_\nu V^\alpha + \Gamma_{\nu c}^\alpha V^c) \\ &= V^\nu\partial_\nu W^\alpha - W^\nu\partial_\nu V^\alpha +(V^\nu\Gamma_{\nu c}^\alpha W^c - W^\nu\Gamma_{\nu c}^\alpha V^c) \\ &= V^\nu\partial_\nu W^\alpha - W^\nu\partial_\nu V^\alpha +(V^\nu\Gamma_{\nu c}^\alpha W^c - W^c\Gamma_{c \nu}^\alpha V^\nu) \\ &= V^\nu\partial_\nu W^\alpha - W^\nu\partial_\nu V^\alpha +V^\nu W^c (\Gamma_{\nu c}^\alpha - \Gamma_{c \nu}^\alpha) \\ &= V^\nu\partial_\nu W^\alpha - W^\nu\partial_\nu V^\alpha. \end{align*}

Answer 2

It seems relevant to mention that the torsion tensor $$T(V,W) ~:=~\nabla_VW - \nabla_WV -[V,W]\tag{T}$$ is precisely defined as the difference between the antisymmetric combination of the covariant derivative and the Lie bracket for vector fields, so OP's eq. (1) simply states that the connection $\nabla$ is torsionfree.