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I am having troubling interpreting a particular expression in differential geometry. It arose in computing the Lie derivative along a unit normal, $n$, of the extrinsic curvature of a sub-manifold embedded in spacetime. Specifically,

$$\mathcal{L}_n K_{bd} = \underbrace{\nabla_n K_{bd}}+(\nabla_b n^c)K_{cd} + (\nabla_d n^c)K_{bc} = \underbrace{\nabla_n K_{bd}} + 2K_{bc}K^c_d$$

I do not know how to make sense of what looks like a covariant derivative $\nabla_n$. Normally, the subscript is an index, but in this case $n$ is a vector. So I can only assume that $\nabla_n$ is compact notation for something else.

Source: The notation was encountered at roughly 25:00 minutes on the Perimeter Institute lecture available at: http://pirsa.org/displayFlash.php?id=12020017.

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This means $$\nabla_n T_{\dots} = n^a \nabla_a T_{\dots}$$

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  • $\begingroup$ Just a quick question, is there a physical interpretation of $\nabla_n$, i.e. a covariant derivative contracted with a vector? Is it similar, intuitively, to a Lie derivative? $\endgroup$ – JamalS Jun 4 '14 at 17:29
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    $\begingroup$ @user1997744 It is better to say that the covariant derivative in the direction of a vector is a connection contracted to this vector, and this is what the formula in my answer really means. The point is that a connection is an additional piece of information, whereas the Lie derivative is defined on any smooth manifold along any smooth vector field without making additional choices. $\endgroup$ – Yuri Vyatkin Jun 5 '14 at 0:02

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