Covariant derivative with respect to commutator

I have some confusion with the notion of $$\nabla_{[A, B]}\bf{v}$$, that expression, with a commutator of vector fields as the subindex of the connection appears for instance in the definition of the Riemann tensor as

$$R(u, v) w=\nabla_{u} \nabla_{v} w-\nabla_{v} \nabla_{u} w-\nabla_{[u, v]} w.$$

what I know is the covariant derivative with respect to a vector, but here we have

$$\nabla_{[u,v]}f = \nabla_{uv - vu}f = \nabla_{uv}f - \nabla_{vu}f = u^av^b\nabla_{ab}f - v^au^b\nabla_{ab}f$$

but what is this

$$\nabla_{ab} = \nabla_{\partial_a \partial_b}$$

and how does it act? Thx.

• $[u,v]$ is not a commutator per se, it is a Lie bracket, i.e., a Lie derivative. In any case, this is a math question, not a physics one. – AccidentalFourierTransform Jul 14 at 1:04
• Unfortunately if $u, v$ are vector fields, "$uv$" is not a vector field. It fails the Leibniz property as $u(v(fg)) = u(v(f)g + fv(g)) = u(v(f))g + v(f)u(g) + u(f)v(g) + fu(v(g))$. However, "$uv-vu$" is a vector field because if you work out the algebra, you'll find the Leibniz property is restored. When it comes to covariant differentiation, $\nabla_{uv}$ is not defined. Instead, you need to work out $[u, v]$ first, and then apply $\nabla_{[u, v]}$ because $[u, v]$ is a vector field. – SpiralRain Jul 14 at 1:06
• If you want an explicit way to work with this, consider the equation $[X, Y] = (X^{i}\partial_{i}(Y^{j}) - Y^{i}\partial_{i}(X^{j}))\partial_{j}$ in local coordinates, and use that to take apart $\nabla_{[X, Y]}$. I wonder if anything interesting can be found. – SpiralRain Jul 14 at 1:10
• Your comment about “the subindex of the connection” indicates confusion about the (confusing!) notation. For example, the $u$ in $\nabla_u$ is not an index; it is a vector field indicating in which direction the directional derivative is being taken. Similarly, $[u,v]$ is a vector field, not an index. But sometimes, as in the $\nabla_\mu$ in $\nabla_u=u^\mu\nabla_\mu$,, the subscript is an index! Sometimes the distinction is made more apparent by using different fonts. – G. Smith Jul 14 at 2:13
• Thank you, everyone. It seems clear to me that what I have when I see [u, v] as a subindex is the Lie derivative, so then it's a vector field. I think this pretty much solves it. – David Jul 15 at 9:08