Checking derivation of the curvature tensor

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The discussion revolves around deriving the curvature tensor by calculating the commutator of two covariant derivatives. The user presents their work, showing the steps taken to express the curvature tensor in terms of the covariant derivatives and Christoffel symbols. They arrive at the conclusion that the difference between the two covariant derivatives yields an expression that resembles the Riemann tensor. The final result indicates that the derived expression aligns with standard formulations found in differential geometry literature. The user expresses confidence that their derivation is correct, referencing its commonality in textbooks.
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Homework Statement


I am trying to derive the curvature tensor by finding the commutator of two covariant derivatives. I think I've got it, but my head is spinning with Nablas and indices. Would anyone be willing to check my work? Thanks

Homework Equations


I am trying to derive the curvature tensor by finding the commutator of two covariant derivatives: $$\nabla_{c}\nabla_{b}V_{a}-\nabla_{b}\nabla_{c}V_{a}.$$
The covariant derivative of ##V_{a}## is given by$$\nabla_{b}V_{a}=\partial_{b}V_{a}-\Gamma_{ab}^{d}V_{d}.$$

3. The Attempt at a Solution

Start by taking another covariant derivative: $$\nabla_{c}\nabla_{b}V_{a}=\partial_{c}\left(\nabla_{b}V_{a}\right)-\Gamma_{ac}^{e}\nabla_{b}V_{e}-\Gamma_{bc}^{e}\nabla_{e}V_{a}.$$
For the first right-hand side term $$\partial_{c}\left(\nabla_{b}V_{a}\right)=\partial_{c}\partial_{b}V_{a}-\partial_{c}\left(\Gamma_{ab}^{d}V_{d}\right).$$
Then using the product rule I get $$\partial_{c}\left(\nabla_{b}V_{a}\right)=\partial_{c}\partial_{b}V_{a}-\partial_{c}\left(\Gamma_{ab}^{d}\right)V_{d}-\partial_{c}\left(V_{d}\right)\Gamma_{ab}^{d}.$$
For the second right-hand side term
$$-\Gamma_{ac}^{e}\nabla_{b}V_{e}=-\Gamma_{ac}^{e}\left(\partial_{b}V_{e}-\Gamma_{eb}^{d}V_{d}\right).$$
And for the third right-hand side term $$-\Gamma_{bc}^{e}\nabla_{e}V_{a}=-\Gamma_{bc}^{e}\left(\partial_{e}V_{a}-\Gamma_{ae}^{d}V_{d}\right).$$
Put all these together to get $$\nabla_{c}\nabla_{b}V_{a}=\partial_{c}\partial_{b}V_{a}-\partial_{c}\left(\Gamma_{ab}^{d}\right)V_{d}-\partial_{c}\left(V_{d}\right)\Gamma_{ab}^{d}-\Gamma_{ac}^{e}\left(\partial_{b}V_{e}-\Gamma_{eb}^{d}V_{d}\right)-\Gamma_{bc}^{e}\left(\partial_{e}V_{a}-\Gamma_{ae}^{d}V_{d}\right).$$
Interchanging ##b## and ##c##
gives $$\nabla_{b}\nabla_{c}V_{a}=\partial_{b}\partial_{c}V_{a}-\partial_{b}\left(\Gamma_{ac}^{d}\right)V_{d}-\partial_{b}\left(V_{d}\right)\Gamma_{ac}^{d}-\Gamma_{ab}^{e}\left(\partial_{c}V_{e}-\Gamma_{ec}^{d}V_{d}\right)-\Gamma_{cb}^{e}\left(\partial_{e}V_{a}-\Gamma_{ae}^{d}V_{d}\right).$$
Subtract one from the other: $$\nabla_{c}\nabla_{b}V_{a}-\nabla_{b}\nabla_{c}V_{a}=\partial_{c}\partial_{b}V_{a}-\partial_{c}\left(\Gamma_{ab}^{d}\right)V_{d}-\partial_{c}\left(V_{d}\right)\Gamma_{ab}^{d}-\Gamma_{ac}^{e}\left(\partial_{b}V_{e}-\Gamma_{eb}^{d}V_{d}\right)-\Gamma_{bc}^{e}\left(\partial_{e}V_{a}-\Gamma_{ae}^{d}V_{d}\right)-
\left(\partial_{b}\partial_{c}V_{a}-\partial_{b}\left(\Gamma_{ac}^{d}\right)V_{d}-\partial_{b}\left(V_{d}\right)\Gamma_{ac}^{d}-\Gamma_{ab}^{e}\left(\partial_{c}V_{e}-\Gamma_{ec}^{d}V_{d}\right)-\Gamma_{cb}^{e}\left(\partial_{e}V_{a}-\Gamma_{ae}^{d}V_{d}\right)\right).$$
Multiplying out the brackets in the second line gives $$\partial_{c}\partial_{b}V_{a}-\partial_{c}\left(\Gamma_{ab}^{d}\right)V_{d}-\partial_{c}\left(V_{d}\right)\Gamma_{ab}^{d}-\Gamma_{ac}^{e}\partial_{b}V_{e}+\Gamma_{ac}^{e}\Gamma_{eb}^{d}V_{d}-\Gamma_{bc}^{e}\partial_{e}V_{a}+\Gamma_{bc}^{e}\Gamma_{ae}^{d}V_{d}-
\left(\partial_{b}\partial_{c}V_{a}-\partial_{b}\left(\Gamma_{ac}^{d}\right)V_{d}-\partial_{b}\left(V_{d}\right)\Gamma_{ac}^{d}-\Gamma_{ab}^{e}\partial_{c}V_{e}+\Gamma_{ab}^{e}\Gamma_{ec}^{d}V_{d}-\Gamma_{cb}^{e}\partial_{e}V_{a}+\Gamma_{cb}^{e}\Gamma_{ae}^{d}V_{d}\right).$$
Note that ##\Gamma_{ac}^{e}\partial_{b}V_{e}=\Gamma_{ab}^{e}\partial_{c}V_{e}## and ##\partial_{c}\left(V_{d}\right)\Gamma_{ab}^{d}=\partial_{b}\left(V_{d}\right)\Gamma_{ac}^{d}##
(because of the definition of ##\frac{\partial e_{x}}{\partial x^{z}}=\Gamma_{xz}^{y}e_{y}).##
I then end up with $$\nabla_{c}\nabla_{b}V_{a}-\nabla_{b}\nabla_{c}V_{a}=\left(\partial_{b}\left(\Gamma_{ac}^{d}\right)-\partial_{c}\left(\Gamma_{ab}^{d}\right)+\Gamma_{ac}^{e}\Gamma_{eb}^{d}-\Gamma_{ab}^{e}\Gamma_{ec}^{d}\right)V_{d}.$$
The expression inside the brackets on the rhs is the Riemann tensor, meaning $$\nabla_{c}\nabla_{b}V_{a}-\nabla_{b}\nabla_{c}V_{a}=R_{\phantom{\mu}abc}^{d}V_{d}.$$
 
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It looks like more or less the right thing. I'm not going to go through the details. This is "textbook" stuff that you can find in lots of different differential geometry texts.
 

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