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peter46464
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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 $$\nabla_{c}\nabla_{b}V_{a}-\nabla_{b}\nabla_{c}V_{a}=R_{\phantom{\mu}abc}^{d}V_{d}.$$