Second Covariant Derivative Leibniz Rule

Question

I’ve been trying to understand why

$$\nabla_{u,v}^2 w := (\nabla(\nabla w) )(u, v) = \nabla_u{(\nabla_v w)}-\nabla_{\nabla_u v}w$$

Should be true. Is there some easy way to prove this identity? I'd like to see a proof that doesn't expand out the product in terms of it's coordinates if possible as well. Thanks in advanced!

Answer 1

Remember that we define the covariant derivative of arbitrary tensors so that it satisfies the Leibniz rule and commutes with contractions.

The term $\nabla W(V)$ is the contraction of $\nabla W\otimes V.$ The Leibniz rule tells us that $$\nabla_U (\nabla W \otimes V) = (\nabla_U(\nabla W)) \otimes V +\nabla W\otimes\nabla_UV.$$

Contracting this equation yields

$$\nabla_U (\nabla W (V)) = (\nabla_U(\nabla W))(V) +(\nabla W)(\nabla_UV).$$

Recalling the definition $(\nabla W)(X) = \nabla_X W,$ we can rewrite this as the desired

$$ \nabla_U(\nabla_V W) = (\nabla (\nabla W))(U,V)+ \nabla_{\nabla_U V} W.$$

Answer 2

Use the semilatus theorem that states $\delta^i_j=\delta^j_i$ if and only if $i=j$, followed by $$\Gamma_i\oplus \Gamma_j=\sum_{i=1}^{k}\hat{\Gamma}_i$$ where $\hat{\Gamma}$ is a half-metric tensor under a morphic dilational phase. If we use this on gradient fields, we have $$\nabla \Gamma_i\oplus \Gamma_j=\sum_{i=1}^{k}\nabla_i \hat{\Gamma}_i$$ Hence, use the same for your argument. If you combine each psuedoterm correctly, you'll get $$\nabla_{u,v}^2 w := (\nabla(\nabla w) )(u, v) = \nabla_u{(\nabla_v w)}-\nabla_{\nabla_u v}w$$ as you wanted to prove!

That being said, there are alternative methods. This is the method I learnt when I did tensor field theory. Hope this helps!