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Question: How to simplify the following local expression: \begin{equation}\tag{*} g^{ij} \Big( \partial_j (\partial_i V^k + \Gamma_{im}^k V^m) - \Gamma_{ij}^l (\partial_l V^k + \Gamma_{lm}^k V^m) \Big), \end{equation} where $V$ is a vector field on a Riemannian manifold $(M,g)$, $\Gamma$ is the Christoffel symbols.

Motivation: I am trying to get through the old paper of Dohrn and Guerra, which has the following quantity in its eqaution (12): \begin{equation}\tag{**} g^{ij} \nabla_{\partial_j} \nabla_{\partial_i} V, \end{equation} where $\nabla$ is the Riemannian covariant derivative. According to my derivation from the preceding text of the paper, the quantity $(**)$ should coorespond to the local expression $(*)$. However, a simple application of definitions gives the local expression of $(**)$ as follows, \begin{equation} g^{ij} \nabla_{\partial_j} \nabla_{\partial_i} V = g^{ij} \Big( \partial_j (\partial_i V^k + \Gamma_{im}^k V^m) + \Gamma_{jm}^k (\partial_i V^m + \Gamma_{il}^m V^l) \Big) \partial_k, \end{equation} which does not coincide with (*). So I strongly suspect that the expression (**) in the paper is not correct.

But I still want to know if it is possible to simplify the local expression (*) to a quantity with a global expression, which may be similar to (**) ? TIA...

Dreamer
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    The usual (somewhat unfortunate) convention in physics (and many math) texts is that the expression $\nabla_i \nabla_j V$ represents the components of $\nabla^2_{\partial_i, \partial_j} V$ (where $\nabla^2$ is the second covariant derivative) and not the components of $\nabla_{\partial_i} \left( \nabla_{\partial_j} V \right)$ (which is not even a tensor in the sense that it does not give you a well-defined global tensor). – levap Aug 20 '21 at 23:42
  • So when $V$ is fixed, $\nabla^2_{\partial_i, \partial_j} V$ is a $(2,1)$-tensor and you can use the metric to contract both inputs and get a vector. The resulting operator is called the "connection Laplacian". – levap Aug 20 '21 at 23:49

2 Answers2

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Expanding the outer covariant operator gives us: $$ g^{ij} \nabla_{\partial_j} (\nabla_{\partial_i} V^k) = \\g^{ij} \Big( \partial_j (\nabla_{\partial_i} V^k) + \Gamma_{jm}^k (\nabla_{\partial_i} V^m) - \Gamma_{jk}^i (\nabla_{\partial_i} V^k) \Big) $$ This has a positive signed Christoffel symbol for the inner contravariant index $k$ and a negative signed one for the covariant inner index $i$. This leads to the full expansion: $$ g^{ij} \nabla_{\partial_j} (\nabla_{\partial_i} V^k) = \\g^{ij} \Big( \partial_j (\partial_i V^k + \Gamma_{im}^k V^m) + \Gamma_{jm}^k (\partial_i V^m + \Gamma_{il}^m V^l) - \Gamma_{jk}^i (\partial_i V^k + \Gamma_{im}^k V^m) \Big) $$ which looks like a combination of the two expressions in your post.

user3257842
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  • Thank you. But what is the usual meaning of $\nabla_{\partial_j} \nabla_{\partial_i} V$? It is $\nabla_{\partial_j} (\nabla_{\partial_i} V)$ or $\langle \nabla_{\partial_j} \nabla V, \partial_i \rangle$? – Dreamer Aug 20 '21 at 22:01
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    I don’t think there is any usual meaning. You have to figure out each time. – Deane Aug 21 '21 at 02:31
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    Given that in $g^{ij} \nabla_{\partial_j} \nabla_{\partial_i} V$ both the indices $i$ and $j$ appear twice (once in $g^{ij}$ as contravariant indices and once in $\nabla_{\partial_j}\nabla_{\partial_i}$ as covariant indices, indicating contraction, I think $\nabla_{\partial_j} \nabla_{\partial_i} V$ means that $\partial_i$ refers to a free index and not a pre-specified direction vector. I tend to write all my formulas in Ricci notation to avoid such ambiguities. Eg the formula would be $g^{ij}{V^k}_{;ij}$. – user3257842 Aug 21 '21 at 05:22
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Expanding out $\nabla_{\partial_j} \nabla_{\partial_i} V$ gives an additional term compared to what you wrote from the outer covariant derivative acting on the Christoffel symbol. More precisely, this gives an additional $(\partial_j \Gamma_{i\ell}^k) V^\ell \partial_k$. Which does not exactly help, but it still notable.

Michael Heins
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  • Thank you. Could you make it more clear? – Dreamer Aug 20 '21 at 23:25
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    Never mind, you have this term yourself: multiplying out your $\partial_j(\partial_i V^k + \Gamma^k_{im} V^m)$ produces it via the Leibniz rule from the second term. Sorry for the confusion. – Michael Heins Aug 21 '21 at 07:30