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I was studying about the divergence of a smooth vector field in the book "Calculus of variations and harmonic maps"by Urakawa. For a smooth vector field $X$, the divergence is defined by $div(X)(p) := g(e_i,\nabla_{e_i}X)(p) ; p \in M$ wher $\{e_i\}_{i=1}^n$ is a local orthonormal frame field.

Further, it was written that the definition does not depend on the choice of $\{e_i\}$but i am not able to prove that $g(e_i,\nabla_{e_i}X)(p)= g(f_i,\nabla_{f_i}X)(p)$ if $\{e_i\}$ and $\{f_i\}$ are two local orthonormal frame fields.Can someone tell me how to prove this?

Thanks!

user103005
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    you might notice my proof does not really rely on Riemannian geometry, it would also apply if we replace $\delta_{ij}$ with $\eta_{ij}$ where $\eta = diag(-1,1,1,\dots, 1)$. As it stands, the matrix of $P$ is a rotation whereas in the case of $\eta$ it would be a Lorentz transformation (they preserve $\eta$)... – James S. Cook Mar 11 '14 at 04:54

2 Answers2

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The proof is based on the assumed orthonormality of the frames.

We assume there exists an isometry $P(e_i)=f_i$ thus as $g(f_i,f_j)=g(e_i,e_j) = \delta_{ij}$ (assuming Riemannian geometry here) $g(P(e_i),P(e_j))=g(e_i,e_j)$. In fact, in general, $g(P(v),P(w))=g(v,w)$ for the isometry $P$.

Note, the covariant derivative has $$ \nabla_{P(e_i)} X = P(\nabla_{e_i} X). $$ Collecting our thoughts, $$ g(f_i, \nabla_{f_i} X)(p) = g(P(e_i),P(\nabla_{e_i} X))(p) = g(e_i, \nabla_{e_i} X)(p). $$

James S. Cook
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Let $a_{ij}$ be a matrix such that $f_i = a_{ij} e_j$. Let both $\{e_i\}$ are $\{f_j\}$ are orthonormal basis, we have $\sum_{i=1}^n a_{ij}a_{ik} = \delta_{jk}$ (That is, $A = (a_{ij})\in O(n)$). thus

$$g(f_i , \nabla_{f_j} X) = a_{ij}a_{ik}\ g(e_j, \nabla_{e_k} X) = \delta_{jk}\ g(e_j, \nabla_{e_k} X) = g(e_j , \nabla_{e_j}X)$$

The answer given by Cook is good (and are just the same), but I think this answer should also be recorded as these arguments (writing out as $a_{ij}$) show up all the time.