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I am attempting to prove that divergence of a smooth vector field $X$ over a $n$-dimensional Riemannian manifold $(M,g)$ is invariant under change of coordinates (or invariant of the choice of frame).

I am stuck in the proof and here is my attempt.

I am starting from the formula given in [p. 33, 1]. So, we have for $$X=X^i \frac{\partial}{\partial x^i}$$ the divergence is $$ \text{div}\left( X^i \frac{\partial}{\partial x^i}\right) = \frac{1}{\sqrt{\det g_{ij}} }\frac{\partial}{\partial x^i}\big( X^i \sqrt{\det g_{ij}}\big). $$

As a first step I am trying to prove that the RHS of that formula is the same for two different orthonormal frames.

So, I assume an orthonormal frame $\{E_i,E_2,...,E_n\}$ with dual frame $\{e^1,e^2,...,e^n\}$. First of all in this frame we have $$X^i = e^i(X)$$ and due to orthonormality we obtain $$ \det g_{ij} = 1.$$ I shall use $\text{div}(X)_{Ee}$ to emphasize the dependence on the frame. So we obtain $$\text{div}(X)_{Ee} = E_i(e^i(X))$$

Now I consider another orthonormal frame $\{F_i,F_2,...,F_n\}$ with dual frame $\{f^1,f^2,...,f^n\}$. For this frame we have $$\text{div}(X)_{Ff} = F_i(f^i(X))$$

Now, to prove invariance under the two frames, I need to prove $$\text{div}(X)_{Ff} = \text{div}(X)_{Ee}$$

I expand the second frame in terms of the first as $$\begin{eqnarray} F_i &=& F_i^l E_l \\ f^i &=& f^i_m e^m\end{eqnarray}$$

Then we have $$\begin{eqnarray} \text{div}(X)_{Ff} &=& F_i(f^i(X)) \nonumber \\ &=& F_i^lE_l(f^i_m e^m(X)) \nonumber \\ &=& F_i^lE_l(f^i_m) e^m(X) + F_i^lf^i_m E_l(e^m(X)) \\ &=& F_i^lE_l(f^i_m) e^m(X) + \delta^l_m E_l(e^m(X)) \\ &=& F_i^lE_l(f^i_m) e^m(X) + \text{div}(X)_{Ee} \end{eqnarray}$$

Where the third step is from Leibniz's rule and the fourth step is because the matrices $[f_m^i]$ and $[F_i^l]$ are inverses of each other.

So essentially I need to prove $$F_i^lE_l(f^i_m) e^m(X) = 0$$ and I am not able to do this, can someone help?

[1] Lee, John M. Introduction to Riemannian manifolds. Vol. 176. Springer, 2018.

P.S. I did go through Divergence of a smooth vector field and that answer uses the covariant derivative. I am attempting to avoid using it.

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I don't understand what you're trying to do here. The formula you're using ($\text{div}(X)_{Ee} = E_i(e^i(X))$) doesn't appear in my book, and it's not correct, except in a flat Riemannian manifold with a coordinate frame.

Page 32 of my book gives a definition of $\text{div}\, X$ that's manifestly coordinate-invariant: it's the unique function that satisfies $$ d(X\mathrel{\lrcorner} dV_g) = (\text{div}\, X) dV_g.$$ Then on page 33, there's a coordinate formula that works in any local coordinates: $$ \text{div}\left( X^i \frac{\partial}{\partial x^i}\right) = \frac{1}{\sqrt{\det g} }\frac{\partial}{\partial x^i}\big( X^i \sqrt{\det g}\big). $$ This reduces to the familiar vector calculus formula in the special case of standard coordinates on Euclidean space.

It's an interesting exercise to prove that the general coordinate formula follows from the coordinate-free definition. You can also try to prove directly that the coordinate formula transforms correctly when you change coordinates, but that's a lot messier and not necessary.

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  • $\begingroup$ You are right, my question is vague. I have added more steps now showing how I arrived at my formula for divergence. I hope I have made my question clear and hopefully you can spot the error in my attempt. $\endgroup$
    – Coniferous
    Oct 16 '19 at 17:19
  • $\begingroup$ Your mistake is in assuming that the two frames are both orthonormal and coordinate frames. That is possible only on a flat Riemannian manifold. $\endgroup$
    – Jack Lee
    Oct 16 '19 at 18:25
  • $\begingroup$ @Coniferous: (Look at the warning near the bottom of page 14.) $\endgroup$
    – Jack Lee
    Oct 16 '19 at 18:29
  • $\begingroup$ Yes, I had mistakenly wrote that the orthonormal frames are coordinate frames. I have corrected that in the text now. None of the two frames I use are coordinate frames. In fact what I want to do is what you mentioned: "You can also try to prove directly that the coordinate formula transforms correctly when you change coordinates". Am I doing it right? I kind of feel I am making a mistake somewhere, but cannot spot where. $\endgroup$
    – Coniferous
    Oct 17 '19 at 16:36
  • $\begingroup$ @Coniferous: You wrote "None of the two frames I use are coordinate frames." But that formula you're working with applies only to coordinate frames. $\endgroup$
    – Jack Lee
    Oct 18 '19 at 14:05
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Say, $E_i = \dfrac{\partial}{\partial u^i}$ and $F_i = \dfrac{\partial}{\partial v^i}$, where $\mathbf{u}$ and $\mathbf{v}$ are the respective coordinate systems.

Notice that $F_i(f^i_m) =0$ because this would involve a term of the form$\dfrac{\partial^2 v^1}{\partial v^1 \partial u^{\{1,2\} }}$ which is zero. \begin{align} F_i(f^i(X))= F_i(f_m^i e^m(X))& = F_if^i_m e^m(X) + f^i_m F_i(e^m(X)) \\ &= 0+ E_m(e^m(X)) \end{align}

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  • $\begingroup$ Can you pleas explain the notation $\dfrac{\partial^2 v^1}{\partial v^1 \partial u^{\{1,2\} }}$? I am not faimliar with the curly bracket notation $\{1,2\}$ yet! $\endgroup$
    – Coniferous
    Oct 16 '19 at 20:53
  • $\begingroup$ $\dfrac{\partial^2 v^1}{\partial v^1 \partial u^{\{1,2\} }} = 0$ means that both $\dfrac{\partial^2 v^1}{\partial v^1 \partial u^{1 }} = 0$ and $\dfrac{\partial^2 v^1}{\partial v^1 \partial u^{2 }} = 0$ $\endgroup$ Oct 17 '19 at 3:41
  • $\begingroup$ If $(u^1,\dots,u^n)$ and $(v^1,\dots, v^n)$ are different coordinate systems, then there's no reason to expect $\partial v^1/\partial u^1$ to be zero, and thus $\partial^2 v^1/\partial v^1\partial u^1$ won't be zero either. $\endgroup$
    – Jack Lee
    Oct 17 '19 at 14:46
  • $\begingroup$ @JackLee: My understanding is that $\dfrac{\partial v^1}{\partial v^1} = 1$, hence, $\dfrac{\partial^2 v^1}{\partial u^1 \partial v^1} = 0$. Also, order of partial derivatives does not matter. Please correct me, if I am wrong. $\endgroup$ Oct 18 '19 at 1:13
  • $\begingroup$ Order of partial derivatives doesn't matter if you're taking derivatives with respect to different variables from the same coordinate chart. There's no reason to expect partial derivatives from different charts to commute with each other. $\endgroup$
    – Jack Lee
    Oct 18 '19 at 13:59

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