2

We know that for a vector field $X$, $\operatorname{div}(X)$ is defined as $\nabla_I X^I$. This is not the same as $\partial_i X^i$ right?

I would assume that $\nabla_I X^I=\partial_i X^i-X^j\Gamma_{ij}^i $.

Logos
  • 644

1 Answers1

3

Yes, you are correct.

Another way to expand the divergence on a Riemannian manifold in local coordinates is to take the expression similar to what you find in the Laplace-Beltrami operator:

$$ X^i \partial_i \mapsto \frac{1}{\sqrt{|g|}} \partial_i \left( \sqrt{|g|} X^i \right) $$

and you see that $\nabla_I X^I$ and $\partial_i X^i$ are equal only when either the volume element relative to the local coordinate chosen has a critical point, or that $X^i$ itself vanishes at that point.

Willie Wong
  • 73,139
  • Which leads me to the question, is $\partial _i V^j +\partial _j V^i=\nabla_I V^J + \nabla_J V^I$? I am getting the former in some calculation (of the Lie derivative of a metric), but I want the latter –  Jul 31 '20 at 13:19
  • The expression as you have written makes no sense. You are adding a term with a lowered $I$ index to a term with a raised $I$ index: you cannot add two things of two different types. – Willie Wong Jul 31 '20 at 13:35
  • If you post your computation of the metric Lie derivative as a separate question, and ping me here, I can take a look. – Willie Wong Jul 31 '20 at 13:37
  • Doesn't $\nabla_I V^J$ just mean $(\nabla_I V)^J$, which is the $Jth $ components of the vector $\nabla_I V$? Why do I have to add something? –  Jul 31 '20 at 13:38
  • I see. But $\partial _i V^j$ I suppose still makes sense, because I am differentiating the $jth$ component of $V$, which can be thought of as a function, in the $i$th direction? Sorry I'm just trying to find out where I am making a mistake so I can correct it –  Jul 31 '20 at 13:42
  • Oh I see. Your point is that the right hand side is a tensor, while the left hand side seems to be a function. I meant for the left hand side to be multiplied with $\partial _j\otimes dx^i$. I suppose I should have been clearer. –  Jul 31 '20 at 13:44
  • Notation aside, I am sure you made a mistake somewhere. If you want someone to look over your work computing the Lie derivative of the metric, as I said above: please post a new question (and ping me here if you wish). – Willie Wong Jul 31 '20 at 13:47
  • Thanks so much, I'll do that in a bit –  Jul 31 '20 at 13:48
  • Hi, I think I finally got the calculation correctly. I was trying to prove that $(L_X g)_{IJ}=\nabla_I X_J+\nabla_J X_I$. I have now realized that this is only true when $X$ is a Killing vector (I am also working in an orthonormal basis. I suppose this doesn't matter). –  Jul 31 '20 at 18:30
  • Wait, that's not right. When $X$ is Killing, $L_Xg \equiv 0$. The formula $(L_Xg)_{IJ} = \nabla_I X_j + \nabla_J X_I$ holds in general. – Willie Wong Jul 31 '20 at 18:56
  • Yes sorry, you're right. That was a careless remark. This is true for all vector fields. –  Jul 31 '20 at 18:57
  • I thought $\nabla_X g=0$ is true for $X$ a Killing vector field, so I was confusing $\nabla$ with Lie derivative. –  Jul 31 '20 at 18:58