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I thought I'd try my luck with another question, so here it goes:

I have to show that if $(y^1,...,y^n):\mathbb R^n \rightarrow \mathbb R^n$ is a diffeomorphism and $f\in C^\infty(\mathbb R^n)$. Then the Laplacian in $\mathbb R^n$ $\nabla_{\mathbb R^n} = \sum_{j=1}^n(\frac{\partial}{\partial x^j})^2$ on the composite function $f$ is

$\nabla_{\mathbb R^n} f(y^1(x^1,...,x^n),...y^n(x^1,...x^n)) = (\widetilde{\nabla}f)(y^1(x^1,...,x^n),...,y^n(x^1,...,x^n))$

where $\widetilde{\nabla} = \sum_{i,j=1}^n g^{ij}((\frac{\partial}{\partial y^i} \frac{\partial}{\partial y^j}) - \sum_{k=1}^n \Gamma_{ij}^k \frac{\partial}{\partial y^k})$

and $g^{ij}=\sum_{l=1}^n \frac{\partial y^i}{\partial x^l}\frac{\partial y^j}{\partial x^l}$

and $\Gamma_{ij}^k = \sum_{l=1}^n \frac{\partial y^k}{\partial x^l} \frac{\partial ^2 x^l}{\partial y^i \partial y^j}$ are the Christoffel symbols

So far I've tried to just write out the Laplacian but I got stuck on the second derivative.

My Calculations:

$\nabla_{\mathbb R^n} f(y^1(x^1,...,x^n),...y^n(x^1,...x^n)) = \sum_{j=1}^n (\frac{\partial}{\partial x^j})^2(f) = \sum_{j=1}^n \frac{\partial}{\partial x^j}(\frac{\partial f}{\partial y^j}\frac{\partial y^j}{\partial x^j}) = \sum_{j=1}^n \frac{\partial ^2 f}{\partial (y^j)^2}(\frac{\partial y^j}{\partial x^j})^2 + \frac{\partial f}{\partial y^j}\frac{\partial ^x y^j}{\partial (x^j)^2}$

but then I'm not quite sure what I should do next. Can anyone help me with this? Thanks in advance!

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You are basically done. I think you copied down $\tilde\nabla$ wrong though: the first term is a "scalar" while the second term has two free indices. – Willie Wong Jan 3 '13 at 15:14
@WillieWong oh, sorry about that, forgot a pair of brackets, other than that it's what I've got in my notes. I'm afraid I still don't quite see the connection between the two expressions, especially because the right hand side seems to sum over so many different indices. – Longeyes Jan 3 '13 at 15:27
I still don't think you have put in the brackets correctly. Look again. The $g^{ij}$ should also act on the second term. Also, the evaluation of $$\frac{\partial}{\partial x^j} \left( \frac{\partial f}{\partial y^i} \frac{\partial y^i}{\partial x^j} \right) $$ is not strictly correct, now that I look at it more carefully. You need to do your chain/product rules paying more attentions to summations and indices. – Willie Wong Jan 3 '13 at 15:57
@WillieWong Sorry again! You're right of course XD However now I'm completely lost as I don't really understand how to apply $\widetilde{\nabla}$ to $f$ – Longeyes Jan 3 '13 at 16:10
up vote 3 down vote accepted

This is a computation in which you can largely follow your nose.

A direct evaluation by chain and product rules gives

$$ \nabla_{\mathbb{R}^n}f(y^1(\ldots), \ldots , y^n(\ldots)) = \sum_{i,j ,k = 1}^{n} \frac{\partial^2 f}{\partial y^i \partial y^j} \frac{\partial y^j}{\partial x^k} \frac{\partial y^i}{\partial x^k} + \sum_{i,j = 1}^n \frac{\partial f}{\partial y^j} \frac{\partial^2 y^j}{\partial x^i \partial x^i} $$

The first term in the expression is clearly equal to $ \sum_{i,j} g^{ij} \frac{\partial^2 f}{\partial y^i \partial y^j}$ as desired. To read off the second term we use the change of variables formula for Christoffel symbols. Using that the Christoffel symbol for the Euclidean metric in the $x$ (standard) coordinate system vanishes, the change of variable formula (look at here but switching the $y$ and $x$ coordinates) gives $$ 0 = \sum_{ij} \frac{\partial y^i}{\partial x^p} \frac{\partial y^j}{\partial x^q} \Gamma^k_{ij} + \frac{\partial^2 y^k}{\partial x^p \partial x^q}$$ which when substituted into the second term gives the correct Christoffel symbol expresion.

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So the second term I managed to set equal to<br> $-\sum \frac{\partial y}{\partial x^i} \frac{\partial y}{\partial x^i} \Gamma ^j \frac{\partial f}{\partial y^k}$<br> But over which indices do I do the summation and which y components do derive in the $\frac{\partial y}{\partial x^i}$ expressions? – Longeyes Jan 3 '13 at 17:13
Also thank you so much for your help!! – Longeyes Jan 3 '13 at 17:15
The second equation I wrote is meant to be parsed as written: the second term inside has no summations. To plug it in you have to replace both $p$ and $q$ by $i$ and sum over it. – Willie Wong Jan 3 '13 at 17:23
Oh! So then if in your first equation I replace $i$ by $l$ and also replace both $p$ and $q$ by $l$ I get the expression with the $l$ from the the defintion of $g^{ij}$ right? I can't thank you enough! XD – Longeyes Jan 3 '13 at 19:08
Yes. That's right. Good luck. – Willie Wong Jan 4 '13 at 8:22

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