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I am trying to understand how mixed partials are defined for a function $\gamma : \mathbb R^m \rightarrow M$, where $M$ is an $n$ dimensional manifold, from Peter Petersen's "Riemannian Geometry" (Page 112). Please refer to the book.

Let $\gamma\colon \mathbb{R}^m \to M$. We wish to define the second partials so that they lie in $TM$ as opposed to $TTM$.

Lemma 6 (Uniqueness of mixed partials): There is at most one way of defining mixed partials so that (1) $\frac{\partial^2 \gamma}{\partial t^i \partial t^j} = \frac{\partial^2 \gamma}{\partial t^j \partial t^i}$ and (2) $\frac{\partial}{\partial t^k}g(\frac{\partial\gamma}{{\partial t^i}}, \frac{\partial \gamma}{\partial t^j}) = g(\frac{\partial^2\gamma}{\partial t^k \partial t^i}, \frac{\partial \gamma}{\partial t^j}) + g(\frac{\partial \gamma}{\partial t^j}, \frac{\partial^2\gamma}{\partial t^k \partial t^j})$ both hold.

My question is about lemma 6 (Page 112). I understand how he proves the Koszul type formula and makes an extension of $\gamma$ to $\overline{\gamma}$, but why is that $\frac{\partial^2 \gamma}{\partial t^i \partial t^j} = \frac{\partial^2 \overline{\gamma}}{\partial t^i \partial t^j}$? More specifically, what does the last line of the proof mean and how does he conclude the proof with this last statement (please refer to the link provided by Anthony below)

Thanks!

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April
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1 Answers1

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As set up in the proof, suppose $\gamma: \Omega \to M$, $\Omega \subset \mathbb R^m$ is a $C^2$ map into the Riemannian manifold $M$, represented in a coordinate system by $(\gamma^1, \ldots, \gamma^n)$. We have the freedom to chose our extension, so define $$\overline \gamma : \Omega \times (-\epsilon, \epsilon) \to M$$ satisfying the conditions $\overline \gamma (x,0) = \gamma(x)$, $x \in \Omega,$ and $$\frac{\partial \overline \gamma}{\partial t^{n+1}}|_p = v \in T_p M,$$ where $p = \overline \gamma(x_0, 0)$ and $v$ are chosen arbitrarily. As demonstrated earlier in the proof, there is a Koszul-type formula which separately applies to both $\gamma$ and $\overline \gamma$. Since $\gamma(x) = \overline \gamma(x, 0),$ it follows that $\frac{\partial \overline \gamma}{\partial t^{k}}|_p = \frac{\partial \gamma}{\partial t^{k}}|_p$, $k < n+1$. The Koszul-type formula then gives the uniqueness statement provided that we show that the second partials of $\overline \gamma$ are equal to the second partials of $\gamma$, provided we only differentiate in the $\frac{\partial}{\partial t^k}$ direction, $k < n +1$. But $\gamma$ and $\overline \gamma$ are identically equal on the restricted domain $\Omega := \Omega \times { 0 }$, so any fixed but arbitrary definition of second partial derivative has to give the same answer on $\gamma$ and $\overline \gamma$. This is what Petersen means when he says the definition is independent of the extension.

treble
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    If you look at the development in the book, there is no apriori geometrical assumption of what the second derivative might mean. All that we know is it is a vector in the corresponding tangent space. So the conclusion that the second partials of $\gamma$ and $\overline{\gamma}$ must be equal because both these functions are equal on $\Omega$ doesn't seem right to me. – April Sep 05 '13 at 04:23
  • If the functions are equal on $\Omega$ then they are the same function, so all I'm asking for is that you believe the concept to be well-defined, and that $\Omega$ and $\Omega \times 0$ be regarded as the same set. If you don't believe these things, then I think Lemma 6 is not justified by Petersen's argument, as you say. But he is being a little informal. – treble Sep 05 '13 at 16:27
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    Well, the whole point of the section is to define the second derivative/mixed partials. I have no problem with looking at the sets $\Omega$ and $\Omega \times 0 $ as the same and the corresponding functions as the same, hence I do not have a problem of considering $\frac{\partial \gamma}{\partial t^i} =\frac{\partial \overline{\gamma}}{\partial t^i}$ when $i < n+1$, but the second derivatives is a totally new concept which he hasn't defined and all this work is for defining it. I hope you understand my point. – April Sep 05 '13 at 17:39
  • But if two functions are the same, then their second partials, whatever they are, must be the same as well (Lemma 6 is of course just a uniqueness statement, so we do assume there is at least one way to define second partial derivatives). Otherwise the concept is not well defined. – treble Sep 05 '13 at 18:47
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    But why is it that $\frac{\partial^2 \overline{\gamma}}{\partial t^i \partial t^ j}$ depends only on the behavior of $\overline{\gamma}$ on $\Omega \times 0$ ? The only thing we are assuming about the second partial derivative is that it satisfies the 2 conditions mentioned in the lemma above and you can't conclude the statement above from these 2 conditions. – April Sep 05 '13 at 19:32