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I was trying to work out a integration by parts formula $S^2$ of the form $$\int_{S^2}f(x)\frac{\partial g}{\partial x_1}dx \tag{1}$$ where $f,g:\mathbb{R}^3\rightarrow\mathbb{R}$. Given $g$ and $\tilde{g}$ that agree on $S^2$, I am not sure if (1) independent from which of the two we choose.

Having realized this I tried two work purely on the manifolds. Does anyone know a good reference that explains how to use Stokes Theorem and an the volume form associated with the Riemanian metric to derive a integration by parts formula.

Thank you,

warsaga

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What is $\widetilde{g}$? –  Neal Nov 4 '12 at 16:55
    
$\tilde{g}$ is another function $\tilde{g}:\mathbb{R}\rightarrow\mathbb{R}$. $g=\tilde{g}$ on $S^2$. Is (1) dependent on the particular choice of $g$? –  warsaga Nov 4 '12 at 17:48

2 Answers 2

Just to give an alternative way of looking at things, here's a look at integration by parts through Clifford (geometric) algebra.

$$\oint_{\partial V} A(x) \; dS \; B(x) = \int_V \dot A(x) \dot \nabla \cdot dV \; B(x) + \int_V A(x) \dot \nabla \cdot dV \; \dot B(x)$$

The overdots denote which function the vector derivative $\nabla$ acts on, even though its vector character has a certain position within the expression. Note that $dS, dV$ are not scalars but rather multivectors appropriate to the type of object being integrated over. All products are the "geometric" product, combining the inner and outer products, except where others are explicitly denoted by $\cdot$ or $\wedge$.

An integration by parts formula follows from isolating one of the "volume" integrals on the right. Nevertheless, if you have this integral over a sphere and you want to use integration by parts, I don't think you're going to find much help if you want to consider the sphere $V$ in this problem, as a sphere has no boundary. If you consider the sphere $\partial V$, the boundary of the volume, then you will get terms involving further derivatives of $\partial g/\partial x_1$. Furthermore, you said that $g,\tilde g$ agree on the sphere, but nothing about their derivatives. It seems to me that there's no need for your (1) to be the same between the two because you've explicitly written it with a derivative.

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Here's a general integration-by-parts using Stokes' theorem. I saw this in class, so I don't recall reading this in a text, but I'm sure it's covered in any text that covers differential forms --- try Principles of Mathematical Analysis (Rudin), or Differential Topology (Guillemin-Pollack).

Remember that if $\alpha,\beta$ are respectively $k-1$ and $n-k$-forms on $M$, then $$d(\alpha\wedge\beta) = (d\alpha)\wedge\beta + (-1)^k\alpha\wedge (d\beta).$$ We have required that $\alpha$ and $\beta$ be $k-1$ and $n-k$ forms so that $d(\alpha\wedge\beta)$ is a multiple of the volume form and so can be integrated against $M$ instead of a submanifold of $M$.

Now we can use this formula and Stokes' theorem: \begin{align*}\int_M d\alpha\wedge\beta &= \int_M \bigg(d(\alpha\wedge\beta) - (1)^k\alpha\wedge (d\beta)\bigg) \\ &= \int_{\partial M}\alpha\wedge\beta + (-1)^{k+1} \int_M \alpha\wedge (d\beta) \end{align*} Note that if $M$ is closed, the term integrating over the boundary vanishes.

This formula recovers integration by parts on the real line. Let $M = [a,b]\subset \mathbb{R}$, so that $\partial M = \{a\}\cup\{b\}$. If $f$ is a function, then $df = \frac{df}{dx}dx$. Integrating by parts is expressed by: $$\int_a^b \frac{df}{dx}gdx = \int_a^b df\wedge g = \int_{\partial M} fg - \int_a^b f\wedge dg = (fg)\bigg|_a^b - \int_a^bf\frac{dg}{dx}dx.$$ Integration of the zero-form $fg$ over $\partial M$ is signed evaluation.


In response to the question in comments below, integrating a top-dimensional form is exactly the same as integrating a function against the measure induced by the volume form. On an orientable manifold $M$, the top exterior bundle $\Lambda^nM$ is a trivializable rank $1$ bundle. Choice of a volume form $Vol$ trivializes the bundle, hence induces an isomorphism ($C^\infty(M)$-modules) $\Omega^n(M)\cong C^\infty(M)$, where $1\leftrightarrow Vol$. So integrating a function against the volume measure is the same as integrating its corresponding $n$-form.

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Thanks for the good answer. How does this relate to integrating functions (not forms) against the measure associated with the riemanian metric. –  warsaga Nov 4 '12 at 17:52
    
I'll try to find time tomorrow morning to edit additional information in to my answer. –  Neal Nov 5 '12 at 3:36
    
I appreciate your help. –  warsaga Nov 6 '12 at 15:10
    
@warsaga I'm sorry to not get to the edit - I'm having an extremely busy week. Just letting you know I haven't forgotten. –  Neal Nov 8 '12 at 12:53
    
It would be great if you find time to explain it. –  warsaga Nov 15 '12 at 10:30

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