# Gauss-Green Theorem from generalized Stoke's Theorem.

I am trying to deduce the next identity (Green-Gauss theorem) $$\int_\Omega \dfrac{\partial u}{\partial x_i} dx = \int_{\partial \Omega} uv_i dS$$ from the generalized Stoke's theorem for manifolds. Here $u$ is a $C^\infty$ function on an open bounded set $\Omega$ with $C^1$ boundary and $v_i$ denotes the i-th component of the unit normal vector field.

I've been trying to define $n-1$ forms on $\Omega$ whose derivatives are $\dfrac{\partial u}{\partial x_i} dx$ to apply the theorem but I cannot prove that those forms have the same integral as $uv_i$ on $\partial \Omega$.

Any ideas?

Also, I would really appreciate references of the proofs of classical calculus integration results using the generalized Stoke's theorem.

• As stated, this does not make sense, since the integrand on the LHS is a $1$-form, which you can only integrate over a curve. Actually, if your coordinates are $x_i$, I do not know what $dx$ all by itself means. Did you mean $dV$ here? – Steven Gubkin Oct 30 '14 at 18:09
• Is just a matter of notation, $dx$ means the volume form on $\mathbb{R} ^{n}$, $dx = dx_1 dx_2 ... dx_n$ if you want to see it that way. – Frank Trujillo Oct 30 '14 at 19:30
• Ah, then I can answer your question. – Steven Gubkin Oct 30 '14 at 19:48

First of all, you need to understand what $dS$ is.

$dS$ is the form $dV$ contracted by the vector field $\nu$ on $\partial \Omega$. In other words, to find $dS(v_2,v_3,...,v_{n})$ for tangent vectors $v_2,v_3,...,v_{n}$ to $\partial \Omega$, you just evaluate $dV(\nu, v_2,v_3,...,v_{n})$. The motivation for this definition is that the area of a parallelogram $P$ is the same as the volume of the parallelepiped with base $P$ and height $1$.

So now note that,

$\frac{\partial f}{\partial x_1} dV = d(f dx_2 \wedge dx_3 \wedge ...\wedge dx_n)$

Letting $\widehat{dx_1} = dx_2 \wedge dx_3 \wedge ...\wedge dx_n$

Let $p \in \partial \Omega$ and $v_2,v_3,...,v_n \in T_p( \partial \Omega)$ then

\begin{align*} f(p) \widehat{dx_1}(v_2,v_3,...,v_n) &= f(p) dV(e_1,v_2,...,v_n)\\ &=f(p) dV(\textrm{proj}_{\nu(p)}(e_1),v_2,...,v_n)\\ &=f(p) dV(\nu_1(p)\nu(p),v_2,...,v_n)\\ &=f(p)\nu_1(p) dV(\nu(p),v_2,...,v_n)\\ &=f(p)\nu_1(p) dS(v_2,...,v_n) \end{align*}

so as forms on $\partial S$, we have $f\widehat{dx_1} = f\nu_1dS$. The big step is the second line in the equality above, which is justified by writing $e_1$ as a linear combination of $\nu$ and $v_i$'s. All of the $v_i$ terms die because of the alternating property of forms.

Now we conclude that

\begin{align*} \int_\Omega \frac{\partial f}{\partial x_1} dV &= \int_\Omega d(f \widehat{dx_i})\\ &=\int_{\partial \Omega} f \widehat{dx_i} \text{by Stokes' theorem}\\ &=\int_{\partial \Omega} f \nu_1 dS \end{align*}

If any of this does not make sense, please ask for follow up in the comments. Notation in differential geometry is horrible, and everyone makes it up for themselves I think. So I understand if some part of this is hard to parse. I am also no differential geometer, but I do a lot of calculus in high dimensional spaces (I work in several complex variables). This kind of thing used to frustrate me a lot.

• There is some care needed with signs if you look at $x_i$ instead of just $x_1$ as I did here. But I think we get a $(-1)^i$ once when rewritting the original form as the exterior differential of something else, and we get another $(-1)^i$ when we contract by the vector field $e_i$ later on, so the conclusion you seek is still true without any sign worries. Sticking to $x_1$ let me glide over this issue, which I thought would be easier as a first pass. – Steven Gubkin Oct 30 '14 at 20:22
• Also I realized that you may not have the same perspective on forms as I do. I know in some presentations they are "just symbols of the form...", with various definitions of integration. My definition of a form on a vector space is an alternating multilinear map to the base field. A form on a manifold is then a smooth section of a form bundle. – Steven Gubkin Oct 30 '14 at 20:53
• Don't worry, I was aware of the sign and I also understand the notation. That was exactly the form that I was trying, thank you. You were right, I was not seeing the second equality. – Frank Trujillo Oct 30 '14 at 21:21

Steven's answer is good, let me reformulate it in more formal flavor. By Stokes, it suffices to prove $$l^*(dx^2 \wedge...\wedge dx^n) = dS := l^*(i_v(dx^1\wedge...\wedge dx^n))$$ Where $l:\partial U \to U$ the inclusion map, $i_v$ is the contraction operator with respect to $v$. Let $i_{\frac{\partial}{\partial x^1}}$ be the contrition operator with respect to $\frac{\partial}{\partial x^1}$, then we have $l^*(dx^2 \wedge...\wedge dx^n) = l^*i_{\frac{\partial}{\partial x^1}}(dx^1,...,dx^n)$. Then we just need to prove $l^*i_{\frac{\partial}{\partial x^1}}(dx^1,...,dx^n) = \langle \frac{\partial}{\partial x^1}, v \rangle l^*(i_v(dx^1\wedge...\wedge dx^n))$. To see this, decompose $\frac{\partial}{\partial x^1} = \langle \frac{\partial}{\partial x^1}, v \rangle v + \alpha$, where $\alpha$ is tangent to $\partial U$.(essentially we used Lemma 16.30 in Lee's introduction to smooth manifold.), then we have $$l^*i_{\frac{\partial}{\partial x^1}}(dx^1,...,dx^n) = \langle \frac{\partial}{\partial x^1}, v \rangle l^*(i_v(dx^1\wedge...\wedge dx^n))+l^*i_\alpha (dx^1\wedge...\wedge dx^n)$$ Now it remains to prove $l^*i_\alpha (dx^1\wedge...\wedge dx^n) = 0$, which is because choosing $v_2,...,v_n$ tangent to $\partial U$, then $\alpha$ and $v_2,...,v_n$ must be linear dependent thus have zero determinant, so $l^*i_\alpha (dx^1\wedge...\wedge dx^n)(v_2,...,v_n)=(dx^1 \wedge...\wedge dx^n)(\alpha, v_2,...,v_n)=0$