# Differential Forms and Applications by do Carmo - Divergence theorem

I read the proof of Stokes theorem for manifolds by do Carmo's book and I'm trying understand an example (the Divergence theorem) given after the proof of Stoke's theorem, but I didn't understand. A preliminary definition of star Hodge operator $$*$$ given by do Carmo which is used in the example can be found here. The example is this. The map $$i$$ is the inclusion map $$i: \partial M \longrightarrow M$$.

I would like to know why $$i^* *\omega(e_1,e_2) = \omega (N)$$.

I thought that $$i^* *\omega(e_1,e_2) = *\omega(di(e_1),di(e_2)) = *\omega(e_1,e_2) = \omega(N)$$, but this seems incorrect to me, because we can indentify $$\partial M$$ as $$\mathbb{R}^2$$ and $$M$$ as $$\mathbb{R}^3$$ and see the inclusion map as

$$i: \mathbb{R}^2 \longrightarrow \mathbb{R}^3$$

$$(x_2, x_3) \mapsto (0,x_2,x_3)$$

due to this proposition of do Carmo's book, then

$$di_p = \begin{pmatrix} 0 & 0\\ 1 & 0\\ 0 & 1 \end{pmatrix}$$, $$di_p (e_1) = (0,1,0) = a_2$$ and $$di_p (e_2) = (0,0,1) = a_3$$, therefore

$$i^* *\omega(e_1,e_2) = *\omega(di(e_1),di(e_2)) = *\omega(a_2,a_3) = \omega(a_1),$$

where $$a_1 = (1,0,0)$$, but don't know what to do now, because I don't know if $$a_1 = N$$.

I will appreciate any comment in order to elucidate the reason why $$i^* *\omega(e_1,e_2) = \omega (N)$$ is valid. Thanks in advance!

A mistake I noticed is that you say you can identify $$\partial M$$ with $$\mathbb{R}^{2}$$, when really it should be $$T_{p}\partial M$$ with $$\mathbb{R}^{2}$$ for any point $$p \in \partial M$$ (and similarly for $$T_{p}M$$ with $$\mathbb{R}^{3}$$).
So with that in mind, let the inclusion map $$i:\partial M \rightarrow M$$ be given by $$i(x_{1}, x_{2}) = (x_{1}, x_{2}, 0),$$ then in letting $$p = (x_{1}, x_{2})$$, in a neighbourhood $$U \subset \partial M$$ with $$p \in U$$, let $$\{e_{1}, e_{2}\}$$ be an orthonormal frame for $$T_{p}\partial M$$. Then $$d_{p}i:T_{p}\partial M \longrightarrow T_{i(p)}M,\qquad (e_{1}, e_{2}) \longmapsto (e_{1}, e_{2}, 0).$$ The tangent space $$T_{i(p)}M$$ can be seen to decompose as $$T_{i(p)}M = T_{p}\partial M\ \oplus\ N_{i(p)}M,$$ where $$N_{i(p)}M$$ is the normal space to $$M$$ at the point $$i(p)$$, i.e. the orthogonal complement to $$T_{p}\partial M$$, which is one-dimensional and spanned by $$N$$, say. Then $$e_{1}, e_{2}, N$$ forms an orthonormal frame for $$T_{i(p)}M$$.
So then, in a calculation similar to what you have already done, $$i^{\ast} \star \omega(e_{1},e_{2}) = \star \omega \big(d_{p}i(e_{1}), d_{p}i(e_{2})\big) = \star \omega (e_{1}, e_{2}) = \omega(N),$$ in considering $$\omega$$ to be a 1-form on $$T_{i(p)}M \cong \mathbb{R}^{3}$$, $$\star \omega$$ for be a 2-form on $$T_{i(p)}M \cong \mathbb{R}^{3}$$, and finally $$i^{\ast}\star\omega$$ to be a 2-form on $$T_{p}\partial M \cong \mathbb{R}^{2}$$.
• thanks for correct me! Why $di_p$ maps $(e_1,e_2)$ to $(e_1,e_2,N)$? Are you assuming that the map $i$ is as I defined? – George Nov 8 '18 at 21:57
• When do Carmo prove the Stokes theorem, he states that "the inclusion map $i$ can be written as: $x_1 = 0$, $x_j = x_j$" for the case that a coordinate neighborhood intersects $\partial M$, so $i$ is defined as I defined in the OP, but I can't see why $di_p$ will be exactly what you said, because I think $e_1 \mapsto (0,1,0)$, $e_2 \mapsto (0,0,1)$ as I computed in the OP. – George Nov 8 '18 at 22:51
• I think you wanted other thing where you wrote $(e_1,e_2) \mapsto (e_1,e_2,0)$, because this is not make sense since we identify $T_p \partial M \cong \mathbb{R}^2$ and $T_p M \cong \mathbb{R}^3$ and, in this sense, $e_1, e_2 \in \mathbb{R}^2$ and not coordinates as you put when you wrote $(e_1,e_2) \mapsto (e_1,e_2,0)$ – George Nov 8 '18 at 23:02
• You will get the same result whether the inclusion map has $x_{1} = 0$ or $x_{3} = 0$, since the orientation of the orthonormal frame will be the same in either case. I have added more clarification to the inclusion map as its derivative now, hopefully that differentiates between what is a coordinate point and what is a tangent vector. – BenCWBrown Nov 8 '18 at 23:19