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!

up vote 1 down vote accepted

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 at 21:57
  • Sorry, just realised I had made a mistake regarding that part! I have edited it so hopefully it makes sense now. I had a look at Do Carmo's book, so it should be the same as what you have. – BenCWBrown Nov 8 at 22:33
  • 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 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 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 at 23:19

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