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I have a problems doing exercise 16 of chapter 3 (p.98 in my edition) of Spivak's book. The problem is very simple. Let $M$ be a manifold with boundary, and choose a point $p\in\delta(M)$. Now consider an element $v\in T_p M$ which is not spanned by the vectors on $T_p\delta(M)$, that is, it's last coordinate is non-zero (after good identifications). We say that $v$ is inward pointing if there is a chart $\phi: U\rightarrow \mathbb{H}^n$ ($p\in U$) such that $d_p\phi(v)=(v_1,\dots,v_n)$ where $v_n>0$.

It is asked to show that this is independent on the choice of coordinates (on the chart).

I think that Spivak's idea is to realize first that the subespace of vectors in $T_p\delta M$ is independent on the chart, which can be seen noticing that if $i:\delta (M)\rightarrow M$ then $d_pi(v_1,\dots,v_{n-1})=(v_1,\dots,v_{n-1},0)\in T_p \mathbb{H}^n$

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"Spivak's book" ? Spivak has writtten more than one book, and as far as I know he defines in all of them "outward pointing" by $v_n\lt 0$ ... –  Georges Elencwajg Jan 10 '13 at 19:34
    
The book is, as I said in the title of the question, A comprehensive introduction to differential geometry vol1. And I don't think that the sign of the last coordinate affects the core of the question, take $v_n<0$ as definition of outward pointing if you want. –  Miguel Jan 11 '13 at 2:42
    
It is preferred on this site that the body of the question be independent of its title. Also, although the core of a question is indeed not affected by the convention, words have an accepted meaning in mathematics and you are not allowed (here or elsewhere) to change a universally adopted convention nor to misquote an author. –  Georges Elencwajg Jan 11 '13 at 8:22
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Ok, I thought this place was meant to be more informal!, anyway I will be more careful in the future, post the whole references and not to misquote things. Sorry! –  Miguel Jan 11 '13 at 12:31
    
No problem Miguel, and welcome to our site! –  Georges Elencwajg Jan 11 '13 at 12:36

1 Answer 1

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A change of coordinates between charts for the manifold with boundary $M$ has the form $x=(x_1, \cdots,x_n) \mapsto (\phi_1(x), \cdots,\phi_n(x))$, with $x_n, \phi_n(x)\geq0$ since $x_n,\phi_n(x)\in \mathbb H_n$.

The last line of the Jacobian $Jac_a(\phi)$ at a point $a\in \partial \mathbb H_n$ has the form $(0,\cdots , 0,\frac { \partial \phi_n}{\partial x_n}(a))$ :
Indeed, for $1\leq i\leq n-1$ we have $\frac { \partial \phi_n}{\partial x_i}(a)=0$ by the definition of partial derivatives since $\phi(\partial \mathbb H_n)\subset \partial \mathbb H_n$ and thus $\:\frac {\phi_n(a+he_i)-\phi_n(a)}{h}=\frac {0-0}{h}=0$.
Similarly $\frac { \partial \phi_n}{\partial x_n}(a)\geq 0$ because $\:\frac {\phi_n(a+he_n)-\phi_n(a)}{h}=\frac {\phi_n(a+he_n)-0}{h}\gt 0$.
Actually, we must have $\frac { \partial \phi_n}{\partial x_n}(a)\gt 0$ because the Jacobian is invertible.

The above proves that, given a tangent vector $v\in T_a(\mathbb H_n)$, its image $w=Jac_a(\phi)(v)$ satisfies $w_n=\frac { \partial \phi_n}{\partial x_n}(a)\cdot v_n$ with $\frac { \partial \phi_n}{\partial x_n}(a)\gt 0$, which shows that outward pointing vectors are preserved by the Jacobian of a change of coordinates for $M$ and thus that the notion of outward pointing vector is well-defined at a boundary point of a manifold with boundary .

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Ok, I assume that $\frac{\phi(a+he_n)}{h}\geq 0$ because $\phi(\mathbb{H}^n)\subseteq\mathbb{H}^n$? Thanks for the answer! –  Miguel Jan 11 '13 at 13:24
    
@ Miguel: yes, exactly. Beware also that you have to assume $h\geq 0$ for $\phi (a+he_n)$ to be defined (so that the partial derivative $\frac { \partial \phi_n}{\partial x_n}(a)$ is actually a limit from the right, i.e. obtained for $h\gt 0$ tending to zero). And you are welcome! –  Georges Elencwajg Jan 11 '13 at 15:07

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