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I am trying to understand how the orientation is induced on the boundary of a differentiable manifold with boundary. Here is what I have worked out so far:

Let $M$ be a differentiable manifold and let $\partial M$ denote its boundary. Let $U$ be an open set in $M$ with $U \cap \partial M \neq 0$. Similarly let $V$ be an open set in $M$ with $V \cap \partial M \neq 0$. Let $x_i$ and $y_i$ be local coordinates in $U$ and $V$ respectively. On the boundary $x_1 = 0 = y_1$. Furthermore we know that the function $y_i = y_i (x_1,\dots, x_m)$ is differentiable. Then the orientation on $\partial M$ will be the orientation given by $x_i$ and $y_i$. It remains to be shown that this is well-defined that is the same on $\partial M \cap U \cap V$. Assuming what I wrote so far is correct my question would be how to do this final step?

Student
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1 Answers1

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Note that at $x_1=0$ we have $\tfrac{\partial y_1}{\partial x_1}>0$ and $\tfrac{\partial y_1}{\partial x_i}=0$ for $i>1$. Therefore, if $x_1=0$ then the Jacobian of $(x_1,x_2,\dots,x_n)\mapsto (y_1,y_2,\dots,y_n)$ equals to $\tfrac{\partial y_1}{\partial x_1}$ times the Jacobian of $(0,x_2,\dots,x_n)\mapsto (0,y_2,\dots,y_n)$. In particular these Jacobians have the same sign. Hence everything follows.

Anton Petrunin
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  • Thank you, @AntonPetrunin, may I ask two questions about your answer? One question is, why is ${\partial y_1 \over \partial x_1} > 0$ at $x_1 = 0$? I hope it is not a stupid question. – Student Jul 23 '13 at 14:08
  • Both $x_1$ and $y_1$ are non-negative and if $x_1=0$ then $y_1=0$, therefore $\tfrac{\partial y_1}{\partial x_1}\ge 0$ if $x_1=0. Further, if "=" then Jacobian vanishes. The later can not happen since the map is regular. – Anton Petrunin Jul 23 '13 at 20:43
  • Why are $x_1$ and $y_1$ non-negative? – Student Jul 27 '13 at 15:35
  • A chart of manifold with boundary is defined on an open set of half-space; you may assume that the half-space is defined by $x_1\ge 0$. – Anton Petrunin Jul 30 '13 at 20:32
  • Thank you. What happens if one chooses the half plane $x_1 \le 0$? Does it still hold that ${\partial y_1 \over \partial x_1} \ge 0$ or would it then be $\le 0$? – Student Aug 03 '13 at 08:31
  • I guess I really don't understand what's going on. To me the expression $\tfrac{\partial y_1}{\partial x_1}$ is just the derivative of a function. And I don't understand why, if the function is positive and the derivative is taken on a positive half plane that it follows that the derivative is also positive. If the function is the constant function $-1$ then it seems to be false. – Student Aug 03 '13 at 10:21
  • Lets fix $x_2,\dots, x_n$ and consider the function $f\colon x_1\mapsto y_1$. Note that $f(0)=0$ and $f(x)>0$ if $x>0$, hence $f'(0)\ge 0$. In our case $f'(0)\ne 0$; hence everything follows. – Anton Petrunin Aug 03 '13 at 18:00
  • What about $f(x)=\sqrt{x}+\sin x$ or $5x - 9x^2 + 5x^3$? They are such that $f(0)=0$ and $f(x)>0$ for $x>0$ but $f'(x)$ is not positive for all $x>0$. – Student Aug 04 '13 at 06:57
  • It is hard for me to guess where is the spot which you do not see. Well the first function is not differentiable at 0, the second is okey, and $f'(0)>0$. I think you should find and read the definition of manifold with boundary carefully, – Anton Petrunin Aug 04 '13 at 09:43
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    Space is short! If ∂y/∂x<0 for some scalar-value entry in your Jacobian, then if you apply this matrix to any standard basis vector (entries all zero except one with 1), then you will get a point in the second patch having one of coordinates a negative value, which is not in the upper half plane. A contradiction. In other words, if you have a bijection between patches, partial derivatives approximate your bijection function. All unit vectors from point x on the boundary direct into upper half plane, but $g(x+e)=Dg*e+\epsilon$ has one negative coordinate for your negative entry in Dg, ie ∂y/∂x. – Mikhail D Jan 31 '18 at 21:37