1

In $\mathbb{R}^3$ consider the the 2-dim manifold $$ M:=\left\{(x,y,z)\in\mathbb{R}^3:z=xy\right\}. $$ Let the orientation of $M$ be in such a manner that in the Point $(0,0,0)$ the vector $e_3=(0,0,1)$ is the positive orientated normal vector. How is the orientation of M given?

Hello, i do not know if I understand it right. Do I have to find an orientated Atlas of $M$ that gives an orientation $\sigma$ to which the normal field is positive orientated?

Or something else?

Adriano
  • 41,576

1 Answers1

2

If $z=xy$ then the natural choice for parameters is just $(x,y)$ and $R(x,y) = (x,y,xy)$ parametrizes the surfaces. The normal vector field is given by $N(x,y) = R_x \times R_y$, $$ R_x \times R_y = (1,0,y) \times (0,1,x) = \langle -y, -x, 1 \rangle $$ hence clearly $N(0,0)= \langle 0,0,1 \rangle$ which shows the standard parametrization does in fact have the desired orientation for $z=xy$. No need for atlas here, as you can easily see the normal vector field is well-behaved on the $xy$-plane so the current choice of coordinates will work well for whatever problem we face.

Notice that if we calculate the determinant in your viewpoint we obtain: $$ det \left[ \begin{array}{ccc} -y & 1 & 0 \\ -x & 0 & 1 \\ 1 & y & x \end{array} \right] =y^2+x^2+1 \neq 0$$ for all $(x,y) \in \mathbb{R}^2$. Of course, for $z=xy$ there should be a symmetry for exchange of $x,y$ in equations, this allowed my brother to immediately deduce that my original normal vector was calculated wrong. Now our view-points agree, the normal vector field $N(x,y)$ is defined on the whole surface $z=xy$ and the determinant maintains its positive sign throughout. Incidentally, we should normalize $N(x,y)$ if we want it to be a unit vector field, I didn't bother since that is not needed for our interest here.

James S. Cook
  • 16,755
  • I understand your map and the normal vector calculation. But why does this show that the normal field is positive orientated related the orientation that is given by the global map? –  Jul 31 '13 at 09:49
  • To my knowledge the normal field is positive orientated related to the orientation that is given by an orientated atlas (and here your map is indeed an orientated atlas) if $det(\nu(a),\frac{\partial \phi(c)}{\partial t_1},\frac{\partial\phi(c)}{\partial t_2})>0$ where $\nu$ is the normal field, $a\in M, c\in\mathbb{R}^2, \phi(c)=a$. And now the question here is if this is fullfilled for $a=(0,0,0)$, right? Or for all a? –  Jul 31 '13 at 10:04
  • Then I think in common it is $\mbox{det}\left(\begin{pmatrix}-y & 1 & 0\x & 0 & 1\1 & y & x\end{pmatrix}\right)=-x^2+y^2+1$ and so in the special case that $a=(0,0,0)$ then the determinant is to my opinion 1>0. So this shows the positive orientation of the normal field at least in the zero point? –  Jul 31 '13 at 10:13
  • @math12 given three vectors $v_1,v_2,v_3$ if $det(v_1|v_2|v_3)>0$ then this means that $v_3$ is on the same side of the plane spanned by $v_1,v_2$ as $v_1 \times v_2$. So, I think we agree at the origin. My approach was purely geometric, I just observed that $N(0,0)=\langle 0,0,1 \rangle$. For points on $z=xy$ where $1=x^2-y^2$ your calculation troubles me for it seems the deteminant is zero, yet $N(x,y) = \langle -y,x,1 \rangle$ is still above the plane spanned by $R_x$ and $R_y$. I am missing something here. – James S. Cook Jul 31 '13 at 14:34
  • But does the determinant has to be larger than zero for points that are not the zeros point? The task only says that the normal vektor of the zero points is positive orientated related to the orientation. To my opinion that means that for the other points it is not necessary. –  Jul 31 '13 at 15:09
  • @math12 right, for your question, what we found suffices, it just is bothering me immensely that I cannot cipher the meaning of your determinant being zero! Of course the determinant can be negative or zero, but that signals a linear dependence at the zero point and then a change in handedness for where it is negative. So, I don't see why that determinant is zero since the coordinates $(x,y)$ look good for all values, in other words, the coordinate chart seems regular from the normal vector, but seems otherwise from your determinant. Curses. – James S. Cook Jul 31 '13 at 15:17
  • Can I ask you another question here concerning this manifold and this global map? Or should I open a new thread for that? It's about calculating an integral of a 2-form. –  Jul 31 '13 at 16:16
  • @math12 yes ask a new question. That's how we do it here. I'm working on some things so I'll leave your question to someone else, don't worry, it'll get answered... – James S. Cook Jul 31 '13 at 16:40
  • 1
    @math12 the error is found and I updated the answer so now our views are not in any logical tension. – James S. Cook Jul 31 '13 at 17:53
  • Okay, thanks a lot. So the unit normal field is positive orientated related to the orientation $\sigma$ that is given by the global map $\phi$ (you called it $R$). –  Jul 31 '13 at 17:57
  • 1
    @math12 yes, the normal vector field defines an orientation of the surface and it is naturally induced from $R$. From a higher-dimensional perspective, it is a happy accident of three dimensions that this normal vector field defines an orientation, it is the Hodge duality that connects the more natural two-form surface area form. Generally, for $p$-dimensional manifold you'll want a non-vanishing $p$-form to orient the space. But, I must cease talking here now since the website begins to grumble. – James S. Cook Jul 31 '13 at 18:13