differential geometry help

**marianne** · November 27th 2008, 01:13 AM

I'm not sure this is the right forum, I hope the moderator will move it if necessary.

(1) I have to sketch level sets $f^{-1}(c)$ , c=1, 0, -1, for the function $f(x_1, x_2, x_3)=x_1 x_2 -x_3^2$

I know level sets are $\{(x_1, x_2, x_3) | x_1 x_2 -x_3^2=1, 0, -1\}$ but how do i sketch this?

(2) Let $\phi:\mathbb{R}^3 \rightarrow S^3$ be the inverse of stereographic projection from $S^3-\{(0, 0, 0, 1)\}$ to the equatorial hiperplane $x_4=0$ . Show that for every vector $p \in \mathbb{R}^3$ exists $\lambda(p) \in \mathbb{R}^3$ such that
$||d\phi(v)||= \lambda (p) ||v||, \forall v \in {\mathbb{R}^3}_p$ .

I don't even know how to start..

Thank you for your time and help!

**Andreas Goebel** · November 27th 2008, 03:22 AM

Hi,

those level-sets are surfaces in R3. You can draw them with most math-packages. The command usually is called "implicitplot" or similar.

Do you have maple or mupad?

Regards,

Andreas

**Laurent** · November 27th 2008, 01:31 PM

Originally Posted by marianne

I'm not sure this is the right forum, I hope the moderator will move it if necessary.

(1) I have to sketch level sets $f^{-1}(c)$ , c=1, 0, -1, for the function $f(x_1, x_2, x_3)=x_1 x_2 -x_3^2$

I know level sets are $\{(x_1, x_2, x_3) | x_1 x_2 -x_3^2=1, 0, -1\}$ but how do i sketch this?

(1) You can use a software, but you can also sketch it yourself.
For our convenience, let us denote the coordinates by $(x,y,z)$ , so that the equations are $z^2=xy-c$ for $c=-1,\,0,\, 1$ .
These are equations of quadrics. A usual trick in this case is to introduce the new coordinates $X=\frac{1}{\sqrt{2}}(x+y)$ and $Y=\frac{1}{\sqrt{2}}(x-y)$ (this just corresponds to rotating the axes of the x-y plane by an angle $\frac{\pi}{4}$ ), for then we obtain $X^2-Y^2=xy$ , and the equations are now $X^2=z^2+Y^2+c$ .

First remark is that it depends only on $r^2=z^2+Y^2$ , so that the surface has a "revolution symmetry" around the axis $(OX)$ . So it suffices to find the sketch of a slice (containing $(0X)$ ), for instance $z=0$ , and rotate it around $(OX)$ to generate the whole surface.

Consider $c=0$ . Then for $z=0$ we have $X^2=Y^2$ , hence $Y=X$ or $Y=-X$ . The graph is nothing else but a cone with axis $(OX)$ and angle $\frac{\pi}{2}$ .

Consider $c=-1$ . Then for $z=0$ we have $X^2=Y^2-1$ , hence $Y^2=X^2+1$ . This is the equation of a hyperbola centered at 0, with asymptots $Y=\pm X$ (one piece above the axis, one piece below). The surface is a hyperboloid of one sheet (i.e. a connected one).

Consider $c=1$ . This time you have $Y^2=X^2-1$ , this has no solution when $|X|<1$ , this is the equation the same hyperbola as above but rotated by $\frac{\pi}{2}$ . Hence we get a hyperboloid of two sheets (i.e. with two connected parts).

**Laurent** · November 27th 2008, 02:18 PM

Originally Posted by marianne

(2) Let $\phi:\mathbb{R}^3 \rightarrow S^3$ be the inverse of stereographic projection from $S^3-\{(0, 0, 0, 1)\}$ to the equatorial hiperplane $x_4=0$ . Show that for every vector $p \in \mathbb{R}^3$ exists $\lambda(p) \in \mathbb{R}$ such that
$||d\phi(v)||= \lambda (p) ||v||, \forall v \in {\mathbb{R}^3}_p$ .

I don't even know how to start..

Thank you for your time and help!

A first step is to find an expression for $\phi$ .

A sketch is pretty helpful. Of course not in $\mathbb{R}^4$ , but the situation (and the result) is pretty much the same in any dimension, so you can draw in $\mathbb{R}^2$ : draw the unit circle, the line $y=0$ (this is the hyperplane $\{x_4=0\}$ of the text), pick a point $X=(x,0)$ on the line, draw the line from $N=(0,1)$ (the "north pole") to $X$ . It intersects the circle at one point, which is $M=\phi(X)$ . In my explanation, I will think at $X$ as a point outside the circle (or the sphere), so perhaps it will be clearer if you draw your sketch accordingly.

In the following, I will denote the north pole by $N=(0,\ldots,0,1)$ .

In a geometric way: remark that $ON=1=OM$ (remember $M=\phi(X)$ where $X$ is in the hyperplane $\{x_n=0\}$ of $\mathbb{R}^n$ ), hence the triangle $OMN$ is isocele at $O$ , so that the point $O$ projects orthogonally to the middle of the segment $[M,N]$ . Subsequently, $\overrightarrow{NM}$ is twice the projection of $\overrightarrow{NO}$ to the line $(NX)$ , that is to say: $\overrightarrow{NM}=2\left(\overrightarrow{NO}\cdo t\frac{\overrightarrow{NX}}{NX}\right)\frac{\overr ightarrow{NX}}{NX}$ . Notice that $\overrightarrow{NO}\cdot\overrightarrow{NX}=\overr ightarrow{NO}\cdot(\overrightarrow{NO}+\overrighta rrow{OX})=1+0=1$ since $\overrightarrow{OX}\perp\overrightarrow{ON}$ and $ON=1$ . Finally, we have:

$\phi(X)=M=N+2\frac{\overrightarrow{NX}}{\|\overrig htarrow{NX}\|^2}$ .

If you prefer not to look at the picture, you can write $\overrightarrow{NM}=\lambda\overrightarrow{NX}$ for some $\lambda\in\mathbb{R}$ (because $M$ lies on the line $(NX)$ ), and find $\lambda$ by writing $\|\overrightarrow{OM}\|^2=1$ (because $M$ lies on the unit sphere). You'll end up with the same expression of course.

--

Now, we must differentiate. It would have been tedious with coordinates, but this is easy with vectors, and the conclusion will be very nice.

The differential of $\overrightarrow{NX}=\overrightarrow{NO}+\overright arrow{OX}$ (at $X$ , evaluated at $v\in\mathbb{R}^n_X$ ) is just $\vec{v}$ .
Then the differential of $\|\overrightarrow{NX}\|^2=\overrightarrow{NX}\cdot \overrightarrow{NX}$ is $2\vec{v}\cdot\overrightarrow{NX}$ .
So that the differential of $\frac{1}{\|\overrightarrow{NX}\|^2}$ is $-\frac{2\vec{v}\cdot\overrightarrow{NX}}{\|\overrig htarrow{NX}\|^4}$ . (by the chain rule)

Finally, $d\phi_X(\vec{v})=2\frac{1}{\|\overrightarrow{NX}\| ^2}\vec{v}-\frac{2\vec{v}\cdot\overrightarrow{NX}}{\|\overrig htarrow{NX}\|^4}\overrightarrow{NX}$ .

Let me write is slightly differently: $d\phi_X(\vec{v})=\frac{2}{\|\overrightarrow{NX}\|^ 2}\left(\vec{v}-2\left(\vec{v}\cdot\frac{\overrightarrow{NX}}{\|\o verrightarrow{NX}\|}\right)\frac{\overrightarrow{N X}}{\|\overrightarrow{NX}\|}\right)$ .

--

What do you recognize in the bigger parentheses? This is $\vec{v}$ minus twice its projection on $(NX)$ . In other words, this is the symmetric of $\vec{v}$ with respect to the vectorial hyperplane orthogonal to $(NX)$ .

A symmetry preserves the length, so that $\|d\phi_X(\vec{v})\|=\frac{2}{\|\overrightarrow{NX }\|^2}\|\vec{v}\|$ . This is what you need, with an explicit $\lambda$ . That means that $d\phi$ multiplies lengths by a constant factor: it is a similitude (if my translation is correct). In particular, this shows that $\phi$ preserves the angles. But this is another story...

**whipflip15** · November 27th 2008, 06:43 PM

Love the question (and the solution)!

**marianne** · November 30th 2008, 02:31 AM

Thank you so very much!!!
I was offline for two days and until now I didn't see you reply.
I'm sure it took a lot of time and efford and I am extremely grateful!

I will study it in detail today, and I hope it's okay to post some further questions, if I have some.

Thank you!

**marianne** · November 30th 2008, 05:38 AM

I can't seem to get the graphs right.

I don't have the necessary software, could someone please post the output photo?

Thank you again.

**Andreas Goebel** · November 30th 2008, 11:16 AM

Hi,

I have plotted those surfaces with MuPAD.

It seems that there is a transition from "two sheets" to "one sheet".

I don´t have time to look into the math itself, you will learn more from Laurent here.

Regards,

Andreas

**marianne** · November 30th 2008, 12:50 PM

Originally Posted by Laurent

First remark is that it depends only on $r^2=z^2+Y^2$ , so that the surface has a "revolution symmetry" around the axis $(OX)$ . So it suffices to find the sketch of a slice (containing $(0X)$ ), for instance $z=0$ , and rotate it around $(OX)$ to generate the whole surface.

I understood everything but this "revolution symmetry" part, and why does it suffice to sketch a slice and rotate it around OX.
I've googled it, but couldn't understand it.
Could you please explain a bit more?

And thank you, Andreas!

**Laurent** · December 1st 2008, 01:51 AM

Originally Posted by marianne

I understood everything but this "revolution symmetry" part, and why does it suffice to sketch a slice and rotate it around OX.
I've googled it, but couldn't understand it.
Could you please explain a bit more?

And thank you, Andreas!

Perhaps the "revolution symmetry" is not the proper term in English, I'm not sure. Anyway, here is what it means:

We noticed that the equation depends only on $X$ and on the distance from the $X$ -axis to the point $(X,Y,z)$ (this distance equals $r=\sqrt{z^2+Y^2}$ ). So if a point $(X,Y,z)$ is on the surface, then any point $(X, Y', z')$ where $Y^2+z^2=Y'^2+z'^2$ will lie on the surface as well. And how can we go from the first point to the second one? By rotating around the axis $(OX)$ .
Or I could say that the circle of center $(X,0,0)$ , of radius $r$ , and orthogonal to $(OX)$ is a subset of the surface.

Now if we know the intersection of the surface with a slice containing the axis, like the slice $\{z=0\}$ , then the whole surface is the union of the previous circles, centered on the axis, orthogonal to the axis, and going through the points of the section.

In other words, the whole surface is obtained by rotating the slice around this axis. You can see this on Andreas'plots, except that the scale is not the same on the three axes, so what we can see on the pictures is not really a revolution symmetry (the sections are ellipses rather than circles).

I hope it was clearer...

**marianne** · December 6th 2008, 11:20 AM

Much clearer, thank you. :-)
I am now required to solve the second one using a given hint, but I will post it in another thread.

Thank you for all your help!

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