1. ## Unit normal

The map
F : R³ \ (0,0,0) → R³ \ (0,0,0):v→(v/((v⋅v)))
is called inversion with respect to S². Let S be a surface that does not pass
through the origin, and let S' = F(S). Show that, if S is orientable, then so is
S' . Do this by showing that if N is the unit normal of S at a point p, then the
unit normal N' of S' at F(p) is
N'=[((2p⋅N)p/(|p|²)]-N

I have tried using a general parametrisation in terms of u and v, and three arbitrary functions f, g, and h for each component.I have then constructed the normal by the cross product of the partial derivatives.Now since S is an orientable surface this implies that this cross product is never zero. Then to obtain the unit normal i divide by the magnitude. At this point i am completely stuck

2. Originally Posted by ulysses123
The map
F : R³ \ (0,0,0) → R³ \ (0,0,0):v→(v/((v⋅v)))
is called inversion with respect to S². Let S be a surface that does not pass
through the origin, and let S' = F(S). Show that, if S is orientable, then so is
S' . Do this by showing that if N is the unit normal of S at a point p, then the
unit normal N' of S' at F(p) is
N'=[((2p⋅N)p/(|p|²)]-N
To take the easy part first, you can soon check that N' is a unit vector. In fact,

$N'\cdot N' = \bigl(\frac{2(p\cdot N)p}{|p|^2} - N\bigr)\cdot\bigl(\frac{2(p\cdot N)p}{|p|^2} - N\bigr) = \frac{4(p\cdot N)^2(p\cdot p)}{|p|^4} - \frac{4(p\cdot N)(p\cdot N)}{|p|^2} + (N\cdot N) = 1$

(because $N\cdot N = 1$ and the other two terms cancel).

The normal vector N can be characterised as the unique unit vector with the property that ${\displaystyle\lim_{q\to p}} \frac{(q-p)\cdot N}{|q-p|} = 0$, where the limit is taken along any path in the surface S. We want to show that the vector N' has this property for the surface F(S). So we need to show that ${\displaystyle\lim_{q\to p}} \frac{(F(q)-F(p))\cdot N'}{|F(q)-F(p)|} = 0$. The numerator of that fraction is $(F(q)-F(p))\cdot N' = \bigl(\frac q{|q|^2} - \frac p{|p|^2}\bigr)\cdot\bigl(2(p\cdot N)\frac p{|p|^2} - N\bigr),$ which with a bit of juggling you can rearrange as

$\frac{(p\cdot N)}{|p|^2|q|^2}(2(p\cdot q) - |q|^2-|p|^2) + \frac1{|q|^2}\bigl((p-q)\cdot N\bigl) =-\frac{(p\cdot N)|p-q|^2}{|p|^2|q|^2} + \frac1{|q|^2}\bigl((p-q)\cdot N\bigl).$

It follows that $\frac{(F(q)-F(p))\cdot N'}{|F(q)-F(p)|} =- \frac{(p\cdot N)}{|p|^2|q|^2}|p-q| + \frac1{|q|^2}\frac{(q-p)\cdot N}{|q-p|}.$ As $q\to p$, both of those terms go to zero. That shows that N' is normal to F(S) at F(p).

3. Thanks, you made it look so easy!
I have never actually seen this property before, we have been shown that if we have a parametrised surface in terms of two parameters u and v, and we take the partial derivatives with respect to u and v, if the cross product never equals zero, they will be linearly independant.Then by this fact the cross product will be a vector perpendicular to both, and the unit normal will be defined as this cross product divided by its magnitude. Firstly i tried this approach for a point p and then a point F(p) which was a very long equation, why wouldn't this method work?
And where can i find some information on the method you used i would like to look into it so i can fully understand it.

4. Originally Posted by ulysses123
I have never actually seen this property before, we have been shown that if we have a parametrised surface in terms of two parameters u and v, and we take the partial derivatives with respect to u and v, if the cross product never equals zero, they will be linearly independant.Then by this fact the cross product will be a vector perpendicular to both, and the unit normal will be defined as this cross product divided by its magnitude. Firstly i tried this approach for a point p and then a point F(p) which was a very long equation, why wouldn't this method work?
And where can i find some information on the method you used i would like to look into it so i can fully understand it.
The definition of a unit normal to a surface S at a point p is that it is a unit vector perpendicular to every tangent vector to S at p. If S is given in terms of a parametrisation then the partial derivatives with respect to the parameters give tangent vectors in the directions of the parameters. Their cross product is perpendicular to both of them and therefore gives the direction of the normal vector. That is a very useful way to find the normal, but it can only be used if you know the parametrisation. In this problem, the surface S is not given a parametric description, so you can't use that method. The only alternative is to go back to the definition of a tangent vector at p, namely the limit, as q tends to p, of the line segment from p to q.

I said in my previous comment that the unit normal is unique. Of course it isn't, because there are always two unit normals, an "outward" normal and an "inward" normal. The surface S is said to be orientable if there is a way of selecting one of these normals for each point of the surface, in a continuous way. In a non-orientable surface, such as a Möbius strip, that is not possible. On a Möbius strip, if you start by choosing a unit normal at one point of the strip, and then go round the strip choosing (at each point) a unit normal that varies continuously from point to point, then by the time you get back to the point you started from, you will find that the normal is pointing in the opposite direction from the one that you started with.

The aim of the problem in this thread is to show that the unit normal N' of the inverted surface S' = F(S) depends continuously on the normal N of S. So if S is oriented then so is S'.

5. thanks you explain things in a good way.