Orientation of Surfaces

Question

I'm having a little trouble understanding how to orient a surface in $\mathbb{R}^3$ For example, how would I orient the ellipsoid given by:

$$x^2+y^2+z^2+xy+xz+yz=\frac12$$

for $(x,y,z) \in \mathbb{R}^3$ ?

score 9 · Answer 1 · answered Feb 12 '13 at 12:45

In elementary differential geometry of surfaces in $\Bbb R^3$ orienting a surface $S$ means defining a unit normal vector $n(p)$ at each point $p\in S$ such that $n$ variies continuously with $p$. The surface $S$ is called orientable if such an assignement is possible.

In your case $S$ is presented as level surface of the function $$f(x,y,z):=x^2+y^2+z^2+xy+yz+zx\ .$$ In this case the vector $\nabla f(p)$ is normal to the tangent plane at $S$ and can be used for our purposes if it is $\ne0$. In order to check whether this condition is fulfilled we compute $$\nabla f(x,y,z)=( 2x+y+z,\ x+2y+z,\ x+y+2z)$$ and easily check that the only point in $\Bbb R^3$ where $\nabla f$ vanishes is the origin $O$, and this is not a point of $S$. Therefore the unit vector $$n(p):={\nabla f(p)\over|\nabla f(p)|}\qquad(p\in S)$$ provides an orientation of your ellipsoid $S$.

The chosen orientation points outward. If for some reason you want to work with the opposite orientation you have to replace the above $n(p)$ by $-n(p)$.

score 1 · Answer 2 · answered Feb 12 '13 at 19:24

As the OP put in a tag for differential forms, I suspect this final ingredient is desired: from Guillemin and Pollack Differential Topology pages 173-174, exercises 13 and 14. In particular 14 has $$ \omega = f_1 dy \wedge dz + f_2 dz \wedge dx +f_3 dx \wedge dy, $$ with a triple $$ F = (f_1, f_2, f_3). $$ Then the restriction of $\omega$ to the given surface $S$ is $$F \cdot N \; dA$$ where $S$ is oriented and $N$ is the (outward) unit normal, as in Christian's answer. This should feel familiar, as it is exactly this setup in Stokes' Theorem for integrating a vector field over a surface.

As a technical note, as long as the surface is smooth and strictly star-shaped around the origin (the radial vector is never tangent to the surface, never orthogonal to the normal), one may take $$ \omega = x dy \wedge dz + y dz \wedge dx + z dx \wedge dy, $$ as a 2-form that never vanishes when restricted to the surface.

Your feeling is indeed correct: I'm doing an exercise that asks to compute the integral of the form $\omega=xdy\wedge dz-2zf(y)dx\wedge dy+yf(y)dz\wedge dx$ over the ellipsoid, for an $f(y)$ such that $d\omega = dx\wedge dy\wedge dz$ and $f(1)=1$. So I needed the orientation. — user61974, Feb 12 '13 at 21:29

score 0 · Answer 3 · answered Feb 12 '13 at 11:34

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The general form of an ellipsoid is:

$$({\bf x-v})^T A^{-1} ({\bf x-v})=1$$ From the form you presented, we can write the matrix $A^{-1}$ for your example as: $$A^{-1}= \left( \begin{array}{ccc} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array} \right)$$ And you may find $A$: $$A = \frac{1}{4}\left( \begin{array}{ccc} 3 & -1 & -1 \\ -1 & 3 & -1 \\ -1 & -1 & 3 \\ \end{array} \right)$$ The eigenvectors of $A$ give you the axes of the ellipsoid: $$(-1, 0, 1), (-1, 1, 0), (1, 1, 1)$$ And the eigenvalues $(1,1,1/4)$ give you the squares of the radii.

answered Feb 12 '13 at 11:34

Nathaniel Bubis

33,018

2

I believe the OP is asking for an orientation as studied in e.g. differential topology, much like "the mobius strip is not orientable". In that case I don't think this answer is what user61974 is looking for. Of course, I'm not sure. – Daan Michiels Feb 12 '13 at 11:46
@DaanMichiels - If that was the case, what's the question? All ellipsoids are quite clearly orientable surfaces. – Nathaniel Bubis Feb 12 '13 at 12:10
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I believe the OP is asking for a consistent choice of normals, an atlas with only orientation preserving transitions, a volume form or maybe something else depending on the definition of orientation that (s)he uses. Of course, graphically it is indeed clear that ellipsoids are orientable, but I believe the OP is looking for technical justification. – Daan Michiels Feb 12 '13 at 14:01

score 0 · Answer 4 · answered Jul 26 '14 at 16:43

A simple way to do your integral is to use Stokes' theorem: the integral of $\omega$ over the surface of the ellipsoid is equal to the integral of $d\omega$ over the enclosed volume. Since you were given that $d\omega=dx\wedge dy\wedge dz$, which is the standard volume form in $R^3$, the integral is simply plus or minus the volume enclosed in the ellipsoid. To choose the sign, you must choose the orientation. It is ``standard" to choose the orientation determined by $dx\wedge dy\wedge dz$ in $R^3$. If you do this and if you then choose the induced orientation on the ellipsoid using the outward direction, the sign is positive. I wouldn't be surprised if the problem posed to you neglected to specify explicitly which orientation it wanted you to use.

Orientation of Surfaces

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