# How to get the normal bivector to a surface?

1. Apr 18, 2007

### HolyPhia

How to get the normal bivector to a surface??

Given a surface $$f(x)=0$$ in a manifold, for examle $$R^n$$, its normal vector can be constructed as $${\partial}^{\mu}f$$.

But i don't know how to construct the so-called normal bivector (or binormal) to this surface. It has the form as $${\epsilon}_{\mu \nu}$$...

2. Apr 18, 2007

### Chris Hillman

You need to be more specific, I think. Is this in the context of Frenet-Serret theory in E^3? Or in say the context of minimal surfaces in E^4?

3. Apr 19, 2007

### HolyPhia

Thank you, i think i've got it.

For example, given a D-dimensional Lorentzian(or Riemann) manifold as $$(M,g_{ab})$$ as
$$ds^2 = g_{tt}dt^2 + g_{rr}dr^2 + f(r)d{\Omega}^2_{D-2}$$,
then the hypersurface with constant t and r is a (D-2)-dimensional sphere. The point is that this sphere is of co-dimension 2. So its normal is a bivector.

We may get its binormal as follows:
First we choose the coordinates as $$\{ t,r,x^1,\cdots,x^{D-2} \}$$, in which $$\{x^1,\cdots,x^{D-2}\}$$ are coordinates on $$S^{D-2}$$. Then we have the nature volume element on the (D-2)-sphere as
$$\tilde{\epsilon}=\frac{1}{(D-2)!}\sqrt{h}{\epsilon}_{a_1 \cdots a_{D-2}}dx^{a_1}\wedge \cdots \wedge dx^{a_{D-2}}$$.
Here h is the determinant of metric on the sphere.
Then we can use the standard "Hodge dual" to get the binormal as
$$*(\tilde{\epsilon}) = *\left( \frac{1}{(D-2)!}\sqrt{h}{\epsilon}_{a_1 \cdots a_{D-2}}dx^{a_1} \wedge \cdots \wedge dx^{a_{D-2}} \right)= \cdots = \frac{1}{2} {\omega}_{\mu \nu} dx^{\mu} \wedge dx^{\nu}$$
For $$x^{a_1},\cdots,x^{a_{D-1}}$$ is limited on the sphere, so the non-zero component of the above equation is $$\mu,\nu = t,r$$.
Finally we can get a 2-form as
$$\omega \equiv \frac{1}{2} {\omega}_{\mu \nu}dx^{\mu} \wedge dx^{nu} \equiv {\omega}_{t r} dt \wedge dr$$.
This is just the so-called binormal to the sphere $$S^{D-2}$$. Of course, some normalization may be made.

Last edited: Apr 19, 2007
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