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.
Chris Hillman said:
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?
