# Differential Geometery: 3-forms on a surface

3-forms on a surface. They are equal to 0. i know this. What i dont know is how to prove it.

This is not a homework question, but my final is tomorrow, and this is a fair question he might ask.

All i can think of is the following

Let dx and dy be one forms

dx^dy would be two form

if we take the differential of that it would be a 3 form, so we get the following

d(dx^dy) = d(dx)^dy + dx^d(dy) by leibniz rule, which is 0 +0.

I know this proof is not good enough cause it is only a special case, and doesnt even take into account that we are on a surface.

Dick
Homework Helper
A 3-form is a totally antisymmetric linear form on vectors. E.g. f(x,y,z)=-f(y,x,z) and etc for all transpositions of the vectors x,y,z. If x,y,z are from a two dimensional space then at most two of them are linearly independent. Why does this mean f must be 0?

Well if at most 2 of them are linearly independent, then this implies that in 2d space

x=u
y=v
z= av + bu

Thus f(u,v, av+bu) = f(u,v,av) + f(u,v,bu) ...

actually thats probably not right, and im stuck.

Dick
Homework Helper
What would f(u,v,v) and f(u,v,u) be if f is totally antisymmetric? Remember f is linear, you can pull the constants out.

If it is anti-symetric then

f(u,v,v)= - f(u,v,v) (i switched the v's there)

Thus f must be 0

That good enough?

Dick