3-forms on a surface. They are equal to 0. i know this. What i dont know is how to prove it.(adsbygoogle = window.adsbygoogle || []).push({});

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.

So could somebody please help me outline a proof?

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# Homework Help: Differential Geometery: 3-forms on a surface

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