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. So could somebody please help me outline a proof?