Differential form

[daishin]

1. The problem statement, all variables and given/known data
Let w be the form w= xdydz in R^3. Let S^2 be the unit sphere in R^3.
If we restrict w on S^2, is w exact?


2. Relevant equations



3. The attempt at a solution
My guess is w is not exact on S^2.
Suppose w is exact on S^2. Then w=da for some 1-form a=fdx+gdy+hdz.
Then by definition of exterior derivative, we get
w=(-df/dy+dg/dx)(dx^dy)+(-df/dz+dh/dx)(dx^dz)+(-dg/dz+dh/dy)(dy^dz)
So we get the conditions:
df/dy=dg/dx, df/dz=dh/dx, x=-dg/dz+dh/dy.
I think I should use a fact that I am working on a unit sphere. Could anybody help me?

[matness]

I did not tried but it would be useful to write everything in spherical coordinates and to take radius as 1.

[daishin]

Using stereographic projection,(say (s,t)) I attained follwoing condition if I assume xdydz is exact on S^2,

for some smooth function g and f,
-df/dt+dg/ds = (-24(s^2)(t^2)-8(s^4)-8s)/(1+s^2+t^2)^4.
Now in order to show xdydz is not exact, it suffices to show such f and g does not exist. How can I show it?

[matness]

Different way: I tried to use the thm saying if a differential form is exact then
its closed.
i.e.if you can show dw is not equal to zero(meaning not closed)
you can conclude w is not exact.If i didn't make a mistake dw=dx dy dz
But still don't now how it is related to S^2

[Dick]

Different way: I tried to use the thm saying if a differential form is exact then
its closed.
i.e.if you can show dw is not equal to zero(meaning not closed)
you can conclude w is not exact.If i didn't make a mistake dw=dx dy dz
But still don't now how it is related to S^2

dw IS zero when restricted to S^2, it's a three form on a two manifold. But you can't apply the Poincare lemma to show it's exact since S^2 isn't contractible.

[Dick]

Try it in spherical coordinates, as matness suggested. It's pretty straight forward to find a solution to w=da, I think.