Differential form

1. Aug 6, 2007

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?

2. Aug 6, 2007

matness

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

3. Aug 6, 2007

daishin

streographic projection

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?

4. Aug 7, 2007

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

5. Aug 7, 2007

Dick

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.

6. Aug 7, 2007

Dick

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