# A question about differential form

1. Dec 9, 2015

### 1591238460

Suppose x ∈ Ω^(n−1)(Rn \{0}) is closed and the integral of x on S^(n-1) equals to 1. I am stuck on how to show
there does not exist an n − 1 form y ∈ Ω(n−1)(R^n) with y|R^n\{0} = x.

2. Dec 9, 2015

### lavinia

Stokes Theorem says that the extension of this form to all of $R^n$ integrated over the sphere equals the integral of its exterior derivative over the interior region bounded by the sphere. Since it is closed on $R^n - 0$, its exterior derivative integrates to zero on the interior of the sphere.

Last edited: Dec 9, 2015
3. Dec 10, 2015

### 1591238460

Dear Lavinia, thank you！

4. Dec 10, 2015

### mathwonk

some restrictions like "smooth" belong in some of these sentences or else there does exist an extension that does not satisfy stokes' thorem. (I see now it depends on your definition of the symbol Omega.)

Last edited: Dec 11, 2015
5. Dec 11, 2015

### lavinia

Right. By extension to a differential form, smooth is implicit.

Last edited: Dec 11, 2015
6. Dec 11, 2015

### mathwonk

Aha!, I have finally re - read the OP's question carefully enough to see he did not use the word "form", but rather a symbolic omega notation for them. And no doubt in his book this notation was reserved for smooth forms. So you guys are right. Before answering I should have asked for the definition of big Omega.

But I still like hypotheses in my theorems. Let me ilustrate by precising the OP's question: assuming the original form was smooth, 1) does there exist any extension? 2) does there exist a smooth extension? 3) does there exist a continuous (but not necessarily smooth) extension?

Last edited: Dec 11, 2015
7. Dec 13, 2015

### lavinia

-There is always a discontinuous extension. Give it any value at the origin.
-By Stokes Theorem, there is no smooth extension. There is not even a continuously differentiable extension.

-One cannot extend the form to a continuous form on all of $R^n$
Since the form integrates to 1 on the unit sphere and is closed, it integrates to 1 on every sphere centered at the origin. This means that it is of the form
$π^{*}(ω) + dφ$ where $ω$ is the volume element of the unit sphere and $π$ is radial projection. $dφ$ is the exterior derivative of an $n-2$ form. $π^{*}(ω)$ can not be extended to the origin continuously. If ,for instance,one evaluates the form on a constant (n-1)-tuple of tangent vectors to $R^n$ - e.g. in the plane, the constant vector field that points parallel to the x-axis - one sees that there is no limit at the origin. One still needs to show that one can not fix things with $dφ$.

Last edited: Dec 13, 2015
8. Dec 13, 2015

### mathwonk

nice. iw as thinking of using stokes on a ball with small ball removed about the origin, then taking a limit. since the integral over the boundary of the small ball goes to zero (by continuity of the extension), you get the same result as if the extension were smooth at the origin.

9. Dec 13, 2015

### lavinia

Lovely idea.

Last edited: Dec 13, 2015