Extending a volume form on $S^k$ to $\mathbb R^{k+1}$ such that the exterior derivative is zero.

Question

As we know that $S^k$ is a properly embedded submanifold of $R^{k+1}$. Let $w$ be a volume for of $S^k$(a non vanishing top form). Then $dw = 0$ on $S^k$. Is it possible to extend $w$ to $R^{k+1}$ such that $dw = 0$? We can extend the form for sure by first extending it locally then patch it together. But it does not seem to be the case that the extension should always have exterior differentiation equal to zero.

Answer 1

If that were possible, then by Stokes, $$\int_{B^{k+1}}dw=\int_{S^{k}}w$$ where $B^{k+1}$ is the unit ball in $\Bbb R^{k+1}$. But the LHS is zero, and the RHS is positive.

Answer 2

I would say no, because then, from Stokes’ theorem, $$0 = \int_{B^{k+1}}{dw}=\int_{S^k}{w},$$ whoch is clearly false.

Answer 3

Using Stokes' theorem is probably the best argument. But here is another perspective, using de Rham cohomology: The inclusion $\iota: S^k\rightarrow \mathbb{R}^{k+1}$ induces a map $$ \iota^*: H^k_{dR}(\mathbb{R}^{k+1}) \rightarrow H_{dR}^k(S^k) . $$ Since the space on the right hand side is spanned by the cohomology class $[w]$, being able to find a closed extension of $w$ would imply surjectivity of $\iota^*$. But this is impossible since $H^k_{dR}(\mathbb{R}^{k+1}) = 0$.