Differential Forms on submanifolds

Question

Say I take an embedded submanifold of $\Bbb R^n$, like the sphere. Any differential form on $\Bbb R^n$ can be restricted to the sphere. My question is this: is any differential form on the sphere (or any embedded submanifold) the restriction of a form on $\Bbb R^n$? My gut says no, but I can't think of any counter examples.

Answer 1

Yes. Since the product of a form and a smooth function is again a form, extend the form to a small tubular neighborhood of your submanifold and multiply by a bump function.

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Georges Elencwajg has suggested I explain further. To extend a $p$-form $\omega\in\Omega^p(M)$ to a $p$-form on $\mathbb{R}^n$, we need to first define a $p$-form on $T\mathbb{R}^n|_{M}$, the tangent bundle of $\mathbb{R}^n$ on $M$.

We define such a form by splitting $T\mathbb{R}^n|_M$ into $TM\oplus \nu(M)$, the sum of $TM$ and the normal bundle of $M$ in $\mathbb{R}^n$. With this splitting, any vector $T_x\mathbb{R}^n|_M$ decomposes into tangent and normal components: $v = v^t + v^n$.

Now simply define the extension $\hat{\omega}$ to be $\omega$ on $TM$ and $0$ on $\nu(M)$. Precisely, $$\hat{\omega}(v_1,\ldots,v_p) = \omega(v_1^t,\ldots,v_p^t).$$

The normal bundle $\nu(M)$ is diffeomorphic to a small neighborhood $U$ of $M$ by traveling along the normal directions a short distance. Define the projection $\pi: U\to M$ by mapping a point $p$ to the closest point in $M$. Extend $\hat{\omega}$ from $T\mathbb{R}^n|_M$ to $TU$ by defining it, again, to be $\omega$ applied to vectors parallel to $M$ and zero on vectors parallel to the normal direction. (That is, the pullback of $\hat{\omega}$ by $\pi$.)

Finally, multiply $\hat{\omega}$ on $TU$ by a smooth cutoff function supported on $U$ to obtain a smooth form on $\mathbb{R}^n$.

(The Riemannian metric on $\mathbb{R}^n$ gives natural choices of bundle splittings. We also assume that $M$ is embedded, so that the construction of $U$ is possible.)