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I'm studying differential geometry right now. I've encountered a problem considering the definition of a function/differential forms on a surface $S \in \mathbb{R}^3$.

Suppose we're given a $C^1$ function $f$ on $S$, can we say that there must be a restriction of some function $F$ on $\mathbb{R}^3$ such that $F|_{S} = f$. And, does this result hold for differential forms as well?

Chris Jing
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    I assume you are requiring some kind of smoothness of $F$? Otherwise the result is trivial: just consider $1_S f$. If you require smoothness, the answer is yes, but higly non trivial. See, for example, https://math.stackexchange.com/questions/131642/the-extension-of-smooth-function –  Apr 20 '20 at 10:06
  • Thank you!! Does this result also hold for differential forms? – Chris Jing Apr 20 '20 at 10:12
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    Cover a neighborhood of $S$ with charts in which $S$ is represented by the $x$-$y$ plane. Extend your function to the entire chart by making it constant in $z$ direction. Then use a smooth partition of unity to stitch the extensions together. – Conifold Apr 20 '20 at 10:15
  • Yes, what Conifold said is the way to do it — and if $S\subset\Bbb R^3$ is an embedded submanifold, then it has a tubular neighborhood, so you can just use that. Works for differential forms just fine, as well, with the same idea. – Ted Shifrin Apr 20 '20 at 18:47

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