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In Guillemin&Pollack, the notions of smooth maps between manifolds and smooth maps from an arbitrary subset to a manifold are defined. However in the problem sets, the term "smooth on $\mathbb{R}^{k}$" is used. What does this mean? In other words, I am unsure what the domain and codomain are of such a map.

More context: We have the definitions

  1. A mapping $f$ of an open set $U\subset \mathbb{R}^{n}$ into $\mathbb{R}^{m}$ is called smooth if it has continuous partial derivatives of all orders.
  2. A map $f:X \to \mathbb{R}^{m}$ defined on an arbitrary subset $X$ of $\mathbb{R}^{n}$ is called smooth if it may be locally extended to a smooth map on open sets.

And the problem in question: If $k$<$l$ we can consider $\mathbb{R}^k$ to be the subset {$(a_{1},...,a_{k},0,...,0)$} in $\mathbb{R}^l$. Show that smooth functions on $\mathbb{R}^k$, considered as a subset of $\mathbb{R}^l$, are the same as usual.

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    $\begingroup$ Can you give a complete context? This is unclear. $\endgroup$
    – EQJ
    Oct 8, 2017 at 19:59
  • $\begingroup$ The definitions are well-known; the additional context that would be useful is the statements in which "smooth on $\mathbb{R}^k$" is used. $\endgroup$ Oct 8, 2017 at 20:07

1 Answer 1

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A map which is "smooth on $\mathbb{R}^k$" is a smooth map with domain $\mathbb{R}^k$. More generally, a map "on" a set is a map whose domain is that set. The codomain is not specified if you say "smooth on $\mathbb{R}^k$" and nothing else. If you refer to a "smooth function on $\mathbb{R}^k$", that would typically mean a smooth map $\mathbb{R}^k\to\mathbb{R}$ (more generally, "functions" on a space are usually maps to the "base field" which may be $\mathbb{R}$ or $\mathbb{C}$ or something else depending on context). Otherwise, you'll have to figure out the codomain from context or other words that are used.

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