From: Jules Bean Subject: Re: extension property of smooth functions between manifolds Date: Thu, 18 Nov 1999 19:10:00 +0000 Newsgroups: sci.math Keywords: extending maps on submanifolds of R^k to open neighborhoods of R^k bkaster@cs.vu.nl wrote: > > I wonder if the following is true and if it is, how to prove it. > > Let M be an m-dimensional manifold in R^k > N an n-dimensional manifold in R^s > f: M -> N smooth (all partial derivatives and iterates exist). > > For all p in M there exists U \subset R^k open p in U and there exists > F: U -> R^s smooth, with F restricted to M \cap U = f. > > [M \cap U is the intersection of M and U] Let me try to write what you've asked in words. M is an m-manifold in k-dimensional euclidean space, and N is n-dimensional embedded in s-dimensional space. f is a smooth map from M to N. You ask whether or not it's true that around each point p in M there is an open subset U in k-space and F a smooth map from U to s-space such that F is a local extension of f. The answer is certainly yes. I will give a sketch argument: Choose U such that U n M [using n for \cap] is contained in a coordinate neighbourhood, and hence identify U n M with R^m. Similarly find a coordinate neighbourhood around f(p) and identify it with R^n. Now we can think of f as (up to a diffeomorphism) being a smooth map from R^m to R^n. [This is exactly what we mean when call f smooth in the first place]. In a sufficiently small such pair of neighbourhoods, we can identify U diffeomorphically as (U n M) x R^(k-m). Similarly on the other side. So the question really boils down to the simpler euclidean question, given a map f: R^m to R^n, can we extend it to a map from R^k to R^s, agreeing on the natural inclusion R^m in R^k and R^n in R^s. The answer to this is certainly yes, by setting F(x_1,...x_k) = f(x_1,...x_k), i.e. ignore the extra coordinates. This map fails to 'fill' R^s in the way which you think it might. That's not hard to change, though. Five-second summary: A map between submanifolds in larger manifolds, locally, is simply a map between euclidean spaces embedded in larger euclidean spaces, and these are trivially extendable (and can be extended in a variety of fashions to satisfy many possible conditions - like maximal rank - if desired). Does that help? Jules -- /----------------+-------------------------------+---------------------\ | Jelibean aka | jules@jellybean.co.uk | 6 Evelyn Rd | | Jules aka | | Richmond, Surrey | | Julian Bean | jmlb2@hermes.cam.ac.uk | TW9 2TF *UK* | +----------------+-------------------------------+---------------------+ | War doesn't demonstrate who's right... just who's left. | | When privacy is outlawed... only the outlaws have privacy. | \----------------------------------------------------------------------/ ============================================================================== From: foltinek@math.utexas.edu (Kevin Foltinek) Subject: Re: extension property of smooth functions between manifolds Date: 23 Nov 1999 10:45:10 -0600 Newsgroups: sci.math In article <810mnl$md@cs.vu.nl> bkaster@cs.vu.nl writes: > I wonder if the following is true and if it is, how to prove it. > > Let M be an m-dimensional manifold in R^k > N an n-dimensional manifold in R^s > f: M -> N smooth (all partial derivatives and iterates exist). > > For all p in M there exists U \subset R^k open p in U and there exists > F: U -> R^s smooth, with F restricted to M \cap U = f. The answer depends on what you mean by "manifold in R^k". If you mean that, for every point p in M, there is a neighbourhood V in R^k of p and coordinates (x^1,...,x^k) on V such that M \cap V is defined by 0=x^{m+1}=...=x^k, then the answer is yes, and the proof is easy: let (y^1,...,y^s) be coordinates on R^s so that (locally) N is defined by 0=y^{n+1}=...=y^s; then the function f:M->N can be written as (y^1,...,y^n) = f(x^1,...,x^m), and you can define F:R^k->R^s by F(x^1,...,x^m,x^{m+1},...,x^k) = (f(x^1,...,x^m); 0,...,0) where the f(...) is the original n-component vector, and the remaining components are zero so that the image is entirely in N. (If k>m and s>n, then you can modify this by replacing one of the zeros by (x^{m+1})^2+...+(x^k)^2, which gives you the added property that f(x) is in N precisely when x is in M\cap U.) On the other hand, if you mean that, for every point p in M, there is a neighbourhood V in M of p and coordinates around p in R^k such that V is defined by 0=x^{m+1}=...=x^k, then the answer is no, and the following is a counterexample: there is a well-known way to put R into R^2 in the shape of a figure-eight, and I'll try to describe it here without drawing it. First, map R bijectively to a finite-length open interval (say, by the arctan function); then imagine this finite interval to be a piece of string. Use this string to draw a figure-eight, such that the two ends of the string just barely touch the "crossing point", which is at "x=0" in R. As these are really the open ends of the interval, they don't actually intersect, so this gives you an injection of R into R^2, with the property I described above (there is a neighbourhood V...). Let N be R sitting in R as itself. Let f:M->N be defined by f(x)=arctan(x). f(0)=0, but as x-> infinity or -infinity, f(x)->pi/2 or -pi/2, but these three "points" are all "the same point" in R^2 (the crossing point of the figure-eight), so there is no extension to a smooth F:R^2->R. (I'll let you make this rigorous.) I believe the first case is what is usually meant by "manifold in R^k". (Usually an embedded submanifold has the induced topology.) Kevin.