Newsgroups: sci.math From: jrioux@act.ulaval.ca (Jacques Rioux) Subject: A question about functions on differentiable manifolds Date: Thu, 1 Dec 1994 23:24:35 GMT Hi, I think the following statement may be true, would anybody in the know comment on possibly wrong or insufficient assumptions or references on the subject in the bibliography. Theorem? If F and G are two analytical real valued functions defined on a closed differentiable manifold M such that F(x_i)=G(x_i) for an infinite collection of points {x_i | i=1...infinity} in M, then F(x)=G(x) for all x in M. ============================================================================== From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: A question about functions on differentiable manifolds Date: 2 Dec 1994 17:52:01 GMT In article , Jacques Rioux wrote: >Theorem? > >If F and G are two analytical real valued functions defined on a closed >differentiable manifold M such that F(x_i)=G(x_i) for an infinite collection >of points {x_i | i=1...infinity} in M, then F(x)=G(x) for all x in M. Your theorem holds. Recall 'closed' manifolds are compact and connected. you'll need this, otherwise the statement is false. Let E = { x in M | F(x)-G(x)=0 } and C=E', the set of limit points of E. A priori, E may have isolated points, so C may be smaller than E, but in fact I will show inclusions M \subset C \subset E \subset M. First observe that C is not empty: since M is compact and E is infinite, E must have at least one limit point. As F-G is continuous, E is closed, and thus C is contained in E and is closed as well. On the other hand, for any point c in C we may choose a neighborhood of c which is diffeomorphic to Euclidean space. Since c is in C, there are other points of E arbitrarily close to c, almost all of which lie in this neighborhood. You know from analysis in R^n that since F-G is analytic in this Euclidean open neighborhood and is zero on this converging sequence, that F-G is zero on the whole neighborhood. In particular, the neighborhood of M around c lies in E (and in fact in C). Since this holds at each c in C, we see C is open. Well, then, C is a non-empty, open and closed subset of a connected space M, hence has to be all of M. This forces E=M too, so that F(x)=G(x) for all x. dave ============================================================================== From: [Permission pending] To: rusin@washington.math.niu.edu Subject: Re: A question about functions on differentiable manifolds Date: Fri, 2 Dec 1994 11:00:44 -0800 >Recall 'closed' manifolds are compact and connected. >you'll need this, otherwise the statement is false. Of course you are rigth that the statement is false for noncompact manifolds. But are you aure this is how the word "closed" is usually used by topologists? I thought maybe it is used just to refer to a submanifold N of a manifold M such that Cl(N) = N (i.e., either compact or properly imbedded noncompact). Can you find a reference? (I can't find one easily myself.) --[Permission pending] ============================================================================== Date: Fri, 2 Dec 94 14:16:45 CST From: rusin (Dave Rusin) To: [Permission pending] Subject: Re: A question about functions on differentiable manifolds (Blush) I was relying on _you_ to have the right definition of closed! I put that definition in just to cover my tail because that's what I need in the proof; really I can't remember quite how it's used. But it's insufficient to have N be closed in some other manifold -- what's to keep one from using N itself as the other manifold in that definition? And if you say N is closed in another manifold which is _comapct_, then N is compact too. Anyway, as I noted, compact and connected are sufficient for your result. Connected is clearly necessary, and probably compactness is too. For example, if M = R, F(x) = sin(x) has an infinite set of zeros but isn't identically zero. (I don't know if in general you can find such an F on every noncompact manifold and keep F analytic.) [deletia - djr] ============================================================================== Date: Fri, 2 Dec 1994 12:21:37 -0800 From: [Permission pending] To: rusin@math.niu.edu (Dave Rusin) Subject: Re: A question about functions on differentiable manifolds Thanks for your comments. I'm almost certain that compact and connected are not only sufficient but also necessary for this theorem to be true. [deletia - djr] ============================================================================== From: x151509@rubc.rz.ruhr-uni-bochum.de (Joerg Winkelmann) Newsgroups: sci.math Subject: Re: A question about functions on differentiable manifolds Date: 7 Dec 1994 10:40:01 GMT Dave Rusin (rusin@washington.math.niu.edu) wrote: : In article , : Jacques Rioux wrote: : >Theorem? : > : >If F and G are two analytical real valued functions defined on a closed : >differentiable manifold M such that F(x_i)=G(x_i) for an infinite collection : >of points {x_i | i=1...infinity} in M, then F(x)=G(x) for all x in M. : Your theorem holds. Recall 'closed' manifolds are compact and connected. : you'll need this, otherwise the statement is false. No, this is not enough. Counterexample: Let M1,M2 be two compact (=closed) real-analytic manifolds, dim(M1)>0, f a non-constant real-analytic function on M2 such that f(y)=0 for some y on M2. Let M=M1 x M2. Now let F be the function which is identically zero, and G be the function on M=M1 x M2 defined by G(a,b)= f(b). Now F(a,b)=G(a,b) whenever f(b)=0, hence F=G at infinitely many points on M1 x M2. Joerg ============================================================================== Date: Wed, 7 Dec 94 12:34:51 CST From: rusin (Dave Rusin) To: x151509@rubc.rz.ruhr-uni-bochum.de Subject: Re: A question about functions on differentiable manifolds In article <3c43e1$khi@rubb.rz.ruhr-uni-bochum.de> you write: >No, this is not enough. >Counterexample: >Let M1,M2 be two compact (=closed) real-analytic manifolds, >dim(M1)>0, >f a non-constant real-analytic function on M2 such that f(y)=0 for some >y on M2. Let M=M1 x M2. Oops, I think. Your point is well taken: I posted after having thought only about analytic functions of one variable, whereas clearly for example f(x,y)=x and f(x,y)=point are distinct analytic functions from M1xM2 to M1 which agree on {point}xM2. However, the example I just gave was not a _real-valued_ analytic function. I can't produce an easy example because I'm not sure if there are any non-constant analytic functions at all from a compact manifold to R^1. Certainly there are none if the manifold is one-dimensional (i.e. = S^1). Do you know if there can be any at all for higher-dimensional manifolds? dave ============================================================================== Date: Thu, 8 Dec 1994 13:29:25 +0100 From: Joerg Winkelmann To: rusin@math.niu.edu, x151509@rubc.rz.ruhr-uni-bochum.de Subject: Re: A question about functions on differentiable manifolds By a (highly non-trivial) theorem of Whitney every real-analytic manifold admits a real-analytic embedding into some R^n. For this reason, every real-analytic manifold admits many non-constant real-analytic functions. This is also the case for S^1 which is embedded in R^2. Thus you get real-analytic functions on S^1 by restricting real-analytic functions on R^2 to S^1. Maybe you mixed this up with the complex case. On a complex-analytic compact manifold every complex-analytic function is constant. But for a real-analytic manifold the theorem of Whitney guarantees the existence of non-constant real-analytic functions. Joerg ============================================================================== Date: Thu, 8 Dec 94 23:34:57 CST From: rusin (Dave Rusin) To: Joerg.Winkelmann@rz.ruhr-uni-bochum.de, rusin@math.niu.edu, Subject: Re: A question about functions on differentiable manifolds Yes, of course you are right. I'm not sure how I let myself get so confused! Thanks for calling my error to my attention. dave