From: asabharw@magnus.acs.ohio-state.edu (Ashutosh Sabharwal) Newsgroups: sci.math Subject: Simple differential geometry condition Date: 7 Feb 1996 21:32:11 GMT Hi everyone, I have a differential geometry question. f : A \subset R^k --> R^m , m > k > 0 A is open, f is a C^\infinity(A) function but f is not full rank everywhere on A (actually it is full rank everywhere except on sets of measure zero). Or f is many-to-one on some sets of measure zero. I am interested in the geometry of sets f(A). So, Is f(A) a manifold ? OR since f is not guaranteed to be full rank everywhere, does it mean that I am out of the realm of differential geometry ? I checked several books and none seem to refer to such a situation. What books are recommended if the above can be handled by diff. geom. ? Thanks a lot in advance. -- Ashu ============================================================================== From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Simple differential geometry condition Date: 8 Feb 1996 02:09:16 GMT In article <4fb5or$s9s@charm.magnus.acs.ohio-state.edu>, Ashutosh Sabharwal wrote: > > f : A \subset R^k --> R^m , m > k > 0 > > A is open, > > f is a C^\infinity(A) function > but f is not full rank everywhere on A (actually > it is full rank everywhere except on sets of measure zero). > Or f is many-to-one on some sets of measure zero. > >I am interested in the geometry of sets f(A). So, > >Is f(A) a manifold ? No. Certainly the existence of any singular points is likely to cause trouble. Consider f: R^2 -> R^2 given by f(x,y) = (xy,y); it's of full rank except where y=0. The image is all of R^2, less the horizontal axis, plus the origin. At this last point, the image fails to have the structure of a manifold. More interestingly, f(A) can fail to be a manifold even if nonsingular. The fact that f has full rank at a means that f'(a) maps the tangent space at a injectively to the (m-dimensional) tangent space at f(a). Now, if you want f(A) to be a submanifold at least near f(a), you need to know that the set of tangent vectors forms a subspace (presumably you intend it to have dimension k). You know that the image of f'(a) is such a subspace, but you don't know there are no other tangent vectors. For example, consider f: R^1 -> R^2 given by f(t) = (cos(t), sin(2t)). This traces out a figure-8 (the curve y^2=4x^2(1-x^2) ), which has a singularity at the origin. This f is globally of (maximal) rank 1. Another example is the function f: R^2 -> R^3 which I will not attempt to express in symbols but whose image is the classic "picture" of a Klein bottle. (Cognoscenti: I mean to compose the universal covering R^2 --> K with an embedding K--> R^4 and a projection R^4 --> R^3 which results in an immersion of K-->R^3) The image f(A) is not reallly a manifold because of the points of "self-intersection". >OR since f is not guaranteed to be full rank everywhere, >does it mean that I am out of the realm of differential >geometry ? I checked several books and none seem to >refer to such a situation. Look up "immersions". You may also find Sard's theorem useful (the _image_ of the singular set has measure zero). If you can stomach algebraic geometry (or perhaps its analytic cousins) you may be able to describe your image as a variety, if f is in the appropriate category. By the way, this isn't really differential geometry (to me) -- the question has nothing to do with a metric. I would classify this as a topic of differential topology. A very pleasant book is John Milnor's "Topology from the differentiable viewpoint". dave ============================================================================== From: Ashutosh Sabharwal Newsgroups: sci.math Subject: Re: Simple differential geometry condition Date: Thu, 08 Feb 1996 17:39:13 -0500 Dave Rusin wrote: > > In article <4fb5or$s9s@charm.magnus.acs.ohio-state.edu>, > Ashutosh Sabharwal wrote: > > > > f : A \subset R^k --> R^m , m > k > 0 > > > > A is open, > > > > f is a C^\infinity(A) function > > but f is not full rank everywhere on A (actually > > it is full rank everywhere except on sets of measure zero). > > Or f is many-to-one on some sets of measure zero. > > > >I am interested in the geometry of sets f(A). So, > > > >Is f(A) a manifold ? Thanks for the reply. What if I decide to remove all the points at which f is not full rank. Since these point form a set of measure zero (here), A'=A-set_of_measure_zero is again open. Can f(A) be a manifold now ? BTW, are manifolds required to be connected ? Looks like f(A') can be disconnected. What are the sufficient conditions on f and A for f(A) to be a manifold ? > More interestingly, f(A) can fail to be a manifold even if nonsingular. > The fact that f has full rank at a means that f'(a) maps the > tangent space at a injectively to the (m-dimensional) tangent space at f(a). > Now, if you want f(A) to be a submanifold at least near f(a), you need > to know that the set of tangent vectors forms a subspace (presumably you > intend it to have dimension k). You know that the image of f'(a) is > such a subspace, but you don't know there are no other tangent vectors. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ I am sorry but I did not get this point. > Look up "immersions". You may also find Sard's theorem useful (the _image_ > of the singular set has measure zero). If you can stomach algebraic > geometry (or perhaps its analytic cousins) you may be able to describe > your image as a variety, if f is in the appropriate category. > > By the way, this isn't really differential geometry (to me) -- the question > has nothing to do with a metric. I would classify this as a topic of > differential topology. A very pleasant book is John Milnor's > "Topology from the differentiable viewpoint". (Unfortunately, the book is not available in the library right now) I was hoping to show that I have a differential manifold and then define a suitable metric. But it appears that I am stuck at the first stage. If I stick to the my previous formulation (with A not A'), is there a natural way of defining a metric on f(A) without going through the ? Thanks -- Ashu ============================================================================== From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Simple differential geometry condition Date: 9 Feb 1996 21:47:34 GMT In article <4fb5or$s9s@charm.magnus.acs.ohio-state.edu>, Ashutosh Sabharwal wrote: [Given] > f : A \subset R^k --> R^m , m > k > 0 [and] > A is open, [f was given to be smooth at some points] > >Is f(A) a manifold ? I responded: not at singular values (except by accident, of course), and not necessarily at regular values either. In article <311A7B91.167EB0E7@ee.eng.ohio-state.edu>, Ashutosh Sabharwal wrote: > >What if I decide to remove all the points at which f is >not full rank. Since these point form a set of measure >zero (here), A'=A-set_of_measure_zero is again open. Can f(A) be >a manifold now ? BTW, are manifolds required to be connected ? Ahem: A - (set of measure zero) is not in general open; it's A-(closed set) which is still open. But this does in fact apply in your situation. f(A) _can_ be a manifold, but it need not be, as I'll explain below. Manifolds are not required to be connected (e.g. GL(n,R) is a popular disconnected manifold) but in practice one studies their connected components. >Looks like f(A') can be disconnected. Yes. Actually, it probably won't be. If k < m, then the only way p can be a singular point is for the rank of f'(p) to be less than k. This requires that all k x k minors have determinant zero. If, for example, f is a polynomial map, this would require that many polynomials vanish simultaneously, so that the set of simgular points is an algebraic set (e.g. a variety), usually of rather large codimension. In particular, its complement A' will still be connected if A was. For example, a generic polynomial map f: R^2 -> R^3 is nonsingular except on a discrete set in R^2, so that A' is connected. >What are the sufficient conditions on f and A for f(A) to be a manifold ? The answer is perhaps messier than you'd like. You've already discounted the problems which can arise from singular points. But you still have the phenomenon I mentioned in a previous post: > More interestingly, f(A) can fail to be a manifold even if nonsingular. > The fact that f has full rank at a means that f'(a) maps the > tangent space at a injectively to the (m-dimensional) tangent space at f(a). > Now, if you want f(A) to be a submanifold at least near f(a), you need > to know that the set of tangent vectors forms a subspace (presumably you > intend it to have dimension k). You know that the image of f'(a) is > such a subspace, but you don't know there are no other tangent vectors. The original poster found this unclear, and I suppose it was. The problem is that if f is not one-to-one, say if f(a)=f(b), then you only know that the images of f'(a) and f'(b) are k-dimensional subspaces passing through f(a) [=f(b)]. They could be _different_ subspaces, so that f(A) does not have a linear tangent space at this point. The simplest example comes from composing a nice map like f1(t) = (cos(t), sin(t)) with a singular map like f2(x,y)=(xy,y). The composite f(t)=(cos(t) sin(t), sin(t)) is nonsingular, but at f(0)=f(pi)=(0,0), we have two different lines of tangent vectors: the line y=x is the image of the linear map f'(0), and the line y=-x is the image of the linear map f'(pi). I had better add another condition to your list of worries. Suppose you further restrict your domain A so that f | A is now not only non-singular but one-to-one. Is f(A) guaranteed to be a manifold now? The answer is _still_ no. An instructive example is that old reliable f(t) = (t, sin(1/t)) on A=(0,1) (the open interval, I mean) Extend this (in a smooth way) so that f is defined on (0,2) and the image of (1,2) is a graceful arc which just happens to include a portion of the vertical axis near the origin. This is not a manifold: there is no neighborhood of the origin which is homeomorphic to Euclidean space. (All the neighborhoods include infinitely many _copies_ of R^1.) What went wrong this time? Well, the given conditions now make f a smooth (hence continuous), 1-to-1, onto map from A onto its image f(A). Unfortunately, in topology, that's not enough for "isomorphism" (homeomorphism). You must also know that f^(-1) is continuous. Equivalently, f establishes a one-to-one correspondence between A and f(A), but there can be more open sets in the former than in the latter. If this situation does _not_ arise, then f is a homeomorphism of A onto f(A). Since A is open in R^k, it's a manifold, and so anything homeomorphic to it can also be given the structure of a manifold. So you should add a condition that f^(-1) be continuous, i.e., that f carry closed sets to closed sets. There's one more little bugaboo here. If all the other conditions are met, then f(A) is a manifold. But is it a submanifold of R^m? (You didn't ask that but I suspect you had it in mind.) For example, the function f(x) = (x, |x|) gives a homeomorphism between the real line and a "V" shape in R^2. Well, via this homeomorphism, the image of f can be viewed as a manifold, but it's not a submanifold of R^2 (that is, it is impossible to choose local coordinates at (0,0) smoothly equivalent to the coordinates already in use at (0,0) in such a way that the subset is the set of points where the first few coordinates [in this case, the first 1 coordinate] vanishes). The problem looks artificial because |x| is not smooth, but the same image results from the _smooth_ map f(x) = (x^3, |x| x^2), and you can even make a C-infinity map with this image. But, despair not. For if you maintain the assumption of _nonsingularity_ as well, then this can't happen. If f'(p) has rank k, then choose coordinates in R^m so that f'(p) has a non-singular matrix in the first k columns. Then the determinant of this matrix of derivatives doesn't vanish at p, hence not in a neighborhood of p, hence in that neighborhood of p, we can trivially modified f to give a function f~ : R^k x R^(m-k) --> R^m, now a (local) diffeomorphism, with the image of f being the image under f~ of the subspace R^k x {0}. In summary: if f : A --> R^m is a differentiable, nonsingular, one-to-one map defined on an open subset of R^k which carries closed sets to closed sets, then f(A) is a manifold. But I'll bet you were hoping for something nicer :-). >I was hoping to show that I have a differential manifold >and then define a suitable metric. But it appears that I >am stuck at the first stage. If I stick to the my previous >formulation (with A not A'), is there a natural way of defining >a metric on f(A) without going through the ? Well, you're welcome to treat the image f(A) as a metric space. It certainly inherits a metric from R^m. But if you need to discuss geodesics, for example, then this is unsatisfactory (e.g., a point near the origin in the topologists' sine curve is "far" if you must measure distances along the curve). You'll have to decide how you feel about the figure-8 and the topologists' sine curve. Do you think they are irrelevant? Then add conditions as above and you will prevent them from arising. If they might legitimately occur in what you wish to study, you can't wish them away; you'll have to decide what you had hoped your differential-geometric-structure to do for you when they did arise. dave ============================================================================== From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Simple differential geometry condition Date: 10 Feb 1996 19:10:14 GMT In article , Laura Helen wrote: >> Now, if you want f(A) to be a submanifold at least near f(a), you need >> to know that the set of tangent vectors forms a subspace (presumably you >> intend it to have dimension k). You know that the image of f'(a) is >> such a subspace, but you don't know there are no other tangent vectors. > >It can fail to be a manifold even if the tangent space is unambiguous, >can't it -- take f: R^1 --> R^2 : f(t) = (-sin (t), -cos(3t) + cos(t)), >where the point (0,0) is reached at t = 0 and pi, and the tangent vector >is (1,0) at both t-values. Isn't the crucial part that f is not >1-1, rather than having extra tangent vectors? Yes indeedy. The possibility of having more tangent vectors than you thought you'd have at f(a) alerts you to the need to require something like f being one-to-one. One might imagine that something weaker would suffice (like insisting that all the points a in f^(-1) (p) have the same image for f'(a) ) but Laura's example shows that you'd then have other worries which you hadn't prevented yet. I don't really know what the minimal conditions would be on a map f to guarantee that its image be a manifold. There are lots of highly-singular, wildly non-one-to-one functions with very nice images (e.g. f=constant). >Dave Rusin, the mathematical Energizer Bunny >writes: Can I quote you one that? dave ============================================================================== From: asimov@nas.nasa.gov (Daniel A. Asimov) Newsgroups: sci.math Subject: Re: Simple differential geometry condition Date: 13 Feb 1996 19:59:25 GMT In article Laura Helen writes: > Ashutosh Sabharwal wrote [in essence]: > >> Suppose f: A -> R^m is C^oo, where A is open in of R^k and m > k > 0 >> but Df is not full rank everywhere on A. >> >> Must f(A) be a manifold ? > > [...] > Isn't the crucial part that f is not 1-1, rather than having extra tangent > vectors? ---------------------------------------------------------------------------- Even if Df: T(R^k) -> T(R^m) is of full rank everywhere, AND f is globally one-to-one, the image f(A) can still fail to be a manifold. E.g., consider the bifolium given by f: (-pi/2, pi/2) -> R^2 via 2 2 f(t) = ((cos t) (sin t), (sin t) (cos t)). --Dan Asimov