Questions in topology/geometry

1. May 13, 2005

kleinwolf

I have summarized 2 questions about topology that seems interesting to me, but are quite vague still :

1) Are there surfaces (parametrized by 2 free variables), that are possible in R^4, but not in R^3 ?

2) Are there differences between the 2 following parametrizations of the 3-sphere :

w=cos(a)sin(b)
x=sin(a)sin(b)
y=cos(c)cos(b)
z=sin(c)cos(b)

w'=cos(a')
x'=sin(a')cos(b')
y'=sin(a')sin(b')cos(c')
z'=sin(a')sin(b')sin(c')

?

2. May 13, 2005

matt grime

1) Like the Klein bottle

2) What does 'different' mean in this context?

3. May 14, 2005

kleinwolf

I thought there were representations of the Klein bottle in 3 dimensions, except that the surface has to intersect itself ??

For the second, I meant

a) are they homeomorphic, for example ?...but I think the answer is no.

I tried this : I suppose the manifold are giving the same points (the 3-sphere)...so that i can try to find which relationship link the free parameters so that the points in the embedded space are the same : w'=w,....

this gave : c'=c....but then the relationships are no more solvable, so that I thought it was allowed to deduce the representations are equivalent but they define the same manifold (set of points)....I don't understand this....or is there a mistake in my resolution ?

Last edited: May 14, 2005
4. May 14, 2005

mathwonk

a smooth compact connected n manifold can of course be mapped into any space at all, the question is when can the mapping be an embedding, i.e. injective with injective derivative.

in general all n manifolds can be embedded into space of dimension 2n+1, and whitney showed smooth amnifolds can even be embbedded in space of dimension 2n.

the usual method is to embed in some high diemnsional space and then start projecting down into lower dimensional spaces. if the point one projects from is on netiher a secant nor a tangent to the manifold, then the projection is still an embedding.

since secants to an n manifold are parametrized by pairs of points on the manifold there is a 2n dimensional family of them, and each is a line, so they form a 2n+1 dimensional solid containing the original n manifold.

if one is in a space of dimenjsion higher than 2n+1, hence one can one avoid any point of this solid, but in general not otherwise.

this proof works also for complex manifolds and algebraic vieties, but whitneys proof for the smooth case lowering the number to 2n is more special and difficult.

so all real curves can be embedded in the plane but not all complex riemann "surfaces" i.e. complex curves, can be embedded in the complex projective plane.

and not all complex surfaces can be embedded in spoaces of dimension less than 5.

a kelin bottle cannot be embedded in three space but can be smoothly embedded in 4 space.

in general a compact non orientable manifold cannot be embedded as a hypersurface in an orientable manifold since the distinction between "inside and outside" would orient the hypersurface.

5. May 14, 2005

Hurkyl

Staff Emeritus
Oh, so that's the key step? I think I understand that. I'll have to figure out that proof sometime then! (Of course, I still need to figure out how to produce the original embedding, but one step at a time, eh?)

6. May 15, 2005

mathwonk

well the original embedding is probably done first locally, then by taking a product of the local ones, which is finite by compactness.

i.e. a map into a product space has injective derivative if true as a map into some factor. then you have to make it globally injective.

7. May 15, 2005

mathwonk

the theory of non embeddability is actually one of the main uses of the theory of characteristic classes, since an embedding implies restrictions on the characteristic classes, which sometimes can be shown to be false.

so this is a good place to start learning characteristic classes, i.e. natural homology classes associated to a manifold.

e.g. take any compact manifold, consider its only natural invariant, its tangent bundle.

then choose any n general tangent vector "fields", and consider the subset of the manifold consisting of points where these n vectors are dependent.

that depends of course on the choice if vector fields, but its homology class does not. so this is a natural homologu calss associated to the manifold called a characteristic class.

or for one vector field just consider the (usually) finite set where it is zero. for a general vector field the numebr of these points is the same, called the euler number of the manifold. so this characteristic class is called the euler class.

now by the jordan hypersurface theorem, a compact connected hypersurface separates R^n into an inside and an outside. so any hypersurface has a no where zero normal vector field.

this implies that the tangent bundle of the hypwersurface, added to the one dimensional normal bundle, is a product bundle, and the whitney product theorem for characteristic classes implies an equation on these classes equalling 1.

but more directly, the outward normal chooses a "side" for the htypersurface which is thus "two sided", whereas a non orientable surface cannot be embedded as a 2 sided hypersurface in any orientable manifold. (I think?)

I am far from expert on this stuff and am partly just recalling lectures i heard over 30 years ago from experts like ioan james.

8. May 15, 2005

Hurkyl

Staff Emeritus
I'm having trouble trying to figure out how to piece together the embedding of two open sets into an embedding of their union.

Though I've figured out an alternate approach -- for each of the open set U homeomorphic to Euclidean n-space of the manifold M, I can find a finite collection of smooth functions such that:

(1) Each function is zero everywhere on M - U
(2) The mapping whose components are those functions is injective on U

Then, for a finite cover of M, I can simply take the collection of all of these functions to be the coordinate functions of a smooth map into Euclidean space.

Of course, now I'm curious about an analytic map, or an isometric map. But, one step at a time!

Last edited: May 15, 2005
9. May 15, 2005

mathwonk

for analytic maps, at least complex analytic ones, you can use line bundles, and divisors. i.e. given an appropriately positive divisor the riemenn roch theorem implies there exists a large number of analytic sections of the line bundle, and enough to separate bioth points and tangent vectors.

see the proof of kodaira's embedding theorem.

of course in the case of compact complex analytic manifolds you meet the situation on non embeddable manifolds, first occurring in complex dimension 2.

i.e. not embeddable in complex projective space of any dimension.

Last edited: May 15, 2005
10. Jun 27, 2005

mathwonk

i seem to have plunged off the shallow end.

11. Jun 28, 2005

HackaB

Don't worry. Your insights are much appreciated, even if most of them go way over my head.

12. Jun 28, 2005

mathwonk

here is the standard "riemann roch" proof of embeddability of a riemann surface of genus g.

the theorem says that if we choose d points and e more points, then whenever d-e > 2g-2, the dimension of the space of meromorphic functions with poles at most at the d points and zeroes at least at the e points, is 1+d-e-g.

thus if we take any d points with d > 2g, then whenever we require one or two more zeroes, we lose one then two dimensions.

thus for any points pq, the set of functions vanishing at p is not all the functions, and the set vanishing at both p and q, is not even all those.

so this set of functions eparates points and tangent vectors (take p=q).

thus we can embed every compact connecetd riemann surface of genus g in projective space of dimension at most g+1, and with degree at most 2g.

this is just riemanns theorem. if we use roch's enhancement of it, then

in fact they can be represented either as degree 2g-2 curves in g-1 space (canonical embedding), or as double covers of the projective line (hyperelliptic case).

Last edited: Jun 28, 2005
13. Jun 28, 2005

Doodle Bob

All right, I'm confused now: a compact Riemann surface of genus g is a sphere with g handles attached to it, which means that topologically it's automatically embeddable in R^3, although clearly not holomorphically -- the above construction, I believe, gives the holomorphic embedding. Is that right? I think I'm misremembering the connection between the genus and the number of handles.

If so, then my question is: is there a noncompact orientable surface that cannot be embedded (homeomorphically) in R^3? Clearly, any nonorientable surface will not be embeddable (I'm not allowing self-intersections).

If not, then just ignore me.

14. Jun 28, 2005

mathwonk

that is right, the above is the holomorphic embedding.

to see that a holomorphic structure can differ from a topological structure picture a torus, i.e. riemann surface of genus one, mapped two to one noto the projective line, i.e. the complex "plane".

fropm topology it follows that the 2 to 1 map is branched at exactly 4 points. i.e. the derivative of the map vanishes at 4 points of the torus, or equivalently, there are 4 points of the plane over which the map has only one preimage.

this map gives the torus T its holomorphic structure. i.e. function from T to the complex plane is holomorphic if and only if everywhere locally the corresponding fucntion from a small disc in the plane is holomorphic.

then you can imagine another torus branched over 4 different points. in general these 4 points cannot be brought into the same positions as the first 4 by a n mobius transformation. so there is no holomorphic 1-1 map from one torus to the other.

i do not know whether there are non compact (second countable) orientable surfaces that do not embed in R^3, but of course there are some topologically pathological ones, non second countable i guess.

it is also not at all clear that a non orientable, non compact, surface will not embed, since a mobius strip does embed in R^3.

15. Jun 28, 2005

Doodle Bob

Ah, I was making an unspoken assumption: that the surface is compact without boundary, so that there is a well-defined normal and thus a well-defined complex structure and thus a well-defined orientation. Although now that I think about it, I'm not quite sure that the lack of a boundary would give a well-defined normal. Isn't there some sort of 3-dimensional Jordan separation theorem for compact surfaces w/o boundary?

16. Jun 28, 2005

mathwonk

yes every compact connected smooth hypersurface in R^n separates R^n into two connected components, an inside and an outside. in particular the hypersurface cannot be one - sided.

17. Jun 29, 2005

HackaB

Is that the jordan curve theorem?

18. Jun 29, 2005

mathwonk

in the plane yes, but in general it is called the jordan brouwer separation theorem. actually this is the smooth case, for the continuous case i need to check the cohomology theory (lefschetz duality?) in higher dimensions.

Jordan-Brouwer Separation Theorem Any imbedding of the n-1 dimensional sphere into n-dimensional Euclidean space, separates the Euclidean space into two disjoint regions.

Brouwer was unable to prove the analog of the Jordan-Schönflies theorem, that the inside and outside of such an imbedded sphere are homeomorphic to the inside and outside of the standard sphere in Euclidean space (ie. the sphere of unit vectors). In 1921 J. W. Alexander announced that he had a proof of this result. However before he published his paper, he discovered a mistake in his proof. Then in 1924 he discovered a counterexample to the analog of the Jordan-Schönflies theorem in 3-dimensions, what came to be known as "Alexander's horned sphere":

The outside of the horned sphere is not simply connected, unlike the outside of the standard sphere: a loop around one of the horns of the horned sphere can't be continuously deformed to a point without passing through the horned sphere

Last edited: Jun 30, 2005
19. Jun 30, 2005

mathwonk

ok here is the continuous case. it follows from alexander duality (which follows from lefschtetz duality)

if a compact n-1 manifold A is embedded in R^n, then the complement of A has one more connected component than A has.

20. Jun 30, 2005

mathwonk

if it seems intuitive that the outside of any embedded sphere should be connected, what about simply connected? is that not just as intuitive?

i guess i would need to examine the proof or the counterexample to get better intuition on that.