topologically different 4d objects

[bo198214]

I have to say that I am not satisfied with all these arbitrary construction mechanism that are invented here.

What would interest me, what topologically *different* 4d objects can we construct. There should be made some notes:
1. The 1dim closed surfaces are the loops/circles, i.e. S₁.
2. There is a topological classification of all 2dim closed surfaces: they can either be a sphere with attached handles or a sphere with attached cross caps. For example the torus is a sphere with one attached handle and the Klein bottle is a cross cap with one attached handle which is in turn a sphere with two attached cross caps (proof this!).
3. There is no classification of the 3dim closed surfaces yet.

These classifications are without regard to the embedding space. I.e. two objects are regarded equal if there is a homeomorphism between them (as space) but not whether they can continuously deformed into each other in a sourrounding space. For example the 1d-knots are all S₁ but two embeddings in R³ may not continuously deformable into each other. In 4d however they can.
So similarely there are homeomorphic 2d-knots (closed 2d-surfaces embedded) in 4d so that they can not continuously deformed into each other (but maybe in higher dimensions).
Also similarly there would be 3d closed surfaces that are not embeddable into 4d (like the crosscap and Klein bottle in 3d).

So my question now is what different embeddings of 2d closed surfaces (sphere with handles or sphere with cross caps/projective planes), i.e. what different 2d-knots can we construct in 4d.
And what different 3d closed surfaces can we construct in 4d (S₃ is clear, but as other candidates I only see the tapered 2d-torus, or what was a 3d-torus?)

[moonlord]

It seems I've got to read some more, I haven't understood much from your post...

[PWrong]

Unfortunately, I have a feeling that Houserichichi is the only one here who's actually learnt topology formally.

Maybe we could extend the "squares with arrows" model to a "cube with arrows" model. Unfortunately, I don't understand how the square with arrows model works.

I.e. two objects are regarded equal if there is a homeomorphism between them (as space) but not whether they can continuously deformed into each other in a sourrounding space.

Suppose you have the parametric equations for two objects, A and B. Now suppose you find a set of continuous equations with an additional parameter 't', such that when t=0, the equations describe A, and when t=1, they describe B. Is that what you mean by a continuous deformation?

what different 2d-knots can we construct in 4d

The most obvious knots would be of the form circle # knot or knot # circle, using the torus product. Note that knot # circle is just a 3D shape.

[bo198214]

Unfortunately, I have a feeling that Houserichichi is the only one here who's actually learnt topology formally.

So my 1 semester topology does not count?! *pouts*

Maybe we could extend the "squares with arrows" model to a "cube with arrows" model. Unfortunately, I don't understand how the square with arrows model works.

Actually there is already the model of simplicial complexes, though I am not in this theory. The "square with arrows" model (as you call it) works as follows. Given a 2d closed manifold. Then you can always cut it off so that you get some patches (which are all homeomorphic to the disc). Simply give each cut a letter write this letter together with the direction of the cut at both corresponding edges of the patch(es).
I mean algebraic topology is all about compound of patches. *gg*

Is that what you mean by a continuous deformation?

Ok, I strike out a bit more. A continuous map f is a map where for each environment U of f(x) there is an environment V of x such that f(V) c U (subset). Its quite similar to the definition of a continuous map in analysis.
An embedding is an injective continuous map (in algebra it is an injective homomorphism).
If we for example embed S₁ into R³. Then can we do it by giving a continuous injective map f from [0,1] to R³ except that f(0)=f(1). Because S₁ can be regarded as [0,1) where the environments of 0 "wrap around to 1". And with the demand of f(0)=f(1) we assure that for each environment U of f(0) there is an environment V of 0 with f(V) c U. If f(0)!=f(1) then we could find an environment U around f(0) such that f(1) is not in U. But every environment of 0 contains elements near 1 (wrapped around). So we can not find V with f(V) c U.

Ok, maybe everyone is now still more confused. But it was a try.
If we have now two embeddings F and G of S₁ into R³ (i.e. knots) then we can ask whether there is H(x,t) (where t is of [0,1]) with H(x,0)=F(x) and H(x,1)=G(x) and each H(.,t) is an embedding of S₁ into R³ and H is continuous.
This I mean by continuosly deform.

what different 2d-knots can we construct in 4d

The most obvious knots would be of the form circle # knot or knot # circle, using the torus product. Note that knot # circle is just a 3D shape.

hm

It seems I've got to read some more, I haven't understood much from your post...

Because it seems we will anyway not get any significant results here, we can use the thread also as Q&A about topology. I will give my best :wink:

[pat]

Take a 3-cube. Identify each face with its opposite. Assuming the cube with corners at (+/-1, +/-1, +/-1), then a point (x,y,+1) is identified with (x,y,-1)... and similarly for the other dimensions. You have created the 3d torus. This is not homeomorphic to a sphere. It's easy to show it's not homotopic to a sphere. And, it can't be homeomorphic if it's not homotopic.

If you again start with a 3-cube and do two dimensions as above, but put a mirroring into the other identification... so that (x,y,+1) is identified with (-x,-y,-1), the resulting manifold is not orientable. As such, it cannot be homeomorphic to either the 3-torus or the S³.

But, as for classifying them... I don't know. It wouldn't have surprised me at all if it was again something like... spheres with handles... or spheres with crosscaps... where by crosscaps, I mean projective 3-space.

Because I'm still working on getting comfortable with the tapered 2-torus, I am pondering the the 3-torus versus an extrusion of a 2-torus. I think that if I took an extrusion of a 2-torus, this is what I would consider a sphere with one handle. There is a way to add a second handle so that the result is homeomorphic to a 3-torus. And, I am pretty sure that any way you add a second handle makes it homemorphic to a 3-torus.

The tapered torus is (obviously?) orientable. Thus, it can't have crosscaps. But, if it tapers to a single point, then it's not a smooth manifold.... so I'm not sure it fits with what you're trying to classify. You said "There is a topological classification of all 2dim closed surfaces". I think "smooth" was kinda tied up in the definition of surfaces there. Otherwise, you could take a non-trivial circle on a 2-torus, scrunch it down to a single point, and come up with something that isn't a sphere with handles or a sphere with crosscaps.

[PWrong]

So my 1 semester topology does not count?! *pouts*

Sorry, I didn't know you'd taken topology. I'll be taking something like it next year, hopefully.

Ok, I strike out a bit more. A continuous map f is a map where for each environment U of f(x) there is an environment V of x such that f(V) c U (subset). Its quite similar to the definition of a continuous map in analysis.

Are you talking about a function from a vector to a vector? Like the transformation T(r, th) = {r cos(theta), r sin(theta)} that transforms a rectangle into a disk? Could you give me a few more examples of continuous and discontinuous maps?

An embedding is an injective continuous map (in algebra it is an injective homomorphism).

Injective meaning "one-to-one", but not "onto", right?

If we have now two embeddings F and G of S1 into R3 (i.e. knots) then we can ask whether there is H(x,t) (where t is of [0,1]) with H(x,0)=F(x) and H(x,1)=G(x) and each H(.,t) is an embedding of S1 into R3 and H is continuous.
This I mean by continuosly deform.

That looks similar to what I said, except you're not using parametric equations. Was I essentially right after all, or did I miss some conditions?

Because I'm still working on getting comfortable with the tapered 2-torus

Which one? There are four: torus-> point, torus->circle, torus-> sphere, and circle->sphere via torus.

But, if it tapers to a single point, then it's not a smooth manifold....

Does smooth mean "continuous" or "infinitely differentiable"? I think it depends how you taper it. It's possible to taper a torus to a point without it intersecting itself.

I think I finally understand homeomorphic and homotopic now . They both mean "continuously deformable", but homotopic only goes one way, and homeomorphic has to go both ways.

[bo198214]

@pat
Thanks for saving this thread

Take a 3-cube.

Maybe I should add that in homology theory, each face, regardless of its dimension, of a (simplicial) complex can only have two orientations (that by a certain algorithm can be computed). So two faces can only be identified for equal orientation or inverted orientation.

Finishing the identifications with the cube I would add the cube with two pairs of opposite faces identified with equal orientation and the other pair identified with opposite orientation. And the cube with one pair identified with equal orientation and the other both pairs identified with opposite orientation. They all should be non-homeomorphic.

Though it is not quite clear to me what is embeddable into 4D (the 3-torus at least seems to be). For 2-manifolds each orientable could be embedded into 3d and each non-orientable into 4D. Maybe Wendy can help here with her inborn-4d visualization capabilities According to the whitney embedding theorem it would be possible that they are not embeddable neither into 4d nor 5d. At least the projective 3-space needs at least dimension 6 for embedding.

However, pat, there was a blunt: If two objects are homotopic then they are homeomorphic and not vice versa. As the example with the knots embedded in 3d showed.

I think I finally understand homeomorphic and homotopic now . They both mean "continuously deformable", but homotopic only goes one way, and homeomorphic has to go both ways.

No!!! Absolutely not! Homotopic means deformable into each other in a surrounding space. Homoemorphic means (very roughly) "in the same way connected".
Let us take another example: Consider an open disk with a hole in it.
If we have one closed curve around the hole and one closed curve apart from the hole, than both curves are not homotopic. Though they are both S₁ and so homeomorphic.

But, as for classifying them... I don't know.

I mean the classification of 3-manifolds is an important major mathematical problem of this century, I would be really surprised if youd know the answer Though a major step towards this classification was the Poincare conjecture which now seems to be proved by some chinese mathematicians.

But, if it tapers to a single point, then it's not a smooth manifold.... so I'm not sure it fits with what you're trying to classify. You said "There is a topological classification of all 2dim closed surfaces". I think "smooth" was kinda tied up in the definition of surfaces there.

Does smooth mean "continuous" or "infinitely differentiable"? I think it depends how you taper it. It's possible to taper a torus to a point without it intersecting itself.

Smooth does indeed mean (arbitrarlily times) differentiable (as in real life, where smoth means having no edges, and tips.). Continuous is always assumed for a manifold.

Otherwise, you could take a non-trivial circle on a 2-torus, scrunch it down to a single point, and come up with something that isn't a sphere with handles or a sphere with crosscaps.

But a circle is not homeomorphic to a point and the deformation of a circle into a point is not continuous. And the so constructed object is no closed 2-manifold because it overlaps/forks at this point (i.e. each embedding wouldnt be injective or wouldnt be continuous).

Injective meaning "one-to-one", but not "onto", right?

Exactly. (Bijective means "one-to-one" and "onto".)

That looks similar to what I said, except you're not using parametric equations. Was I essentially right after all, or did I miss some conditions?

Essentially right. We have to take special care about (self-)intersections. If two *embeddings* are homotopic, you have to regard that the intermediate steps also have to be embeddings (and must not be self-intersecting), as with the knots. Also the intermediate steps must completly be located in the given space (in a space like the disk with a hole, they must not pass the hole.)

Are you talking about a function from a vector to a vector?

Generally I talk about a function from one topological space to another topological space. Where you can regard a topological space as a set together with the environments of each point.
In a metric space for example the basic environments of the point p are given by {x: d(p,x)<eps} for all eps>0. I.e. by the balls around the point. (I say "basic" because all environments of p are then all sets that contain such a ball. There is also a formal definition of an environment system, which I omit here...) So a metric space is also in a natural way a topological space, same for a vector space with norm.

Like the transformation T(r, th) = {r cos(theta), r sin(theta)} that transforms a rectangle into a disk?

This is especially not continuous if you take R₊x[0,2pi) as domain, or it is not injective if you take R₊x[0,2pi] as domain. But if you would take R₊xS₁ as domain (i.e. you topologically identifiy 0 and 2pi in having the same environments) then the transformation would be continuous. If you take for example (1,2)xS₁ you get a disc with a hole, if you take (0,1)xS₁ you get a perforated disc, but if you take [0,1)xS₁, yielding the whole unit disc, it is no more injective though continuous.

[PWrong]

No!!! Absolutely not! Homotopic means deformable into each other in a surrounding space. Homoemorphic means (very roughly) "in the same way connected".

Oh, ok. I think I get it. I think I need to do some exercises to understand all this though.

Generally I talk about a function from one topological space to another topological space. Where you can regard a topological space as a set together with the environments of each point.

Ok, now I need a definition for "environment". I couldn't find it on mathworld.

Perhaps you could add some stuff to the wiki about which objects are homotopic/homeomorphic to what. If we had some data in 2D and 3D, the rest of us might be able to develop an intuition and help with the rest.

[houserichichi]

Read this book: Allen Hatcher's Algebraic Topology.

[bo198214]

Ok, now I need a definition for "environment". I couldn't find it on mathworld.

Oops perhaps I used the wrong word (a direct translation from german). As I now see the word is not environment but neighborhood.
The definition is as follows:
We assign each point the set B(x) of its neighborhoods (B is bold to denote a set of sets) such that the following axioms are valid:
1. For each neighborhood V of x is x in V.
2. For each two neighborhoods V and W of x is VnW again a neighborhood of x
3. Each superset of a neighborhood of x is again a neighborhood of x.
4. For each point y in a neighborhood V of x there is a neighborhood of y that is completely contained in V.

If we want to restrict to a base, i.e. a system such that all environments are gained taking all supersets of sets in this base system. The conditions are slightly modified:

1. as above
2. for V, W there is a neighborhood U of x such that U c VnW
3. discarded
4. as above

For example are all epsilon-balls around x in an euclidean space a base of all neighborhoods of that point x. Even all 1/n-balls around x are a base.

On the other hand this is general topology. The definitions are barely needed in algebraic topology, which is already indicated by the lack of this basic definition in the book introduced by Houserichichi. For the intuitive understanding of continuous deformation, general topology is not needed.

PS: I am away till Sunday.

[wendy]

A criss-cap in 4d is like a knot in 3d. You can make them, but not as parts of solids or something.

Remember first that in 4d, 2d does not divide space, and so one can freely twist it around as one twists 1d in 3d. You can even tie a hedrix or 2-space into a reef-knot prism, ie a space wz appears in xyz as the line z, which you can then knot.

The only real topologicals in 4d is two kinds of hole: tunnel and bridge.

A tunnel is what you get if you stick a line through a solid, and leave the hole behind. A bridge is where you stick a tunnel through the outside.

Apart from that, you can sort of knot things etc, as in 3d, but you have to deal with 4d limitations and freedoms.


W

[bo198214]

Yes but what about the embeddability of the basic identifications of the 3d-cube:
+++ (3d-torus)
++- (?)
+-- (?)
--- (?)

Where the 3 positions stand for the 3 opposite face pairs of the cube and + means to identify them in the same orientation and - means to identify them in the opposite orientation ...
And how would you build them with tunnels and bridges?

[Marek14]

How about identifying two opposite faces with a 90-degree twist, and working from there? Or identifying two faces through a mirror?

[Keiji]

PWrong: Not sure if someone managed to explain the square with arrows thing to you or not, because I didn't make much of this topic apart from the first post, so here's my explanation:

For the arrows:
+ means "up" or "right"
- means "down" or "left"
order of signs is left, top, right, bottom.

Combinations possible:
"++++" or "----" or "+-+-" or "-+-+": The torus.
"++--" or "--++" or "-++-" or "+--+": The real projective plane, which seems to be impossible in any dimension.
"+---" or "-+--" or "--+-" or "---+" or "-+++" or "+-++" or "++-+" or "+++-": The Klein bottle.

How you get there: Take the square, and attach each side to the opposite side in three or more dimensions, making sure that the arrows correspond to each other. So for the torus, it's obvious as you simply curve it up to attach top and bottom, and then curve the uncapped cylinder you created round to attach the sides. For the Klein bottle, you start with the uncapped cylinder and then twist round this object in the fourth dimension so when attached, the arrows correspond properly. This object is of course made similarly to the Möbius strip, where you twist round a rectangle.

I like the idea of your "cube with arrows", but I can't see how it would work. I imagine the arrow would have to point along a diagonal of a face, rather than on the edges?

[bo198214]

Marek, as I already told it seems to me that in homology two faces (of any dimension, here for the case d=2 as faces of the cube) can only be connected in two ways that is equally oriented or oppositely oriented. To be really sure perhaps one have to read the book suggested by houserichichi. But we can take at look at how we can connect those the cube faces

Code: Select all

  4      4
  /|3    /|3
1|/    1|/
   2      2


I imagine it as pulling out a hose from each of the opposite faces of the cube and now asking how to connect them.
If we for example connect 11 then it is clear that 44 and 22 and hence 33 must be connected to get a closed manifold. Not matter with what identification of edges we start the other identifications are then already determined to get a closed manifold. And now each connection where we start identifying x and y in the same direction leads to a topological equal identification, and all identification where we start identifiying x and y in opposite direction leads also to topologically equal identifications. These are the both ways we can connect the opposit faces of a cube. (I know the description is a bit blurry.)

Rob, the real projective plane is embeddable into 4d space. In general every k-dim manifold can be embedded into 2k-dim space (this is the already mentioned Whitney embedding theorem).
These "arrows" are indeed the orientations of homology. Each face has an either positive or negative orientation according to the order of the vertices. The 1dim edges are only a special case.