# Mobius strip and 4D

1. Sep 12, 2014

### volcanolam

What is the relationship with the Mobius strip (or loop) and the 4 dimension? Is Mobius strip a four dimensional object?

2. Sep 12, 2014

3. Sep 12, 2014

### HallsofIvy

Staff Emeritus
Where did you see anything that would lead you to think that the Moebius strip is a four dimensional object?

The Moebius strip is a two-dimensional object.

4. Sep 13, 2014

### volcanolam

From here:
http://stickrpg2.wikia.com/wiki/Fourth_Dimensional_Objects

Also putting two of them together in some sort of ways will result as a Klein bottle, which is truely a 4 dimensional object. This makes me believe that the moebius strip has to have relationship with the 4 dimension.

5. Sep 13, 2014

### HallsofIvy

Staff Emeritus
Putting two of them together will give you the boundary of a Klein bottle just as a putting several two dimensional squares together will give you the boundary of a three dimensional cube.

6. Sep 14, 2014

### volcanolam

7. Sep 14, 2014

### Terandol

Yes, the above link is misleading. The only actual four dimensional object in that link is the hypercube.

At least according to all the definitions I've come across, the Klein Bottle is a 2-dimensional surface which can be obtained by gluing two closed mobius bands together along their boundary or alternatively by identifying the edges of a square in a certain way. There is also the notion of a solid Klein bottle which can be obtained by taking a quotient of $D^2\times I$ which identifies the upper and lower disks via a reflection. In this case the boundary of the solid Klein bottle is the two dimensional surface usually called the Klein bottle and this might be what a previous poster was referring to when he said gluing two mobius bands together gives the boundary of the Klein bottle. In any case, the solid Klein bottle is still only a three dimensional manifold not a four dimensional one.

There is a relation with $\mathbb{R}^4$ though in that the Klein bottle can be embedded in $\mathbb{R}^4$ without self intersections which is not possible in $\mathbb{R}^3$. You've probably noticed that whenever someone draws a picture of the Klein bottle the bottle intersects itself since a portion of the tube passes through another portion to get to the "inside". In reality the Klein bottle does not intersect itself, this is just because we are trying to visualize it as a two dimensional surface sitting inside $\mathbb{R}^3$ which is impossible to do accurately. If we have one extra dimension to play with, then instead of passing the tube through itself to get to the "inside" of another portion of the tube, it is possible to get inside by moving in the fourth dimension. Hence, the Klein bottle sits inside $\mathbb{R}^4$ as an embedded 2-dimensional submanifold but it isn't an embedded submanifold of any lower dimensional Euclidean space. The Klein bottle itself is still two dimensional though.

I have no idea why they claim the mobius band is four dimensional though. The usual picture everyone draws of it shows that it can be embedded as a submanifold of $\mathbb{R}^3$ so unlike the Klein bottle, you don't even need four dimensions to embed it in Euclidean space.

Last edited: Sep 14, 2014
8. Sep 17, 2014

### lavinia

The general theorem is that no non orientable closed surface without boundary is embeddable in 3 space. The Klein bottle is one example,the projective plane another. The Mobius band is non orientable but it has a boundary so the theorem doesn't apply to it.