# B How a Klein bottle has a 0 volume?

1. Dec 15, 2016

### parshyaa

How a klein bottle has a 0 volume?
I am a 12th grade student and i don't know anything about topology,klein bottles but i am curious to know how something which exist has a 0 volume. I just watched clifford stoll's klein bottles video on numberphile and i came here to know how a klein bottle has a 0 volume, please explain it in layman's term if possible, thanks

2. Dec 16, 2016

### andrewkirk

A klein bottle is a surface, which is a two dimensional manifold. It has a surface area, but not a volume.

The same applies to the unit sphere $S^1\triangleq \{(x,y,z)\in\mathbb R^3\ :\ x^2+y^2+z^2=1\}$. That sphere encloses a volume, because it is a closed manifold embedded in $\mathbb R^3$. But it does not have a volume (the volume is not a measure of any part of the manifold).

The Klein bottle cannot be embedded in $\mathbb R^3$, so it cannot even be said to enclose a volume (it does not have an inside or an outside).

I haven't watched your video so I don't know whether the talker was referring to 'having' a volume or 'enclosing' a volume.

3. Dec 16, 2016

### parshyaa

4. Dec 17, 2016

### Logical Dog

Yeah its just a surface thats been moulded to look kinda 3d I guess. Like a surface of a cube vs the cube itself.

I think. You can take the one surface away and transform it into something that looks 3d I guess , but it wont have any volume, its still one surface. It has lines and length but no width. You are still manipulating its points but not extruding any of its points along any direction.this bottle is more confusing because you cant come across its edge if you "walk" on it.

In school we were taught The formula for a volume of a sphere is 4/3 * PI * r3.

Last edited: Dec 17, 2016
5. Dec 19, 2016

### weirdoguy

Ball, not sphere.

6. Dec 19, 2016

### FactChecker

The idea of a Klein bottle is that the "inside" bends around and becomes the "outside". Look at the picture and follow the "inside" around to see what I mean. So the "bottle" really has no inside as we usually think of it. With no inside, it is a "bottle" with zero volume.

7. Dec 19, 2016

### dkotschessaa

You could buy one and try to fill it. Makes a mess. Trust me.

8. Dec 19, 2016

### parshyaa

Nicely explained thanks

9. Dec 19, 2016

### parshyaa

Not in india

10. Dec 19, 2016

### dkotschessaa

http://www.kleinbottle.com/

Well, he didn't say he DOESN'T ship to India.

Maybe you should start making them. :)

-Dave K

11. Jan 6, 2017

### lavinia

It is a little bit misleading to say that the Klein bottle has no inside or outside. If one means that it does not enclose a volume in $R^3$ that is true but sort of besides the point since it can not be embedded in $R^3$ in the first place. One might say that to create a Klein bottle one must pass a cylinder through itself so that the inside and the outside become connected. That is also true but the resulting surface is not a Klein bottle. In fact it is not even an surface.

The Klein bottle is in fact the boundary of a 3 dimensional manifold so in some sense it does enclose a volume. To see this take a solid cylinder and bend it around to bring the two disks on its two ends together. Instead of gluing them together directly to make a solid torus, glue them by a reflection through an axis through their centers. This makes a 3 dimensional manifold whose boundary is the Klein bottle.

More generally, there are compact closed manifolds in every dimension that are boundaries but can not be embedded as hyper-surfaces of Euclidean space. For instance, all compact flat Riemannian manifolds - of which the flat Klein bottle is an example - are boundaries.

There are closed surfaces that are not boundaries of 3 dimensional solids. The simplest example is the projective plane.

12. Jan 6, 2017

### FactChecker

Does that mean that it partitions a (higher dimensional?) space into three pieces -- the Klein bottle itself, an inside and an outside?

Last edited: Jan 6, 2017
13. Jan 7, 2017

### lavinia

No. The dimension of the solid Klein bottle would be too low. For instance in four dimensions the inside would have to be four dimensional.

14. Jan 7, 2017

### andrewkirk

Just musing a bit further on that. According to the Jordan-Brouwer Separation Theorem, $\mathbb R^4$ can be partitioned into three parts by an embedded 3-sphere, which is a 3D manifold (without boundary).

I would guess that any 3D manifold that has an embedding in $\mathbb R^4$ that partitions the latter into three components must be homeomorphic to the 3-sphere, and presumably also its embedding is homotopic to that of a 3-sphere. But I have not seen statements or proofs of either of those. If it's true, that would be sufficient to prove that the solid Klein bottle interior does not partition $\mathbb R^4$, because it is a manifold with boundary, which cannot be homeomorphic to a manifold without boundary.

15. Jan 8, 2017

### lavinia

You bring up an interesting line of thought.

The Jordan-Brouwer Separation Theorem is more general. It says that any compact connected hyper-surface ( a hyper-surface is assumed to have no boundary) of Euclidean space of any dimension separates Euclidean space into two disjoint components whose common boundary is the hyper-surface itself. For $R^4$ the 3- sphere is an example but there are many others. For instance the three dimensional torus can be embedded in $R^4$.

One might ask what the topological restrictions are for a compact connected hyper-surface of Euclidean space. One of them is that the manifold must be orientable. For instance, the Klein bottle is not orientable and can not be embedded as a hyper-surface of 3 space.

One can prove that a hyper-surface must be orientable in several ways:

By Alexander Duality $H_{n-1}(M^{n-1}) = H^{0}(R^{n}-M^{n-1})$ where $H$ is reduced homology and $R^{n}-M^{n-1}$ is the complement of the manifold in $R^{n}$. (Reduced homology is the same as singular homology except in dimension zero where the number of generators is one less than the number of connected components. ) If the complement of the manifold in $R^{n}$ has two components then $H_{n-1}(M^{n-1})$ has one generator and is thus isomorphic to $Z$. A general theorem on compact manifolds without boundary is that the top integer homology is $Z$ if the manifold is orientable and $0$ if it is not.

Another proof involves showing that the hyper-surface has a well defined unit normal field. This follows from the Jordan-Brouwer separation Theorem. This is an interesting exercise in itself.

In conclusion, no non-orientable compact smooth manifold of any dimension can ever be a hyper-surface of Euclidean space.

Back to the solid Klein bottle:

If the complement of the solid Klein bottle in $R^4$ had two components then the Alexander duality argument would imply that $H_{3}($SolidKB$) ≈Z$. But the top integer homology of a compact manifold with boundary is always zero so by Alexander Duality, its complement in $R^4$ must have a single component.

Why is the top integer homology of a smooth n-manifold with boundary always zero? The idea is that for any triangulation of the manifold( a smooth manifold is always triangulable) pairs of n-simples can share at most a single n-1 face. Therefore in order to get an n-cycle, these faces must cancel in pairs under the simplicial boundary operator. But the n-1 faces of the boundary lie on only one n-simplex so they cannot cancel out. Therefore there is no n-cycle and the top $Z$-homology is zero.

For the solid Klein bottle one does not need this general homology theorem. Like the solid torus, the solid Klein bottle can be continuously shrunk onto its central equatorial circle. This means that it has the homology of a circle so its third homology is zero.

The Projective Plane:

Another topological restriction for a hyper-surface of Euclidean space is that is must have even Euler characteristic. So any manifold with odd Euler characteristic can not be a hyper-surface. The Euler characteristic of the Projective Plane is 1 and this gives another proof that it can not be embedded in $R^3$. One might ask whether the Projective Plane can be a boundary of some manifold other than a bounded region of $R^3$. The answer is no. In fact, any manifold that is a boundary must have even Euler characteristic. ( Note that the Euler characteristic of the Klein bottle is zero which is even.)

It is remarkable that there could be a smooth compact surface with no edges that still can not be solidified into a manifold with boundary. It would be interesting to try to see why this is true geometrically, maybe from a parameterized embedding of the Projective Plane in $R^4$.

Last edited: Jan 8, 2017