How a Klein bottle has a 0 volume?
andrewkirk said:
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.
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.
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##.
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.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.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.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##.
