On the possible forms of our space
What does the space we live in look like? Such a question appears
quite frequently in various journals and an answer is usually
provided by an architect, an ecologist or a geographer. What can
a mathematician add to this?
For instance, you can ask him about the geometric properties of
the surrounding space. Naturally, what we are interested in is
the space itself, and not the objects it contains. Therefore,
to make things easy, we will simply neglect the latter.
So, imagine you are standing in the middle of a huge and dusky
empty hangar. What geometric features of the space are you liable
to observe? It seems that the only reasonable observation is that
you can freely move around. If you focus your attention on one
particular point, e.g. on the tip N of your nose, then
you can displace it in any direction over a practically arbitrary
distance. Moreover, even if there is an infinity of directions
to choose of, any translation of the point N can be obtained
by successive displacements in three (only!) "basic"
directions: front - rear, right - left and top - down. As for
the distances, we'll say that a translation of 10 m to the rear
is the same as a translation of -10 m to the front, and similarly
for the other two distinguished directions. Then any displacement
in each of the basic directions can be described by one real number,
while any arbitrary displacement is uniquely determined by a triple
(x, y, z) of real numbers.
Fig. 1. Let No
be a fixed arbitrary point of our space. The position of any other
point N can now be expressed by a triple of real numbers
describing the displacement that carries No
onto N. This approach makes possible an identification
of the space with the set of all triples of real numbers.
Now we have at our disposal a nice arithmetical model of the space. The points correspond to number triples and the distance between the points N1=(x1, y1, z1) and N2=(x2, y2, z2) can be derived from Pythagoras's theorem:
By treating points as number triples we can describe even complicated subsets of our space by equations or inequalities, easily subject them to various kinds of transformations etc.
The above model of the space has one more advantage: it can be easily carried over to "other dimensions". Indeed, why not consider n-tuples (x1, ..., xn) of real numbers instead of triples and use the obvious generalisation of Pythagoras's formula to measure distances between them? Thus we arrive at an n-dimensional Cartesian space , a simple model for phenomena determined by n independent parameters.
Does this really mean that we live in the three-dimensional Cartesian space ? Remember that our description originated in the hangar, i.e., it relies on the observation of a very small fragment of the space only. Thus it cannot be excluded that by hastily generalising the results of our local investigation to the entire space we have committed some important errors.
You can easily imagine such errors in the two-dimensional case: a sphere with large radius does indeed look like a fragment of a plane when observed at a very close distance, although it certainly is not one. This is why the notion of a flat Earth once had so many fervent supporters.
What certain knowledge do we have, then? The point No has a neighbourhood that looks like a part of . Besides, there is no reason to believe that the part of the space we live in is anyhow distinguished, so most probably each point N has a similar neighbourhood.
This brings us to the notion of a manifold. An n-dimensional manifold is a space with the following property: each point has a neighbourhood that looks like a fragment of the Cartesian space . Hence, we live in some three-dimensional manifold. Which one?
The answer to this question requires some physical experiments. There is, however, something left for the mathematician. namely, he should provide a list of all possible worlds together with a description of their geometric properties. Only then will the physicist be able to decide which of the proposed models corresponds to the physical reality.
Thus our problem consists in creating a complete list of three-dimensional manifolds.
How can we cope with this problem, if it's not even obvious that there are three-dimensional manifolds at all which are not itself? If we are to look for some new one, it should "bend" somehow, but it's rather difficult to imagine how this can occur.
The reason for this difficulty lies in our habit to situate things in our nearest neighbourhood, i.e., in precisely. Therefore we must learn to think of spaces in a different way.
For instance, we can imagine that the space has been divided into small polyhedral solids, which fill it completely. Rather then "seeing" the entire space in one glimpse, we may think of it as a collection of solids glued one to another by a pair of faces. Clearly, the gluing should be performed so that any point on the weld has a neighbourhood of the type.
The solids can be tetrahedrons. If so, we say that the space has been triangulated. Moise's theorem of 1952 states that there is a triangulation for every manifold of dimension three.
If you come down to dimension two, the analogue of our situation would be to use triangular block to construct a surface which resembles in the neighbourhood of each of its points. It is not hard to produce in this way an infinite series of cracknels (see Fig. 2) with an increasing number of holes.
Cracknels are compact and connected surfaces. Compactness
means that they have been glued up out of finitely many triangles,
whereas connectedness is the property of being one piece, i.e.,
any two triangles in a triangulation can be connected by a chain
of triangles such that any two successive triangles in the chain
have a common side.
Fig. 2. All these are cracknels: with no holes (a
sphere), one hole (a torus), two holes and three holes.
Let's return to dimension three. How can a compact and connected three-dimensional manifold be glued up of tetrahedrons? Obviously, the essence of the problem is to describe the gluing. This can be done quite easily due to a simple trick.
Assume for a while that we have been given such a manifold with
a triangulation. Then we can associate with it a system of points
and of segments between the points, which will be called the dual
graph. We proceed in the following way: choose a point within
each of the tetrahedrons. Two such points are joined by a segment
if and only if their corresponding tetrahedrons have a common
face (Fig. 3).
It can be easily proved by induction that every finite connected
graph contains a maximal tree, i.e., a subgraph that includes
all the vertices of the graph and has no loops (Fig. 4).
Now break the manifold up into separate tetrahedral blocks and reassemble them gluing only those faces that correspond to the edges of the dual graph which belong to some fixed maximal tree. The result will be a polyhedron with triangular faces, which can be easily embedded into . Observe that it contains all the blocks of the original manifold; if you want to reconstruct the manifold, you just have to complete the gluing process and that consists in gluing pairs of triangular faces of our new solid. We have thus proved that every compact and connected three-dimensional manifold can be obtained from some polyhedron by an appropriate identification of its faces.
Consider, for example, the rectangular parallelepiped ABCDA'B'C'D'
Fig. 5. An impossible picture: by successively gluing
pairs of opposite faces of the rectangular parallelepiped we get
a three-dimensional torus.
By gluing pairs of opposite faces we get a manifold called the three-dimensional torus. Being compact, this manifold is clearly not , since is not compact.
If we change the polyhedron and the way its faces are identified, new examples of manifolds may arise. We know by now that all compact and connected manifolds can be produced by this procedure. But then, how many of them are there? Suppose two very patient students have been given a polyhedron each and asked to show all possible ways of gluing pairs of faces that yield a manifold. Assume moreover that the task has been completed. How can we be sure that the manifolds thus obtained are indeed different?
The problem of distinguishing objects is solved in mathematics by indicating an appropriate invariant. Let be the set of all n-dimensional manifolds with triangulation. To distinguish two manifolds M1 and M2 we may find a function a function which assigns to each manifold X an integer f(X) so that if X and Y are the same (i.e. homeomorphic), then f(X)=f(Y), and moreover, .
This observation is, of course, completely trivial. All the art lies in finding good functions f.
First of all, let's have a look at compact and connected manifolds of dimension two. To make things simpler, we restrict our considerations to two-sided surfaces, i.e. those that do not contain the Möbius strip.
By copying the argument of the three-dimensional case you may check for yourself that every such surface can be derived from some polygon by appropriately gluing pairs of its sides. Some not too sophisticated manipulation with the polygon further proves that every two-sided surface is a cracknel Pg with g holes, for some non-negative g. To have a complete classification of such surfaces it only remains to prove that the cracknels Pg are different for different values of g.
Let's start with P0, i.e. the two-dimensional sphere. You can imagine a sphere divided into triangles as the border of a convex polyhedron with triangular faces. If V, E and F represent the number of vertices, edges and faces respectively, then the celebrated Euler's theorem states that independently of the partition of the sphere, always V-E+F=2.
So, consider a function with integer values given by the formula . As for the sphere, it can be proved that the number does not depend on the triangulation of the surface X, i.e. is an invariant, called the Euler-Poincaré characteristic of the manifold X. Thus it is sufficient to compute for non-negative g. If we find that the values are different for different g, we will have proved that the surfaces Pg are pairwise topologically different.
We know already that . Observe that the cracknel Pg+1 can be obtained from Pg by performing successively two operations (Fig. 6):
Fig. 6. From g=1 to g=2.
Note also that the cracknel Pg+1 thus obtained has 2 triangles, 3 edges and 3 vertices less than the original cracknel Pg. If , then . An easy inductive proof yields the formula .
Let's try to carry our experiments over to the three-dimensional case. If the triangulation of a manifold X consists of T tetrahedrons, which together have V vertices, E edges and F faces (common faces are only counted once!), then we may consider the number .
The beginning seems to be promising: it can be proved that also in this case the number does not depend on the triangulation of X. Alas, there is an unpleasant surprise awaiting: this time the function will not satisfy our expectations. In fact, the following theorem holds: for any compact three-dimensional manifold X, .
The theorem has a simple and yet very ingenious proof. Once again we'll make use of the cracknels. Observe that every one of them, considered as a subset of , is the border of some three-dimensional figure which we will call a full cracknel and denote by Qg. For example, Qo is the good, old sphere.
What is Euler-Poincaré's characteristic of the set Qg? The sphere Qo has a natural triangulation consisting of one tetrahedron, i.e. V=4, E=6, F=4, T=1, hence . For the subsequent full cracknels, as well as for the empty ones, an inductive computation yields .
Now we note that each compact connected manifold X of dimension three can be represented as the join of two equal full cracknels Qg which meet in the common border Pg. The first of the two Qg's can be produced by making all the edges of some triangulation X thicker. The fact that what remains of X is again a full cracknel may require some perceptiveness. Let's recall the dual graph corresponding to the given triangulation once more. Its thicker version is obviously a full cracknel.
Thus we have placed two disjoint full cracknels within X.
Imagine now that you inflate them and watch what is going on with
one of the tetrahedrons of the triangulation. It is easy to see
that after some time both cracknels will meet border to border,
filling tightly the entire X. Both of them having a common
- and therefore identical - border, they are both of the same
form Qg. The decomposition of X
thus obtained is known as the Heegaard decomposition of
This decomposition will be of much help in the computation of the characteristic of X. Note that if a polyhedron X is represented as the join of its two subpolyhedrons M and L, then . The formula would be obvious, were just the number of vertices (edges, faces etc.). Now you can just observe that the formula we intend to justify is but an alternating sum of such obvious formulas.
Apply the formula to Heegaard's decomposition , where M=L=Qg, . We now have , which is precisely what we intended to prove.
Summing up: Euler-Poincaré's characteristic is useless for distinguishing three-dimensional manifolds. Other, more efficient invariants are needed. Maybe you have some good ideas in this respect?
At the moment of this writing (i.e. December 1996), the classification
problem for three-dimensional manifolds, i.e. the problem of describing
all the possible forms of our space, remained open.