Singular Homology, The Homology of the Spheres

A More General Approach

In order to derive h(Sⁿ), we will solve a much more general problem. If Sⁿ is all one wants it is not necessary to do this much work, but cellular homology uses h(Dⁿ/S^n-1), and other applications involve h(X*Sⁿ) for various spaces X. Therefore, we will derive a formula for h(X*Dⁿ/X*S^n-1) in terms of h(X), and use this to construct a formula for h(X*Sⁿ) in terms of h(X). Set X to a single point to obtain h(Sⁿ).

Notation Review

Remember that Sⁿ is the n dimensional sphere, which is often pictured in n+1 dimensions, and Dⁿ is the n dimensional disk or ball. Thus S^n-1 is the topological boundary of Dⁿ, and it is reasonable to talk about the homology of the disk relative to its boundary. Beyond this, we may take the topological product of these two spaces with another space X, and talk about the relative homology of X*Dⁿ relative to X*S^n-1.

The Homology of the Disk Relative to its Spherical Boundary

Consider the three spaces X*Dⁿ, X*S^n-1, and X*U^n-1, where U^n-1 is the norther hemisphere, i.e. the northern half of S^n-1. Take these in the reverse order, so that X*U^n-1 embeds in X*S^n-1, embeds in X*Dⁿ. The corresponding chain complexes embed as well. The first is a subchain of the second, is a subchain of the third. Take the triple homology sequence, and the quotient groups become exact at every dimension.

0 → C_i(X*S^n-1/X*U^n-1) → C_i(X*Dⁿ/X*U^n-1) → C_i(X*Dⁿ/X*S^n-1) → 0

Take the long exact sequence to build a chain of relative homologies through all the dimensions.

… → h_i(X*S^n-1/X*U^n-1) → h_i(X*Dⁿ/X*U^n-1) → h_i(X*Dⁿ/X*S^n-1) → …

Notice that the disk retracts continuously onto its northern hemisphere. Therefore the middle column consists of a space relative to its retract. The relative homology is 0. This means the first and third columns are equal, with a phase shift. We are interested in the third column - so determine the first column, and we're done.

Use the excision theorem to delete the north pole. This does not change the relative homology. Now pull the punctured northern hemisphere down to the equator, leaving the southern hemisphere alone. The result is a disk and its equator, one dimension lower. Since the two spaces are homotopically equivalent, the relative homologies are the same. The homology of the first column equals the homology of the third column in another table, based on the disk and sphere at a lower dimension.

With a middle column of 0, the first and third columns correspond, with a phase shift, thanks to the wrap around of the long exact sequence. And the first column corresponds to the third column in another table. Put this all together and write the following equation.

h_i(X*Dⁿ/X*S^n-1) = h_i-1(X*D^n-1/X*S^n-2).

That's great, but we need a place to start. Consider D¹ containing S⁰ containing q, where q is one of the two points in S⁰. This yields another triple homology sequence in which the first column, h(X*S⁰)/h(X*q), yields h(X). the second column, h(X*D¹/X*q) gives 0 by homotopy equivalence. The third column equals the first column with a phase shift.

h_i+1(X*D¹/X*S⁰) = h_i(X)

With this as a base, we can write the following formula.

h_n+i(X*Dⁿ/X*S^n-1) = h_i(X)

Set X to a point, which has no homology beyond h[0]. Thus the reduced homology h_n(Dⁿ/S^n-1) = 1.

The Homology of the Spheres

I think this is the music of the spheres - don't you?

We wish to find the homology h(X*Sⁿ) in terms of h(X). Write the long exact sequence for the inclusion of X*Sⁿ into X*Dⁿ⁺¹ to get the known homology of X*Dⁿ⁺¹ relative to X*Sⁿ. The middle column is known to be h(X), since X is homotopy equivalent to X*Dⁿ⁺¹. The third column can be mapped back to h(X). The first column is what we seek.

We will prove the first column is split exact - the direct product of the second column and the third column one row up. This makes h(X*Sⁿ) the direct product of two homology groups from X.

Look at row n+i. The middle column is h_n+i(X). Up one row and to the right is h_n+i+1(X*Dⁿ⁺¹/X*Sⁿ), which is h_i(X) by the previous theorem. When we're done, h_n+i(X*Sⁿ) will be the direct product of these two groups.

Let c be a cycle in X*Dⁿ⁺¹ representing a homology class. Since the inclusion of X into X*Dⁿ⁺¹ is a function homotopic to the identity map, inducing a homology isomorphism, we may pick cycles in X to represent the homology classes of X*Dⁿ⁺¹. These cycles are also present in X*Sⁿ. Moreover, if two such cycles were homologous in X*Sⁿ they would be homologous in X*Dⁿ⁺¹, hence the correspondence is 1-1. The map from the first column to the second is surjective, and there is a reverse map which, when composed with the forward map, gives the identity map on the homology groups. We are one step away from being split exact. The map from the third column to the first must be injective, or equivalently, the map from the second column to the third must be trivial.

Again let cycles in X represent the homology classes of X*Dⁿ⁺¹. All of these cycles wind up in X*Sⁿ, and are homologous, since X*Sⁿ is modded out in the relative homology group. That completes the proof, and establishes the following formula.

h_n+i(X*Sⁿ) = h_n+i(X) * h_i(X).

Make X a point to find the homology of the sphere. Since X is convex, its only homology is h₀. This comes into play when i = 0 or -n. The reduced homology of the sphere is 1 at dimension 0, as expected, since Sⁿ is path connected, and 1 at dimension n. If n = 0, the reduced homology is 2 at dimension 0, and 0 elsewhere. This is consistent with the results we found earlier, for h₀ and h₁.

The Spheres are Topologically Distinct

It's practically a corollary. Homology is a topological invariant, and each Sⁿ has its own homology, hence each Sⁿ is distinct. The circle really is different from the sphere.

The Homology of the Torus

Set X to the circle, which has reduced homology h₀ = h₁ = 1. The torus is now X*S¹. Apply the formula, and get the following reduced homology.

h₀ = 1
h₁ = 2
h₂ = 1

Let X = T² and cross with S¹ to get T³. The above numbers are shifted and added to themselves. Continue this process and build pascal's triangle. Thus reduced h_i(Tⁿ) = n choose i.

Finding the Generators

Exactly what are the simplexes that generate the relative homologies h_n(Dⁿ/S^n-1)?

Recall earlier, we said we needed a place to start, and we built a triple homology sequence for D¹ containing S⁰ containing q. You know this better as an interval, containing its endpoints, containing one of its endpoints. The boundary operator induces a relative homology isomorphism from the interval mod its endpoints to the endpoints mod q. The latter only has a 0 homology, and its generator is the trivial map from the 0 simplex onto p, which is the point opposite to q. A 1 simplex, mapping into the interval, is a generator of its homology iff the boundary operator maps it onto ±p, and possibly q, which we don't care about. This hapens iff the two ends of our 1 simplex map to p and q. Any continuous function from one end of the interval to the other will do. The linear map is the simplest example. This is a generator for h₁(D¹/S⁰).

A linear combination of simplexes in our unit interval can be broken up into disjoint cycles and chains of red arcs, as described in the 1 homology. An open ended chain can still be a cycle, if its endpoints are p and q. Add up the chains that end in p, and subtract the chains that end in q. The result generates h₁ iff the sum is ±1.

Now step up to the disk, with the circle as boundary. Place a triangle inside the disk, and let c be a point inside the triangle. Draw a ray in any direction, starting at c, through the triangle, and out to the circle. Use these rays to push the triangle out to the circle. This defines a map, in fact a homeomorphism, from our 2 simplex onto the disk. The boundary maps to the circle, so it is a cycle, and represents an element in the relative homology h₂(D²/S¹). Apply the boundary operator and find the boundary of the triangle embedded in the circle. We are interested in the relative homology of the circle mod the upper semicircle, as described above. To get there, we used the excision theorem, and that means we may need to chop the three arcs into smaller pieces, so that each piece fits in the upper semicircle, or in the circle without V, which was removed from the top. The result is a chain of arcs from one side of the upper semicircle, down around the bottom, and reaching into the other side of the upper semicircle. Apply a homotopy, and this becomes a chain of arcs connecting the two endpoints of an interval. This generates the 1 homology, as described above. Therefore our original function was a generator for the 2 homology of the disc relative to its boundary.

Many other functions are possible. They need not be neat clean radial maps from a point c in the disk. Any continuous function will do, as long as it wraps the border of the simplex once (net) around the circle. (Remember that we have reversed the blue path and made it red, so the singular boundary marches counterclockwise around the topological boundary.) Place a start point s in V, which was the arc of the upper semicircle that was excised. The border of the simplex leaves V, and returns, and leaves, and returns, perhaps many times, but it always returns. The sides of V become p and q in the interval. Leaving and returning on the same side creates a path whose boundary is p-p, or q-q; basically nothing. A path that moves from one side to the other gives p-q, or q-p. Add these up and the coefficient on p corresponds to the winding number of the path, which is the number of times the boundary of the simplex wraps around the circle. This has to be once, either clockwise or counterclockwise. This is a necessary and sufficient condition for a simplex to be a generator of h₂ on the disk relative to its boundary.

What if the simplex maps to the disk, which then maps to itself via the complex function z². This has a winding number of 2, and is not the generator of the homology. Twice the original map should be homologous to the z² map. The boundary of some chain of 3 simplexes has two of the original z maps and - a z² map. It's only 3 dimensions, but I can't picture such a chain of simplexes right now. Perhaps you can send a description along.

Return two our original generator, the 2 simplex mapped homeomorphically onto the disk, and take a subdivision. The boundary of the subdivision is the subdivision of the boundary, which lies on the circle. It has the same winding number; hence any subdivision of our generator is still a generator for h₂.

Beyond this, we can merge adjacent triangles together and still get the same boundary. For instance, the first subdivision of a triangle produces 6 smaller triangles; But we can merge adjacent triangles together to get three triangles. This is a nonstandard subdivision, but it has the same boundary, and is a generator of h₂.

Moving up to the 3 dimensional ball relative to its surface, embed the 3 simplex in the ball, place a point c inside, and push the simplex outward towards the surface of the ball, like blowing up a balloon. This is a 3 simplex that generates h₃(D³/S²). Excise V from the sphere, and the boundary of the 3 simplex becomes three triangles of the tetrahedron; the fourth triangle being contained in the northern hemisphere. This is the aforementioned nonstandard subdivision of the triangle that generates h₂ on the disk relative to its boundary. Thus our original simplex, and any subdivision thereof, generates h₃.

This reasoning continues, by induction, up through higher dimensions. The simplest radial map you can construct from an n simplex onto the n ball, or anything homotopic thereto, provided the homotopy carries the disk into itself and keeps the sphere within the sphere, generates the relative homology.

Moving on, what are the generators for the homology of the spheres? Refer again to the long exact sequence that equates the homology of a disk relative to its boundary with the homology of the sphere in the next row down. The sphere is the boundary, and the map is the boundary operator. Thus the boundary of a simplex, one dimension up, wrapped around a sphere, generates the homology for that sphere.

Anything homotopic will do as well, provided the homotopy is derived from a homotopy of the sphere within itself. Consider the 2 sphere, and expand one of the triangles outward until it squashes the other three triangles onto the north pole. These degenerate simplexes are - + and -, so only one is left. This leaves one simplex covering the sphere, minus the trivial simplex at the north pole. If the dimension of the sphere is odd, the trivial simplexes cancel completely. This suggests a fundamental difference between even and odd dimensional spheres.