The goal of this post is to compute the cohomology of the -torus in as many ways as I can think of. Below, if no coefficient ring is specified then the coefficient ring is by default. At the end we will interpret this computation in terms of cohomology operations.
Method 1: Mayer-Vietoris
For this particular method write . We can compute the cohomology of inductively by regarding it as the union of two copies of with intersection and using Mayer-Vietoris. The cohomological version of Mayer-Vietoris is a long exact sequence of the form
The maps are induced by pulling back along the inclusion , whereas the maps are induced by the difference between the pullbacks along the inclusions . Because these maps are homotopic to the identity map , we can think of as being given by
where , and we can think of as being given by two copies of a single map , which we’ll denote by . It follows that is the antidiagonal copy of in , hence factors through the map from to and contains a copy of given by .
It also follows that is the diagonal copy of , hence that is surjective. Finally, is the kernel of , hence the quotient of by is . In other words, we have short exact sequences
But inductively it will turn out that all the groups involved are free abelian so all of these exact sequences split. In fact, inducting on the above relation it follows that the Poincaré polynomials
satisfy and , hence
So by induction we conclude that . Note that we have not computed the cup product structure.
Method 2: the Künneth formula
This method will compute the cup product structure. is the product of copies of , whose cohomology as a ring is ; there are no interesting cup products. By the Künneth formula, the cohomology of is the graded tensor product, as algebras, of copies of (since all of the cohomology groups involved are free). This is precisely the exterior algebra , with each generator in degree . In particular, naturally, and under this isomorphism the cup product corresponds to the wedge product.
Method 3: de Rham cohomology
This method will compute the cohomology over by computing the de Rham cohomology of . One particularly nice way to do this is to use the following.
Theorem: Let be a compact connected Lie group acting on a smooth manifold . The inclusion of invariant differential forms into differential forms is a quasi-isomorphism (induces an isomorphism on cohomology).
The idea behind this result is that, since is compact, there is an averaging operator given by averaging over the action of with respect to normalized Haar measure on . But since is connected, the action of any individual element of is homotopic to the identity, so this average is also homotopic to the identity.
In particular, letting act on itself by translation, we conclude that we can compute its de Rham cohomology using translationally invariant differential forms on , or equivalently on its universal cover . But these are precisely the differential forms obtained by wedging together the -forms . The exterior derivative vanishes on all such forms, so we conclude that the de Rham cohomology of is the exterior algebra on .
Method 4: Hopf algebras
This method will compute the cohomology over . Since is a topological group, it’s equipped with a product operation . The induced map in cohomology has the form
by the Künneth formula. This map is coassociative and compatible with cup product, so equips with the structure of a bialgebra. Together with the map induced by the inversion map and the identity , the cohomology of acquires the structure of a Hopf algebra, and in fact this was Hopf’s motivation for introducing Hopf algebras. Hopf algebras arising in this way satisfy the following very stringent structure theorem.
Theorem (Hopf): Let be a finite-dimensional graded commutative and cocommutative Hopf algebra over a field of characteristic zero such that (the Hopf algebra is connected). Then is the exterior algebra on a finite collection of generators of odd degrees.
The comultiplication sends each generator to , the antipode sends each generator to , and the counit sends each generator to .
To compute the cohomology of it therefore suffices to determine what the possible generators of the exterior algebra are. For starters, let’s write more abstractly as where is a finite-dimensional real vector space of dimension and is a lattice in of full rank (the subgroup generated by a basis of ). Covering space theory gives us that . By the Hurewicz theorem, , so by the universal coefficient theorem,
This gives us generators of degree , one for each element of a basis of , and so at the very least contains the exterior algebra . But now we’re done: the cohomology can’t contain any generators of higher degree because wedging them with the generators we’ve already found would produce nonzero elements of the cohomology of in degrees higher than , and no such elements exist (either because admits a CW-decomposition involving cells of dimension at most or because the de Rham complex only extends up to dimension for a smooth manifold of dimension ).
Method 5: suspension
Recall that cohomology is a stable invariant in the sense that
where is the (reduced) suspension of (here a pointed space). Recall also that for nice pointed spaces the suspension of a product has homotopy type
Two spaces are said to be stably homotopy equivalent if for some ; in particular, stably homotopy equivalent spaces have isomorphic cohomology. The above result tells us that is stably homotopy equivalent to (once we know that suspension commutes with wedge sums). More generally, by induction we conclude that a product is stably homotopy equivalent to a wedge obtained formally by expanding
where denotes the unit of the smash product, and removing the unit. It follows that is stably homotopy equivalent to a wedge of copies of the -sphere, , and by a simple application of Mayer-Vietoris (for wedge sums), the cohomology of such a wedge is the same as what we’ve computed before.
This argument does not get us the cup product structure, since the cup product is an unstable phenomenon; after suspension, all cup products are trivial. However, it does describe the stable homotopy type of , which contains information that cohomology doesn’t (e.g. about stable homotopy groups).
Method 6: cellular homology
To compute the cohomology of it suffices to compute the homology and apply either universal coefficients or Poincaré duality. It is possible to describe fairly concretely what the homology of looks like using cellular homology. Recall that cellular homology describes a chain complex computing the homology of a CW-complex which in degree is free abelian on the -cells in a cell decomposition of . Our particular admits a cell decomposition with cells of dimension given by starting with the minimal cell decomposition of into two cells (a -cell and a -cell connecting the -cell to itself) and taking products, where we’re thinking of cubical -cells here. Equivalently, we can think of as being with opposite -faces identified, and then our cells are the faces of up to this identification.
The boundary maps in the cellular complex are as follows. If is a -cell and its attaching map (where here denotes the -skeleton of ), then the differential is
where runs over an enumeration of all -cells , denotes the degree, and is the map induced by collapsing all of except the cell to a point.
In this particular case all of the boundary maps in the cellular complex are trivial, so the homology is free abelian on cells. To see this, note that if is not surjective, then it necessarily has degree since it is null-homotopic, so we reduce to the surjective case. In this case the -cell must be a face of the -cell , and since we’ve collapsed everything else we can reduce to the case that , so that is the top-dimensional cell. At this point we will cheat a little: if in this case, then we would have , but is a compact orientable manifold and therefore must satisfy .
In particular, the cell decomposition we gave above is minimal: it is not possible to give a cell decomposition with fewer cells. In addition, by Poincaré duality the cohomology can also be thought of as free abelian on cells, and moreover we can describe the cup product in terms of transverse intersections of submanifolds representing homology classes. We can do this by explicitly intersecting the cells above, but the following description is perhaps more elegant: if we think of as , then a subspace represents a homology class if it is translation-invariant (given by the pushforward of a fundamental class). The images of two such subspaces intersect transversely if , and then their intersection represents a homology class which Poincaré dualizes to the cup product of the Poincaré duals of . In particular, note that the short exact sequence
implies that . Its Poincaré dual is therefore a class in , which has the correct degree.
Method ???: the Lefschetz fixed point theorem
This method is not numbered because the argument is incomplete. Consider the map
where each is a positive integer equal to at least . This map has fixed points, since in each coordinate the fixed points of are precisely the th roots of unity. Each fixed point has index . By the Lefschetz fixed point theorem it follows that
Knowing what we already know about the cohomology, it is tempting to identify a monomial on the LHS with a cohomology class on the RHS on which acts by multiplication by that monomial. We can do this as follows. For any subset of indices we have a projection map . Since is a compact orientable manifold, it has a fundamental class generating its top cohomology. The map induces a map on such that any point has preimages, hence has degree as a map on , so acts on the fundamental class by multiplication by . This action induces an action on the pullback of the fundamental class of to which is also by multiplication by .
As the vary this argument shows that the cohomology classes arising in this way are all linearly independent, hence all contribute to the RHS of the Lefschetz fixed point theorem. The sum of the corresponding contributions to the RHS exhaust all terms on the LHS, so if there is any more cohomology to be found then it isn’t being detected by .
Method ???: Morse theory
There is a convenient choice of Morse function on given by
The gradient of this function is , and in particular it vanishes iff for all . There are therefore critical points , organized in batches of critical points such that coordinates are equal to and coordinates are equal to . At such a point of the second derivatives of each term are equal to and are equal to , with no other contributions to the second-order Taylor series expansion of , so all critical points are nondegenerate (hence we do in fact have a Morse function) with index . Morse theory then guarantees that has the homotopy type of a CW-complex with cells of dimension .
This argument should be placed in the context of a Morse-theoretic proof of the Künneth formula; more generally, if are manifolds with Morse functions , then is a Morse function on the product , and critical points of are precisely products of critical points on the , and so forth.
With more effort Morse theory even provides a complex computing the homology, but I wasn’t able to easily compute the differentials in it (they should all vanish in this case).
Our computations admit the following interpretation. Recall that is the Eilenberg-MacLane space representing integral cohomology in the sense that there is a natural isomorphism , where denotes the space of homotopy classes of maps (or weak homotopy classes if are not CW-complexes) . It follows that represents -tuples of cohomology classes in . By the Yoneda lemma, cohomology classes in , or equivalently homotopy classes of maps , can naturally be identified with natural transformations
Such natural transformations between cohomology functors are called cohomology operations, and the computations we did above imply that the only cohomology operations of this form are generated by wedge products under addition. (“Interesting” cohomology operations over , not generated by addition and the wedge product, require higher cohomology classes as input. The smallest one is a cohomology operation ; see this math.SE question.)