The cohomology of the n-torus

October 12, 2013 by Qiaochu Yuan

The goal of this post is to compute the cohomology of the $n$ -torus $X = (S^1)^n \cong \mathbb{R}^n/\mathbb{Z}^n$ in as many ways as I can think of. Below, if no coefficient ring is specified then the coefficient ring is $\mathbb{Z}$ by default. At the end we will interpret this computation in terms of cohomology operations.

Method 1: Mayer-Vietoris

For this particular method write $X^n = (S^1)^n$ . We can compute the cohomology of $X^n$ inductively by regarding it as the union of two copies of $I \times X^{n-1}$ with intersection $X^{n-1} \sqcup X^{n-1}$ and using Mayer-Vietoris. The cohomological version of Mayer-Vietoris is a long exact sequence of the form

$\displaystyle \cdots \to H^k(X^n) \xrightarrow{a_k} H^k(X^{n-1}) \oplus H^k(X^{n-1}) \xrightarrow{b_k} H^k(X^{n-1} \sqcup X^{n-1}) \\ \xrightarrow{d_k} H^{k+1}(X^n) \to \cdots$ .

The maps $a_k$ are induced by pulling back along the inclusion $(I \times X^{n-1}) \sqcup (I \times X^{n-1}) \to S^1 \times X^{n-1}$ , whereas the maps $b_k$ are induced by the difference between the pullbacks along the inclusions $\{ 0 \} \times X^{n-1} \sqcup \{ 1 \} \times X^{n-1} \to I \times X^{n-1}$ . Because these maps are homotopic to the identity map $X^{n-1} \to X^{n-1}$ , we can think of $b_k$ as being given by

$\displaystyle a \oplus b \mapsto (a-b) \oplus (b-a)$

where $a, b \in H^k(X^{n-1})$ , and we can think of $a_k$ as being given by two copies of a single map $H^k(X^n) \to H^k(X^{n-1})$ , which we’ll denote by $a_k'$ . It follows that $\text{im}(b_k) = \text{ker}(d_k)$ is the antidiagonal copy $a \oplus (-a)$ of $H^k(X^{n-1})$ in $H^k(X^{n-1}) \oplus H^k(X^{n-1})$ , hence $d_k$ factors through the map $a \oplus b \mapsto (a + b)$ from $H^k(X^{n-1}) \oplus H^k(X^{n-1})$ to $H^k(X^{n-1})$ and $H^{k+1}(X^n)$ contains a copy of $H^k(X^{n-1})$ given by $\text{im}(d_k)$ .

It also follows that $\text{im}(a_k) = \text{ker}(b_k)$ is the diagonal copy $a \oplus a$ of $H^k(X^{n-1}) \oplus H^k(X^{n-1})$ , hence that $a_k' : H^k(X^n) \to H^k(X^{n-1})$ is surjective. Finally, $\text{ker}(a_{k+1}) = \text{im}(d_k)$ is the kernel of $a_{k+1}'$ , hence the quotient of $H^{k+1}(X^n)$ by $\text{im}(d_k) \cong H^k(X^{n-1})$ is $H^{k+1}(X^{n-1})$ . In other words, we have short exact sequences

$\displaystyle 0 \to H^k(X^{n-1}) \to H^{k+1}(X^n) \to H^{k+1}(X^{n-1}) \to 0$ .

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

$\displaystyle p_n(t) = \sum_{k=0}^n (\dim H^k(X^n)) t^k$

satisfy $p_0(t) = 1$ and $p_n(t) = (1 + t) p_{n-1}(t)$ , hence

$\displaystyle p_n(t) = (1 + t)^n = \sum_{k=0}^n {n \choose k} t^k$ .

So by induction we conclude that $H^k(X^n) \cong \mathbb{Z}^{{n \choose k}}$ . Note that we have not computed the cup product structure.

Method 2: the Künneth formula

This method will compute the cup product structure. $X$ is the product of $n$ copies of $S^1$ , whose cohomology as a ring is $\mathbb{Z}[x]/x^2$ ; there are no interesting cup products. By the Künneth formula, the cohomology of $X$ is the graded tensor product, as algebras, of $n$ copies of $\mathbb{Z}[x]/x^2$ (since all of the cohomology groups involved are free). This is precisely the exterior algebra $\mathbb{Z}[x_1, ... x_n]/(x_i^2, x_i x_j + x_j x_i)$ , with each generator in degree $1$ . In particular, $H^k(X) \cong \Lambda^k(H^1(X))$ 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 $\mathbb{R}$ by computing the de Rham cohomology of $X$ . One particularly nice way to do this is to use the following.

Theorem: Let $G$ be a compact connected Lie group acting on a smooth manifold $M$ . The inclusion $\Omega^{\bullet}(M)^G \to \Omega^{\bullet}(M)$ of invariant differential forms into differential forms is a quasi-isomorphism (induces an isomorphism on cohomology).

The idea behind this result is that, since $G$ is compact, there is an averaging operator $\Omega^{\bullet}(M) \to \Omega^{\bullet}(M)^G$ given by averaging over the action of $G$ with respect to normalized Haar measure on $G$ . But since $G$ is connected, the action of any individual element of $G$ is homotopic to the identity, so this average is also homotopic to the identity.

In particular, letting $X$ act on itself by translation, we conclude that we can compute its de Rham cohomology using translationally invariant differential forms on $X$ , or equivalently on its universal cover $\mathbb{R}^n$ . But these are precisely the differential forms obtained by wedging together the $1$ -forms $dx_1, ..., dx_n$ . The exterior derivative vanishes on all such forms, so we conclude that the de Rham cohomology of $X$ is the exterior algebra on $dx_1, ..., dx_n$ .

Method 4: Hopf algebras

This method will compute the cohomology over $\mathbb{Q}$ . Since $X$ is a topological group, it’s equipped with a product operation $X \times X \to X$ . The induced map in cohomology has the form

$\displaystyle H^{\bullet}(X, \mathbb{Q}) \to H^{\bullet}(X, \mathbb{Q}) \otimes H^{\bullet}(X, \mathbb{Q})$

by the Künneth formula. This map is coassociative and compatible with cup product, so equips $H^{\bullet}(X, \mathbb{Q})$ with the structure of a bialgebra. Together with the map induced by the inversion map $X \to X$ and the identity $\text{pt} \to X$ , the cohomology of $X$ 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 $A^{\bullet}$ be a finite-dimensional graded commutative and cocommutative Hopf algebra over a field $k$ of characteristic zero such that $A^0 \cong k$ (the Hopf algebra is connected). Then $A^{\bullet}$ is the exterior algebra on a finite collection of generators of odd degrees.

The comultiplication sends each generator $x$ to $x \otimes 1 + 1 \otimes x$ , the antipode sends each generator $x$ to $-x$ , and the counit sends each generator $x$ to $0$ .

To compute the cohomology of $X$ it therefore suffices to determine what the possible generators of the exterior algebra $H^{\bullet}(X, \mathbb{Q})$ are. For starters, let’s write $X$ more abstractly as $V/\Gamma$ where $V$ is a finite-dimensional real vector space of dimension $n$ and $\Gamma$ is a lattice in $V$ of full rank (the subgroup generated by a basis of $V$ ). Covering space theory gives us that $\pi_1(X) \cong \Gamma$ . By the Hurewicz theorem, $H_1(X) \cong \Gamma$ , so by the universal coefficient theorem,

$\displaystyle H^1(X, \mathbb{Q}) \cong \text{Hom}(\Gamma, \mathbb{Q})$ .

This gives us $n$ generators of degree $1$ , one for each element of a basis of $\text{Hom}(\Gamma, \mathbb{Q})$ , and so $H^{\bullet}(X, \mathbb{Q})$ at the very least contains the exterior algebra $\Lambda^{\bullet}(\text{Hom}(\pi_1(X), \mathbb{Q}))$ . But now we’re done: the cohomology can’t contain any generators of higher degree because wedging them with the $n$ generators we’ve already found would produce nonzero elements of the cohomology of $X$ in degrees higher than $n$ , and no such elements exist (either because $X$ admits a CW-decomposition involving cells of dimension at most $n$ or because the de Rham complex only extends up to dimension $n$ for a smooth manifold of dimension $n$ ).

Method 5: suspension

Recall that cohomology is a stable invariant in the sense that

$\displaystyle H^{n+1}(\Sigma X) \cong H^n(X)$

where $\Sigma X$ is the (reduced) suspension of $X$ (here a pointed space). Recall also that for nice pointed spaces the suspension of a product has homotopy type

$\displaystyle \Sigma(X \times Y) \simeq \Sigma X \vee \Sigma Y \vee \Sigma(X \wedge Y)$

where $\vee$ is the wedge sum and $\wedge$ is the smash product. Finally, recall that $S^1 \wedge X \simeq \Sigma X$ and that $\Sigma S^n \simeq S^{n+1}$ , so $S^1 \wedge S^n \simeq S^{n+1}$ .

Two spaces $X, Y$ are said to be stably homotopy equivalent if $\Sigma^k X \simeq \Sigma^k Y$ for some $k$ ; in particular, stably homotopy equivalent spaces have isomorphic cohomology. The above result tells us that $X \times Y$ is stably homotopy equivalent to $X \vee Y \vee (X \wedge Y)$ (once we know that suspension commutes with wedge sums). More generally, by induction we conclude that a product $X_1 \times ... \times X_n$ is stably homotopy equivalent to a wedge obtained formally by expanding

$\displaystyle \bigwedge_{i=1}^n (1 \vee X_i)$ ,

where $1$ denotes the unit of the smash product, and removing the unit. It follows that $X$ is stably homotopy equivalent to a wedge of ${n \choose k}$ copies of the $k$ -sphere, $1 \le k \le n$ , 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 $X$ , which contains information that cohomology doesn’t (e.g. about stable homotopy groups).

Method 6: cellular homology

To compute the cohomology of $X$ 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 $X$ looks like using cellular homology. Recall that cellular homology describes a chain complex $C_{\bullet}(X)$ computing the homology of a CW-complex which in degree $k$ is free abelian on the $k$ -cells in a cell decomposition of $X$ . Our particular $X$ admits a cell decomposition with ${n \choose k}$ cells of dimension $k$ given by starting with the minimal cell decomposition of $S^1$ into two cells (a $0$ -cell and a $1$ -cell connecting the $0$ -cell to itself) and taking products, where we’re thinking of cubical $k$ -cells $(I^k, \partial I^k)$ here. Equivalently, we can think of $X$ as being $[0, 1]^n$ with opposite $(n-1)$ -faces identified, and then our cells are the faces of $[0, 1]^n$ up to this identification.

The boundary maps in the cellular complex are as follows. If $e_k^{\alpha}$ is a $k$ -cell and $\chi_k^{\alpha} : \partial e_k^{\alpha} \cong S^{k-1} \to X_{k-1}$ its attaching map (where $X_{k-1}$ here denotes the $k-1$ -skeleton of $X$ ), then the differential is

$\displaystyle d(e_k^{\alpha}) = \sum_{\beta} \deg(\chi^{\alpha \beta}_k) e_{k-1}^{\beta}$

where $\beta$ runs over an enumeration of all $k-1$ -cells $e_{k-1}^{\beta}$ , $\deg$ denotes the degree, and $\chi^{\alpha \beta}_k$ is the map $S^{k-1} \to X_{k-1} \to S^{k-1}$ induced by collapsing all of $X_{k-1}$ except the cell $e_{k-1}^{\beta}$ to a point.

In this particular case all of the boundary maps in the cellular complex are trivial, so the homology $H_{\bullet}(X)$ is free abelian on cells. To see this, note that if $\chi^{\alpha \beta}_k$ is not surjective, then it necessarily has degree $0$ since it is null-homotopic, so we reduce to the surjective case. In this case the $k-1$ -cell $e_{k-1}^{\beta}$ must be a face of the $k$ -cell $e_k^{\alpha}$ , and since we’ve collapsed everything else we can reduce to the case that $k = n$ , so that $e_n^{\alpha}$ is the top-dimensional cell. At this point we will cheat a little: if $\deg(\chi^{\alpha \beta}_k) \neq 0$ in this case, then we would have $H_n(X) \cong 0$ , but $X$ is a compact orientable manifold and therefore must satisfy $H_n(X) \cong \mathbb{Z}$ .

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 $H^{\bullet}(X)$ 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 $X$ as $\mathbb{R}^n/\mathbb{Z}^n$ , then a subspace $V \subseteq \mathbb{R}^n$ represents a homology class if it is translation-invariant (given by the pushforward of a fundamental class). The images of two such subspaces $V, W$ intersect transversely if $V + W = \mathbb{R}^n$ , and then their intersection $V \cap W$ represents a homology class which Poincaré dualizes to the cup product of the Poincaré duals of $V, W$ . In particular, note that the short exact sequence

$\displaystyle 0 \to V \cap W \to V \oplus W \to \mathbb{R}^n \to 0$

implies that $\dim (V \cap W) = \dim V + \dim W - n$ . Its Poincaré dual is therefore a class in $H^{(n - \dim V) + (n - \dim W)}(X)$ , which has the correct degree.

Method ???: the Lefschetz fixed point theorem

This method is not numbered because the argument is incomplete. Consider the map

$\displaystyle f : X \ni (z_1, ... z_n) \to (z_1^{m_1}, ..., z_n^{m_n}) \in X$

where each $m_i$ is a positive integer equal to at least $2$ . This map has $\prod (m_i - 1)$ fixed points, since in each coordinate the fixed points of $z_i \mapsto z_i^{m_i}$ are precisely the $(m_i - 1)$ th roots of unity. Each fixed point has index $(-1)^k$ . By the Lefschetz fixed point theorem it follows that

$\displaystyle \prod_{i=1}^n (1 - m_i) = \sum_{i=0}^n \text{tr}(H^{\ast}(f) : H^{\ast}(X, \mathbb{Q}) \to H^{\ast}(X, \mathbb{Q}))$ .

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 $H^{\ast}(f)$ acts by multiplication by that monomial. We can do this as follows. For any subset $I \subseteq \{ 1, 2, ... n \}$ of indices we have a projection map $\pi_I : X \to (S_1)^I$ . Since $(S_1)^I$ is a compact orientable manifold, it has a fundamental class generating its top cohomology. The map $f$ induces a map on $(S_1)^I$ such that any point has $\prod_{i \in I} m_i$ preimages, hence $f$ has degree $\prod_{i \in I} (m_i - 1)$ as a map on $(S_1)^I$ , so acts on the fundamental class by multiplication by $\prod_{i \in I} m_i$ . This action induces an action on the pullback of the fundamental class of $(S_1)^I$ to $X$ which is also by multiplication by $\prod_{i \in I} m_i$ .

As the $m_i$ 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 $(-1)^{|I|} \prod_{i \in I} m_i$ 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 $f$ .

Method ???: Morse theory

There is a convenient choice of Morse function on $X$ given by

$\displaystyle f : X \ni (e^{i \theta_1}, ..., e^{i \theta_n}) \mapsto \cos \theta_1 + ... + \cos \theta_n \in \mathbb{R}$ .

The gradient of this function is $(-\sin \theta_1, ..., -\sin \theta_n)$ , and in particular it vanishes iff $e^{i \theta_k} = \pm 1$ for all $1 \le k \le n$ . There are therefore $2^n$ critical points $(\pm 1, ..., \pm 1)$ , organized in batches of ${n \choose k}$ critical points such that $k$ coordinates are equal to $1$ and $n-k$ coordinates are equal to $-1$ . At such a point $k$ of the second derivatives of each term $\cos \theta_k$ are equal to $-1$ and $n-k$ are equal to $1$ , with no other contributions to the second-order Taylor series expansion of $f$ , so all critical points are nondegenerate (hence we do in fact have a Morse function) with index $k$ . Morse theory then guarantees that $X$ has the homotopy type of a CW-complex with ${n \choose k}$ cells of dimension $k$ .

This argument should be placed in the context of a Morse-theoretic proof of the Künneth formula; more generally, if $M_1, ..., M_n$ are manifolds with Morse functions $f_1, ..., f_n$ , then $f = \sum f_i$ is a Morse function on the product $M_1 \times ... \times M_n$ , and critical points of $f$ are precisely products of critical points on the $M_i$ , 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).

Interpretation

Our computations admit the following interpretation. Recall that $S^1$ is the Eilenberg-MacLane space $K(\mathbb{Z}, 1)$ representing integral cohomology in the sense that there is a natural isomorphism $[X, K(\mathbb{Z}, 1)] \cong H^1(X)$ , where $[X, Y]$ denotes the space of homotopy classes of maps (or weak homotopy classes if $X, Y$ are not CW-complexes) $X \to Y$ . It follows that $K(\mathbb{Z}, 1)^n$ represents $n$ -tuples of cohomology classes in $H^1(X)$ . By the Yoneda lemma, cohomology classes in $H^k( K(\mathbb{Z}, 1)^n )$ , or equivalently homotopy classes of maps $K(\mathbb{Z}, 1)^n \to K(\mathbb{Z}, k)$ , can naturally be identified with natural transformations

$\displaystyle H^1(-) \times ... \times H^1(-) \to H^k(-)$ .

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 $\mathbb{Z}$ , not generated by addition and the wedge product, require higher cohomology classes as input. The smallest one is a cohomology operation $H^3(-) \to H^8(-)$ ; see this math.SE question.)

Posted in math.AT | Tagged fixed point theorems | 13 Comments

13 Responses

on March 3, 2016 at 10:51 am | Reply JKryton

You could also compute this using group cohomology theory – in fact, it is essentially a trivial example of the Hochschild-Serre spectral sequence.
- on March 3, 2016 at 3:59 pm | Reply Qiaochu Yuan
  
  Yes, I’ve since learned several other methods to do this computation!
on December 5, 2014 at 6:42 am | Reply buckdedalus

“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).” more on this.
on January 19, 2014 at 2:41 am | Reply Fahime Alibabaei

Would you please give me a reference for method 3?
- on January 19, 2014 at 11:05 pm | Reply Qiaochu Yuan
  
  I don’t know a reference.
on December 9, 2013 at 6:43 am | Reply Joe Hannon

The unit of the smash product is the two point space, S^0
- on December 9, 2013 at 10:55 am | Reply Qiaochu Yuan
  
  Whoops; thanks.
on December 8, 2013 at 6:48 pm | Reply jmracek

This may be killing an ant with a flamethrower, but proceed by induction by applying the Leray-Serre spectral sequence to the fibre bundle $S^{1}\hookrightarrow T^{n} \to T^{n-1}$. Always collapses at $E_{2}^{p,q}$ for degree reasons. Morally, this is the same as the Kunneth formula solution.
on October 29, 2013 at 11:38 am | Reply David Speyer

(Real) Hodge theory: Take the obvious metric on $S^1$ . The laplacian operator takes $f(\theta)$ to $f''(\theta)$ and takes $g(\theta) d \theta)$ to $g''(\theta) d \theta$ . The only periodic solution to $f''(\theta)=0$ is taking $f$ constant, so the only harmonic functions are the constant functions and the only harmonic 1-forms are multiples of $d \theta$ .

Switching from $S^1$ to $\mathbb{C}^{\ast}$ :

On any affine (or Stein) space, Cartan’s Theorem B implies that using holomorphic differential
forms in the de Rham complex computes topological cohomology. A differential form $g(z) dz$ on $\mathbb{C}^{\ast}$, with $g$ holomorphic, has a holomorphic anti-derivative if and only if it has residue zero at 0, so $H^1$ is $1$ again.

By a theorem of Grothendieck, we can replace holomorphic functions with polynomial functions in this setting. Sure enough, a Laurent polynomial $g(z) dz$ has a Laurent polynomial anti-derivative if and only if the coefficient of $z^{-1} dz$ is 0.

Write $\mathbb{C}^{\ast}$ as $\mathbb{P}^1 \setminus \{ 0, \infty \}$ . I’ll abbreviate $\mathbb{P}^1$ to $X$ and $\{ 0, \infty \}$ to $D$ . Then Deligne computes the Hodge filtration on $H^1(X \setminus D)$ using a spectral sequence which collapses at $E_1$ and gives us $H^1(X \setminus D) = H^1(X, \mathcal{O}) \oplus H^0(X, \Omega^1(D)) = 0 \oplus \mathbb{C}$ .

One could probably write down several more good computations by treating $S^1 \times S^1$ as an elliptic curve, but I’m out of time for now.
on October 18, 2013 at 10:32 pm | Reply citedcorpse

Reblogged this on citedcorpse and commented:
I like this concept a lot.
on October 17, 2013 at 7:47 am | Reply Qiao Zhou

How about Lie algebra cohomology? The n-torus is a compact Lie group, so the corresponding Lie algebra cohomology computes its cohomology.
- on October 17, 2013 at 10:26 am | Reply Qiaochu Yuan
  
  That’s what I’m using to compute the de Rham cohomology. But it didn’t seem worth being explicit about it in this example because of all the differentials are trivial.
  - on November 27, 2017 at 10:37 am | Reply Allen Knutson
    
    The thing that’s messed up/wonderful about this is that it’s defined for Lie algebras of noncompact groups (like $\mathbb R^n$ ), but when they’re also Lie algebras of compact groups, it’ll find the cohomology of the compact group. How does it know!!

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