De Rham Cohomology Ring of \(S^{2}\times S^{1}\)

Since I didn’t find this question answered anywhere on the Internet I thought It would be a good idea to write an article explaining how to do it, so that Math Stackexchange lurkers who are learning about De Rham Cohomology can have a nice example to use as a model. I’ll be drawing some results from John M. Lee’s "Introduction to Smooth Manifolds" (which I’ll call "Lee" from now on), but this shouldn’t be a problem, since everything is standard course material for De Rham Cohomology groups.

Applying the Mayer-Vietoris Theorem

The statement of the Mayer-Vietoris Theorem for De Rham Cohmology groups is given in Lee - Theorem 17.20. Let us call \(M:=S^{2}\times S^{1}\), the product of a unit sphere and a unit circle. We shall find an open cover of \(M\) by two open subsets \(U\) and \(V\) that by the Mayer-Vietoris Theorem will induce a long exact sequence in the Cohomology groups of \(M\) \[\cdots\to H^{p}(M)\to H^{p}(U)\oplus H^{p}(V)\to H^{p}(U\cap V)\to H^{p+1}(M)\to\cdots .\] Here we need to be a bit careful to select a convenient open cover. For example, one may be tempted to choose \(U=S^{2}\times B^{1}_{\varepsilon}\) and \(V=B^{2}_{\varepsilon}\times S^{1}\), where \(B^{2}_{\varepsilon},B^{1}_{\varepsilon}\) are two balls of sufficiently small radius \(\varepsilon >0\), centered at a common point on \(S^{2}\times S^{1}\) and of dimensions 2 and 1, respectively.

The problem with the previous approach is that, since \(V\) is homotopically equivalent to a circle and de Rham cohmology groups are invariant under homotopical equivalence, we are left with the following sequence in the beggining of the de Rham Cochain: \[0\to\underbrace{H^{0}(M)}_{\cong \mathbb{R}}\to\underbrace{H^{0}(U)}_{\cong\mathbb{R}}\oplus\underbrace{H^{0}(V)}_{\cong\mathbb{R}}\to\underbrace{H^{0}(U\cap V)}_{\cong\mathbb{R}}\to H^{1}(M)\to\underbrace{H^{1}(U)}_{\cong 0}\oplus \underbrace{H^{1}(V)}_{\cong \mathbb{R}}\to\cdots .\] (Recall from Lee - Theorem 17.21, that the de Rham cohomology groups of the spheres \(S^{n}\) are: \[H^{p}(S^{n})\cong \begin{cases} \mathbb{R}& \text{if } p=0 \text{ or } p=n \\ 0 & \text{if } 0<p<n \end{cases} .\] ) And since this part of the cochain doesn’t end in \(0\) we can’t deduce the dimension of \(H^{1}(M)\) as a real vector space from it.

However, the trick of using the first cohomology groups would work if we could choose \(U\) and \(V\) in such a way that \(H^{1}(U)\) and \(H^{1}(V)\) were both zero. For example, if both were homotopically equivalent to the unit sphere.

One way to do this is to simply cut the circle into two semicircles and consider the product of each of these with the unit sphere. In other words, to take \[\begin{aligned} U&=(S^{1}\backslash \{(0,1)\})\times S^{2}\simeq \mathbb{R}\times S^{2} \simeq \{\ast\} \times S^{2} \simeq S^{2} \\ V&=(S^{1}\backslash{(0,-1)})\times S^{2} \simeq \mathbb{R}\times S^{2} \simeq \{\ast\} \times S^{2} \simeq S^{2} \end{aligned}\] where \(\simeq\) denotes homotpical equivalence, \(\{\ast\}\) is the topological space of only one point and we’ve used the obvious homotopies.

Of course, \(U\cup V = M\) and \(U\cap V \simeq S^{2}\sqcup S^{2}\) a disjoint union of two unit spheres, which has two connected components and therefore \(H^{0}(U\cap V)\cong \mathbb{R}^{2}\).

Now, since \(U,V\) and \(M\) are all connected, we have that the zeroeth cohomology groups of these three spaces are all \(\cong \mathbb{R}\). Thus, we are left with the following in the \(p=0\) section of the sequence \[0\to \underbrace{H^{0}(M)}_{\cong \mathbb{R}}\to \underbrace{H^{0}(U)}_{\cong \mathbb{R}}\oplus\underbrace{H^{0}(V)}_{\cong \mathbb{R}}\to\underbrace{H^{0}(U\cap V)}_{\cong \mathbb{R}^{2}}\to H^{1}(M) \to \underbrace{H^{1}(V)\oplus H^{1}(V)}_{\cong 0}\to \cdots .\] And since it starts and ends with a zero we can calculate \[\begin{aligned} \mathrm{dim}_{\mathbb{R}}(H^{1}(M)) &= \mathrm{dim}_{\mathbb{R}}(H^{0}(U\cap V))-\mathrm{dim}_{\mathbb{R}}(H^{0}(U)\oplus H^{0}(V))+\mathrm{dim}_{\mathbb{R}}(H^{0}(M)) \\ &= 2 - 2 + 1 \\ &= 1 \end{aligned}\] And from this we get that \(H^{1}(M)\cong \mathbb{R}\).

To calculate the cohmology groups of the intersection we use that the de Rham cohomology of a disjoint union is the direct sum of the cohmologies of the summands (this is Lee - Proposition 17.5). \[H^{p}(U\cap V)\cong H^{p}(S^{2}\sqcup S^{2}) \cong H^{p}(S^{2})\oplus H^{p}(S^{2})\cong \begin{cases} \mathbb{R}^{2} & \text{if } p=0,2 \\ 0 & \text{if } p\neq 0,2 \end{cases} .\]

The rest of the de Rham cohmology.

We have that \(M\) is a compact, connected (since both \(S^{2}\) and \(S^{1}\) are compact and connected) oriented (since both \(S^{2}\) and \(S^{1}\) are oriented and we may equip \(M\) with the product orientation) manifold of dimension \(3\). Therefore \(H^{3}(M)\cong \mathbb{R}\) by Lee - Theorem 17.31 and \(H^{p}(M)\cong 0\) for all \(p>3\) since there are no \(p\)-differentiable forms on a three-dimensional manifold for all such \(p\).

To calculate the last cohomology group we have two options: we either use the exactness of the sequence to do a direct calculation of \(H^{2}(M)\), or we use the Poincaré Duality. I’ll explain both of these approaches below.

For the former approach, let us write again the relevant part of the sequence: \[\begin{aligned} \underbrace{H^{1}(U\cap V)}_{\cong 0}\xrightarrow{\ell}H^{2}(M)\xrightarrow{k}\underbrace{H^{2}(U)}_{\cong \mathbb{R}}\oplus\underbrace{H^{2}(V)}_{\cong \mathbb{R}}\xrightarrow{h}\underbrace{H^{2}(U\cap V)}_{\cong \mathbb{R}^{2}}\xrightarrow{g} \underbrace{H^{3}(M)}_{\cong \mathbb{R}} \xrightarrow{f} \underbrace{H^{3}(U)}_{\cong 0}\oplus\underbrace{H^{3}(V)}_{\cong 0} \end{aligned}\]

We have that \(\mathbb{R}\cong H^{3}(M)=\mathrm{ker}(f)=\mathrm{ran}(g)\), so \(\mathrm{ker}(g)\cong\mathbb{R}\).

Therefore, \(\mathbb{R}\cong\mathrm{ker}(g)=\mathrm{ran}(h)\implies \mathbb{R}\cong\mathrm{ker}(h)=\mathrm{ran}(k)\).

Lastly, we note that \(k\) is injective because \(\mathrm{ker}(k)=\mathrm{ran}(\ell)=0\). Thus: \[H^{2}(M)\cong \frac{\mathrm{ran}(k)}{\mathrm{ker}(k)}\cong \mathbb{R} .\] In conclusion, \[H^{p}(M)\cong \begin{cases} \mathbb{R}& \text{if } p=0,1,2,3 \\ 0 & \text{for all other values of } p \end{cases} .\] For the latter approach, we first recall the statement of Poincaré Duality for the de Rham cohomology.

Where \([\omega]\) denotes the equivalence class of the \(k\)-form \(\omega\) in \(H^{k}(M)\) and \(H^{n-k}_{c}(M)\) is the \(n-k\)-th compactly supported de Rham chomology group as described in chapter 17 of Lee’s book. Having the definition of this isomorphism at hand will prove useful later, as we’ll see.

Since \(M\) is already a compact manifold, we get that \[H^{2}(M) = H^{3-1}(M)\cong \left( H^{1}_{c}(M) \right) ^{\ast} = \left( H^{1}(M) \right) ^{\ast} = \mathbb{R}^{\ast} = \mathbb{R} .\]

Calculating the ring structure of \(H^{\bullet}(M)\).

Thus far, we’ve seen that \(H^{\bullet}(M)\cong \mathbb{R}\times \mathbb{R}\times \mathbb{R}\times \mathbb{R}\), where each copy of \(\mathbb{R}\) comes from each of the de Rham cohomology groups \(H^{p}(M)\) with \(p=0,\ldots,3\).

It’s not difficult to see that the wedge product \(\wedge\) of differential froms in \(M\) induces a ring structure on the de Rham cochain complex. Our intuition tells us that if we let \(X\) be a generator of the cohomology group \(H^{1}(M)\), and \(Y\) be a generator of \(H^{2}(M)\), then \(XY\) (the wedge product of both generators) will be a generator of \(H^{3}(M)\). Therefore the cohomology ring of \(M\) will be a ring of polynomials in the variables \(X,Y\) ; with \(X^{2}\) and \(Y^{2}\) being equal to \(0\) because of the anticommutativity of differential forms and the fact that \(M\) is three-dimensional. I.e., we expect that \(H^{\bullet}(M)\cong \frac{\mathbb{R}[X,Y]}{\left<X^{2},Y^{2} \right>}\). This is indeed the case, but we need to be a bit careful to make the correct calculations.

Let \([\omega]\) be a generator of \(H^{2}(M)\). Since \(M\) is compact, \(H^{1}_{c}(M)=H^{1}(M)\) and since \(M\) is oriented and connected we get by Poincaré Duality that \[H^{2}(M)\to\left( H^{1}(M) \right) ^{\ast},\quad [\omega]\to\left( [\eta]\to \int_{M}\omega\wedge\eta \right) .\] is a linear isomorphism. Since both cohomology groups are \(\cong\mathbb{R}\), we have that \([\eta]\to\int_{M}\omega\wedge\eta\) is not the zero linear map.

Therefore, there exists \(\eta\in \Omega^{1}(M)\) such that \(\int_{M}\omega\wedge\eta \neq 0\). We have that \(\eta\) is not exact, for if it were exact then there would exist some \(f\in \Omega^{0}(M)\) such that \(\eta=df\), the exterior derivative of \(f\). Then \[\begin{aligned} 0 \neq \int_M\omega\wedge\eta = \int_M\omega\wedge &df = \int_M (d(\omega\wedge f) - d\omega\wedge f) \\ &\stackrel{Stokes}{=} \int_{\partial M}\omega\wedge f - \int_M d\omega\wedge f \end{aligned}\] Since \(\omega\) is closed, \(d\omega = 0\). And since neither \(S^{2}\) nor \(S^{1}\) have any boundary we get that \(\partial M = \emptyset\). Therefore both integrals vanish and we get a contradiction. Consequently, \([\eta]\neq 0\) is a generator of \(H^{1}(M)\).

Lastly, we have that \([\omega\wedge\eta]\) is a generator of \(H^{3}(M)\). To see this, we assume by way of contradiction that \(\omega\wedge\eta\) is exact, therefore there exists \(\alpha\in \Omega^{2}(M)\) such that \(\omega\wedge\eta = d\alpha\). Once again using Stoke’s Theorem we get that \[0 \neq \int_M\omega\wedge\eta=\int_Md\alpha=\int_{\partial M}\alpha = 0 .\] Which is again a contradiction.

Defining an isomorphism between \(H^{\bullet}(M)\) and \(\frac{\mathbb{R}[X,Y]}{\left<X^{2},Y^{2} \right>}\)

Let us consider the linear morphism: \[\mathbb{R}[X,Y]\xrightarrow{\Phi}H^{\bullet}(M); \quad X\to[\omega], \, Y\to[\eta] .\] Note that \(XY\stackrel{\Phi}{\longmapsto}[\omega]\wedge[\eta]=[\omega\wedge\eta]\). This shows that \(\Phi\) is surjective, given that \(H^{\bullet}(M)=\left<1,[\omega],[\eta],[\omega\wedge\eta] \right>_{\mathbb{R}}\) and all the generators lie in the rank of \(\Phi\).

Now, \[[\omega]\in H^{2}(M) \implies \Phi(X^{2})=[\omega]^{2}=[\omega\wedge\omega]\in H^{4}(M)=\{0\} .\] Therefore, \(\Phi(X^{2})=0\) and \(\Phi\) can be factored by \(\left<X^{2} \right>\). On the other hand, by the anticommutativity of \(\wedge\) we have that \[\eta\wedge\eta = (-1)^{1.1}\eta\wedge\eta=-\eta\wedge\eta \implies \eta\wedge\eta = 0 .\] So \(\Phi(Y^{2})=0\) and \(\Phi\) can be factored by \(\left<Y^{2} \right>\)

Therefore there is a surjective linear morphism:

\[\frac{\mathbb{R}[X,Y]}{\left<X^{2},Y^{2} \right>} \xrightarrow{\widetilde{\Phi}} H^{\bullet}(M) .\] And it is injective by the Dimension Theorem because both its domain and rank have dimension \(4\) as real vector spaces.