Cohomology rings

One of the great wonders of cohomology, as opposed to homology, is that when the coefficients carry a ring structure, the cohomology groups inherit the structure of a graded ring. This lets one say much more. For example, when $n \geq 3$ and $k\geq 2$, any map $S^n \to \mathbb{CP}^k$ induces the zero map on cohomology in positive degrees. In particular, there are no degree 1 maps, which contrasts with the fact that an $n$-manifold always admits a degree 1 map to $S^n$ (squash the complement of an open ball). (I initially used the example of tori instead of projective space before realising that this was a bad example since tori are aspherical: their universal covers are $\mathbb{R}^n$ and hence contractible.) The particularly pleasing cohomology ring structures of $\mathbb{RP}^{\infty}$ and $\mathbb{CP}^{\infty}$, as the classifying spaces of line bundles, are also responsible for the well-behavedness of Stiefel-Whitney and Chern classes.

However, it is not so obvious what sort of rings can appear, for any choice of non-trivial coefficients, and I believe this is an open problem. Nonetheless, there are already interesting things to be said. For instance, for any CW complex $X$ and $x \in H^1(X, \mathbb{Z})$, we have that $x \cup x =0$. A priori, it is only 2-torsion, but $S^1$ happens to be a $K(\mathbb{Z},1)$ and since continuous maps induce ring homomorphisms, we get the vanishing from $S^1$, where it is obvious. I think this is really neat.

Analysing special cases can also give interesting and important results. One can ask whether, for a given ring $R$, degree $d$, and positive integer $k$ (or $\infty)$, whether there is a space $X$ and a cohomology class graded in degree $d$ such that $H*(X; R) \cong R[\alpha]/(\alpha^k)$. When $R$ is an integral domain of characteristic not 2 graded commutativity implies $d$ must be even. By $K$-theoretic methods one can show that when $R=\mathbb{F}_2$, $d=1,2, 4, 8$, and  when $R=\mathbb{F}_p$ for an odd prime $p$, $d$ must be an even divisor of $2(p-1)$. Combining these implies that when $R=\mathbb{Z}$ and $k>3$, we must have that $d=2$ or $4$ (reduce mod 2 and mod 3), and these correspond to complex and quaternionic projective spaces. This gives a conceptual explanation for the failure to build octonionic projective spaces: the lack of associativity means that the usual projectivisation construction involving lines in $\mathbb{O}^n$ doesn't work, and it isn't that there is some cleverer construction that does: there is a genuine topological obstruction.

One can also ask in some sense the dual question: given spaces we sort of know, what rings do we get? The simplest spaces are just products of those that we know well, where the Kunneth formula gives rise to tensor products of rings in some cases. For instance tori give rise to exterior algebras, and . In principle one can use spectral sequences to extract information about more general fibre bundles, but this tends to be very difficult. The next class of important manifolds are those of dimensions 1 and 2, where compact manifolds have a relatively straightforward classification. The natural question is now what happens for 3-manifolds, and here too much is known.

Let $M$ be a closed 3-manifold and let $\pi=\pi_1(M)$. Let $R=\mathbb{Z}$ or a field of characteristic $\not=2$. We will concentrate on the orientable case with torsion-free integral cohomology. We leave the reader to look up the case of $\mathbb{F_2}$-coefficients, proved by similar surgery methods to what is discussed below, and the relatively simple non-orientable case in Hillman's paper (which we partially follow).

We identify $H_1(M, \mathbb{Z})=\pi^{ab}$ and let $H^1(M, R)=Hom(H_1(M, \mathbb{Z}),R)$. If $M$ is orientable then $H_3(M;R)\cong{R}$, and cap product with a fundamental class $[M]$ defines Poincar\'e duality isomorphisms $D_2:H^2(M;R)\to{R\otimes{H}}$ and $D_3:H^3(M;R)\to{R}$, while cup product and duality define homomorphisms $\gamma:\wedge_2H^1\to{H^2(M;R)}$ and $\mu=\mu_M:\wedge_3H^1\to{R}$.
(Alternating trilinear functions such as $\mu$ may be identified with elements of $\wedge_3H$.)
These satisfy the equations
\[
D_3(a\smile{D_2^{-1}(h)})=a(h)\quad\forall~a\in{H^*}~and~h\in{H}
\]
and
\[
\mu(a\wedge{b}\wedge{c})=a(D_2\gamma(b\wedge{c}))\quad
\forall~a,b,c\in{H^*}.
\]
If $R$ is a field of with characteristic $\not=2$) or if $R=\mathbb{Z}$ and $H$ is torsion-free then the cohomology ring $H^*(M;R)$ is determined by $H$, $\mu$ and the duality isomorphisms $D_2$ and $D_3$, via these equations.
Let $\mathcal{H}^0=R$, $\mathcal{H}^1=H^*$, $\mathcal{H}^2=H$ and $\mathcal{H}^3=R\varepsilon_3$, and let $rs$ be the unique solution of  ``$t(rs)=\mu(r\wedge{s}\wedge{t})\varepsilon_3~\forall~t\in{R^1}$", and $rh=r(h)\varepsilon_3$, for $r,s\in\mathcal{H}^1$ and $h\in {H}^2$.Then $rst=\mu(r\wedge{s}\wedge{t})\varepsilon_3$,  for all $r,s,t\in{H^*}$,and $\mathcal{H}^*(H,\mu)=\oplus_{i=0}^3\mathcal{H}^i$ is a graded ring. We may use $D_2$ and $D_3$ to determine an isomorphism  $H^*(M;R)\cong\mathcal{H}^*(H,\mu)$.) If $H^1\not=0$ then $D_2$ determines $D_3$, by the first equation above.

Theorem (Sullivan): If $R=\mathbb{Z}$ then every such pair $(H,\mu)$ is realisable by some closed 
orientable $3$-manifold with torsion-free homology.

 We want to construct a 3-manifold \( M \) with \( H_1(M) = H \) and \( \mu_M = \mu \) with \( (H, \mu) \) arbitrary, and motivated by the fact that any closed 3-manifold has a handlebody decomposition we try to do this by gluing a suitable choice of handlebodies. Let \( \beta = \dim H \) and consider the handlebody \( S \) of genus \( \beta \) and boundary \( \partial S \) the surface of genus \( \beta \).

By half-lives half-dies, we identify \( H \) with the kernel of the inclusion map \( H_1(\partial S) \to H_1(S) \) and let \( G \) be the subgroup of homeomorphisms \( \sigma \) of \( \partial S \) satisfying \( \sigma|_H = \text{identity} \), i.e. act trivially on that subspace of the homology group.

We can form for each such homeomorphism \( \sigma \) the three manifold \( M_\sigma = S \cup_\sigma S \) by pasting two copies of \( S \) together along \( \partial S \) using \( \sigma \). It is easy to check that \( H_1(M_\sigma) \) is isomorphic to \( H \) but is not quite so easy to identify the \( \mu \) invariant, \( \mu(M_\sigma) \), which will turn out to be the general \( \mu \). Now we have the following additivity proposition:

if \( \sigma, \sigma' \in G \), then
\[
\mu(M_{\sigma' \cdot \sigma}) = \mu(M_\sigma) + \mu(M_{\sigma'}).
\]

The additivity proposition can be proved directly using homologies in a collar neighborhood of \( \partial S \) between \( \{x_i\} \), the standard basis of \( H \), and \( \{\sigma(x_i)\} \) on the cycle level. The additivity follows because we can calculate independently in separate collars for \( \sigma \) and \( \sigma' \) to obtain the sum \( \mu \) for \( \sigma \cdot \sigma' \). This proof of the additivity is due to François Laudenbach.

With this in hand, we can give the proof of Sullivan's theorem.

Proof: The proof proceeds by analyzing a Heegaard decomposition of the three torus \( T = S^1 \times S^1 \times S^1 \). We will write \( T = M_\sigma \) for \( \sigma \in G \) and since \( \mu(T) = x_1 \wedge x_2 \wedge x_3 \) we will be done by additivity. Consider \( T \) as the quotient of Euclidean space \( \mathbb{R}^3 \) by the integral translations in three orthogonal directions with the standard cube in \( \mathbb{R}^3 \) as the fundamental domain. Consider the two systems of lines \( L_1 \) and \( L_2 \) obtained by translating the edges of the basic cube on the one hand and the perpendicular axes of symmetry through the faces of the cube on the other hand. If we carefully thicken \( L_1 \) and \( L_2 \) to tubular neighborhoods we obtain an infinite equivariant Heegaard decomposition of \( \mathbb{R}^3 \) which yields upon identification a decomposition, \( S^1 \times S^1 \times S^1 = S \cup_\sigma S \), where \( \partial S \) has genus 3 and \( \sigma \colon \partial S \to \partial S \) induces the identity on \( H = \mathrm{ker}(H_1(\partial S) \to H_1(S)) \).

One calculates visually that \( \sigma \) is given by the matrix
\[
\left(\begin{array}{ccc|ccc}
1 & 0 & 0 & 0 & 1 & 1 \\
0 & 1 & 0 & 1 & 0 & 1 \\
0 & 0 & 1 & 1 & 1 & 0 \\
\hline
 &  &  & 1 & 0 & 0 \\
 & * &  & 0 & 1 & 0 \\
 &  &  & 0 & 0 & 1
\end{array}\right)
\]
in the natural basis, after reflection.

Now consider the general case of \( H \) with basis \( x_1, \ldots, x_\beta \) and \( \mu = \sum_{i<j<k} a_{ijk} \, x_i \wedge x_j \wedge x_k \) in \( \Lambda^3 H \), \( a_{ijk} \in \Z \). Let \( S_\beta \) denote the solid torus with \( \beta \) one-handles \( h_1, \ldots, h_\beta \) attached. For a moment think of \( h_i \), \( h_j \), and \( h_k \) attached to the upper hemisphere of the boundary of the ball and the others to the lower. Construct \( \sigma_{ijk} \), a homeomorphism of \( \partial S_\beta \) using the \( \sigma \) considered before in the decomposition of \( S^1 \times S^1 \times S^1 \) on the upper part of \( \partial S_\beta \) and the identity on the lower part. It is clear that \( M_{\sigma_{ijk}} \equiv (S^1 \times S^1 \times S^1) \# (\beta - 3 \text{ copies of } S^1 \times S^2) \), so \( \mu(M_{\sigma_{ijk}}) = x_i \wedge x_j \wedge x_k \).

Now since \( \sigma_{ijk} \) is in the subgroup fixing \( \mathrm{ker}(H_1(\partial S_\beta) \to H_1(S_\beta)) \), we can calculate \( \mu(M_\tau) \) where
\[
\tau = \prod_{i<j<k} (\sigma_{ijk})^{a_{ijk}}
\]
by the additivity proposition to be the desired
\[
\mu = \sum_{i<j<k} a_{ijk} \, x_i \wedge x_j \wedge x_k.
\]
Note that by varying the order of the \( \sigma_{ijk} \)'s in \( \tau \) we obtain many different manifolds with invariant \( \mu \). Each one however has no torsion in \( H_1 \) and its genus (the minimal genus in a Heegaard decomposition) equals the first Betti number \( \beta \). $\blacksquare$

Later works shows that we may take the 3-manifold to be irreducible.

 The case with torsion

When $R=\mathbb{Z}$ and $H_1(M;\mathbb{Z})$ has nontrivial torsion then $\gamma$ is no longer determined by $\mu_M$. The torsion subgroup has a complementary direct summand, but the splitting is not canonical. Cup products with coefficients in other rings and their compatibilities with integral cup product must also be considered.

For each $n>1$ let $\nu_n=\nu_{nM}:\wedge_3H^1(M;\mathbb{Z}/n\mathbb{Z})\to\mathbb{Z}/n\mathbb{Z}$ be defined by
\[
\nu_n(X,Y,Z)=(X\cup{Y}\cup{Z})\cap[M], \quad\mathrm{for~all}~X,Y,Z\in{H^1(M;\mathbb{Z}/n\mathbb{Z})}.
\]
Then $\nu_{nM}$ and Poincar\'e duality together determine the ring $H^*(M;\mathbb{Z}/n\mathbb{Z})$.
Every 3-form $\nu:\wedge_3(\mathbb{Z}/n\mathbb{Z})^\beta \to\mathbb{Z}/n\mathbb{Z}$ lifts to a 3-form $\widehat{\nu}:\wedge_3\mathbb{Z}^\beta\to\mathbb{Z}$.
Hence it is an immediate consequence of Sullivan's construction that every such 3-form $\nu$ can be realized as $\nu_{nM}$ for some closed orientable 3-manifold $M$ with $H_1(M;\mathbb{Z})\cong\mathbb{Z}^\beta$. If $p$ is an odd prime it follows that every finite graded-commutative graded $\mathbb{F}_p$-algebra satisfying 3-dimensional Poincar\'e duality is the $\mathbb{F}_p$ cohomology ring of such a 3-manifold.

The Bockstein homomorphism
$\beta_{\mathbb{Q}/\mathbb{Z}}:Hom(H,\mathbb{Q}/\mathbb{Z})\to{H^2(M;\mathbb{Z})}$ has image $Ext(H,\mathbb{Z})$,  the torsion subgroup of $H^2(M;\mathbb{Z})$. We may use this to define the torsion linking  pairing $\ell:tH\times{tH}\to\mathbb{Q}/\mathbb{Z}$ by 
\[
\ell(u,v)=(D_2\circ\beta_{\mathbb{Q}/\mathbb{Z}})^{-1}(v)(u)\quad\forall~u,v\in{tH}.
\]
This pairing is nonsingular and symmetric, and $\ell$ and $\beta_{\mathbb{Q}/\mathbb{Z}}$ determine each other (given $D_2$). It is a theorem of Kawauchi and Kojima that every such pairing is realisable by some $\mathbb{Q}$-homology 3-sphere. Taking connected sums shows that every such triple $(H, \mu, \ell)$ is realisable by some 3-manifold. On the other hand,  $\mu$ and $\ell$ are independent invariants.

It is easily verified that if $x\in{H^1(M;\mathbb{Z}/n\mathbb{Z})}$ then $x^2=\frac{n}2\beta_{\mathbb{Q}/\mathbb{Z}}(x)$. (It suffices to check this for $x=id_{\mathbb{Z}/n\mathbb{Z}}$, considered as an element of $H^1(\mathbb{Z}/n\mathbb{Z};\mathbb{Q}/\mathbb{Z})= H^1(\mathbb{Z}/n\mathbb{Z};\mathbb{Z}/n\mathbb{Z})$.)
Let $\psi_n:\mathbb{Z}/n\mathbb{Z}\to\mathbb{Q}/\mathbb{Z}$ be the standard inclusion. If $x\in{H^{*n}}$ let $\hat{x}$ be the element of $tH$ such that $\ell(\hat{x},a)=\psi_n(x(a))$ for all $a\in{tH}$. Turaev showed that
\[
\psi_n(\nu_n(x,x,y))=\frac{n}2\ell(\hat{x},\hat{y})~\mathrm{for~all}
~x,y\in{H^{*n}}.
\]
When $n=2$ this condition implies the orientable case of the Postnikov-Wu identity. (Note that if $n$ is odd then both sides are 0.)

Let $A$ be a finitely generated abelian group, let $\ell:tA\times{tA}\to\mathbb{Q}/\mathbb{Z}$ be a nonsingular symmetric pairing, and suppose that $\nu:\wedge_3A^*\to\mathbb{Z}$ and $\nu_n:\wedge_3A^{*n}\to\mathbb{Z}/n\mathbb{Z}$ is a system of alternating trilinear functions which are compatible under reduction {\it mod}-$n$.

Then Turaev showed that such a group $A$, pairing $\ell$ and system of trilinear functions may be realisable by the homology and cohomology of a closed orientable 3-manifold if and only if the above condition deriving from the interaction of the Bockstein $\beta_{\mathbb{Q}/\mathbb{Z}}$ with the cup-square holds for all $n$ dividing the order of $tA$.

4-manifolds

Since 4 is the next number after 3, one might reasonably ask what happens in the case of 4-manifolds. As all people in topology know, however, 4 is the dimension where all hell breaks loose. Here's what is known. The homotopy type of a simply connected compact 4-manifold only depends on the intersection form on the middle dimensional homology. A famous theorem of Michael Freedman implies that the homeomorphism type of the manifold only depends on this intersection form and ${\displaystyle \mathbb{Z}}$the Kirby–Siebenmann invariant, which implies the topological Poincare conjecture, and moreover that every combination of unimodular form and Kirby–Siebenmann invariant can arise, except that if the form is even, then the Kirby–Siebenmann invariant must be the signature/8 (mod 2).  Freedman invents an analogue of the 4-dimensional Whitney trick, and this extends to some manifolds whose fundamental group is non-trivial. Groups/manifolds for which this extends are called 'good in the sense of Freedman', and conjecturally this is precisely the class of amenable groups. In general, however, any finitely presented group is the fundamental group of a closed 4-manifold, so there is a limited extent to which one can hope for as complete a picture as in the lower dimensional cases. The same is true in higher dimensions, but in higher dimensions one has additional extremely powerful tools from surgery theory, so the description is in some sense better if one takes as input the extra data (on top of the cohomology ring) that surgery theory needs.