Homology of the $\mathbb {E}_d$ operadCohomology

3 Cohomology

We now produce some cohomology classes of $H_*(\operatorname{Conf}_n(\mathbb {R}^d))$ in degree $d - 1$, which will generate the whole cohomology.

Definition 3.1

For $i, j \in \{ 1, \ldots , n\}$, let $\alpha _{ij}: \operatorname{Conf}_n(\mathbb {R}^d) \to \operatorname{Conf}_2(\mathbb {R}^d) \simeq S^{d - 1}$ send $(x_1, \ldots , x_n)$ to $(x_i, x_j)$. Let $a_{ij} \in H^{d - 1}(\operatorname{Conf}_n(\mathbb {R}^d))$ be the pullback of the fundamental class along $\alpha _{ij}$.

Note that $a_{ij} = (-1)^d a_{ji}$, and $a_{ij}^2 = 0$, since the square of the fundamental class on $S^{d - 1}$ vanishes.

The first claim is that these $a_{ij}$ are dual to the $P_{ij}$.

Lemma 3.2

Under the homology-cohomology pairing, if $i < j$ and $k < \ell$, then

$\langle a_{ij}, P_{k\ell }\rangle = \delta _{ik} \delta _{j\ell }.$

In particular, the set $\{ a_{ij}: i < j\}$ is linearly independent and forms a basis of $H^{d - 1}(\operatorname{Conf}_n(\mathbb {R}^d))$, since we know this group is free of rank $\binom {n}{2}$.

Proof
Consider the composite

$S^{d - 1} \to \operatorname{Conf}_n(\mathbb {R}^d) \to S^{d - 1}.$

where the first map picks out $P_{ij}$ and the second $\alpha _{ij}$. The pairing above is given by the degree of this map.

If $i = k$ and $j = \ell$, then this is the identity map, which has degree $1$. Otherwise, taking $\varepsilon \to 0$ gives a homotopy to the constant map.

Proof

We now know that as a ring, the cohomology of $\operatorname{Conf}_n(\mathbb {R}^d)$ is generated by the $a_{ij}$. We introduce a graphical way to depict products of the $a_{ij}$. Such a product is represented by a graph with vertices $\{ 1, \ldots , n\}$. Each edge is oriented, and the set of edges is ordered. Multiple edges between vertices is allowed but loops are not.

To depict $a_{ij}$, we take the graph whose only edge is an edge from $i$ to $j$. Products are then represented by unions. For example, $a_{42} a_{43} a_{13}$ is represented by the graph

The numbers on the edges denote the ordering of the edges.

If we swap the ordering of two adjacent edges or reverse an arrow, the class represented picks up a sign of $(-1)^{d - 1}$. Moreover, since $a_{ij}^2 = 0$, any graph with a repeated edge represents $0$.

There is one further relation, which morally is dual to the Jacobi identity.

Lemma 3.3 (Arnold)

Proof
Without loss of generality, we assume $n = 3$ and $(i, j, k) = (1, 2, 3)$. We have to show that

$a_{12} a_{23} + a_{23} a_{31} + a_{31} a_{12} = 0.$

We will prove this using the intersection product.

Recall that if $M^n$ is a compact manifold, then Poincaré duality tells us

$H^k(M^n) \cong H_{n - k}(M^n).$

Under this isomorphism, the cup product on $H^*(M^n)$ induces a product on $H_*(M^n)$. If $x, y \in H_*(M^n)$ are represented by submanifolds $X, Y$ that intersect transversely, then $x \cdot y$ is represented by $X \cap Y$. This allows for a geometric way to compute the cup product structure of a compact manifold.

In the case of a non-compact manifold, there are two ways to fix Poincaré duality. One is to replace cohomology with compactly supported cohomology, but this is not what we want. Instead, we can replace homology with “locally finite homology”, also known as Borel–Moore homology, which is given by the homology of the complex of locally finite chains. In particular, arbitrary closed submanifolds of $M^n$ represent a Borel—More homology class.

Let $\mathrm{col}_i$ be the submanifold of $\operatorname{Conf}_3(\mathbb {R}^d)$ where $x_1, x_2, x_3$ are colinear and $x_i$ is in the middle. This is a submanifold of codimension $d - 1$, and represents a class in $H^{d - 1}(\operatorname{Conf}_3(\mathbb {R}^d))$. We claim that up to a sign, it is $a_{ij} - a_{ik}$ (where $\{ j, k\} = \{ 1, 2, 3\} \setminus \{ i\}$).

To compute the relevant class, we evaluate it on $P_{ij}$, $P_{jk}$ and $P_{ik}$, which is computed by intersecting the relevant submanifolds. It does not intersect $P_{jk}$, because in $P_{jk}$, the point $i$ is at “infinity”, so cannot be in between $j$ and $k$.

On the other hand, in $P_{ij}$ and $P_{jk}$, there is exact one point where $i, j, k$ are colinear and $i$ is in the middle, and these points come with opposite signs.

Now $\mathrm{col}_1$ and $\mathrm{col}_2$ are disjoint, so the represented cohomology classes have zero product. So

$0 = (a_{12} - a_{13})(a_{23} - a_{21}) = a_{12} a_{23} + a_{23} a_{31} + a_{31} a_{12}.$

Proof

Definition 3.4

We let $\mathrm{Siop}^d(n)$ be the abelian group generated by graphs on $\mathbf{n}$ subject to the previous relations. This is a ring by taking unions.

There is then a map $\mathrm{Siop}^d(n) \to H^*(\operatorname{Conf}_n(\mathbb {R}^d))$, which we will show to be an isomorphism.