Homology of the Ed\mathbb {E}_d operadCohomology

3 Cohomology

We now produce some cohomology classes of H(Confn(Rd))H_*(\operatorname{Conf}_n(\mathbb {R}^d)) in degree d1d - 1, which will generate the whole cohomology.

Definition 3.1

For i,j{1,,n}i, j \in \{ 1, \ldots , n\} , let αij:Confn(Rd)Conf2(Rd)Sd1\alpha _{ij}: \operatorname{Conf}_n(\mathbb {R}^d) \to \operatorname{Conf}_2(\mathbb {R}^d) \simeq S^{d - 1} send (x1,,xn)(x_1, \ldots , x_n) to (xi,xj)(x_i, x_j). Let aijHd1(Confn(Rd))a_{ij} \in H^{d - 1}(\operatorname{Conf}_n(\mathbb {R}^d)) be the pullback of the fundamental class along αij\alpha _{ij}.

Note that aij=(1)dajia_{ij} = (-1)^d a_{ji}, and aij2=0a_{ij}^2 = 0, since the square of the fundamental class on Sd1S^{d - 1} vanishes.

The first claim is that these aija_{ij} are dual to the PijP_{ij}.

Lemma 3.2

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

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

In particular, the set {aij:i<j}\{ a_{ij}: i < j\} is linearly independent and forms a basis of Hd1(Confn(Rd))H^{d - 1}(\operatorname{Conf}_n(\mathbb {R}^d)), since we know this group is free of rank (n2)\binom {n}{2}.

Consider the composite

Sd1Confn(Rd)Sd1. S^{d - 1} \to \operatorname{Conf}_n(\mathbb {R}^d) \to S^{d - 1}.

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

If i=ki = k and j=j = \ell , then this is the identity map, which has degree 11. Otherwise, taking ε0\varepsilon \to 0 gives a homotopy to the constant map.


We now know that as a ring, the cohomology of Confn(Rd)\operatorname{Conf}_n(\mathbb {R}^d) is generated by the aija_{ij}. We introduce a graphical way to depict products of the aija_{ij}. Such a product is represented by a graph with vertices {1,,n}\{ 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 aija_{ij}, we take the graph whose only edge is an edge from ii to jj. Products are then represented by unions. For example, a42a43a13a_{42} a_{43} a_{13} is represented by the graph

    \node (1) {1};
    \node (2) at (1, 0) {2};
    \node (3) at (0, -1) {3};
    \node (4) at (1, -1) {4};
    \draw [->] (4) -- (2) node [pos=0.5, right] {\scriptsize \color{red} 1};
    \draw [->] (4) -- (3) node [pos=0.5, below] {\scriptsize \color{red} 2};
    \draw [->] (1) -- (3) node [pos=0.5, left] {\scriptsize \color{red} 3};

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)d1(-1)^{d - 1}. Moreover, since aij2=0a_{ij}^2 = 0, any graph with a repeated edge represents 00.

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

Lemma 3.3 (Arnold)
      \begin{tikzpicture}[scale=0.5, baseline={([yshift=-.5ex]current bounding box.center)}]
        \node (i) at (-1, 0) {$i$};
        \node (k) at (1, 0) {$k$};
        \node (j) at (0, 1.732) {$j$};

        \draw [->] (i) -- (j) node [pos=0.7, left] {\scriptsize \color{red} 1};
        \draw [->] (j) -- (k) node [pos=0.3, right] {\scriptsize \color{red} 2};
      \begin{tikzpicture}[scale=0.5, baseline={([yshift=-.5ex]current bounding box.center)}]
        \node (i) at (-1, 0) {$i$};
        \node (k) at (1, 0) {$k$};
        \node (j) at (0, 1.732) {$j$};

        \draw [->] (j) -- (k) node [pos=0.3, right] {\scriptsize \color{red} 1};
        \draw [->] (k) -- (i) node [pos=0.5, below] {\scriptsize \color{red} 2};
      \begin{tikzpicture}[scale=0.5, baseline={([yshift=-.5ex]current bounding box.center)}]
        \node (i) at (-1, 0) {$i$};
        \node (k) at (1, 0) {$k$};
        \node (j) at (0, 1.732) {$j$};

        \draw [->] (k) -- (i) node [pos=0.5, below] {\scriptsize \color{red} 1};
        \draw [->] (i) -- (j) node [pos=0.7, left] {\scriptsize \color{red} 2};
      = 0.

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

a12a23+a23a31+a31a12=0. a_{12} a_{23} + a_{23} a_{31} + a_{31} a_{12} = 0.

We will prove this using the intersection product.

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

Hk(Mn)Hnk(Mn). H^k(M^n) \cong H_{n - k}(M^n).

Under this isomorphism, the cup product on H(Mn)H^*(M^n) induces a product on H(Mn)H_*(M^n). If x,yH(Mn)x, y \in H_*(M^n) are represented by submanifolds X,YX, Y that intersect transversely, then xyx \cdot y is represented by XYX \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 MnM^n represent a Borel—More homology class.

Let coli\mathrm{col}_i be the submanifold of Conf3(Rd)\operatorname{Conf}_3(\mathbb {R}^d) where x1,x2,x3x_1, x_2, x_3 are colinear and xix_i is in the middle. This is a submanifold of codimension d1d - 1, and represents a class in Hd1(Conf3(Rd))H^{d - 1}(\operatorname{Conf}_3(\mathbb {R}^d)). We claim that up to a sign, it is aijaika_{ij} - a_{ik} (where {j,k}={1,2,3}{i}\{ j, k\} = \{ 1, 2, 3\} \setminus \{ i\} ).

To compute the relevant class, we evaluate it on PijP_{ij}, PjkP_{jk} and PikP_{ik}, which is computed by intersecting the relevant submanifolds. It does not intersect PjkP_{jk}, because in PjkP_{jk}, the point ii is at “infinity”, so cannot be in between jj and kk.

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

Now col1\mathrm{col}_1 and col2\mathrm{col}_2 are disjoint, so the represented cohomology classes have zero product. So

0=(a12a13)(a23a21)=a12a23+a23a31+a31a12. 0 = (a_{12} - a_{13})(a_{23} - a_{21}) = a_{12} a_{23} + a_{23} a_{31} + a_{31} a_{12}.


Definition 3.4

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

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