Homology of the Ed\mathbb {E}_d operadHomology

2 Homology

We produce some homology classes associated to trees.

Definition 2.1

Let S{1,,n}=nS \subseteq \{ 1, \ldots , n\} = \mathbf{n}. An SS-tree is a binary tree with a root vertex and leaves labelled by elements of SS (each label is used exactly once). We also distinguish between the left and right branches of the binary tree.

By convention, the term “vertex” does not refer to the leaves. Thus, the number of vertices is the number of leaves 1 - 1. We write T|T| for the number of vertices of TT.

Example 2.2

The following is are trees with 1 and 3 vertices respectively:

\begin{tikzpicture} [scale=0.3]
      \draw (0, 0) -- (-0.5, 0.5) node [above=-0.05] {\tiny $i$};
      \draw (0, 0) -- (0.5, 0.5) node [above=-0.05] {\tiny $j$};
    \end{tikzpicture}
\begin{tikzpicture} [scale=0.3]
      \draw (0, 0) -- (1.5, 1.5) node [above=-0.05] {\tiny $\ell$};
      \draw (0, 0) -- (-0.5, 0.5) node [above=-0.05] {\tiny $i$};
      \draw (0.5, 0.5) -- (0, 1) node [above=-0.05] {\tiny $j$};
      \draw (1, 1) -- (0.5, 1.5) node [above=-0.05] {\tiny $k$};
    \end{tikzpicture}

Fix 0<ε<13 0 < \varepsilon < \frac{1}{3}. Given an SS-tree TT, define a map PT:(Sd1)×TConfn(Rd)P_T: (S^{d - 1})^{\times |T|} \to \operatorname{Conf}_n(\mathbb {R}^d) as follows. If inSi \in \mathbf{n} \setminus S, put xix_i at some fixed point “at infinity”. For the rest, we define this by example.

If TT is

\begin{tikzpicture} [scale=0.3]
  \draw (0, 0) -- (-0.5, 0.5) node [above=-0.05] {\tiny $i$};
  \draw (0, 0) -- (0.5, 0.5) node [above=-0.05] {\tiny $j$};
\end{tikzpicture}

, we send vSd1\mathbf{v} \in S^{d - 1} to the configuration

\begin{tikzpicture} 
    \draw [dashed] circle (2);
    \node [anchor = north east] {$0$};

    \draw [->] (0, 0) -- (2 / 1.414, 2 / 1.414) node [pos=0.5, anchor = south east, yshift=-2, xshift=2] {$\varepsilon \mathbf{v}$};

    \node [hollow] at (0, 0) {};
    \node [circ] at (2 / 1.414, 2 / 1.414) {};
    \node [anchor = south west] at (2 / 1.414, 2 / 1.414) {$x_i$};
    \node [circ] at (-2 / 1.414, -2 / 1.414) {};
    \node [anchor = north east] at (-2 / 1.414, -2 / 1.414) {$x_j$};
  \end{tikzpicture}

We abbreviate this as PijP_{ij}.

If TT is

\begin{tikzpicture} [scale=0.3]
  \draw (0, 0) -- (1, 1) node [above=-0.05] {\tiny $k$};
  \draw (0, 0) -- (-0.5, 0.5) node [above=-0.05] {\tiny $i$};
  \draw (0.5, 0.5) -- (0, 1) node [above=-0.05] {\tiny $j$};
\end{tikzpicture}

we send (v1,v2)(\mathbf{v}_1, \mathbf{v}_2) to

\begin{tikzpicture} 
    \draw [dashed] circle (2);
    \node [anchor = north east] {$0$};

    \draw [->] (0, 0) -- (2 / 1.414, 2 / 1.414) node [pos=0.5, anchor = south east, yshift=-2, xshift=2] {\small $\varepsilon \mathbf{v}_1$};

    \node [hollow] at (0, 0) {};
    \node [circ] at (2 / 1.414, 2 / 1.414) {};
    \node [anchor = south west] at (2 / 1.414, 2 / 1.414) {$x_i$};
    \draw [dashed] (-2 / 1.414, -2 / 1.414) circle (0.7);
    \draw [->] (-2 / 1.414, -2 / 1.414) -- +(0.7, 0) node [above, pos=0.5] {\tiny $\varepsilon^2 \mathbf{v}_2$};
    \node [hollow] at (-2 / 1.414, -2 / 1.414) {};

    \node [circ] at (-2 / 1.414 + 0.7, -2 / 1.414) {};
    \node [right] at (-2 / 1.414 + 0.7, -2 / 1.414) {$x_j$};

    \node [circ] at (-2 / 1.414 - 0.7, -2 / 1.414) {};
    \node [left] at (-2 / 1.414 - 0.7, -2 / 1.414) {$x_k$};
  \end{tikzpicture}

The fundamental class of (Sd1)×T(S^{d - 1})^{\times |T|} gives us a corresponding homology class in Confn(Rd)\operatorname{Conf}_n(\mathbb {R}^d), which we still call PTP_T.

Theorem 2.3

The classes in H(Confn(Rd))H_*(\operatorname{Conf}_n(\mathbb {R}^d)) satisfy the relations

\begin{useimager} 
    \[
      \begin{tikzpicture}[scale=0.3, baseline={([yshift=-.5ex]current bounding box.center)}]
        \draw (0, 0) -- (-0.5, 0.5) node [above=-0.05] {\tiny $T_1$};
        \draw (0, 0) -- (0.5, 0.5) node [above=-0.05] {\tiny $T_2$};
        \draw (0, 0) -- (0, -0.707) node [below=-0.05] {\tiny $R$};
      \end{tikzpicture}
      = (-1)^{d + |T_1| |T_2| (d - 1)}
      \begin{tikzpicture}[scale=0.3, baseline={([yshift=-.5ex]current bounding box.center)}]
        \draw (0, 0) -- (-0.5, 0.5) node [above=-0.05] {\tiny $T_2$};
        \draw (0, 0) -- (0.5, 0.5) node [above=-0.05] {\tiny $T_1$};
        \draw (0, 0) -- (0, -0.707) node [below=-0.05] {\tiny $R$};
      \end{tikzpicture}
    \]
  \end{useimager}
\begin{useimager} 
    \[
      \begin{tikzpicture}[scale=0.3, baseline={([yshift=-.5ex]current bounding box.center)}]
        \draw (0, 0) -- (-1, 1) node [above=-0.05] {\tiny $T_1$};
        \draw (-0.5, 0.5) -- (0, 1) node [above=-0.05] {\tiny $T_2$};
        \draw (0, 0) -- (1, 1) node [above=-0.05] {\tiny $T_3$};
        \draw (0, 0) -- (0, -0.707) node [below=-0.05] {\tiny $R$};
      \end{tikzpicture}
      +
      \begin{tikzpicture}[scale=0.3, baseline={([yshift=-.5ex]current bounding box.center)}]
        \draw (0, 0) -- (-1, 1) node [above=-0.05] {\tiny $T_2$};
        \draw (-0.5, 0.5) -- (0, 1) node [above=-0.05] {\tiny $T_3$};
        \draw (0, 0) -- (1, 1) node [above=-0.05] {\tiny $T_1$};
        \draw (0, 0) -- (0, -0.707) node [below=-0.05] {\tiny $R$};
      \end{tikzpicture}
      +
      \begin{tikzpicture}[scale=0.3, baseline={([yshift=-.5ex]current bounding box.center)}]
        \draw (0, 0) -- (-1, 1) node [above=-0.05] {\tiny $T_3$};
        \draw (-0.5, 0.5) -- (0, 1) node [above=-0.05] {\tiny $T_1$};
        \draw (0, 0) -- (1, 1) node [above=-0.05] {\tiny $T_2$};
        \draw (0, 0) -- (0, -0.707) node [below=-0.05] {\tiny $R$};
      \end{tikzpicture}
      = 0
    \]
  \end{useimager}

for any (sub)trees T1,T2,T3,RT_1, T_2, T_3, R

The second identity is, of course, the Jacobi identity.

Proof
The first identity comes from changing the orientations of (Sd1)×T(S^{d - 1})^{\times |T|}.

To simplify notation, we only prove the second identity in the case where T1,T2,T3,RT_1, T_2, T_3, R are trivial and n=3n = 3, i.e. we show that

\begin{useimager} 
    \[
      \begin{tikzpicture}[scale=0.3, baseline={([yshift=-.5ex]current bounding box.center)}]
        \draw (0, 0) -- (-1, 1) node [above=-0.05] {\tiny 1};
        \draw (-0.5, 0.5) -- (0, 1) node [above=-0.05] {\tiny 2};
        \draw (0, 0) -- (1, 1) node [above=-0.05] {\tiny 3};
      \end{tikzpicture}
      +
      \begin{tikzpicture}[scale=0.3, baseline={([yshift=-.5ex]current bounding box.center)}]
        \draw (0, 0) -- (-1, 1) node [above=-0.05] {\tiny 2};
        \draw (-0.5, 0.5) -- (0, 1) node [above=-0.05] {\tiny 3};
        \draw (0, 0) -- (1, 1) node [above=-0.05] {\tiny 1};
      \end{tikzpicture}
      +
      \begin{tikzpicture}[scale=0.3, baseline={([yshift=-.5ex]current bounding box.center)}]
        \draw (0, 0) -- (-1, 1) node [above=-0.05] {\tiny 3};
        \draw (-0.5, 0.5) -- (0, 1) node [above=-0.05] {\tiny 1};
        \draw (0, 0) -- (1, 1) node [above=-0.05] {\tiny 2};
      \end{tikzpicture}
      = 0
    \]
  \end{useimager}

The general case admits the exact same proof.

Consider the submanifold of Conf3(Rd)\operatorname{Conf}_3(\mathbb {R}^d) consisting of points satisfying:

  1. xi=0\sum x_i = 0

  2. The perimeter of the triangle spanned by x1,x2,x3x_1, x_2, x_3 is 4ε+2ε24\varepsilon + 2 \varepsilon ^2.

  3. The sides of the triangle have length at least 2ε22\varepsilon ^2.

One observes that this manifold has three boundary components, achieved when one of the sides have length exactly 2ε22 \varepsilon ^2. One then sees that this component is homotopic to PTP_T for some TT in the identity. For example, the component where the x1x_1x2x_2 side has length 2ε22 \varepsilon ^2 is homotopic to the first term.

Proof

Definition 2.4

An nn-forest is a collection of SS-trees where each element in n\mathbf{n} is used exactly once.

If F=TiF = \bigcup T_i is an FF-tree, we define PF:(Sd1)×FConfn(Rd)P_F: (S^{d - 1})^{\times |F|} \to \operatorname{Conf}_n(\mathbb {R}^d) in the same way as PTP_T, where the position of xkx_k is specified by the tree that contains kk, and the iith tree is translated by (i,0,0,,0)(i, 0, 0, \ldots , 0).

Definition 2.5

Let Poisd(n)\mathrm{Pois}^d(n) be the abelian group generated by nn-forest subject to the (anti)commutativity and the Jacobi identity.

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