2 Homology

We produce some homology classes associated to trees.

Definition 2.1

Let $S \subseteq \{ 1, \ldots , n\} = \mathbf{n}$ . An $S$ -tree is a binary tree with a root vertex and leaves labelled by elements of $S$ (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 $- 1$ . We write $|T|$ for the number of vertices of $T$ .

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 $0 < \varepsilon < \frac{1}{3}$ . Given an $S$ -tree $T$ , define a map $P_T: (S^{d - 1})^{\times |T|} \to \operatorname{Conf}_n(\mathbb {R}^d)$ as follows. If $i \in \mathbf{n} \setminus S$ , put $x_i$ at some fixed point “at infinity”. For the rest, we define this by example.

If $T$ 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 $\mathbf{v} \in S^{d - 1}$ to the configuration

We abbreviate this as $P_{ij}$ .

If $T$ 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 $(\mathbf{v}_1, \mathbf{v}_2)$ 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 $(S^{d - 1})^{\times |T|}$ gives us a corresponding homology class in $\operatorname{Conf}_n(\mathbb {R}^d)$ , which we still call $P_T$ .

Theorem 2.3

The classes in $H_*(\operatorname{Conf}_n(\mathbb {R}^d))$ 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 $T_1, T_2, T_3, R$

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

Proof

□

The first identity comes from changing the orientations of

(S^{d - 1})^{\times |T|}

To simplify notation, we only prove the second identity in the case where $T_1, T_2, T_3, R$ are trivial and $n = 3$ , 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 $\operatorname{Conf}_3(\mathbb {R}^d)$ consisting of points satisfying:

$\sum x_i = 0$
The perimeter of the triangle spanned by $x_1, x_2, x_3$ is $4\varepsilon + 2 \varepsilon ^2$ .
The sides of the triangle have length at least $2\varepsilon ^2$ .

One observes that this manifold has three boundary components, achieved when one of the sides have length exactly $2 \varepsilon ^2$ . One then sees that this component is homotopic to $P_T$ for some $T$ in the identity. For example, the component where the $x_1$ – $x_2$ side has length $2 \varepsilon ^2$ is homotopic to the first term.

Proof

□

Definition 2.4

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

If $F = \bigcup T_i$ is an $F$ -tree, we define $P_F: (S^{d - 1})^{\times |F|} \to \operatorname{Conf}_n(\mathbb {R}^d)$ in the same way as $P_T$ , where the position of $x_k$ is specified by the tree that contains $k$ , and the $i$ th tree is translated by $(i, 0, 0, \ldots , 0)$ .

Definition 2.5

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

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