# The two-sided Bar Construction on Orbifolds

Submitted by jeff on Sun, 20/07/2014 - 05:24

The bar construction is a way to turn an algebraic object into a topological space.

References:

• Two sided Bar construction: http://math.stanford.edu/~carym/bar.pdf
• General Orbifold constructions: Adem, Leida & Ruan - Orbifolds and Stringy Topology

Background.

Without going into too much detail, we recall that a groupoid $\mathcal{G}$ is made up of a set of objects $G_0$ and a set of arrows $G_1$.  Each arrow $g \in G_1$ has a source and target map $(s, t) : G_1 \rightarrow G_0 \times G_0$.  Along with other structure maps.  Orbifolds make $G_0$ and $G_1$ a space, adds smoothness etc.

Our use of the two-sided Bar Construction is an attempt to describe the preimage of a simplex $(Y_0 \rightarrow \cdots \rightarrow Y_n, t)$ under an orbifold morphism $f$.  The two-sided bar construction involves having a space $X$ with a right $\mathcal{G}$ action, and a space $Y$ with a left $\mathcal{G}$ action to construct a complex $B (X, \mathcal{G}, Y)$.  We now define left/right $\mathcal{G}$-spaces.

Definition(Definition 2.14 - Adem, Leida & Ruan): Let $\mathcal{G}$ be an orbifold groupoid.  A left $G$-space is a manifold $E$ with an action of $\mathcal{G}$.  The action comes in two parts:

1. an anchor map $\pi : E \rightarrow G_0$; and
2. an action map $\mu : G_1 \times_{G_0} E \rightarrow E$, which is defined on $(g, e)$ when $\pi(e) = s(g)$ and is written $\mu(g, e) = g \cdot e$.

These must satisfy $\pi (g \cdot e) = t(g)$, $1_x \cdot e = e$ and $g \cdot (h \cdot e) = (g h) \cdot e$ when $x \xrightarrow{h} y \xrightarrow{g} z$ in $G_1$ and $e \in E$ with $\pi(e) = x$.

With a slight modification, we define a right $G$-space.

Definition: Let $\mathcal{G}$ be an orbifold groupoid.  A right $G$-space is a manifold $E$ with an action of $\mathcal{G}$.  The action comes in two parts:

1. an anchor map $\pi : E \rightarrow G_0$; and
2. an action map $\mu : E \times_{G_0} G_1 \rightarrow E$, which is defined on $(e, g)$ when $\pi(e) = t(g)$ and is written $\mu(e, g) = e \cdot g$.

With similar conditions to above.

Construction.

Let $X$ be a right $\mathcal{G}$-space and let $Y$ be a left $\mathcal{G}$-space with anchor and action maps \begin{align*} \pi_X : X &\rightarrow G_0,\\ x &\mapsto s(x),\\ \pi_Y : Y &\rightarrow G_0,\\ y &\mapsto t(y),\\ \mu_X : X \times_{G_0} G_1 &\rightarrow X,\\ (x, g) &\mapsto x \circ g,\\ \mu_Y : G_1 \times_{G_0} Y &\rightarrow Y,\\ (g, y) &\mapsto g \circ y.\end{align*}  The two-sided Bar construction $B (X, \mathcal{G}, Y)$ is now, $$B (X, \mathcal{G}, Y) = \bigsqcup_{n \geq 0} (X \times_{G_0} G_1 \times_{G_0} G_1 \times_{G_0} \cdots \times_{G_0} G_1 \times_{G_0} Y) \times \Delta^n / \sim,$$ where elements look like $((x, g_1, g_2, \ldots, g_n, y), (t_0, \ldots, t_n))$ with $s(g_i) = t(g_{i + 1})$, $\pi_X (x) = t(g_1)$, $\pi_Y (y) = s(g_n)$, $\sum t_i = 1$ and equivalence relation $\sim$ generated by:

1. \begin{align*} &((x, g_1, \ldots, g_{i - 1}, 1, g_{i + 1}, \ldots, g_n, y), (t_0, \ldots, t_n))\\ &\sim ((x, g_1, \ldots, g_{i - 1}, g_{i + 1}, \ldots, g_n, y), (t_0, \ldots, t_{i - 2}, t_{i - 1} + t_{i}, t_{i + 1}, \ldots, t_{n})); \end{align*}
2. \begin{align*} &((x, g_1, \ldots, g_n, y), (0, t_1, \ldots, t_n))\\ &\sim ((\mu_X (x, g_1), g_2 \ldots, g_n, y), (t_1, \ldots, t_{n})); \end{align*}
3. \begin{align*} &((x, g_1, \ldots, g_n, y), (t_0, \ldots, t_{i - 1}, 0, t_{i + 1}, \ldots, t_n))\\ &\sim ((x, g_1, \ldots, g_{i - 1}, g_{i} g_{i + 1}, \ldots, g_n, y), (t_0, \ldots, t_{i - 1}, t_{i + 1}, t_{i + 2}, \ldots, t_{n})); \end{align*}
4. \begin{align*} &((x, g_1, \ldots, g_n, y), (t_0, \ldots, t_{n - 1}, 0))\\ &\sim ((x, g_1, \ldots, g_{n - 1}, \mu_Y (g_n, y)), (t_0, \ldots, t_{n - 1})). \end{align*}

A good way to think of the last three equivalences, is to think of a simplex as $$\xleftarrow{x} z_0 \xleftarrow{g_1} z_1 \xleftarrow{g_2} \cdots \xleftarrow{g_n} z_n \xleftarrow{y}.$$  Then if $t_i = 0$, we drop $z_i$ and compose the surrounding arrows giving $$\xleftarrow{x} z_0 \xleftarrow{g_1} \cdots \xleftarrow{g_{i - 1}} z_{i - 1} \xleftarrow{g_{i} \circ g_{i + 1}} z_{i + 1} \xleftarrow{g_{i + 2}} \cdots \xleftarrow{g_n} z_n \xleftarrow{y},$$ and similarly for $i = 0$ and $i = n$ bringing in the right/left actions or $\mathcal{G}$ on $X$ and $Y$.

Also notice that if $X = Y = *$, we get the usual bar construction $B \mathcal{G}$, as our action maps $\mu_X : * \times_{G_0} G_1 \rightarrow *$ and $\mu_Y : G_1 \times_{G_0} * \rightarrow *$ are now trivial.