# Claire: Classical and homological perspectives on scissors congruence (Monday, June 13)

## Classical scissors congruence

:::{.remark}
We'll be thinking about polytopes in $X = E^n, S^n, H^n$.
We saw that in $\RR^2$, area determines SC.
Outline:

- Defining $P(X, G)$ and $I(X) \da \Isom(X)$.
- Defining the Dehn invariant $D$
- Hilbert's 3rd problem
- The Dehn-Sedler theorem
- $\ker D$ and $\coker D$.
- Recasting everything in terms of group cohomology
- Action on the Steinberg module
- The Tits building.

:::

:::{.remark}
Let $X=\RR^n$, note $I(\RR^n) = T(n) \semidirect \Orth_n$.
:::

:::{.definition title="Simplices and polytopes"}
A **geometric simplex** in $X$ is the convex hull of $(n+1)$ points, and we write $\sigma = \abs{(a_0,\cdots, a_n)}$.
A **polytope** is a finite union of simplices
:::

:::{.definition title="Scissors congruence"}
Write $P = P' \disjoint P''$ if $P = P' \union P''$ and the interiors of $P', P''$ are disjoint.
For $G\leq I(X)$, we say $P$ and $Q$ are **$G\dash$scissors congruent** if $P = \disjoint P_i$, $Q = \disjoint Q_i$, and $P_i = g Q_i$ for some $g\in G$.
We write $P\sim_G Q$.
Define $P(G, X)$ as the free abelian group generated by $[P]$ for polytopes $P$ with relations

1. $[P] = [P'] + [P'']$ if $P = P' \disjoint P''$
2. $[P] = [gP']$ for some $g\in G$.

:::

## The Dehn Invariant

:::{.remark}
As seen in Inna's talk, volume alone is not enough to determine $\SC$ in $\RR^3$.
:::

:::{.theorem title="Dehn-Sydler"}
If $\vol(P) = \vol(P')$ and $D(P) = D(P')$ then $P\sim_g P'$, i.e. $P$ and $P'$ are $\SC$.
The invariant is iven by
\[
D: \mcp \to \RR \tensor_\ZZ \RR/\ZZ \\
P &\mapsto \sum_{e\in \mathrm{ edges of } P }\ell(p) \tensor \theta(e)/\pi
,\]

:::

:::{.remark}
Note that $\ker(D) \mapsvia{\vol} \RR$ is an isomorphism.
The theorem can be recast as injectivity of $D$, and surjectivity is via scaling polytopes.
One can also identify the cokernel:
\[
P \mapsvia{D} \RR\tensor_\ZZ \RR/\ZZ \surjects \coker(D) \cong \Omega_{\RR/\ZZ} \to 0$.
.\]
:::

## Scissors congruence in homological terms

:::{.proposition title="Trivial scissors congruence group as homology"}
Note that $P(X, 1) \cong H_n(\complex{C}(X)/\complex{C}^{n-1})$ where $\complex{C}(X)$ is the simplicial chain complex and this quotient can be written as the set of *proper* $n\dash$simplices.
:::

:::{.proof title="?"}
Let $\eps_\sigma = \pm 1$ be the orientation of a simplex $\sigma$.
Define a map
\[
\phi: C_n(X) &\to P(X, 1) \\
\sigma &\mapsto

\begin{cases}
\eps_\sigma [\abs{\sigma}] & \text{ if $\sigma$ is proper}  
\\
0 & \text{otherwise};
\end{cases}
\]
this is what induces the desired isomorphism after quotienting and taking homology.
:::

:::{.proposition title="General scissors congruence as homology"}
If $[P] = [gP]$ for some $g$ then $P(X, G) \cong H_0(; P(X, 1))$.
:::

:::{.proof title="?"}
Group homology is the left-derived functor of $M \to M_G \da M/\gens{m-gm}$, or equivalently of tensoring in $\ZZ[G]$, and $H_0$ is precisely the coinvariants.
:::

:::{.definition title="Steinberg"}
The **Steinberg module** is defined as 
\[
\St(X) = H_n (\complex{C}(X)/ \complex{C}(X)^{n-1})
.\]
:::

:::{.remark}
Note that the induced map $\phi_*$ is not quite an isomorphism of $I(X)\dash$modules, the following square doesn't quite commute:

\begin{tikzcd}
	{C_n(X)} && {P(X,1)} \\
	\\
	{C_n(X)} && {P(X,1)}
	\arrow["\phi", from=1-1, to=1-3]
	\arrow["\phi", from=3-1, to=3-3]
	\arrow["{g^*_{\text{twist}}}", from=1-1, to=3-1]
	\arrow["{g^*}", from=1-3, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJDX24oWCkiXSxbMiwwLCJQKFgsMSkiXSxbMCwyLCJDX24oWCkiXSxbMiwyLCJQKFgsMSkiXSxbMCwxLCJcXHBoaSJdLFsyLDMsIlxccGhpIl0sWzAsMiwiZ14qX3tcXHRleHR7dHdpc3R9fSJdLFsxLDMsImdeKiJdXQ==)

One instead needs to twist by $\det(g) = \pm 1$, which yields the twist $P(X, 1) = \St(X)^t$.
Thus
\[
P(X, G) \cong H_0(G; \St(X)^t)
,\]
which realizes $P$ as Lie group homology made discrete, i.e. we ignore the topology of $\Orth_n$ and more generally of $G$.
:::

## Steinberg as the homology of its Tits building

:::{.remark}
In the euclidean case, $\St(X)$ has a homological origin:
:::

:::{.definition title="Tits Building"}
For any vector space $X$, define the **Tits building** $\tau(X)$ as the geometric realization of the poset of proper nonzero subspaces of $X$ ordered by inclusion.
Concretely, as a simplicial set this has

- Vertices: proper nonzero subspaces, and
- Simplices: flags in $X$.

:::

:::{.definition title="Link of a vertex"}
For $T$ a poset and $t$ a vertex, the **link** of $t$ is defined as
\[
L_t \da \ts{s\in T \st s < t \text{ or } s > t}
.\]

:::

:::{.exercise title="?"}
Draw $\tau(\FF_2^3)$; it has 21 1-simplices!
:::

:::{.theorem title="Homotopy type of the Tits building for $\RR^n$"}
For $X = \RR^n$, Quillen showed $\tau(X) \homotopic \bigvee S^{n-2}$.
:::

:::{.remark}
$\tau(X)$ has one nontrivial homology group, and Dupont shows $H_{n-?}(\tau(X)) \cong \St(X)$.
:::