# Oliver: Translation Scissors Congruence and Rational Structures (Monday, June 13)

:::{.remark}
Setup: $X= E^n$ where $n\da \dim(X)$ is fixed.
Noting that $\Isom(E^n) \cong \RR^n\semidirect \Orth_n$, consider $P(X, \Isom(E^n))$.
We can realize this as iterated coinvariants:
\[
\RR^n\semidirect \Orth_n 
&\cong H_0(\Orth_n; P(X, \RR^n))\\
&\cong H_0\qty{ \Orth_n; H_0\qty{  \RR^n; {\complex{C}(X) \over \complex{C}(X)^{n-1}}  }}
,\]
where all Lie groups in sight are discrete.
Note that 

- $C_q(X) = \ZZ\adjoin{(q+1)\dash\text{tuples of points in } X}$,
- $C_q(X)^{n-1} = \ZZ\adjoin{(q+1)\dash\text{tuples of points in a proper linear subspace of } X}$,
- $C_q(X) = C_q(X)$ when $q\leq n-1$

:::

:::{.remark}
By left-exactness we can commute homologies:
\[
H_0\qty{ \Orth_n; H_0\qty{ \RR^n; {\complex{C}(X) \over \complex{C}(X)^{n-1}}  } } \cong H_n\qty{ \complex{C}(X)/\RR^n \over  \complex{C}(X)^{n-1}/\RR^n}
.\]

:::

:::{.definition title="Translation scissors congruence"}
Let $\FF$ be a characteristic zero field and $V\in \Vect\slice\FF^{\dim =n}$, the **translational scissors congruence group of $V$** is
\[
P_T(V) \da H_n\qty{ \complex{C}(V)/V \over \complex{C}(V)^{n-1}/V }
.\]
:::

:::{.question}
What is the structure of $P+T(V)$?
:::

:::{.remark}
Recall that $C_q(G) \da \ZZ\adjoin{G^{q+1}}$ defines a complex resolving $\ZZ[G]$.
Note that $G\actson \ZZ[G^{q}]$ diagonally, so define
\[
H_*(G; \ZZ) = H_*(G) \da H_*(\complex{C}(G)/G)
.\]

:::

:::{.definition title="Eilenberg-Zilber map"}
Consider the following bar notation:
\[
g[g_1 | g_2 | \cdots | g_q] \da \qty{ g, g g_1, g g_1 g_2,\cdots, g \prod_{1\leq i\leq q} g_i}
.\]

With this, we define the **Eilenberg-Zilber map** as follows:
\[
\EZ: \complex{C}(G)\tensorpower{\ZZ}{2} &\to \complex{C}(G\cartpower{2}) \\
g[g_1| \cdots | g_q] \tensor g'[g'_1 | \cdots | g'_p] &\mapsto \sum_{\text{shuffles }\sigma} \sign(\sigma) (g, g') [h_{\sigma(1)} | \cdots | h_{\sigma(p+q)} ]
.\]
:::

:::{.example title="?"}
Let $G = \RR$, then
\[
0[1] \tensor 0[1] \mapsto (0,0)[(1,0), (0, 1)] - (0,0)[(0, 1), (1,0)]
.\]
:::

:::{.slogan}
$\EZ$ triangulates products.

\todo{Missing simplicial interpretation!}
:::

:::{.remark}
If $G$ is abelian then addition is a group morphism.
Composing with $\EZ$ yields the Pontryagin product, making $H_*(G)$ into a graded commutative ring:

\begin{tikzcd}
	{H_*(G)\tensorpower{\ZZ}{2}} && {H_*(G\cartpower{2})} && {H_*(G)}
	\arrow["\EZ", from=1-1, to=1-3]
	\arrow["{+_{\scriptscriptstyle G}}", from=1-3, to=1-5]
	\arrow["\Lambda", curve={height=-30pt}, from=1-1, to=1-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJIXyooRylcXHRlbnNvcnBvd2Vye1xcWlp9ezJ9Il0sWzIsMCwiSF8qKEdcXGNhcnRwb3dlcnsyfSkiXSxbNCwwLCJIXyooRykiXSxbMCwxLCJcXEVaIl0sWzEsMiwiKyJdLFswLDIsIlxcTGFtYmRhIiwwLHsiY3VydmUiOi01fV1d)

:::

:::{.observation}
Note:

- $H_0(G) \cong \ZZ$.
- $H_1(G)\cong G$ is $G$ is abelian.
- If $\tors(G) = 0$ then there is an isomorphism $\Extalg_\ZZ H_1(G) \to H_*(G)$.
- If $V\in \qmod$ then there is an isomorphism $\Extalg_\QQ V\to H_*(V)$.

:::

:::{.remark}
The main takeaway: for $U\in \qmod$, $H_*(\complex{C}(U)/U) \cong \Extalg_\QQ U$.
:::

:::{.remark}
There is a giant double complex $\bicomplex{A}$; taking homology one way yields
\[
H_k(\bicomplex{A}) \iso H_{k+1}(\tau(X))
,\]
noting the degree shift.
Taking homology the other way yields
\[
E_{p, q}^2 = \tilde H_p\qty{ \tau(V); \Extprod_\QQ^2\lieg} \quad q = ?, \qquad 0 \text{ otherwise}
,\]
where the coefficients are in a local system.

> Todo: what is this local system?

:::

:::{.remark}
Consider the dilation operator $\mu_a: V \mapsvia{\cdot a} V$, multiplication by an integer $a > 1$ on $V$.
For any $U \subseteq V$, this induces a chain endomorphism
\(
\complex{C}(U)/U \selfmap
\)
and thus a morphism of spectral sequences.
On $E^1_{p, q} = \bigoplus \Extalg^q U_p$, this induces $x\mapsto xa^q$.
Suppose for $r>1$ this produces a diagram commuting with the differentials which forces $d^r = 0$.
:::

:::{.remark}
Summary:

- \(H_*(
\complex{C}/V
/ \complex{C}^{n-1}/V
)\) is in $\qmod$ and thus so is $P_T(V)$.
- There is a splitting 
\[
P_T(V) \cong \bigoplus _{1\leq q\leq n} \tilde H_{n-q-1}\qty{ \tau(V); \Extprod_\QQ^q \lieg}
\]
into $a^q$ eigenspaces.
- These summands are zero when $p+q < n$.

:::

:::{.example title="?"}
What are these eigenspaces?
The unit square is a $a^q$ eigenvector for $a=q=2$:

<!-- Xournal file: /home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Projects/Talbot2022/sections/figures/2022-06-13_15-46.xoj -->

![](figures/2022-06-13_15-47-56.png)

:::

:::{.theorem title="?"}
There is a map $\Extalg^q_\QQ \lieg\to \Extalg_F^q \lieg$ which induces an isomorphism 
\[
\tilde H_{n-q-1}\qty{ \tau(V); \Extprod_\QQ^q }
\iso
\tilde H_{n-q-1} \qty{ \tau(V); \Extprod^q}
.\]
This yields a containment $P_T(V) \subseteq \bigoplus F$, and **Hadwiger invariants** $P_T(V) \to F$.
:::


\todo[inline]{todo}