# Talk 0: ?? (Monday, June 13) :::{.remark} Numbers were historically defined as lengths, but an area is not quite a number! A modern formulation of a definition of area by Euclid: ::: :::{.definition title="Scissors congruence"} If $P, Q$ are polygons then $P$ and $Q$ are **scissors congruent** if $P = \uplus P_i, Q = \uplus Q_i$ where $A\uplus B$ means $\area(A \intersect B) = 0$. ::: :::{.theorem title="?"} $P$ is scissors congruent to $Q$ iff $\area(P) = \area(Q)$. ::: :::{.proof title="?"} Some reductions: - STS $P$ is SC to a 1 by $\area(P)$ rectangle. - STS this holds if $P$ is a triangle, by folding the triangle; ![](figures/2022-06-13_10-14-37.png) - STS any rectangle is SC to a $1\times \area$ rectangle. ::: :::{.exercise title="?"} Inna collects proofs, send one to her! ::: :::{.conjecture} Hilbert's third problem: is this a well-defined notion of volume in higher dimensions? ::: :::{.remark} \envlist - 1901, Dehn shows that a cube and a regular tetrahedron are not SC. There is a Dehn invariant: \[ D: P \to \sum_{\mathrm{edges } e} \len(e) \tensor \mathrm{angle}(e) \in \RR\tensor_\ZZ \RR/\ZZ ,\] and he shows $D(\mathrm{cube}) = 0$ but $D(\mathrm{tetrahedron}) \neq 0$. Note that the tensor product wasn't even defined until 1938! - 1965: Sedler shows that if $\vol(Q) = \vol(Q')$ and $D(Q) = D(Q')$ then $Q,Q'$ are SC. - 1968: Jessen defines the following. For $X$ a geometry and $G$ a group of isometries, define \[ P(X, G) = \ZZ[\text{polytopes in }X] \modulo \gens{ [P \uplus Q] = [P] + [Q], [g.P] = [P] } .\]. ::: :::{.theorem title="?"} If $X = E^n, S^n, H^n$ then $P,Q$ are SC iff $[P] = [Q]\in P(X, G)$. ::: :::{.proof title="?"} Take the upper half-plane model for $X = \HH^2$: ![](figures/2022-06-13_10-29-10.png) Note that red $= 1 +3$ and blue $=2 + 3$ and they are translates and thus equal, but $[2] = [3]$ while they are not scissors congruent (they move an ideal vertex at $\infty$). ::: :::{.problem title="?"} What is the structure of $P(X, G)$ for $X=E^n, S^n, H^n$ and $G\leq \Isom(X)$? ::: :::{.conjecture} Goncharov: there are higher Dehn invariant $D_i$, consider $H^{2n+1}$. Is the volume map \[ \intersect_i \ker D_i \mapsvia{\vol} \RR \] injective? Is the Borel regulator \[ (\gr^{\gamma}_{n+1} \K_{2n+1}(\CC) \tensor \eps(n+1))^- \mapsvia{\mathrm{BR}} \RR \] injective? Note that the regulator *is* injective for number fields. Here the negative is the $-1$ eigenspace of complex conjugation, and $\gr$ comes from the gamma filtration, which is related to the rank filtration and amounts to choosing a dimension. Moreover, is there a commutative diagram: \begin{tikzcd} {\intersect_i \ker D_i} && {(\gr^{\gamma}_{n+1} \K_{2n+1}(\CC) \tensor \eps(n+1))^- \to \RR} \\ \\ & \RR \arrow[hook, from=1-1, to=3-2] \arrow["{\text{Borel regulator}}", hook, from=1-3, to=3-2] \arrow["{\exists ?}"', dashed, from=1-1, to=1-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMiwwLCIoXFxncl57XFxnYW1tYX1fe24rMX0gXFxLX3sybisxfShcXENDKSBcXHRlbnNvciBcXGVwcyhuKzEpKV4tIFxcdG8gXFxSUiJdLFswLDAsIlxcaW50ZXJzZWN0X2kgXFxrZXIgRF9pIl0sWzEsMiwiXFxSUiJdLFsxLDIsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzAsMiwiXFx0ZXh0e0JvcmVsIHJlZ3VsYXRvcn0iLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFsxLDAsIlxcZXhpc3RzID8iLDIseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=) ::: # Algebraic K :::{.definition title="?"} Define \[ \K_0(R) = \ZZ\freeon{\rmod^{\fg, \proj}} \modulo \gens{ [B] = [A] + [C] \st A\injects B \surjects C } .\] Note that $A\cong B\implies [A] = [B]$. ::: :::{.remark} Note that for $F$ a field, $\K_0(F) = \ZZ$ since vector spaces of the same dimension are isomorphic, so all fields look the same to $\K_0$. But $\K_1(F) = F\units$, which can be very different for different fields! Why do $\K$ groups seem to appear everywhere? $\K_0$ encodes ways of breaking a problem into sub-problems, and a spectral sequence measures how higher $\K_i$ determine how these pieces can be reassembled. ::: :::{.remark} $\K_0$ has a 3-term relation $[B] = [A] + [C]$, which can't be encoded by just an arc -- so use a triangle, i.e. a loop. So in general, $\K = \Loop(?)$ to move the homotopy groups down: we want $\K_0(?) = \pi_1(?)$. Encoding this: ![](figures/2022-06-13_10-51-41.png) The Q construction: let $\cat{C} = \rmod^{\fg, \proj}$. Let $Q\cat{C}$ denote that category whose objects are $\mathrm{ob}(\cat C)$ and is generated by monics and op-epics, modulo the relation \begin{tikzcd} & {X\fiberprod{Z} Y} \\ X & {=} & Y \\ & Z \arrow[two heads, from=1-2, to=2-1] \arrow[two heads, from=2-3, to=3-2] \arrow[hook, from=1-2, to=2-3] \arrow[hook, from=2-1, to=3-2] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMCwxLCJYIl0sWzEsMCwiWFxcZmliZXJwcm9ke1p9Il0sWzIsMSwiWSJdLFsxLDIsIloiXSxbMSwxLCI9Il0sWzEsMCwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoiZXBpIn19fV0sWzIsMywiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoiZXBpIn19fV0sWzEsMiwiIiwxLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMCwzLCIiLDEseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dXQ==) Then define $\K(R) \da \Loop \B Q\cat C$ -- all other variants of $\K$ are extra data to take into account extra details (e.g. weak equivalences, non-existence of pullbacks). ::: :::{.remark} For SC, suppose there are 2 different ways for $P$ to be a subobject of $Q$, $P\injects_1 Q$ and $P\injects_2 Q$. We need a way of commuting them, so take: \begin{tikzcd} {P\intersect Q} && P \\ \\ Q && { Q \union P} \arrow["{{}_2}", hook, from=1-1, to=3-1] \arrow["{{}_2}", hook, from=1-3, to=3-3] \arrow[hook, from=1-1, to=1-3] \arrow[hook, from=3-1, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJQXFxpbnRlcnNlY3QgUSJdLFsyLDAsIlAiXSxbMCwyLCJRIl0sWzIsMiwiIFEgXFx1bmlvbiBQIl0sWzAsMiwie31fMiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzEsMywie31fMiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzAsMSwiIiwxLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMiwzLCIiLDEseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dXQ==) > Todo: missed some stuff at the end. :::