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Stream: deprecated: discrete geometry and entanglement

Topic: Update: 2021-21-01

Eric Forgy (Jan 21 2021 at 23:04):

Ok! Another update...

With John's help via

• Relationship between spans and bimodules

and working through the puzzles and their generalizations in

• Bimodules vs Spans: General solution

${}$
I now know an important relationship between spans and bimodules. Namely, if $C$ is a cartesian monoidal category with pullbacks (so it is finitely complete), then
${}$
$\mathbf{Span}(C,\times)\cong \mathbf{Bim}(C^\mathsf{op},\times).$
${}$
This is huge progress :rocket:

Eric Forgy (Jan 21 2021 at 23:19):

However, in the process of working that :point_of_information: out, I was convinced (thanks again John!) to not try treating these as 1-categories, but accept the fact they are both bicategories. Accepting they are bicategories forced me to rethink my favorite functor:
${}$
$G: \mathbb{N}\to\mathbf{Span(Set)}.$
${}$
An obvious candidate replacement is
${}$
$\mathbf{Span}(\mathbb{N},\times),$
${}$
but it didn't take me long to figure out that a better replacement is actually
${}$
$\mathbf{Cospan}(\mathbb{N},+)\cong \mathbf{Span}(\mathbb{N}^\mathsf{op},+)\cong \mathbf{Bim}(\mathbb{N},+)$
${}$
so my new favorite functor (and I love this one even more :heart:) is
${}$
$G: \mathbf{Span}(\mathbb{N}^\mathsf{op},+)\to \mathbf{Span(Set,\times)}$
${}$
that I introduced here. It is worth repeating the highlight:
${}$

• $G(\bullet) = V\quad$ (set of vertices)
• $G(\bullet\overset{0}\rightarrow \bullet \overset{0}\leftarrow \bullet) = V: V\to V\quad$ (identity span)
• $G(\bullet\overset{0}\rightarrow\bullet \overset{1}\leftarrow \bullet) = E: V\to V\quad$ (directed graph)
• $G(\bullet\overset{1}\rightarrow\bullet \overset{0}\leftarrow \bullet) = E^\dagger: V\to V\quad$ (same directed graph with reversed arrows, i.e. the adjoint graph)
• $G(\bullet\overset{1}\rightarrow\bullet \overset{1}\leftarrow \bullet) = E^\dagger\circ E = E\circ E^\dagger: V\to V\quad$ (directed graph composed with its adjoint)
• $G(\bullet\overset{m}\rightarrow\bullet \overset{n}\leftarrow \bullet) = {E^\dagger}^{\times_V m}\circ E^{\times_V n} = E^{\times_V n}\circ {E^\dagger}^{\times_V m}: V\to V\quad$ (adjoint paths of length $m$ composed with paths of length $n$)

${}$
This is very cool because the adjoint graphs, which are important for discrete differential geometry appear so naturally :heart:

Eric Forgy (Jan 21 2021 at 23:26):

I then introduced a new shorthand notation
${}$
$E^{m,n} := {E^\dagger}^{\times_V m}\circ E^{\times_V n} = E^{\times_V n}\circ {E^\dagger}^{\times_V m}$
${}$
so that
${}$
$E^{0,0} = V,$
${}$
$E^{0,1} = E,$
${}$
and
${}$
$E^{1,0} = E^\dagger$
${}$
together with a new set of sets
${}$
$\begin{aligned} E^{(r)} = \coprod_{r=n-m} E^{m,n} = \coprod_{r=n-m} {E^\dagger}^{\times_V m}\circ E^{\times_V (m-r)}\end{aligned}$
${}$
which, I think (still need to verify, but its looking good) gets mapped to bimodules
${}$
$\begin{aligned} \Omega^{(r)} = \bigoplus_{r=n-m} \Omega^{m,n} = \bigoplus_{r=n-m} {\Omega^\dagger}^{\times_V m}\circ \Omega^{\times_V (m-r)}.\end{aligned}$
${}$
Assuming that pans out (I'm sure it will), we have a space of graded bimodules
${}$
$\begin{aligned}\Omega = \bigoplus_{r\in\mathbb{Z}} \Omega^{(r)}.\end{aligned}$

Eric Forgy (Jan 21 2021 at 23:42):

In particular,
${}$
$\Omega^{(0)} = A\oplus\Omega^{1,1}\oplus\Omega^{2,2}\oplus\Omega^{3,3}\oplus\cdots,$
${}$
where all $\Omega^{m,n}$ are $A$-bimodules ($A=\Omega^{0,0}$).

This helps explain our metric operator (3.5 Metrics on cubic graphs) and more :+1:

Eric Forgy (Oct 24 2023 at 16:20):

Looking at this almost 3 years later, I don't even remember what I was doing here :sweat_smile: :older_man:

All of these topics in this stream all have the same goal: Re-express more elegantly and possibly generalize this:

• Discrete Differential Geometry on Causal Graphs, Eric Forgy & Urs Schreiber

All of the stuff in the Bourbaki topic provides the background into the universal differential envelope, which can be used to turn any directed graph into a formal first order differential algebra.

All the stuff about Spans and Bimodules is my attempt to generalize this to higher order. In particular, I noticed that a slight modification of the definition of the Moore complex provide a nice way to express what we meant by "diamonds", but I have so far failed to enunciate that clearly :sweat_smile:

Eric Forgy (Oct 24 2023 at 16:29):

The, admittedly weak, connection to entanglement came in when I learned about the fascinating relationship between entanglement and spacetime topology. For example see this:

• Quantum Reality: Space, Time, and Entanglement
  Creating a discrete form of differential geometry involves starting with a complete graph and removing edges.
  Creating spacetime from entanglement involves removing connections.
  Could those two things be related?
  

Eric Forgy (Oct 24 2023 at 16:45):

Why am I talking about this here? :sweat_smile:
I suspect there is some deep connection between these two things. To unravel any such connection, if it exists, I am pretty sure the right explanation / intuition will require (higher) categorical thinking. I wish I was smarter and could figure this out myself, but it wouldn't take long exploring my rambling to see I am not equipped mathematically to carry this out, but I'm trying :sweat_smile: