Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: deprecated: discrete geometry and entanglement

Topic: Update: 2021-08-01

Eric Forgy (Jan 09 2021 at 07:58):

I've been working on this for a while now (too long!) and have made some progress. I thought it might be a good idea to take stock and summarize what I have so far.

As mentioned above, I now have a pretty good understanding of the first order discrete calculus $(A,\Omega,d:A\to\Omega)$.

Our algebra $A$ is constructed from a set of vertices of a directed graph, i.e. $A := K^V,$ and has basis elements
${}$
$e^i(j) = \delta^i_j,\quad\text{for}\quad i,j\in V$
${}$
and unit element
${}$
$1 = \sum_{i\in V} e^i.$
${}$
Our $A$-bimodule $\Omega$ is constructed from a set of directed edges, i.e. $\Omega := K^E,$ and has basis elements
${}$
$e^\epsilon(\epsilon') = \delta^\epsilon_{\epsilon'},\quad\text{for}\quad \epsilon,\epsilon'\in E.$
${}$
However, now that I write this, I think there is some restriction on the allowed directed graphs that I've thought about before but hadn't formalized (until now?).

The $A$-bimodule $\Omega$ is actually a sub-bimodule of the universal one appearing in $\tilde d: A\to \tilde\Omega$, which is the kernel of the multiplication $m: A\otimes_K A\to A.$ If $\Omega$ is a sub-bimodule of $\tilde\Omega$ which is a sub-bimodule of $A\otimes_K A,$ then obviously

$\Omega$ is a sub-bimodule of $A\otimes_K A.$

If $\Omega = K^E$ and $\Omega$ is a sub-bimodule of $A\otimes_K A$, I believe this means that $E$ should be a subset of $V\times V.$ Moreover, since $\Omega$ is a sub-bimodule of $\tilde\Omega = \mathsf{ker}(m)$, then $E$ cannot contain elements $(v,v)\in V\times V$ in the diagonal either.

In other words, I think the directed graph
${}$
$V\leftarrow E\rightarrow V$
${}$
has to be an "irreflexive binary relation". As a directed graph, this means there can be at most one directed edge connecting any two vertices and there cannot be any loops, i.e. we are dealing with digraphs.

If this is correct, it simplifies things a lot. Intuitively, I suspected I wanted to be working with digraphs, but I couldn't find a justification until realizing the above argument. I still need to think about it, but it seems reasonable at the moment :thinking:

Eric Forgy (Jan 09 2021 at 08:50):

Another MAJOR step was when I discovered a way to extend the first order calculus to higher orders (I think it happened around Dec 21).

I'm not ready to talk about that in detail yet, but it involved constructing tensor products (over $A$) of $\Omega$ with itself, i.e.
${}$
$\Omega^{\otimes_A n}$
${}$
and then introducing maps
${}$
$\Omega^{\otimes_A n}\to A^{\otimes_K n}$
${}$
so that finally my higher order elements $\Omega^n$ are special kernels of these maps resulting in a cochain complex with
${}$
$d:\Omega^n\to\Omega^{n+1}\quad\text{and}\quad d^2 = 0$
${}$
with $d$ satisfying the graded product rule (on cochains) so I end up with an associative differential graded algebra (with unit).

I had never thought much about spans, but at some point (can find the exact point because it happened on this Zulip), I learned that "tensors over $A$" might secretly imply I am working with spans somehow. In the process, I also learned that directed graphs can be thought of as endomorphisms in $\mathsf{Span(Set)}$ with composition of a directed graph (as an endomorphism) with itself $n$ times gives a span
${}$
$E^{\times_V n}: V\to V,$
${}$
where $E^{\times_V n}$ consists of directed paths of length $n$.
${}$
I got excited by this. It really felt (and still feels) like I was / am on the right track.

Since then, I've been studying $\mathsf{Span(Set)}$ and my article contains a pretty decent "Introduction to $\mathsf{Span(Set)}$ for Engineers".

Then on Dec 24, Amar suggested looking at the functor:
${}$
$\mathbb{N}\to\mathsf{Span(Set)}.$
${}$
This functor provides an alternate way to think about directed graphs because, given a functor
${}$
$G:\mathbb{N}\to\mathsf{Span(Set)},$
${}$
then
${}$
$G(\bullet) = V\quad\text{and}\quad G(1) = E: V\to V,$
${}$
is a directed graph. Also $G(n) = G(1)^{\times_V n}:V\to V$, i.e. the directed graph (as an endomorphism) composed with itself $n$ times gives directed paths of length $n$.

This all feels good and I feel I'm on the right track, but I also feel my progress is too slow.

Eric Forgy (Jan 09 2021 at 08:56):

On Dec 27, Dan made a helpful comment pointing out that maybe the maps I mentioned above, i.e.
${}$
$\Omega^{\otimes_A n}\to A^{\otimes_K n}$
${}$
looked like they might involve a natural transformation
${}$
$\alpha: G\Rightarrow G',\quad G,G': \mathbb{N}\to\mathsf{Span(Set)}.$
${}$
Thinking about this led me to realize that given spans $X:*\to *$ and $Y:*\to *$, where $*$ is the terminal (one-element) set, if you compose them, you get a span
${}$
$X\times_* Y: *\to *,$
${}$
where $X\times_* Y$ is just the usual Cartesian product $X\times Y.$ That is kind of neat :blush:

The point being that given two functors $G,G': \mathbb{N}\to\mathsf{Span(Set)}$ with
${}$
$G(\bullet) = V,\quad G(1) = E:V\to V,\quad G(n) = E^{\times_V n}:V\to V$
${}$
and
${}$
$G'(\bullet) =$,\quad G'(1) = V:\to *,\quad G'(n) = V^{\times n}: *\to *,
${}$
then (borrowing from Dan's suggestion) a natural transformation $\alpha: G\Rightarrow G'$ involves maps
${}$
$E^{\times_V n}\to V^{\times n}.$
${}$
Now, if only I could move this to algebras and bimodules so that we get the similar looking
${}$
$(K^E)^{\otimes_{K^V} n}\to (K^V)^{\otimes_K n}$
${}$
or (equivalently)
${}$
$\Omega^{\otimes_A n}\to A^{\otimes_K n},$
${}$
which is what I want.

Eric Forgy (Jan 09 2021 at 09:15):

That brings me to where I am now. I'm currently trying to understand how to get algebras and bimodules from sets and spans.

John kindly pointed out that starting with a span of sets $V\leftarrow E\rightarrow V$ and hitting it with a functor $K^-: \mathsf{Set\to Alg}$ with $V\mapsto K^V$, then we don't get a span, we get a cospan $A\rightarrow \Omega\leftarrow A$ so we have a map
${}$
$\mathsf{Span(Set)\to Cospan(Alg)}.$
${}$
In hindsight, this seems obvious, but I'm in unfamiliar waters with this stuff and I tend to panic and forget things I already know in such situations.

I'm still working things out. For example, I think a span $K^V: K^E\to K^E$ should push out to a span $K^{E\times_V E}: K^E\to K^E$, but intuitively, I also think $K^{E\times_V E}$ should be equivalent to $K^E\otimes_{K^V} K^E$, but I'm struggling a bit to confirm / reject the idea.

Just a note on acknowledgements. Several people have helped me out (and one great thing about Zulip is that I'll be able to record everyone). Above, I highlighted some important comments from Amar and Dan, but I can't name names without pointing out that I would be nowhere without the help of John Baez. Thank you so much :pray:

Eric Forgy (Jan 09 2021 at 09:29):

Maybe one final note on motivation.

I started out already knowing that I have algebras and bimodules (in fact differential graded algebras) and knowing that these algebras and bimodules are related to directed graphs somehow, but I didn't know the precise relationship.

For example, I know I have an algebra $A$ with basis elements $e^i$ described above so that a general element $f\in A$ can be expressed as
${}$
$f = \sum_{i\in V} f(i) e^i.$
${}$
I also know I have an $A$-bimodule $\Omega$ with basis elements $e^{i,j}$ described above so that a general element $\alpha\in\Omega$ can be expressed as
${}$
$\alpha = \sum_{(i,j)\in E} \alpha(i,j) e^{i,j}.$
${}$
The left and right actions can be defined on basis elements by
${}$
$e^i\,e^{j,k} = \delta^{i,j}\,e^{j,k}\quad\text{and}\quad e^{i,j}\,e^k = \delta^{j,k}\,e^{i,j},$
${}$
where $\delta^{i,j}$ is the Kronecker delta functions. This is super remiscent of a path algebra. Now, I could just declare I have a product like that, but I'm trying to construct a setting where the actions are natural consequences, e.g. functorial, in nature. I don't feel satisfied just declaring rules for a multiplication without some better justification. That is why I was happy to stumble on spans. The composition of spans gives the exact structure I want, but it is a challenge for me to work out the details.

As usual, any help, questions, comments are more than welcome :pray:

Matteo Capucci (he/him) (Jan 10 2021 at 09:58):

Hi Eric! Nice to see your progress :) In reading I stumbled upon two questions which are maybe sorted out, actually, so I'll just leave them here for future reference:

1. What's the $A$-bimodule structure of $\Omega$ (aka $K^E$)? In other words, how does $A$ act on the left/right of $\Omega$? My guess is that on the left $e^i \cdot e^\epsilon(\epsilon') = \delta^{s(\epsilon)}_i \delta^{\epsilon}_{\epsilon'}$ and similarly on the right by replacing $s$ with $t$ (here I'm assuming there are two maps $E \to V$ which assign to each edge the source and target vertex, if $E$ is just $V \times V$ then those maps are the projections)
2. This was a question but I solved it: Why is $\Omega$ a sub-bimodule of $\tilde \Omega$? I see $\Omega$ can be embedded in $A \otimes_K A$ by sending $e^\epsilon$ to $s(\epsilon) \otimes t(\epsilon)$, but why should $s(\epsilon) \cdot t(\epsilon) = 0$? If we assume $s(\epsilon) \neq t(\epsilon)$ then $\delta^{s(\epsilon)}_j$ and $\delta^{t(\epsilon)}$ will never be $1$ at the same time, thus their product will always be zero. This also explains why the graph can't have edges $v \to v$. (my categorey sense tickle when one dispenses with identities though :laughing: )

Eric Forgy (Jan 10 2021 at 21:05):

(Note: I moved our subsequent discussion about this latest update to a new stream here . Thanks again :pray: )

Eric Forgy (Jan 22 2021 at 00:26):

Update: 2021-21-01