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Stream: deprecated: discrete geometry and entanglement

Topic: Invitation (Discussion on Jan 8 update)

Eric Forgy (Jan 10 2021 at 10:44):

(Note: This stream was split from the original Invitation stream.)

Thanks for your comment Matteo :blush:

Matteo Capucci said:

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)

Yep. You got it :+1:

The challenge for me at the moment is to arrive at that functorially rather than just declare it. It really is just like concatenating paths.

It is probably fine to just take a functor

$G:\mathbb{N}\to\mathsf{Span(Set)}$

and define a category algebra $K[G]$ in an obvious way. I'm not sure though. Still thinking about it.

Matteo Capucci said:

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: )

Ah, but we are not dispensing with identities :blush:

By analogy (and making the details precise is work-in-progress), take my favorite functor:

$G:\mathbb{N}\to\mathsf{Span(Set)}$

There are three key pieces and the rest are derived from them:
${}$

1. $G(\bullet) = V$
2. $G(0) = V: V\to V$ (the identity span also written $V\leftarrow V\rightarrow V$)
3. $G(1) = E: V\to V$ (the directed graph as an "endospan" also written $V\leftarrow E\rightarrow V$ )
   ${}$

There is an important distinction between 1. (a set) and 2. (a trivial span, i.e. a path of length 0).

By analogy, we have three pieces:
${}$

1. An algebra $A$
2. An $A$-bimodule $A$ (I believe this can be expressed as an identity cospan)
3. An $A$-bimodule $\Omega$ (I believe this can be expressed as an "endo-cospan")
   ${}$

The $A$-bimodule $A$ is the identity :blush:

Details need to be worked out, but that is morally the way it should work :blush:

Another way to think about why we want $\tilde\Omega = \mathsf{ker}(m)$ comes from the other stream:

• Universal problem for derivations

We can show that the derivation $\tilde d: A\to\tilde\Omega$ is universal and that any other derivation $d:A\to\Omega$ can be written as $d = \phi\circ\tilde d$ for some bimodule morphism $\phi: \tilde\Omega\to\Omega.$ It is all pretty cool :blush:

When you construct a derivation from a graph, $\tilde\Omega$ comes from a complete graph (mod some hand waving about loops) and $\phi$ essentially amounts to removing some edges from the complete graph, i.e. setting them equal to 0.

The hand wavy argument goes roughly like, you can keep the loops in your graph, but the derivation $d$ makes them disappear so whether you keep them and they disapear or you never start with them doesn't really matter. I go back and forth on this one. Sometimes I like to keep them in because then the universal derivation $\tilde d:A\to\tilde\Omega$ can be written as a (graded) commutator:
${}$
$\begin{aligned}\tilde d f &= 1\otimes_K f - f\otimes_K 1 \\ &= (1\otimes 1)f - f(1\otimes 1) \\ &= [\tilde G, f],\end{aligned}$
${}$
where
${}$
$\begin{aligned}\tilde G &= 1\otimes_K 1 \\ &= \sum_{(i,j)\in V\times V} e^i\otimes_K e^j \\ &= \sum_{\epsilon\in\tilde E} e^\epsilon\end{aligned}$
${}$
and
${}$
$e^\epsilon = e^{s(\epsilon)}\otimes_K e^{t(\epsilon)}$
${}$
and $\tilde E = V\times V$, i.e. the edge set for a complete graph.

Eric Forgy (Jan 10 2021 at 11:00):

If you work with digraphs (no loops, no parallel edges), then you can't consider
${}$
$\tilde G = 1\otimes_K 1$
${}$
to be the sum of all edges because that sum includes the diagonals / loops $e^i\otimes_K e^i.$

Eric Forgy (Jan 10 2021 at 11:09):

Actually, the reasoning above kind of tilts me toward having identity loops because they would correspond to the identity span $V\leftarrow V\rightarrow V$ and likely also to the identity $A$-bimodule $A$ (which probably turns out to be an indentity cospan).

Eric Forgy (Jan 10 2021 at 11:13):

I still don't want parallel edges though because then we can't say
${}$
$e^\epsilon = e^{s(\epsilon)}\otimes_K e^{t(\epsilon)}.$
${}$
Since $e^\epsilon$ is a basis for $\Omega$ and $\Omega$ is a sub-bimodule of $A\otimes_K A,$ I think this is ok :+1:

Matteo Capucci (he/him) (Jan 10 2021 at 11:17):

Eric Forgy said:

The challenge for me at the moment is to arrive at that functorially rather than just declare it. It really is just like concatenating paths.

It is probably fine to just take a functor

$G:\mathbb{N}\to\mathsf{Span(Set)}$

and define a category algebra $K[G]$ in an obvious way. I'm not sure though. Still thinking about it.

Oh I see :thinking: though since $G$ gives you paths of all lengths I would expect $K[G]$ to be a complex of algebras, if not a dg-algebra already.

The whole setup of $G$ makes me think of the nerve construction... could this be its version for graphs? In the nerve construction, you associate a simplicial set (= a presheaf on the simplex category) to a small category (= a graph + composition). You do so by considering paths of composable edges of length 0, 1, 2, and so on. Since a simplicial set also has face (they go $n \to n+1$)/degeneracies (they go $n+1 \to n$) maps, one has also to specify those. It's combinatorially fiddly, but basically they amount to either add identities somewhere in the path (these are face maps, since they lengthen the path), forget the first/last morphism (these are degeneracies $d_0$ and $d_n$) or compose the two consecutive middle maps (these are degeneracies between $d_1$ and $d_{n-1}$).
Don't take my word for it, check the nlab where everything is worked out in more detail (so I won't repeat everything here).
The moral is, to each category one can associate a simplicial set which track the way you can move around it.
The cool thing is, there's a canonical way to turn simplicial things into complexes, the Dold-Kan correspondence. It's easiest when you have simplicial abelian groups (or any good abelian category, I guess $K$-algebras are good too) instead of simplicial sets, i.e. if the presheaf on the simplex category actually lands in $Ab$ (or, again, your preferred abelian cat). Then by waving a magic wand you automagically get a complex out of this.
I think it's not unreasonable to think that your construction amounts to take a graph $G$, build its free category $F(G)$, build its nerve $N(F(G))$, then take free $K$-algebras to get a simplicial $K$-algebra $K^{N(F(G))}$ and run the Dold-Kan construction.

Matteo Capucci (he/him) (Jan 10 2021 at 11:19):

The good thing about this approach is that it resuses a lot of well-known tools in categorical algebras, so (1) you don't have to reinvent the wheel, (2) (some) people will be ready to read your paper and (3) you get some giant shoulders to stand on.

Eric Forgy (Jan 10 2021 at 11:19):

Yes. I believe what you just outlined is what John has been telling me all along. I call it the treasure map :blush:

Matteo Capucci (he/him) (Jan 10 2021 at 11:20):

Oh, wow

Eric Forgy (Jan 10 2021 at 11:21):

I believe this is what Sullivan-Wilson did and it is slightly different than what Urs and I did. For sure, the two are related (we both end up with Navier-Stokes as a tautology), but the construction is slightly different. If you do as you suggest, you end up with a DGA that is not associative on cochains.

Matteo Capucci (he/him) (Jan 10 2021 at 11:23):

I don't know enough to understand where the culprit is

Eric Forgy (Jan 10 2021 at 11:24):

I believe there is a "No Go" theorem along the lines that a DGA derived from a countable set cannot be both associative and graded commutative. Sullivan-Wilson come up with something that is nonassociative yet graded commutative. Urs and I came up with something that is associative yet noncommutative. I feel like giving up associativity is more of a sin than giving up commutativity and, as we discussed elsewhere, the way the noncommutativity comes in is pretty cool and leads to things like quantum mechanics and stochastic calculus.

Matteo Capucci (he/him) (Jan 10 2021 at 11:25):

Though it seems to me you're revolving around this construction. I would be surprised if having the algebra structure so close to the 'compositional' structure of the graph doesn't give you something closely related to the 'treasure map' construction.

Eric Forgy (Jan 10 2021 at 11:25):

Sure. It will be very similar :+1:

Matteo Capucci (he/him) (Jan 10 2021 at 11:26):

Eric Forgy said:

I believe there is a "No Go" theorem along the lines that a DGA derived from a countable set cannot be both associative and graded commutative. Sullivan-Wilson come up with something that is nonassociative yet graded commutative. Urs and I came up with something that is associative yet noncommutative. I feel like giving up associativity is more of a sin than giving up commutativity and, as we discussed elsewhere, the way the noncommutativity comes in is pretty cool and leads to things like quantum mechanics and stochastic calculus.

Oh I see!

Matteo Capucci (he/him) (Jan 10 2021 at 11:26):

So you're probably after the non-abelian version of Dold-Kan

Eric Forgy (Jan 10 2021 at 11:26):

But instead of simplices, we have diamonds :blush: :large_blue_diamond: So it is a shape thing.

Eric Forgy (Jan 10 2021 at 11:27):

Diamonds are constructed from directed $n$-paths and John says directed $n$-paths are secretly simplices, so both are related to simplices and my favorite functor obviously gives something with the same data as the nerve.

Eric Forgy (Jan 10 2021 at 11:29):

Matteo Capucci said:

So you're probably after the non-abelian version of Dold-Kan

I already have the answer. I am only stuck on writing it up. I can write 20 pages of exposition with the primary motivation being "It works", but that is unsatisfying. I want to express it cleanly / functorially.

Eric Forgy (Jan 10 2021 at 11:33):

Matteo Capucci said:

Eric Forgy said:

I believe there is a "No Go" theorem along the lines that a DGA derived from a countable set cannot be both associative and graded commutative. Sullivan-Wilson come up with something that is nonassociative yet graded commutative. Urs and I came up with something that is associative yet noncommutative. I feel like giving up associativity is more of a sin than giving up commutativity and, as we discussed elsewhere, the way the noncommutativity comes in is pretty cool and leads to things like quantum mechanics and stochastic calculus.

Oh I see!

Plus, you can always start with a DGA that is associative and noncommutative and construct a DGA that is nonassociative and commutative by (anti)symmetrizing, e.g. defining a product (similar to wedge product of forms)
${}$
$a\wedge b := ab + (-1)^{deg(a)*deg(b)} ba$
${}$
but this will no longer be associative (except in the continuum limit).
${}$
So I do suspect what I am doing is closely related to the treasure map, but maybe slightly cooler :sunglasses:

Matteo Capucci (he/him) (Jan 10 2021 at 11:38):

I see. Maybe diamonds are really different? Are the different from cubes? Cubes and simplices are 'equivalent' for homological algebra, so that wouldn't warrant a complete change of foundations.
Eric Forgy said:

I already have the answer. I am only stuck on writing it up. I can write 20 pages of exposition with the primary motivation being "It works", but that is unsatisfying. I want to express it cleanly / functorially.

What I want to say is, expressing it functorially it's very likely going to reproduce well-known homological algebra. So I don't see the point of starting from scratch again :thinking: at the very least I would try to see where the beaten path brings me. (I apologize in advance if this comes across as rude, that's not intended! :blush: )
[Continuing to beat my drum, I came across this: semi-abelian Dold-Kan correspondence (associative algebras form a semi-abelian category, according to the nLab)]

Eric Forgy (Jan 10 2021 at 11:38):

Hint: The face map is not a bimodule morphism unless it doesn't act on the leading or trailing vertex.

Eric Forgy (Jan 10 2021 at 11:40):

Matteo Capucci said:

I see. Maybe diamonds are really different? Are the different from cubes? Cubes and simplices are 'equivalent' for homological algebra, so that wouldn't warrant a complete change of foundations.

Diamonds are different than cubes. That is my new result. In my paper with Urs, we only consider cubical diamonds, but I found (by accident) the class of diamonds is larger than cubes and are more closely related to "directed spheres".

Eric Forgy (Jan 10 2021 at 11:45):

You can combine $n$-paths from one pole of an $m$-sphere to the other in a special way. That is a diamond. As I mentioned before, a diamond is a special intersection of the future and past of two points in spacetime.

So I don't know if it is a "non-abelian version of Dold-Kan". More like a "directed version of Dold-Kan" maybe.

Eric Forgy (Jan 10 2021 at 11:56):

Matteo Capucci said:

What I want to say is, expressing it functorially it's very likely going to reproduce well-known homological algebra. So I don't see the point of starting from scratch again :thinking: at the very least I would try to see where the beaten path brings me. (I apologize in advance if this comes across as rude, that's not intended! :blush: )

You are right and I can (and try to) learn from the beaten path, but there is more to life than homological algebra. What if I want to design a medical imaging device through simulation? What if I want to simulate the earth's climate? You will need more than homological algebra to do that (I think).

Eric Forgy (Jan 10 2021 at 11:59):

My endgame is to build numerical algorithms for applied scientific computation.

Eric Forgy (Jan 10 2021 at 12:03):

Another nice thing about using spans is that the 1-category $\mathsf{Span}(C)$ (as opposed to bicategory) is a $\dagger$-category and will facilitate the definition of an inner product. A DGA
${}$
$\begin{aligned}\Omega = \bigoplus_{r\ge 0} \Omega^r\end{aligned}$
${}$
with an adjoint operator / inner product is sufficient to do a lot of physics.

Eric Forgy (Jan 10 2021 at 12:06):

Matteo Capucci said:

[Continuing to beat my drum, I came across this: semi-abelian Dold-Kan correspondence (associative algebras form a semi-abelian category, according to the nLab)]

Cool. Thanks. Will have a look :+1: :blush:

Eric Forgy (Jan 10 2021 at 12:12):

Matteo Capucci said:

Eric Forgy said:

The challenge for me at the moment is to arrive at that functorially rather than just declare it. It really is just like concatenating paths.

It is probably fine to just take a functor

$G:\mathbb{N}\to\mathsf{Span(Set)}$

and define a category algebra $K[G]$ in an obvious way. I'm not sure though. Still thinking about it.

Oh I see :thinking: though since $G$ gives you paths of all lengths I would expect $K[G]$ to be a complex of algebras, if not a dg-algebra already.

Right. $K[G]$ would be a graded algebra, but not yet differential (unless you follow the beaten path which leads to nonassociativity).

There is one more step that is similar in spirit to the trodden path, but gives you an associative DGA.

Eric Forgy (Jan 10 2021 at 12:24):

Btw, when I say "nonassociative", I should stress that I mean "nonassociative on cochains". When you pass to cohomology classes, I think it might be associative, but physics happens on cochains.

Eric Forgy (Jan 10 2021 at 21:14):

I'm starting my day a little late here (was up until 5am working on this stuff :zzz: ) and looking at Dold-Kan correspondence. I remember reading about this stuff back in grad school (1998-2002), but I wasn't mathematically mature enough to absorb it. Today, I'm older and wiser and maybe a little more mature, but that is countered by the fact I am so rusty :sweat_smile:

Matteo Capucci (he/him) (Jan 10 2021 at 21:15):

It's pretty advanced stuff, I'm myself far from acquainted with it, I just know enough to grasp the idea and be comfortable with it

Eric Forgy (Jan 10 2021 at 21:15):

I think maybe what I'm doing constitutes a Kan complex, but not sure. Still waking up :coffee: