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Stream: event: ACT20

Topic: July 7: David Myers' talk

Paolo Perrone (Jul 02 2020 at 02:17):

Hello all! This is the thread of discussion for the talk of David Jaz Myers, "Double Categories of Open Dynamical Systems".
Date and time: Tuesday July 7, 20:00 UTC.
Zoom meeting: https://mit.zoom.us/j/7055345747
YouTube live stream: https://www.youtube.com/watch?v=5Rs_7ZCH1Wc&list=PLCOXjXDLt3pZDHGYOIqtg1m1lLOURjl1Q

Paolo Perrone (Jul 07 2020 at 19:45):

15 minutes!

Maria (Jul 07 2020 at 20:41):

Thanks! Are the slides for this talk available?

James Fairbanks (Jul 07 2020 at 20:42):

So you if we are just talking about composing systems by matrix multiply on their steady states, you need to "solve the whole system" you mean that you actually build the relation between parameters and trajectories. and then you can compose those things by matrix multiply

James Fairbanks (Jul 07 2020 at 20:42):

is that a correct understanding?

Bruno Gavranović (Jul 07 2020 at 20:42):

So to check my understanding, I have a basic question. the functor $T$ can be the tangent bundle functor (i.e. $\mathcal{C} \to \mathcal{C}$), but also something more general, right? Is $\mathcal{C} \to \mathcal{C}$ the most general form of this functor? EDIT: Ah, so after seeing the slides,$T$ actually has the type $T : \mathcal{C} \to \int Bun$
The other question I asked was, is it accurate to say that $TS : Bun(S)$, for some $S : \mathcal{C}$?

David Jaz (Jul 07 2020 at 20:45):

Hi! I've attached the slides. dyn.pdf

There is also a slightly expanded version on my website.

Maria (Jul 07 2020 at 21:30):

Thank you so much! :)

David Jaz (Jul 07 2020 at 21:41):

James Fairbanks said:

So you if we are just talking about composing systems by matrix multiply on their steady states, you need to "solve the whole system" you mean that you actually build the relation between parameters and trajectories. and then you can compose those things by matrix multiply

So if you are using wiring diagrams, then it is computationally trivial to build the matrix; you can effectively "read it off" from the diagram. Then the theorem says if you have a vector which contains the number of steady states (indexed by choice of parameter and output) for each internal system, you multiply it by this matrix to get the vector which contains the number of steady states for the combined system.

Mike Shulman (Jul 07 2020 at 22:01):

What are the companions and conjoints in the double category of interfaces?

Mike Shulman (Jul 07 2020 at 22:02):

Also I wonder whether your double functors are related to the Yoneda embedding of http://www.tac.mta.ca/tac/volumes/25/17/25-17abs.html?

David Jaz (Jul 07 2020 at 22:24):

Great question! The companions are the vertical morphisms and the conjoints are the cartesian morphisms! (or maybe I have that backward, the covariant ones are the vertical ones and the contravariant ones are the cartesian ones)

David Jaz (Jul 07 2020 at 22:34):

Mike Shulman said:

Also I wonder whether your double functors are related to the Yoneda embedding of http://www.tac.mta.ca/tac/volumes/25/17/25-17abs.html?

So, I'm not quite sure. If you take a double Grothendeick construction of the representable indexed double functors, the codomain is now a little bit weirder than the double category of spans. I think it is Pare's representable on the base, however, though I'd have to be a bit more careful in checking to be sure. That is, in the base it is representable in the sense of Pare, but on the whole it is something more like "indexed representable".

Sophie Libkind (Jul 07 2020 at 23:15):

@David Jaz , can you walk me through the bottom half of this diagram on an interface ${I'} \choose {O'}$. Along the top I get a category whose objects are dynamical systems with interface ${I'} \choose {O'}$. Then the transformation maps such a dynamical system to a set (call it $X$) with a map (call it $p$) into the set of bundle maps from ${I} \choose {S}$ to ${I'} \choose {O'}$. Is that correct?

Sophie Libkind (Jul 07 2020 at 23:15):

If it is correct, I'm missing the intuition about what the set $X$ and the fibers of $p$ represent.

Paolo Perrone (Jul 08 2020 at 01:52):

Here's the video!
https://www.youtube.com/watch?v=f9fjf9lo2_M&list=PLCOXjXDLt3pYot9VNdLlZqGajHyZUywdI

David Jaz (Jul 08 2020 at 11:02):

Sophie Libkind said:

David Jaz , can you walk me through the bottom half of this diagram on an interface ${I'} \choose {O'}$. Along the top I get a category whose objects are dynamical systems with interface ${I'} \choose {O'}$. Then the transformation maps such a dynamical system to a set (call it $X$) with a map (call it $p$) into the set of bundle maps from ${I} \choose {S}$ to ${I'} \choose {O'}$. Is that correct?

That's correct, yes. The specific map into the set of bundle maps from ${I \choose S}$ to ${I' \choose O'}$ is the set of horizontal maps of dynamical systems projecting down their bottom component. For any such map ${I \choose S} \to {I' \choose O'}$, with our dynamical system in mind, we get a set from the profunctor that the indexed double category of dynamical systems assigns to that map; we take the union of all such sets, projecting down.

Sophie Libkind (Jul 08 2020 at 15:14):

Oh right! I see. A system ${v \choose r}: {{TS} \choose {S}} \leftrightarrows {{I'}\choose {O'}}$ maps to the set of commuting squares like the one below(taken from your talk) where the left "system" is $u \choose {id}$. The projection maps this square on to the bottom bundle map.

This square represents a trajectory through the system $v \choose r$ using the clock $u \choose {id}$. Thanks!

David Jaz (Jul 08 2020 at 18:22):

Sophie Libkind said:

Oh right! I see. A system ${v \choose r}: {{TS} \choose {S}} \leftrightarrows {{I'}\choose {O'}}$ maps to the set of commuting squares like the one below(taken from your talk) where the left "system" is $u \choose {id}$. The projection maps this square on to the bottom bundle map.

This square represents a trajectory through the system $v \choose r$ using the clock $u \choose {id}$. Thanks!

Yes exactly!