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11:25 AM
@SaalHardali FFES = fully faithful and essentially surjective
@SaalHardali Yes, and I would say this is known, though the general topos version might need some slight variants of what's actually in the literature. (Note these are internal operads though, not enriched)
For a general topos, I think you want to localize w.r.t. something like the product of the Segal inclusions with a set of generating objects
 
 
2 hours later…
1:19 PM
@RuneHaugseng When you emphasis these are internal as opposed to enriched do you mean that this definition might work in the generality of a Topos but not necessarily for arbitrary symmetric monoidal categories?
 
Internal (multi)categories and enriched (multi)categories are different things in general
 
And Operad could mean both things?
 
Operad == symmetric multicategory (at least in Lurie-speak)
 
So aren't internal multi categories a special case of enriched multi categories?
If not why not?
 
Why should they be the same thing?
 
1:22 PM
Ah wait I have things confused
this is already wrong for categories
Sorry.
I'm glad you pointed this distinction out.
 
I think in a topos enriched categories are the same as internal category with constant space of objects, but I don't know a reference off the top of my head. No idea of what happens for operads
 
So when someone says "let O be an operad in spectra/chain complexes etc..." does this mean an enriched operad or internal operad?
 
Usually enriched. Also, usually they assume it has only one object
(since this is the original meaning of the word operad)
 
And an $\infty$-operad enriched in Spaces is the same as just a plain Operad corret?
The infinity should be on the second operad
 
Exactly
 
1:29 PM
So how does one model enriched operads in the Lurie framework? I only just touched this part in HA but from the beggining it seems like hes developing "plain" $\infty$-operads...
 
I don't think people have a well-developed theory of that yet. You can pretty much work out the theory for enriched operads with one object by mimicking the classical definition, and that's been enough so far
Although there are people in this room that know much more than me on this
 
So do I understand correctly that the content about Operads in HA is in the framework of symmetric $\infty$ multi-categories and it doesn't directly talk about stuff like "operad in spectra" for example?
 
Good to know. Thanks!
Could you bootstrap his theory to do stuff like "operad in spectra" or is there a real need for some outside model dependent input?
 
My feeling is that it should be at least as hard as doing enriched ∞-category theory
 
1:39 PM
And do I have it correctly that (implicitly at least) enriched $\infty$-category theory is not a walk in the park even when building on Lurie?
 
Is it even known, for example, that Spectra is a symmetric monoidal spectrally enriched infinity-category?
 
@SaalHardali It's not trivial, although it's definitely not as bad as it looks when you open the paper for the first time
@EspenNielsen I don't think that symmetric monoidal spectrally enriched ∞-categories have even been defined
 
@DenisNardin I mean, the ∞-category of spectrally enriched ∞-categories is symmetric monoidal by Gepner-Haugseng, so we can take $\mathbb{E}_{\infty}$-objects in there?
 
Hmm.. in that case it's probably true and known but you're probably better off asking Rune for a reference :)
 
Not an uncommon situation to be in for these matters. :P
 
2:16 PM
Slightly unrelated question. What's the stabilization of the category of (single coloured) $\infty$-Operads? Actually I don't even know what's the stabilization of $A_{\infty}$-Spaces so maybe this first?
(I need to add pointed everywhere)
 
user131753
Probably a naive question, dut does anyone know the rationale behind the name of "commutative diagrams"? To be precise, what is(are) the reason(s) for calling those diagrams "commutative" and in what sense?
 
@user170039 Because a commutative square is a diagrammatic depiction of an identity like fg = g'f', for example...
 
@user170039 I assume MacLane was thinking of natural transformations when designing his terminology. The naturality square can be thought of as a commutation relation for the natural transformation.
 
 
1 hour later…
3:33 PM
From Eilenberg & Steenrod, Foundations of Algebraic Topology, preface:
The reader will observe the presence of numerous diagrams in the text. Each diagram is a network or linear graph in which each vertex represents a group, and each oriented edge represents a homomorphism connecting thegroups at its two ends. //
A directed path in the network represents the homomorphism which is the composition of the homomorphisms assigned to its edges. Two paths connecting the same pair of vertices usually give the same homomorphism. This is called a commutativity relation. The combinatorially minded individual can regard it as a homology relation due to the presence of 2-dimensional cells adjoined to the graph.
(It's still not clear to me why commutative is the word chosen for this.)
 
3:47 PM
I am with @TomBachmann: It expresses the fact that the order of doing things does not matter. If $f:A\rightarrow A$ and $g:A\rightarrow A$ then there is one obvious diagram one might write down. If it is commutative then $fg=gf$. If one interprets a group as a category (The group as the single object: morphisms are left multiplication by group elements) then every diagram in the category is commutative iff the group is commutative.
 
But a commutative diagram can also express something like $fg=hk$, which is not in any sense a matter of "order of doing things does not matter". It's just an identity between two different expressions.
 
@DenisNardin @EspenNielsen I just found this: arxiv.org/abs/1707.08049 in case youre interested
It looks really neat and not that technical from a first glance
 
4:23 PM
@EspenNielsen I feel like this might be doable using Hinich's results, which give a really nice way of getting (say) spectra as a spectral infinity-category, but haven't thought about it.
 
 
2 hours later…
user131753
6:06 PM
@CharlesRezk I had some discussion with one of my friends regarding this. He said that maybe the idea that is intended to be conveyed is something like the following, namely that the operation done first along a "row" and then along a "column" is equal to the operation done first along a "column" and then along a "row" (for a commutative square, of course).
 
user131753
I have asked a question regarding this in MO here.
 

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