of course, to avoid the eilenberg swindle i think we're only supposed to take K-theory of subcategories of suitably bounded objects, anyways. this leads me to wonder if the function {t-structures on C} --bounded objects--> {subcategories of C} is injective. (i suppose there's an evident guess for a morphism of t-structures which would make this into a functor of posets, though i think my question is almost equivalently whether there are any interesting nonidentity morphisms.)

@ArunDebray haha that's pretty good

@AaronMazel-Gee Neat paper! but this sentence was triggering: "complexes will be cohomologically graded"

can we not use mental illness language like internet trolls do? that would be great

why so defensive? @PeterNelson

internet trolls are made, not born

troll elsewhere

Following up, the researchers reviewed 16 million posts on CNN.com, noting which posts were flagged by moderators, and applying computer text analysis and human review of samples to confirm that these qualified as trolling.

Now, that's some serious data analysis.

@PeterNelson I agree with making this request.

And support avoiding ableist language in general.

Anyone know if there are any results saying something about how many E_n-ring structures an Eilenberg-MacLane spectrum HA can have?

the endomorphism operad of an eilenberg-maclane spectrum HR in spectra is discrete, and i think this implies that the space of E_n-structures on HR is contractible

Hm... what's the argument precisely? I'm pretty sure I believe connected, but contractible?

i found a reference for this in the E_oo-case, it's at lemma 5.5 in arxiv.org/pdf/1411.7988.pdf

@skd Yeah, I knew about that case.

In particular would be sort of interested to know if, say, the E_2-structure on HF_p coming from the Thom spectrum construction is unique, and how its related to the E_\infty structure you referenced.

@JonathanBeardsley The argument referenced by @skd works for arbitrary $n$, as what they show is that the endomorphism operad of $HA$ in spectra is equivalent to the endomorphism operad of $A$ in abelian groups. That is an $E_{n}$-algebra structure on $HA$ is the same as a ring structure on $A$, commutative if $n \geq 2$.

@PiotrPstrągowski oh, i see! wonderful!

should have gone and read the proof

@_@

Anyone know of a model structure on Hausdorff spaces? I.e. NOT on compactly generated Hausdorff spaces?

I mean, I get that there are probably relatively trivial ways to produce such a model structure, I'm just wondering if there's like... a "canonical" one? This is related to a undergraduate thesis project one of my students is doing.

@JonathanBeardsley I don't know about the Hausdorff condition, but this: ncatlab.org/nlab/show/model+structure+on+topological+spaces seems to think that there is a model structure on all topological spaces (albeit possibly not a monoidal one)

@DenisNardin Right, I wonder... is that a questionable statement, about there being a model structure on all topological spaces?

Hmm, I guess it depends on what you take as equivalences

Apparently there's even one where the equivalences are the homotopy equivalences. I have no idea what it presents

It seems at least reasonable that you can construct one by transferring the model structure along the Sing/| - | adjunction, although I haven't thought it through

Oh, this is in Hovey

I wonder if there's a model structure where the w.e. are the shape equivalences, and if so whether it presents all of Pro(Space)

Yeah so okay... this naturally induces a model structure on Haus, but this doesn't seem to be written down anywhere.

@PeterNelson @JonathanBeardsley sorry, are you guys referring to my movie clip?

no

@DenisNardin super-basic question: do homotopy equivalences of topological spaces induce shape equivalences of $\infty$-topoi?

@DenisNardin if you're transferring the standard model structure on sSet, you'll just get the usual weak homotopy equivalences though right?

I'm pretty sure yes. The shape of X is the functor sending a Kan complex Y to the Kan complex Map(X,|Y|) whose n-simplices are continous maps X×|Δ^n|→|Y|. A homotopy equivalence X→X' should induce a homotopy equivalence Map(X',|Y|)→Map(X,|Y|) more or less by a simple combinatorial argument

(and the answer to the second question is yes, I was a bit rambling above)

Uh, I guess by shape I mean the strong shape (so the object of Pro(Spaces), not of the localization at π_*-equivalences)

cool, i didn't know that definition of shape -- it's way simpler than the one passing through $\infty$-topoi!

It's the same, I'm just using the fact that $p_*p^*Y=Map(X,|Y|)$ where $p:X\to *$ is the obvious map and $X$ is a paracompact Hausdorff space

gotcha, i see that now. but i wouldn'tve noticed to unpack it this way

@PeterNelson ok. just for context, what were you referring to then? has it since been deleted?

the comment right above mine

@AaronMazel-Gee I'm pretty sure he's referring to the usage of the word "triggering". I had to reread the exchange several times to get it, though (it's not a usage of the word I encounter frequently).

...the word "triggering" is triggering?

aaron.

i'm certainly willing to listen if you feel that this is in poor taste (perhaps a discussion by email would be better than in a public forum?), but at present i'm having a hard time seeing how a reference to the phenomenon of trigger warnings is itself offensive

no, it's the extreme minimization and corruption of what the term actually means and what content warnings are actually for, and also the usage of the term by people in bad faith in ways that are extremely similar to how you just used it

okay, i wasn't aware that the topic of trigger warnings was sensitive. (certainly i know that trigger warnings themselves are meant to indicate sensitive content.)

yeah, me too

I urge you to learn more about why they exist.

i apologize for my intrusion

yeah at some point "triggered" became a casual slang term for being really annoyed by something.

do not ever use it that way

I don't use it that way. I do think it's obnoxious to phrase that statement as a command

:-D

@skull ah ok, I see

welcome @G.Bergeron

@skull I guess what I had in mind can be remotely seen as algebraic geometry :)

askaway

In the context of quantum algebras, is defining a braiding relation on the tensor product always equivalent to deforming the coproduct?

@G.Bergeron this sounds cool, and plausible. could you be more specific about what data you're starting with, and what further sorts of data you're hoping are equivalent? first of all, by "quantum algebra" do you mean "hopf algebra" (or maybe "bialgebra")?

i at least know the slogan that representations of a hopf algebra form a braided-monoidal (a.k.a. E_2-monoidal) category

@AaronMazel-Gee This is coming from a mathematical physics context. By quantum algebra, I do mean possibly non-co-commutative Hopf algebras.

A q-deformed version of the Schwinger construction for Lie algebras exists and relies on q-oscillators algebra see for example iopscience.iop.org/article/10.1088/0305-4470/22/21/001/meta for such a construction for $U_q(su(n))$

haha you're gonna lose me fast if you try to explain the context. here's a super basic question, just to orient myself: given a not-necessarily-cocommutative hopf algebra, do its representations carry a tensor product?

@AaronMazel-Gee However, in cases I have seen, they seem to not define a Hopf algebra structure on this q-oscillator algebra..

i think this requires the ambient category (of vector spaces or whatever) to have a braiding, but that's all

@AaronMazel-Gee I would think so

okay, great. and so, what do you mean by a deformation of the coproduct here?

@AaronMazel-Gee Not a symmetric coproduct, for instance, not $\Delta(X) = X\otimes 1 + 1 \otimes X$.

sorry, i'm confused. we're already starting out with a hopf algebra that has a non-cocommutative coproduct. is this the structure that you're deforming?

@AaronMazel-Gee deformed with respect to the non-deformed co-commutative case.

oh, okay. here's another basic question: how do you describe the braiding on the category of representations?

@AaronMazel-Gee As relations of the form $e_i \otimes e_j = q_{ij} e_j \otimes e_i$.

haha okay. where are those $q_{ij}$'s coming from, do they arise as structure constants of the hopf algebra?

[sorry, in case it's not clear i don't know if i know the answer to your question -- i'm just curious!]

@AaronMazel-Gee In a way: for $k_i$ a generator of the Cartan sub-algebra, one has $k_i e_j k_i^{-1} = q_{ij} e_j$

My question is somewhat related to this question mathoverflow.net/questions/20683/… where the coproduct appears symmetric and the deformation is now in the braiding data.

@PeterNelson well anyways, i'm sorry if i offended you. i do think i know why they exist: it's so that people can avoid content that would cause a negative reaction for them, especially based on their past experiences. or maybe i'm misinformed? in any case, let me know if there is something further i should be looking out for in learning about them, and feel free to share any reading material you think would be particularly enlightening. again, i propose that we move this conversation to email.

More precisely, I'm looking at inducing the entire Hopf algebra structure of $U_q(su(3))$ through a Schwinger-like construction by defining a Hopf algebra structure on the q-oscillator algebra. All so that I can derive a recurrence relation for the q-polynomials I derived as the matrix elements of the symmetric $U_q(su(3))$ representations.

@AaronMazel-Gee At least I know the fit seems ok for those questions! Thank you for your attention.