@skd @DexterChua @TylerLawson thank you all for weighing in! I'm not actually 100% sure what the map I want is. I think the Stolz-Teichner conjecture predicts the existence of a map TMF -> KO given by dimensional reduction (a 2d SQFT, compactified on a circle, gives you a 1d theory; 2d SQFTs hopefully describe TMF, and 1d SQFTs describe KO)

my understanding of Stolz-Teichner is not as strong as I'd like, so if the actual conjecture means using the map Tmf -> KO or something else instead then that's great. But I think the conjecture involves periodic TMF and periodic KO

If it's not clear what the map is (or if there is no nice map TMF -> KO) that is in itself interesting/useful. basically, whatever you have to say will be helpful!

and indeed, already has been

@TimCampion I think one of the big things that occurs to me is just being a bit more careful with out language about things being "clear" or "obvious." Maybe recognizing that a lot of people are interested in the sort of stuff we're doing without having much background.

I think we should normalize asking people "Are you familiar with ____?" I know I am often nervous about asking such questions because I don't want to "insult" someone, but the idea that any particular person will automatically know such and such facts is, in my opinion, a source of a lot of anxiety, especially among younger researchers.

I think we can also point out when we're trying to give answers or suggestions that are NOT precise, or that we CANNOT produce specific references for, and be careful of language like "I'm pretty sure it follows from..." or "It's immediate as a result of...".

@TimCampion I've been missing MSRI a lot too, recently

I know some people will disagree with this, but that language, in my experience, is rarely helpful. I think it's more useful to say things like "I think I could prove it using _____ but I don't have the time to think about it right now."

Because when someone says "It follows from ______" to me, my thought is usually "I can't see how, and I don't want to ask, and this person seems to think it's obvious."

On another note, I have a math question...

@JonathanBeardsley To my ears, "I'm pretty sure it follows from..." sounds more polite than "I think I could prove it using _____ but I don't have the time to think about it right now.". Perhaps this is a cultural difference.

@JonathanBeardsley @TimCampion I'm really glad this was brought up. As a "young researcher" myself and I found that I experience a lot of anxiety due to this tension between the massive amounts of mathematical content I feel like I must to know and the bounded amount of avaliable time I have for learning it.

I feel like there are at least two reasons for this which are not entirely my own fault. The first is not very specific to our field and it's the tendency of some mathematician to have less respect for complicated theorems once theyve internalized. This lack of respect can come in different forms. Sometimes it's just answering concisely "yes, that's true" without further elaboration to a question that requires some non-trivial arguments to resolve.

Sometimes it's saying "this easily follows from ----" or even just "it's easy" where "easy" is almost always in the eyes (and mind) of the beholder. We all are guilty of doing this sometimes simply because when we've internalized something it does often feel "easy" to us because we already have a very clear path in our mind which didn't exist before. I think it's important to keep this in mind, especially when talking to reaserchers younger than yourself.

The second reason is more specific to our field and it's the lack of textbooks covering the fundamentals of our field. It feels like no matter what subfield of HT/AT you're going to focus on there's a huge gap between what can be found in textbooks and what's expected of a young researcher in the field to know. This causes the already amorphous division between "fundamentals" and "extracurricular" to be even murkier.

As a young researcher it can sometimes feel like the choice is between on the one hand giving up on learning the fundamentals or on the other being in a constant race to fill all the percieved holes in your background (probably learning a lot more than "fundamentals" in the process). Both options can lead to anxiety.

@SaalHardali I think I am guilty of this particular fault. I try to watch my language but I was brought up (mathematically) to consider "easy" as synonym with "you can work it out in an hour or so" or "when you read the proof, it doesn't take a lot of effort to understand it", which might sound as patronizing

If F:C→D and G:D→E are left Quillen functors between model categories, then L(GF)=LG LF, right?

Yes, see Prop. 7.5.29 of Cisinski's Higher Categories and Higher Algebra

@DenisNardin If Q denotes a generic cofibrant replacement functor, then L(GF) = GFQ and LG LF = GQFQ, but F preserves cofibrant objects and G preserves weak equivalences between cofibrant objects, so the natural map GQFQ -> GFQ is a weak equivalence

@ArunDebray Not sure if this is helpful, but I believe if u is a unit, then the cofiber of u - j: TMF -> TMF is equivalent to KO

(j as in the j invariant)

Thank you! @AdrianClough @DexterChua

@ArunDebray dimensional reduction in the Stolz-Teichner program is a bit subtle because of the rigid Euclidean super-geometries that the "spacetime" manifolds are equipped with. To reduce from 1d to 0d, one can indeed compactify over a circle bundle since there is a quotient map from the 1d to the 0d super Poincare group (cf arXiv:0711.3862). But from 2d to 1d, such a map just doesn't exist. Dan Berwick-Evans has some ideas about this in arXiv:1311.6836.

@TimCampion Hinich shows that localization is compatible with cocartesian fibrations (in his paper called something like "Dwyer-Kan revisited"), in the sense that if you have a functor to relative categories you can either (A) localize to get a functor to infinity-categories or (B) form a 1-categorical fibration, then localize at the fiberwise equivalences; then the output of (B) is a cocartesian fibration corresponding to the functor from (A).

Here we'd need a generalization of that to locally cocartesian fibrations though...

@AlexanderCampbell yeah maybe "I think I could prove it..." is not the best example. I just sort of mean that there are shorthands that people use (e.g. Denis' "easy" that he mentioned above) that are not universal, and these result in a lot of exclusion. I think it'd be better if people were explicit about what they meant.

I feel like at some point someone explained to me why "truncation" in an ∞-topos should probably commute with "the category of modules," in the case that the algebra we're taking modules over is already, say, n-truncated. Does anyone have a sense of why this is true? I keep trying to prove it and am going in circles.

In other words, why is it the case (is it the case?) that LMod_A(τ_{n}X)=τ_{n}LMod_A(X), if A is an algebra whose underlying object of X is n-truncated.

Where I shouldn't use the same τ for both sides probably since the on the right hand side we're considering the full subcategory of objects which are truncated as objects of LMod_A(X), rather than those whose underlying objects of X are n-truncated.

I'm trying to leverage the equivalence X_{/BA} = LMod_{A}(X), obvi, and I can relatively easily show that τ_nLMod_A(X) is a full subcategory of LMod_A(τ_{n}X), but the other direction is tripping me up.

I thought I had a proof of this fact many months ago, but now I am quite certain that my past self was extremely confused.

@JonathanBeardsley I think I'm confused somehow. Like what if n=0 and I take A=*, a point. Then shouldn't LMod_\tau_0A(Spaces)=LMod_A(Spaces) = Spaces, but \tau_0Spaces = Sets?

Or I'm possibly just wrong... let me see...

Uhhh, wait, so on both sides you should be taking the truncation of an entire category.

Like, I think what I would want, in your case, is that the τ_0 LMod_A(Spaces) = LMod_A(τ_0Spaces)

In which case both are Sets?

oh, so like: \tau_n(LMod_{\tau_nA}(X)) vs. \tau_nLMod_A(X)? I think I was just missing a \tau_n somewhere

oh whoop

Well I was skipping applying the τ_n to A, since A was already n-truncated

**whoops, missed it again, I should just ready what you wrote!

**read

... and learn how to type

I probably should have just used the ChatJax thing instead of trying to be fancy with my custom keyboard Greek letters, haha.

okay, now I'm caught up I think, but don't necessarily have anything helpful to say yet :)

I somehow genuinely forgot that I could just type LaTeX in this room.

I THINK I have a proof... I think it's one of those things where like, it's almost so "trivially obvious" that somehow I can't, like, say it precisely.

(this seems to happen to me a lot with ∞-categories, because I'm just so bad at saying things with, say, universal properties rather than, say, set theoretic ideas)

Also my proof, in an attempt to say things using universal properties, now has a bunch of goofy diagrams, so I don't think I can post it here,.

hmm. Okay so is it right that it'd be useful to find just some general way to compute "\tau_nX_{/y}" in general, and then, as you said, just S/U over?

I think that's right. And I think that's sort of what I'm doing.

like are you hoping there's just some nice fact like "$\tau_n(X_{/y})\simeq (\tau_n X)_{/\tau_n y}$"?

cuz that seems pretty reasonable... except maybe one of my n's should be \pm 1

Right so I actually sort of misspoke. A should be n-1-truncated, so that BA is n-truncated

In which case τ_n(Y_{/BA}) = (τ_n(Y))/BA

@JonathanBeardsley Maybe I'm misunderstanding you, but is what you're looking the statement that an A-module is n-truncated (in the infinity-category of A-modules) if and only if the underlying object is n-truncated in the underlying infinity-category?

@PiotrPstrągowski yeah i believe so.

Does that seem false to you?

I think it's true. If M is an A-module, and suppose first it is n-truncated as such. Then, for any $F \in X$, where $X$ is the underlying $\infty$-category, $map_{A}(A \otimes F, M) \simeq map(F, M)$ is an $n$-truncated space, so we see that the underlying object of $M$ is n-truncated.

Ah I see, okay, so you use the fact that mapping into M, in A-modules, always gives you an n-truncated space, and then specifically use the free A-modules on X-objects.

So I have a proof of that direction, although it's not nearly so elegant, haha. Is the other direction implied by your argument?

from the S/U point of view, maybe we could use HTT.5.5.6.14?

I THINK that that sort of gets you the same direction that Piotr just pointed out.

but it's an if and only if, right?

I didn't actually realize that that was in HTT, but I was basically making the LES argument when I said that $\tau_n (X_{/BA})\simeq (\tau_n X)_{/BA}$

oh sorry, I'm behind again! I thought that's the statement we wanted to prove. but you're saying unstraightening after doing that statement doesn't work out or is unclear? I guess I didn't think about that yet

Conversely, the same adjunction tells you that if $M$ is an $A$-module whose underlying object is n-truncated, then $map_{A}(N, M)$ is $n$-truncated whenever $N$ is free. But any module $N$ is a colimit of free modules (for example, you can take the bar resolution), and $n$-truncated spaces are closed under limits.

well at this point I like Piotr's direct argument better anyway :)

Haha, yeah Piotr's argument is much nicer.

At least, nicer than the godforsaken mess of notes and diagrams on my desk right now.

Yeah thinking about it as I'm walking to my car. Goddamn that's nice Piotr.

I feel like there's a sort of "motto" one should always remember which is something like "Try it with free objects first," but that I rarely remember.

@JonathanBeardsley Happy to be of help :)

@PiotrPstrągowski is it clear that some part of the above breaks when we try to do the same for Algebras? I am quite sure the statement is false for algebras.

@RuneHaugseng Oh wow, that's great! I might comment on the question with links to some of this discussion.

Is it the part about every algebra being a colimit of free Algebras?

@TimCampion Sure - but just to be clear I have no idea about how to prove the hypothetical generalization to locally cocartesian fibrations!

@JonathanBeardsley That's true for algebras over any monad, though (without claiming that applies here, as I haven't followed this discussion)

@JonathanBeardsley I think it's as Rune says, and just to add a reference, the fact that free objects generating everything in a monadic $\infty$-category is 4.7.3.14 in HA. (It's the bar resolution, again.)

@PiotrPstrągowski hm okay... So less clear why it may not hold for Algebras. For some reason my phone wants to capitalize Algebras

Right, yeah monadic bar resolution for any algebra. Hm...

The example I have in mind is just algebras in groupoids vs the 2-truncation of Algebras in Spaces

What does algebras here mean? Monoids / E_{1}-objects?

Yeah

I think the former includes things which are loops on 2-types, but the latter does not...

What do you mean by 2-truncation of algebras in spaces?

The subcategory of 1-truncated objects?

Yeah

One way to be explicit about it is to pass to pointed connected spaces, truncate to a 1-type, and then cobar back to a monoid

Oh and I'm assuming group like here

So since bar/cobar is an equivalence, it truncates to an equivalence on subcategories. This means that everything in \tau_{\leq 1} of monoids in Spaces must be of the form loops on a 1-type, therefore a 0-type

Whereas the monoids in \tau_{\leq 1}Spaces, i.e. groupoids, include loops on 2-types

I think ??

I think loops $\Omega X$ on a $2$-truncated pointed space $X$ are in fact $1$-truncated as a monoid in spaces.

You're right that this would imply in particular that $\Omega X$ is $1$-truncated in group-like monoids in spaces, in which case you can deloop and this becomes equivalent to saying that $X$ itself is a $1$-truncated object in pointed, connected spaces.

This sounds counterintuitive, I agree, since as a space $X$ is only assumed to be $2$-truncated. However, any pointed connected space $Y$ has a $1$-skeleton $Y_{1}$ which is a wedge of circles, and then $map_{*}(Y_{1}, X)$ is a product of $\Omega X$ which is actually $1$-truncated.

I'm not certain I follow. I guess my assertion is that \tau_{\leq n-1}Mon(Spaces) = Mon(\tau_{n-2}Spaces) but you're saying this is not the case?

Then to pass from the mapping space with source $Y_{1}$ to $Y$ you will need to consider the effect of attaching higher-dimensional cells, but this will not decrease the truncatedness of this mapping space.

Alright, maybe this is going to be hard to think about on my phone, haha.

@JonathanBeardsley I don't really know much about bar/cobar equivalence, so it's hard to say. But I think instead we have that n-truncated monoids in spaces are the same as monoids in n-truncated spaces.

Hm ok. I mean this fits with your earlier proof for the module case, but yeah feels very counterintuitive... Maybe I've been thinking about bar/cobar incorrectly

I'm beginning to think I may have at various points in my thinking mixed up the "truncate as an algebra" and "truncate the underlying object" functors. But this will have to wait until tomorrow to think more about it seems.