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00:01
@BenSteffan Gotcha here--thank you!
welcome :)
somebody publicly shared his personal notes for the algebraic geometry class this term, which is very kind
however, they are... bilingual, with spanish and english mixed haphazardly (?!)
sometimes I wish I could peek into the minds of other people to see what's going on in there
00:50
a major point of confusion for me has been that the "iterated Bar construction" does not appear to be the result of iterating the Bar construction
that would be too obvious wouldn't it
@Thorgott see page 842, right before the proposition
I kid you not
I just alt-tabbed into HA and that was the page I had already open
that said, I don't see where he actually makes the thing precise he claims to be making precise
yeah, I was looking for that as well
I guess it's Example 5.2.3.14?
See also the discussion on page 848
ah yes
this almost sorta kinda makes sense
"under good circumstances the iterated bar construction is an iterated bar construction" is a perfectly normal and sensible statement to have
I'm sad that we skipped everything about the $\mathbb{E}_k$-operads in our HA seminar now :(
much fun could've been had
01:07
the operads themselves are pretty reasonable, the stuff that's being done with them...
speaking of, do you have any intuition for why colim E_k = E_infty? the proof is surprisingly easy, but not particularly enlightening
I wager it's some sort of 'infinitary Eckman-Hilton principle', but that's also kind of just saying words
one way to characterize $\mathbb{E}_\infty$ is as the unique operad (let's say classical for the moment, and then with values in $\mathrm{Top}$) with each operation space contractible
(up to homotopy)
and then a computation shows that the $\mathbb{E}_k(r)$ "stabilize" to something contractible when $k \to \infty$, q.e.d.
right, that's the geometric perspective, but what's the algebraic one?
an E_k-algebra consists of k compatible associative algebra structures
why are infinitely many compatible associative algebra structures the same as one commutative associative algebra structure?
hmm
I think one should think of "$k$ compatible associative algebra structures" as being coherently commutative up to "level" $k$ (insert something about being $k$-truncated)
the translation happening via some form of Eckmann-Hilton
a grouplike $\mathbb{E}_1$-algebra is a loop space, so in particular a monoid in the homotopy category. a grouplike $\mathbb{E}_2$-algebra is a two-fold loop space, so by Eckman-Hilton in particular a commutative monoid in the homotopy category
in particular there is some square in the homotopy category witnessing this commutativity, with the $\mathbb{E}_2$-algebra structure providing the homotopy necessary to make it commute
an $\mathbb{E}_3$-algebra structure should then correspond to additional coherence data making certain cubes commute, ect.
01:27
I think the different algebra structures themselves also ought to agree up to some "level", but I can't quite make this precise
so e.g. in 1-categories, an E_2-algebra actually gives two copies of the same associative algebra structure + a braiding
what's, say, an E_2-algebra in 2-categories?
I presume the two associative algebra structures don't need to fully agree, but perhaps on objects and 1-morphisms?
not sure
In particular I don't know what model of $\mathbb{E}_2$ you'd use to make sense of that in the first place
01:45
theres multiple?
there's a bunch: cubes, disks, ... :)
but that's not what I mean. What I mean is that, to the best of my knowledge, to have an algebra over some operad $O$ in a category $C$ this category $C$ should be... enriched over the category $O$ takes values in (I think)?
but $\mathbb{E}_k$ is an operad in $\mathrm{Top}$
perhaps we just forget to $\mathrm{Set}$
I guess that's what people mean when they talk about $\mathbb{E}_k$-algebras in things like $\mathrm{Cat}$ (?)
02:01
@BenSteffan you don't need that
an O-algebra in C should just be a map of operads O -> C
(C being a symmetric monoidal category here)
you're implicitly forgetting to $\mathrm{Set}$, but then it's the same thing, yes :)
where does Set enter the picture?
maybe I'm misunderstanding what you mean by map of operads, but if you view $\mathbb{E}_k$ as some form of colored operad a la Lurie, then you will have spaces $\mathbb{E}_k(r)$, and if you package this up into a category of operators this will be naturally $\mathrm{Top}$-enriched.
Now either a map of operads (in the first description) consists of giving maps $\mathbb{E}_k(r) \to \mathrm{Mul}_C(...)$, but the domain is a space and the codomain a set, or it consists of giving a functor with certain properties from $\mathbb{E}_k^\otimes$ to $C^\otimes$, but the first category is $\mathrm{Top}$-enriched and the second isn't.
In either case you have to explain to me how to do that, and in both cases the obvious thing to do is to simply forget the topological structure on the domain :)
02:22
sorry, I meant to view it as an infty-operad
ah ok
then I don't know how to model 2-categories
(nevertheless this discussion should have the unfortunate upshot that an $\mathbb{E}_k$-algebra in $NC$ may not be the same thing as an $\mathbb{E}_k$-algebra in $C$ when $C$ is a 1-category, unless $k = 1, \infty$.)
let's say (2,1)-categories
i.e. (infty,1)-categories with 1-truncated mapping spaces
and I believe those can all be strictified actually
I will take your word for it
 
3 hours later…
05:12
there is a concept of "categroid" presumably a generalization of category.
 
2 hours later…
07:22
Cleo's mystery has been finally solved!
3
@PM2Ring interesting patterns
07:56
Can anyone provide references to why addition and multiplication over the surreal numbers are defined in the way they are? I am trying to get an intuitive feeling about them, but can’t get any foothold.
08:29
Would it be accurate to consider the total differential of a function as the directinal derivative in some arbitrary direction of an infinitesimal vector dx ?
 
5 hours later…
13:11
Is the unoriented S^2 bundle over S^1 diffeomorphic to unoriented S^1 bundle over S^2?
13:51
I wonder how people study non-oriented manifolds.
14:29
@hbghlyj what unoriented S^1-bundle over S^2?
15:14
Is it circular to use L'Hospital's rule when computing $$\lim_{x\to0}\frac{\log(x+1)}{x}?$$
@psie Probably.
It depends. What is your definition of $\log(x)$?
I would also note that L'H is not so much of a computational tool (though it is generally taught that way) as it is a useful tool for proving theorems.
There are a lot of applications of L'H in, for example, the study of differential equations.
@XanderHenderson the logarithm is defined to be the inverse of the exponential function $E$, which in turn is defined in terms of its power series. So the logarithm $L$ is the function that satisfies $$E(L(y))=y,\quad y>0.$$This is Rudin's definition. He then derives through the chain rule the identity $$L(y)=\int_1^y\frac{dx}{x}$$and comments that sometimes this is taken to be the starting point of the theory of the logarithm.
In that case, it seems like the power series expansion is much more useful than L'H.
15:51
Chat will be unavailable for ~15 minutes "shortly", while the database is migrated to the Cloud.
11
Q: Chat maintenance - Tuesday February 18th @ 14:45 UTC (Tuesday February 18th @ 9:45 AM ET)

Cesar MWe have planned maintenance that will impact our chat servers. The window is Tuesday February 18th @ 14:45 UTC (Tuesday February 18th @ 9:45 AM ET) (Aka: nowish). During the maintenance window, chat will be unusable and likely throw several errors. The sites will continue to function normally and...

@PM2Ring that was over an hour ago!
hmm I guess the wording in that post could be clearer
> Update: Things are taking a little longer than expected, and we're still working on chat.meta.stackexchange.com. Once this one is finished, the other chat servers should still only take up to 15 minutes.
@BenSteffan Clarity in communication isn't their strong suit... ;)
I like the "this is basically now btw."
"isn't this a little late to be 'planned'? is this some emergency update?"
"nope, totally planned :)"
We've known it was going to happen for a couple of months, but they forgot to give us a definite time.
16:24
Why do we bother with the $\delta$-functor approach to cohomology?
cause we don't want to explain model categories in a homological algebra class
@BenSteffan Because category theorists are jealous of analysts, and are stretching to find a way to shoehorn epsilons and deltas into their work. It's kind of sad, really. :P
@XanderHenderson lol
@Thorgott ...but you could still just pick a resolution and go to town? you need to do that anyways
16:43
yeah, but that is not properly motivated
and it does not tell you the universal property
but is the universal property something you ever use?
@BenSteffan Ugh... I don't want to go to town... it's, like, a 90 minute drive!
@BenSteffan yeah
even the universal property among non-exact delta-functors
see this old question of mine for example math.stackexchange.com/questions/4374334/…
17:14
But if I want to represent 20 elements with 3 symbols, the length of the words must be log_3 (20) ≈3, so I have 27 different strings right?
I don't understand how 33 strings can be ....
18:09
@Ben I love making notationally clear definitions
You missed a $= \mathrm{Alg}_{\mathrm{Comm}}(\mathbf{Cat}_{(\infty, \infty)})$ there :)
please no
or the variant writing Alg_/N(Fin)
18:48
Related to this question: math.stackexchange.com/questions/2816786/… I have a question: Let Σ_g be a genus g compact connected oriented surface, for a finite group G, is the set {g | Σ_g has an action by G that is trivial on H_2 but no element of G acts trivially} the same as the natural numbers that are $1\mod n_G$?
Is the number n_G the order of G?
Take $G = \mathbb{Z} / 2$ and test this conjecture for plausibility.
@Ben "We now partition the collection of new simplices  of E into eleven groups"...
?
I think that reference missed me
it's not a reference, just a ridiculous proof in HA
oh,
oh no :)
19:00
imagine needing 11 cases
group theorists laughing in the background
perhaps they'll find a better proof this century
it is our duty to clean up this mess so future generations don't have to suffer as much
something something
Joe
Joe
I don't know if I am going crazy but I don't understand this definition of generic point. What exactly does "unique" mean here?
what a bizarre definition
19:07
I... what?
what they should be saying is that it is the unique point specializing to every point
Joe
Joe
Apparently, this definition is meant to be a generalization of the definition where we say that $x$ is a generic point of $X$ is the closure of $\{x\}$ equals $X$.
9
Q: Explaining the motivation behind two different definitions of a generic point

RankeyaThis question is primarily regarding the definition of a generic point of a topological space that I came across in Qing Liu's Algebraic Geometry and Arithmetic Curves. First I will give the definition of a generic point that I am used to: Definition I am used to: Let $X$ be a topological space....

But it seems that it was awfully garbled.
garbled?
I suppose they want uniqueness (which will be automatic if $X$ is sober)
cause a general space can have multiple dense points, and I'm not sure if you would want to call those generic or not
but I am sure you don't want to look at non-sober spaces anyway
the use case will be schemes so
Joe
Joe
19:23
Hmmm, I think I am still confused about what this definition is saying. I understand that if $x,y\in X$ then saying that $x$ specializes to $y$ means $y\in\overline{\{x\}}$. But is there is a mistake in the next sentence? What does it mean for $x\in X$ to be a generic point?
it means that $x \in \overline{\{y\}}$ implies that $x = y$.
Clearly $x$ specializes to itself, and if $y$ is another point that specializes to $x$, then $y = x$
"$x$ is not in the closure of any point different from $x$"
Joe
Joe
Oh, I see what it is saying now, thanks. I think I was getting the quantifiers in my head mixed up.
19:45
I need a sanity check
the category of symmetric monoidal categories should have biproducts given by the cartesian product with the product symmetric monoidal structure, so it shouldn't be a cartesian closed category since products don't distribute over products, right?
(this phrasing is a bit deliberately vague, it won't work $1$-categorically, but I think you can interpret this $2$-categorically or $\infty$-categorically and the logic should be the same)
where would you need to distribute products over products?
Any y'all algebraic geometry people?
have a quick q but don't wanna pollute the chat if it's not the right venue
Ok, generally, if two multivariate polynomials share the same solution space, can I say that they share part of their factorization, and that the shared portion contains that solution space?
It feels right but I'd rather just cite it than prove it myself
@BenSteffan cartesian closed implies $C\times(D\sqcup E)=(C\times D)\sqcup(C\times E)$, but if $\sqcup=\times$, this doesn't work
@ChristopherD'Arcy what does 'sharing a part of their factorization' mean
20:01
So suppose I had (x-3)(y+4)(z+2) and (x-3)(y+9)(z-1), both share (x-3) as a factor. I guess more generally, do the real roots explicitly define the real solution space for multivariate polynomials? I guess yeah
now i'm wondering what "share the same solution space" means
Yeah of course, I don't even know why I asked
null space? IDK I'm not an algebra guy
you likely have an area specific term
well for any polynomial f in d variables in some field k you have {x in k^d: f(x) = 0} which people sometimes denote by things like "V(f)"
but if "solution space" means "zero set" in that sense, then the two examples you just gave don't have the same "solution space"
there's stuff like, if V(f) [as defined above] is a subset of V(g) and f is irreducible and k is nice enough, then f divides g. so there are contexts in which containment of zero sets says something about factorization
but i think you need to begin adding in hypotheses about f and g and k to get anything useful
almost anyone in this chat would be closer to AG than i am though, i am just spitting out some relevant definitions or definitional concepts, and memory fragments
you're making sense, leslie
@Thorgott on the other hand, it feels like the subcategory of the functor category on strong monoidal functors should inherit a pointwise symmetric monoidal structure...
I mean, it makes sense but I don't understand why you're asking for restatement? I could only imagine a real solution space to mean the set of real values a polynomial can take to be 0. Which is also what you guessed it to mean
20:10
@Thorgott ok, but does it have biproducts?
the thing is, if you put enough of the usual adjectives you find yourself talking about $\mathrm{Alg}_{\mathrm{Comm}}(\mathrm{Pr}^L)$ as your category of "symmetric monoidal (presentable) $\infty$-categories," and there the coproduct is the tensor product
at least if you don't mind the ${}^L$
christopher: a couple of things there, you're interested in zero sets of polynomials in real variables? by "share the same solution space" do you mean that their zero sets are the same, or just that they have nonempty intersection, or something else? or is this all up in the air pending further input and maybe it's part of the question
the L is a bit weird, why would you demand the tensor product $\mathcal{C}\times\mathcal{C}\rightarrow\mathcal{C}$ on a monoidal category to be a left adjoint?
it depends on what you want out of it; generally this is a fairly common assumption ime
but yeah, it'll change how the category behaves
I'm fine with assuming presentability, but demanding the tensor product to be a left adjoint is odd IMO
it's not even always the case for cartesian monoidal categories
christopher: as one example that may be of interest, with f(x,y,z) = xyz and g(x,y,z) = x^2 + y^2 - 1, the intersection of the real zero sets includes "the unit circle in the plane z = 0," i.e. the set {(x,y,0): x^2 + y^2 = 1}, but the polynomials f and g do not share any common factors. i took this example from sean haight's answer in math.stackexchange.com/questions/3256880/…
20:18
@ChristopherD'Arcy real solution spaces behave poorly. however, if two multivariate polynomials have the same complex solution space, then they have the same irreducible factors (though potentially with different multiplicities).
@Ben I want to look at, say, Alg(Cat_infty) (with the cartesian monoidal structure on Cat_infty), which is also what Lurie calls the infty-category of symmetric monoidal infty-categories
I genuinely don't know how I can possibly get more specific than "Do the real roots explicitly define the real solution space for multivariate polynomials"
now this inherits a pointwise symmetric monoidal structure from Cat_infty, but limits in Alg of something are always computed pointwise, so this is again a cartesian monoidal structure, but Prop. 3.2.4.7 in HA suggests this is also the coproduct
@ChristopherD'Arcy your original question mentioned factorization and this formulation does not? what is the difference between "the real roots" and "the real solution space"
@Thorgott ah, alright
I followed up at 15:01
Honestly it feels like you're taking the piss and I solved it myself 5 minutes ago anyway. Thanks for the time but I'm bouncing
20:23
Then yes, the fact that products don't distribute over products should imply that it cannot be closed
christopher it is pretty common for people to inquire about definitions, both here and on the main site. it isn't "taking the piss," it's, not everybody is using the same terminology and when a question is underspecified it can be more difficult to answer (or impossible to answer) than when it is not
it wasn't like anybody was sitting there knowing exactly what you meant and just playing games to get you to ask it the right way. the way you phrased it genuinely left a lot open to interpretation
There's also the question of the internal hom
my answer was not even acknowledged
@BenSteffan ok thanks
but I'm still missing as to why the internal Hom doesn't work, hmm
@Thorgott I just didn't see how the behavior of the space would apply to what I was trying to solve and didn't want to misuse the time
another example which is maybe too degenerate to be interesting is that in any number of variables the real zero sets of x^2 + 1 and x^2 + 2 (imagine as many other variables as you like just not appearing in these particular polynomials) are the same, although the polynomials do not share any common factors, whether you mean real or complex polynomials in your factors
20:35
@Thorgott something fishy is going on mathoverflow.net/questions/319580/…
that would suggest that it is closed
Maybe some context, I'm saying solution space because it's on the top of mind as I'm working with ODEs. I care about what happens to a high-order ODE when it is reduced to a first-order system. This produces a pair of polynomials that admit a non-trivial solution iff they have real solutions.

I particularly care about situations in which two polynomials have the same space of possible real solutions because this equates to a set of distinct differential equations that admit the same behaviors.
Hi all: I've got an announcement for interested parties. (I promise i really don't do this often.)

https://databaseofringtheory.wordpress.com/2025/02/15/help-choose-darts-200th-ring/
6
Hi
20:50
Hello
@BenSteffan they only claim this for the target V itself, not for any V-enriched category, though it's still od
they also use a Day convolution (which I never really understood) instead of the pointwise monoidal structure, hmm
ah, I see now that the pointwise monoidal structure on the category of strong monoidal functors does certainly not provide an internal Hom
ah, yeah
it does not define a strong monoidal evaluation functor cause $(F\otimes G)(c\otimes d)\neq F(c)\otimes G(d)$
you want Day convolution
I suppose that makes sense
but I think you can't define Day convolution on a functor category where the target is not just the base of enrichment (spaces in the infty-setting)
so I'd say there is no contradiction left
20:59
you absolutely can :)
there's nothing requiring the target to be the base of the enrichment
section 2.2.6 of HA is about just that
but perhaps the tensor-hom adjunction only holds when the target is the base of enrichment, hmm
hmm, this is troubling
a quick scan of the Section seems to suggest to me that Fun(C,D) with the Day convolution is not always a symmetric monoidal category, even if C,D are (though it is always an infty-operad)
see 2.2.6.17
@Thorgott that example says that it is, though?
if you enforce the smallness conditions
and the preservation of colimits condition (there it is again)
exactly, but you don't have preservation of colimits in general
so I would expect this to fail without that assumption
not that I have a counter-example, but if the condition weren't necessary, Lurie would tell us :)
I suppose this suggests that the infty-category of closed symmetric monoidal infty-categories is again closed?
and I believe this subcategory is stable under products, but probably not under colimits
but that should suit your situation well: this makes it very unlikely that $\mathrm{CAlg}(\mathrm{Cat}_\infty)$ has internal homs
@Thorgott not sure, but it should at least be closer to it than that of non-closed ones
you still need to add something like "presentable" to get a shot at it
perhaps, but I'm ok with everything being presentable
anyway, I think I'm probably not going insane
21:16
praise be

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