Mathematics - 2025-01-15

Homesick Iguana

00:29

0

Q: singularity types of a geometric flow with specific Dirichlet boundary conditions?

Let $p_1,p_2$ be antipodal points on $S^{n}$. Assume $p_1,p_2$ are fixed points under any smooth geometric flow which strictly decreases the enclosed volume $\mathrm{vol}(S^{n})$, for all time. Assume any flow is unnormalized. Does any flow always result in exactly two possibilities? $a)$ result...

Any hints?

does a sphere always become a line segment if you fix 2 antipodal points througouth the smooth deformation

leslie townes

01:47

iguana it may help to be clearer about what you mean by 'become a line segment', particularly if this is something you are getting out of physics it would not surprise me if people aren't as clear about that as one might like

if you just do the mean curvature flow of something like a circle in R^2 it stays a circle the entire (finite) duration of the flow. it never becomes anything other than that (and i guess guess at that last instance it's a single point, not a segment)

same result for an ellipse although the eccentricity is not preserved, it becomes "more like a circle" (in a sense that poeple usually make sense of via some kind of rescaling) as it evolves

there are probably flows by something other than the usual curvature that preserve or even increase the eccentricity and maybe get closer to an idea of something becoming more like a line segment but idk

this is something where its very easy for a physicist to conjure up all kinds of words about subsets of R^n converging to other subsets of R^n without paying really any attention to what any of it "actually" means

if something stays a simple closed curve the entire duration of its flow, except maybe at the last minute, a lot of people would struggle to say how that's "becoming" a line segment or even converging to one in some easily definable sense

it might help to make clearer what you mean by 'geometric flow' with fixed points, a lot of the most commonly studied examples i can think of wouldn't have fixed points without some kind of rescaling

BAYMAX

02:10

Hi chat,
Any thoughts on this would be much helpful
https://math.stackexchange.com/questions/5023366/computing-the-most-common-overlapping-region-and-the-number-of-vertices-formed

Oceans Bleed

05:56

@Jakobian Well it just hit me that I have been ignoring set-theoretic issues in a place where for all I know there become genuine issues. I was going to write that the formation of orthogonal subcategories is an order-reversing map, but it occurred to me that the underlying ZFC set of my category is a $\mathfrak{U}$-class (in my convention), and I've been writing

$\mathcal{S}^\perp$ is the full subcategory spanned by the objects $\{x\in \mathcal{T}\mid \operatorname{Hom}_{\mathcal{T}}(S,x)\cong 0,\text{ for all }S\in\mathcal{S}\}$, where I would like to say that $(-)^\perp$ is an order-reversing map. But order reversing on what? I'd be talking about endomorphisms on the power-set of the object class, which I suppose in the case where I have a Von-Neumann universe for an inaccessible cardinal, this is still a $\mathcal{U}$-class

I will note that my lack of background makes the last sentence there essentially empty words. Basically a phrase, so it could very well be wrong

Jakobian

I don't know much category theory, and especially when I'm sleep deprived

Oceans Bleed

If there was no time pressure in life, I'd set aside a year straight to just purely sort all this out. (Set theory, formal logic, model theory, etc)

Well, the category theoretic part of that is not important. You could just pretend that $\operatorname{Hom}_{\mathcal{T}}(S,x)$ is referring to an inner product of $S$ and $x$, if you wish, that's where the name is motivated from. The relevant part is that one is using the axiom of specification on a $\mathcal{U}$-class, and I was worried this was not allowed

Sorry I changed from $\mathfrak{U}$ to $\mathcal{U}$, please identify them in your mind

@Jakobian I suppose my concern is what the axiom schema of separation actually says - I think I'm concerned about what a ZFC set even means. If my von Neumann universe is a model for ZFC, it sits inside larger models for ZFC, but what even is a ZFC set? Does a ZFC set only make sense in the context of picking a model?

So you say I need not pick the larger universe $\mathfrak{V}$, but I don't actually know what it means to deploy the axiom of specification without picking a universe first

Jakobian

A 'set' is really just an object that ZFC talks about

Oceans Bleed

In short, my possibly incorrect way of handling this is the following:

1) Pick massive cardinal,
2) Form universe
3) Use set theory in the universe - where by this I mean that the axioms of ZFC are allowed to be used on $\mathfrak{U}$-sets, but nothing else

I suppose you'll say that
4) Set-theoretic operations, like forming the powerset and such, on $\mathfrak{U}$-sets are still perfectly fine sets, and we don't care where they live. So you can use ZFC axiom schema with respect to them

@Jakobian This doesn't really mean anything to me unfortunately. It kind of moves my concern to the word 'object' in that sentence. Not trying to be annoying/pedantic

Jakobian

06:14

For a better explanation you should read about what ZFC is

Oceans Bleed

Is this a correct reading of what you're saying: ZFC prescribes the existence of two different sets (empty, and infinite [if I'm remembering correctly from long ago]), and in all other cases tells you that you can build new sets from old ones via prescribed operations

Jakobian

ZFC doesn't talk about anything else other than sets

everywhere you see a quantifier, all those things, we just call them sets, that's just the name

Oceans Bleed

But the prescribed operations are relevant to what is my concern. One talks about them being impermissible for classes, but presumably by 'classes' they mean something very different from $\mathfrak{U}$-classes (which are just subsets of the VN universe)

Okay, how about this: what is a class? in an absolute sense [Rather than relative to a universe]

Jakobian

what does absolute sense mean

are you asking what a class is in ZFC

Oceans Bleed

I'm not 100% sure to be honest. People always write words about 'class' issues, and as far as I can tell, there are actually no $\mathfrak{U}$-class issues at all, is what we're saying above(?), so they are referring to something else.

Does saying "what is a class in ZFC" even make sense? In ZFC as you say, everything is a set, and presumably a class is not a set

Jakobian

06:19

a class, as an object, doesn't exist in ZFC, we only informally talk about classes, those are really just some logical sentences $\varphi$ and we say $x$ belongs to the class $\{x : \varphi(x)\}$ when $\varphi(x)$ holds

so classes are defined in an informal way in ZFC

Oceans Bleed

So you can still use the axiom of specification, and obtain a class, that is not a set - you just cannot then make use of set-theoretic operations on it?

Wait you can't though, sorry

Jakobian

I am saying what I just said

Oceans Bleed

Axiom of specification picks out subsets, so I don't know what that is. is set-builder notation something named, when not picking out subsets

@Jakobian You are indeed

I'm just trying to understand, not critiquing

Jakobian

axiom schema of specificaton (because it's not just a single axiom), says that given some logical sentence $\varphi$ and a set $A$, we can form a subset of elements of $A$ which satisfy $\varphi$

so it can be understood that, if you intersect a set and a class, you get a set

Oceans Bleed

Whereas your 'class construction' didn't construct it as a subset, unless $\varphi(-)$ had as part of its definition a condition that $x\in A$

Jakobian

06:23

construct what as a subset

Oceans Bleed

This class: $\{x : \varphi(x)\}$. It was not constructed as a set, but one can simply take every set satisfying $\varphi(-)$, apparently

Jakobian

this class is not an object of ZFC, but we can still write shorthand sentences using it

$y\in \{x : \varphi(x)\}$ just means the sentence $\varphi(y)$

that's how it should be interpreted in ZFC, every occurrence you just replace it with that

it's more about the language than concrete objects in our theory

Oceans Bleed

That's what I was trying to clarify above with the
"So you can still use the axiom [schema] of specification, and obtain a class, that is not a set - you just cannot then make use of set-theoretic operations on it?"
I meant you can construct these classes, but can now not make use of the power-set of them, etc

[I thought your set-builder notation, to construct a class, was axiom schema of specification, but realised I was mistaken because it does not pick out a subset of anything]

Alright

Jakobian

well you can do anything that you can with logical sentences

so you can take their (finite) intersections, unions

I guess you can't take powersets, I don't know what would that mean

$\{x : \varphi(x)\}$ is just notation, there is no actual objects like this in your theory, it's just a shorthand which given some logical sentence involving sets, we will always substitute for it

Oceans Bleed

Can you clarify why I would not need to pick $\mathfrak{V}$ as before? It seems the only way to be 'safe' is to make sure you make clear what sets you are taking operations on, at any time. So sometimes you would be forced to move up a universe, and you must actually track that?

Jakobian

06:31

you don't need to

and the reason is because it works without it. What else can I say

there is nothing forcing you to take some $\mathfrak{V}$

I think you would benefit from reading a chapter or two out of Jech's Set theory

Oceans Bleed

I swear I remember reading some warnings over the last 5 years of the form: Note, you cannot do this operation twice, for set theoretic reasons. Free cocompleting a free cocompletion comes to mind. Also some of the collections of objects I am interested in are not sets in my universe

@Jakobian I do have that book sitting somewhere here. I suppose I'll have to read it in a couple weeks, sadly can't at the moment

I think there was some set-theoretic error in showing some specific algebraic stack was a stack, which had nothing to do with algebraic geometry, so I am a little worried I might make errors by ignoring foundations

Jakobian

@OceansBleed well sure, depending on the problem you might have to, but there you were just asking if axiom schema of specification holds for your $V_\kappa$

Oceans Bleed

Oh right, sorry. Gotcha

in that case I don't have to

I guess I'm generating XY-problems by accident. In my mind I suppose the meta-question is "Can I literally ignore set-theoretic issues again, by working with my two-universe method, or are there still potential issues", where by that I mean are there any operations I care about that would still break. I suppose I'm being dumb though: the short version is that by definition, my $\mathfrak{U}$-classes are sets, so I can do set operations, and

my fear about doing set operations on classes is unrelated, since $\mathfrak{U}$-classes have nothing to do with classes, except the name

And if I was to make precise what my set-theoretic operations are on these $\mathfrak{U}$-classes, I could work out their cardinalities, and that would suffice to know they live in some other larger universe (if they do), but that wouldn't matter at all, so it doesn't actually need to be mentioned

I think that is the conclusion

Actually the deeper conclusion is that I have broken my potential understanding of ZFC by implementing this universe setup

Since I'm realising this is completely pointless phrasing, just now

You do at least need to do all this crap when you define the categories you care about. There will be a different category of abelian groups for whichever universe you choose to work in, right? Bigger universe means bigger groups. Or is even that mistaken? How far back does my error go

Jech gives the universal class on page 3, I see

Okay, I see, as you said, a formula can be used to pick out a class. which can only contain sets. Some classes are not sets. I wonder if he'll tell me how you can tell if the class is not a set

Okay page 5 gives me intersection of class and a set is a set, as you claimed, that makes sense

Page 85 Lemma 10.2 gives me that $V_\kappa$ is a model of ZFC.

Is your $\varphi$ above only defined after picking a model, as part of an interpretation as on page 80?

Oceans Bleed

07:12

Seems like I need to start foundations again from scratch. I wouldn't even normally need to ask anyone any questions, except I can't even tell where I've corrupted my understanding

Homesick Iguana

14:21

baseball is just too easy to use a dynamic martingale strategy. Bet on in play balls bet low on first pitch ---> increase bet on successive pitches not necessarily a strict double of money.

walk away with 1k every game (they last about 2 hours)

thank me later!

not even kidding

problem is the fun won't last long cause betting apps will start restricting your bets if you are too succesfull - so maybe don't try it

example: bryce harper comes up to the plate. he swings more often on first pitch than any other guy on the team. This trends even higher as game progresses (assuming game is close or within reach and not a blowout). So bet 25 on a in play ball right before first pitch. say he swings and its strike. place 50 on next pitch etc.

Jakobian

Remember kids, gambling is fun

Homesick Iguana

its not gambling if you know how to win consistently :)

for those folks it just called easy money

but sadly we are a rare breed nowadays

one potato two potato

14:36

what would be a usage of Grassmann bundle in differential geometry? I mean "naturally" arises in some sense. The only example I know is during the proof of a given continuous section is a smooth subbundle.

Homesick Iguana

15:02

@onepotatotwopotato The Gauss map can be viewed as a section of the Grassmann bundle $\mathrm{Gr_2}(T \Bbb R^3)$ where each point on the surface is mapped to the corresponding tangent plane (a point in $\mathrm{Gr_2}$).

this is how it classically arose which is pretty natural

psie

Let $p,q$ be conjugate pairs, $\nu$ a $\sigma$-finite measure. I'm reading about the theorem that if $\Phi: L^p(E,\mathcal A,\nu)\to\mathbb R$ is a continuous linear form, there exists a unique $g\in L^q(E,\mathcal A,\nu)$ such that for every $f\in L^p(E,\mathcal A,\nu)$, $\Phi(f)=\int fg\,\mathrm{d}\nu$.

The author has already derived that the statement is true for $\nu$ finite and $f\in L^p(\nu)$ and $g$ integrable, i.e. that we have $\Phi(f)=\int fg\,\mathrm{d}\nu$ where $g\in L^1(\nu)$. Now he wants to show that if $p=1$, then $g\in L^\infty(E,\mathcal A,\nu)$.

What he does is this; $$\left|\int_A g\,\mathrm{d}\nu\right|=|\Phi(\mathbf1_A)|\leq \|\Phi\| \|1_A\|_1=\|\Phi\|\nu(A),$$and says this easily implies $|g|\leq\|\Phi\|$ by applying the previous display to $A=\{g>\|\Phi\|+\epsilon\}$ or $A=\{g<-\|\Phi\|-\epsilon\}$. I don't understand how we reach $|g|\leq\|\Phi\|$. I'm just getting weird inequalities by applying the display to the sets he suggests.

Oceans Bleed

@Jakobian Here's an actual specific question of the sort that concerns me. I can write down a specific way of associating to a subcategory of a given category, some subset of a topological space, and I can give a map going the other way. Both sides have a poset structure by inclusion. I want to say I have a Galois connection between the power-set of the object class of the category, and the power-set of the topological space. But I can't take the power-set.

Then what the heck are these 'completely obviously well defined' maps I am writing down, which easily satisfy the defining conditions of a Galois connection

Or in short, any 'element of' $\mathcal{P}(\mathcal{C})$ where $\mathcal{C}$ is a proper class, I can easily assign an element of $\mathcal{P}(X)$ where $X$ is some topological space, and I have a map going the other way, so I should have $F:\mathcal{P}(\mathcal{C})\leftrightarrow \mathcal{P}(X):G$ where $F,G$ are set maps that preserve the inclusion order on both sides, But it seems I must already be talking nonsense, since I am taking the power-set of a class

I guess I have to take the 'power class' of the class, and define this as a class function

mo-_-

15:34

hi

Oceans Bleed

hi

psie

@psie for instance, applying the display to $A=\{g>\|\Phi\|+\epsilon\}$, I get $$\nu(A)(\|\Phi\|+\epsilon)<\left|\int_A g\,\mathrm{d}\nu\right|\leq \|\Phi\|\nu(A).$$Does this somehow imply $|g|\leq\|\Phi\|$?

Thorgott

@OceansBleed sure, that works

Jakobian

15:56

@psie display?

psie

@Jakobian yes, the author uses this word when equations have double dollar signs, i.e. when they are displayed. By display I meant $$\left|\int_A g\,\mathrm{d}\nu\right|=|\Phi(\mathbf1_A)|\leq \|\Phi\| \|1_A\|_1=\|\Phi\|\nu(A),$$in one of my first messages.

Jakobian

They want to show that $|g|\leq |\Phi|$ a.e., so they need to show the converse inequality is a set of measure zero

psie

ahh, makes sense

Jakobian

@psie here you don't know that a strict inequality holds. You have $\nu(A)(\|\Phi\|+\varepsilon)\leq \|\Phi\| \nu (A)$ which if $\nu(A) > 0$ would give you a contradiction

psie

@Jakobian hmm, ok. But the inequality in the set $A=\{g>\|\Phi\|+\epsilon\}$ is strict. If I use then the monotinicity of integrals, I get $$\nu(A)(\|\Phi\|+\epsilon)<\left|\int_A g\,\mathrm{d}\nu\right|\leq \|\Phi\|\nu(A).$$

Jakobian

16:09

@psie no

psie

ok, what did I do wrong? Also, I don't see the role $\epsilon$ plays here. Why is it needed?

oh wait

for the contradiction, we need the $\epsilon$.

I don't see how you got $\nu(A)(\|\Phi\|+\varepsilon)\leq \|\Phi\| \nu (A)$...

Jakobian

inequality is a transitive relation

psie

16:27

hmm, I still don't see what I did wrong. I applied $\left|\int_A g\,\mathrm{d}\nu\right|\leq\|\Phi\|\nu(A)$ with $A=\{g>\|\Phi\|+\epsilon\}$ just as was hinted at.

Oceans Bleed

16:38

@Thorgott If someone is 'using ZFC', does that mean explicitly that they are working in a model for ZFC? Is there a meaning for 'ZFC set' outside the context of working in a model for ZFC?

psie

@Jakobian I'd be grateful if you could elaborate on this "no".

Oceans Bleed

"The axiom of foundation (or regularity) demands that every set be well founded and hence in V"

Oh, maybe my confusion is totally resolved

Claudio

is there a rule to establish if integrals such as $\int_{0}^{2\pi} \cos^m(x)\sin^n(x)dx, n,m \in \mathbb{N}$ are zero or not?

$\int_{[0,2pi]}\cos^n(x)/ \sin^n(x)dx$ are zero when n is odd for example

I only need to know which of those are zero, not the general expression of the solution

Sine of the Time

try $u=\pi-x$

Claudio

16:56

no ok, I meant is there a rule like $m+n = odd, m+n = even$ or something like this

because integrals where either one of the two exponents $m$ or $n$ is 1 are zero

Sine of the Time

If you perform the substitution I suggested you integrate over $[-\pi,\pi]$

Which is a symmetric domain

Claudio

Oh I can use the parity of the overall function

Sine of the Time

yes, it's easier to see when the integral is $0$

Claudio

@SineoftheTime if you have sin^2x cosx, then you get $\sin^2u \cos u $ which is even but the integral should be 0

there's a minus sign, my bad it is indeed odd

yeah it works, @SineoftheTime thanks my guy

Sine of the Time

:)

Claudio

17:06

@SineoftheTime no wait hahah even with the minus sign it is still even

this is weird

Sine of the Time

let me see

It's even but it's not always positive

So it can be $0$

$-\int_{-\pi}^{\pi}\sin^2 x \cos x dx=\frac13\sin^3 x \big|^{-\pi}_{\pi}=0$

Vuk Stojiljkovic

17:28

Is there someone I can ask a question in functional analysis?

Jakobian

> Just ask; don't ask to ask.

there is no "someone" that's not how it works, you ask and someone might or might not answer

sure you can try "hey @leslietownes solve my problem about bornological spaces" but that might or might not work

there's lots of people who know lots of different things - just ask

"functional analysis" is a broad topic in itself

Thorgott

@OceansBleed I don't know what a 'model' is (in the formal sense of model theory). I live life by just working with the axioms of ZFC and assuming the existence of sufficiently many inaccessible cardinals, which I then use to talk about small/large/very large (usually, that's all that's needed) sets. This is ultimately equivalent to using nested Grothendieck universes, but more accessible IMO.

@VukStojiljkovic you've been taught about this like three times within the past couple days already

Jakobian

@psie "no you don't know that a strict inequality holds"

Thorgott

you're not entitled to receiving answers to your question. you can always feel free to ask a question in this chat and then you may or may not receive an answer and that's life.

psie

@Jakobian well, how do we know the non-strict inequality holds then?

Jakobian

17:36

@psie well this is basically Chebyshev inequality

psie

you mean $\mu(\{x\in E: f(x)\geq a\})\leq \frac1{a}\int f\,\mathrm{d}\mu$?

Jakobian

I'm sure you know how to prove Chebyshev inequality, $\int_A |g| \geq \int_A (\|\Phi\|+\varepsilon) = \nu(A)(\|\Phi\|+\varepsilon)$

@psie yes, it's basically this inequality, more or less

the difference is so negligible that I'll just call it Chebyshev inequality, this one follows as a corollary anyway

leslie townes

vuk please ask here, with questions of general interest you are likely to get higher quality feedback if the full chat can see/respond

Jakobian

unless it's a question about bornological spaces, then please ping leslie directly

leslie townes

yes also my desktop client periodically loses contact with the chat server, so i sometimes don't see requests to join other rooms until hours after they happen

i'll just give you my personal cell number if you need to reach me about bornological spaces

Jakobian

17:45

lmao

Thorgott

give me your address, I'll fly over and we can host a seminar on bornological spaces

Xander Henderson

@Thorgott You aren't going to make them pay for the flight?

How nice of you.

psie

@Jakobian ok, I think I understand now. And in Chebyshev's inequality (I'm referring to the one I posted, which I think is Markov's inequality), it's not necessarily true that a strict inequality in the set gives a strict inequality between the measure and the integral I think, see for instance here

thanks, I have to hit the road now. One for the road!

Jakobian

@psie yeah, and it's not necessarily the case in monotonicity of integral either

Oceans Bleed

18:19

@Thorgott I feel like I'm being driven insane by there being some presumably basic fact I am missing, that noone is stating. I just can't tell what a 'proper class is', whether it has an actual non-relative meaning or not. Is there such a thing as a proper class that is not a set in some bigger universe?

Xander Henderson

Isn't a proper class just a class that isn't a set?

Oceans Bleed

But then what is a class. In ZFC it's just an informal notion it seems, but then what is a large category?

Can large categories simply not actually be discussed in terms of ZFC

Xander Henderson

I really don't do class theory---I just know some basic examples. For example, an example of a proper class is the class of all sets.

Oceans Bleed

@XanderHenderson As far as I can tell, that might literally be the only class (from the perspective of ZFC)

Xander Henderson

@OceansBleed Isn't the class of all ordinals also a proper class?

4

Q: What is the exact reason why the class of all ordinals, $\textbf{ORD}$ is not a set?

In this question, some explanations about why the class of all ordinals is not a set are given. I can also find similar explanations in many books. However, all those explanations and proofs just outline how we can reach a contradiction by assuming $\textbf{ORD}$ is a set, without saying where ex...

set-theory ordinals

Oceans Bleed

18:25

Maybe I should amend what I said, I think proper classes might only come from taking some sequence of sets with increasing rank, such that it climbs the Von Neumann hierarchy forever, and then the union of these will be a proper class

Xander Henderson

@OceansBleed Again, I don't know this area of mathematics well, but... so what if they do?

Oceans Bleed

Or just starts as all of V at the first step

It says this on wiki:

V is not "the set of all (naive) sets" for two reasons. First, it is not a set; although each individual stage Vα is a set, their union V is a proper class. Second, the sets in V are only the well-founded sets. The axiom of foundation (or regularity) demands that every set be well founded and hence in V, and thus in ZFC every set is in V

Von Neumann universe

In set theory and related branches of mathematics, the von Neumann universe, or von Neumann hierarchy of sets, denoted by V, is the class of hereditary well-founded sets. This collection, which is formalized by Zermelo–Fraenkel set theory (ZFC), is often used to provide an interpretation or motivation of the axioms of ZFC. The concept is named after John von Neumann, although it was first published by Ernst Zermelo in 1930. The rank of a well-founded set is defined inductively as the smallest ordinal number greater than the ranks of all members of the set. In particular, the rank of the empty set...

@XanderHenderson Actually I am just trying to make sure something I'm writing isn't wrong for set-theoretic reasons

It's interesting to me that set-theoretic stuff seems to be becoming an issue in mathematics again, at least I am seeing people far from set theory having to deal with set-theoretic issues more

Thorgott

@OceansBleed 'proper class' does not have any meaning in ZFC, it does in other theories like NBG

Oceans Bleed

@Thorgott So what exactly is a large category?

Thorgott

depends on whom you ask

Oceans Bleed

18:29

Maybe this is the thing I am 'missing'. I skim over the sentence that says "In ZF you do X" and "in NBHGHZGK you do Y", and try to synthesise information from incompatible contexts, and then find myself confused

Thorgott

as I said, my preferred way of doing things is by working in ZFC + assuming a sufficient supply of inaccessible cardinals (which is consistent), but there are other ways of doing things

Oceans Bleed

So to be clear, those inaccessible cardinals are just to pick out all sets of lower rank than each of them, so you can stratify sets into small, large, Large, LARGE, etc?

Thorgott

yes, exactly

Oceans Bleed

In which case I never need to use the term 'power class', since the power-set of a large-class (A subset of the large-universe, as opposed to an element of the large-universe) would be a Large set, etc?

Jakobian

@Thorgott I don't think that's consistent with ZFC, actually?

not provably anyway

it's not provable that existence of inaccessible cardinals in ZFC is possible

Thorgott

18:45

@OceansBleed yes, if $A\subseteq V_{\kappa}$, then $\mathcal{P}(A)\subseteq V_{\kappa+1}$, hence $\mathcal{P}(A)\in V_{\kappa+2}$ and this is contained in $V_{\kappa^{\prime}}$ whenever $\kappa^{\prime}$ is a limit cardinal greater than $\kappa$

@Jakobian true, not from ZFC itself, sorry

Oceans Bleed

@Thorgott So the category of groups, say, is itself actually secretly defined relative to picking some background cardinal, to pick out some stage in the hierarchy, even though this is not mentioned

You read "has the class of all groups for objects", and immediately interpret that as actually meaning, class of all groups definable in $V_\kappa$

Thorgott

the way I like to do things, yes

people who use other foundations may do it differently, but the punchline is that the difference doesn't really matter most of the time

even if you don't wanna assume any additional foundational axioms, working in the small category of $\kappa$-small groups relative to some large enough regular cardinal $\kappa$ suffices for most applications

Oceans Bleed

And so transfinite induction/recursion, for you, actually caps off, then. I guess one only bothers to inductively prove up to $\kappa$

Thorgott

yeah

I think the NBG-like approaches via classes are more popular in classical ordinary category theory, but the approach using inaccessible cardinals is pretty standard in the world of (weak) higher categories, where it was popularized by Lurie's HTT

Oceans Bleed

I suppose it was vague memories from the start of HTT or HA that made me understand immediately what you meant about the chain of inaccessible cardinals

Thorgott

19:00

nice

Vuk Stojiljkovic

Well, I'll just ask here, can someone explain to me why does an operator $T\in B(H_1\oplus H_2)$ as the matrix representation $a,b,c,d$, why does a act here in $H_1$ $d$ in $H_2$ and $b,c$ in $H_1\to H_{2} $ and $H_{2}\to H_{1}$, I mean, why do they map in such way.

Oceans Bleed

I'm still somewhat confused. Take the following: https://arxiv.org/pdf/math/060225 Corollary 8.4

One shows that there is a proper class of t-structures on the derived category of the integers. Certainly he isn't working in some fixed universe. He is taking groups as large as he wants, and he is showing there are at least a power-sets worth of things that produces, and so the collection of objects he can construct is a proper class. Would you interpret that in your [cont]

[cont] framework as merely saying that a classification of such things must take place in a larger universe, but you would only be classifying the $\kappa$-rank objects, now inside the $\kappa'$-rank universe

I would then not even know how to interpret his result. He is clearly talking about t-structures on the mega-biggest possible D(Z)

I think you just simply wouldn't even have a statement of this form? Are all propositions now labelled with the rank in the hierarchy you are intending all objects/categories/etc that appear in the statement to defined relative to

leslie townes

vuk this is done in the comments to your question in the case H_1 = H_2 = H. the maybe trickiest part is understanding how a linear operator from H_1 + H_2 into some other space W can be thought of as determined by a pair of linear operators, one from H_1 into W and the other from H_2 into W. this is basically just T(x,y) = T(x,0) + T(0,y)

Oceans Bleed

Possibly I just have to read the referenced shelah.logic.at/files/95858/44.pdf to check what set-theoretic formalism Shelah chose

leslie townes

the other part is how an operator from a space W into H_1 + H_2 (or indeed any map a from a set W into H_1 + H_2) can be thought of as a pair of maps, one from W into H_1 and the other from W into H_2. this and the previous observation give you the "matrix entries" of an operator from H_1 + H_2 to itself

again, this is done in the comments to your question on main already. it might help to dispense with generalities start with an example T and H_1 and H_2 if you are still stuck

Thorgott

19:08

your arxiv link doesn't work

Oceans Bleed

arxiv.org/pdf/math/0602252

Oh I deleted the 2 at the end by accident

You can just black box the terms in Corollary 8.4, if you don't think about these things, but the proof should be black-box-readable

It's disturbing that one can publish papers in highly abstract areas of math, and not even have an undergraduate level understanding of set-theory

leslie townes

vuk if you want to think purely algebraically (not that this is a good idea) if you let P_1 and P_2 denote the orthogonal projections from H_1 + H_2 onto H_1 + 0 and 0 + H_2 respectively, and I the identity on H_1 + H_2, use the fact that I = P_1 + P_2 and expand any operator T as (P_1 + P_2) T (P_1 + P_2) = P_1 T P_1 + P_1 T P_2 + P_2 T P_1 + P_2 T P_2

Vladimir Lysikov

@Vuk That is basically by definition
What does it mean that an operator $T \in B(H_1 \oplus H_2)$ is represented by a matrix $\begin{bmatrix}a & b \\ c& d\end{bmatrix}$?
It means that $T (x_1, x_2) = (y_1, y_2)$ is equivalent to $y_1 = a x_1 + b x_2$, $y_2 = c x_1 + d x_2$ (in matrix form $\begin{bmatrix}y_1 \\ y_2\end{bmatrix} = \begin{bmatrix}a&b\\c&d\end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix}$)
For this expression to make sense, we need $a \colon H_1 \to H_1$, $b \colon H_2 \to H_1$, $c \colon H_1 \to H_2$, and $d \colon H_2 \to H_2$

leslie townes

this is another view of the "matrix entries" of T although in this guise they are regarded as operators on all of H_1 + H_2, and they just have kernels and ranges that make it appropriate to think of them as operators among the subspaces H_1 + 0 and 0 + H_2

Oceans Bleed

@Thorgott Meaning I think his proof probably shows that if you are working with $D(\Bbb{Z})$ relative to $V_\kappa$, then the collection of $V_\kappa$-relative t-structures is not in $V_\kappa$, and this is true for all $\kappa$. Or more precisely, the collection of $V_\kappa$ t-structures is in $V_{\kappa +1}$ (?)

But the fundamental take away is that you cannot classify t-structures on D(Z), and yet you can, just in a larger universe

But his result is about classifying 'them all', and you can't have that statement, since no stage of the hierarchy is ever about 'them all'

So your way of dealing with set-theoretic issues doesn't actually capture seemingly important mathematical statements

Unless I'm confused

Thorgott

19:20

yeah, I think this is all fine. he appears to be working in an NBG-like setting, as is not uncommon in ordinary category theory setting. in the setting we have just discussed, you would define a version D(Z) via chain complexes that consist of $\kappa$-small abelian groups (relative to the inaccessible cardinal $\kappa$) has a set of $t$-structures that is not $\kappa$-small (and it should not be hard to see that the set of $t$-structure is small relative to the next inaccessible cardinal)

@OceansBleed I think that is more of a philosophical discrepancy

Oceans Bleed

Well the conclusion is that you 'cannot classify all t-structures', which is something quoted in like 30+ papers

I'm not sure if I could consider it a philosophical discrepancy for that reason. You cannot classify them, so you move to the collection of 'compactly-generated t-structures', which do form a set, so you can classify them

Thorgott

what it means to 'classify' something does not (at least in most cases) have a concrete mathematical meaning anyhow, it's more of a moral statement

Oceans Bleed

I just worry that these things are just catchphrases, and noone is doing their due diligence (including me until now)

Thorgott

the general philosophy (which you may recall from HTT) is that "how big" something is in absolute terms is usually not meaningful, only "how big" things are relative to another

Oceans Bleed

@Thorgott The classification theorems here have a very specific meaning. You can construct a bijection to some other set you care about, and the bijection is nicely controlled

Thorgott

19:26

and the Theorem there tells you the collection of $t$-structures is bigger than the abelian groups you consider, irrespective how big either of those are in absolute terms

Oceans Bleed

See 3.4, 3.5 for some 'flavour' to the classification type things: people.ucsc.edu/~beren/pdfs/mackey-stratification.pdf

@Thorgott In which case you have a cardinal indexed family of propositions, all of which tell you that you can classify, in the relative to \kappa stage, inside the in \kappa+1 stage, but none of those statements is the statement that is 'important' presumably

Thorgott

well, what is 'important' is the philosophical discrepancy I was alluding to

cause what sets do you care about that are not contained in $V_{\kappa}$ for $\kappa$ the first uncountable inaccessible cardinal (assuming such a thing exists)?

every mathematical object you have ever written down explicitly is in there

Oceans Bleed

Depends what explicitly means

Xander Henderson

@OceansBleed The dude whose website you are linking looks like such a UC Santa Cruz prof.

people.ucsc.edu/~beren

Oceans Bleed

@XanderHenderson hahaha

Very true

Xander Henderson

19:29

I can just hear him saying "Bruh, let's hit the beach and ride some gnarly waves!"

Oceans Bleed

hahaha, I can see that for sure

Thorgott

my point is that $V_{\kappa}$ is itself a model of ZFC, it is no better or worse than the model you started with

you worry about these statements not being about 'them all', but in the quoted version, this theorem also only addresses sets and not what happens if we allow class-sized abelian groups

Vuk Stojiljkovic

@leslietownes Can we move this conversation to a private chat, many messages flood the chat and I can't properly read nor respond

Thorgott

there's always an arbitrary cut-off point

all that really matters is the relative size: what you get is bigger than what you put into it

Oceans Bleed

@Thorgott I don't think there is. I think when someone talks about the category of groups, they have all of them

I think they mean a collection that is not contained in the hierarchy at any stage

Think of the proper class obtained by taking the union of all $V_\kappa$ defined groups as you vary $\kappa$, a union of sets indexed by the ordinals

Is that a legal operation?

I think it should be, assuming something @Jakobian said

Thorgott

19:34

it's not a legal operation in ZFC, but if you extend to something like NBG, this will be a (proper) class and it's called the von Neumann universe

@OceansBleed they have all of them that are sets, but the Theorem as stated is in a setting where classes exist, too, so you still have a cut-off point

Oceans Bleed

Wait maybe it's not permissible from what he said. I guess when s/he said I can still take a class defined by formula, the first order language only has symbols for the sets, so I can't index the union over a non-set

Thorgott

there needs to be a bigger size that your theory is able to address than what you allow for your groups

Oceans Bleed

Goddamnit, I have to go learn another set theory to understand a statement properly

hahaha

Thorgott

you don't really need to learn NBG IMO

the philosophy of the Theorem is the same no matter how you formalize it

Oceans Bleed

I guess the theorem is beyond insanely stronger than needed. The classifications people care about are usually based on support, giving certain subsets, or filtrations of subsets, on Spec(R), or D_{qc}(X), and it would be rather difficult for subsets of Spec(Z), or filtrations on Spec(Z), to be put in bijection with the \kappa-class of t-structures

I wonder what insane concoctions one would come up with for support spaces for the \kappa-class of t-structures

Sorry, I meant on Spec(R) when working with D(R) for R a commutative ring, and on X, in the case of D_{qc}(X) for X a scheme (or algebraic space/stack),

Anyway, these spaces atleast seem to not depend on \kappa at all, so I guess some things stay the same upwards in the hierarchy

Thorgott

19:48

I think being compactly generated is much stronger than just a size restriction

Oceans Bleed

Well the compact-object subcategory in these cases is essentially small, so that would be a size restriction I suppose

A set of isomorphism classes of compact objects

Compactness in triangulated categories, stable \infty-categories, means that Hom(k,-) commutes with small coproducts

Compact-generation of the t-structure means it is determined by a set of compact objects in a very precise way

But it is a 'set' of compact objects, and commuting with 'small' coproducts, which we must take as \kappa-small

Thorgott

yeah, that makes sense

Oceans Bleed

Thanks for chatting, I think you've substantially cleared up my confusion! Probably the 'philosophical' bit needs further refinement, but I'll have to do that later (probably after next sleeping at least)

Thorgott

no problem, glad it helped!

Sine of the Time

20:36

@Claudio you may find this useful

Vuk Stojiljkovic

21:37

@VladimirLysikov Thank you for this analogy!

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$Mathematics$ Mathematics