Mathematics - 2024-12-01

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Silly Goose

01:54

user image

is there a name for this theorem?

oh oops i have a typo. $H_0$ is the identity component of $G$ in $H$.

leslie townes

one data point: i personally don't know of a name for that theorem

from my extremely limited knowledge of algebraic topology, i would not expect there to be a name for something like that, but i dunno, maybe it comes up often enough to have a name

usually i am accustomed to see somewhat more general results (or families of results) being given names, or prose descriptors arising from ways of seeing how they come about (e.g. "the long exact sequence of homotopy groups associated with a fibration")

Silly Goose

hm i see

do you know if this result is some direct corollary of a more general result?

leslie townes

but i know some stuff in e.g. matrix groups often has people's names or other names associated to where it first arose in that context so who knows

Silly Goose

this is from a physics paper, so i am not sure if it is really of relevance to mathematicians as you say to bear its own name.

i am trying to see if i can just prove it using the first isomorphism theorem or something...because the proof is a bit long in the paper i took the theorem from

leslie townes

i haven't checked but i'm guessing it comes out of considering a long exact sequence associated with the pair G and H, which exists for technical reasons that are baked into the hypotheses of what a "continuous group" is. H might need to be nice e.g. maybe normal subgroup or at least a closed subgroup or a subgroup where the quotient G/H has a nice structure

Silly Goose

02:06

but the hypotheses of connected, simply-connected suggest that there is more to it

hm okay i will try to look into that direction

leslie townes

yes that's the kind of kind of stuff you would plug into a more general long exact sequence and results like that would fall out

usually this channel is littered with topologists who would be all over this by now

but that's just your bad luck

Silly Goose

hehe

Ben Steffan

What's $H_0$? It's the identity component of $H$, right?

Silly Goose

hm well the data available seems to be the covering map $p: G \to G/H$, the surjective group homomorphism $q: H \to H/H_0$ with $\ker q = H_0$.

@BenSteffan yes that is correct

Ben Steffan

Why would $p$ be a covering map

Silly Goose

02:13

since $G$ is simply-connected i thought one can construct a covering map for $G/H$

separately, $H$ is not necessarily a normal subgroup, so $G/H$ is not necessarily a group.

Ben Steffan

@SillyGoose Take $H = G$...

Silly Goose

i guess to leslie's point: mathoverflow.net/questions/143734/…

Silly Goose

02:33

user image

so then the proof is something like this

Silly Goose

03:27

er actually more like this

user image

1 hour later…

copper.hat

04:44

its all a Lie

2

6 hours later…

nickbros123

10:21

ive never in my life written an SOP. any tips an tricks? for context im thinking of applying to this program: phd.pages.ista.ac.at/isternship

1 hour later…

Thomas Finley

11:43

Let $\ell_2=\{(x_n)\subseteq \Bbb C: \sum_{n=1}^{\infty}|x_n|^2\lt \infty\}$ and $A=\{(x_n):|x_n|\leq \frac 1n,n\in \Bbb N\}\subseteq X=\ell_2.$ Show that $A$ is closed.

My solution goes like this:

Let $x\in A'$, where $A'$ is the derived set of $A.$ This means, that $\exists$ a sequence $(x^{(k)})$ that converges to $x\in A',$ where $x^{(k)}=\{x_1^{(k)},x_2^{(k)},...\}$

So, for $\epsilon>0$ $\exists N\in\Bbb N$ such that $d(x^{(k)},y)<\epsilon,\forall k\geq N.$ Hence, $\sum_{i=1}^{\infty}|x_i^{(k)}-x_i|^2\lt \epsilon,\forall k\geq N,\forall i\in \Bbb N.$

@XanderHenderson Can you please make some comment on this?

4 hours later…

Xander Henderson

15:26

@ThomasFinley Honestly, I'm out at "derived set". I don't know what that means, and the next sentence has an unnecessary $\exists$ in the middle of the text for no reason, and no explanation about what the $x^{(k)}$ are. I don't want to read further to see what you are doing.

Sine of the Time

it's the set of all limit points

Thorgott

clearly it means working in some sort of derived category

Jakobian

16:18

@XanderHenderson you can always google it

@ThomasFinley $A$ is the so called Hilbert cube. Its not only closed, its actually compact

Consider a sequence $x^{(k)}$ in $A$. Take convergent subsequence so that $x^{(k)}_1$ converges, then so that $x_2^{(k)}$ converges and so on, consider the diagonal argument by taking subsequence $k_n$ so that $x_m^{(k_n)}$ converges to some $x_m$ for all $m$. The claim is now that $x^{(k_n)}$ converges to $x = (x_m)$.

To see this note that $|x_m|\leq \frac{1}{m}$ so that indeed $x\in \ell^2$ and $$\|x^{(k_n)} - x\| \leq \|(x_1^{(k_n)}-x_1, ..., x_N^{(k_n)}-x_N)\|_2 + \sqrt{\sum_{m=N+1}^\infty \frac{1}{m^2}}$$

Claudio

The sets in which a series converges uniformly and *totally*(which, as I've mentioned some time ago, basically means the series satisfies Weierstrass M-test) are closed. I wrote this down in my notes without any explanation, and I wonder why such statement holds

Jakobian

now the second term on the right will be smaller than epsilon for big enough $N$, and first term will be smaller than epsilon for big enough $n$ since $x_m^{(k_n)}$ converges to $x_m$

so this proves that $A$ is compact

Lukas Heger

16:55

here's a different proof. For any $k$, the function $\pi_k:\ell_2 \to \Bbb C,(x_n)_{n \in \Bbb N} \mapsto x_k$ is continuous (even Lipschitz). Now the set $\{z \in \Bbb C \mid |z| \leq \frac{1}{k}\}\subset \Bbb C$ is closed and hence the preimage $\pi_k^{-1}(\{z \in \Bbb C \mid |z| \leq \frac{1}{k}\})=\{(x_n) \in \ell_2 \mid |x_k| \leq \frac{1}{k}\}$ is closed.

As an intersection of closed subsets, the set $A=\bigcap_{k \in \Bbb N} \{(x_n) \in \ell_2 \mid |x_k| \leq \frac{1}{k}\}$ is thus also closed

Xander Henderson

17:39

@Jakobian Sure, but if someone is asking me for feedback, it is incumbent upon them to make sure that they are communicating effectively. I am going to stop reading when what I am reading is no longer clear.

leslie townes

pushing back is also a good way (maybe the only way) to find out if the asker has ever tried googling it :)

it may also reveal that rare beast, a choice of definition

Xander Henderson

@leslietownes Indeed.

leslie townes

and helpfully reinforces the notion that math is not just: terms are floating around out there and figuring out what they mean from partial context, and whatever sources we can grab is The Job of Doing Math

Xander Henderson

Though my more serious critique as about the habitat of the $x^{(k)}$. I assume that they live in $A$, but that isn't made explicit.

leslie townes

well, figure it out, math man

Jakobian

18:39

@XanderHenderson I see, yeah, I missed that they asked you specifically

pie

19:08

@think_meaning_buildß Is this your channel?

@leslietownes How do they not have a mental breakdown when they try to read a book that's too advanced for them? I tried this once, and my thoughts were: 'Am I really so stupid that I can't understand this?' 'Maybe I'm just not good enough for math.' 'I bet everyone else understood this on their first try.' 'I said I wanted to be a mathematician, but I can't even understand this... (sarcastic voice) Great job, me.'

Jakobian

19:35

Because they approach things differently from you

Jakobian

19:47

its not really complicated that different people have different experiences, and its the fundamental of what things like empathy is built upon

Thorgott

I mean, most math books have an Introduction in which they specify their target audience and the prerequisites

leslie townes

yeah pie as jakobian mentions a lot of it is just differences in how people are built. it really helps if you don't conceive of math as a kind of battle or test of general intelligence. also, to adjust expectations downward. why would anybody expect to be able to instantly understand a subject they had maybe never read anything about

a math book is just a book. it isn't a personal challenge that you have to rise to, or be the equal of. it's just there on the shelf and there are lots of other ones

same as any book

Silly Goose

Is there a generic statement about $(A/B)/(B/C) \cong A/C$ where $\cong$ is homotopic or homeomorphic equivalence? I.e., an analogue of the third isomorphism theorem for topological spaces.

leslie townes

if i picked up a book written in arabic i wouldn't be able to read that either, and i wouldn't think "oh my god, i'm such an idiot, i will never accomplish anything." i'd just need to study arabic if i cared about reading that book. a lot of math books in one's native language are maybe closer to this analogy than most people might think

Ben Steffan

@SillyGoose That statement does not parse

Thorgott

19:57

yeah, you switched the order up

Silly Goose

Oh oops. you're right. I mean $(A/B)/(C/B) \cong A/B$

Ben Steffan

(also "homeomorphic equivalence" is not a term)

Thorgott

anyway, if we generously assume that $A/B$ denotes the pushout of $\ast\leftarrow B\rightarrow A$ for some implicitly assumed map $B\rightarrow A$ and $\ast$ the terminal object, then this is true for any composite $C\rightarrow B\rightarrow A$ in any category as long as the required pushouts existence

Silly Goose

homotopic equivalent or homemorphic :P

Ben Steffan

still no

homotopy equivalent or homeomorphic :)

leslie townes

19:59

sometimes people react to math stuff like "omg i would never be able to understand that," leading me to wonder, do they walk up to people who speak languages that they don't, and share that observation with them? why would you expect to understand a language you haven't studied? where does the expectation come from that if it's math and not a language, anybody who is "just smart enough" should be able to just pick it up and run with it

Silly Goose

ah

Thorgott

sorry, $B\rightarrow C\rightarrow A$ the way you wrote it

leslie townes

i guess popular notions around a correlation between math aptitude and some kind of general intelligence might lead you to that conclusion

Ben Steffan

@Thorgott now the pushout in $\mathrm{hTop}$ just has to exist :)

Thorgott

the statement about homotopy is a bit confusing cause it's true up to homeomorphism, unless you actually meant to take a homotopy pushout (a mapping cone) instead

in which my previous statement still applies, but you should work $\infty$-categorically instead

at the end of the day, rather than an abelian phenomenon or anything like that, I think the third isomorphism theorem is just a particular consequence of the pasting lemma

Silly Goose

20:01

ah no im good with taking it up to homeomorphism. there is just wiggle room to allow for homotopy equivalence

since the statement i am ultimately looking to show is about isomorphic fundamental groups of the two spaces

but if the wiggle room is not necessary then it's even nicer

leslie townes

why is "work infinity-categorically instead" always the answer

Ben Steffan

because $\infty$-categories exist to solve this particular kind of problem, among other things

leslie townes

fiddlesticks

Ben Steffan

the $\infty$-category of spaces is a gimmick that lets you do homotopy theory categorically

It is, among other things, a setting in which limits are homotopy limits and colimits are homotopy colimits, by design

Thorgott

user image

this is the relevant diagram

left square and upper composite square are pushouts, hence right upper square is pushout, so if the right lower square is a pushout (meaning the bottom right term is $(A/C)/(B/C)$), then the right composite square is a pushout (meaning the bottom right term is $A/B$)

(this back-and-forth chase with the pasting lemma is a very common type of argument)

Jakobian

20:08

@leslietownes What's interesting is that almost everyone not experienced with math does that

the issue I think is an idealization of mathematics as this hard, unapproachable subject

in the eyes of the general population

leslie townes

jakobian: well that, paired with the weird expectation that special people can do it instantly if they are smart enough, which maybe they wouldn't have about foreign languages

Jakobian

I find it interesting how many things boil down to - this is just the culture of how people approach things

Silly Goose

is it implicit that a long exact sequence begins and terminates with a $1$?

leslie townes

omg i never would have thought of being able to speak polish. like, how do you even say the words. i guess i'm not a "polish speaker."

copper.hat

i have the same feeling about Irish.

Jakobian

20:12

I never would have thought of being able to speak Polish, and yet, here I am...

Ben Steffan

in all fairness you might have had some time to polish it

Jakobian

I wonder how much can I get away with by acting superior for speaking my native language

Thorgott

@SillyGoose if it has a beginning and an end, it does not deserve to be called a long exact sequence

it can have one of them, though

Silly Goose

oh okay

Thorgott

in which case there is no harm in adding a $1$

Jakobian

20:18

on the other hand, treating reading a book like a competition might add to some motivation, for better or worse

everything within reason though, I suppose

Ben Steffan

@Thorgott but the sequence need not end in an injective / surjective map?

ending in 0 or 1 is an extra condition

Jakobian

I think there is indeed some kind of relationship between being motivated to do something and being competitive

Thorgott

sorry, you also add the kernel/cokernel

and don't get started on LESs that end in pointed sets

Ben Steffan

@Thorgott I was about to say something lol

Thorgott

takes an algebraic topologist to know one

Daniel Donnelly

21:00

0

Q: Roughly speaking can a Proof Tree always be written as $((A \implies B) \implies (C \implies D)) \implies \dots \implies E$ for example?

I'm in the process of coding a web-based proof engine namely for the purpose of storing diagram chases in a database. For the purpose of display, I have opted for the nice properties of Right-associative implies $\implies$ arrows to display both Definitions and Statements. But for Proofs, things...

logic soft-question trees proof-theory type-theory

Gonna need some expert input on this

@Room

I know it's Sunday, but where is everyone in the math world...

Everyone's at church or somethin 😂

Sh0w m3h th3 c0d35, p1s. :)

Xander Henderson

@DanielDonnelly probably more on topic elsewhere. proofassistants.stackexchange.com

Daniel Donnelly

@XanderHenderson good idea! Thankyou!

0

Q: Roughly speaking can a Proof Tree always be written as $((A \implies B) \implies (C \implies D)) \implies \dots \implies E$ for example?

I'm in the process of coding a web-based proof engine namely for the purpose of storing diagram chases in a database. For the purpose of display, I have opted for the nice properties of Right-associative implies $\implies$ arrows to display both Definitions and Statements. But for Proofs, things...

proof-assistant soft-question software-application proof-tree meta-logical-framework

I put in on PA exchange, it's there if anyone here goes there too

Sine of the Time

21:16

please don't ping random users @Daniel

Daniel Donnelly

@SineoftheTime k, noted

I would show screenshots in the PA post but my recent edits broke the site. I have diagram search up to variable subst working, but just works with single diagram at the moment

Got it working last night because I recently finished the bulk of the text language part where I extract the varables and replace with a Neo4j regex standard form.

This standard form is important because each user has a particular way they like to or know of entering something in LaTeX

But my definition of variable is very non-standard, and includes stuff like $\mathscr{A}$ is a variable!

that's on the user I/O side, but it gets ASCII-tized to go into the database

So each variable in the database is the regex [a-zA-Z][0-9]* in order to offer an unbounded set of them, but on the UX side you get even larger set of fancy vars.

It's second-nature to mathematicians to substitute variables, but to get a computer to do it the right way, it takes months of coding / re-design

It wouldn't be write if I said the only variables are A..Z, a..z with a number of ''' (primes) attached. So I spent a lot of time on this part

copper.hat

21:52

i find combinatorics very unintuitive. i came across a simple such problem but could not follow the answers and my own answer was different. rather than being mature and walking away i choose to avoid dealing with my real life issues to waste many hours, including some coding to get some closure. mathematics is some kind of sick porn. no doubt Gaetz is lurking around somewhere.

Jakobian

22:13

@copper.hat combinatorics* is some kind of sick porn

copper.hat

indeed. then there's abstract nonsense.

1 hour later…

Jakobian

23:29

@BenSteffan I was asking some stuff about covers before. I'm not sure if that one is true, but you can still prove that fully normal implies strongly collectionwise normal - if you take a neighbourhood $U$ of the diagonal then you can take open $W_x$ with $x\in W_x$ and $W_x\times W_x\subseteq U$ for all $x$, then star-refinement of $\{W_x : x\in X\}$, say $\mathcal{V}$, and then $V = \bigcup\{A\times A : A\in\mathcal{V}\}$ is such that $V\circ V\subseteq U$.

So the result I wanted to hold is still true, I just had a bad reference I guess

Ben Steffan

I see :)

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