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15:00
what's with this brand new neo-nazi trend
@BalarkaSen I dunno. When I first heard about this, I figured people were blowing it out of proportion
But no, their manifesto is explicitly anti-semitic, homophobic, empirialistic, socialist (lol), etc.
Yeah, I was genuinely surprised to see there's a nontrivial support for this movement.
also lots of posters with "national socialism" and eagles. I think it's an elaborate troll
I'll send you a link on discord. It gives me the creeps
Dear god.
Re:troll, I initially thought that it's an alt-right troll movement like the cult of kek.
@BalarkaSen I don't know how they're not self-aware enough to know that poster goes way over the line
15:14
It's kind of logical that a fraction of the white nationalism in US today would try to conflate themselves with the Nazi propaganda
I wouldn't deny that
But
15:37
You're on discord too Balarka?
Yeah I am.
Do you have one?
Have I talked you on there? On 'Proper mathematics'
Hmm no
Oh right, Daminark is in that chat room too though
btw you're Alex Clark right? I never got around to asking
15:38
I am
kk
good to see you again :P
Did you stalk me at last :')
Ah no it just struck me a few days ago when you mentioned you're from Australia
who is alex clark
I've been on here for ages under a different name
15:41
i had an answer involving "algebraist fuck" and the c-word but I don't want to get banned
2
Oh god
You have to be Australian to not be banned
When you see over 10k users enter the room, you best be deleting those messages
rip
gotcha
Serious business mathz in the most part
Did you join before I deleted the link, I couldn't tell from the names
15:49
I joined, yep
$\mathcal{F}(X,Y)$ is the set of all mappings $X\toY$ that are provided with the operations of pointwise addition and scalar multiplication. Am I supposed to take it for granted that continuous functions are a subset of this set?
That the collection of all set theoretic maps contains the continuous maps?
or that they don't form a proper class?
I am really just asking, whether given f,g as continuous functions, can you assume that $(f+g)(x)=f(x)+g(x)$.
In the context of introductory functional analysis, if that is relevant
You shouldn't really take anything for granted, and should instead try to prove/disprove it
especially the axiom of choice
15:54
exactly
I think I located you @BalarkaSen lmao
I guess the name trivially gives it away after the fact
@Narcissus I posted something on the algebraic geometry room
I know that two continuous functions add up to a continuous function. I just don't know what definition of addition of functions I should use over the linear space of continuous functions.
@Narcissusjewel is there a PDE room
there seems to be
15:56
@0celo7 There is. They took every arxiv section and made a room for it
I'm the famous mathematician Morita
got it
is this server dead
PDE is
Nah, they just kick people who troll and whatnot
most activity is in hub
AT room is pretty active I think
Or it was
AT is actually cancer tho
@Daminark
what about him
16:00
He's our AT guy right?
I haven't heard him talk about AT
He did a project under JP May
He's into J Peter May
In algebraic topology, the Dold–Thom theorem, proved by Albrecht Dold and René Thom (1956, 1958), states that the homotopy group πi(SP(X)) of the infinite symmetric product SP(X) of a connected CW complex X is the i-th singular reduced homology group of X, usually denoted by H ~ i ( X ; Z ) {\displaystyle {\widetilde {H}}_{i}(X;\mathbb {Z} )} ...
I still don't understand that theorem
16:04
@BalarkaSen I am tempted to invoke the softness of $C^\infty_M$
Or I could just say "because we can extend $g$ off of a closed set"
Please just say that
hmm
16:22
"You shouldn't really take anything for granted, and should instead try to prove/disprove it" i like that attitude
16:49
Just had an odd thought: Does anyone stream their mathematics research?
I don't think it could possibly be made interesting
So you want to stream someone sitting at a desk /standing in front of a blackboard who thinks really hard and scribbles occasionally?
Well, like many good streamers, they'd have to discuss what they're doing
If they're focused on working, trying to be entertaining for stream viewers is distracting
16:54
Maybe? It depends how much they like to think out loud
Like, you're really trying to see if this lemma you'd need is true and then you remember you're on twitch, turn to the camera and say "Oh so what I'm doing is..."
Well I mean, I talk to myself when I do my research. And talking through things might sort out ideas (like rubber duck programming, or whatever that's called)
I'm just imagining some nerd scribbling stuff with some dope rap playing on the background
2
and it cracks me up
Basically, I think it could be done in a way I'd want to watch, and I also think I could do it if I thought anyone would want to watch, but it seems likely no one is doing it.
"Be sure to tune in next thursday where we're going to check if some spectral sequences degenerate to give us some exact sequences we need, that's going to be thrilling! If I get enough subscribers, I'll make a special that redoes the same calculations in the derived category.", yeah, I'd imagine that would be really popular on Twitch
16:58
But like, imagine being there when someone has a big breakthrough
MatheinBoulo exactly! I could watch that
I imagine it being like a talk, except instead of stuff they figured out a while ago, you get to watch it happen in real time
people certainly watch talks
The punchline though is that research is slow, random, and rarely rewarding
So is a lot of stuff on Twitch...
But you can only give a talk after you figured everything out and know what works and what doesn't
I mean it's better than a stream of figdet spinners spinning... fidget for 5 hours
Is it though? At least that happens continuously
17:03
Yeah what Mark S is suggesting someone just continuously babbles on about what they are thinking on stream
That's not how it works
"Now I'm going to facebook my buddy Kevin and see if he knows how to do this"
Rubberducking works for very specific things (in particular, precisely when you have an idea and you want to talk it out loud to see if it works)
"He didn't and I'm bored so I'm going to scroll through the timeline"
(I'm not trying to pick on you - this is what I do when I work :) )
Also who could actually follow that? Only researchers in the same subarea would have a chance to get what's going on, and they probably have more important things to do
I think there is a potential merit to the idea if someone is streaming when they are reading a textbook or whatever
17:05
@MatheinBoulomenos That is the primary objection that comes to my mind, too
@MatheinBoulomenos I will never watch you stream, Mathei
This is my promise to you
But problem-solving is way too dull and happens in "aha" moments to stream
@MikeMiller haha
My aha moments are usually just mad I didn't see it earlier lol
@BalarkaSen There's a lot of streaming that's mostly "dull" with random success/completion; I don't think that's actually a significant issue - just one that might cut down on the popularity
17:07
@MarkS. Well that's why I don't watch most (in fact any) Twitch streams.
I'm thinking of speedrunners right now and I don't think it's boring to watch them when they don't make a new record or something - you're watching someone show off a skill
Specifically when someone is trying for a new record it can actually be pretty boring with the resets haha
But that's uncommon
@BalarkaSen I pinged you elswhere about 15 ago
playthroughs are at least a little gripping
Oh I wasn't checking
I figure if no one's tried/done this, then we don't know how gripping it can be.
I can make a guess and say that I won't watch your stream ;)
Maybe someone will
That's fine; I'd rather watch someone better than me stream anyway
17:09
@MarkS. I think the main point of the majorities of streams is the entertainment value. If it's entertaining to some people, then it doesn't really matter if it's random. I have a hard time imagining that it's possible to do research and entertain people at the same time
Have you ever worked on a problem together with someone in real time?
Was it fun to see the new ideas/perspective they were trying compared to what comes to your mind?
that were easy things, problem that are designed to solved in reasonable time
ah, that's a bit different then
Collaboration on unsolved problems is similar, you just might not get a solution as fast.
At the very least, maybe a camera at a workshop might be interesting for people who couldn't attend...
17:12
Also it's not like you can just watch someone do research and follow it. You have to spend years working through books and papers and talk to people who know more than you before you begin to comprehend what the current research is even about
Well yeah, in the same sense not just anyone could attend a workshop on blah blah subtheory
But the people who are giving the workshop don't try to figure out new things at the same time!
The people who are attending do; isn't that what makes it a workshop?
Someone who gives a talk at a workshop has time to prepare the talk. They're not developing the theory in real time on the blackboard
Right, but then after the talk people might work together on things. Maybe "workshop" is used differently in different subfields?
17:17
But working together is also quite different from watching another person work alone
streams of discussions at workshops would be valuable, just like streams of talks at workshops, yeah
ok, so maybe math research can hypothetically be streamed, but only if there are multiple people there
I'd agree with that
17:40
@0celo7 OP responded by saying my answer was interesting
17:50
@BalarkaSen did he say it wasn't rigorous enough?
no lol
I smell a topologist in the OP
I might have to pull a topologist
I can't write down a diffeomorphism 100% correctly
but I can sure as shit say that I want to shove stuff radially outwards
I have some spare time. Should I think about my answer deeply?
Maybe I should learn Sobolev spaces from Brezis
I think @EricSilva should think about your answer deeply.
17:58
He's a busy man
this summer I mean
jesus god is my answer that valuable
i should write more glib answers like that
no, it's trash
but it could be good
kind of like Fermat trolling future generations while taking a shit
y u hurt me like this
@BalarkaSen I compared you to fermat
18:15
Hi @Ted
Hi @Balarka
@BalarkaSen I think I can actually avoid the Morse theory altogether
I didn't really use any Morse theory
Just that the gradient flow is relevant to Morse theory
in my thesis
18:31
@BalarkaSen I have the Morse argument written up but I like the idea of not having to use Morse theory for this...
18:43
alternating group is subgroup of special linear group and subgroup of special linear group is subgroup of general linear group, is this right?
18:56
I've been banging my head against the wall for a while now. I am trying to show

$\square (p \rightarrow \square q) \rightarrow \square (\diamond p \rightarrow q)$

with KT45 modal logic.

I get to the point that I have a dashed box (working in all neighbouring worlds) where I have $p \rightarrow \square q$. I then know I need to make the assumption $\neg \square \neg p$ and somehow within the scope of that assumption show that $p$ holds that I can use the premise. Any ideas on how I can get $p$ in that scope?
19:12
Hi Guys
Guys, is a nuclear weapon a machine?
When people say, Saturn V is the most powerful machine made by man.
Why do they disregard a nuclear bomb?
given a range 0 .. N and a number M
I am doing a loop (i) from 0 .. to N
each time I test i modulo M = 0 here I print the result
else (here I am finding a formula that can return the previous number was modulo M)

example 0..5(=N) M=2

at the moment i = 3 I want to print 2 (because 2 is the most approximate modulo M that less than 3)
I musn't use variables I want to do anything with formula
I can rexplain if it's not clear
What exactly is the question?
19:36
lets take an exemple:
i = 0 if()
oh sorry, hold on
lets take an exemple:
N= 5 , M= 2
i = 0 if( 0 modulo 2) the condition is true so print 0
i = 1 if( 1 modulo 2) the condition is false so print the approximate previous number which is modulo 2 ,thats 0 then print 0
i = 2 if( 2 modulo 2) the condition is true so print 2
i = 3 if( 3 modulo 2) the condition is false so print the approximate previous number which is modulo 2 ,thats 2 then print 2
The question I am finding a formula that can return the approximate previous number which is modulo M for a number not modulo M
I hope it's clear
for example if I say 3 and M=2 the formula must result to 2
mathematic formula
if it's not clear again I will re-explain
20:04
@AliGajani they could be called war machines.
Define: machine
A device designed to perform a task
A machine uses power to apply forces and control movement to perform an intended action. Machines can be driven by animals and people, by natural forces such as wind and water, and by chemical, thermal, or electrical power, and include a system of mechanisms that shape the actuator input to achieve a specific application of output forces and movement. They can also include computers and sensors that monitor performance and plan movement, often called mechanical systems. Renaissance natural philosophers identified six simple machines which were the elementary devices that put a load into motion...
20:45
Is Cantor-Schröder-Bernstein (|A|=|B| iff. |A| ≤ |B| and |B| ≤ |A|) equivalent to the axiom of choice? My Linear Algebra textbook says so but I thought you could prove it without referring to the axiom of choice?
No, Schröder-Bernstein is a theorem in ZF
(What definition does that book use for |A|<|B|?)
Isn't that usually just used to mean that there's an injection from $A$ to $B$ but no injection from $B$ to $A$?
As I understand it the author defines it as |A| ≤ |B| if there is a injection from A to B.
Which is the standard definition right?
Yes
(I was asking because if you do it with surjections for some weird reason then you actually need AC)
Okay that's interesting
So what happens in general to the comparison of the cardinalities of infinte sets without AC?
Because you need AC to say that if f:A→B is surjective than there is a g:B→A which is injective
@JannikPitt A mess. Cardinalities are not necessarily linearly ordered, that is you can have sets A and B with no injections from A into B and no injections from B into A
And a lot of cardinal arithmetic stops working. You can't say $|A\times A|=|A|$ for example
Anyway that claim about Schröder-Bernstein is really weird, which book is it from?
Just read the passaga again and I'm just really dumb. The author states two theorems in succession and then remarks that one of them is equivalent to the axiom of choice and I missread and thought he meant Schröder-Bernstein-Cantor, sorry haha
21:08
Hello...
Which theorem is the other one?
I've got what I think is a kindof complicated, question; at least, it's beyond my math knowledge. Would it be okay to ask?
Just ask; don't ask to ask ;)
@Alessandro The comparison theorem which you mentioned earlier
Ah, yeah, that's definitely equivalent
There's really a lot of things which are equivalent to AC over ZF
21:11
@AlessandroCodenotti I never really understood the phrase "equivalent to AC", does it just mean "requires AC to be true" or what?
It means that the theorem is true iff AC is true
An assertion A is equivalent to AC (wrt ZF, this is assumed if not specified) if ZF+AC proves A and ZF+A proves AC
Ah right lol
For example "every set admits a group structure" is equivalent to AC, while Hahn-Banach is strictly weaker (meaning that ZF+AC proves HB, but ZF+HB doesn't prove AC)
Interesting, I know absolutely no logic or set theory beyond basic stuff that everyone knows
21:16
Thanks. Suppose I have 4 particles travelling through space. Their speeds can be measured in relation to each other, that is for example that the speed of particle A to B may be in the ratio 1.8, B to C 0.4, etc. The ratios are constantly changing due to acceleration or deceleration from outside forces which we don't need to know about.
Is it possible to say which particle is the fastest at any point in time?
What made me think about it in my slow-witted way is the current cryptocurrency craze. If I could track 4 currencies against the dollar then I wonder if it's possible to figure out where the greatest value lies at any point in time.
It seems like calculus should have something to say about it, rate of change...
@ÍgjøgnumMeg they're not very popular areas, but they are the ones I'm most interested in
@Alessandro takes pride in being unpopular (and dangerous)!
Hi Alessandro :)
Well if they were more popular maybe there would also be more undergrad courses about them
21:23
@AlessandroCodenotti I did a small amount of Model Theory at a summer school, which I found VERY interesting, especially as there were objects called Ultrapowers and Ultraproducts and Ultrafilters
@GG de la DGSE : alors toujours avec ton esprit grincheux ?
I've only heard about those, I wanted to go to the model theory course lectures next semester but it overlaps with my courses
@Alessandro: It's sort of a vicious circle. Most departments don't want to hire logicians because they don't have people working in that area and they'd rather build up those areas (and be of greater support to the new person hired).
Hi all, this is written in the book : ''If X ⊆ End(V ) and W ≤ V , then W is called X-invariant if, for any A ∈ X and any w ∈ W, one has Aw ∈ W, i.e., XW ⊆ W.'' First of all, what is X here? and secondly, endomorphism is a homomorphism from one object to itself, how is X included in a map? I am talking about X ⊆ End(V )
$X$ is a collection of maps.
I'm not fond of the notation, but what you wrote is pretty clear. $X$ is a subset of the set of all maps from $V$ to $V$.
21:25
Hey @Ted!
For every $T\in X$, $T$ maps $W$ to $W$.
@AlessandroCodenotti It was used to prove the compactness theorem, I found it most interesting because a lot of constructions I've heard of from other areas of mathematics could be expressed by ultraproducts
dopey question, I guess.
hi Demonark
Hi @Ted
21:26
Yo @Balarka
@Gerard: Are you worrying about relativistic issues and what an observer can or cannot observe? Or are you asking a straightforward math question?
Find $p$ prime number with : $P(x)=x^{\frac{p-1}{2}} \mod p$, $P(1)=P(2)=P(3)=...=P(22)$, and with :
$p\neq 11013658661829071$ and $p \neq 197521$.
Eyo @Daminark
rehi @Balarka
I'm thinking about Sobolev spaces
Or rather thonking
21:26
I understand the reasoning behing that, but it's still rather disappointing. My uni is rather small and we have a grand total of two logicians (and one is actually a type theorist)
@Balarka: Are they thinking back?
@GG de la DGSE : tu connais mon prix, on se tient aux jus. Tchuss
@TedShifrin You mean End(V) is the set of all maps from V to V? now makes sense, thanks
@Alessandro: At UGA the only courses on mathematical logic were taught in the philosophy department, in fact.
Yup @Leyla.
Ted, I think it must be a straightforward math question.
21:27
@Ted not yet ;3
@AlessandroCodenotti Thats unfortunate, I'm in the same position; I'm at a uni with ZERO number theorists, and I'm writing my undergrad dissertation on Algebraic Number Theory, with a supervisor whose research is mainly in computer algebra
@ÍgjøgnumMeg oh, yeah, I've only seen it proved via completeness, but I've been told that it can be proved directly
Sobolev spaces are cruel folk, they rarely think about you
I doubt Eric would agree, Demonark.
21:28
Number theory isn't very big here either, we only have one ANT course in the undergrad curriculum
Do most places have more?
Sobolev spaces are a pain in the ass
Well, the undergraduate curriculum in Europe is quite a bit more advanced than most curricula in the US ...
@BalarkaSen are you thinking about anything good?
@AlessandroCodenotti That sucks, we have no number theory courses on our curriculum except for a weird 4 week course at the start of the first year where we were taught about RSA, but never in the context of number theory, the course was just called "intro to modern mathematics" so nobody knew it was number theory hahaha
21:30
@0celo7 I'm peeking at Evans
Analysis is hard to read for us mortal men dawg
false modesty from a topologist is really quite annoying
It's always a chunk of baggage
It's easier for me to read than most of the topology you do.
@TedShifrin That's because we take only math or related courses instead of having humanities requirements and stuff, I think
@Alessandro: I think more is expected of students, too, generically.
In the US the best undergraduates take a number of graduate courses (if they're at universities, rather than small colleges).
21:32
@TedShifrin I think the US graduate programs are much nicer than the standalone masters degrees we have here though, if it weren't so far away I'd quite like to go to a US style grad school
@TedShifrin Having never been in a US school (or in the US) I can't comment on this, but I trust your experience
If you trust me so much, why are you always trying to run me over? :D
its pretty late and i feel dumb for not knowing how to dissolve "x + 3.5x^0.9 = 100" to x
Let me put it another way. At time T1 if A is worth 1.2B and B is worth 0.4C and C is worth 3D, then at time T2 if B is worth 0.6C and C is worth 2D then how much is A worth in relation to each of B, C and D. It seems like it should be reasonably easy, maybe I just have to do more thinking about it...
21:34
@Yannik: You mean solve, not dissolve :)
@BalarkaSen what is wrong with evans?
yeah solve sorry
@Yannik: I think the only hope is to do it numerically (Newton's method).
@Gerard, you can relate everything to D at time 1. At time 2, you omitted to say anything about A. Is it still worth 1.2B?
21:36
@TedShifrin "fidarsi è bene ma non fidarsi è meglio" says an Italian proverb, trusting is good but not trusting is better :P
really ... maybe then i made a mistake because this is what i was getting when trying to solve a simple lagrange multiplier exercise
Doesn't sound right, @Yannik.
Although I do recall that in one textbook we used such things occasionally did happen in exercises.
@Alessandro: Just as I suspected.
@AlessandroCodenotti Is "fidarsi" a noun or reflexive verb? lol
In infinitive form, you mean?
Yeah, something like the "infinitive + se" you would have in Spanish
21:41
how do i write LaTeX in here ? just staring with $ and then the code ?
top right corner
Writing in LaTeX and seeing it compiled are two different things. Yes, enclose math in dollar signs and use the appropriate syntax.
Sup chat
$ \frac{7x^{0.2}y^{-0.3}}{10} = 2x^{-0.8}y^{0.7} $
yes i know, there is rendering involved ^^
Oh, you did your algebra wrong with exponents.
This turns out nicely.
21:43
@ÍgjøgnumMeg it's a pronominal verbs which takes a reflexive pronoun, something like sich schämen in German I suppose, it's constructed lie a reflexive verb but it doesn't have a reflexive meaning
Multiply both sides by $x^{0.8}y^{0.3}$.
@Ted had a fantastic ten course tasting menu yesterday. There was this udon with this cod roe sauce that made me transcend
@EricSilva: You're also enamored of "this" ... where's poor ol' "that"? :P
Sounds interesting :)
@TedShifrin oh yeah because those are negative right
@AlessandroCodenotti That's cool, you can also say "sich trauen" in German which is something like "to trust oneself" but it means more like "to dare" as in "I daren't ask him" or something
21:46
well that explains that my approach was wrong
I love it when this room disintegrates into language and linguistics exchange :P
do you prefer the meme exchange mode
Sorry! lol
No, @ÍgjøgnumMeg, I wasn't being sarcastic.
21:47
@0celo7 Nothing, per se. I like it so far
@TedShifrin Oh good, I rarely get to talk about it with anyone else
@TedShifrin Darn, and I was ready too
I don't understand his explanation of what the closure of $C^\infty_c(U)$ in $W^{k, p}(U)$ is though
But that's okay
You need traces to understand that
21:50
@Balarka disappears without a trace.
I see, we have some verbs that support both a reflexive and a not reflexive construction, like preparare/prepararsi: sto preparando una torta =I'm making a cake, mi sto preparando una torta =I'm making myself a cake, while some verbs are always constructed reflexively without having a reflexive meaning, like impegnarsi =to put effort into something
Morally $W_0^{k,p}$ functions "vanish on the boundary"
And for many domains that "don't have a boundary", $W_0=W$
@0celo7: So it doesn't indicate vanishing at infinity in the non-compact case?
such as $\Bbb R^n$ or complete manifolds with curvature bounds (and bounds on $k$ and $p$ for manifolds)
@TedShifrin I don't have a good answer for that. Does it?
21:53
I always thought it did.
There's Morrey's inequality, but I think you need a bit more to get decay.
In a lot of mathematics, the 0 subscript indicates compact support, in fact.
I know.
They certainly don't always have compact support in the noncompact case.
But I haven't thought about Sobolev stuff in many a year.
@TedShifrin From my hyperbolic PDE notes, if $s>n/2+1$, then $H^s$ functions vanish at $\infty$
21:55
Balarka learning sobolev theory is the best thing that's happened to this chat
That's because you've abandoned Bryant, @EricSilva. Otherwise that would win.
Hey I'm only trying it because I want to understand the intricacies of the shitty answer I posted
@EricSilva I have a feeling your opinion here is biased
I haven't! It's gonna be one of my primary summer projects
@BalarkaSen You only need single variable stuff for that
That's why I recommended Brezis
21:56
Which shitty answer this time, @Balarka?
@AlessandroCodenotti pssssssssshhhhhhhhhhhhhhh nah
@Ted Oh I haven't told you?
Somewhere in all that, did someone tell him to think about the proof of Hopf-Rinow?
@Ted in fact, on Bryant, I now have a nicely printed and bound copy of EDS to fawn over this summer
I didn't
21:58
That's way heavier than the paper we were supposed to be working on.
Is that a constructive way to get a geodesic? I guess so
that paper is so hard :(
@BalarkaSen There isssss and I will write it up one day
i need to get back to it when im better
Remind me when I retire
just heat flow it
21:59
@Balarka: If you know how to measure distance (i.e., the metric space structure), yeah.
i was trying to GRADIENT FLOW CONDDAMIT

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