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00:54
@pie That may be, but whoever is local to you is going to be able to give you much better advice than some random strangers on the internet. They are going to be much more familiar with your current situation, the situation on the ground vis-a-vis the academic environment, and so on.
Xander dropping wisdom
 
2 hours later…
03:11
What are some nocode solutions for writing extensive math docs?
03:22
I think that I am going to spend $70 on a piece of cardboard...
Hey @XanderHenderson
Hay is for horses.
What're you working on?
And why does this cardboard box cost $70
@Michael Because, apparently, that is the price the market has determined is fair and just.
The particular bit of cardboard that I am looking at is an alpha edition Prodigal Sorcerer, which is a gamepiece I enjoyed quite a bit in high school. But I don't know where all the cardboard I had in high school ran off to, so... :/
03:57
what do you mean by nocode? who writes math documents with code?
04:28
"Every closed, oriented 3-manifold $M$ contains a link $L\subset M$ such that $M-L$ is homeomorphic to a finite sheeted covering space of the Whitehead link complement"
@BenSteffan Surprising, did you know this?
i have become quite adapt at answering trivial questions.
 
6 hours later…
10:14
Consider $A_rf(x):=\frac1{m(B(r,x))}\int_{B(r,x)}f(y)\,dy$ where $B(r,x)$ is a open ball of radius $r$ and center $x$. Let $Hf(x):=\sup_{r>0}A_r|f|(x)$ (the Hardy-Littlewood maximal function). It is claimed that $$\limsup_{r\to0} |A_r(f-g)(x)|\leq H(f-g)(x).$$How? We have that $|A_r(f-g)(x)|\leq A_r|f-g|(x)$, but is $\limsup_{r\to0}A_r|f-g|(x)=H(f-g)(x)$?
10:34
I'm not sure if it makes sense to argue like this; $$|A_r(f-g)(x)|\leq A_r|f-g|(x)\leq H(f-g)(x),$$ so $\limsup_{r\to0} |A_r(f-g)(x)|\leq H(f-g)(x)$.
@BenSteffan Alright, thanks
"The map $aH \to bH$ with $ah \mapsto bh$ is a bijection since the mapping corresponds to left-mutiplying by $ba^{-1}$ and the inverse to left-multiplying by $ab^{-1}$." Why is this a valid argument?
10:55
Could anyone be so kind to explain me how to reach the final expression in here?

https://math.stackexchange.com/questions/4059325/how-to-understand-the-reduced-cost-in-simplex-method#:~:text=This%20has%20a,).

In pariticular, I'm referring to "We might as well extend the variable component to..."
Thanks in advance!
@ILikeMathematics why is not valid?
11:12
@SineoftheTime Where does it show injectivity and surjectivity?
11:27
@ILikeMathematics to prove a map is bijective, you can prove it's invertible
11:42
Oh, essentially we are finding the inverse map like this..
Thanks
How do we know that if we have a finite group $G$ and $a \in G$, then $\operatorname{ord}(a) = \begin{cases} \min\{n \in \mathbb Z_{> 0} \mid a^n = 1\} & \text{if such $n$ exists} \\ \infty & \text{otherwise}\end{cases}$ will always be finite?
Maybe by taking powers, we go some cycle and end up at a again, without having covered 1?
11:56
@ILikeMathematics if $a^n = a^m$ then $a^{m-n} = 1$
And such $m$ and $n$ must exist, since $G$ is finite
12:11
Distinct
12:48
@onepotatotwopotato I did not, but then I also don't know much about links and knots :)
 
1 hour later…
13:51
Hi guys! Guys, I have an angle $\alpha$ which lies in $[0,\frac{\pi}{2}]$. In my context, $\alpha=0$ is called horizontal and $\alpha = \frac{\pi}{2}$ is vertical. All angles in between are called "oblique", I guess ( oblíquo in Portuguese). But the cases $0<\alpha <\frac{\pi}{4}$, $\alpha =\frac{\pi}{4}$ and $\frac{\pi}{4}<\alpha < \frac{\pi}{2}$ also represent different cases. How should I call them?
derso that is interesting. i don't know of a common english term for this. surely someone (perhaps many people) have invented one but i have never heard it
"horizontal-oblique", "right oblique" and "vertical-oblique"? I don't like these :'(
the usual english term for an angle between 0 and 90 degrees is an 'acute' angle. even there i am not sure that there is established practice around the 0 and 90 endpoints
I wish I could use as cases of oblique angles something like: "acute", "right" and "obtuse", but these are misleading lol
@leslietownes And the context is, I'm working in a space which looks like the interior of a cylinder. Imagine it like a vertical cylinder.
Then, we have the fibers as vertical geodesics.
We also have horizontal one (in the direction of the basis space)
And then, things in between...
The "$45^\circ$" geodesics will have a kind of asympotic behavior, those with a higher angle than that, but not vertical, will go "more" in the vertical direction and those with a lesser angle "more" in the basis direction...
I don't know how to call them
The space is $\widetilde{SL_2(\mathbb{R})}$
it is traditional to include only one s in the word asymptotic :)
14:00
lol thanks
ordinarily i would not correct spelling, but occurrences of the substring "ass" are occasionally hilarious
core.ac.uk/download/pdf/14416005.pdf In here, they call these cases "$\mathbb{H}^2$-like", "separating light" and "fiber-like" directions, but I don't like them either lol
@leslietownes hahahahaah
i think you have identified a gap in the english language
There are also all the claims about Uranus
@SineoftheTime How should I call these angles?
"pointy" and "somewhat less pointy"
14:12
What about hoblique, roblique and voblique?
I hate it.
subacute and superacute
(joking)
LOL
supercute names
Well, if we look at $2\alpha$... It lies on $[0,\pi]$, with the separating case right in $\frac{\pi}{2}$...
less than 45 should be 'too cute' because when you multiply them by '2' they are still a'cute'
2
So, the cases I want to distinguish can be said in terms of $2\alpha$, using the names acute, obtuse and right angles
14:39
@leslietownes I fully endorse this proposal.
 
1 hour later…
15:58
Let $A_rf(x):=\frac1{m(B(r,x))}\int_{B(r,x)}f(y)\,dy$. In Theorem 3.18 in Folland's, which says that if $f$ is locally integrable then $\lim_{r\to0} A_rf(x)=f(x)$ a.e., it says that the result holds true for a continuous function $g$ everywhere. What I don't understand is, the inequality $$|A_rg(x)-g(x)|<\delta$$holds for $|y-x|<r$. Why does this mean that $A_rg(x)\to g(x)$ as $r\to0$? I don't see $r$ tending to zero in the statement $|y-x|<r$.
16:20
This is one of those Hardy-Littlewood things, right?
indeed
It's been a while since I've read Folland... I never get tired of the terseness of his exposition...
haha yeah, he is concise :)
Basically, if you an find some that does the job, you can find a smaller $r$ that still does the job.
If $|x-y|<r$ implies that $|g(x)-g(y)| < \delta$, then you will still have $|g(x)-g(y)|<\delta$ for any smaller $r$.
ah yeah!
and that is what $A_rg(x)\to g(x)$ as $r\to 0$ means anyway
thank you
 
1 hour later…
17:50
@leslietownes About the angles: one should call them space/light/time-like geodesics. Their model justify this terminology
$\alpha=\frac{\pi}{4}$ is the separating light-like cone direction...
mathematician discovers general relativity
lol @BenSteffan Not yet, for me it's just $\widetilde{SL_2(\mathbb{R})}$ geometry, I don't care for any physical interpretation (for now)
"one should call them space/light/time-like geodesics"
"I don't care for any physical interpretation"
(thinking emoji)
hahahahaha
I just needed to give them some names...
@Thorgott I've read and reread your answer about the relative colimits thing, and I've come to the following conclusion: "a $f$-colimit $\bar{p}: K^{\triangleright} \to A$ is initial among the cones $\alpha$ under $p: K \to A$ such that $f \circ \alpha$ factors through $f \circ \bar{p}$ (seen as a cone under $f \circ p$)"
I've wanted to say this because this agree with the answer you gave me, but I don't think this is the same as the last comment you've left under your question
17:57
I really liked @leslietownes suggestion of naming them "too cute", but maybe space/light/time like is a bit better
Because if I just use the last comment you have left, I don't quite get coCartesian lifts as an example (I think)
(sorry for still bringing this up, but I thought you'd like to know, also because this thing is driving me crazy)
@BenSteffan 'sup
not much, just understanding what I don't understand of higher category theory
sounds like progress
small progress is still progress
18:00
currently wrestling with the "hard" AT1 exercises and having, uh, fun
Have fun lol
it's good when you google the exercise statement and you find a proof for a stronger setting together with a comment by somebody saying that they don't think it generalizes to your setting right :^)
but maybe I'm misreading things
18:44
Hi
PD Is coming 2.5, not 7.5
Help pls
I fixed it
@BenSteffan AT1?
Algebraic Topology 1
(course name)
19:08
@BenSteffan whats in that course
this time around basic homotopy theory
this time around?
basic homotopy theory?
@Jakobian next year it probably will be something else
@SineoftheTime yes
as in homotopy groups, fibrations, cofibrations, homotopy excision, ...
@BenSteffan by next year do you mean the exact same course?
19:10
yes
why is that though
in the sense of "course offered under the same id"
because lecturers have different opinion on what to do when during the algebraic topology cycle
to be clear, what's being done right now is technically what's supposed to be taught in the course
are you some kind of supervisor for this course, or are you taking it rn
The former. I'm teaching one of the exercise groups.
@BenSteffan that makes sense with a field like that, full of novel approaches I imagine
19:12
yeah. people rush through the material to get closer to modern topics
what are the modern topics?
probably some category theory horrors
$\infty$-categories & stable theory, chromatic homotopy theory
@Jakobian correct :^)
equivariant stuff as well, maybe
19:16
@SoumikMukherjee do you like category theory?
no
not yet, maybe in the future I will
btw sine, Ding or Gukesh?
category theory for me is a necessary evil, but an evil nonetheless
$\infty$-category theory is (not) category theory
joking, probably Gukesh
19:18
@SineoftheTime lol, someone asked me Ding or Gukesh and I also replied with Magnus
Gukesh seems more psychologically stable, despite his age
@BenSteffan is it infinity minus category theory?:P
@SineoftheTime yeah
well it is also category theory but compared to category theory (which is easy) it is actually quite difficult :^)
...and it's topology at the same time so
If Ding wins, we need a new cheating scandal to make chess more popular :D
@BenSteffan ah okay
@SineoftheTime "let's start the procedure"-you know who
19:24
If Gukesh wins, do you think Magnus will take part in the candidates?
no I guess
I feel like Magnus only considers Alireza to be the next big name
In terms of surpassing Magnus
and imo he won't be 100% sure he'll beat Gukesh in a WC match, so he'll not risk
eh I don't think so
WC is a totally different chess format, I don't think Gukesh has the strength to dominate Magnus in WC format
19:32
not this year
but Gukesh can only improve, and Magnus is getting old :D
Bro is so good that he can casually call a world championship challenger lot weaker than himself
X4J
X4J
Let G be a group. Does Z(G/Z(G)) = {eZ(G)} necessarily hold?
My intuition is that it is false, trying to simplify it by looking for a counter-example within Inn(G) where a conjugation C_g conjugates with each other but isn't the identity
@SoumikMukherjee and no one disagree
X4J
X4J
Also the group cannot be abelian
Struggle to come up with counter example
@X4J look up upper central series
19:44
X4J if |G| = p^2 (p a prime) and G is nonabelian there will be order reasons why Z(G/Z(G)) isn't trivial
oh maybe there is no such group
anyway consider the case of p groups
the center could be trivial in that case, at least a priori
ah, nevermind
i was just thinking, G/Z(G) is also going to be a p group and probably have lots of normal subgroups
19:56
okay, so any non abelian p group would be an example, right? because p groups have nontrivial center?
X4J
X4J
20:11
I struggle to find concrete examples for non abelian p groups
I havent introduced to it yet
I've seen D4, D3
as non abelian
and I follow my intuition for other examples
try the heisenberg group over Z_3
X4J
X4J
oh
like 3x3 matrices
matrix groups are a really good source of examples for this kind of thing
but here $D_4$ would suffice, as would $Q_8$
yes heisenberg groups over Z_p are just a readily identifiable infinite family of examples, shoudl you want one
20:40
Let $R$ be a subring of a number field $K$. Assume $R$ is finitely generated as a $\mathbb Z$ module. One can argue that the rank of $R$ as a $\mathbb Z$-module is equal too $[Q(R):\mathbb Q]$, where $Q(R)$ is the quotient field of $R$, from
$$
R\otimes_{\mathbb Z}\mathbb Q=Q(R).
$$
I would like to know why this equality holds. The inclusion $\subset$ is clear, but the inclusion $\supset$ not yet. If I can argue that $R\otimes_\mathbb Z\mathbb Q$ is a field, then I'd be done
 
1 hour later…
21:51
For anyone waiting with bated (or, for the fish lovers, baited) breath, I did pull the trigger, and spent $70 on a piece of cardboard. It is not the most expensive piece of cardboard I own, but it is the most I have ever actually spend to acquire a single piece of cardboard.
(The most expensive pieces of cardboard I own are the 1955 Sandy Kofax and Hank Aaron cards I inherited from my father---both PSA 6s).
22:03
financially responsible mathematicians would have spent the $70 on chalk instead
it's much tastier than cardboard anyways :^)
@BenSteffan Have I ever told you about the piece of cardboard I ate in 1997?
I played a lot of Magic: The Gathering in high school, but before I left for Russia, I tried to sell a large part of my collection---I wasn't really playing much anymore, and some of the cards were, theoretically, worth a lot of money.
So I tried to sell one of these. Scry Magazine said that I should be able to get around \$300 for it. I couldn't get anyone to pay that much. I couldn't get anyone to pay half that much. So, at an event, I announced that if no one would give me \$150 for it, I would eat it. I think people really wanted to see someone eat a Black Lotus, so that's what ended up happening.
I kind of regret that, now. On the other hand, I think that my ex-wife ran off with all of the other Magic cards that I had left over from high school, so I'm not really out any money that I wouldn't otherwise not have.
22:19
you ate a black lotus???
Yes.
It did not taste good.
I am impressed
22:34
@ShaVuklia Let $r$ be in $R$. Then there is a polynomial equation $P(r)=0$ with $P(0)\ne 0$
In particular $\frac{P(r)-P(0)}{r P(0)} \in R\otimes_{\mathbb Z}\mathbb Q$ and it is an inverse of $r$
@XanderHenderson "Black Lotus is toxic; consuming it can cause nausea, vomiting, and abdominal pain."
I am not an expert, but it sounds dangerous
@Jakobian I drank dishwashing detergent once (not so long ago)
fantastic
@Jakobian The packaging clearly said that it was non-toxic.
Came back home after a jog, thirsty, and took a huge gulp from the first gatorade bottle I've seen, which seemed to have water in it...
I forgot I left the bottle with detergent in it :P
22:49
happens to the best of us
@Derso how much
hopefully it can't destroy your organs
@Jakobian Only one gulp, but it was fast and huge
I don't recommend it. Maybe black lotus tastes better
dish soap isn't so bad
after all you consume tiny amounts of dish soap every time you eat off a plate, or using cutlery, etc.
true, it should be somewhat digestible
@BenSteffan What? Don't we rinse these things after wash?
23:01
rinsing is a process of dilution
also some people don't, apparently
and manufacturers have to account for those people
23:56
@BenSteffan how different are AR's for normal and metrizable spaces?
absolute retracts
I heard that the theory is the most satisfactory for metrizable spaces, but there also exist ones for normal spaces, and I am wondering

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