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11:00 PM
Unless I dreamed it, I received a text message that then disappeared.
Stay safe, guys. There's frightening technology about these days.
 
From the first equation I can go to the second equation using Fubini, correct?
actually is not even Fubini is just, taking constants out of the integral
 
Yes.
 
Sure, you can delete text messages, @Symposium. What doesn't work is sending one, realizing you sent it to the wrong person, and then trying to delete it on your end to "unsend" it.
 
@TedShifrin that works in whatsapp
 
@TedShifrin I worded that wrong. I meant to send one, then delete it.
 
11:10 PM
interesting, Leaky. I don't use that. It works provided the person hasn't already received and read it?
You said you received one, @Symposium.
But never mind.
 
@TedShifrin it works regardless, but it says "this message has been deleted"
 
Yeah, I received one, then it disappeared without me deleting it.
 
Oh, so is this on Whatsapp? It won't work with SMS or iMessage.
 
right, this is a relatively new feature of whatsapp
 
@LeakyNun Because WhatsApp is basically iMessage service (wifi/internet based). I'm pretty sure they use different wave signals or something to normal texts!
 
11:13 PM
Definitely can't do it on iMessage.
 
I was very sleepy. Probably a dream. Weird because my dreams are usually about invading neighbouring countries.
My standards are falling!
 
LOL
 
xD
 
Oh, interesting what I just found by googling. Tiger text is an app that allows you to unsend a text, but both parties must have it installed.
 
@TedShifrin so we all know what $GL_n(R)$ is, for a given commutative ring $R$
but the question is, what is $GL_n$ without the $R$?
well, you can define $GL_n(R)$ as a subset of $R^{n^2+1}$ satisfying some polynomial equations
 
11:17 PM
Oh, that depends on the context, but usually working over $\Bbb R$ or $\Bbb C$.
Why the +1?
Oh, right, to get an equation.
Sorry.
 
if you define it as a subset of $R^{n^2}$, it would be those not satisfying a polynomial equation instead
that's an open embedding
 
Yeah, yeah, I got it.
 
in the $n^2+1$ case, it's a closed embedding
ok
so let the determinant polynomial be $p \in \Bbb Z[X_{11}, \cdots, X_{nn}]$
 
@Leaky I think I sorted that thing earlier; I can send $f$ to $\varphi \circ f \circ \varphi^{-1}$ and this should be an isomorphism with an obvious inverse
 
@ÍgjøgnumMeg right
 
11:19 PM
woo
 
@TedShifrin then $GL_n(R)$ can be identified with $\operatorname{Hom}(\Bbb Z[X_{11}, \cdots, X_{nn}, T]/(pT-1), R)$ by some universal properties of quotients and polynomial rings
 
@TedShifrin Interesting. That sounds like the app Signal; the one where all the criminals moved to after it turned out the feds had a backdoor to WhatsApp.
 
I don't usually think about algebraic geometry over non-fields, but, sure, Leaky. Why are you telling me all this? It's not like I don't know it.
 
oh you know algebraic geometry :o :D
sorry
I thought you do differential geometry
 
The student testing the master! xD
 
11:23 PM
well, I'm going to have to stand in front of my poster explaining stuff on friday
 
Forgive me, I've seen too many Karate/Kung Fu films.
 
how do you think I sounded @Ted
 
Depends if you're explaining to experienced experts or to noobs.
For people who haven't done a lot of algebra and know a lot of notation, it's hopeless.
 
I'll be explaining to two professors who know (know as in research on) the thing I'm going to present
but I'm not supposed to be doing something that hard, so there's that factor
 
Did they tell you what level to give the presentation? For them? For non-experts?
 
11:25 PM
it's mainly them asking me questions
it's really a coincidence that they know what i'm talking about
 
So did you just summarize the whole thing, or was that just the beginning?
 
well that's something half related
as in, I should have mentioned GL_n once in my poster
 
I guess you should state stuff in a sophisticated manner, but be prepared to back up and fill in details. Also, as I said to Alessandro, I'm a big believer that you should have interesting examples, not just abstraction.
 
I do have examples
do you think I should talk about the examples?
 
Generally, showing command of examples you've figured out is way more impressive than regurgitating abstract proofs.
 
11:27 PM
no, I won't be proving anything
 
Hi @Ted @LeakyNun and everyone else
 
@MatheinBoulomenos git'n tug
 
hi @Mathein
 
hey @Mathein
 
yeah it was a quite eye-opening experience for me
 
11:31 PM
Sad that I got only one star for saying we need more insanity checks.
 
Hi @ÍgjøgnumMeg
 
"what is $GL_n(R)$ without the $R$"
15 mins ago, by Leaky Nun
but the question is, what is $GL_n$ without the $R$?
 
So are you talking about a functor?
 
mind = blown
@TedShifrin right
 
Given who's asking the question, that has to be the answer. shrug
 
11:32 PM
I never recognized it as a functor
 
@LeakyNun yeah I always said that you should mention in LA that the determinant is a morphism of group schemes
 
it blew my mind
 
smacks Mathein
 
protects Mathein
@MatheinBoulomenos can you explain me the three types of sheaf induced by a map?
 
@LeakyNun I'm not in the mood, I'm going through the stuff for my presentation tomorrow (I'm already done with preparations, but I want to be sure)
 
11:34 PM
oh ok
 
And Alessandro presents Friday. ... This is fingernail-biting chat time.
 
vielen Erfolg @MatheinBoulomenos
 
@LeakyNun dankeschön, dir auch!
 
and I present Friday
 
Hey everyone!
 
11:40 PM
Hey @Daminark
 
hi @Daminark
 
How's it going?
 
everyone has a presentation now
MB presents tomorrow, Alessandro and I present Friday
 
@LeakyNun Is this part of your course or summer thing that you're doing?
 
part of my course
 
11:41 PM
Hi @Daminark
 
well good luck to me, I can't stand for more than a minute
or sit, for that matter
 
Hi Demonark. Did you see I referenced you in a ping earlier?
 
how do I see that $\Bbb Z$ is not profinite-complete?
I mean, the intersection of the finite index normal subgroups is (0)...
 
@LeakyNun Haar measure or Baire, pick one
just show it for countable infinite groups in general
 
what was the criterion again?
 
11:44 PM
profinite groups that are not finite are uncountable
 
no, I mean my criterion
41 secs ago, by Leaky Nun
I mean, the intersection of the finite index normal subgroups is (0)...
surely this is a half-truth
 
That gives you an injection into the profinite completion
 
is mine not an injection?
 
oh, what's the remaining half of what I'm quoting?
 
11:46 PM
@LeakyNun You're a first year right?
 
yes, as stated in my profile
 
@Symposium: We have a number of advanced students in this chat ...
 
my knowledge is nothing compared to the other people here
 
Including a couple high schoolers who just graduated and are already graduate students :P
 
@LeakyNun I don't know which criterion you're refering. The half that you mentioned is just the definition of the profinite completion, basically
 
11:47 PM
Oh, that reminds me ... @MeowMix: You were supposed to email me?
 
@MatheinBoulomenos oh, if the intersection isn't {0} then we can't complete it?
 
@TedShifrin I'm convinced @MatheinBoulomenos is a seasoned expert, not an undergrad.
 
@LeakyNun we can complete every group
 
xD
 
@MatheinBoulomenos was ist die Komplezion des Q?
 
11:48 PM
@LeakyNun 0
 
ajskdfajsldfjakjakdsjfal????
 
Symposium: We've had this issue with about a half dozen chatters over the past 4 years.
 
the canonical morphism into the completion is injective iff the intersection of all finite index normal subgroups is 0
 
I'm only joking, of course.
 
are you telling me that there is no finite index proper subgroup of Q so it's 0?
 
11:49 PM
@Symposium I'm an honest-to-goodness undergrad
@LeakyNun yes
 
ok I pulled that out of my gut
need to clean it a bit
 
@Ted I'll check it out now!
 
It's algebra/analysis, Demonark.
 
@LeakyNun just show that any morphism from $\Bbb Q$ to a finite group is $0$
 
@MatheinBoulomenos oh, done
indeed, the categorical point of view is always better
 
11:51 PM
Even I can do that one, Mathein :P
 
I was going to consider cosets, you know
 
@LeakyNun always? I'm not sure I agree with that. It's nice to have it, yeah, but
@TedShifrin I would've never doubted you could
 
I can't do most of the stuff you guys talk about.
 
:o
 
I like to believe there are still a few things I know you don't, though. :)
 
11:53 PM
of course there are!
 
@TedShifrin for sure
 
@MatheinBoulomenos what next, Q/Z and Z-hat are duals?
 
@LeakyNun that's a really fun exercise
 
@MatheinBoulomenos Hehe! I guess by now you know what area you want to get into eventually?
 
Hmm, I may think about that at some point. At the moment I'm a bit busy because I'm trying to read around a bunch of stuff and ideally have a topic idea (at least to start with) next week
 
11:54 PM
@Daminark about Q/Z and Z-hat being dual?
 
try even to prove that $\widehat{\Bbb Z}$ is the endomorphism ring of $\Bbb Q/\Bbb Z$
@LeakyNun yes
 
OK, Demonark. I thought you might have run into that.
 
@MatheinBoulomenos yeah let's use Pontryagin agin
 
When I hear categorical, the categorical imperative comes to mind. That's how you know I'm more of a philosopher than a maths person.
 
there's a categorical and a concrete approach, I even prefer the concrete in this case
 
11:55 PM
DogAteMy!!!
You semi-permanently moved across the ocean?
I'm stunned, @Mathein.
 
@MatheinBoulomenos whaddayaknow, Q=limZ, so Q/Z = lim Z/nZ, done
 
We didn't talk much about commutators in functional, sadly
 
@Daminark yeah, the commutator of Q is 0
 
I suspect this relates to traceless and operators of trace class or somethin' ...
 
@Symposium I'm not sure actually. I would like number theory, representation theory or noncommutative algebra. Of those three, number theory seems the most probable right now, as that is what a lot of pros here in Heidelberg do.
 
11:57 PM
@MatheinBoulomenos not AG?
it's treason then
 
@LeakyNun lol I'm talking about a different question at the moment
 
@LeakyNun I see it more as a tool for number theory :P
 
:P
 
You saw my comment, Demonark (since I don't ping you) :P
Any luck finding your laptop?
 
@MatheinBoulomenos amirite
 
11:58 PM
@LeakyNun Q=lim Z? you can do that, but you have to be careful with the transition maps and also with quotients and limits
 
Sadly, no, reported it to the police so we'll see what comes of it, but yeah for now I'll probably be working mostly off phone and library computers
 
ok, Q = lim (1/n)Z, how about that
 
I'm sorry, Demonark.
 
so Q/Z = lim [(1/n)Z]/Z
 
Thanks
 
11:59 PM
@LeakyNun why are allowed to pull a quotient into a colimit?
 
just check everything
 
Is this definition correct? $B_n$ is the unit open ball in $\mathbb{R}^n,$ with its center at $\vec{x}$, defined as: $B_n := \{\vec{y} : |\vec{y} - \vec{x}| < 1 \}$
 

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