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12:00 AM
yes
 
:)
and closed ball just changes so it has its limit points which means we just have \leq instead of <, correct?
 
@LeakyNun do you want to see the concrete approach?
 
@MatheinBoulomenos I knew some that learned ridiculous amounts of maths like you as an undergrad. Similar interests too. I think they are studying the Langlands program at the moment, whatever that's.
 
@MatheinBoulomenos go ahead
24 secs ago, by Maximus
and closed ball just changes so it has its limit points which means we just have \leq instead of <, correct?
that's the right conclusion with the wrong reason
@Symposium i'm doing a poster on local langland for tori on friday
but ok i don't know much about langland in general
 
what do you mean wrong reason
 
12:02 AM
@LeakyNun but that's not much more than local class field theory + some Galois descent, isn't it?
 
indeed
so ok what i'm doing is nothing
so, that's that
@Maximus it means wrong reason
 
nah, that's really impressive and I think you know that :P
But I think most people think more of $GL_n$ or something when they hear Langlands
 
indeed
 
@LeakyNun Sounds interesting!
 
@Maximus it is not true that the closure of the open ball is the closed ball
@MatheinBoulomenos could you tell me what commutative, connected, and reductive mean?
 
12:06 AM
I see, it it because you can define random limit points?
along with the open ball?
 
not really
 
@LeakyNun all three together mean a torus :P
 
@MatheinBoulomenos vielen Dank
das ist sehr benuztlich
und hilflich
ich habe viel gelernt
 
@LeakyNun I know, you're welcome
 
@Maximus (I believe that) the closure of the open ball must be contained inside the closed ball
since the norm should be continuous
right, of course it is continuous
@MatheinBoulomenos is the metric continuous?
 
12:10 AM
@LeakyNun yes, reverse triangle inequality tells you that
 
ok
 
@LeakyNun I'm going to prove the thing with the endomorphism ring. Let $\phi \in \operatorname{End}(\Bbb Q/\Bbb Z)$.
Then for any $n \in \Bbb N$, we can look at the image $\phi(\frac{1}{n})$. Since that will be $n$-torsion again, and every $n$-torsion element can be uniquely written as $\frac{a}{n}$ with $a \in \Bbb Z/n\Bbb Z$, we get an element $a_n \in \Bbb Z/n\Bbb Z$. We can do this for every $n$. If $kd=n$, then $\frac{d}{n}=\frac{1}{k}$, so this gives us some compatability between $a_n$ and $a_k$ if $k \mid n$. If you check the details, this is exactly the compatability you need that $
The proof that $\Bbb Z_p$ is the endomorphism ring of $\Bbb Z[1/p]/\Bbb Z$ is basically the same
 
what is the topology for torsion groups?
 
@LeakyNun discrete
 
und was ist der Funktor that sends Z to Q/Z?
 
12:12 AM
I'm just talking about the endomorphism ring as abelian groups
 
are you telling me that Q/Z has the discrete topology :o
yeah i'm just asking it in general
 
@LeakyNun I'm not following. There are many functors that do that
 
I mean, can I torsion a group like I profinite a group?
 
@LeakyNun you can just take the torsion part, that would be the analog
 
I mean, take its normal subgroups, do something
 
12:14 AM
No, I don't think anything like that works
 
ok
 
Torsion groups are direct limits of finite groups, so basically unions
 
ok I'm convinced that it will work
provided that sending Z to Z-hat is a functor
 
the quotients don't form a direct limit
 
is it?
 
12:15 AM
@LeakyNun you mean profinite completion? You can't just describe a functor by saying what it does on one object
 
-2
Q: How can I do the last question?

JoeI have done the first 5 parts, but I can't seem to get the last one right. Can someone help me figure that one out? Thank You!

which 5 parts? lol
@MatheinBoulomenos well yes that's what I mean
 
The analogy between torsion and profinite is pretty complete.

Torsion groups are direct limits of finite groups, profinite groups are inverse limits
The finite subgroups form a direct system, the finite quotients form an inverse system.
So what you can do is you take the direct limit over the finite subgroups, that's just the torsion part of a group, that's the right analog for the inverse limit over all finite quotients
@LeakyNun you can do that just by using the definition of limits (multiple times)
 
ok, now take that functor
and then just Pontryagin it
then it should do what I want it does
sending Z to Q/Z and whatnot
 
ah, sure you can do that
 
@MatheinBoulomenos are you sure they don't
 
12:20 AM
pretty sure, yes
For Z they do, but I think that's more of a coincidence
 
interesting
I thought finite abelian groups are noncanonically isomorphic to their duals
 
that's true of course
I wouldn't call that "taking the colimit over the finite quotients" though
if you do that, you just get the Pontryagin dual of the profinite completion again
 
I see
 
@LeakyNun taking the Pontryagin dual of the profinite completion may also be described as just taking all homomorphisms to $\Bbb Q/\Bbb Z$ with finite image
 
finite because locally compact hausdorff -> discrete?
 
12:29 AM
yes
 
I also need totally disconnected I suppose
 
compact -> discrete is enough
 
to be safe :P
 
morphisms from the group with finite image are the same as morphisms from the profinite completion with finite image
 
wait what
 
12:32 AM
having finite image means that it factors over a finite quotient
 
and then use the projection?
 
for one direction, you just compose with the canonical morphism into the profinite completion
@LeakyNun yeah, exactly
 
genau
 
There are other ways to get $\Bbb Q/\Bbb Z$ from $\Bbb z$, you could also say that it's the cokernel of the canonical morhism into the injective hull
 
the injective hull?
 
12:36 AM
it's the "smallest" injective module containing a given module in some well-defined sense
but I need to sleep now
See you!
 
Schlaf gut, @Mathein.
 
@TedShifrin danke!
 
12:53 AM
See you!
 
1:09 AM
God it's gonna be hard to decide what to do as a summer project
 
1:29 AM
Good evening all
 
 
1 hour later…
2:30 AM
Me: "oh yeah I study math"
My friends: "Great, you can help me build my D&D character!"
 
lol
I've actually thought about some of the distributions and whether they have nice closed forms
in particular, things like
"roll 2 20-sided dice and keep the highest"
2d20kh
or roll 4d6 keep highest 3 , 4d6kh3
 
That seems like it comes up quite a lot more in 5E than 3.5 because of advantage/disadvantage.
 
yes exactly
I play 5e :)
i'm not even sure how to find things like E[2d20kh]
certainly $\langle 1d20 \rangle < \langle 2d20kh \rangle < 20$
but we can certainly do better than that
 
@Fargle lmao
@jonem hello!
Also hey @GFauxPas!
 
DAMI!
 
2:42 AM
How are you, @Daminark?
 
hm
 
Doing alright, how about you?
 
what do you think about those dice rolling variables
 
Alright as well. Now that I'm finally done helping someone build their D&D character (and in their defense, I'm their DM, so...), I'll probably do some G-P
@GFauxPas I'm not sure. My probability is far from fresh.
 
G-P?
geometric progression?
 
2:49 AM
Guillemin-Pollack
Differential topology
 
my answer didn't make sense so I'm glad it wasnt right
what campaign setting?
 
D&D, nice.
 
mine is custom made by the dm
 
quick: what's the average roll of 20d1 is?
 
$\pi$
 
2:50 AM
snap
 
@GFauxPas Forgotten Realms. I've been familiar with it since childhood so it was a natural choice.
 
nice
i've read a ton of Salvatore's books
 
a d1 is in the shape of a sphere, of course.
 
a d1 is a point mass
or maybe that's a d0
 
1 dg?
haha
what would be the shape of a d0, do you suppose?
nvrmind
so what dice-rolling variables were you looking for?
 
2:56 AM
4
A: How does rolling two dice and taking the higher affect the average outcome?

Drunk CynicProbability The answers provided effectively cover the probability for every result, 1 through 20, for advantage/disadvantage with 2d20. For completeness, the probabilities follow: Expected Value When rolling 2d20, and keeping the Maximum value from each of the 400 permutations, the expect...

 
i was wondering about that
 
Sorry I was out for a sec, and @Fargle G-P sounds like a good plan!
Out of curiosity, what prompted the Hatcher -> G-P switch? Was it just wanting to change things up a bit?
 
3:12 AM
Yeah. I wouldn't call it a total switch--I'm just going to go back and forth whenever I get too terribly stuck.
 
Good plan, taking a break and doing something else has helped for sure
 
What about $\langle 4d6\text{kh}3\rangle$?
roll 4 6-sided dice, pick the highest 3 dice
guess i have to read that question in the rpg.se and see how it applies to other situations
 
Mmm... D&D...
I haven't done that is... 20 years...
I hear that they got rid of thac0, which makes me sad :9
 
thac0 was dumb
 
on that note, as tired as I am of hearing about Alexa and all that, you have this: amazon.com/kp-Roll-Dice/dp/B07456M6WB
 
3:24 AM
I would agree with you, @GFauxPas
 
which is kinda fun
 
but then we would both be wrong
 
lol at the second review semi
 
haaah
i played a little D&D when I was younger, but I never really had the community for it
so my knowledge is all pretty second hand and non-mechanical
 
Same, until recently.
 
3:28 AM
(though somehow I know a crapton of WH40k lore from way too much wiki diving, despite never having touched an actual set)
 
Long ago I played like 2 sessions of D&D in high school and that's it. Much more recently I played a Pathfinder campaign with some friends, and now I'm a DM.
 
well thanks to the internet you can always find a community for it now :)
 
though the 'if the emperor had text-to-speech' video series helps as well
 
ew pathfinder
 
I dunno, I found Pathfinder pretty fun.
 
3:42 AM
so @Rudi_Birnbaum does this look good ?

Let $H \subset G$ be finite groups,
let $V$ be an irreducible representation of $G$, such that
when restricted to $H$, $V = \oplus V_i$.

Conjecture 1
If $W$ is an irreducible representation of $G$ that doesn't branch when restricted to $H$,
then the number of occurences of $W$ in the decomposition of $V \otimes V$ into $G$-irreducibles
is less than or equal to the number of occurences of $W$ in the decompositions of $V_i \otimes V_j$ with $i \neq j$ into $H$-irreducibles.
 
4:42 AM
I looked at S3 and those conjectures are wrong anyway
and still wrong even if you only look at the antisymmetric part
 
Isn't this maximum function periodic with period 1? My book says its not.
Is the thing in my book wrong?
Just need verification,I believe its wrong.
Someone please verify
 
it looks like it's only defined on $[-2;2]$
 
yes
 
then it can't be periodic
 
whyy
why should we check periodicity beyond domain
in its domain its periodic
 
4:55 AM
because then you could say it's also periodic with period $\pi^e$
and then you are going to get everyone confused
 
 
1 hour later…
6:11 AM
Seems I've got a bit of a back problem :(
 
D:
 
It's missing
 
Wait what? Where'd it go?
 
Oh wait nvm
I got it back
 
o_o
 
6:18 AM
Lmao
 
 
2 hours later…
8:36 AM
Laplace-Beltrami Kissing Prank (GONE WRONG!!!) — Balarka Sen 59 secs ago
 
@Daminark I saw it :P
 
rEmOvEd
 
[LEAKY EXPUNGED]
 
information got leaked out
 
Zee
Hello fellow human beings
 
8:43 AM
How rude
 
Yo
 
Zee
Shouldn’t you be sleeping Chicago boy ?
 
I should be, this is true
 
Zee
Guys guys , I just realized something
It’s probably trivial
 
Daminark is not a Cali boi. He's from Chicago tho, white boi
 
Zee
8:44 AM
But it blew my mind
Also , I don’t have a proof but it seems true
So you know how when you take the total derivative of a function
You get F_x dx + F_y dy
For example
Those dx and dy are the same as the columns in the Jacobson matrix , acting as linear functional in the sense of differential forms
Am I making sense ?
God I had too much beer
 
What you said is correct
But you meant Jacobian not Jacobson
 
Zee
Is that trivial ? Nobody ever told me
I just found it today while doing some problem and I could not believe it
Man , math is crazy
 
Most multivariable calculus courses don't teach you about tangent spaces
In that formalism this becomes clear
 
Zee
Well I don’t even remember calculus but I took a smooth manifolds course and I don’t remember this
Maybe couse the professor was a big pde guy and just used the course as an excuse to teach us a bunch of pde theory
Blach
Anyway , I’ll let you all be
 
9:00 AM
Hi everyone
 
HARRO EBURYNYAN
 
Slowly backs away from the chat
 
@BalarkaSen I finally finished chem.
And even if I failed I'm never doing it again.
I'll take the U.
 
@AlessandroCodenotti hey! Also lol
 
9:40 AM
"Morning" all
 
Morning
 
Want to help me with an abstract algebra thing that should be easy but I'm not seeing? @ÍgjøgnumMeg
 
o..o
 
@Alessandro I probably can't but I can try lol
else there are others here who know more!
 
9:51 AM
So I have $\mathcal O_K$ a Dedekind domain, $K=\operatorname{Frac(\mathcal{O_k}})$, $L/K$ a Galois extension and $\mathcal{O}_L$ the integral closure of $\mathcal{O_k}$ in $L$ (which is another Dedekind domain)
As soon as one mentions ANT a wild @mercio appears! Hi
 
hello
 
Anyway I have $\mathfrak p$ a prime ideal of $\mathcal{O}_K$, I want to prove that $\operatorname{Gal}(L/K)$ acts transitively on the set of ideals of $\mathcal{O}_L$ lying over $\mathfrak p$
 
Ah okey
i wrote about this in my dissertation
 
Oh, nice, I knew I was asking the right person :D
 
9:55 AM
if it wasn't you would be able to make a product of some of them
 
(This is proposition 9.1 in Neukirch, chapter 1, which I'm reading it from)
 
hmmm
then hopefully by restricting that product to $O_K$ that gives you an ideal of $O_K$
and uuuh
i don't know
 
I guess you take two primes of $L$ dividing $\mathfrak{p}$
 
and that product ideal should contain $\mathfrak p$ strictly
 
and suppose that $\sigma(\mathfrak{P}_i) \neq \mathfrak{P}_j$ for all $\sigma \in \operatorname{Gal}(L/K)$
 
9:57 AM
assume that $\mathfrak B$ and $\mathfrak B'$ are two prime ideals of $\mathcal{O}_L$ lying above $\mathfrak p$ and suppose that they are not conjugate, then we can find $x\in\mathcal{O}_L$ such that $x\equiv 0$ mod $\mathfrak B'$ and $x\equiv 1$ mod $\sigma(\mathfrak B)$ for all $\sigma$ in the Galois group, by the Chinese remainder theorem
 
right
and then set $b = N_{L/K}(x)$
(product over all $\sigma \in \operatorname{Gal}(L/K)$)
 
Right, but I'm missing why such an $x$ exists
 
well you said it, by the chinese remainder theorem
 
(Also I just discovered that what I always assumed to be mathfrak Bs are actually Ps)
 
hahah
 
10:00 AM
@ÍgjøgnumMeg well Neukirch said it! I'm not seeing it, don't I need coprime ideals to use CRT? Or is it some form of CRT I'm not familiar with?
 
The ideals are coprime by assumption
because they're not equal
and this is a dedekind domain so they're both maximal ideals
 
@ÍgjøgnumMeg ohhhh, of course, that's what I was missing!
feels obvious in hindsight
 
haha yis
 
I was stuck on the fact that prime ideals are not necessarily coprime in general forgetting that Dedekind domains are super nice
 
exactly
so you have $b \in \mathfrak{P}^\prime \cap \Bbb Z = \mathfrak{p}$
so now you have
 
10:04 AM
$\mathcal{O}_K$ isn't necessarily $\Bbb Z$ in Neukirch, but yeah, I see how to conclude from there, I was only missing that step!
 
Ahhh okay
that's cool
and yeah
I'm actually ad hoc translating from my dissertation
I meant to write $\mathcal{O}_K$
lol
the proof is exactly the same though
 
What's the topic of your dissertation exactly?
 
Fermat's Last Theorem for Regular Primes
+ Hilbert Class Fields stuck onto the end because why not
 
Sounds cool
 
yeah it is! FLT is a cool application of all the ideas in a first ANT course
The only bit I actually assumed without proof is "Kummer's Lemma" which requires some p-adic stuff that I didn't learn
also the group of lecturers marking it gave me an absurdly high grade (mostly because they expect a lot less of the pure mathematicians at my uni due to it being mostly applied)
so that makes me happy.
hahah
 
10:10 AM
that's great!
 
yeahhhh thanks man :)
 
I'm still finishing to write my dissertation, but it's in model theory
 
that's cool, I only did a little bit of model theory at a summer school (compactness theorem?)
but I reaaaally liked it
via Łoś's theorem I think
 
That's the usual proof if you want to avoid completeness
And model theorists usually want to remain on the semantic side
 
I think this was stated informally at the beginning as something like "if it's true in the product then it's true in each of the factors" or something and then made more precise when he introduced ultraproducts properly
 
10:15 AM
It's true in the product iff it's true in "almost all" the factors
 
if and only if the set of indices for which it's true is in the ultrafilter?
 
cool
what's the topic of your dissertation? Just model theory in general?
 
Quantifier elimination
Focussing on algebraically closed fields and real closed fields
 
sounds like magic
 
10:24 AM
Some results are pretty magical indeed
 
11:00 AM
ok so how can I prove more rigorously than stating that it follows from $n \geq \pi(n)$, that $\lfloor \sqrt{n} \rfloor \geq \pi(\sqrt{n})$
 
11:30 AM
if $x \ge 0$, $\lfloor x \rfloor \ge \pi (\lfloor x \rfloor) = \pi (x)$
 
11:45 AM
yeah ok I concede I don't quite know what I am looking for other than the obvious
 
11:59 AM
Rehi Alessandro, hey mercio and Ígjøgnum!
 

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