00:00 - 19:0019:00 - 00:00

12:00 AM
The Picard group is super beautiful, and I guess you could do it in special cases without the formalism of line bundles if you just read about Dedekind domains
Yes! I remember being very confused about the point at infinity in S-T
I am still not great with projective (algebraic) geometry, tbh. Hoping it's something that improves in the next few years

What does addition mean in the definition of a ring? e.g. a ring is an abelian group under addition.

Good luck! Hopefully that can be worked in what seems to be a rather busy time :P
@Nebulae it's just the name of one of the operations
You define a ring to be a set with two operations, called addition and multiplication, satisfying blah blah blah

I love sets with two operations that satisfy blah blah blah
Haha, thanks! Same to you - it sounds like you've got a big year ahead. Are you applying to graduate schools?

Yup!
Once I finish the general GRE and paper revisions I'm gonna have to relearn how to do calc quickly, and learn multi at all

Nice! That stuff sucks

12:10 AM
(Literally a few days ago I learned about integration in polar coordinates)

You've got the right approach though. There's so much bad advice on the internet on how to do well on the subject GRE, IMO

Hmm, what do they tend to recommend?

@Daminark Thanks. I'm confused by the whole new addition vs my good old addition.

Well, I guess I shouldn't speak for everyone on the internet. Some people will say you need to do every single problem in Rudin, Munkres, D&F

Good Lord

12:11 AM
That you need to know all about every subject on there, including probability, logic, etc
Indeed, my reaction as well

If that were true then Soug would've actually helped quite for it

I'm not trying to dissuade anyone from learning that stuff for grad school
But like, that's ridiculous prep
50% of the stupid test is dumb tricky calc problems

Yeah rip

IMO, best use of time is: be able to to do every calc problem (including multivariable), every linear algebra problem, do every practice test available on the internet, and know the major theorems that tend to pop up (e.g., classification of f.g. abelian groups, Cauchy-Riemann equations, Great Picard)
I mean, I'm sure you actually know these theorems, but the point is, one doesn't (unfortunately) have to actually know anything deep about them, just their statements

I'll keep that in mind, thanks!

12:17 AM
Lol, sorry, I realize you didn't ask for advice. But I do see a lot of crappy advice out there, which I'm mainly reacting to

Lately I've been thinking of asking on here every single exercise of Magnus et al's book on combinatorial group theory. There's a good few hundred of them and they're very detailed, but I'm doing a PhD in the area and I'd like a firm grasp of the subject. It'd take a long time and I've already asked a handful. What do you think?
I don't care how difficult I find them: if it's too easy for me at first sight, perhaps I'm fooling myself as to the rigour of my solution, the details of which I can always post in a proof-verification; if it's too difficult for me, then a question is in order anyway! I'd ask on meta about it but I want some preliminary reactions first.

No no, actually thank you, it's very helpful since I'm taking the test in a bit over a month :P

The definition of a field is nice. I've seen its axioms before, but defined w.r.t. ring, it's nice.

@Shaun you should probably try the exercises yourself first. Maybe for the first few easy ones if it's good to ask around after you do them and make sure you have the right idea, but it seems to me like at some point you'd want to gain confidence in your own ability to answer questions and believe you have them right without relying too much on others

Good luck! I'm sure you'll do very well. If you don't mind my asking, are you applying to UCLA?

12:24 AM
I have considered UCLA, and I think it's moderately likely I'll apply there. For the most part my criteria at the moment is that there's at minimum good representation in the general sphere of algebra, number theory, and hopefully topology
Which UCLA seems to fit from what I hear, though at some point I'll have to check the faculty interests page and ask around

@Daminark Don't worry: I'll definitely try each if them first. If I'm confident in my answer, perhaps a proof-verification would be appropriate anyway to iron out any kinks.

:)

In (the?) subring test, we show (i) $a+b$, (ii) $-a$, and (iii) $a \cdot b$ and (iv) $1$; belong to the subset. How come we don't show that $0$ belongs to the subset? Is it because (i) and (ii) give this, and (iv) ensures that it's nonempty, so no need to show that?

12:50 AM
Yup
(1)+(2)+(4) => 0

Thanks.
I can't wrap my head around the fact that R = {0} is a ring.

1:34 AM
Could somebody explain why the order of summation for a double sum doesn’t matter?
E.g in matrix multiplication

It's a finite sum so addition is commutative

1:57 AM
@loch yo
here ?

2:12 AM
Exercise 1.1.1. It took ages to type on my phone.

@rschwieb. You around?

2:31 AM
@Mason Yes... still waiting for a response to my email!

Ah. Yes.

You mentioned a draft?

I meant literally my email draft

ahh, heh

I want some feedback on the most recent question I wrote
If you have time.

2:34 AM
Very fast

Head to bed. I'll catch you later.

Yeah.

First impression: kind of all over the place. I believe it's interesting, but it might take some more persuasion.
in terms of tricking people into seeing the interest if they're not inclined to be interested.
But that's just a superficial assessemtn
Will look more tomorrow. Goodnight

Night.

2:57 AM
@Mason There's a chatroom specifically for constructive feedback, called constructive feedback, funnily enough :)

Yup yup
Noted.

Ah: I just noticed you posted in there recently. My bad m . . . It is 4am here.

It's all good. I am clearly a tad overenthusiastic about getting feedback on this question.

4 hours later…
7:28 AM
So the ICM still have not uploaded any of the plenary lectures (hopefully they will be up soon). But at least University of Bonn has uploaded the talk by Scholze.

7:42 AM
@geocalc33 Cool! What country do you live in? Sounds like chemistry or physics to me. :)

1 hour later…
8:54 AM

9:09 AM
Hey @loch

hey @AlexClark

@loch do you know about conductor etc?

Hey @LeakyNun

for a short period of time in my life yes but not now lol

9:58 AM
@Daminark You are just 'denmark' now?

10:08 AM
You know a talk is in pure math when the final part is called "applications" and is followed by a slide titled "no torsion in cohomology of Shimura varieties"
7

Hahahaha

### 2018 Moderator Election Chatroom

This is where users and candidates can interact in a construct...
3

@user2646 Any particular reason you linked that?

9 hours left to vote

Oh

10:42 AM
:-)

most have their votes maxed out, thus it seemed settled to me

11:00 AM
@Tobias What was the talk about?

Hey
@OskarTegby I live in Maryland

@OskarTegby It was Peter Scholze's plenary lecture at ICM. It was about period maps in $p$-adic geometry
(and I know this because that was the title :) )

@MatheinBoulomenos Is the easiest way to see a local ring with nilpotent radical is complete to think of the completion as Cauchy sequences which are just eventually constant? In fact, sequences which are constant after some fixed n?
That seems to make it clear all the Cauchy sequences have unique limits.

sorry I am probably going to need to alter and or edit a post ive made with the assistance of another how do I properly reference the assistance of that person into my SE question/poost

@geocalc33 Cool! I'm going on an exchange year in the US after next summer. I haven't decided on where yet.
@Tobias Sweet! Is it possible to explain $p$-adic geometry in a sentence or two? I'm not very familiar with it.

11:08 AM
3

Crazy Turing Machine is the same as Turing machine with one stripe , except of the fact that after each ten steps the head jumps back to the beginning of the stripe. Lunatic Turing Machine is the same as Turing machine with one stripe , except of the fact that after each 10, 100, 1000, ... steps...

Is the family $(1-x)^n +y^n$ a non Abelian group under composition with identity element $y=x$, and inverse elements $(1-x)^{(1/n)} +y^{(1/n)}$

@OskarTegby Neither am I, so I at least can't
@geocalc33 What do you mean by composition?

lolwut, I wonder what will a lunatic infinite time turing machine look like

@OskarTegby But Scholze has written a very nice answer on MO about what perfectoid spaces are, which are the main ingredient in a lot of his work (winning him the Fields medal this year).
Also, his ICM talk starts with an overview of the main ideas
(the talk is on the uni-bonn youtube channel, since the ICM have not yet gotten around to uploading the things)

Cool!

11:14 AM
@TobiasKildetoft I mean that composition is the group operation. Such that $f$ composed with $f$ inverse equals the identity function

@geocalc33 But how is composition of those things supposed to be defined? What sort of objects are they even?

For $x,y \in (0,1)$ in the reals

@rschwieb I think the easiest way seems to be $A/(0) \cong A$, so $\varprojlim A/\mathfrak m^n \cong A$, since the inverse system is eventually constant

@geocalc33 so you did not actually mean to compose such things. You mean that composing $x$ and $y$ should be given by that formula

I find it easier to think about categorical limits than Cauchy sequences, but that's probably just a matter of taste

11:20 AM
Hey @Mathein, I'm probably gonna come to Heidelberg in the summer semester instead -_-

I was plugging in f inside f inverse for x and y

@ÍgjøgnumMeg wow, why? :(

@Mathein I am too poor to finance studying immediately in the winter semester so I will probably work for 6 months to get some money together and then just pray that I can get a scholarship from the DAAD (I have quite a strong application I think so hopefully this works out)

@geocalc33 What is $f$ now?

f is a function element
Defined above
Yeah in just not sure if the operation I'm doing is actually valid

11:24 AM
@geocalc33 So far you have not managed to define anything precise enough for me to understand

Sorry I'm typing on my phone

Hi, if I have a normalized vector, is there a quick and easy way to denormalize it? I am working with python right now and there I get only the normalized eigenvectors of a matrix...

@ÍgjøgnumMeg did you check that it is possible to start in the summer semester? there are subjects where you can only start in the winter semester (but it maybe applies more to bachelor students)

@Mathein yeah i checked that you can start in the summer semester (at least I think I did!) but scholarships can only be given from the winter semester, so I would have to finance one full semester by myself I think

@philmcole There is no such thing as a "denormalization"

11:27 AM
I mean once a vector is normalized, can I figure out a scalar multiple with which multiplied the components become integers again?

@philmcole You mean assuming such a scalar exists?
Just make one of the entries $1$, then check if the rest are rational and multiply by their denominators

@Mathein Also I guess that there is also a chance that I would not be accepted in the summer semester (although I don't see why this would happen if they already accepted me now)

@LeakyNun Hi

11:30 AM

Like on trains?

Like wires?

nope

merry christmas

11:31 AM
he knows everything

yis

froehliche weinnachten

I've found his blurbs very useful

I think I once read something involving conductors, but I don't recall what (possibly something related to automorphic representations)

I want to ask a question on dynamics of a rigid body. But I can't find a tag for it
What should I do ? 🤔🤔🤔

11:39 AM
That moment when you want to study 200% pace just to manage to take all the cool math courses.

@TobiasKildetoft what I was doing was taking let's say $(1-x)^2+y^2=1$ and $(1-x)^.5+y^.5=1$ and compositioning them (substituting in the second one for the first one for x and y) and the result is y=x

That's the story of my life since moving away from home.

I forgot to add the equals 1 to the RHS!

@geocalc33 Still no idea what you are doing or why

@anyone else know what I'm doing

11:44 AM
I mean something chat appropriate

@Adam how is that chat "appropriate"?
thank you for removing that

12:11 PM

12:25 PM
Can the identity element of a group be whatever you want it to be
Nvm

How could that be the case?

I'm just making no sense don't mind me

It's okay. That's me half of the time.
Did I advise you to listen to the Ben, Ben, and Blue podcast? I listened to the first one and it was really nice.

12:40 PM
No I'll check it out though
Is it on Spotify

Cool

You can choose any of the alternatives in the menu bar.

Can the identity element of a group be asymmetrical with respect to the elements and inverse elements

What do you mean by asymmetrical?

12:43 PM
The integers have 0 as identity and this perfectly splits the positive side and the negatives side
Yeah I had a feeling it wouldn't work out

You cannot do that for a finite group that has some linear ordering
actually... maybe not, if you have involutive elements (elements that are self inverses)

Well, the reals have 1 as multiplicative identity and it doesn't split the positive and the negative side.

the coxetar groups came into mind

In what context is this property important now?

12:51 PM
I believe I'm talking about reflection groups

Okay.

I don't know how important they are

Okay. Well, I think the answer is yes if my example was valid.
Are reflection groups necessarily only additative or multiplicative?

Not sure. I was just wondering if you could take functions and reflect them to form a group

Interesting idea.

1:02 PM
With y=x as identity function

I think it should work. Right?

And what I was considering was a family of super ellipses $x^s+y^s=1$
Which is asymmetrical on either side of the identity element, $y=x$
So that's why I was asking
About an asymmetrical group sorta thing
For the super ellipse family the inverses are themselves
Oops, I made a mistake. The family is actually asymmetrical about y=1-x
This is so confusingly subtle
Like why did the right handed majority enact these mathematical conventions such as the identity function Hass to be y=x
Done.
I guess convention provides structure, clarity and a stepping stone for progress. So that's always good

1:21 PM
Do you have time to look through a proof that I made. Something feels wrong.

Sure but I'm probably not the best person to ask about proofs lol
I'm more of a proof by intuition person

depends on what proof, I can handle some topology ones

Which does not suffice for mathematicians in terms of rigor

It just seems way too easy.

@MatheinBoulomenos Ugh, I feel really dense, but I have a hard time seeing that isomorphism based on what little I know about direct/inverse limits. I can tell the "tails" look like elements of $A$ since they're constant... but it seems like the initial segment can vary a lot. How does the tail being constant choke it down to just $A$?
Thinking in terms of categorical limits is completely foreign to me, but I'd like to hear what the reasoning sounds like.

1:31 PM
I felt like this line is not strong enough, but I cannot think of anything more precise
Also, before you say $f(x_0)=L$ you probably can quote intermediate value theorm as $f$ is continuous

I'm leaving for a while, so if someone wants to comment the solution please tag my name in that comment! :)

@OskarTegby And I do felt like you need to consider the case where f(x) is positive in some values and negative in others. Perhaps there is an $f(x)$ that grew in just a way so that it get cancelled out by $x^{-n}$ thus getting it to zero

@secret take the segment of a unit circle in unit square, reflect it about$y=1-x$. For what value of s does the reflected circle equal $x^s+y^s$
I love this question
The RHS should be equal to 1 my bad

so something about $x^{-n}$ prevent that equality from satisfying, thus ruling out all nonzero $f(x)$

Irrational, rational,...
Can't be an integers I'm sure

1:44 PM
$x^s + y^s \mapsto x^s + (1-x)^s$

$x^2 + y^2 \mapsto x^2 + (1-x)^2 = 2x^2 - 2x +1$
$x^s + y^s = x^2 + (1-x)^2$
$r^s = r^2 \cos^2 x + (1- r\cos x)^2$

Shouldn't there be a y on the RHS

I reflected the circle about $y=1-x$, thus all points on $y$ should map to $1-x$?
o wait, no

I did an approximate solution and got 1.77
For s
Trying to make it exact

$x^2 + y^2$
Let $x' = 1 - x \implies x = 1 - x'$ Then $y=1-x \implies y=x'$

$(1-x')^2 + y^2 \mapsto (1-y)^2 + x'^2$
$(1-x')^s + y^s = (1-y)^2 + x'^2$
$x^s + y^s = (1-y)^2 + (1-x)^2$
$x^s + y^s = 1 -2y + y^2 + 1 - 2x + x^2$
$x^s + y^s = x^2 + y^2 - 2 (x + y) + 2$
$x^{s-2} + y^{s - 2} = -2(x + y -1)$
$r^{s-2} = -2r (\cos \theta + \sin \theta) + 2$
$(s-2) \ln r = \ln (2 - 2r (\cos \theta + \sin \theta))$
$s = \frac{\ln (2 - 2r (\cos \theta + \sin \theta))}{\ln r} + 2$
If $\ln r = 0$ instead, then:
$0 = \ln (2 - \cos \theta - \sin \theta)$
$0 = 1 - \cos \theta - \sin \theta$
$\cos \theta + \sin \theta = -1$

fine I apologise for my inappropriate behaviour tonight any will make an effort to make less socially awkward remarks in the future

1:57 PM
($r = 0$ is probably irrelevent, since you will be dealing with superellipses of zero radii)
But yeah. there is a relationship between $s$ and $\theta, r$
a polar plot will be useful to find all the $s$

Thanks so much @Secret

but actually that equation is a bit strange, since you are starting with a circle thus $r$ should be constant, then it makes no sense why $s$ has to vary with angle
I think I am not very sure, unless I actually need to deal with 2 sets of (x,y) independently

Yeah

Silent!

2:04 PM
@LeakyNun, above is a theorem from Rudin PMA. The proof still holds if we take radius exactly equal to $\frac12 d(p,q)$, right?
* radius of $V_q$ and $W_q$.
@geocalc33 hi!

yes
more generally, a compact subspace of a T2 space is closed

Thank you very much!
T2=Hausdorff?

yes

yep.

Pretty visual theorem, basically compactness ensure every net will hit the boundary of the set or some points in its interior outside, hence guarentee closure
Now that makes me curious, how does that spillover occur for spaces that are non T2...
Ah, T1 spaces allow intersecting neighbourhoods, thus a net within the compact set can spillover to these neighbourhoods outside and hence landed in the neighbourhood of points outside the boudnary, thus preventing closure

2:19 PM
@Secret I want s and theta held constant and r to vary. I think it would make sense then

ok then

@LeakyNun Yes. I think I figured it out. If $(X,\tau)$ be the topological space then we define $\text{Cl}_\tau$ as described here (this is also probably the one you mentioned. Let $\text{Cl}$ be any other closure operator on $X$ giving the same topology $\tau$.
Since $\text{Cl}$ is a closure operator giving the same topology $\tau$, so $\text{Cl}(A)$ is $\tau$-closed. Also since it is a closure operator we have $$A\subseteq \text{Cl}(A)$$ for all $A\subseteq X$. Then from the definition of $\text{Cl}_\tau$ we get, $$\text{Cl}_\tau(A)\subseteq \text{Cl}(A)$$To prove the converse we simply note that, $$A\subseteq \text{Cl}_\tau(A)\implies \text{Cl}(A)\subseteq \text{Cl}(\text{Cl}_\tau(A))\implies \text{Cl}(A)\subseteq \text{Cl}_\tau(A)$$
Did you mean something along this line @LeakyNun?

21 hours ago, by Leaky Nun
@user170039 yes, because you can recover the closure operator from the topology
this is what I meant.

@LeakyNun So, if I understand you correctly, you were suggesting the existence of such an operator but nothing about its uniqueness, right?

2:34 PM
I am sorry but I don't understand you then. Can you clarify?

I think "recover" is a standard term used in mathematics (or maybe more in category theory)
Given a closure operator Cl on a set X, I can form a topology (X,τ).
What I mean when I said you can recover the closure operator, is that you can produce a closure operator Cl from the topology (X,τ) that is isomorphic to the original closure operator
where "isomorphic" really means "equal"
so I'm describing an equivalence between { closure operators on X } and { topologies on X } by giving a function in both ways
you already have the function in ---> way
when I said "recover", I'm referring to a function in <--- way

I see. Now it is much clearer.
I think I have essentially worked out the same details above.

Thanks @LeakyNun.

maybe you should think about why "isomorphic" means "equal"

2:42 PM
@LeakyNun Maybe it has something to do with the skeleton of the category $\mathbf{TCSp}$ but I have not gone that far in reading my category theory book.

well it's because a closure operator is really a function P(X)->P(X)
and functions have extensionality

3:05 PM
2

Okay, I am going to do my best articulating my question. Suppose that $P = [p_{ij}] \in M_{n \times n} (\mathbb{C})$ is a positive semi-definite matrix. Then its off-diagonal entries satisfy $|p_{ij}| \le \sqrt{ p_{ii} p_{jj}}$, which says that the the entry $p_{ij}$ lies in the closed disc of ra...

3:33 PM
2

How to show that multiplication of matrices, when defined, is associative and distributive over addition. Let $R$ be a ring, and the set of all $n \times m$ matrices over $R$.

Could somebody explain to me why you’re allowed to swap which limit you do first?
@LeakyNun I think I’ve asked you a similar question before and you gave me a good answer xxxxxxx

4:11 PM
0

Take the arc of the unit circle $(r=1)$, inside the square $[0,1]$ by $[0,1]$. Reflect this arc about $y=1-x$. Show or disprove that an approximation can be made arbitrarily precise to the reflected circle using the form $x^s+y^s=1$.

Why is the set of matrices with determinant $ad-bc$ s.t. $c=0$ a subring of $M_2(\mathrm{R})$?
Wouldn't we have problem with the multiplicative inverse when $a=0$ or $d=0$?
Set of 2 by 2 matrices that should say, with first row {a, b} and second row {c,d} such that c=0.

4:28 PM
Rings don't have multiplicative inverses for most elements

Most rings ;P

Then do I misunderstand the following from the definition:
There is an element 1 in R such that a · 1 = a and 1 · a = a for all a in R
Oh, not an inverse.
It's the multiplicative identity.

5:15 PM
It turns out that even on the easiest mode, Pandemic is really hard as a 2-player game when the other player is 4 years old.
(still fun though)

5:28 PM
haha

@Mike This question is probably too vague but can something interesting be said about topological spaces for which invariance of domain holds? (So spaces $X$ such that whenever $U\subset X^m$ and $V\subset X^n$ are open and homeomorphic we have $m=n$)

@TobiasKildetoft I really want lp space to be symmetric about $y=1-x$, but it isn't...

@geocalc33 What space?

$L_p$ space

5:31 PM
rather, $L^p$ space
like $L^2$ is a Hilbert space

@geocalc33 If your measure space is finite...

How can something be symmetric around an equation? Or did you mean the function $x\mapsto 1-x$?

but it's just so annoying how it isn't symmetric
@robjohn but if you use the reals and a probability measure it can be symmetric?
probability measure in the unit square that is
okay I got it I just need an infinite measure.
...make it continuous not discrete...
@robjohn can you explain more ?
four intersecting $L^p$ spaces
I don't want to know what will happen to me if I keep asking too many questions in such an academic setting. Hopefully no one will put a hit on me

6:36 PM
If $V$ is a normed linear space, and $W$ is a linear subspace with basis $B$, is $\overline{W}$ spanned by $\overline{B}$? Does $\overline{B}$ form a basis?

the adherence of $B$ ??

$\overline{B}$ denotes the closure of $B$ with respect to the norm topology.

@user193319 So presumably, the space is infinite dimensional here?

I see no reason why that would hold

6:41 PM
Under a discrete measure, Lp space, specifically, the part only located in the unit square, is not symmetric on either side of the function ϕ=1−x. So how exactly do you fill in the space with an infinite measure, say a probability measure in the unit square, such that the space becomes symmetric?

.....

............
.....

what is a measure, what is a discrete measure, what is a Lp space, are you saying an Lp space is a measure, what is an infinite measure, what is a probability measure, and when is a measure symmetric ?

0

Prove that $I^2/ \partial I^2$ is homeomorphic to $\mathbb{S}^2$ This was the idea I came up with to prove the above. I was thinking that I could prove the above by showing that $I^2/ \partial I^2$ was homeomorphic to the one-point compactification of $(0, 1) \times (0, 1)$ which itself is h...

Any topologists around?

@Mercio I just made all of that up

6:53 PM
then you should make up an answer

okay

with pixie dust and fairies and unicorns

okay i'm writing
I think I'm going to add funicorns though
the composition of fairies and unicorns

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