01:00 - 18:0018:00 - 00:00

1:10 AM
@Thorgott you can try to read Herstein's books :P

2 hours later…
3:09 AM
1

Say we have a machine and we give it a task to do (vision task, language task, game, etc.), how can one prove that a machine actually know's what's going on/happening in that specific task? To narrow it down, some examples: Conversation - How would one prove that a machine actually knows what i...

@TedShifrin Hello!
If we are asked: Can we find an appropriate subset of $$\{0, \pm 1, \pm 2, \cdots , \pm n-1\}$$ such that sum of its members is divisible by $n$?
My answers is yes, for example take $\{2, n-2\}$ the sum is of course divisible by $n$. Or we can take $-(n-1)$ and $-1$ then also we have the sum divisible by $n$. But I know this is not a proof, but is it okay or can we prove it more rigorously ?

3:55 AM
But your original question, @Knight, was about $n$ distinct integers, right? First, when you look at residues mod $n$, you don't have $\pm$; just have $0,\dots,n-1$. Second, the integers might give the same answers mod $n$, so we don't know we have $n$ different residues mod $n$. Or maybe I'm remembering someone else's question, not yours.

4:10 AM
@TedShifrin I tried that original question and found that we can reduce it to this one.
The reduced question is: Given $n$ different integers (all of them are less than $n$) and repetition is allowed. Can we choose members from that set such that their sum is divisible by $n$?
Of course, we can but do I need to prove it? I can prove it through verbal logic.

4:31 AM
@Knight: No, it's not this simple. When you start with $n$ distinct integers, you are likely to get repetitions when you reduce mod $n$. Of course, if you don't, the probably is trivial, as you're saying, because you must have one number $\equiv 0\pmod n$ (or two whose sum is $0\pmod n$). But what if I start with numbers some of which reduce to the same thing mod $n$? Then it gets very complicated.

5:06 AM
@AlessandroCodenotti I see

5:45 AM
@Joanna are you willing to learn a programming language?
re: this

6:16 AM
@robjohn Robbie sir, can you please tell me what does this quote means in simple words
> Life lived in the absence of the psychedelic experience that primordial shamanism is based on is life trivialized, life denied, life enslaved to the ego.
@AFURIOUSMIND What was your previous user name :-) ?

4 hours later…
10:12 AM
What's an example of a countable topological space which is not second countable? Is it totally obvious?
I guess we can even break first countability, $\Bbb N$ with a topology such that neighborhoods of $0$ are $\{0\} \cup A$ where $A$ is a nonempty subset of $\Bbb N \setminus \{0\}$
Discrete elsewhere.
That seems to do the trick, right?
Let's see, so the topological space is $X = \Bbb N \cup \{0\}$ where $\Bbb N$ is given the discrete topology, and neighborhoods of $\{0\}$ are of the form $\{0\} \cup A$ where $A \subset \Bbb N$, $A \neq \emptyset$. Suppose $U_1, U_2, \cdots$ is a countable nbhd basis of $\{0\}$.
Then each $U_i = \{0\} \cup A_i$ for some $A_i \neq \emptyset$. $\{A_i\}$ is a countable subset of $P(\Bbb N)$, and I want to argue that we can definitely choose some element $S$ which does not contain any of the $A_i$'s. Then $\{0\} \cup S$ will be an open nbhd of $0$ in $X$ which does not contain any of the basic nbhds around $0$, and that will be a contradiction.

10:33 AM
then $\{0,1\}$ and $\{0,2\}$ would be open sets, but their intersection, $\{0\}$ would not be, no?
or am I missing sth stupid

Yeah you're right
I dunno

Remind me what second xountable means

this is annoying
countable basis @Mike

Note that every countable topological space is obviously Lindelof

How about one point compactificafion of Q? Neighborhoods of infinity should be sparse

10:42 AM
Yeah that probably works I was trying something like this

I don't know how to characterize compact subsets of Q

they need to be complete
so all sequences in their should have rational limits

19

$\newcommand{\ms}{\mathscr}$The compact subsets of $\Bbb Q$ are a little awkward to describe unless you’re familiar with infinite ordinals, but here’s a rough, intuitive description. Let $\mathscr{C}_0=\big\{\{q\}:q\in\Bbb Q\big\}$; clearly every member of $\ms{C}_0$ and every finite union of m...

doesn't seem nice

I suspect you can fiddle with this to show that you don't have a countable base at infinity

10:44 AM
ah, you can do it recursively, that's nice-ish

This describes a family of compact sets $C_\alpha$ for each ordinal $\alpha < \omega_1$. If I am not mistaken, $X \in C_\alpha$ and $Y \in C_\beta$ lie in $C_\kappa$ where $\kappa \leq \alpha, \beta$
In particular, thinking of complements, we get families $U_\alpha$ whose unions have ordinal less than the ordinals you're taking the union of
If you had a countable base you could only construct open sets in a countable number of these $U_\alpha$

I think the way it is defined, the sets have to be more complex if you go up the hierarchy

Sorry Sorry

Given an integer $n$, we have a set $$\{\underbrace{3,3 \cdots 3}_{n-1~terms}, n-22 \}$$ Can we find some members from the set above such that their sum is divisible by $n$ ? (please don’t consider my previous question related to this).

at least that's what he's implying by "for each α<ω1 the sets in Cα are homeomorphic to the ordinal α with the order topology.", I think

10:47 AM
I meant to say that X,Y have $X \cap Y$ in ...

Given an integer $n$, we have a set $$\{\underbrace{3,3 \cdots 3}_{n-1~terms}, n-2\}$$ Can we find some members from the set above such that their sum is divisible by $n$ ? (please don’t consider my previous question related to this).

I want to turn this into a union property for opens to show no countable base at infinity, by the uncountability of this collection of ordinals

ah, that does sound promising

Remarkably hard lol

the intersection of two compact subspaces in a Hausdorff space is a compact subspace
and the order type only decreases when taking intersection

10:50 AM

so that should be true

@LeakyNun Cheater

Hola Leaky

when will you spell my name correctly...

The Arens-Fort space looks nice

10:50 AM
at least that affirms the desired conclusion

I corrected it

I think the argument I gave is more or less correct. I'm kinda proud

$\Bbb{Z}_{\geq 0} \times \Bbb Z_{\geq 0}$, discrete away from $(0, 0)$ and open nbhds of $(0, 0)$ are sets containing all but a finite number of points from all but a finite number of columns

Oh sure that's built for this to be true isn't it

Easy to argue that $(0, 0)$ does not have a countable basis

10:51 AM
nice

"An anti-Hausdorff Fréchet-Urysohn US space"

Yeah I was trying something with $\Bbb Z_{\geq 0}$
But couldn't manage to build a topology
Obviously you can break first countability at a point I mean, there had to be an example

Q cup {infty} was just a more convoluted version of this example

It's a good one, I should have thought of that. Seems annoying to argue is all

should've just looked into Steen-Seebach

11:09 AM
Can someone please suggest me a site where I find the all possible $x$ such that $$x \equiv 156 ~~~~~~~(mod 153)$$

(0, 0) is really a weird point, because there's no sequence away from (0, 0) which converges to (0, 0)
But there's a net converging to (0, 0) namely the open neighborhoods of (0, 0)

That means I want the list of all numbers which when divided by 153 gives us the remainder 3

I need a some thing like a calculator for that

@Balarka wouldn't you expect that to happen?
since there's no countable neighborhood base
well, I guess you can still have a sequence converging to a point which doesn't have a countable neighborhood base
@Knight there will be countably infinitely many such integers and you should be able to write them down yourself

11:12 AM
Yes, just wedge a bunch of [0, 1]'s at 0

@Thorgott Thorgott, I’m not joking but I really need only 152 of them
That is I need 152 numbers such that when they are divided by 153 we get the remainder 3.
Am I too demanding ?

9847922864743813118262264566390547551333516894250128770747655765000487989029680512375968717425176429561137255044253770284826215035060231325229055185039385580694823912850244466633167037055051383025141949
12233128709506429261872689053383605493937737899149445221808654785792005533748045424614957415776728323140548890760243404186765106982611096893019573778480846177461443950219652703480727397604441716509669744
1482854938021814270963761116943424667796900722488370036576310080802053636899394204789105038722710302010935332663388038625188001904681765922354452058229821950592552076871968483586449684793185…
@Knight there you go
import random
for _ in range(152):
print(153*random.randrange(10**100,10**200)+3)

I have no clue why you would want that
but if you can list 152 of them, you can list all of them and vice versa

Is that a single number or collection of numbers? Leakay

there's a very clear pattern that you will be able to recognize after just writing down a couple examples (or directly from definition)

11:20 AM
each line is one number

@Thorgott That’s a good question. Well, I think the set of 153 numbers such that 152 of them leaves the remainder 3 and the last one is 151. We cannot find a subset from it such that their sum is divisible by 153
@Thorgott I know it will be 153n +3 or 153n -3
@LeakyNun Don’t you think you have given quite big numbers?

the former, yes

you didn't specify the size

so you're done then

@Knight Nice

11:22 AM
@Thorgott Which former ?

unless you want to manually plug in 153 different values for n
in which case, good luck

oh no
you misspelled my name again

the former of the two

@LeakyNun Yes I accept my fault

I like how he's creating infinitely many variations on "Leaky"
It's humanly impossible

11:23 AM
@Thorgott I think you got me wrong.
Okay.

11:41 AM
You could just spell it right instead of accepting your fault, I think

Hi Maik

Hmaic
Hi chat

Yo Astex

Aztecs ?

Ok, let $G,H$ be groups. We define $G\coprod H$ as the set of all alternating words $g_1h_1....g_nh_n$ or $h_1...g_nh_n$ or $g_1h_1...g_n$ with $g_1,...,g_n\in G$ and $h_1,...,h_n\in H$ and none of them the respective identity. Composition is the concatenation of words, where two adjacent elements of the same group get replaced by their composition within that group and any instance of the identity gets removed altogether (this reduction process terminates in finitely many steps by finite length). The identity in this new group is the empty word and the inverse of an element is given by the

11:45 AM
Asterix

@Astyx Beluga Sen is spelling names wrong.

Beluga-San would be an japanese manga character right ?

robjhon I think you spelled my name incorrectly

12:10 PM
@Knight That quote, by Terence McKenna, essentially says that life without drugs is unfulfilled.

12:21 PM
@Thorgott This construction is also known as free product of two groups, by the way. It's easier to describe by writing $G = \langle S_G | R_G \rangle$, $H = \langle S_H | R_H \rangle$, $G * H = \langle S_G \cup S_H | R_G \cup R_H \rangle$, presentation-wise
Where remember $\langle S | R \rangle$ is the group $F(S)/\overline{\langle R \rangle}$, quotient of the free group on the generators $S$ by the normal closure of the subgroup generated by the relator words $R$

12:46 PM
@robjohn Okay... thanks by the way

Now, assume $G,H$ abelian and let $X$ be arbitrary abelian, $f_1,f_2$ the same. Take the defining diagram of $G\coprod H$. Abelianization takes this diagram to a diagram in the category of abelian groups, where $G,H,X,f_1,f_2$ stay the same (up to irrelevant isomorphy). This diagram plus uniqueness then asserts that $G\coprod_{Ab}H=G\coprod H/[G\coprod H,G\coprod H]$ (up to irrelevant isomorphy). There's a map $\varphi\colon G\coprod H\rightarrow G\times H$ sending $g_1h_1...g_nh_n\mapsto(g_1...g_n,h_1...h_n)$. Reduction happens in accordance with the group operations, so this is a homomorp
@Balarka right, that makes sense. you essentially want words freely constructed from $G$ and $H$ that only reduce like $G$ and $H$ do, so this does that

1:33 PM
@Fargle!

where?
how're you doing?

long time we havent see you around in chat

fine, I guess---just relaxing for the summer
been turning my math brain against video games in an attempt to make it more useful :^)

whatchoo playing
i played some games during the pandemic

Factorio, mostly
a game which is just crack for the analytical mind

1:41 PM
aha

Prefer spacechem

never played it, but might do so at some point
I'm generally a sucker for games that are just wrappers for massive interlocking systems

2:00 PM
Hi @Fargle!

heya Alessandro
grats on the offer if I didn't already say so :)

Thanks!

There does not exist connected countable regular space with more than one point. There does exist connected countable Hausdorff space with more than one point.
Random factoids from the void

@LeakyNun Can we write a code such that we input $n$ in the equation $$3x + (n-2)= k~n$$ such Let $x$ vary from 0 to (n-1) and see for which $x$ our $k$ is an integer ?

@BalarkaSen Hm is the idea to argue that otherwise it'd be normal and somehow get a surjective function onto $[0,1]$?

2:10 PM
Only integral $x$ Are allowed

@Alessandro Yeah

I don't see normality though. I wanted to go through regular+second countable implies normal but we don't necessarily have second countable here

Ah regular Lindelof also implies normal
That's what you want to use

Ah I see

$g_1h_1g_2h_2g_3h_3g_4h_4=[g_1,h_1][h_1,g_4^{-1}g_3^{-1}][g_4^{-1}g_3^{-1},h_1h_2][h_1h_2,g_4^{-1}][g_4^{-1},h_4^{-1}]$
I hate myself

2:16 PM
wth are u doing man

That makes sense, in the proof of regular+second countable implies normal you need second countability to reduce to a countable cover, so it should be similar with Lindelöf

Yeah I believe a minor modification of the proof goes through

im showing $(G\ast H)^{ab}\cong G\times H$ via overly convoluted direct computation

Use presentations bro
Direct computation

2:20 PM
Be a true algebraist, show that $G\times H$ has the universal property of $(G\ast H)^{ab}$

that's the easy way

Brackets are too hard

$(G * H)^{ab} = \langle S_G \cup S_H | R_G \cup R_H \cup [S_G, S_G] \cup [S_H, S_H] \cup [S_G, S_H] \rangle$
Definition of $G \times H$
$\langle S|R\rangle^{ab} = \langle S|R \cup [S, S]\rangle$
All of group theory should be done at the level of presentations, 60's combinatorial group theory style

At the level of Cayley graphs*

how did they arrive at that expression underlined in red?
I understand that $e^{h(t)}$ is proportional to $z(t)-a$ from their previous remarks, but where did the $z(\alpha)-a$ come from

2:36 PM
@BalarkaSen OK how do I tell if a group is trivial

$G=\langle S\vert R\rangle$

That's not a presentation
Give me a presentation, with high probability I will be able to give you an algorithm for if it's trivial

Every group is hyperbolic and those have solvable word problem, qed

That's what haha
In fact for a large class of hyperbolic groups you can tell from the presentation, if it has C'(1/6) small cancellation property for example

2:40 PM
(nvm btw, I see why $e^{h(\alpha)}=1$, and that's all I need)

@BalarkaSen I have vague memories of reading something related to this

Yeah I was trying to cheat you but I'm too lazy to find a presentation you can't deal with in short time

I don't know how big the smallest presentations with unsolvable word problem are
I know that they are small enough to actually be written down explicitely

Huh

I think there's some confusion here
Groups have unsolvable word problem, independent of presentation, right? You're just saying the smallest group in the sense of cardinality of presentation
Given a presentation if it's trivial there's an algorithm to reduce it to the trivial group
There shouldn't be a finite presentation for which you can't determine if it's trivial or not
(Possibly I am the one that is confused)

2:47 PM
Is that clear

@MikeMiller Hm there's finite presentations for which the problem of deciding whether two words are equal. By multiplying with an inverse this reduces to checking whether the group is trivial
whether two words are equal is undecidable*

That means there's no algorithm that will determine whether for any two words in those group, those two words are equal, but that doesn't seem to guarantee that there is no word you can show is nonzero

Right, but apparently being trivial is also undecidable
berstein2015.wordpress.com/2015/02/17/… according to this basically anything is undecidable

good morning

3:17 PM
@AlessandroCodenotti That suggests that no algorithm works for every presentation but not that there's a presentation for which it's impossible to determine if it's trivial
That sounds like you're getting to it being independent of set theoreric axioms or something

A question on root field. Is $\mathbb{Q}(\sqrt{-2},\sqrt{-3})$ the root field of the same irreducible polynomial $a(x)=x^4+5x^2+6$ over both $\mathbb{Q}$ and $\mathbb{Q}(i)$ ?

@MikeMiller Ah, I see your point now. However if we enumerate all possible presentations as $\{P_n\}_{n\in\Bbb N}$ and suppose that every $P_n$ has an algorithm $A_n$ deciding whether $P_n$ is trivial, can't we build an algorithm deciding whether an arbitrary presentation is trivial by diagonalizing? Run $A_1$ for one step, then run $A_1$ for two steps and $A_2$ for one step, then run $A_1$ for three steps, $A_2$ for two steps and $A_3$ for one step, and so on
This computability stuff is too tricky for me

3:43 PM
@AFURIOUSMIND Yes, I've coded a lot in Matlab - math modeling code, using the standard ODE solvers, writing data plotting and simulation movie code.
@AFURIOUSMIND A math professor that knows me fairly well says to not go into Pure Math (he does Analysis) and that Applied and Computational Math is my best bet. He also tells this to my classmates. But the no. 1 advice he gives us is to not do a PhD at all, because for most people, it's supposedly not a good choice.

@AlessandroCodenotti Hm this seems reasonable to me. I get really confused by this stuff

3:59 PM
LOOOOL
Good fucking luck
Half of you formalizers will get rekt trying to read Gromov
@LeakyNun More seriously: I like the blueprint. I could try to read it; if I knew Lean I would have considered trying stuff out.

cool

4:48 PM
@LeakyNun Have you seen the video Outside In?

yeah didn't really understand it

really good video
took me years to understand tho

Trying to search for the one about sphere eversion is made much more difficult because there was a 2017 drama made with the same name (and nothing to do with differential geometry).
There is the real one

I once found a really quick, visual proof of Whitney-Graustein theorem that immersed curves in the plane upto isotopy are classified by turning number without using any algebra or analysis
I haven't seen it written down anywhere, most of the proof uses Whitney's convex integration trick or directly deduces it from the general Smale-Hirsch theorem, which reduces to computing a simple homotopy group, $\pi_1(SO(2))$.
The idea in the Outside In video is that one gains flexibility by deforming an immersion $S^1 \to \Bbb R^2$ to an immersion which has many many many cancelling pairs of "loop-de-loops", which makes the Gauss map extremely surjective
The basic algorithmic question boils down to the following: You are given two integers of the form $\sum_i \varepsilon_i$ where $\varepsilon_i = \pm 1$. How do you decide they are the same with minimal number of computations?
You can compute the sum out but that's not efficient
The right idea, I think, is to add a couple (+1 - 1)'s to one expression so that the two expressions become roughly of equal size, and then flip - some sorting algorithm
Maybe @robjohn can confirm if this is indeed correct and more efficient
But I think this is what happens in the Outside In proof of Whitney-Graustein

5:18 PM
every time I watch Outside In, I feel like I understand it just a tiny bit more
some day I will actually understand it

@Thorgott You should intersperse those viewings with Inside Out

A good thing to notice in Outside In is that during the sphere eversion there are multiply belt tricks happening

@BalarkaSen I haven't really thought about it. Ted might know.

watching Outside In suffices WLOG, by duality

@Thorgott So also by duality, understanding Inside Out suffices?

5:21 PM
It feels more like a CS question for Ted

sure
if you can do it forwards, you can do it backwards

I wonder if there are mathematical scenarios where time-irreversibility is visible (and surprising). Maybe the answer lies in monodromy. When you go around counterclockwise in the complex plane, the value of $\log(z)$ jumps $+2\pi$. When you go clockwise, it jumps $-2\pi$.
But that's not really it
Because $+2\pi - 2\pi = 0$

@Alessandro @Balarka algebraic number theory exam and representation theory seminar went well, modular forms did n o t go well hahaha

Congrats on the first two. Just forget about the third one

@EdwardEvans What was the topic of the seminar?

5:27 PM
@Tobias induced representations/induced modules

nice

well
yeah
it was gross though
because we're meant to stick to the text, and the textbook's definition for the induced module was weird
even the prof didn't like it
hahaha

@BalarkaSen a CS question for Ted?

congrats

Haha I meant the question sounds too CS for you to be interesting

5:29 PM
@EdwardEvans What context was it? finite groups and complex reps?

I actually have no idea if the belt trick shows up in sphere eversion.

Finite groups, any old field (except in the cases where you need that $\operatorname{char} K \nmid \lvert G \rvert$)

@EdwardEvans Right, non-dividing characteristic is not as such important for the definition, but for the properties

speaking of groups

(basically whether you can get a biadjoint of restriction or only a one-sided one)

5:31 PM
Oh nice

I just dropped a group of hard edm tracks

the "punchline" of my talk was Frobenius reciprocity
for modules (and then for characters)

@TedShifrin Essentially by Smale-Hirsch theorem, the isotopy class of immersions $S^2 \to \Bbb R^3$ are classified by $\pi_2 SO(3)$. So it's kind of natural that the belt trick will show up, although that's really a demonstration of $\pi_1 SO(3) \cong \Bbb Z_2$

but the proof was very weird without defining the induced module as a tensor product

heya Ted

5:32 PM
Right, Frobenius reciprocity requires non-dividing characteristic (in the usual formulation)

heya @Fargle — long time!

ditto
for several months I was extremely busy, and now I'm doing typical wastrel things

You doing OK in these crazy times?

Hello Ted

I'd say so---having to keep close to family because many of them are older and therefore it's safer for me to do errands than them, but I'm naturally a homebody anyway

5:35 PM
I'm mostly staying in, but after almost 3 months I'm starting to go to doctors and do limited shopping. And then civil war is breaking out .... Crap.
hi @Knight

@EdwardEvans The more general version of Frobenius reciprocity does not require non-dividing characteristic, but comes in two versions depending on which definition of induced representation you use (and they are not the same)

@TedShifrin The remainder problem that I posted in the morning, cannot be proved.

Huh?

We're given integers like these $$0, \pm 1, \pm 2, \cdots , \pm(n-1)$$
We want to prove that if we always take $n$ integers of them (repetition is allowed) we can always find from *the taken ones* whose sum is divisible by $n$

I already corrected your plus/minuses. There should only be $n$ total. You're listing every element (except $0$) twice.

5:47 PM
Repetition is not allowed, otherwise you can go $1 + 1 + \cdots + 1$ (n times)

So, except for the $\pm$ issue, yes, that's the problem.

@BalarkaSen Thats allowed

Repetition is allowed as many times as you're given to work with.

@TedShifrin Okay, we can "prohibit" negative remainders

You're listing $2n-1$ numbers, most twice, @Knight. It's just wrong.
So, yes, that's the problem, and Balarka gave a sketch of the proof yesterday.

5:49 PM
So, you know Ted the main problem is that we need to know something about $n$ before telling if the sum is divisible by $n$

No.
True for every positive integer $n$.

You're rephrasing the problem weirdly. Just work with whatever was given to you initially

Because take this collection, $$3, 3, 3 ... 3 , n-2$$ Can we say any sum of this collection is divisible by $n$ without knowing what $n$ is?

The solution will depend on $n$, of course. There's not a universal formula.

@BalarkaSen :-) Actually, I worked through it and I progressed from "given $n$ different integers" to "given integers are $0, 1, ... (n-1)$

5:51 PM
Its not much of a progress but ok
The given integers aren't 0, 1, ..., (n-1)

@TedShifrin We cannot deduce something in this manner without knowing $n$ is, do you roger that?

The given integers can be all 1's
@LeakyNun Yeah I have seen that one

see it again

Lol
Ok

5:52 PM
this is with the twitch chat

you got me

@BalarkaSen Actually, what I did was: Let's say we have integers $$\{a_1, a_2, \cdots a_n\}$$ then all of them can be written in the form of $$a_i = n~q_i + r_i$$ and any sum will be of the form $$n (q_1 + q_2 \cdots q_m) + (r_1 + r_2 + \cdots r_m)$$
and $r_i$ s can only range from $0$ to $n-1$

I know what you did. Stating the given integers are "0, 1, ..., n-1" is wrong
They are a collection of n numbers chosen with repetition from {0, 1, ..., n-1}

Yes, Balarka is correct. Sloppy English makes for sloppy math.
@Knight No, I do not roger. I can, in fact, make an argument based on cases with $n\pmod 3$.
Remember, you have $n-1$ copies of $3$ listed.

@BalarkaSen Then I said lets focus on the sum of reminders, because the first term of my last equation is divisible by $n$
@TedShifrin Where was my English sloppy?

5:57 PM
@Knight This

Well, thats not sloppiness. All we can say is that communication is tough over the chat.
Okay, let's go off it

I'm not saying it's sloppiness I'm just saying it's wrong

Have you understood my progress? I changed the problem from "given $n$ integers" to "given some known integers"

Communication need not be tough. One has to learn to say things precisely and correctly. That's an important step in becoming a mathematician.
4

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