Mathematics - 2019-06-27

Akiva Weinberger

12:41 AM

@Adam Isn't there like a free PDF of that one online

user193319

Does the word metric on a group always induce the discrete topology?

Oh, of course. It's obvious. Just center $1/2$ ball centered at any group element.

Akiva Weinberger

12:59 AM

@user193319 What's the word metric?

user193319

proofwiki.org/wiki/Definition:Word_Metric

Too many words for me to type.

Adam

1:10 AM

@AkivaWeinberger yeah you might be right, this computer is a mess so wasn't able to find it despite having EKHAD

on a non related note have they managed to catch that serial killer in America? the one that had everyone believe they were suicides convincingly

user76284

1:47 AM

Can the construction of the integers as pairs of naturals be generalized to the ordinals?

Adam

2:15 AM

explain what you mean using algebraic definitions

potential group of serial killers*

more than one would certainly explain the high level of professionalism

Secret

Last night dream: Very weird large cardinal numbers:

Recall that an inaccessible cardinal is a regular cardinal and also it is some $\kappa > 2^{\alpha}$ for any $\alpha < \kappa$

Adam

are you answering user76284's question? if not it's polite to start with "on an unrelated note"

Secret

ok sorry about that.

@user76284 Ordinals are actually encoded as a well ordering of the natural numbers, at least until you reach $\omega_1$

e.g. $\omega + 1$ is encoded as 1,2,3,4,...,0 algorithmically speaking. No pairs needed

(continue on unrelated note)

Last night dream introduce a very weird kind of large cardinals which is made from non well founded sets (i.e. sets that contain itself such as X={X}). These weird cardinals, are "large" and inaccessible, in the sense that they are larger than any self containing sets, which are already large because they contain themselves as elements and hence an infinitely decreasing sequence. Thus let a non well founded set be Y={Y}. Then these large cardinals are T such that $Y \subsetneq T$ for any $T > Y$

Adam

define an encoding with an algebraic lemma, as well as well ordering

Secret

According to discussions with user21820 on this topic, an encoding is basically a function $f : \Bbb{N}^2 \to \{0,1\}$. Thus e.g. $f(x,y)$ outputs $1$ if $x > y$, otherwise outputs $0$

Hence $f$ is a program that tells you how to order the natural numbers, and hence they encode a given countable ordinal

in Mathworks (Not the main chat!), Oct 24 '17 at 8:56, by user21820

We would thus say that f denotes the ordering 1<2<3<...<0.

for details

Adam

2:32 AM

$f : \Bbb{N}^2 \to \{0,1\}$ absolutely does not necessarily imply $f(x,y)=1$ if $x \gt y$, if I am correct in my interpretation of the former statement, $f$ is any two variable arithmetic function that is equal to either $1$ or $0$, for any $x \in \mathbb N$ and $y \in \mathbb N$

Adam

2:50 AM

But I feel as if I should point out, this is not a medium for collaboration between the vast many of us in the sense that our thoughts are going to be respected as having come from us as individuals, it's more a sham set up to benefit a select few, much like all things are really, and providing answers isn't going to lead you anywhere but perpetually being taken for granted

Adam

3:19 AM

But this is pretty funny, when was $0$ ever considered a natural number?

Ultrafinitism

In the philosophy of mathematics, ultrafinitism (also known as ultraintuitionism, strict formalism, strict finitism, actualism, predicativism, and strong finitism) is a form of finitism. There are various philosophies of mathematics that are called ultrafinitism. A major identifying property common among most of these philosophies is their objections to totality of number theoretic functions like exponentiation over natural numbers. == Main ideas == Like other finitists, ultrafinitists deny the existence of the infinite set N of natural numbers. In addition, some ultrafinitists are concerned with...

en.wikipedia.org/wiki/Ultrafinitism

Ultrafinitism

In the philosophy of mathematics, ultrafinitism (also known as ultraintuitionism, strict formalism, strict finitism, actualism, predicativism, and strong finitism) is a form of finitism. There are various philosophies of mathematics that are called ultrafinitism. A major identifying property common among most of these philosophies is their objections to totality of number theoretic functions like exponentiation over natural numbers. == Main ideas == Like other finitists, ultrafinitists deny the existence of the infinite set N of natural numbers. In addition, some ultrafinitists are concerned with...

Adam

3:43 AM

It's very strange, I mean in that yes I have done quite a bit of work in analytic number theory, I cannot see where they have drawn the conclusion of denying the existence of an infinite natural number line, and yet, they seem to agree that there exists a limit of measurability, which is really the same thing as saying that the set is infinite lol

Secret

Depending on who you ask, you can have the naturals forming a proper class or a set

Adam

Depends on how long I can pretend to care about linguistic nonsense

Secret

The notion of a proper class is only a natural consequence of how Peano axioms work but it does not prove there exists a proper class, only that you can define such in logic

It remains open whether the existence of infinite set can be a theorem

Adam

sounds classy to read

MatheinBoulomenos

6:03 AM

@user193319 they have coefficients in the prime field $\Bbb F_p$ where $p$ is the characteristic of $E$ and $F$ (which is necessarily the same)

so you can consider them as they same polynomial in $\Bbb F_p[x]$

Adam

6:40 AM

JFK - The Speech That Killed Him - About Secret Societies

►

user76284

6:56 AM

Can we learn the mean value of an analytic function in the interior of some region by integrating something along the boundary of that region?

ÍgjøgnumMeg

7:33 AM

Morning

MatheinBoulomenos

@ÍgjøgnumMeg Morning!

ÍgjøgnumMeg

How's it going? :)

MatheinBoulomenos

Pretty well, thanks.

And yourself?

ÍgjøgnumMeg

Yeah not bad, had a fun day yesterday and now I'm back at work T_T

just about to stick an i7 into someone's PC

MatheinBoulomenos

if you want to keep up with my series on category theory, I'm at part 4 right now

ÍgjøgnumMeg

7:44 AM

Oh nice! Will take a look

user76284

7:59 AM

What's the analogue of en.wikipedia.org/wiki/Stieltjes_constants for the prime zeta function?

Secret

8:34 AM

who really is 76284, seemed to be piling a bunch of googlable questions

ÍgjøgnumMeg

user76284 is the unique user between user76283 and user76285

Secret

that's a technically correct and useless answer

(and then they said technically correct is the best kind of correct)

Looking at his chat history, it seems most of the questions are googlable, except for the anormalous messages related to producing new functions

ÍgjøgnumMeg

rofl

Secret

Otherwise, I think I started to understand what Adam is warning about earlier:

6 hours ago, by Adam

But I feel as if I should point out, this is not a medium for collaboration between the vast many of us in the sense that our thoughts are going to be respected as having come from us as individuals, it's more a sham set up to benefit a select few, much like all things are really, and providing answers isn't going to lead you anywhere but perpetually being taken for granted

and providing answers isn't going to lead you anywhere but perpetually being taken for granted

I guess combining with the 76284 data, that kinda justify it

Well at least 76284 is not as JEE machine as gateprep, but it really give me the urge to spam about large cardinals

on another note:

Jun 5 at 0:22, by user76284

What’s the order type of the set of all integer polynomials, where $f \leq g \leftrightarrow \exists x \in \mathbb{Z} : \forall y \in \mathbb{Z} : f(y) \leq g(y)$?

Turns out that RudyRucker book I have been reading in the past weeks have an answer to this question

It's $\omega_1$

Adam

9:05 AM

@Secret come on, don't go and think that a username composed of any alpha numeric characters awards any kind of legitimate identity, everyone is just a unit of a certain monetary fraction of the website owner's profits

we are all Stack Exchange lol

@RyanUnger ok that's quite a number of rhetoric @BalarkaSen compliments yes he knows he is good too I can only assume, therefore probably doesn't get any satisfaction except for the first one you give if he is impressed by you, the rest id keep to references to specific proofs he has shown you if I were to give social advice there

Slereah

Do you think we should insult @BalarkaSen more so that the compliments are all the more special

Secret

probably not a good idea unless you want lightning bolt singularities everywhere due to barlaka folding simplices after simplices

in response to something of the nature of that starred comment

Alessandro Codenotti

@Secret I dont believe this

First of all because this order is not a well order, second there's countably many polynomials

Secret

> One last picture of alef-one. Go back to Figure 42. In this picture we saw how various ordinals can be represented as sequences of functions ordered according to steepness (the <bep ordering). How long a sequence of functions can be found with each function steeper than all the ones before? At least alef-one. That is, if S is a set of functions so that for every g (no matter how steep), there will be an f in S that is steeper, then S must have at least alef-one members.

Ok nvm, I made a read error. They are saying all eventually dominanting functions, not just polynomials

Alessandro Codenotti

I still don't believe the book, the set of functions $x\mapsto nx$ for $n\in\Bbb N$ has this property unless I'm missing something

And regardless having uncountably many members and having order type $\omega_1$ are very different

Secret

9:21 AM

They are talking about aleph 1 as the first number that cannot be injected into $\omega$, so they are talking about $\omega_1$. However, when I first read this I do felt strange. Problem is I do not have as good an intuition nor experience to handle the set of all increasing functions ordered by eventual dominance, thus I have not tried to (dis)prove myself on that yet

Leaky Nun

@AlessandroCodenotti here's a measure theory question

α β : Type u,
_inst_1 : measurable_space α,
_inst_2 : measurable_space β,
ν : measure β,
s : set (α × β),
hs : is_measurable s
⊢ measurable (λ (b : α), ⇑ν {y : β | (b, y) ∈ s})

is this true or false?

ÍgjøgnumMeg

Hey @Alessandro @Leaky

Adam

@Slereah lol no I am just saying it probably only feels rewarding when he impresses himself, if I show my math to my family or my cat and the miracle of them saying something nice occurs, I don't think I would be any happier with the math, so if he is the same in that regard, it only feels good when there is mutual appreciation for something he has done

Thomas Shelby

Does anyone know the meaning of "unitary inner product"?

s.harp

9:38 AM

@thomasshelby likely a complex scalar product

Lorenzo

mbari

Alessandro Codenotti

@LeakyNun translate it in English? I'm not that familiar with lean yet, still learning it

Hi @ÍgjøgnumMeg

Actually let me translate it

Suppose that $\alpha$ and $\beta$ are measurable spacces, $\nu$ is the measure on $\beta$ and $s\subseteq \alpha\times\beta$. Then $s$ measurable implies that for all $b\in\alpha$, $\{y\in\beta\mid (b,y)\in s\}$ is $\nu$-measurable?

(If my translation is correct the theorem is false by the way)

Slereah

9:54 AM

@Adam Are you saying we are all as dumb beasts next to @BalarkaSen's majesty

Alessandro Codenotti

Specifically let $V\subseteq\Bbb R$ be the Vitaly set and consider $V\times\{0\}\subseteq\Bbb R^2$. It is measurable having measure $0$, but obviously not all slices are measurable

The correct version is "let $(X,\mu)$ and $(Y,\nu)$ be spaces with measure. Suppose that $S\subseteq X\times Y$ is $\mu\times\nu$-measurable. let $S_x=\{y\in Y\mid (x,y)\in S\}$ and $S^y=\{x\in X\mid (x,y)\in S\}$. Then $S_x$ is measurable for $\mu$-almost every $x$ and $S^y$ is measurable for $\nu$-almost every $y$."

The converse can fail badly, you can ask all slices to be measurable and yet $S$ can be nonmeasurable, you can even ask all slices to be open while keeping $S$ nonmeasurable

@AlessandroCodenotti Vitali*

Leaky Nun

@AlessandroCodenotti eh

(λ (b : α), ⇑ν {y : β | (b, y) ∈ s}) is a function from alpha to [0,infty], and measurable means that the preimage of measurable sets is measurable

also I don't think the Lebesgue sigma algebra on R^2 is the product of the Lebesgue sigma algberas on R

Alessandro Codenotti

No, it's the completion of the product

Leaky Nun

right, so your counter-example wouldn't fit

because we're using the product sigma algebra

Alessandro Codenotti

Which is why some people define the product algebra as the completion of the product, because you really want to say that every subset of a measure zero set is measurable with measure zero

Leaky Nun

10:06 AM

but we don't have a measure on alpha

Alessandro Codenotti

I don't see how's that relevant, build a counterexample with a measure on alpha and then forget about the measure?

Leaky Nun

well, again, we are not using the completion of the product sigma algebra

so your counter-example wouldn't work

Alessandro Codenotti

Sure then it works just because the inclusion maps $X\to X\times Y$ are measurable

Leaky Nun

eh, what's your counter-example?

Alessandro Codenotti

@AlessandroCodenotti I have none, it is true with the product $\sigma$-algebra

Leaky Nun

10:19 AM

I'm confused

Alessandro Codenotti

If you use the product $\sigma$-algebra on $X\times Y$ then $S\subseteq X\times Y$ measurable implies that $S_x$ and $S^y$ are measurable for all $x,y$

Leaky Nun

but does that mean the measure is measurable

18 mins ago, by Leaky Nun

(λ (b : α), ⇑ν {y : β | (b, y) ∈ s}) is a function from alpha to [0,infty], and measurable means that the preimage of measurable sets is measurable

Alessandro Codenotti

Oh, you mean the function $b\mapsto\nu(S_b)$?

Yes that is also measurable

But it's not as straightforward

Look at @Ryan, summoned by measure theory

Leaky Nun

eh... how should I prove it?

Alessandro Codenotti

Hmm actually maybe you need $\sigma$-finite measure spaces, because this is usually proved as part of Fubini

Leaky Nun

10:26 AM

right, but I don't even have a measure now

and the measure on beta isn't sigma-finite

Alessandro Codenotti

10:43 AM

I'm pretty sure there's counterexamples with sigma infinite stuff

Adam

@Slereah well no I didn't say that at all, that interpretation indicates it isn't worth me paraphrasing, but no I never implied he was from a royal family either. I don't know what the social environment is like in India, I doubt they have a kings etc, if they do they are probably like ours, totally unimportant super rich reality tv stars

also everyone should look at the way @LeakyNun writes questions this makes it far easier for people not necessarily in that specific field of mathematics to attempt it and or learn from the experience, asking in natural language with a f%$# ton of field specific terminology is just silly

Leaky Nun

11:00 AM

@AlessandroCodenotti I'm pretty sure the sigma-infinite stuff either gives you non-uniqueness of measure or failure of equality in Fubini

rather than non-measurability

user193319

11:10 AM

In the context of local compactness, what is the definition of a compact neighborhood?

Alessandro Codenotti

@LeakyNun Cohn states your theorem for $\sigma$-finite spaces

I'm trying to figure out whether this is to make the proof easier or to make the proof possible

(fun fact: for non sigma-finite spaces Fubini can fail for all possible product measures)

MatheinBoulomenos

11:25 AM

@user193319 a compact subset that contains an open set containing the point $x$ is a compact neighborhood of $x$

user193319

@MatheinBoulomenos Ah, I see. Thanks!

Adam

11:51 AM

odd asf

Adam

1:24 PM

It's a little strange that some people can have a career and declare themselves to be ultrafinitists. Sure ok academics is in all likelihood like any other industry now, it isn't what you know it's who you know. Actually it was always like that wasn't it

oh well

Adam

1:36 PM

psychiatry is by far the most corrupt industry by a long shot tho

in my own personal experience I mean, of course there are banks that have stolen billions from people and never spent a single day in a cell, so round that up to all authorities and we have a winner

cchr.org/cchr-reports/eroding-justice/introduction.html

anakhro

@Adam ultrafinitism is a philosophical view. It's really no different from being religious.

It's not like it makes their research in mathematics any less compelling.

Adam

I am aware of that but I'm just curious as to how Doron can have a career and still go in opposition to classical view, sure yes his algorithms are very impressive we study very similar areas of number theory, that's the only reason I came across them, but we live in a world where majority rules, popular opinion is equal to correct opinion in an institutionalized mind set, so, I don't make any bold assertions like that, only state what i know to be true, its just interesting that's all

anakhro

1:52 PM

@Adam Why would being an ultrafinitist preclude having a career? I don't follow.

Adam

Because it's a minority

anakhro

Why does that preclude having a career?

Adam

I already explained why i think this

anakhro

I don't see why any of that has to do with having a career as a mathematician.

Adam

I guess already being important might explain it so if i were to guess i would have to go with nepotism. Yes it shouldn't have anything to do with it, but it does. \

anakhro

1:59 PM

It's amusing when people try to reduce the careers and success of others down to unfair advantage.

Adam

well no I wasn't talking about the success and already stated I hold his work in very high regard, I only implied this for the career of the individual

do you truly believe we are all given equal opportunity @anakhro ? That's very amusing and defensive of you

anakhro

His career is his success.

I never said everyone was given equal opportunity.

But I merely find it amusing when people reduce success down to unfair advantage.

Adam

no they are not the same thing. So in the hypothetical scenario of what I implied being true, which you have agreed, does happen, you find this amusing?

anakhro

I never said they were the same thing. Imagine it is an inclusion, not an equality.

Ryan Unger

@AlessandroCodenotti that was my phone doing mysterious things while I was asleep

anakhro

2:05 PM

$\text{career}\subseteq\text{success}$

Ryan Unger

@Adam I've known Balarka for years

Adam

no it's just a subset of factors that contribute to the person's standard of living. success isn't measurable

anakhro

I don't know what it being measurable has to do with the conversation.

MatheinBoulomenos

is success a vitali set?

Secret

that will be a very bad idea

In fact, I think success is beyond measurable

(Think berstein set but more inconceivable by picking a suitable large cardinal axiom)

on another note

Can we have a $\eta_f$?

MatheinBoulomenos

2:14 PM

@Secret you mean a morphism going between morphisms? Concepts like these exist, and the lead to higher category

Ryan Unger

higher nonsense

MatheinBoulomenos

in a 2-category for example, you have objects, which are 0-morphisms and 1-morphisms going from objects to objects and you have 2-morphisms going between 1-morphisms

Ryan Unger

I hear Jacob Lurie is going to be here soon

Wonder if he'll teach courses

Adam

well, a career is a good measure of how comfortably the individual lives, and how much appreciation for their work they have received from others. This is not in any sense a measure of success

MatheinBoulomenos

The category of small categories is such a 2-category, with objects being small categories, 1-morphisms being functors, and 2-morphisms being natural transformations

another example is the category of topological spaces with 2-morphisms being homotopies between continuous maps

Secret

2:17 PM

I see. I just had this thought because it will look more like an interpolation if $\eta$ literally translate the top row of arrows to the bottom row, and then for some reason I started to imagine the whole thing to behave like a homotopy where the top row of the commutative diagram looks like a directed curve to me

snipped

MatheinBoulomenos

that's a good intuition to have

in fact, I have a section on the analogy to homotopies

hello.

hi.

hey.

wagwan.

2:23 PM

sound

Haaaabe d'eeehri

hello there

General Kenobi

behold the masterful princeton landscaping

Simone

lol 39°C right now

I'm sublimating

ÍgjøgnumMeg

2:26 PM

f*ck that

Adam

I've got better puddles at home now that I found a pressure cleaner

Simone

Italy is supposed to be temperate

Ryan Unger

29 here

Adam

my mother was in Italy recently actually

Simone

visiting?

Adam

2:28 PM

yeah i think it was Venice? I wasn't really listening but yeah her brother's wife is Italian so they have family over there enabling her to stay for over a month which was great

Simone

nice

I've been to Venice only once.

ÍgjøgnumMeg

Austria is super hot in the summer :(

Simone

I live in Tuscany

Adam

Oh ok yeah I did plan at one stage to do one of those semesters abroad things in Budapest, but I literally don't want to travel unless I can live in that place for an extended period, I just don't see the point in vacations personally

Simone

@ÍgjøgnumMeg given the name I'd imagined you were icelandic.

icelandian?

Viking!

ÍgjøgnumMeg

2:32 PM

@Simone the name is Faroese :) But actually I'm English

Adam

My ancestors are French lol apparently they fled from the French revolution to England and then well by no wrong doings ill assume one generation ended up here in Australia

Simone

old blighty eh?

Akiva Weinberger

@RyanUnger Hey you don't happen to know a guy named Clem Brown, do you?

Ryan Unger

@AkivaWeinberger no, what makes you ask

ÍgjøgnumMeg

Screw the UK

Awful, awful place

Ryan Unger

2:34 PM

I agree

Akiva Weinberger

He's also in Princeton

Ryan Unger

an undergrad?

Simone

@RyanUnger Homonymy I presume?

Akiva Weinberger

Yeah

Ryan Unger

@Simone wot

Simone

2:35 PM

@RyanUnger This

I forgot to tag that

Adam

I'd like to go to America but not die in a new York apartment owned by one of the Olsen twins. I mean ketamine vacuums

same thing really

user193319

Let $X$ be some compact Hausdorff space, let $C(X)$ denote all of the continuous complex valued functions on $X$, and let $A \subseteq C(X)$ be a subalgebra. If $f \in A$, does it follow that $|f| \in A$?

$|f| = \max \{f,-f\}$, but I don't think that is helpful...

Leaky Nun

@user193319 that's false

Adam

this guy would be really interesting to meet wow what an extraordinary life to have lived en.wikipedia.org/wiki/William_Breeze#Gnostic_Catholic_Church

Ryan Unger

@AkivaWeinberger I don't know any undergrads here

@Simone lmao my friend just used this word on the NTY spelling bee

nice one

Leaky Nun

2:47 PM

I feel like there is no finite-dimensional counter-example

Ryan Unger

@LeakyNun ?

Leaky Nun

9 mins ago, by user193319

Let $X$ be some compact Hausdorff space, let $C(X)$ denote all of the continuous complex valued functions on $X$, and let $A \subseteq C(X)$ be a subalgebra. If $f \in A$, does it follow that $|f| \in A$?

Ryan Unger

oh

Leaky Nun

I was trying to think of a counter-example

Ryan Unger

he said complex valued

and then wrote max

:thonk:

Leaky Nun

2:48 PM

1 min ago, by Leaky Nun

I feel like there is no finite-dimensional counter-example

because $f^n$ are going to be generally linearly independent

unless $f$ consists of roots of unity or zero

in which case the magnitude contains only $1$ and $0$

maybe $(i,0)$ will work

no it won't

no it won't work

because $f^N$ would be $|f|$ for some big $N$

and I'm not going to generate infinite-dimensional counter-example

MatheinBoulomenos

$(-2,0)$?

Leaky Nun

(1,1) and (-2,0) are already linearly independent

I'm interpreting subalgebra as $\Bbb C$-subalgebra

so we want a function $f$ such that there is no polynomial $p$ with $p(f) = |f|$

thinking about it universally... $|z|$ is not a polynomial in $z$ because of complex analysis

so let $X$ be the closed unit disc (so that it's compact and we have the identity theorem)

and let $f(z) = z$

then this should be a counter-example

@user193319 there you go

$|f|$ is not in the $\Bbb C$-subalgebra generated by $f$

user193319

Ah, very nice! Thanks!

MatheinBoulomenos

note that if the $\Bbb C$-subalgebra is closed under conjugation and topologically closed, then it will satisfy $f \in A \Rightarrow |f| \in A$

Leaky Nun

how?

you only have $|f|^2 \in A$ right

MatheinBoulomenos

2:59 PM

yeah, but you can use some trickery with the taylor series of the square root function

Leaky Nun

hmm

I'm not very convinced

user193319

The subalgebra generated by $f(z) = z$ is not topologically closed, right?

Ryan Unger

@MatheinBoulomenos the question didn’t say that though, did it?

You do that for the proof of Stone-Weierstrass

MatheinBoulomenos

@RyanUnger no, it didn't, I just thought it was interesting

@RyanUnger exactly, that's where I saw it

Leaky Nun

@user193319 right

the subalgebra is $\Bbb C[z]$

the closure is a subalgebra of $\Bbb C[[z]]$

user193319

3:09 PM

Topological closure?

Leaky Nun

right

MatheinBoulomenos

I think the topological closure is convergent power serires, not formal ones

Leaky Nun

so I said subalgebra

MatheinBoulomenos

a right

flowian

3:36 PM

Please recommend me good book to learn undergrad algebra. I took algebra course at uni which covered speedily topics of basic structures, vector fields, transformations, polynomials, characteristic polynomials, residue classes and matrices.
I didn't understand well many things so I wish to re-learn by myself things.

don't really care for uni content, I'm rather interested in learning these

Adam

4:23 PM

no.

flowian

4:37 PM

:(

Adam

Read this instead en.wikipedia.org/wiki/Psychological_operations_(United_States)

flowian

the link does not seem to lead to algebra topics

Thorgott

I will go so far as to dare claim that this was Adam's intent all along

flowian

I suspect the same thing

Adam

5:07 PM

lol how could you have guessed

ok well you can learn about multisets before you learn about vectors I guess

Simone

it would be nice if senior users were to put up a reading list.

Adam

definitely and demonstrating your knowledge of said reading list is guaranteed to lead us all to career opportunities

flowian

5:28 PM

who cares about career opportunities

But would be nice to have such list indeed, like in reddit.

Adam

flowian

https://www.reddit.com/r/learnmath/comments/8p922p/list_of_websites_ebooks_downloads_etc_for_mobile/

https://www.reddit.com/r/math/comments/7i9t5y/book_recommendation_thread/

https://www.reddit.com/r/math/wiki/faq#wiki_what_are_some_good_books_on_topic_x.3F

ÍgjøgnumMeg

5:58 PM

*Research*
My research is on number theory, arithmetic geometry, topology and ways to avoid administrative work.

From Ted Chinburg's homepage

MatheinBoulomenos

lmao

ÍgjøgnumMeg

I've just discovered that he has a 4 semester course on ANT on youtube

MatheinBoulomenos

oh wow

4 semesters is a lot. We did ANT up to local and global CFT and some stuff beyond that in 3

ÍgjøgnumMeg

Was just looking at the math reddit that @Flowian posted and it links to an MO post that is essentially "Let's compile a list of all the full courses on various subjects that have been made publicly available"

So thanks a lot indeed, @Flowian

99% of Semester 4 focusses on Kato's Class Field Theory

whatever that is

Daminark

@ÍgjøgnumMeg it's ordinary CFT but the mathematical objects are all personified

ÍgjøgnumMeg

6:06 PM

@Daminark wtf man

MatheinBoulomenos

I guess that's higher-dimensional CFT

as Kato worked on that

ÍgjøgnumMeg

I see

MatheinBoulomenos

that's some pretty deep stuff

ÍgjøgnumMeg

what "dimension" does the higher refer to?

MatheinBoulomenos

for example in the local case, you can define dimension inductively: a 0-dimensional local field is a finite field and then a n+1-dimensional local field is the quotient field of a complete DVR whose residue field is a n-dimensional local field

ÍgjøgnumMeg

6:08 PM

Ah nice

MatheinBoulomenos

for example $\Bbb F_q((x_1, \dots, x_n))$ (formal Laurent series in n variables) has dimension $n$

I actually don't know the definition in the global case

another example would be $\Bbb Q_p((x))$ which has dimension $2$

ordinary local fields are those of dimension 1

ÍgjøgnumMeg

Right

So which complete DVR is $\Bbb Q_p$ a quotient of?

MatheinBoulomenos

$\Bbb Z_p$

ÍgjøgnumMeg

and it's residue field is just $\Bbb F_p$ again so you get a $0$-dimensional local field?

its*

Leaky Nun

$\Bbb Q_p[[X]]$?

he said quotient

MatheinBoulomenos

6:10 PM

oh right

yeah

ÍgjøgnumMeg

I see

Leaky Nun

oh, quotient field

then it's $\Bbb Z_p$

MatheinBoulomenos

there's also some examples with Witt vectors

ÍgjøgnumMeg

@Leaky No you got what I meant the first time

MatheinBoulomenos

the $p$-typic Witt vectors of $\Bbb Q_p$ are a complete DVR with quotient $\Bbb Q_p$ and you can take the quotient field of that

basically you have two ways to go up a dimension: either take formal power series or Witt vectors

all higher dimensional local fields are constructed from finite fields like that

ÍgjøgnumMeg

6:13 PM

Cool :)

user193319

If $\mu$ is a complex measure on a locally compact space $X$, what does $|\mu|(E)$ denote, where $E$ is some subset of $X$?

ÍgjøgnumMeg

Fesenko was working in Nottingham (where I almost went for my Masters) and he does a bunch of higher dimensional weird mumbo jumbo

MatheinBoulomenos

@ÍgjøgnumMeg yeah I know, he's another big name in the field

ÍgjøgnumMeg

that's where the Nottingham group comes from!

Tobias Kildetoft

@ÍgjøgnumMeg Isn' Fesenko the guy who keeps claiming that people who don't understand Mochuzuki's work on IIUT have just not spend enough time on it?

ÍgjøgnumMeg

6:16 PM

@Tobias yeah something like that

he runs a seminar on IIUT in Nottingham I believe

IUTT

lol

Tobias Kildetoft

sure, whatever

He is also being really aggressive in his defense of the theory

ÍgjøgnumMeg

I gotta jump in the shower, cy'all later

s.harp

is mochuzuki's work culturally like the crazy shit deligne did in the 70-80s?

the comparison is probably stupid for a million obvious reasons i guess

Adam

Shinichi Mochizuki's father was Fellow of Center for International Affairs and Center for Middle Eastern Studies at Harvard University interesting stuff

s.harp

everybody's father was somebody, sometimes even somebody else

Adam

6:27 PM

My Dad was a farmer

s.harp

there is a tongue twister there with further, father, farther, farmer and falter

MatheinBoulomenos

6:57 PM

Hi @Ted

Ted Shifrin

Hi @Mathein

MatheinBoulomenos

How are doing? Did you have fun with your friends?

Ted Shifrin

Yes, fun and some tremendous food. I took them to my favorite restaurant last night.

MatheinBoulomenos

Sounds nice. I applied for intro algebra TA for next semester and got accepted. I went over my past evaluations for the application (since I was asked for them) and noticed how I improved over time (I've been a TA 3 thrice)

Ted Shifrin

Félicitations! ... Most people have a rocky start in teaching and those who care do improve. We've commented before on your improvement in explanations and patience in the chatroom :)

Daminark

7:06 PM

Hello

MatheinBoulomenos

@Daminark happy birthday!

Ted Shifrin

Howdy, nerd Demonark. Oh oh, are you a year older and allegedly wiser?

ÍgjøgnumMeg

Happy birthday @Daminark

assuming that's true

ÍgjøgnumMeg

7:27 PM

Congrats @Mathein

on the TA position

Daminark

Hey sorry I was out for a bit, thanks guys!

@TedShifrin yeah aging a year in one day is quite a feeling :P

MatheinBoulomenos

@ÍgjøgnumMeg thanks

Daminark

Good thing you've been improving! Does TAing in your case give you the chance to do stuff like problem sessions?

cdt

@BalarkaSen Remeber I was telling you of meeting an ISI M Math interview guy who admitted that he had no fucking idea of compactness ? Well now that results are out I have seriously no fucking idea (I remember he telling to someone near me after the interview that he couldn't even prove that if a, b are nilpotent then a+b need not to be nilpotent even after crapload of hints and time) how he got selected in M Math that too with a single digit rank...

MatheinBoulomenos

@Daminark yes, TAing means grading and doing problem sessions

at least here

user76284

8:47 PM

Is there a Schwartz function $f$ such that $\int_0^\infty \frac{f(x)}{x} dx = 0$ and $\lim_{x \rightarrow 0} f(x) = 1$?

user76284

8:59 PM

Or more directly, is there a function $f$ such that $\lim_{\varepsilon \rightarrow 0^+} \sum_{n=0}^\infty \frac{1}{n} f(n,\varepsilon)$ exists and $\lim_{\varepsilon \rightarrow 0^+} f(n,\varepsilon) = 1$?

Such an $f$ would be a "regulator" for the harmonic series.

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$Mathematics$ Mathematics