Mathematics - 2018-12-08 (page 1 of 3)

Lucas Henrique

00:08

Hi chat.

I was thinking: in FOL we can instantiate the unique object that satisfies a property by defining a new constant that satisfies and using that “satisfies property $\implies$ equals to constant”. What about usual notation (can be thought of notation for the element of a singleton)? If $f$ is a function and $c$ a variable within its range, how do we define $f(c)$? Actually, how do we formally define $f(c)$ in FOL?

Semiclassical

@Rick what’s unclear about the definition? A graph is bipartite if you can split the vertices in two groups such that there’s no edges between vertices in the same group

Leaky Nun

@LucasHenrique in FOL we just say here are a bunch of function symbols, constant symbols, and relation symbols

Robert Frost

@Fargle this is a good place to jot ideas: stackedit.io/app#

2

Secret

00:25

Looks cool

Maybe I can migrate all my ramblings there

hmm... does it support mobile?

Robert Frost

@Secret not sure - never tried.

Learner

00:56

Could someone help me with a triple integral? Cylindrical coordinates

Symposium

@CaptainAmerica16 I used Ethan D. Bloch's analysis text. It was amazing. I didn't even know he had abstract algebra text.

Secret

01:12

Gaussian 16 quote of the day:

> "MATHEMATICS IS THE ART OF GIVING THE SAME NAME
TO DIFFERENT THINGS."
- H. POINCARE

Holo

@Secret it=this site? If so then yes

> "mathematician's reputation rests on the number of bad proofs he has
given." - A.S. Besicovich

Secret

@Holo ah I mean stackedit

Holo

@Secret So no idea

Secret

In other news, I am currently reading an arxiv that is a set of notes on finite sets, and currently experimenting on defining something I called a "subfinite set"

arxiv.org/abs/1509.02747

In particular, the notes give a way to define finite sets without referencing to the natural numbers, and then proved that they are equivalent

Holo

@Secret I think "the power set of $A$ is D-finite"="$A$ is finite"

Or the power set of the power set? Something like this

Secret

01:26

That I don't remember, though I recall seeing a theorem that mentioned about taking power sets of D-finite sets twice and that is not necessary D-finite or something
(cont. on arxiv) Basically, given any set $E$, you can find some subset $\mathcal{S}\subset \mathcal{P}(E)$ such that $\mathcal{S}$ is an inductive-$E$ system. This is something which satisfies the following:

$\varnothing \in \mathcal{S}$

$B \cup \{a\} \in \mathcal{S}$ for all $B \in \mathcal{S}^p$ and for all $a \in E\setminus B$

where $\mathcal{S}^p$ is the set of subsets of $\mathcal{S}$ that are proper subsets of $E$

Rick

@Semiclassical is there a more visual yet correct way of stating that to a person who knows nothing about graphs . I need to be able to explain this to a retarded person. By retarded I mean someone who likes to make sense of things with their gut

Secret

If you run this procedure on any finite ordinal for example, you will notice that the definition of inductive-$E$ system enumerates the elements in the set in some systematic fashion that does not need to take account of order

Holo

Yes, "the powerset of the powerset of A is Dedekind-finite"="A is finite"

Just checked

Secret

The author than defines $E$ as finite iff $\mathcal{P}(E)$ is the only inductive-$E$ system

Holo

I see

I never heard about this definition

Wait, $\cal S$ is in ${\cal P(P(}E))$, so how subset of $\cal S$ be a subset of $E$?

Secret

01:51

Let's take an example: e.g. E={0,1,2}

Using the definition of inductive E systems, we have:

$\mathcal{P}(E) =$ {0,{0},{1},{2},{0,1},{0,2},{1,2},{0,1,2}}

Let's say $\mathcal{S}$ = {0,{0},{0,1},{0,1,2}} then $S^p \in \mathcal{P}(S) = \mathcal{PP}(E)$ will be {0,{{0}},{0,{0}},{{0},{0,1}}, too messy to write down}

Observe that the element {0,{0}}={0,1} in $S^p$ is a proper subset of $E$

Thus I think the author's definition of $S^p$ allows us to keep track on which stage of induction we are in

Holo

So we start with the definition of ordinals?

With ordinals we can define the natural numbers tho...

Secret

No I just use ordinals as an example. The author in that arxiv linked above actually does it abstractly for any set (because I am terrible with nesting notations), which is why in the past 2 weeks I have trouble understanding it

He will eventually proved that this is equivalent to the finite ordinals

Holo

Something in this definition seems wrong:
$\mathcal{S}\subset \mathcal{P}(E)$, $\mathcal{S}$ is an inductive-$E$ system. This is something which satisfies the following:
$\varnothing \in \mathcal{S}$
$B \cup \{a\} \in \mathcal{S}$ for all $B \in \mathcal{S}^p$ and for all $a \in E\setminus B$
where $\mathcal{S}^p$ is the set of subsets of $\mathcal{S}$ that are proper subsets of $E$

Secret

> The power set of a set E, i.e., the set of all its subsets, will be denoted by P(E) and the set of non-empty subsets by P0(E). If E is a set and S is a subset of P(E) then Sp will be used to denote the set of subsets in S which are proper subsets of E.
Finite sets will be defined here in terms of what is known as an inductive system, where a subset S of P(E) is called an inductive E-system if ∅ ∈ S and F ∪{e} ∈ S for all F ∈ Sp, e ∈ E \ F. In particular, P(E) is itself an inductive E-system.

Holo

"the set of subsets in S which are proper subsets of E", not of

Eran

02:00

How can i find an example of a group G where Z(G) and [G,G] don't have any containment relation , i.e neither Z(G) <= [G,G] , or [G,G] <= Z(G)

Holo

Okay, now it makes sense

Secret

So $\mathcal{S}^p \subset \mathcal{PP}(E)$

Eran

Where [G,G] is the commutator subgroup and Z(G) is the center of G

Holo

No, it still does not make sense. $S^p\in PP(E)$ means that $x\in S^p\implies x\subseteq P(E)$, not $x\subseteq E$...

Secret

hmm... In that case I don't really understood what "then Sp will be used to denote the set of subsets in S which are proper subsets of E"

Trying to comprehend that statement alone took me 2 weeks

Holo

02:07

I think that it should be $S\in P(E)$, not $S\subset P(E)$, otherwise I don't understand it either

Resend the link I will read the whole part, not only the definition. Maybe this will help

Secret

arxiv.org/abs/1509.02747

It's very early in the page in the introduction

page 3 to be exact

Holo

Okay, after reading a bit about Kuratowski-finiteness I think that he mean:

The set $P(E)$ is inductive E-system if for all $A\subseteq P(E)$ we have ($\emptyset\in A$ and $X\in A\implies X\cup\{x\}\in A$ for $x\in E$) impies $E\in A$

@Secret

But I got this from the reference, not from the paper

Secret

02:31

hmm... seems like the usual induction but applied to all the subsets of $E$

Holo

@Secret Yes, it is saying: induction on any subset of E will eventually reach $E$ itself

Secret

ok that is a lot easier to work with

(than that nested mess of $S^p$ in that paper)

Holo

02:46

Yea, I don't get what the paper :/

Daniel ML

05:01

@Daminark wooow that's beauty ❤️

CaptainAmerica16

@Symposium How was his book on analysis? I might look into it after I finish my current one - assuming it's written at a higher level.

Learner

08:09

0

Q: With $\alpha>0$, how to calculate the volume of $E_\alpha=\{(x,y,z)\in\Bbb R^3: x^2+y^2-\alpha^2\le\alpha z\le\alpha(3\alpha-2x)\}$

With $\alpha>0$ let $E_\alpha=\left\{(x,y,z)\in\Bbb R^3:\frac1\alpha(x^2+y^2-\alpha^2)\le z\le 3\alpha-2x \right\}.$ I'm asked to find the volume of this set, which is $V_\alpha=\iiint_{E_\alpha}dxdydz.$ Now, I think I should use the cylindrical coordinates $x=r\cos\theta, y=r\sin\theta, z=z$. ...

Any help?

Alessandro Codenotti

08:21

@LucasHenrique We don't, a function is already defined (in the metatheory)

Leaky Nun

09:11

@AlessandroCodenotti I looked at the answer

I'm like "wow this is dank"

Adam

09:25

Hey if anyone is interested I found a good paper on Gaussian binomial coefficients. Q-BINOMIALS AND THE GREATEST COMMON DIVISOR, Keith R. Slavin,INTEGERS:ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY Volume 8 (2008).

It primary is to do with the greatest common divisor function oc, but Gaussian Binomial Coefficients count the number of subspaces in a $n$ dimensional vector space I think, so yeah it should have to do with all the bizarre objects and large wordy math that seems popular

Nobody recognizeable

09:52

Can someone help how to solve $\frac{d^2f}{dr^2} + \frac{2}{r} \frac{df}{dr}=0$

Secret

11:52

@Holo ok using the definition in the reference, I think $\mathcal{S}^p$ is actually the set, whose elements are proper subset of E, and $\mathcal{S}^p \subseteq \mathcal{P}(E)$

Then the two definitions are equivalent

Holo

12:59

@Secret make sense, but, the way it is written is so confusing

@Nobodyrecognizeable multiply by $r^2$ and use this method

Secret

13:19

Yeah, he seems to like to use a lot of nesting in his definitions

In other news, the longer I look at the statement "The empty set is a subset of any set" and how $\varnothing \in \mathcal{P}(\varnothing)$ and $\varnothing \subset \mathcal{P}(\varnothing)$ are both true, the more I have this thought that any set has uncountably many "empty string elements"

I mean e.g. the subsets of the singleton {a} depends on where you put some partition | through that set. If you do it like this: {|a}, then the left side is $\{\} = \varnothing$ and the right side is $\{a\}$

And you can imagine computing $\mathcal{PP}(\{a\})$ which basically does the following:

$\{|a\} = \{\varnothing, \{a\}\}$, and then:
$\{|\varnothing, \{a\}\} = \{\varnothing,\{\varnothing, \{a\}\}\}$
$\{\varnothing,| \{a\}\} = \{\{\varnothing\},\{\{a\}\}\}$
thus we have in total: $\{\varnothing,\{\varnothing, \{a\}\},\{\varnothing\},\{\{a\}\}\}$

So in theory:

$\{|||...||a\} = \{|a\} = \{a\}$

Holo

@Secret I would say that this is extremely bad way to look at this(as it helps only for singletons) and would say that there aren't uncountably many, there are a proper class many

Secret

yeah, it is really weird to have a humongous amount of nothingness in something

Nobody recognizeable

13:35

@Holo thanks for the help.

Holo

@Nobodyrecognizeable No problems

Ultradark

13:48

hi

christmas_light

hi

Ultradark

anyone care to discuss persistant homologies

Alessandro Codenotti

14:07

@LeakyNun Where? Which proof did you see?

Secret

14:47

[Random]

A cosmology of ZFC:

In the beginning, there is nothing $\varnothing$ and God (Proper classes)

From nothing, comes something $\{\varnothing\}$ and from something comes more things 2,3,4,5,...

Then God commanded: Let there be the axiom of infinity, and then the cosmology populated by nothing, something and God splits to accomodate a new entity, the countable $\Bbb{N}$

Then God diced $\Bbb{N}$ into many pieces. The first uncountable came into existence

And as his final task before going to rest, he commanded: Let there be the axiom of choice. And the infinite cosmos is brought into well order. The creation of ZFC is completed

Holo

You are missing some axioms tho

Secret

that's true, hmm...

Trouble is, the axiom of empty set means that in the beginning there isn't really nothing, which is very strange to conceptualise

I guess we can probably get away with in the beginning, there is only God (because ZFC make no explicit reference to proper classes)

Holo

In ZF(and ZFC) there are no proper classes

There are sets and that it

If something is not a set then ZF(C) does not see it

Secret

Trying to conceptualise a state of being before there is even nothingness is going to be challenging...

Let me think about how to rewrite this when I check my latest batch of chemistry calculations...

Abhas Kumar Sinha

15:14

hello

can any one here can help me understanding contour integral

and how it is different from regular integral?

@Secret hello

what is that?

$$\oint$$

^ this thing

For example if an object goes and comes back then: $$\oint F.dx = 0$$

How?

Fargle

So a regular integral can be thought of as being over a "path" in the 1-dimensional real line. But those paths simply look like intervals, so you can speak of an integral "from a to b".

Abhas Kumar Sinha

I didn't understood, I'm very very new to calculus

:P

@Fargle Can this be used for circle's eqs

?

Fargle

I'm not sure what you mean by the question.

Abhas Kumar Sinha

$$x^2 + y^2 = 36$$

@Fargle You mean that for a closed curve, it should be zero?

Fargle

@AbhasKumarSinha It depends---in the real numbers, that's certainly true, because $\int_a^b f dx + \int_b^a f dx = 0$, and it's like we're integrating over a closed curve in 1 dimension, starting at $a$, going to $b$, and ending at $a$

Abhas Kumar Sinha

15:26

@Fargle Can you give me an equation for example, I think that I can understand better with it

Aidan Rocke

Does anybody here have a strong interest in statistical mechanics?

I have a specific question: mathoverflow.net/questions/317192/…

Secret

@AbhasKumarSinha That means integrating along some closed curve

It's an integral symbol that appears frequently in many domains of calculus as well in physics and engineering

Abhas Kumar Sinha

@Secret can you give me example of such curve? plz

Fargle

Abhas Kumar Sinha

wow

Fargle

15:30

This is what a line integral looks like over a curve in a scalar field. You literally just do a regular integral, but by thinking of the curve as your interval.

Abhas Kumar Sinha

can any mathematical eqs be used as an example?

like circle? it is also a closed curve?

Fargle

Consider $z = x^2 + y^2$, and the curve $C: x^2 + y^2 = 1$.

Abhas Kumar Sinha

okay

then how to take intervals?

how do you know what intervals are to be taken?

Fargle

Parametrize the curve $C$ with the vector function $\vec{r}(t) = (\cos t, \sin t)$.

Abhas Kumar Sinha

what is parametrize?

Is it necessary to convert every eqs into vectors? like ellipse too?

Fargle

15:35

With no offense meant whatsoever, it's probably best that you wait to tackle this until you've seen parametric equations and multivariable calculus/vector calculus in a bit more completeness.

Abhas Kumar Sinha

Is there something like multivariable calculus?

I know how to do things for single variable or convert the variable into a single dependent variable and integrate

Fargle

Multivariable calc (along with parametric stuff, vector calc, and line/surface integrals even) can usually be found in standard calculus textbooks

Off the top of my head, the "complete" versions of Rogawski and Larson both feature several chapters on multi

Abhas Kumar Sinha

which standard, I'm just in high school

I'll start the IA Maron's Calc in single variable next month

Fargle

It looks like that book won't cover multivariable at all...hmm.

I'd personally recommend Rogawski's Calculus: Early Transcendentals as a book that covers basically everything you'll need, both single- and multivariable.

Abhas Kumar Sinha

will that help in jee?

Fargle

15:42

I'm not sure, I have no experience with the JEE

Secret

Let me repeat:

Abhas Kumar Sinha

how do you do that?

i see

@Fargle how can I cover that 1200 page book? is that the correct one I've?

Pandya

don't use offensive words.

Secret

So yeah, just wait until you get on uni and then multivariable calculus will become clear

Abhas Kumar Sinha

@Secret i'll try to wait'

Secret

15:46

It is an interesting and important field as you will get to see in a few years

Fargle

@AbhasKumarSinha Sounds like the right one. It will take a while. Most college students in the US take about a year and a half to go through the whole thing, over 3 courses

I of course can't speak to how things are done anywhere else.

Abhas Kumar Sinha

@Fargle What can I be if I take maths and physics as a major?

dun worry I'll be stanford after next year

Fargle

What do you mean? Like career-wise?

Abhas Kumar Sinha

yes

Mike Miller

Those books are intended to be as clear as physically possible, so hopefully one doesn't struggle on a given page as much as one might on other books. But practice is important.

Fargle

15:48

I dunno, mathematician, physicist, data analyst, teacher, uhh...

...barista...

Mike Miller

That assumes we can predict what a career will look like in 10-20 years

Albas

@AbhasKumarSinha For JEE purposes, apostol volume 1 kind of does the job I think.

Abhas Kumar Sinha

@MikeMiller oh, thanks for that advice :)

Mike Miller

Albas would know about the JEE better than us

Abhas Kumar Sinha

@Fargle Thanks for your time :)

Fargle

15:49

No problem.

Alessandro Codenotti

So everyone states that a closed immersion of schemes is obviously a morphism of finite type, but I'm not seeing why this holds so I think I'm missing something obvious. Any hint in the right direction?

Mike Miller

Can you tell me what everything means very carefully

Just keep in mind I am slow

Abhas Kumar Sinha

"what everything means very carefully"!!? You mean something where this phase can be used or what?

Mike Miller

I was writing to Alessandro.

Alessandro Codenotti

A morphism of schemes $f:X\to Y$ is a closed immersion if it is a homeomorphism onto a closed subspace of $Y$ on the topological spaces level and on the sheaves level the map $f^\#:O_Y\to f_\ast O_X$ is surjective

Mike Miller

15:53

And surjective means stalkwise I guess

Alessandro Codenotti

Yes

Mike Miller

What does sharp mean? Pullback of functions?

Alessandro Codenotti

You can say that surjective means that the sheafification of the image presheaf is the whole of $f_\ast O_X$ but surjective on stalks is usually the most convenient definition

Mike Miller

He's I'm fine with that

Yes*

And finite type?

Also, I think this may be reduced to the affine case

Maybe not, the closed condition is worrying. If it can't be reduced to "locally closed"...

Alessandro Codenotti

The sharp is part of the morphism. A morphism of schemes (ringed spaces really) is a pair $f,f^\#$ where $f$ is a map of the underlying topological spaces and $f^\#$ of the sheaves. The map of top. spaces itself is not enough to determine pullbacks of functions so they need to be specified as part of the morphism

Mike Miller

16:00

Yeah this is too hard for me

I am not (ahem) immersed in this stuff

Alessandro Codenotti

So finite type means that there is a cover of $Y$ by affine open subschemes $\operatorname{Spec} R_i$ such that, for every $i$, $f^{-1}(\operatorname{Spec} R_i)$ can be covered by finitely many affine open subschemes $\operatorname{Spec} A_{ij}$ with $A_{ij}$ f.g. over $B_i$ for every $j$

(I have no idea what finite type should mean geometrically :/ )

@MikeMiller lol

I see a wild @Mathei appeared, summoned by the mention of schemes

Hi @Balarka

Balarka Sen

Hi @Alessandro

Alessandro Codenotti

How is it going?

Balarka Sen

Not bad

Finite type is just the algebraic analogue of branched cover I believe

A morphism of algebraic varieties is of finite type if it's proper and has finite fibers. The schemey definition should translate to that

Mike Miller

I don't think that's true, I think it includes projections of affine space

For isntance I think R -> R[x] is finite type

Abhas Kumar Sinha

16:14

see this - quora.com/What-is-a-complicated-equation-that-equals-80

people are creative than ever!

Mike Miller

I think it's silly

Abhas Kumar Sinha

see those answers!

Mike Miller

I did

Abhas Kumar Sinha

wasn't that fun?

Amit

Please some one help me in understanding math.stackexchange.com/questions/3031296/… I will be very thankful to you . I am trying this for long time but unsuccessful .

Abhas Kumar Sinha

16:18

Abstract algebra is out of my course, there are others out here which can help you

Mike Miller

@Amit Multiply the polynomials. If there is an $x^2$ term, subtract off $x^2 + x +1$ (we may do that because it's defined to be 0). Remember that $1=-1$.

Balarka Sen

@MikeMiller Hm, interesting.

I suppose finite morphisms and morphisms of finite type are different things!

Mike Miller

Those wacky "geometers"

Alessandro Codenotti

There are way too many finiteness conditions around

But I agree with Mike, the projection $\operatorname{Spec} k[x]\to\operatorname{Spec} k$ should be of finite type

Balarka Sen

k[x] is f.g over k after all

Alessandro Codenotti

16:21

Oh, of course it is, the only fiber is $\operatorname{Spec} k[x]$ and $k[x]$ is very much a f.g $k$-algebra

And I got sniped

Balarka Sen

It's more like a bundle with fibers having finite dimension I suppose

Oh, ok, it's precisely like a submersion.

Mike Miller

Something like that. But "bundle" is probably weird too

The definition I saw for rings was: of the form $R \to R[x_1, \cdots, x_n]/(f)$

I guess for some collection of f

Balarka Sen

Fibers are algebraic varieties if the base is $\Bbb A^1$

So for base $\Bbb A^1$ I guess it's a pencil of algebraic varieties

Mike Miller

You wanna see a magic trick?

(shoves pencil of varieties up your nose)

Balarka Sen

I knew you'd do that

@AlessandroCodenotti So surjectivity of $f^\#$ is basically saying that germ of every regular function on $f(X) \subset Y$ (by which I really mean a regular function on $X$, pushedforward by $f$) extends to a germ of regular function on $Y$

$f$ sends $X$ to an "algebraic subset of $Y$", not just a closed subset

Alessandro Codenotti

16:33

@BalarkaSen This looks suspicious to me, I think I can only say that locally a regular function on $X$ (where I mean $f(x)$ really) comes from restricting a regular function on $Y$

Mike Miller

Locally algebraic, right?

Balarka Sen

I mentioned germ, so stalk level

Daminark

I should learn about germs tbh, they seem p sick

3

Mike Miller

Yeah the germ theory revolutionized our understanding of disease

adriana634

Hi people, I'm a bit lost with this proof, I am studying order relations from my algebra book (so I have to use Peano axioms and the addition properties). I have to use induction to complete this proof? Or these types of proofs can be proved without induction? math.stackexchange.com/questions/3030013/a-b-and-cd-imply-ac-bd

Balarka Sen

16:40

@Daminark germs make me quite sick, yes

In fact I am sneezing right now

Mike Miller

Wow was that joke intended @Dami? I am very impressed

Daminark

Yup

Balarka Sen

I didn't realize it until I read Mike's joke, tried to modify it based off of your sentence, and then figured out it might be intentional

Alessandro Codenotti

@BalarkaSen ah, right

MatheinBoulomenos

16:59

@AlessandroCodenotti being of finite type is a local property. Locally a closed immersion looks like $\mathrm{Spec}(R/I) \to \mathrm{Spec}(R)$ which is clearly of finite type

oh no, this only shows locally of finite type

Alessandro Codenotti

@MatheinBoulomenos I don't see why locally a closed immersion locally has that form

Mike Miller

@MatheinBoulomenos Presumably this is why we need the image to be globally closed

MatheinBoulomenos

missed the finite part

this feels like cheating, but: stacks.math.columbia.edu/tag/01QO

condition (2) implies of finite type directly

Alessandro Codenotti

It does

loch

17:33

the analogue of submersion in AG is smooth morphisms

MatheinBoulomenos

@Alessandro have you covered what closed subschemes of an affine scheme look like? If you have that, it's not to bad to show that closed immersion => finite type with the definitions you gave

Alessandro Codenotti

Yes, I know that closed immersions in $\mathrm{Spec} R$ all look like $\mathrm{Spec} R/I\to \mathrm{Spec} R$ for some $I$ (the morphism is induced by $R\to R/I$

We also saw that $f:X\to Y$ is a closed immersion iff there is an open affine cover $Y=\bigcup\mathrm{Spec} R_i$ such that $f^{-1}(\mathrm{Spec}R_i)$ is affine and $O_X(\mathrm{Spec} R_i)\to O_Y(f^{-1}(\mathrm{Spec}R_i))$ is surjective for all $i$ which looks useful here

MatheinBoulomenos

17:55

If $f:X \to Y$ is a closed immersion, then cover $Y$ be affine opens $X=\bigcup U_i=\bigcup \mathrm{Spec}(R_i)$, then $f(X) \cap U_i$ is closed in $U_i$ for all $i$ and $f(f^{-1}(U_i))=f(X) \cap U_i$ since $f$ is a homeomorphism onto its image. Thus $f_{\mid f^{-1}(U_i)}:f^{-1}(U_i) \to U_i$ is a closed immersion, since restricting a sheaf doesn't change the stalks.

Thus $f^{-1}(U_i)=\mathrm{Spec}(R_i/I_i)$

Alessandro Codenotti

And since we can pick $U_i$ such that $f^{-1}(U_i)$ is affine this is enough

MatheinBoulomenos

you don't even need that, it follows that $f^{-1}(U)$ is affine since a closed subscheme of an affine scheme is affine

Alessandro Codenotti

Oh, right

Ted Shifrin

18:37

@MikeMiller Please don't encourage the Demonark. Something about feeding the ... :D

Balarka Sen

Hi @Ted!

Ted Shifrin

Hi, a @Balarka.

Alessandro Codenotti

Hi @Ted

Ted Shifrin

Hi, demonic @Alessandro.

Balarka Sen

@AlessandroCodenotti We should read these notes

By which I mean the whole blog from start to end :3

Alessandro Codenotti

18:47

That's a weird coincidence, I saw the same article earlier this week because Clara Löh has a simplified version of the "universe of groups" picture in her GGT book!

Balarka Sen

Aha

Alessandro Codenotti

I'm mostly thinking about AG at the moment, I'm planning to catch up with GGT during Christmas break

Balarka Sen

Oh nice we can do that togather

Alessandro Codenotti

Sure! That'll be in a couple of weeks for me

Fargle

@AlessandroCodenotti Geometric group theory?

Alessandro Codenotti

18:50

Yep

Ted Shifrin

Hi @Fargle

Fargle

What's that like? It sounds neat by name.

Hey @Ted! How goes it?

CaptainAmerica16

Quick: Who's your favorite mathematician and why?

Balarka Sen

Smale

For many reasons

Ted Shifrin

Not a bad one to pick. Not my choice, unsurprisingly.

Alessandro Codenotti

18:52

@Fargle Unlike algebraic geometry, geometric group theory is actually about geometry!

8

Fargle

Grothendieck. Dude was a boss.

Balarka Sen

@TedShifrin Yours is Chern for sure :)

CaptainAmerica16

ooh

ÍgjøgnumMeg

Jacques Tits

Fargle

@AlessandroCodenotti What text(s) are you using?

Ted Shifrin

18:54

Balarka: If I have to pick just one, it's tough. I have to go with Griffiths, too. And Bryant ... (perhaps too young).

Balarka Sen

Yeah I suppose once you know big swaths of mathematics it's harder to pick one great mathematician

Hm, it's 12:30. I should try to see if I can fix my sleep schedule today

Alessandro Codenotti

@Fargle Clara Löh: Geometric group theory: an introduction and Serre: Trees are the two reference texts for the course. There are some very terse course notes available here

Ted Shifrin

@Alessandro: Algebraic geometry is full of geometry. It's just the way the modern scheme theory people teach it, you never see it.

Balarka Sen

Good night, all. Don't start talking fun mathematics while I'm gone!

Ted Shifrin

Wow, a Balarka is un-unsleeping?

Alessandro Codenotti

18:57

Good night @Balarka

Balarka Sen

Trying to

Fargle

"Fun mathematics" is an oxymoron, math sux

Ted Shifrin

Fare thee well, a @Balarka.

Fargle

(note: I do not hold this opinion in good faith, and refuse to defend it)

Ted Shifrin

I was going to smack you, @Fargle, but decided you weren't serious.

Alessandro Codenotti

18:58

@TedShifrin I know, we actually saw some interesting examples of schemes that are more geometrically motivated

Fargle

Always have a disclaimer ready.

Ted Shifrin

Some day you should read some classic stuff, @Alessandro, particularly since so much of the beautiful 19th century foundations is Italian.

CaptainAmerica16

@Fargle Whatchu mean

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$Mathematics$ Mathematics