Mathematics - 2020-05-28 (page 1 of 2)

Balarka Sen

02:22

@AlessandroCodenotti Id be in

@AlessandroCodenotti Ah strange

@EdwardEvans Check out "Welkin" by Raat, an Indian atmoblack band lol

Ted Shifrin

03:06

Oh, no, we have twin A-Balarkas now.

Balarka Sen

Not sure I follow the joke

Ted Shifrin

You and A are twins.

I thought one had replaced the other, but no.

Balarka Sen

What gave it away that we are not the same person?

Ted Shifrin

Well, I'm working on twin versus identical twin.

Balarka Sen

Haha alright

Bizarre surrealism

Ted Shifrin

03:10

Soon I'll get to existentialism and wonder if either of you actually exists.

Balarka Sen

One morning Gregor Samsa woke up from troubled dreams to find himself transformed in his bed into a

Ted Shifrin

Käfer.

maybe "awoke"

Balarka Sen

Yeah should use awoke

You read it in original German, right?

Ted Shifrin

Centuries ago.

Balarka Sen

Very cool

Ted Shifrin

03:14

So are you enjoying your dual identity?

Balarka Sen

I'm trying to establish them as adjoint to each other

Ted Shifrin

or at least adjunct

Balarka Sen

So there's apparently this notion of a Spivak normal fibration of a PD space $X$. I think one constructs this as follows; let's say $\dim X = n$, embed $X$ in $\Bbb R^{n+k}$ for some large $k$ and take a regular neighborhood $N$ of this CW-complex. Turn $\partial N \hookrightarrow N \simeq X$ into a fibration by the homotopy fiber construction. Because $X$ satisfies Poincare duality, the fibers are $S^{k-1}$'s

Stably this makes sense apparently, and doesn't depend on the choice of the embedding

Ted Shifrin

Oh, yeah, that was his thesis with Milnor.

Balarka Sen

Ah I see

Ted Shifrin

03:17

Unless I am totally forgetful.

This paper might be of interest to you, given that you're thinking about this.

Balarka Sen

Thanks!! I did need a better reference

Ted Shifrin

Not clear this will help, but I came across it while googling to make sure I was correct.

Balarka Sen

I'll look into it. I was interested in a result of Browder which says any H-space is a PD-space (in fact, odd dimensional simply connected H-spaces are always homotopy manifolds)

BAYMAX

03:56

Hi guys!

I was thinking of some free online apps

in which we can deliver online lectures to a group of students?

like group video conferencing

For Zoom, it seems I need paid account

any Stack exchange platform in which I can ask about this?

Alexander Gruber

04:29

@BAYMAX microsoft teams?

BAYMAX

Thanks@AlexanderGruber

how many students at a time does it support for free?

any idea?

seems like 300 participants

@AlexanderGruber seems like video conferencing up to 250 people is not available?

onmsft.com/feature/…

Alexander Gruber

@BAYMAX I'm not sure, I've only ever had 100 people on one of those calls at once.

It can get slow.

BAYMAX

oh cool then!

that will help!

Thanks for looking@AlexanderGruber

Alexander Gruber

No problem, enjoy teaching.

BAYMAX

Thanks!

Joshua Lin

04:59

Is there an unknotting algorithm for 2-knots? I'm having difficulty finding the answer online; and it might just be because I'm searching the wrong words

The Substitute

Hi (off topic). Are there known analogues of the Erdos-Kac Theorem for distributions other than the uniform?

Knight

Does anybody here know about the anecdote of Carl F. Gauss where he got a letter from the school and his mother read it like “Your kid is exceptionally intelligent and school will not be able to develop his intelligence. We suggest you to let him work on his own” while in actuality the letter was written like “Your kid doesn’t have any ability to survive in our school, we have to expel him. We are sorry”

I’m looking for the source of this anecdote

Balarka Sen

05:45

$f : \Bbb R^3 \to \Bbb R^3$, $f(x, y, z) = (xy + x^2 z + x^4, y, z)$

What on earth does this look like

Mike Miller

Trace out a few f(t, y, z) as t varies?

Balarka Sen

If you hold x = t fixed it's just some plane in $\Bbb R^3$, so that doesn't really help, right

y and z are the correct variables to fix

Pig

what are you trying to understand about it though

Balarka Sen

How does it look like?

It's a transformation of the 3-space, so it does something - not clear to me what it does

Should be visualizable

Pig

i don't know, what mike suggests makes sense though, maybe a foliation of the whole thing gives you a sense of what it looks like

Balarka Sen

05:57

yeah

Ah this is interesting

$\det Df = y + 2xz + 4x^3 = 0$ is the following set: grapher.mathpix.com/?latexList=%5B%22y%2B2xz%2B4x%5E3%3D0%22%5D

If you look from bottom-up that set looks like a cubic, whereas if you look from top-down it looks like a line. So the singularity set of $f$ is the movie of a pair of folds converging to a cusp and cancelling each other out

Mike Miller

@BalarkaSen Yeah but you can look at how these planes intersect

Balarka Sen

probably true

I would like to sketch $f(\{\det Df = 0\})$, maybe that's easy to write down

Edward Evans

06:28

@Balarka listening

is cool

Balarka Sen

The tangent space to $\{y + 2xz + 4x^3 = 0\}$ is normal to $(12x^2 + 2z, 1, 2x)$, which contains the direction $(1, 0, 0)$ killed by the projection $\Bbb R^3 \to \{x = 0\}$ iff $12x^2 + 2z = 0$, or $z = -6x^2$. So the "cuspidal curve" in $\{\det Df = 0\}$ is given by $y = -2xz - 4x^3$ and $z = -6x^2$, aka the curve $(t, 8t^3, -6t^2)$

Looks right

So I should study what $f$ does near this curve

Yeah why am I making this complicated. Consider $g : \Bbb R^2 \to \Bbb R^3$, $g(u, v) = (u, -2uv - 4u^3, v)$. The image of $g$ is exactly $\{\det Df = 0\}$. $f \circ g : \Bbb R^2 \to \Bbb R^3$ is $(u, v) \mapsto (-3u^4 - u^2v, -2uv - 4u^3, v)$. Maybe I can plot a 2-parameter surface somewhere

Ah mathpix can

Hot damn

It's a swallowtail

(Put u between -2 and 2 and v between -2 and 2)

@MikeMiller You might like this: This is the description of the swallowtail singularity

My guess (although I have not checked this yet), is that $f$ "creases" along the two fold lines appearing in $\text{det} Df = 0$ that makes the two creases of the swallowtail, and the cusp in $\text{det} Df = 0$ gets mapped to the singularity where those swallowtail creases converge

@Edward dope

Edward Evans

06:53

aye

'twas atmo af

short tho, 19 mins is usually one song

Balarka Sen

yeah very short

Knight

- - - Eoin- - - Morgen

———Eoin———

What is the issue with strike off?

I wanted to greet Eddie in German, but my whole plan got stripped off (I wanted strike off on Eoin)

Edward Evans

Please don't use the name Eddie

lol

Knight

Okay :-)

Edward Evans

tyty

Und guten Morgen :P

Knight

07:04

:D

When will you become an Englishman in Heidelberg?

Edward Evans

wat

Knight

Sting - Englishman In New York (Official Music Video)

►

Okay come to my place and you will become “Englishman in Chicago”

Alessandro Codenotti

10:07

@Balarka Let $X$ be any topological space, let $A$ be any subspace of $X$. How many distinct subspaces of $X$ can you obtain by applying complementation and closure to $A$ in any order? Answer is 14 by a result of Kuratowski wtf

I made a room for the chess group, anybody who is interested please join, we will discuss exactly how to do this in that room! @Balarka @Leaky @Present @Astyx @user21820 chat.stackexchange.com/rooms/108579/1-e4-2-ke2-1-0

Leaky Nun

how to do what?

Alessandro Codenotti

the study group

What to study and how/when

overactor

10:58

Hey everyone, I was wondering if I could ask a very quick question here that I can't seem to find an answer to. (My google fu is letting me down big time)

I want to write a definition for something that can include any of a few operators in a specific place and I need a symbol that is commonly used as a stand-in for an operator.

Something like $\textit{foo}(\odot, n) := \{m \mid m \odot n\}$ for all $\odot \in \{>, <, =\}$

is $\odot$ suitable for this? I have some vague sense that I've seen it used like that, but I can't find any evidence for it. Any help would be greatly appreciated.

topologicalorientablesurface

every manifold with boundary is locally path connected?

what would be the basis?

Alessandro Codenotti

11:14

Locally they look like Euclidean space, so...

topologicalorientablesurface

I mean if it was a manifold without a boundary, then you can use coordinate balls

since every ball is convex and therefore path connected

Alessandro Codenotti

And if it has boundary you can use half balls

topologicalorientablesurface

11:30

half balls?

feynhat

balls in the upper half plane are also path connected.

topologicalorientablesurface

oooohhh

I didn't know that

Is the following true: Let $X$ be a space. $U$ is an open subset of $X$ homeomorphic to an open subset of $\mathbb{H}^n$ iff $U$ is an open subset of $X$ homeomorphic to an open ball in $\mathbb{H}^n$

feynhat

No.

That's like saying all open subsets are balls, which is obviously false.

topologicalorientablesurface

What about: Let $p\in X$. p has a neighborhood homeomorphic to an open ball in $\mathbb{H}^n$ iff $p$ has a nieghborhood homeomorphic to an open subset of $\mathbb{H}^n$? Forward direction clearly holds, what about backwards?

it holds for $\mathbb{R}^n$

feynhat

11:46

Yes, because you can choose a ball inside that open subset of $\Bbb H^n$ and restrict your homeo to the inverse image of this ball.

nbro

13:26

The homogenous representation of a circle is given by x^2 + y^2 + 2*gx + 2*fy + c*z^2 = 0. Now, given 3 points (in homogenous form), we can solve a system of linear equations and retrieve the unknowns f, g and c. But what do these unknowns actually represent?

Yuvraj

13:37

@nbro represent in the sense

nbro

In the sense that what do they represent with respect to the circle? Are they the center's coordinates radius or what?

Yuvraj

yes -g ,-f reprsent the circle centre

nbro

why -g and -f?

why not g and f?

Yuvraj

it has a very simple derivation

nbro

Well, that's what I am asking for

Yuvraj

13:40

do you know how to derive the locus of a point who distance from a fixed point always remains the same?

nbro

I don't know what a "locus" is

is that a germ or something?

lol

I just want to know why -g and -f would be the center and I also want to know what would be the radius

Yuvraj

you have to be cautious before running on this

Locus (mathematics)

In geometry, a locus (plural: loci) (Latin word for "place", "location") is a set of all points (commonly, a line, a line segment, a curve or a surface), whose location satisfies or is determined by one or more specified conditions.In other words, the set of the points that satisfy some property is often called the locus of a point satisfying this property. The use of singular in this formulation is a witness that, until the end of 19th century, mathematicians did not consider infinite sets. Instead of viewing lines and curves as sets of points, they viewed them as places where a point may be located...

please go through it once

nbro

I don't have time for that now. I just want to know what I said above

Can you explain it or not?

If you have a source that also explains it, that would be fine

Yuvraj

sorry I can't :(

@robjohn hi sir are you free?

robjohn

@Yuvraj for something short.

Yuvraj

13:52

on complex numbers

robjohn

"Just ask; don't ask to ask." from the sidebar

Yuvraj

oops

sorry

nbro

that's the most annoying advice ever, btw

"don't ask to ask"

so annoying

People "ask to ask" so that they are polite

And you're asking people not to be polite

lol

Leaky Nun

@nbro but there's also the problem that if someone says "can I ask something" and you say "sure", then you are expected to answer his/her question

nbro

@LeakyNun that's your problem if you say "sure"

Leaky Nun

13:57

and if someone just asks then whoever is interested then answer

robjohn

@nbro person 1: Can I ask a question? person 2: yes: person 1: how does... it takes so much longer and you are waiting for the question.

@LeakyNun precisely

Leaky Nun

@nbro so by asking to ask you are forcing someone to commit to answering your question without knowing what your question will be

3

nbro

@LeakyNun But if nobody knows the answer it's a waste of time to even write anything more

That advice is so annoying and I don't agree with it (no benefit and only discourages people that need help even more if then nobody answers)

One of the most annoying things on these chats on the web

robjohn

@nbro If you don't ask the question, no one knows if they can answer it

Leaky Nun

nobody wants to commit to answering to a question that they don't even know what it is

@nbro you're right, this rule is asking people needing help to value the answerers' time

nbro

14:00

@robjohn You don't understand. First, people need to understand if someone knows anything about the subject

@LeakyNun Nobody should commit to an answer if you don't even know the general subject

That's why people "ask to ask"

Do you know about complex numbers? No, nobody knows, so I will not waste my time here

That's my point

Leaky Nun

yeah but people would say "can I ask a question"

and also there are still so many aspects of complex numbers

if you don't ask the actual question, then nobody can know if they can answer it

nbro

The "can I ask a question" is to be polite. If you are busy, that's a nice way of interrupting the person

robjohn

@nbro That is a question, not "can I ask a question"

nbro

And I am the guy that isn't really polite always, and I find this suggestion of not asking to ask so annoying

Yuvraj

i need help on deriving the orthocenter of a triangle whose vertices are z1 z2 and z3 ?

Leaky Nun

14:02

bingo

robjohn

@Yuvraj the orthocenter is where the altitudes intersect

Yuvraj

yes

i know this ,and i derived this using my straight lines skills

robjohn

@Yuvraj give me a few minutes to see how to put this into $\mathbb{C}$

Yuvraj

sure

Thorgott

you can ask a question politely without asking to ask

that's the difference between people posting genuine questions on MSE and those who literally just copy their homework text "Show that..."

nbro

14:08

That suggestion is stupid. And I am known for having used the term stupid in various loci

But that may offend someone, but it should offend someone that goes straight to the point and ignores redundant and useless words like "stupid" (i.e. people that suggest not to ask to ask)

I hope you can notice my subtle sarcasm

Thorgott

no

nbro

It isn't subtle, right

topologicalorientablesurface

Is every subset of a locally euclidean space locally euclidean?

every open subset is

oh, no, consider the closed ball

unit ball

robjohn

@Yuvraj well, I get $\frac{z_1\left(|z_1-z_2|^2-|z_2-z_3|^2+|z_3-z_1|^2\right)+z_2\left(|z_2-z_3|^2-|z_3-z_1|^2+|z_1-z_2|^2\right)+z_3\left(|z_3-z_1|^2-|z_1-z_2|^2+|z_2-z_3|^2\right)}{|z_1-z_2|^2+|z_2-z_3|^2+|z_3-z_1|^2}$

Thorgott

this would imply every subset of a Euclidean space is a topological manifold

robjohn

14:21

@Yuvraj there might be a way to simplify

Balarka Sen

@AlessandroCodenotti yeah i have seen this before

robjohn

@Yuvraj is that the kind of form you're looking for, or were you thinking of something else?

feynhat

@Pig Cute pic.

robjohn

gotta walk my doggies. bbl

Alessandro Codenotti

@BalarkaSen It's very weird

It makes more sense when you realize that it has nothing to do with topology, but weird nonetheless

Knight

15:46

@robjohn Hello Robbie sir

0

Q: Do you know about this anecdote or its source where the mother reads out the letter to Gauss and made him Gauss, the mathematician?

I remember my brother telling me an anecdote about Carl F. Gauss three-four years ago, I want to know if anybody here also know about it, or can provide the source of it. The anecdote goes like this One day Carl came from school, he was probably a kid of 9 or 10 years, and smilingly handed ...

Ed, can we move things from one side to the other in limit equation

like this $$ \lim_{x\to 0} (f(x)) ^{g(x)} = y \\
\lim_{x\to 0} g(x) ln (f(x)) = ln(y) \\
\lim_{x\to 0 } ln(f(x)) = ln(y)/g(x)$$

Are these things allowed in limits?

Edward Evans

no you can't pull $g(x)$ out of the limit like that

unless $g(x)$ doesn't depend on $x$ lol

Knight

We can't apply $ln$ like that?

Thorgott

that last line doesn't even make sense conceptually, think about it

Knight

Okay, is it allowed if it is given that $\lim_{x\to 0} g(x) \neq 0$ ?

Edward Evans

@Thorgott Now I have to create a quiz for rep theory lol

Ein "angeleiteter Beweis" zur Frobenius-Reziprozität

before the prof asked us to do some exercises and present them, but changed his mind and now we have to create a quiz lol

Thorgott

16:00

lol my entire talk will be based around an angeleiteter Beweis exercise in Serre

@Knight you should rethink what your last line means, because it doesn't make sense

Knight

@Thorgott You mean I should apply limits on both sides?

Alessandro Codenotti

what does angeleiteter mean?

Edward Evans

"More angeleitet"

Nah

it's like a "led" proof

I can't remember how to say it in English lol

basically you do a quiz and each question in the quiz is a step in the proof

Thorgott

instructed?

Edward Evans

maybe

guided

Guided proof

Thorgott

16:03

@Knight I mean the LHS is a limit, which ought to be some number, but the RHS is a function of $x$

Knight

@Thorgott So, how can I fix it?

or is it fundamentally wrong ?

Here comes Robbie sir

Thorgott

I don't know how to fix it, cause I don't know what you want to achieve

linda Oiladali

hello

Knight

Okay, I actually want to solve those kinds of limits where a function is raised to some other function

linda Oiladali

how to prove that $\lim_{x\to-\infty} x^n exp(x)=0$

Thorgott

16:13

the function $(0,\infty)^2\rightarrow\mathbb{R},\,(x,y)\mapsto x^y$ is continuous

continuous functions preserve limits

that's probably the best general result useful in this direction

actually, $(0,\infty)\times\mathbb{R}$ works for the domain

Alessandro Codenotti

@EdwardEvans Ah makes sense

thanks

linda Oiladali

@Astyx hello, please have you an idea o how to prove that $\lim_{x\to-\infty} x^n exp(x)=0$

Astyx

16:56

Yes

But I don't know what you know and don't know

@lindaOiladali

robjohn

17:21

@Knight your ping did not show up in my menu. odd.

Knight

@robjohn no problem.

What you had/ going to have in breakfast ?

Ted Shifrin

heya @robjohn

robjohn

@TedShifrin howdy! Are you doing okay in isolation?

@Knight I had some fresh strawberries. Quite nice!

Ted Shifrin

I suppose. I'm finally going out a little bit. Returning to physical therapy and chiropractor. 2-3 months off is killing me.

robjohn

@TedShifrin My wife started back a while ago. Stopping PT can be bad.

Knight

17:35

@robjohn Were they sweet :-) ?

robjohn

@Knight They were very nice. There is sweet and a lot of strawberry flavor

@TedShifrin she was so glad to get back, but she put it off a couple of weeks after they said they were thinking of opening.

Knight

Wow! It’s very nice to have strawberries in breakfast. I will also try it someday

What is PT? Physical Training ?

robjohn

@Knight if you look at the comment mine was in response to, it says "physical therapy".

Knight

Okay. In school we had PT as something set of exercises, so I thought it was Physical Training

Thorgott

is there a notion of exponentiating categories?

Ted Shifrin

17:44

Gag.

Manolis Lyviakis

id like to discuss about finding topic to write n ssignment on my algebraic topology course

My proff suggested me the reltion between fundemelntl group nd homology group

and maybe Van kampens theorem

Tobias Kildetoft

@Thorgott What properties would that have?

Mike Miller

Sounds like you probably just want functor categories

You'll need some conditions for these to actually be categories (have sets of morphisms Nat(F,G))

Thorgott

I don't know about properties, but here's what I was thinking of: consider the category whose objects are indexed tuples of vector spaces, $(V_i)_{i\in I}$, where $I$ is any set and each $V_i,i\in I$ is a vector space (over some fixed base field). A morphism between $(V_i)_{i\in I}$ and $(W_j)_{j\in J}$ consists of any map $g\colon I\rightarrow J$ and a linear map $f_i\colon V_i\rightarrow W_{g(j)}$ for each $i\in I$.
I feel this should be $\mathbf{Vec}^{\mathbf{Set}}$ in some sense I can't specify.

Mike Miller

Who cares about that category

Thorgott

17:59

I was trying to phrase some manifold stuff categorically and it popped up naturally. The differential can be thought of as a functor from $\mathbf{Diff}$ to $\mathbf{Vec}^{\mathbf{Set}}$.

Transcript for

$Mathematics$ Mathematics