« first day (3736 days earlier)      last day (41 days later) » 
00:00 - 16:0016:00 - 23:00

12:18 AM
that's just the guiding principle of working with smooth manifolds
representation theory is about trying to understand groups by how they can linearly act on vector spaces
 
The Euler characteristic of an annulus is zero, which is fitting for something shaped like a 0
 
12:42 AM
I have a claim here that states $\hat{K}(\xi) = \int_{\mathbb{^n}} (K(x) - K(x - \frac{\xi}{2|\xi|^2}) e^{-2 \pi \xi x}$, but I'm getting
$\int_{\mathbb{R}^n} (K(x) - K(x - \frac{\xi}{2|\xi|^2}) e^{-2 \pi i \xi x}= \int_{\mathbb{R}^n} K(x) e^{-2 \pi i \xi x} - \int_{\mathbb{R}^n} K(x -\frac{\xi}{2|\xi|^2}) e^{-2 \pi i \xi x} =$
$\int_{\mathbb{R}^n} K(x) e^{-2 \pi i \xi x} - \int_{\mathbb{R}^n} K(x - \frac{\xi}{2|\xi|^2}) e^{-2 \pi i \xi x} = \int_{\mathbb{R}^n} K(x) e^{-2 \pi i \xi x} - \int_{\mathbb{R}^n} K(x) e^{-2 \pi i \xi (x + \frac{\xi}{2|\xi|^2})} = \int_{\mathbb{R}^n} K(x) e^{-2 \pi i \xi x} + \int_{\mathbb{R}^n} K(x) e^{-2 \pi i \xi x} = 2\hat{K}(x)$
 
By the way
Elementary Applied Topology by Robert Ghrist
Really weird book
As far as textbooks go, it's on the weird end
but really good
 
TDA guy?
There's a handful of conditions and bounds on $K$, but I don't think they make a difference for this bit
 
There's topological data analysis in the book, yeah
@JoeShmo What's the hat?
 
Fourier Transform
if you were to work it out just a little more you end up with a $e^{-i \pi} = -1$ factor and whoop-de-doo
 
Yeah I dunno
My instinct is to plug in something simple for K but I dunno enough about Fourier transforms to know what that'd be
Like $K(x)=\delta(x)$? Does that work? What's $\hat\delta$
 
12:52 AM
no K is an operator. Vanishes outside an annulus
 
So does $\delta$, no?
 
oh you mean sanity check if that works? I was too lazy for that..
 
Yeah
I mean if you're just hunting for a constant factor
So the topology book
None of the figures have captions
so you're kinda on your own trying to figure out how they relate to the text
which is intentional, apparently
For some it's really clear, but for others you have to think and that can be kinda satisfying
Also
It's more of an overview of the subject, so there's not a whole lot of proofs
Oh and it's available online (but I got a hard copy because I <3 hard copies - it's $20 on Amazon)
 
Elementary Applied Topology by Robert Ghrist , I have heard is super weird.
 
12:57 AM
Nothing there is "elementary" at all.
 
That's the one I'm talking about
I'm about halfway through and I 100% recommend it
He does recommend reading through Hatcher chapter 0 for intuition on homotopy equivalence, though, which is kinda weird
So that's a prereq, as well as calc, some point set, and some linear algebra
The fact that I've already read Hatcher chapters 0-2 in the past probably helps me
Weird thing: he goes over homology before the fundamental group
 
my teacher uses it.
 
Oh wow
For an applied topology course? Or what
 
the book is more proof driven in my eyes
 
Oh you mean Hatcher
That makes a lot more sense
Yeah Hatcher is a standard text for algebraic topology I think
 
1:00 AM
he uses it for convex geometry for his research.
 
Do you mean Hatcher?
 
the one you using, I think person needs to have topology experience.
yeah Hatcher
he has the book when I talk about topology.
the "book" I am using is topology without tears
 
Oh I've heard of that, never really looked at it though
but it's an online text, right?
So - tears? No tears?
 
Has it been tearless for you?
 
1:03 AM
TOPOLOGY = TEARS
 
Howdy, DogAteMy.,
 
the class is more Independent Study
 
Hello Ted
Ted, you were right it was bad mixture to go from diff geo to topology.
 
1:05 AM
I'm reading through Robert Ghrist's Elementary Applied Topology, which is a very weird book but I recommend it anyway
 
Oh, you've changed names.
 
never knew IS is time tolling task.
 
I mean, it kinda hypes itself up to be weird. It's maybe less weird than it wants to be
Oh I don't know who I'm talking to
 
Independent study is very tough for advanced math. A good teacher makes a lot of difference.
 
@TedShifrin, changed my name here and there. The teacher isn't good at all.
the dude is making me to do problems of any course with topology.
he has that mindset "eh I will pass you"
 
1:07 AM
Well, and “independent study” puts a ton of load on a student.
 
Oh I think I found your old name but it's not one I recognize anyway shrug
Nice to meet you
 
@AkivaWeinberger, me?
 
So how far into it are you?
Yes
 
Grades aren’t the point if you actually want to learn
 
I will changed my name after.
 
1:08 AM
Like what are you learning now
 
I know only topology definition and discrete and indiscrete.
 
Anyhow, I said I would help with diff geo, but you haven't asked.
I've forgotten the old name.
 
Ah so you're at the start
 
my old name was EM4 haha...it will return back to it.
Ted, I will after this week exams.
 
But I just got in touch with a guy who took topology and a year of algebra from me 25 years ago.
 
1:10 AM
Any of it stick?
 
OK.
He was brilliant and said I was the best teacher he ever had anywhere :) Now he has high-school age kids and is a free-lance editor for authors.
 
Ah! Nice
 
is he a mathematician?
 
Nope.
 
that's awesome.
Ted, i have seen your lectures in youtube.
 
1:12 AM
Judging by your lectures, I can't say I'm surprised, Ted
 
I got him into grad school, but he didn't persist. It's fine with me :)
blushes
You know the Kuratowski problem in topology?
 
how's grad school for math.
is more proof based.
 
Yes.
 
Is that the one where you want a set closed under a bunch of operations?
> In point-set topology, Kuratowski's closure-complement problem asks for the largest number of distinct sets obtainable by repeatedly applying the set operations of closure and complement to a given starting subset of a topological space.
- Wikipedia
 
never heard of Kuratowski problem haha.
 
1:15 AM
Not "closed" but I was close
 
I know the Kuratowski theorem in graph theory
 
You are allowed closure and complement. What is the largest number of sets you can generate from a starting set?
Ah, too slow on iPad.
 
only definition of ordered pair by Kuratowski
 
This guy was the first student I ever had who worked it out, and beautiful proof.
 
Looking up his biography on Wikipedia
> During World War II, he gave lectures at the underground university in Warsaw, since higher education for Poles was forbidden under German occupation.
I did not know there was such a thing
 
1:16 AM
Subsequently, another student, who did finish his PhD, did it.
 
worked out the kuratowski theorem?
 
is the answer 14....got the answer via online.
have no idea why is 14.
 
Yes, 14.
 
> World War II saw the cultivation of underground education in Poland. Secretly conducted education prepared scholars and workers for the postwar reconstruction of Poland and countered German and Soviet threats to eradicate Polish culture.
 
I didn't know that history, DogAteMy.
 
1:18 AM
I want to see the proof why it is 14 haha.
 
You need to work on it and play with examples.
 
@TedShifrin, I agree
 
Ted, can you take a look at the integral a few comments up?
Either I did something stupid, or I'm right
 
I don’t see it.
 
apparently closure(interior(closure(interior(X)))) equals closure(interior(X))
cici=ci
 
1:20 AM
Don't cheat, DogAteMy. Work on it yourself.
These guys did it pre-internet.
 
Here's a nice fact I learned by the way
 
Given a relation on a pair of finite sets $R\subseteq X\times Y$, you can construct two simplicial complexes in a natural way
 
do you guys gave good reference book for Complex Analysis and Abstract Algebra?
 
An n-simplex is the n-dimensional equivalent of a triangle or tetrahedron
A simplicial complex is a structure made of simplicies glued together
 
1:23 AM
No clue, JoeShmo.
 
hrmp, does it sound right though?
twice the fourier transform
 
No clue.
 
OK.. :-)
 
Let $R_X$ be the simplicial complex with vertex set $X$, where, for every $S\subseteq X$ such that there exists a $y\in Y$ with, for all $x\in S$, $xRy$ (aka $(x,y)\in R$), then you fill in a simplex with vertices $S$
 
Hard to answer, @user1993, depends on level and sophistication. I wrote an algebra book I like. :)
 
1:25 AM
The dimension will be one less than $|S|$. Subsets of $S$ give you faces of $S$
 
what's name of the book @TedShifrin?
 
Similarly, define $R_Y$ the same way
Theorem: $R_X$ and $R_Y$ are always homotopic
 
See my profile. It should be linked. This one isn’t free.
 
(aka they're "equivalent" in a slightly weaker sense than homeomorphism)
 
is the one multi. calc thingy
 
1:27 AM
That's not algebra.
Interesting, DogAteMy. Some well-chosen examples would be good.
 
I see the link of diff geo, mult. calc thingy, and not algebra book
 
Oh, you're right. Only the diff geo is linked. Click on my webpage (on the right).
 
See the image there
That says the homologies are equal, but further down the page he asserts their geometric realizations are homotopic
Apparently you use the nerve lemma
But I don't know how to prove the nerve lemma either lol
 
Well, you're goofing off on sabbatical — get to work!
 
So see how there are 5 columns, and the simplicial complex on the left has 5 vertices
That last column represents the vertex that's not connected to anything
Columns 2, 3, and 4 represent the vertices with the filled in triangle because they're all represented by row 4
Both simplicial complexes look like "two disconnected pieces; one (on top) homotopic to a circle, one (below) homotopic to a point", so you can see how they're homotopic to each other
They each have a piece with a loop and a piece with no loops
 
1:35 AM
your book is Abstract Algebra: A Geometric Approach, so awesome
 
Disappointing how the plural of simplex is simplices but the plural of complex isn't complices
@user1993 A while back we were goofing off with that title
Eventually someone came up with, "Lovecraftian mythos: The Things That Should Not Be Known Approach"
Dec 15 '16 at 2:25, by TheGreatDuck
(for those reading this later: no, we are not drunk)
 
I still get spoofed once in a while.
It was not just that title.
 
@Thorgott you mean representation theory?
 
no
 
1:42 AM
the book looks good.
 
If you're curious about the Ghrist stuff, this video (and the two sequel lectures) is very good
 
good question
 
 
1 hour later…
3:12 AM
just read about representation theory. it's really cool!!
i wonder if there's a way to relate it to topology or something
 
4:11 AM
@LucasHenrique The theory of representation of Compact abelian groups does use topology to a certain extent. But a more geometrical framework can be found in Kirillov theory, where one uses very symplectic geometric techniques of analysing coadjoint orbits to understand group representations. Fascinating ideas!
 
4:22 AM
There's also something called representation of quivers and it has plenty of connections with ideas coming from BPS states in string theory but I know nothing but words here.
 
5:20 AM
I was right about the integral, you'll be happy to know.. and after wasting the entire day figuring out where I went wrong, I will now finally go to sleep
 
5:50 AM
hey joe hows it going
do u know what l.l means?
 
 
1 hour later…
6:55 AM
anyone know this?
 
It looks like the absolute value of a point on the real number line? But that's just a guess. Don't you have a table of symbols?
 
thanks skull
i think u are right
i do but it's not the same as what my professor uses
 
7:13 AM
np
 
7:45 AM
hey is there anyone, who knows a lot about BV spaces?
If I take $f,g \in BV({[0,1]}^2)$ then $fg \in BV([0,1])$. In particular for $f_1,g_1 \in BV([0,1])$, the mapping $h: [0,1]^2 \rightarrow [0,1], f_1(x)g_1(y)$ is in $BV({[0,1]}^2)$. Now I wonder if mappings like $h$ are dense in $BV({[0,1]^2})$ ...
If I take $f,g \in BV({[0,1]}^2)$ then $fg \in BV([0,1])$. In particular for $f_1,g_1 \in BV([0,1])$, the mapping $h: [0,1]^2 \rightarrow \mathbb{},(x,y) \mapsto f_1(x)g_1(y)$ is in $BV({[0,1]}^2)$. Now I wonder if mappings like $h$ are dense in $BV({[0,1]^2})$ ...
fixed that
I don't want a clear answer, unless it is in the literature, but I wonder if someone can give me any tips hints or ideas.
That's why I am asking here as opposed to normal
 
8:34 AM
hi chat
 
Hi @Astyx
 
How are you ?
 
Not bad, just reading about the smooth dual
How about you?
 
doing some AG
 
8:37 AM
yeah
 
9:35 AM
Can anyone provide intuition as to the difference between $f(x)+1$ and $f(x+1)$?
 
if $f(t) =0$ for all $t$, then $f(x+1) =0$, but $f(x)+1 = 0+1 = 1$
 
just spent like an hour trying to work out some stupid fact until I realised I had just forgotten a definition
kms
 
At least you probably won't forget it now
 
probably not
"suuuuuuuuuuurely that's a typo! It doesn't make any sense!"
 
why S has only x_0 as limit point lol
I don't know how this inequality this inequality was derived plus what does for all but finitely many n means
 
9:49 AM
They say S does NOT have limit points
If S had a limit point x, then S would have an infinity of points of norm less than $|x|+1$, which is not true
 
@EdwardEvans the other day I spent 15 minutes trying to figure out why property A implies property B because this was being used implicitly in the paper, only to realize that property A was actually defined as property B+other stuff and I only remembered about the other stuff
 
lool
I had something called $V(K)$ defined as the span of some guys of the form $v - \pi(k)v$ and was wondering why $f(v - \pi(k)v) = 0$ for all such guys means that $f(V(K)) = 0$
having forgotten the definition of $V(K)$
lel
 
lol it happens
 
very disheartening
sitting there like "wtf this should be obvious"
 
An open subset of an affine scheme is not necessarily an affine scheme right ? That's true iff the open subset is distinguished
 
9:58 AM
@Astyx but why he said S has $x_0$ as a limit point?
 
Proof by contradiction I think (I haven't read everything)
He assumes E is closed and shows that leads to S having a limit point which is absurd
 
10:11 AM
@Astyx $\Bbb C^2 \setminus 0$ is not affine yeah
 
oh yeah that's an easy counterexample, cheers
 
quasiprojectives are the best category
closed under image, the absolute best
"open subset of a closed subset" is a weird thing to say
 
sounds dope
I need to go, bye !
 
 
1 hour later…
11:16 AM
The books says that The figure on the left covering space has deck transformation group S_3 But I don't understand why
 
 
2 hours later…
12:58 PM
@love_sodam What is the image in $F_2$ of the covering-induced homomorphism?
Note that this is a normal covering space. So the group of deck transformations will be $\pi_1(S^1 \vee S^1)/\text{im } p_*$, where $p$ is covering map.
Write down, $\text{im } p_*$. That should give you all the relators for $S_3$.
@BalarkaSen I am thinking something very wrong but why isn't the deck group for a normal cover same as the symmetric group of the fiber?
 
What's a normal cover?
 
1:33 PM
pi_1 acts transitively on fibers I think
Galois cover if you like
@feynhat $S^1 \to S^1$, given by $z \mapsto z^n$, is a normal cover. The deck transformation group acts freely and transitively on fibers.
But it's $\Bbb Z/n$ and not $S_n$.
 
So for a normal cover the deck group is never the full symmetric group? Because the deck should have as many elements as the fibers does
 
Why would that be true
Oh yeah deck transformations act freely on fibers?
Whoops
 
Sorry I went afk for a while.
Yeah, that's what I was thinking about.
 
What I forgot is that the deck transformation group isn't the same as the action of $\pi_1$ on fibers, since the deck transformation group corresponds to a proper subgroup of $\pi_1$ acting on the fibers.
There are non-regular covers where $\pi_1$ can act by the full symmetric group, but as Alessandro says as soon as the cover is degree >2 a regular cover cannot act as the symmetric group on the fibers
 
1:49 PM
@MikeMiller Actually Hatcher defines it as deck group acting transitively on the fibers.
 
Yeah that's the correct statement. Sorry.
My memory is a bit hazy.
 
@feynhat This is nonsense. Please ignore, everybody.
 
2:09 PM
Yeah this is a subtle difference. The deck transformation group is quotient of a normalizer of a subgroup of $\pi_1$ if it's not regular in the first place
So the fiber action doesn't match
 
Yeah.
You meant 'normal', right?
 
normal=regular=Galois
 
Oh.
 
Hmm. I am confused.
One sec
Ah yeah nevermind. Take the following beautiful example: Consider the cover of $S^1 \vee S^1$ corresponding to $\langle a \rangle$ (fundamental group is $F_2 = \langle a, b \rangle$). Then my guy above has NO deck transformations; it looks like a two copies of the Cayley graph of $F_2$ coming out of a point, with a self-loop at that point
 
God I hate these terribly nondescriptive adjectives
 
2:19 PM
There are NO automorphisms of the graph which are free (doesn't fix any vertex, in particular, the self-loop guy)
The $\pi_1$ action is perfectly alright though.
 
my favorite descriptive adjective is "good"
 
(Is the $\pi_1$-action alright?)
 
"An adjective is called perfect if it is descriptive"
 
@BalarkaSen Skeptical
I think your first statement was correct
 
thonk_face
 
2:22 PM
Prove or disprove: Let $F$ be a field of characteristic zero. Then $F\left(x^{2}\right) \cap F\left(x^{2}-x\right)=F$
 
First statement being?
 
How to do above problem ?
 
The deck transformation is a quotient etc
 
Ah yeah I think that's correct. But I want a cover with a good $\pi_1$ action but terrible deck action
 
If we think statement is false we try to built counterexample in Q or R field
 
2:23 PM
There are nice pictures of degree 3 covers of surfaces for this
 
Here the deck group is zero visually and also normalizer of $\langle a \rangle$ mod $\langle a \rangle$ is $0$
Dude I only understand graphs
 
Graphs are too hard for me
Find a non-normal index 3 in F_2 then draw that graph
Sounds easier
This infinite junk is too hard
 
"An adjective is called reduced when it consist of one syllable. When it consists of more than one syllable, we call it regular."
 
Let me find a picture, you can tell me yours afterwards
 
Also isn't the cover that you just described actually a line with a circle attached at each integer
 
2:24 PM
Nope
 
Yeah it is
 
That's the one corresponding to normal closure of $\langle a \rangle$
 
Man
Too hard
 
@mathsstudent are those rational function fields?
 
Project to $S^1_a \vee S^1_b$ following the labels/directions
Yeah $\pi_1$-action seems completely transitive
You just have to lift loops corresponding to words starting with $b$
Otherwise you get stuck in that self-loop
 
2:29 PM
I only have a faint idea of what that image you just posted is, Balarka, but it's cool regardless
 
This is actually bad news for me because it means my "proof" of the polynomials puzzle I gave @Thorgott is not a proof. The puzzle might be wrong altogether.
Lol
pUrE tOpLoGy
@MikeMiller Show me the surface pic
 
your brain on topology
 
GIVE A COUNTEREXAMPLE THEN
 
Oh cool.
No deckies
 
idk this stuff man
pOlYnOMiaLs too complicated
 
2:36 PM
@feynhat yeah, and clearly so
 
@Thorgott There are good and bad orbifolds.
Setting up wacom on Arch linux wasn't nearly as painful as I thought it would be.
 
@BalarkaSen This is really offensive to me
 
the existence of good orbifolds is an as-of-currently unproven conjecture
2
 
@MikeMiller The example?
 
Your exampe yes
It's offensive that graph exists
2
 
2:43 PM
Oh that picture is exactly like the standard non-normal cover of wedge of two circles by wedge of four
But thickened
 
Whatever dude I don't care
I'm an algebraist
 
Give me an example of a branched cover $M \to S^2$ of degree $n$ which looks like $z \mapsto z^n$ at every branch point and is not regular
By algebra
 
@MikeMiller Why did he make the middle bit twist ?
 
Ask the topologist
 
Kopologist
I didn't think of thickening that's a cool idea
Covering spaces are really hard
 
2:49 PM
"We say a covering space is hard when...."
 
"We say an idea is cool when it cannot be studied with raw algebra"
6
 
We say an algebraist is a prick if...
3
 
@Balarka do you know who introduced 2-categories
 
is it
MACSHANE
 
Ehresmann
yes, that one
 
2:54 PM
Oh yeah
groupoid guy
 
differential topology and higher category theory, what a combination
 
did he do free yoneda coconut
probably not
 
@BalarkaSen Brown?
 
lool
 
Oh no, another groupoid guy
 
3:00 PM
Brown is groupoid van kampen
 
His fundamental groupoid stuff actually looks useful
 
Yeah man I love setting up the same machine in confusing language to save you some time calculating $\pi_1(S^1)$
Let $X$ be an arbitrary space and $X \ast X$ the join of $X$ with itself. If G-SvK is so useful, tell me what $\pi_1(X \ast X)$ is; I still don't know the answer
 
@MikeMiller I once took a whole homotopy type theory course where the highlight of the last lecture was computing $\pi_1(S^1)$
 
Mike: "I am an algebraist"
"Who are these people I am talking to" - also Mike
after reading Alessandro's message
 
I study group actions, does that make me an algebraist too?
 
3:07 PM
Everyone is an algebraist until proven otherwise
 
What a timing for Tobias to enter the chat
 
An algebraist is never late, nor is he early, he arrives precisely when he means to
 
what's confusing about fundamental groupoids
 
Yeah man I'm sure it was the clearest thing to you when it was the first time you were simultaneously introduced to groupoids and path-homotopies etc
No further questions until you can answer tell me what that group is
 
3:15 PM
I'm confused by why one would use the fundamental groupoid instead of the fundamental $\infty$-groupoid
 
@AlessandroCodenotti what's $S^1$ in HoTT
 
An ugly type
I don't remember the details
 
The circle denoted S^1 is defined as the higher inductive type generated by:

A point `base:S^1`
A path `loop:base=base`
is it this one?
 
I mean, what you need to know about homotopies is the same regardless of whether you do fundamental group or groupoid
it's just a tiny bit more language to keep track of a bit more information
 
3:19 PM
Yes, and you're simultaneously introducing a worthless algebraic notion
Calculate that group
 
3:36 PM
one nice thing about the groupoid perspective is that it allows you to rid yourself of having to work with pointed spaces
 
00:00 - 16:0016:00 - 23:00

« first day (3736 days earlier)      last day (41 days later) »