Mathematics - 2017-06-23 (page 4 of 5)

Ivan Ivanov

11:00

@Secret what do you think of this solution ?

Secret

Still computing, there seemed to be a lot of nested operations whenever i=/=j
s=accdbdab
12345678

Compute
T[1,8]
Check T[1,8]
s[1]=a,s[8]=b=/=s[1]
T[1,8]=min(1+T[1,7],1+T[2,8])
Check T[1,7]
s[1]=s[7]
thus
T[1,7] -> T[2,6]
check T[2,6]
s[2]=/=s[6]
T[2,6] = min(1+T[3,6],1+T[2,5])
Check T[3,6]
s[3]=/=s[6]
T[3,6] = min(1+T[4,6],1+T[3,5])
Check T[4,6]
s[4]=s[6]
T[4,6]=T[5,5]=0
Check T[3,5]

Ivan Ivanov

yeah it is normal because we have to check all cases

Secret

and it seems the number of checks needed grows roughly in powers of 2

Ivan Ivanov

exactly

Abdullah UYU

summer has come here so i can care my hobbies. i am setting my dev. env. up.

Secret

11:14

check T[3,5]
s[3]=/=s[5]
T[3,5]=min(1+T[4,5],1+T[3,4])
Check T[4,5]
s[4]=/=s[5]
T[4,5]=min(1+T[5,5],1+T[4,4])=min(1,1)=1
Check T[3,4]
s[3]=/=s[4]
T[3,4]=min(1+T[4,4],1+T[3,3])=min(1,1)=1
T[3,5]=min(1+T[4,5],1+T[3,4])=min(2,2)=2

T[3,6]=min(1+0,1+2)=min(1,3)=1

check T[2,5]
s[2]=/=s[5]
T[2,5]=min(1+T[3,5],1+T[2,4])=min(1+2,1+T[2,4])

check T[2,4]
s[2]=/=s[4]
T[2,4]=min(1+T[3,4],1+T[2,3])

check T[3,4]
s[3]=/=s[4]

T[3,4]=min(1+T[4,4],1+T[3,3])=min(1,1)=1
check T[2,3]
s[2]=s[3]

T[2,3]=T[3,2]=0

Ivan Ivanov

yeah it is normal,but I think that is the right way :P

Secret

How will you deal with odd pallidromes like abdbdba or abdcdba?, or do you also have a check on the input string on whether it can form a palidrome before the T[i,j] start checking (e.g. a string with all letters distinct cannot from a palidrome for example)?

Ivan Ivanov

I just seek for the number of moves of deleting , I am not interested of the length of the string

Secret

ok I see

Ivan Ivanov

and in such a nice cases where it is palindrome the first thing ofcourse it to chech the start and end character if they are the same we go in depth else we look for the string without the first letter and the string without the last letter

Secret

11:21

Ah so you want to count the minimum number of deletions of an input string such that the leftover is a pallidrome

Ivan Ivanov

yes

Secret

How will your program respond if the user puts in a string that contains no pallidrome e.g. abc, I am guessing T[i,j] will eventually return 1?

but in fact there cannot be any deletion on it that will give you a pallidrome, unless you consider a single letter as a "trivial pallidrome"
or is it irrelevant because of whatever underlying background that the strings came from will guarenteee it contains at least one pallidrome substring?

Balarka Sen

@SohamC ping back

Ivan Ivanov

my solution guarantees because for "abc" it will say 1+1+0 = 2 meaning it will leave a single character which can be considered palindrome

Secret

ok that makes sense

Meanwhile this program does makes me wonder about something: If we are given the following string:

abdcfghrkdfsbcgdjeabdbacdhnfjjfcbdbdbss

we as humans can rather quickly find out where the 3 nontrival pallidromes are, yet a computer will take $2^{|s|}$ steps. I wonder how are we able to achieve that...

Ivan Ivanov

11:31

because we can scan large interval which seem correct and we can even better seek for palindrome in smaller interval of correct interval

computers don't have that sense

Soham Chowdhury

@Balarka email reminder

Balarka Sen

@Soham I do remember :) I just got back from school

Ivan Ivanov

we can find some solutions but it is not guaranteed we will find all while the computer guarantuees

Soham Chowdhury

@BalarkaSen sure.

Secret

I see

Soham Chowdhury

11:34

I cannot believe I am going to be off to ... college ... in a little over a week

Balarka Sen

Neat!

Soham Chowdhury

:(

I mean, more like :/

Daminark

Hey @Balarka

Soham Chowdhury

In any case, colleges have libraries, which is a nice thing.

I really want to look at Eisenbud's comm-alg book

Daminark

@Soham true that, which is convenient

And hope you have a good time in college!

Balarka Sen

11:39

Eisenbud is too huge 4me

(I am writing the email now)

Daminark

(For a shorter one, I've heard Atiyah-Macdonald is good)

Balarka Sen

@Daminark hi

Soham Chowdhury

look. I don't do well with physical math books, it's not how I was brought up :P

I suppose you're the same way to some extent, Balarka.

@Daminark, I've worked through the first couple chapters of that at one point.

Daminark

Nice

Balarka Sen

I can't read pdf's as well as I read physical books actually

Daminark

11:40

How's it going @Balarka?

Soham Chowdhury

@BalarkaSen I like having physical books for the simple reason that I don't have two screens and typing notes is easier without switching back and forth. But I'm more comfortable with pdfs.

Force of habit, I suppose.

Balarka Sen

you techno

i am thinking of refuting a lot of modern technology

cyberpunk is not the greatest post-apocalyptic world to be in.

Daminark

@Soham I've got that Macintosh computer so multiple desktops

Soham Chowdhury

I mean physical screens.

Daminark

I just sorta like the idea of having all my books with me at most times

Soham Chowdhury

11:43

Unless you meant that too?

Daminark

I mean, it's easy to switch between desktops on a Mac is the point, so you may as well have multiple screens because you just slide 4 fingers to the left or right, which is faster

(Also note that I don't really take notes)

@Balarka o lawd

Soham Chowdhury

Virtual desktop-y setups are extremely inadequate. I have an excellent tiling window manager setup (completely keyboard-controlled), but still.

I'm talking about being able to look at something while typing on another screen. You need multiple monitors for that, right?

Balarka Sen

@Daminark there are better options for a post-apocalyptic world.

Soham Chowdhury

@Daminark ahahaha.

Balarka Sen

get used to daminark's puns

and try not to laugh

Daminark

11:45

Oh, yeah likely, they're not simultaneous, I do alternate

Don't listen to @Balarka's advice, let the laughter flow, it's good for the soul

Speaking of souls, when you're done with your email shtick, want to talk a little about algebraic topology?

Soham Chowdhury

@Daminark I hope you don't mean me, I don't know the first thing about topology

I guess you didn't mean me. :)

Secret

people.maths.ox.ac.uk/harrington

Daminark

I was talking about Balarka lmao

Secret

Sorry for some reason, I missread algebraic topology as algebraic biology

Daminark

But nah I demand YOU to help me with this! :P

Soham Chowdhury

11:47

math.berkeley.edu/~lpachter

Balarka Sen

@Daminark Sure, but give me a few minutes

Daminark

Sure

Kek @Secret

Secret

westernsydney.edu.au/__data/assets/pdf_file/0006/513906/…

Turns out that subject does exists in some form...

and there's a springer book

Daminark

Good lord

Secret

Pure mathematics

Broadly speaking, pure mathematics is mathematics that studies entirely abstract concepts. This was a recognizable category of mathematical activity from the 19th century onwards, at variance with the trend towards meeting the needs of navigation, astronomy, physics, economics, engineering, and so on. Another view is that pure mathematics is not necessarily applied mathematics: it is possible to study abstract entities with respect to their intrinsic nature and not be concerned with how they manifest in the real world. Even though the pure and applied viewpoints are distinct philosophical positions...

Looks like nothing is pure nowadays, be it maths, arts etc.

Daminark

11:53

$\infty$-topoi tho

Jk but I mean, I think the point is more, the aim of the study is the abstract object

So many things end up being applicable as a matter of course, but the people working on pure _____ probably ought be defined via taking _____ as an end in itself, as opposed to a means to an other end

Just win baby

What?

Balarka Sen

@Soham Sent.

Secret

I see. For me, I have interest in both fundamental and applied side of ____, which means I can be a pure researcher or an applied researcher.

For example, the recent integral and ordinal mumblings of mine are really more pure driven and applications came as a side product

Whereas for things like dynamical systems, more often there's an application in mind

Due to how I tend to conflate relational existence with physical existence, most abstract entities are tangible to me after I understood them, and I can imagine tossing them around as if they are shapes

Daminark

Hmm

Balarka Sen

@Daminark So, algebraic topology?

Daminark

11:58

I tend to look at subjects with use math more in terms of, wow that's a cool fact about this subject that it involves math somehow. shrugs

Ah, yes, perfect!

Okay so I'm trying to understand what I've now identified to be the Eckmann-Hilton duality

Namely $[\Sigma X,Y] = [X,\Omega Y]$

(equal meaning homeomorphic)

Balarka Sen

Mhm

Daminark

Now, May just sorta wrote that there and possibly some other stuff related to it but I was completely lost

Balarka Sen

So you just want to understand that homeomorphism?

Daminark

For now that's my main goal

So, I remember he wrote something to the effect of how the compact-open topology made one thing work nicely

Balarka Sen

Don't worry about that for now. Do you see how to get a map $X \to \Omega Y$ from $\Sigma X \to Y$?

Let's just try to work out a bijection for now.

Daminark

12:03

Like, that $C(X,C(Y,Z))$ bijects to $C(X\times Y, Z)$, and that in the compact open topology this is a homeomorphism

Balarka Sen

Correct.

Soham Chowdhury

@BalarkaSen Thank you. Do you have any idea how long he takes to reply, usually?

I see you're working with @Daminark, so I'll leave y'all alone for now and go do something useful.

Balarka Sen

@Soham We are talking now. He's on email for 24/7 generically. :P

Daminark

So I think the idea is to use the natural transformation, right?

Balarka Sen

@Daminark do NOT think categorically.

What is $\Sigma X$?

Soham Chowdhury

12:08

@BalarkaSen Oh, I see.

Studentmath

Back again

Daminark

$(X\times I)/(X\times \{0\} \cup X\times \{1\} \cup \{x_0\}\times I)$

Balarka Sen

He agreed to talk, @Soham. I am right now asking whether you should shoot him an email, or vice versa.

Soham Chowdhury

Okay. (He remembers me?)

Balarka Sen

I am not sure. Should he?

I didn't think you knew him.

Soham Chowdhury

12:10

He connected me to SB.

Don't really know him per se.

Just sent him an email, and he said "talk to this prof".

Balarka Sen

@Daminark Right, OK. So what's a way to get a 1-parameter family of maps $X \to Y$ from a map $\Sigma X \to Y$?

Soham Chowdhury

But, no, I think it's likely he doesn't. Ignore that.

Balarka Sen

@SohamC OH. I see. You should mention that to him when you send an email to him.

(He's very forgetful, so it's very possible he does not remember)

Soham Chowdhury

So, what is he saying?

Okay, I'll mention it.

Balarka Sen

hasn't replied to the last email yet. i'll let you know if he wants you to send an email to him

Daminark

12:14

I mean, I think it should be $F_t(x) = F(x,t)$, somehow being careful about the collapsing of $\{x_0\}\times I$

Balarka Sen

@Daminark Good, good. Yes.

Daminark

(Kek Peter May's influence is taking hold on me)

Oh wait hold on, $(x_0,t_1)$ and $(x_0,t_2)$ are the same point so $F$ should basically take the same value on each, meaning that this one parameter family of maps is based properly, I think

Balarka Sen

:)

So, yes, if $F : \Sigma X \to Y$ is a map, then you get a 1-parameter family $f_t : X \to Y$ given by $f_t(x) = F(t, x)$. This one parameter family, as you said, sends $f_t(x_0) = y_0$ for all $t$, where $x_0$ is the "base" of $\Sigma X$, and $y_0 = F(x_0)$

But more importantly, this 1-parameter family $f_t$ "starts" ($f_0$) and "ends" ($f_1$) at the constant map at $y_0$

Hmm, is this not starting to look like a loop of based maps? :)

Daminark

So this should generate a loop somehow, and I imagine this might have something to do with our currying?

Internet...

But yeah awesome, I'll think a bit and make sure to formalize this

Balarka Sen

Ok, let me know if you want me to spell more details out. Remember you need to construct a map $X \to \Omega Y$ out of this.

It's helpful to understand what $\Omega Y$ looks like.

(Not literally, just interpret the definition visually)

Fargle

12:23

Eww, math.

Balarka Sen

In the end it's just $C(X \times I, Y) \cong C(X, C(I, Y))$ with appropriate basing considerations.

(the bijection for that is super-obvious; any map $f : X \times I \to Y$ gives rise to an $X$'s worth of maps $f(x, -) : I \to Y$)

Daminark

So you take that homeomorphism, call it $\phi$, and then you're looking at a precomposition $\phi\circ f$

Or wait hold on not that

$\phi(f)$

Balarka Sen

@SohamC Shoot.

Tell him what you know, what you do, and where you're at.

among other things maybe

Daminark

Well, you're having $f_t(x)$ which we can revert to thinking about as $F(x,t)$ for a bit, then it looks like $\phi(F)$

@Fargle kek

And it works basedly because each $f_t(x_0) = y_0$, so that when you pull it through, you get that the loop fixes $y_0$

Soham Chowdhury

@BalarkaSen I'm supposed to email him?

Balarka Sen

12:29

Yes, you are.

Soham Chowdhury

okay.

Daminark

Okay I think I better understand this, and you were right that it's helpful to not just say that suspension and loop are functors, so then you have natural transformations and then just compose them

At least now I see it, for real

Soham Chowdhury

rkmvu email or tifr one (or something else)?

Balarka Sen

Those are official ones. Do you know the gmail one? If not, send me an email and I'll send it to you.

Soham Chowdhury

Copy it and I'll delete the message.

Daminark

12:32

And now I just need to remember exactly how we linked it to Eilenberg-MacLane

Balarka Sen

delete

@Daminark Give me a few more minutes, I'll see what you have to say.

Daminark

Yeah it's fine, I'll just type it out as you talk

Balarka Sen

@Soham Done.

Daminark

Oh wait I think I see it

Soham Chowdhury

@BalarkaSen got it

Daminark

12:36

So we know that $\Sigma S^{n-1} = S^n$

Balarka Sen

also we should move further discussions about this to email

in my opinion

Daminark

So then $G = [S^n, K(G,n)] = [S^{n-1},\Omega K(G,n)]$, which is suggestive of the identity $\Omega K(G,n) = K(G,n-1)$

Oh wait you could get that identity by directly thinking about loops, I imagine

Balarka Sen

Bingo, @Daminark.

Soham Chowdhury

@BalarkaSen okay.

Daminark

Okay need to make this precise but the pieces are starting to fit together

Balarka Sen

12:39

So just to understand where we were, we understood the loopspace-suspension adjunction?

Daminark

Yeah, I think that part checks out alright for me now

Balarka Sen

Great.

So you want to prove $\Omega K(G, n) = K(G, n -1)$ now?

Daminark

Yup

Balarka Sen

I like your observation. I think that means $\pi_n(K(G, n)) \cong \pi_{n-1}(\Omega K(G, n))$?

Actually, you don't need $S^n$. You can work with $S^k$

Which would tell you $\pi_k(K(G, n)) \cong \pi_{k-1}(\Omega K(G, n))$

Daminark

Well I think that is something that just holds more generally because the suspension of $S^n$ is $S^{n-1}$

Wait what? What's the difference between $S^n$ and $S^k$ here?

Balarka Sen

12:41

you used $n$ in both the dimension of the sphere and K(G, n)

that is not necessary

Daminark

Oh I see

Soham Chowdhury

@BalarkaSen because the groups are 0 in every other dimension, right?

if I remember what K(G,n) means

Balarka Sen

Oh that's true. I was more like, saying, write it out for all homotopy groups than one :)

What I was in fact getting at was $\Omega K(G, n)$ has the same homotopy groups in all dimensions as $K(G, n-1)$

which your comment + Daminark's observation proves

Daminark

Right, nice

Perfect!

Okay now I'm much happier with that

Balarka Sen

Does isomorphic homotopy groups in every dimension imply homotopy equivalent spaces?

Daminark

12:44

And we just need to wait for Peter May to actually construct said spaces and show that their homotopy stuff form groups :P

Balarka Sen

Ponder on that while I go away from keyboard for a while.

Daminark

I know this should give a weak homotopy equivalence

Or at least I think

Perturbative

Hey everyone!

Daminark

My immediate hope for a counterexample would be something which is not contractible but with vanishing homotopy groups

GPhys

seems like my first bounty is going to expire without an answer :(

did manage to bring the question from ~2 to 35 upvotes though and my own work has at least convinced me the question is "hard"

Akiva Weinberger

12:53

Whoa your avatar changed

(Did you get a haircut?)

Daminark

Hey @Akiva and @GPhys!

Akiva Weinberger

@GPhys What's the question?

And hejhej

GPhys

35

Q: Prove that $\sum_{n=1}^\infty \frac{n!}{n^n}$is irrational.

Let $$S= \sum_{n=1}^\infty \frac{n!}{n^n}$$ (Does anybody know of a closed form expression for $S$?) It is easy to show that the series converges. Prove that $S$ is irrational. I tried the sort of technique that works to prove $e$ is irrational, but got bogged down.

Balarka Sen

@Daminark Do you know Whitehead's theorem?

GPhys

I went through the full notes for a lower level grad course on irrationality and transcendence @AkivaWeinberger

I made no progress on this question, but I learned a lot

Daminark

12:56

Nope, that hasn't happened yet

René Descartes

Is Science the knowledge of what is around us? Please give your opinion. I request you.

GPhys

and I proved or empirically showed several of the methods wouldn't work for the series as given

Balarka Sen

It says for CW complexes, a weak homotopy equivalence (a map $f : X \to Y$ such that $\pi_n(f)$ is an isomorphism for all $n$) is a homotopy equivalence.

Daminark

So the main atop course is going a manifoldsy route, but so far what we've done is briefly review the basic topology stuff, then do homotopy, induced homomorphism, weak homotopy equivalence, fundamental group, a bit on higher homotopy groups (including a proof I'm not yet sure about that they're abelian), categories/functors/natural transformations, Eilenberg-MacLane, "the nth cohomology group", and the loop/suspension stuff

Balarka Sen

If what you guess is true, that is, $\pi_n(X) \cong \pi_n(Y)$ for all $n$ implies there is a weak homotopy equivalence $f : X \to Y$, then $X$ and $Y$ in fact should be homotopy equivalent. (assuming everything is a CW complexes). But is it?

Daminark

12:57

(starting from higher homotopy groups, it was the spinoff seminar by May)

We haven't quite defined CW complexes yet

Balarka Sen

I am not liking May's course.

@Daminark It's just "the space is nice enough". Don't worry about it.

Daminark

Alright

So, based on this statement, are weak homotopy equivalence and homotopy equivalence the same in CW complexes?

Balarka Sen

Yes: that is precisely the Whitehead theorem.

Daminark

Oh wait sorry I missed that message

Yeah okay we're good

Balarka Sen

But what I am asking is more like, does $\pi_n(X) \cong \pi_n(Y)$ for all $n$ really mean $X$ and $Y$ are weak homotopy equivalent?

@Daminark It's actually not entirely trivial.

Say, you don't know about covering spaces yet, do you?

Daminark

13:07

Nope

All that's happened is the syllabus you're iffy about above

Balarka Sen

Mmkay.

Ask me the question I asked you after you learn covering spaces and homotopy lifting.

Note it down somewhere meanwhile.

Daminark

How do homotopy groups behave with Cartesian product?

Akiva Weinberger

Every space is weakly homotopy equivalent to a finite space, fun fact

For example, for spheres, you quotient out by equators and stuff

Daminark

Ah, yeah Peter May mentioned that

Balarka Sen

@AkivaWeinberger I don't think this is true for every space.

Daminark

13:09

Maybe every CW complex?

Akiva Weinberger

Like for a 2-sphere, you quotient out the northern hemisphere, the southern hemisphere, the eastern… hemiequator, and the western hemiequator

and you have two points left

so it's a six-point space

Balarka Sen

Should be true for all simplicial complexes.

Daminark

And @Balarka will do

Akiva Weinberger

@Daminark Oh, probably.

@BalarkaSen *Oh, probably.

Balarka Sen

Actually I think something stronger than that is needed.

GPhys

13:11

@AkivaWeinberger as for the question I bounty'd, I suspect you need some clever integral representation

Balarka Sen

@Daminark The answer is "false".

Daminark

Also @Balarka what in particular are you iffy about wrt Peter May's shtick

Steamy was not happy about Peter May's defining $\widetilde{H}^n(X,\pi) = [X,K(\pi,n)]$

Balarka Sen

Ok, no, apparently this is true for all finite simplicial complexes.

@Daminark I am totally not. UGH.

Daminark

It was rather funny when he did it though

Balarka Sen

That is a fantastic theorem I appreciated after I learnt individual definitions of all those.

Daminark

13:13

He was like "So we've got $[X,K(\pi,n)]$, and I wish to give it a name. Let's call it the H stuff!"

(He didn't say the H stuff but I don't feel like TeXing it again)

So that was amusing. But yeah he defined that and said something like "So we're gonna see that this works when we construct E-M spaces, and then we'll see it's dual to something, hence homology exists!"

Balarka Sen

I don't like it.

But I wonder; it might be a fun pedagogy.

lol i can't decide whether i like it or not

Daminark

He's apparently never done it quite this way before, he doesn't like the whole chains stuff used to introduce homology, apparently, and thinks that this is closer to how modern algebraic topologists think about it. Also he likes that you can get other homology theories out of different types of spaces

Balarka Sen

Yes.

Daminark

He wants to introduce the Eilenberg-Steenrod axioms at some point, and that'll explain this

Balarka Sen

$K(G, n)$ is the "Eilenberg-Maclane spectrum" to the homotopy theorists nowadays

You can use different spectrum to get different homologies

Daminark

13:17

I find it rather amusing that he's one of the more modern people around and yet he's one of the oldest people at this school, if not the oldest

Balarka Sen

I don't think any of this is modern. It's more useful from the homotopy theory perspective.

Daminark

No I meant like, his research

Balarka Sen

The chain complex definition is more useful from the geometry perspective.

@Daminark Ah, ok.

Daminark

I think he remarked once that he does research in infinite loop space theory and higher category theory (model categories or smth?)

Balarka Sen

vOv

Daminark

13:21

So I guess you can see why he wants to put a lot of stuff in categorical language

Balarka Sen

I know why. I just don't appreciate it :P

Peter May is a pretty famous name.

Semiclassical

Hi chat

BAYMAX

Hi semi

Daminark

With respect to the homeomorphism between looped and suspended space, he basically said that the two are functors and that this would basically have to do with a natural transformation between them

BAYMAX

classics

semiclassics

Daminark

13:22

He actually said that you straight up can't do math without categorical language, though if you're in analysis you can avoid category theory :P

BAYMAX

Hey you know this one $\dot{x} = 1(mod 1) , \dot{y} = \omega (mod 1)$

Semiclassical

not really

those are independent ODEs, though

BAYMAX

actually I am searching what this type of system is known as ?

Semiclassical

so wouldn't seem like there's anything terribly interesting there

Balarka Sen

@Daminark $[\Sigma X, Y] \cong [X, \Omega Y]$ is a categorical relationship, namely, "Loopspace and suspension are adjoint functors". But that doesn't explain shit about where that isomorphism comes from, unfortunately. :P

BAYMAX

13:23

it is when you plot x vs y

Semiclassical

also, though, what is exactly being taken mod 1 here?

Balarka Sen

I don't believe you... You're a liar!!!

►

Semiclassical

If it's 1, well---1 mod 1 = 0

and omega mod 1 is a constant.

Daminark

Lel @Balarka

Balarka Sen

I love some of the categorical formalism and results, but not the pedagogy.

$H^n(X; G) = [X, K(G, n)]$ is proved (from the Eilenberg-Steenrod axioms of $H^*$) from a categorical theorem, actually.

BAYMAX

13:26

$x ̇ = 1(mod1); y ̇ = ω+αsin(2πx)(mod1)$

a coupled one!

Balarka Sen

The "Yoneda lemma"

BAYMAX

any body can search and can help me in solving this analytically?

Semiclassical

that's a bit different, yeah.

Daminark

I mean if you wanna prove a theorem, Yoneda lemma

BAYMAX

Like just a sample of how to do it?

Semiclassical

13:27

but, again, I don't know what you mean by mod 1 here. Do you mean you only care about the interval [0,1]?

Soham Chowdhury

Infuriatingly trivial if you don't understand it, and unexplainably trivial if you do.

BAYMAX

yeah I think we will be on a unit square

Soham Chowdhury

@Semiclassical often people mean fractional part by that, yeah.

BAYMAX

@Semiclassical

Semiclassical

@baymax sure, but that'd be $x,y$ mod 1

Daminark

13:27

chuckles

Alessandro Codenotti

Every time cathegory theory is mentioned the yoneda's lemma pops up, I'll discover what is says sooner or later

BAYMAX

like i didn't get that?

Daminark

Let's see how long before May teaches us that. Next class is on Tuesday

Also lol I wonder what Ted would think

Balarka Sen

Ted is even more conservative than me about category theory.

(I like it)

Daminark

I know, I kinda wanna see his reaction sitting in on a lecture

Soham Chowdhury

13:30

Well, doesn't it depend on the math you do?

Balarka Sen

Mike has unfortunately become an infinity category theorist

too bad

Daminark

Has he really?

Balarka Sen

not technically, but he's doing lots of categorical algebra

Daminark

I mean given that I've got 8 weeks (hopefully no bootcamp conflict), it May happen to me too

Soham Chowdhury

he's grumpy about it, but characteristically full of good cheer

Daminark

13:32

And ah, well algebra is good

Balarka Sen

no u

Soham Chowdhury

@Daminark have you done/are you doing the Chicago REU?

Daminark

The program I'm officially in is called the bootcamp

It's something my analysis professor set up, we'll do complex, probability, dynamics, and diffgeo

Balarka Sen

I am annoyed. I don't know an entirely elementary proof of $\Omega K(G, n) = K(G, n - 1)$.

I can use ze path space fibration but I don't want to

BAYMAX

$\dot{x} = -x + x^3$ , x(0) = a ;$-1<a< 1$ , does solution exist for all times?

Daminark

13:34

Now, I'm also sitting in on the REU talks, with my top priorities being atop, followed by NT

BAYMAX

@Semiclassical any help?

Daminark

I have done the apprentice last year, which was a 5 week course in linear algebra and a side of graph theory

Akiva Weinberger

What's $[X,Y]$ for spaces?

Balarka Sen

Homotopy classes of maps $X \to Y$

Also used as based maps sometimes

Akiva Weinberger

Ah, right

Wait, so the homology of a space is the homotopy classes of maps from it to $K(\Bbb Z,n)$?

Balarka Sen

13:41

K(Z, n)

and *cohomology with Z coefficients

Akiva Weinberger

Oh, cohomology.

How is that at all useful? I thought $K(\Bbb Z,n)$ was in general a horrible, infinite-dimensional thing

(Also I have no idea how one proves this)

Balarka Sen

Do you know the cellular approximation theorem?

Akiva Weinberger

Vaguely.

Balarka Sen

Ok. If $X$ is an $n$-dimensional CW complex then any map $X \to K(\Bbb Z, n)$ is homotopic to a map $X \to K(\Bbb Z, n)$ which sends all of $X$ to the $n$-skeleton of $K(\Bbb Z, n)$.

This is because of cellular approximation; you can homotope any map between CW complexes so that it sends $i$-skeleton to $i$-skeleton for all $i$.

$K(\Bbb Z, n)$ is a sad space, but it admits a model (by that I mean another space which is homotopy equivalent to it) so that $n$-skeleton of $K(\Bbb Z,n)$ is $S^n$. (you can prove this)

Soham Chowdhury

@Balarka can you come online on Hangouts?

Balarka Sen

13:45

So you end up looking at maps $X \to S^n$

Ah, sure, @Soham.

Akiva Weinberger

Right, I think I remember your construction of $K(\Bbb Z,S^n)$ from a while ago which should work

@BalarkaSen Wouldn't that give you cohomotopy or something?

Balarka Sen

K(Z, n), but yeah.

@Akiva Right, cohomotopy.

Akiva Weinberger

Dammit

That's what I meant

Got to go, be back later

Balarka Sen

As a particular example of what I said, if $M$ is an $n$-dimensional manifolds (do topological manifolds always admit CW structure? I think so) then $H^n(M; \Bbb Z)$ is in bijection with $[M, S^n]$

So $[M, S^n] = \Bbb Z$

With a little more work you can show Hopf degree theorem from this

(you just have to show that the bijection is given by taking preimage of a point and counting the number of preimages; there's a categorical work involved here)

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$Mathematics$ Mathematics