Mathematics - 2017-08-25 (page 2 of 6)

Akiva Weinberger

09:01

@AlessandroCodenotti Wait so the regions are what

for Van Kampen

Heh, apparently Seifert–van Kampen is named after Herbert and Egbert

Alessandro Codenotti

For Van Kampen it's easier if you remove the 2 points from the capping disk

Akiva Weinberger

Can we do that?

It's not a homogenous space

Yeah, no we can't

Alessandro Codenotti

Ah, wait, originally the points were removed from the disk

Akiva Weinberger

Not the torus?

Then that's different, isn't it homotopy equivalent to T^2vS^1 then?

I was thinking you removed them from the torus, that's why H_2 was gonna be zero

If you remove them from the capping disk H_2 is still fine

Alessandro Codenotti

Interesting, I didn't think it'd make a difference

Akiva Weinberger

09:04

@BalarkaSen Double-check me on that

Alessandro Codenotti

Turns out thay I was wrong :P (unsurprisingly)

Will Hunting

@AkivaWeinberger I always find it amazing that the torus is the circle times the circle. =)

Tobias Kildetoft

How can it matter where you remove the points from on a path connected space?

Balarka Sen

@AkivaWeinberger I am not sure why you think they are homotopically different objects.

They shouldn't be

Alessandro Codenotti

@TobiasKildetoft Yeah that was my reasoning too

Akiva Weinberger

09:05

@TobiasKildetoft Take S^1vS^2 for example

Wildly different depending on if you remove the point from the circle bit, the sphere bit, or the intersection bit

Balarka Sen

Oh fuck fair enough

Alessandro Codenotti

@AkivaWeinberger nice, the group become either $F_2$ or trivial

Akiva Weinberger

Homotopy equivalent to the sphere, circle, and pair of points respectively

@AlessandroCodenotti What wait how F_2

Tobias Kildetoft

@AkivaWeinberger Ahh, right. But when removing the points does not change the path connectedness, can it still matter?

Alessandro Codenotti

@AkivaWeinberger sorry, Z, I was still thinking about the earlier example

Removing 2 points

Akiva Weinberger

09:07

@TobiasKildetoft Same example. Removing from the circle bit leaves us with something homotopically equivalent to a sphere. Removing from the sphere bit leaves us with something homotopically equivalent to a circle.

@AlessandroCodenotti Ah fair

Tobias Kildetoft

Ahh, right

Akiva Weinberger

You would want homogeneity for it not to matter where you take it from, I think

Like connected manifolds without boundary are homogeneous

Balarka Sen

That can happen even if the local homology around those points are the same.

Like, T^2 wedge S^2

Alessandro Codenotti

Interesting, that was unexpected

Balarka Sen

It matters if you remove from T^2 or S^2

Akiva Weinberger

09:08

Hm, yeah. T^2 and S^1vS^1vS^2 respectively, right?

Balarka Sen

Yup

Alessandro Codenotti

So how does the group change if we remove from the capping disk?

Balarka Sen

So I don't know a good condition for homogeneity of a space.

Akiva Weinberger

You should get T^2vS^1

Alessandro Codenotti

Which has fundamental group $\Bbb Z^2\ast\Bbb Z$

Akiva Weinberger

09:10

so $(\Bbb Z^2)*\Bbb Z$ I guess?

Merp. That

Daminark

You're saying merp now

My influence grows

Akiva Weinberger

O_O nooOOOOOO

Balarka Sen

Big Smoke starts singing

Akiva Weinberger

Also hi good morning

Vrouvrou

@AkivaWeinberger just to be sure when we work with dx then we work with Lebesgue measure

Akiva Weinberger

09:12

I think so yeah

Alessandro Codenotti

So a presentation would be $\langle \alpha,\beta\gamma|\alpha\beta=\beta\alpha\rangle$

Daminark

Yo

Akiva Weinberger

Yeah I think so

Alessandro Codenotti

Which is still different from what I got with S-vK meh

Balarka Sen

It's one of the examples where drawing it is easier than S-vK

Alessandro Codenotti

09:14

Indeed

A classmate (@lore I summon thee) asked about S-vK though

Balarka Sen

But it's the amalgamated product of $\pi_1(T^2)$ and $\pi_1(D^2 - \{p, q\})$, where you identify one of the generators of $\pi_1(T^2)$ with the commutator curve of $\pi_1(D^2 - \{p, q\})$.

Akiva Weinberger

I guess ideally the thing you got before ($\langle a,b,g|abg=gab\rangle$) should be isomorphic

Vrouvrou

@AkivaWeinberger thank you

Akiva Weinberger

$(a,b,g)=(\alpha,\beta,\gamma)$ clearly

Alessandro Codenotti

Except 500% faster to write :P

Balarka Sen

09:16

Which is <a, b, c, d| ab = ba, a = cdc^-1d^-1> I guess.

SteamyRoot

That smells like an almost-crystallographic group

Daminark

I just realized why I've had a bad time with cycle notation for literally over a year now

I read the product of cycles left-to-right

Balarka Sen

lol

Daminark

So every time I was given a product of two cycles I could never get it right

Alessandro Codenotti

@BalarkaSen uhmm that seems different both from what I got and from $\Bbb Z^2\ast\Bbb Z$

Balarka Sen

09:26

@Alessandro For torus with a double punctured disk capping off the hole? That's what SvK gives me.

Daminark

This is why you don't read Herstein

Alessandro Codenotti

I got <a,b,c|abc=cab>

Akiva Weinberger

@Daminark Neah, you read the product left-to-right, just the cycle notation is backwards. (1 2 3) means 3 goes to 2 goes to 1 :P

(I think that should give the same answer, provided you write the output in cycle notation as well)

Alessandro Codenotti

I guess drawing it until it's S^1 wedge S^2 is just better for this particular space

Daminark

Actually I'm gonna say Herstein is right and every other algebra book is wrong

Balarka Sen

09:28

@AlessandroCodenotti I don't know how you got that but if you we both did it right they should be isomorphic.

Akiva Weinberger

(since it's essentially the idea that $x\mapsto x^{-1}$ is an isomorphism between a group $(G,\circ)$ and its opposite $(G,\circ')$)

Daminark

In fact I'll swtich from $f(x)$ to $xf$

Akiva Weinberger

(where $a\circ'b:=b\circ a$)

Daminark

Even if the context isn't algebra

Akiva Weinberger

@Daminark GOOD idea

:P

Balarka Sen

09:29

NO GOD! PLEASE NO!!! NOOOOOOOOOO

►

Alessandro Codenotti

@BalarkaSen I divided this space into the torus with a bit of the disk (without punctures) and the disk with punctures as open sets so that they intersect in a ring

Balarka Sen

@Alessandro Ok. That seems like my open sets. So how did you get abc = cab?

Tobias Kildetoft

@Daminark So you should write $a+b$ as $ab+$

(or maybe as $ba+$?)

Balarka Sen

$+ab$, rather.

Daminark

$(a,b)+$

Tobias Kildetoft

09:31

@BalarkaSen Not when the functions are to the right

Balarka Sen

Ah yes

Alessandro Codenotti

The group of the intersection has a single generator, which is one of the generators of the torus on one side and is $ab$ in the punctured disk (where $a,b$ generate the groip of the punctured disk)

Akiva Weinberger

RPN, innit

Reverse Polish Notation

Tobias Kildetoft

@Daminark But you did not use brackets there (and one can completely avoid them if needed)

Akiva Weinberger

Where "3 5 + 6 x" means 48 and parentheses are unnecessary

Balarka Sen

09:32

@Alessandro Uh, the boundary circle of the disk with two punctures is not $ab$.

Daminark

Oh I was thinking that as a function, plus takes in the ordered pair $(a,b)\in G^2$

Balarka Sen

It's $aba^{-1}b^{-1}$ is it not

Akiva Weinberger

What

Tobias Kildetoft

@Daminark Once we go there we might as well go all the way to reverse polish. The point is that addition always takes two arguments, so you don't need to group them in any special way

Akiva Weinberger

Doesn't the conjugate have self-intersections and stuff

I don't know what "and stuff" would mean

Just self-intersections

@TobiasKildetoft Yeah I just mentioned it

Daminark

09:34

Fair

Balarka Sen

@AkivaWeinberger Homotopically those self intersections don't really matter do they

Akiva Weinberger

?

I think Ale's right in that it's ab

Alessandro Codenotti

@BalarkaSen wait why? I don't see it, can't you shrink the boundary circle to $ab$?

Balarka Sen

How do you shrink the boundary circle to $ab$? First choose a basepoint on the boundary.

Akiva Weinberger

Go 'round one puncture then the other

In fact, $aba^{-1}b^{-1}$ would be in the kernel of the $\pi_1\to H_1$ abelianization map whereas this thing shouldn't be

Balarka Sen

09:37

Uhhhh

Maybe you're right.

Akiva Weinberger

The boundary circle is where the disk is attached to the torus, yeah?

Alessandro Codenotti

Yep

Balarka Sen

Yeah so my thing should actually have been <a, b, c, d| ab = ba, a = cd>

Alessandro Codenotti

Ok, that's the same as my group

Balarka Sen

Yeah actually the commutator is the Pochhammer curve.

I'm dumb

Tobias Kildetoft

09:39

@BalarkaSen So you can simplify that by removing $a$ as a generator

Balarka Sen

Right, which is what Alessandro did

mago

hi, i'm the code's classmate

Balarka Sen

I was thinking of boundary curve of a torus with boundary for some reason.

Hi @mago

Daminark

Well we gotta make sure you're not an imposter

What's the code word?

Akiva Weinberger

Is "Code" nickname for Alessandro?

Codenotti?

Balarka Sen

09:43

It's derived from Codenotti I think

mago

yup, read his surname and shrink

Akiva Weinberger

Cod.

Co?

How high is the English-speaking rate in Italy

Balarka Sen

I need to wake up a bit more today.

Akiva Weinberger

Preliminary Google says 1 in 3

Daminark

C

Tobias Kildetoft

09:46

@AkivaWeinberger how high of a level does that assume?

Akiva Weinberger

I dunno

Daminark

@TobiasKildetoft About 90

Akiva Weinberger

Denmark has a high English-speaking rate, right? I heard Scandinavia is very English-speaking

(I don't know if that grammars right)

mago

i got lost: which is the space you're "svk"-ing?

Daminark

Sounded grammarful enough to me

Tobias Kildetoft

09:47

@AkivaWeinberger Yeah, we are generally fairly proficient in English

Will Hunting

Hi guys. I don't like Amin's idea of the MSE University, but Heather has sort of a slightly different idea in mind, and there is some discussion of this different idea in that room, and maybe if you are interested you can go and read the transcript and see if you want to say anything or participate in the project. chat.stackexchange.com/rooms/63252/planning-mse-university

2

Tobias Kildetoft

Not sure what the official rate is (I am always surprised when I meet anyone who does not speak at least passable English)

Akiva Weinberger

@mago Uh, take a torus. Take the hole and cap it with a disk. Remove two points from the disk.

(It matters where we remove the points from)

Balarka Sen

@mago There seems to be two spaces we were talking about. One is torus with the hole capped off by disk, and two points removed from the torus. Other is the same thing with two points removed from the disk.

mago

ok, it was the first one i asked to @alessandro

Alessandro Codenotti

09:48

Now the problem is, are $\langle a,b,c|abc=cab\rangle$ and $\langle a,b,c|ab=ba\rangle$ isomorphic

Akiva Weinberger

An interesting example comes from $S^1\vee S^1$, where removing a point leaves you with a space of homotopy type $S^0$, $S^1$, or $S^2$ depending on where the point was

mago

got it, thanks guys

Alessandro Codenotti

yeah he is the source of the problem with the points removed from the disk

the second space was an accident since I believed it was the same

Balarka Sen

So he's the source of our miseries.

Akiva Weinberger

@AlessandroCodenotti Well the one on the left is $\langle ab, c, a\mid abc=cab\rangle$, right?

By which I mean the isomorphic image of $\langle x,c,a\mid xc=cx\rangle$ under the map $x\mapsto ab$

Relable stuffs and you get $\langle a,b,c\mid ab=ba\rangle$

google.com/search?q=hats&tbm=isch

Wait hold on aren't all hats top hats

Where else are you wearing them

Daminark

09:53

Top hats are hats that people in the northern hemisphere wear

Alessandro Codenotti

@AkivaWeinberger hmm, wait, I'm not following this step

Balarka Sen

@Alessandro Without coming up with an isomorphism, you can indirectly see that because the space is homotopy equivalent to torus with a small interval going between two points in the donut hole.

If you shrink that interval you get that it's homotopy eq to T^2 v S^1

Daminark

Anyway I should sleep now, good morning everyone!

Balarka Sen

So the fundamental group is also Z^2 * Z.

Will Hunting

@Daminark See you in your dreams!

Balarka Sen

09:56

Night, @Daminark!

skullpatrol

Cya

mago

night @Daminark, thanks!

Akiva Weinberger

I should go back to sleep

Gnat

G'night

Alessandro Codenotti

night

skullpatrol

later

Alessandro Codenotti

09:57

@BalarkaSen yep, I agree

mago

thanks @AkivaWeinberger too

Alessandro Codenotti

isn't this how we decided it had to be Z^2*Z at first?

Balarka Sen

I guess so?

Alessandro Codenotti

Anyway we (finally) got a group with SvK too so I'll consider this problem done and go back to studying analysis, thanks everyone!

Balarka Sen

I need to figure out how to wake up

Will Hunting

10:02

@BalarkaSen Use two alarm clocks.

Balarka Sen

Good luck with analysis

@WillHunting Doesn't help if your mind is asleep

Alessandro Codenotti

@BalarkaSen don't sleep!

You don't need to solve a problem if the problem never comes up

Balarka Sen

flayless lagic

I guess I could listen to some music. I need to listen to that Scott Walker trilogy in full.

@Alessandro Do you like Pasolini?

Alessandro Codenotti

He's on my list, but I have yet to see any of his movies

I'm surprisingly ignorant concerning Italian cinema

Balarka Sen

Gotcha. I haven't seen anything by him either.

@AlessandroCodenotti lol

I was reminded of him because one of my recent favorite songs is dedicated to him. It's sort of obscure because I don't get all the references but it's very interesting.

Alessandro Codenotti

10:19

Ah, I see

I have Stalker on my computer, but I never find the time to watch it...

SteamyRoot

Do it

Also read Roadside Picnic and maybe even play the games.

FuzzyPixelz

Hello!

Balarka Sen

@Alessandro Stalker is a must watch.

To be honest pretty much any other Tarkovsky is a must watch.

My top 3 are Solyaris, Stalker and Nostalghia

FuzzyPixelz

So.. it's my first time doing a proof by induction on a proposition of the form $P(n,k)$ and I'm confused, I can't decide which is the next proposition and I don't know if I should do induction on $n$ or $k$

It's about the binomial coefficient by the way, and even if you look at Pascal's triangle, I still can't decide what is the "right way to move around the numbers" to see which one is next

Akiva Weinberger

10:35

$$\sum_{n=-\infty}^\infty e^{-(x-n)^2}=\sqrt\pi$$

Is that true??

If so, how would I prove it?

@SimplyBeautifulArt You seem like a knowledgeable person

Similarly, $\sum_{n=-\infty}^\infty e^{-(x-2n)^2}$ seems to outline a cosine curve (i.e. $A\cos(\pi x)+B$ for some constants $A$ and $B$).

Balarka Sen

Looks awfully like $\int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi}$

Akiva Weinberger

Hm, it does

What's $\int_n^{n+1}e^{-x^2}dx$?

Oh that's gonna be error function nonsense

Yep

Balarka Sen

mhm

I am not sure if the same technique that's used to prove the integral thing works

It works in the integral case because of how the area element changes under polar coordinates

Akiva Weinberger

Doubt it. Discrete grids probably don't work well with polar.

Balarka Sen

Right

Akiva Weinberger

10:42

Wolfram seems to suggest it's false, actually

Balarka Sen

Aw

Akiva Weinberger

It's really close, though.

Graphing it on Desmos, it really does look just like a straight line.

Alessandro Codenotti

weren't you 2 both supposed to be asleep?

Balarka Sen

Akiva was

It's 4PM here :P

Akiva Weinberger

Yeah I dunno it's morning now whatever

Sun's up

Balarka Sen

10:48

I was just feeling sleepy. Listening to Tilt helps to wear that off

Alessandro Codenotti

@BalarkaSen Really? I was sure there's a much bigger timezone difference between Italy and India, it's 12:50 here

Balarka Sen

I guess I could also listen to progressive death metal but I don't want my ears to die

Akiva Weinberger

'Bout three hours probably?

4.

Balarka Sen

@Alessandro Yeah our timezones are pretty close actually

Akiva Weinberger

No, 3 and a half

Balarka Sen

10:49

It just seems different because I give ze fingers to my timezone

Akiva Weinberger

Me: 6:50 am. Ale: 12:50pm. Blark: 4:10pm

Blargle

Balarka Sen

4:19 actually, but yep.

Akiva Weinberger

Check again?

Balarka Sen

20

lol

Akiva Weinberger

Blaise it

FuzzyPixelz

10:53

Well dumb questions may actually exist, and I may have asked one.

Akiva Weinberger

@FuzzyPixelz What's the theorem

FuzzyPixelz

The binomial coefficient is always a natural number.

Akiva Weinberger

I'd induct on the row

Which one says how far down the triangle you are, $n$?

Will Hunting

I always am confused about induct and induce, lol.

FuzzyPixelz

The $n$

Akiva Weinberger

10:58

We know it's an integer when $n=0$, for all $k$. Suppose it's true for $n$ (for all $k$), prove it's true for $n+1$ (for all $k$).

We know $P(n,k)+P(n,k+1)=P(n+1,k)$ (I think?)

Also $P(n+1,k+1)=1$

So the first one should take care of when $k$ is between $0$ and $n$, and the last one should take care of when $k$ is $n+1$

which are all the options I think (the second input can't be more than the first input, right?)

FuzzyPixelz

And $k=0$ ?

It's also 1 right ?

Akiva Weinberger

Right

Oh did I mess up the equation

I might've

FuzzyPixelz

I just can't understand how why did we chose $n$

No worries I have it

Akiva Weinberger

Think of it in terms of the triangle

Say we know everything in a certain row is an integer

With the rule where each thing is the sum of the two things above it, we can know that everything in the row after it is an integer

(each entry is the sum of the two things above it, but those two things above it are integers)

(except the edges I guess but they're fine 'cause they're 1)

On the other hand, inducting on $k$ would be like inducting on the column

and I don't see how that would work

Even if we know everything with a certain $k$ is an integer, I don't see how that implies that everything at $k+1$ is an integer

FuzzyPixelz

Makes sense, thank you

Secret

11:13

$$\sum_{n \in \Bbb{Z}}e^{-n^2}$$

The summation chain rule is extremely henious thus I am not sure if it can even be evaluated at all

Leaky Nun

@Secret I don't think it can

Secret

2

Q: Bounds on Gaussian infinite sum

What are some good upper and lower bounds on the following sum? $$S=\sum_{n=-\infty}^{+\infty}\dfrac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{1}{2}\left(\frac{n}{\sigma}\right)^2}$$ I am looking for something better than $1<S<2$.

physicsforums.com/threads/discrete-gaussian-summation.399503

Theta function

In mathematics, theta functions are special functions of several complex variables. They are important in many areas, including the theories of Abelian varieties and moduli spaces, and of quadratic forms. They have also been applied to soliton theory. When generalized to a Grassmann algebra, they also appear in quantum field theory. The most common form of theta function is that occurring in the theory of elliptic functions. With respect to one of the complex variables (conventionally called z), a theta function has a property expressing its behavior with respect to the addition of a period of...

Pretty much the same as saying that it is not evaluated at all

Meanwhile, not sure how to derive the multivariable finite difference chain rule

skullpatrol

11:30

Is your program working now?

Secret

not quite, the 4 blocks are functioning properly, but still trying to get the bash component to work

Mats Granvik

Why is there change in the universe?

skullpatrol

Because without change nothing would change.

Mats Granvik

@skullpatrol Tautology.

skullpatrol

Sometimes that's what it comes down to.

You can't just keep asking "why" forever.

Mats Granvik

11:35

@skullpatrol So the highest form of truth is a tautology?

skullpatrol

Not "highest," but it's a starting point at least.

Secret

$\lim_{n\to \infty}\text{Why} \circ^{n}P = \text{undefined}$

skullpatrol

Perhaps later you may modify it.

Sorta like an assumption. Let's begin by assuming such and such is true...

(and not ask "why")

Mats Granvik

@skullpatrol Let's change the question. Is there change in the universe?

skullpatrol

Why is there "change?" Because without it we can't talk about it.

Leaky Nun

11:41

@Secret because it is only defined for finite numbers?

skullpatrol

Same answer @MatsGranvik Change must be assumed to exist first.

Akiva Weinberger

I feel like I should be able to decompose $\sin^2(x)$ as the sum of infinitely many bump-like functions

like c+$\sum_{n=-\infty}^\infty f(x+\pi n)$ where $f(x)$ roughly looks like $e^{-x^2}$

And by $\sin^2(x)$ I mean $\cos^2(x)$ I guess

Leaky Nun

@AkivaWeinberger the power series of the sum doesn't look good

Akiva Weinberger

I wonder if there's a way to derive what this $f$ function must be, under certain nicety conditions

skullpatrol

Do we measure change? @MatsGranvik

Secret

11:49

The peaks of $\sin^2 x$ are $2 \sin x \cos x = \sin 2x$

skullpatrol

The very definition of "measurement" is circular, but we have to start somewhere.

Secret

one can isolate the peaks by multiplying $\sin^2 x$ with the hat function $2 (x_{i+1}-x_{i})\Pi (\frac{x_{i+1}-x_i}{2})$ where $x_i$ are the roots of $\sin^ 2 x$

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$Mathematics$ Mathematics