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

Ahh, right

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

Like connected manifolds without boundary are homogeneous

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

Like, T^2 wedge S^2

Interesting, that was unexpected

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

9:08 AM

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

Yup

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

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

You should get T^2vS^1

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

9:10 AM

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

Merp. That

You're saying merp now

My influence grows

O_O nooOOOOOO

Big Smoke starts singing

Also hi good morning

Vrouvrou

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

9:12 AM

I think so yeah

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

Yo

Yeah I think so

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

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

9:14 AM

Indeed

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

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 $\pi_1(D^2 - \{p, q\})$.

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

Vrouvrou

@AkivaWeinberger thank you

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

Except 500% faster to write :P

9:16 AM

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

SteamyRoot

That smells like an almost-crystallographic group

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

lol

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

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

9:26 AM

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

This is why you don't read Herstein

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

@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)

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

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

9:28 AM

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

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

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

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

Even if the context isn't algebra

@Daminark GOOD idea

:P

NO GOD! PLEASE NO!!! NOOOOOOOOOO

9:29 AM

►

@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

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

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

(or maybe as $ba+$?)

$+ab$, rather.

$(a,b)+$

9:31 AM

@BalarkaSen Not when the functions are to the right

Ah yes

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)

RPN, innit

Reverse Polish Notation

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

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

9:32 AM

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

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

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

What

@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

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

9:34 AM

Fair

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

?

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

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

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

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

9:37 AM

Uhhhh

Maybe you're right.

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

Yep

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

Ok, that's the same as my group

Yeah actually the commutator is the Pochhammer curve.

I'm dumb

9:39 AM

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

Right, which is what Alessandro did

hi, i'm the code's classmate

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

Hi @mago

Well we gotta make sure you're not an imposter

What's the code word?

Is "Code" nickname for Alessandro?

Codenotti?

9:43 AM

It's derived from Codenotti I think

yup, read his surname and shrink

Cod.

Co?

How high is the English-speaking rate in Italy

I need to wake up a bit more today.

Preliminary Google says 1 in 3

C

9:46 AM

@AkivaWeinberger how high of a level does that assume?

I dunno

@TobiasKildetoft About 90

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

(I don't know if that grammars right)

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

Sounded grammarful enough to me

9:47 AM

@AkivaWeinberger Yeah, we are generally fairly proficient in English

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

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

@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)

@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.

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

9:48 AM

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

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

got it, thanks guys

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

So he's the source of our miseries.

@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

9:53 AM

Top hats are hats that people in the northern hemisphere wear

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

@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

Anyway I should sleep now, good morning everyone!

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

@Daminark See you in your dreams!

9:56 AM

Night, @Daminark!

Cya

night @Daminark, thanks!

I should go back to sleep

Gnat

G'night

night

later

9:57 AM

@BalarkaSen yep, I agree

thanks @AkivaWeinberger too

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

I guess so?

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

I need to figure out how to wake up

10:02 AM

@BalarkaSen Use two alarm clocks.

Good luck with analysis

@WillHunting Doesn't help if your mind is asleep

@BalarkaSen don't sleep!

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

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?

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

I'm surprisingly ignorant concerning Italian cinema

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.

10:19 AM

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.

Hello!

@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

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

10:35 AM

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

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 $B$).

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

Hm, it does

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

Oh that's gonna be error function nonsense

Yep

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

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

Right

10:42 AM

Wolfram seems to suggest it's false, actually

Aw

It's really close, though.

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

weren't you 2 both supposed to be asleep?

Akiva was

It's 4PM here :P

Yeah I dunno it's morning now whatever

Sun's up

10:48 AM

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

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

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

'Bout three hours probably?

4.

@Alessandro Yeah our timezones are pretty close actually

No, 3 and a half

10:49 AM

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

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

Blargle

4:19 actually, but yep.

Check again?

20

lol

Blaise it

10:53 AM

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

@FuzzyPixelz What's the theorem

The binomial coefficient is always a natural number.

I'd induct on the row

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

I always am confused about induct and induce, lol.

The $n$

10:58 AM

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?)

And $k=0$ ?

It's also 1 right ?

Right

Oh did I mess up the equation

I might've

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

No worries I have it

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

Makes sense, thank you

11:13 AM

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

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

Q: Bounds on Gaussian infinite sum

2

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

11:30 AM

Is your program working now?

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

Why is there change in the universe?

Because without change nothing would change.

@skullpatrol Tautology.

Sometimes that's what it comes down to.

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

11:35 AM

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

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

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

Perhaps later you may modify it.

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

(and not ask "why")

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

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

Leaky Nun

11:41 AM

@Secret because it is only defined for finite numbers?

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

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

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

Do we measure change? @MatsGranvik

11:49 AM

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

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