Mathematics - 2017-12-25 (page 1 of 3)

Daminark

00:05

@BalarkaSen it's pure gold

Leaky Nun

00:28

@Daminark could I ask you category?

Daminark

I don't know much but do try?

Leaky Nun

do we have $\pi_1 \circ f = \pi_1 \circ g \land \pi_2 \circ f = \pi_2 \circ g \implies f=g$ for the diagram $\begin{array}{} &&&& A_1 \\ &&& \overset{\pi_1}\nearrow \\ X & \overset f{\underset g \rightrightarrows} & A_1 \times A_2 \\ &&& \underset{\pi_2}\searrow \\ &&&& A_2 \end{array}$?

@Daminark

Daminark

Oh Lord I'm on my phone and cannot read that rn

Leaky Nun

so A1×A2 has two projection maps, π1 to A1 and π2 to A2

I also have two maps from X to A1×A2, one called f and one called g

that's basically my diagram

and I'm wondering if π1∘f=π1∘g and π2∘f=π2∘g imply f=g

@Daminark do you get me?

Daminark

It should be true in, say, sets

Leaky Nun

00:40

sure

it's also true in Grp and Mod and Ring and ...

but is it by construction or by category theory?

Daminark

Actually yeah for a categorical product that's the whole universal property business. So I'm guessing by construction

Bob

I have a new windows machine which version of LaTex should I install?

Daminark

I use sharelatex

Leaky Nun

@Daminark so you're saying that it's true for all categories with products?

Bob

is sharelatex free?

Leaky Nun

00:45

@Bob I use TeXStudio

@Bob I don't think it is

Bob

if I download TexStudio is that all I need?

Leaky Nun

well sure

Bob

I tired installing MikTex but the install halted with the followng error:

MikTex Setup Wizard

Inside the box there is a red circle with an X. The box says: The operation could not be completed for the following reasons:

Permission denied:
path="C:\Users\rsherr\AppData\Local\Temp\mik66364\bidi.tar.lzma"

Details:
path="C:\Users\rsherr\AppData\Local\Temp\mik66364\bidi.tar.lzma"

any comments?

Leaky Nun

0

Q: Permission denied: path="C/program Files/MiKTeX 2.9\doc/bibitex/apalike/apalike.doc

While installing MiKTeX this error is coming: Permission denied: path="C/program Files/MiKTeX 2.9\doc/bibitex/apalike/apalike.doc My OS version is Windows 7 Ultimate and I downloaded complete MiKTeX files from http://miktex.org using 64bit net installer option. Then, during installation, i...

stfw

Bob

I saw that link but it does not say how to fix it

Leaky Nun

00:50

> First: a path for windows 64 bit has to look like c:\Program Files\MiKTeX 2.9\doc\bibtex\apalike\apalike.doc and not yours path="C/program Files/MiKTeX 2.9\doc/bibitex/apalike/apalike.doc. you missed : and mixed ` and /`.

@Daminark should I ask it on main?

Bob

the problem is that I did not set the path

could it be bad installation script?

Daminark

Maybe a good idea

Leaky Nun

@Bob I've no idea; @Daminark might be able to help?

@Daminark I'll ask it now then

Bob

@LeakyNun if you are asking for me, I thank you very much

Daminark

Sharelatex is free and it's online

Just go on sharelatex.com, create an account, and start typing

Leaky Nun

00:54

@Bob no, sorry, I'm asking an unrelated question

Bob

I would prefer something that runs on my machine

Daminark

Ah, well I'm not familiar with those, but chances are any one you find will be alright?

Bob

is there a good spot on any of the stack exchanges to ask about my problem? e.g. why I cannot install miktex

Daminark

And Leaky: do let me know how it goes. I'm not sure exactly which categories contain a product, but with those that do, the universal property should guarantee the uniqueness you seek

Leaky Nun

@Bob just the site I linked you to

Bob

00:59

I think I found the problem. It was my anti-virus software

thank you very much

Merry Chrismas

Good Night

Balarka Sen

I hate writing

robjohn

$\Huge\text{Happy Holidays}\unicode{x1F384}\unicode{x2603}\unicode{x1F54E}$

Leaky Nun

@BalarkaSen maybe you can help me?

Balarka Sen

If it's category theory I can't

Sorry

Leaky Nun

ok

Balarka Sen

01:08

Also I have to keep writing

Leaky Nun

@Daminark there is an answer

1

Q: Morphisms to categorical product completely determined by composition with canonical projections?

Consider the diagram: $$\begin{array}{} &&&& A_1 \\ &&& \overset{\pi_1}\nearrow \\ X & \overset f{\underset g \rightrightarrows} & A_1 \times A_2 \\ &&& \underset{\pi_2}\searrow \\ &&&& A_2 \end{array}$$ where $X$ is any arbitrary object, $(A_1 \times A_2, \pi_1, \pi_2)$ is the category product o...

category-theory

Daminark

@BalarkaSen do I need to repeat my apparent quote?

Balarka Sen

what is your apparent quote

Daminark

""Category theory is an island..."-Amin"-Mike

Balarka Sen

Ohhhhh

Daminark

01:11

@Leaky alright so that works but the question is do you always have products

Balarka Sen

That was golden

Daminark

That's what I don't know

:t h o n k:

Balarka Sen

what about category of fields

sounds like a good candidate to me

Mike Miller

I guess yeah because there's a forgetful functor to Ring

Balarka Sen

Oh hm

Leaky Nun

01:14

@Daminark of course you don’t always have products

Daminark

:blobhyperthinkfast:

Balarka Sen

2

Omae wa mou thonkeru

NANI??

Daminark

Manifolds with boundary won't have a product

Leaky Nun

it seems, in a rather useless manner, that in Sets, $A \times B$ is the categorical coproduct of as many copies of $A$ as $|B|$.

Balarka Sen

@Daminark Well

Smooth manifolds with boundary

I think

'Cuz corners are a smooth phenomenon

Daminark

01:20

Non-smooth manifolds are fake news

Balarka Sen

Kervaire manifold

In mathematics, specifically in differential topology, a Kervaire manifold K4n+2 is a piecewise-linear manifold of dimension 4n+2 constructed by Kervaire (1960) by plumbing together the tangent bundles of two 2n+1-spheres, and then gluing a ball to the result. In 10 dimensions this gives a piecewise-linear manifold with no smooth structure. == See also == Exotic sphere == References == Kervaire, M. (1960), "A manifold which does not admit any differentiable structure", Comm. Math. Helv., 34: 257–270, doi:10.1007/BF02565940, MR 0139172 Shtan'ko, M.A. (2001) [1994], "Kervaire invariant",...

thomas the engine tune plays

DarkRunner

Hey folks

Daminark

Wikipedia is fake news media

Balarka Sen

You're right

Alex Jones is where the true news is

Adeek

I was attending once a physics talk

somebody had a topological space named fock space

haha

Balarka Sen

01:24

well you know what they say

fuck fock

Leaky Nun

"fock" is just "fuck" in a British accent

Narcissus jewel

@LeakyNun I think you dualized it friend.

Balarka Sen

oi thwats focking roight m8

Was British accent always this destabilized? It seems it has gotten thicker and swirlier after the 80's

What happened?

Daminark

I think the northeastern accent in America is closer to how English was spoken in the 1700s than the current British accent

Balarka Sen

Trying to figure out where north-east is in America

Daminark

01:35

New York and Canada basically

Balarka Sen

Hm

Oh yeah Canada speaks decently

Daminark

Oh no I was joking Canada is its own thing

Balarka Sen

I was confused how Canada was north

4/10 prank

Now I have to figure out the proof of the Shadowing lemma

This doesn't look good

Owh. Owww. This doesn't look good at all

Daminark

Which is that?

Oh wait

Is it this?

Balarka Sen

That sounds like it

Daminark

01:47

It's not that bad to prove if you have the monster theorem

Balarka Sen

Is it in your notes?

Daminark

This fucker

Yeah it's there. Also in Brin & Stuck, which may have a less shit explanation than I do

Balarka Sen

Ah, the $\delta$-conjugacy being $\epsilon$-shadowed thing

Thanks, I shall have to check this out

I've been recommended to read the proof by Bowen because Katok-Hasselblatt does it in such severe generality

Daminark

So if you're willing to blackbox this theorem 3

This gives you an open set containing $\Lambda$. Choose some $\delta$ such that $\Lambda_{\delta} \subset O$

Then you want to extend this $x_k$ to a $\delta$-orbit which is infinite on both sides

Balarka Sen

Well, need it be?

Daminark

01:51

A priori no, it can be a finite or one-sided orbit that you start with

Balarka Sen

I was thinking of like the Anosov closing lemma, which says every closed $\delta$-orbit can be shadowed by a periodic orbit

This is a kinda generalization of that I guess

Daminark

But that's easy, wherever the sequence terminates (on either side), choose another point close to the last one and just extend by $f(x_k) = x_{k+1}$, then $f(f(x_k))$, and so forth.

Balarka Sen

Well that doesn't give you a periodic orbit necessarily

Daminark

Not yet

But watch

So you have a delta-orbit, yeah?

And it's doubly infinite

Balarka Sen

Ya

Daminark

01:55

Okay, so in theorem 3 (using my notation)

You let $X = \mathbb{Z}$, $g= f$, $h(k) = k+1$

And $\phi(k) = x_k$

Balarka Sen

oic

Daminark

Because we have a $\delta$-orbit, $d(\phi(h(k)), f(\phi(k))) = d(x_{k+1},f(x_k)) < \delta$

Balarka Sen

Right, so it's a $\delta$-conjugacy between $h$ and $f$

Daminark

Exactly

So you can find some $\psi$ such that $d(\psi(x_k),x_k) < \epsilon$

And $f(\psi(x_k)) = \psi(x_{k+1})$

But then just let your sequence be $(\ldots, \psi(x_{k-1}), \psi(x_k), \psi(x_{k+1}),\ldots)$

(This is why I stretched the sequence out on each side)

Makes sense?

That's just the sequence which shadows your delta-orbit

Balarka Sen

So this is like the boss theorem

Daminark

01:59

This = theorem 3?

Balarka Sen

Yep

Daminark

Oh yeah that theorem is a fuck

Balarka Sen

There should be geometric proofs of this thing

Using the foliation explicitly

Daminark

Like basically the proof amounts to setting up this fixed point problem and then pushing really really fucking hard until you can show that something else is a contraction mapping

Balarka Sen

That's how you prove Anosov closing too

Leaky Nun

02:01

poll: what's your favourite fixed-point theorem?

Balarka Sen

Balarka's fixed point theorem

which says if it's the identity map, every point is a fixed point, and vice versa

Daminark

I only know two: Banach and Brouwer

Balarka Sen

all our names start with B

there's a connection, see

Leaky Nun

conspiracy intensifies

Daminark

@Balarka so I dunno about Foliations, I gave a severely abridged proof of the theorem (Brin and Stuck is also quite terse about it but it's 2 and a half pages there instead of a page in my thing, so maybe that'll be easier to understand)

I'm somewhat skeptical of a way of proving it which isn't pure fuckshit

Though I dunno, people are clever

Balarka Sen

02:05

@Daminark I don't see why this should imply Anosov closing though. Say you have closed $\delta$ orbit i.e. a sequence of points $x_0, x_2, \cdots, x_n$ i.e., $d(x_i, x_{i+1}) < \delta$ (index is modulo n+1). You can $\epsilon$-shadow it by an actual orbit, but why is that periodic son

Daminark

What I proved was shadowing. I could see you doing a similar argument for closing (based on what you just said) by swapping $X = \mathbb{Z}$ in the above with $X = \mathbb{Z}/n\mathbb{Z}$?

Balarka Sen

Hm

That's probably right

But there should be a more direct way to see this than pure algebra

Daminark

Perhaps

Balarka Sen

@Daminark When I say foliations, I of course mean in the case of Anosov diffeomorphisms

There's no foliation for a generic hyperbolic set

That's like a really bad subset of your manifold

You just have a stable-unstable decomposition of the tangent spaces over it

It's fucked

Daminark

I just mean that I don't know anything about foliations

That's my operating definition for a foliation but that definition just makes me say "derp"

Balarka Sen

02:11

the definition of a foliation makes no sense on face value

Daminark

;novathonk

Balarka Sen

you don't get it until you see one

Daminark

Probably can confirm because I just saw that and was like um... so I only use foliations really briefly and not in a way that failure to understand it is a significant conceptual roadblock so I'll just leave it at that

Like I vaguely have it in mind as "Let's cut up a manifold into sub manifolds in a way which isn't completely stupid"

Buuuut... Yea

Balarka Sen

You know

This might follow from the thin waist theorem

The biinfinite shadowing => Anosov closing thing

Hmmmm

Suppose you get your biinfinite shadowing by $f^m(x)$, $m \in \Bbb Z$

Then, starting at $x$, you can find $n$ such that $d(f^n x, x)$ is small.

Because you're shadowing a $\delta$-periodic sequence starting at $y$, something close to $x$

Er

I want to say infinitely many of your biinfinite shadowing sequence hits an $\epsilon$-neighborhood of $x$

Because the shadowing is an $\epsilon$-shadowing

Daminark

02:28

Uh... are you sure?

Oh wait hmm

Balarka Sen

I am not sure but it feels like it

Daminark

So wait you're starting with some $\delta$-periodic sequence

Balarka Sen

Right

Daminark

And then you're $\epsilon$-shadowing it, so yeah, infinitely many points of the shadowing sequence are within $\delta + \epsilon$ of $x$

Balarka Sen

Right, OK

But those points form a sequence inside the closure of the delta+epsilon ball around x

So it converges somewhere

and that somewhere has to be $y$

So the points which are in the shadowing sequence and withing $\delta + \epsilon$ of $x$ are all of the form $g^i(x)$ where $g = f^N$, where $N$ is the length of the $\delta$-periodic sequence.

They loop around $N$ times and hit $B_{\delta + \epsilon}(x)$ each time

Daminark

02:38

Yeah

Leaky Nun

03:01

I have to say, I have no knowledge of category theory, but I am very impressed by and jealous of that LaTeXing. — TreFox 1 min ago

lol

Balarka Sen

@Daminark OK. This about this scenario instead. $x, f(x), f^2(x), \cdots, f^N(x)$ is a $\delta + \epsilon$-periodic orbit. It's an orbit everywhere except when you go from $f^N(x)$ to $x$, and there's a deviation of $\delta + \epsilon$ there. So set up $x_k = f^{k \pmod{N+1}}(x)$ and $\epsilon'$-shadow THIS dude. You get another $x'$ in the game whose orbit shadows this fellow, i.e., $d(x_i, f^i(x')) < \epsilon'$

Akiva Weinberger

math.mit.edu/~dspivak/files/metric_realization.pdf

I have no idea what that paper is talking about

Balarka Sen

Observe: $d(f^i(x'), f^{i+N+1}(x')) < d(f^i(x), x_i) + d(x_i, f^{i+N+1}(x'))$. The first term is bounded by $\epsilon$. The second term is also bounded by $\epsilon$ because $x_i = x_{i+N+1}$.

Akiva Weinberger

But I feel like one of you guys might

So, if it turns out to be good: enjoy

Balarka Sen

So $d(f^i(x'), f^i(f^{N+1}(x')) < 2\epsilon'$ for all $i$

You can of course also choose $\epsilon'$ to be as small as possible, replacing $x'$ by $x''$ if necessary

Daminark

03:04

Good Lord this looks aggressive

Balarka Sen

By the hypothesis of shadowing arbitrary well

Daminark

Fuzzy sets

Balarka Sen

@Daminark This would be much easier if I had pen and paper

to explain the picture

But the general lemma after this is if $V$ is a sufficiently small neighborhood of the hyperbolic set $\Lambda$, there's an $\alpha$ such that if two orbits $O_x$ and $O_y$ are $\alpha$-close (i.e., $d(f^i(x), f^j(y)) < \alpha$) then $x = y$ ie the orbits are the same.

i think it's known as expansiveness or something

This immediately implies $f^{N+1}(x') = x'$ up above, which closes our sequence off

Daminark

I see

Balarka Sen

The lemma is interesting and closely related to the thin waist lemma I was speaking of earlier

You are familiar with $\Bbb H^2$?

How the geodesics look like there?

Daminark

03:19

Not quite

Balarka Sen

Poincare disk model is easier to explain. So you have an open disk $D^2$

Daminark

I do? What a Christmas present

Balarka Sen

lmao

It comes with a metric!

But let us not speak of that horrid metric (it's not that horrid, but whatever)

Let us watch the geodesics!

The geodesics of this model are exactly the semicircles perpendicular to the boundary of this disk (in degenerate cases those are diameters)

$D, D_1, D_2$ are all geodesics there

Daminark

Is that unique? Like, given two points, is there a unique geodesic connecting them?

Balarka Sen

Oh, given any two points, yes.

That is true.

It's a nice exercise in Euclidean geometry

Daminark

03:25

Actually wait a sec, so given two points not on the boundary, a geodesic is gonna be some section along a geodesic of boundary points containing them, right?

Xander Henderson

I want to speak of the horrid metric!

Can we, please?

Daminark

So in that case I'm actually skeptical

Lol, do go for it

Balarka Sen

I am not sure what "section along a geodesic of boundary points containing them means"

The geodesic is simply a hemicircle connecting the two points, perpendicular to the boundary

Daminark

Perpendicular when you keep extending the hemicircle until it hits the boundary of the disk, right?

Balarka Sen

That's right.

Daminark

03:28

Okay

Xander Henderson

I'm not sure that "hemicircle" is quite the right word here; "circular arc" may be better?

Balarka Sen

Yeah

Xander Henderson

or "hyperbolic geodesic"? :P

Balarka Sen

Tautological, but that works :p

Xander Henderson

all the best tautologies are tautological

Balarka Sen

03:30

@Daminark You can either do the geometry or use ... differential geometry of surfaces to prove it! Your nightmare mwahaha

Daminark

:GWChadThonkery:

Xander Henderson

hah

Daminark

@BalarkaSen or I could... Not prove it :P

And go back to quasifibrations

Balarka Sen

What this model fails to have is uniqueness of a parallel passing through a given point

link

Xander Henderson

But that is an advantage of the model, no? Or, at least, motivation for its creation...

Balarka Sen

03:32

Yes, it proves that the fifth axiom of Euclid is independent of the other four.

Xander Henderson

exactly

Daminark

I see

Balarka Sen

@Daminark Notice how any two geodesics diverge off exponentially from each other

If $\gamma$ and $\gamma'$ were two hyperbolic geodesics such that $\|\gamma - \gamma\| < \alpha$ for ANY given $\alpha$, then $\gamma = \gamma'$

27 mins ago, by Balarka Sen

But the general lemma after this is if $V$ is a sufficiently small neighborhood of the hyperbolic set $\Lambda$, there's an $\alpha$ such that if two orbits $O_x$ and $O_y$ are $\alpha$-close (i.e., $d(f^i(x), f^j(y)) < \alpha$) then $x = y$ ie the orbits are the same.

This ^ is an analogue of that for hyperbolic orbits/orbits of a hyperbolic dynamics

Although a much looser form. It just says there is some $\alpha$ such that...

(Also I meant $d(f^i(x), f^i(y)) < \alpha$. No $j$ there)

Also notice that if $\gamma$, $\gamma'$ are two hyperbolic geodesics usually the ends diverge off from each other and the middle bit is the closest. That's the analogue of the thin waist lemma I was speaking of for orbits

There's a lot of close parallels

No pun intended

Xander Henderson

ouch...

Daminark

Kek

Balarka Sen

03:41

Well, I guess thin waists are not that much of an attraction point in 2017 anymore. The taste has, er, um, shifted.

Daminark

Kek

Balarka Sen

Unfortunately, in hyperbolic geometry, that's the best you're gonna get. Savor it!

Daminark

No that's the best you're gonna get, I'm gonna get long exact sequences

Balarka Sen

such depravity.

absolutely vulgar nonsense

Daminark

Lol Xander you're probably just watching our bickering like "Erm..."

Typhon

03:46

@AkivaWeinberger all piecewise constant constant function can be written as h(floor(g(x))). I'm limiting myself to things of a closed form so that my algorithm can say "replace _ with a constant". Technically I could say any expression of the specified sort.

in fact cosine is redundant as it can be written as a function of sine

i believe proving it relies on using the left and right hand derivatives somehow

ill think on it more tonight

perhaps it will work out as a surprise christmas gift.

that would awesome considering Ive been trying to prove it works for 2.5 years.

Leaky Nun

how do you remember that monomorphism is g1of=g2of implies g1=g2 or fog1=fog2 implies g1=g2?

@Daminark

Daminark

I mean I just think about mono as injection and then epi as surjection

Balarka Sen

Not true in general categories, dr. algebraist

Daminark

I know but like, the analog

Balarka Sen

I never think about mono/epis though

Daminark

03:52

Because it's easier to remember that and just say okay, which side of cancellation works for injection? Okay that's the mono one

Balarka Sen

Oh as an intuition

I see

Daminark

Yup

Leaky Nun

but which side is it?

oh wait, it actually tells us that injectivity and surjectivity are dual concepts.. can’t believe this gem is hidden there so long and i didn’t find it

Balarka Sen

OH WAIT. tHe rOcK is DuAl tO ThE tReE!!111

2

Leaky Nun

:p

i do have an obsession with duality

Balarka Sen

03:59

I think it's okay. Categorists are too easily excited by it

Daminark

Duality sounds dank enough, much as I haven't quite gotten deep enough into things to appreciate it beyond the fact that concepts connect

Leaky Nun

so im(phi) where phi:X->Y is B such that X->B epi B->Y mono and triangle commute?

is this a pushout/pullback of some sort?

Daminark

I don't know much about pullbacks in general but... I'm not wit getting such a vibe here

Leaky Nun

04:22

nlab confirms my epi-mono factorization

Daminark

I see

BEHOLD

I am wearing a hat

@Balarka b e h o l d

Balarka Sen

2

Daminark rn

Santathonk

Leaky Nun

nlab even confirms the pushout

Adeek

04:48

@BalarkaSen ....

@BalarkaSen .....

:D

Daminark

04:59

Hey @Adeek, @Faust, and @MeowMix!

Meow Mix

hi

Daminark

How've you been?

Meow Mix

ive been being

Daminark

:blobhyperthinkfast:

Adeek

05:15

hi @Daminark

Daminark

How's it going?

Adeek

good just trying to understand few thigngs

Daminark

Nice, which things?

Leaky Nun

inb4 this thing, that thing, and some other things

Daminark

05:35

Seems like you predicted exactly what Adeek was gonna say and he was just like "Aight peace out

Typhon

07:48

MERRY CHRISTMAS

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$Mathematics$ Mathematics