Mathematics - 2020-08-04 (page 1 of 2)

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12:36 AM

If I have a faithfully flat ring homomorphism $R \to R^\prime$ and a flat $R$-module $M$, why is $M \otimes_R R^\prime$ also flat? Can I just hit an ES $N_1 \to N_2 \to N_3$ of $R^\prime$-modules with $(M \otimes_R R^\prime) \otimes_{R^\prime} -$ and use that $M$ is flat to prove that the resulting sequence is also an exact sequence of $R^\prime$-modules?

If $f\colon R\rightarrow R^{\prime}$ is the faithfully flat hom, we have $f_{\ast}(f^{\ast}M\otimes_{R^{\prime}}N_i)=M\otimes_Rf_{\ast}N_i$ (hooray, projection formula) and exactness of the sequence is independent of whether we look at them as $R$- or $R^{\prime}$-modules, but the sequence $0\rightarrow M\otimes_Rf_{\ast}N_1\rightarrow...\rightarrow0$ is exact, because a) $M$ is flat over $R$ and b) $0\rightarrow f_{\ast}N_1\rightarrow...\rightarrow0$ is exact, because $f$ is faithfully flat.

3 hours later…

AbstractAlgebraLearner

4:05 AM

You know how there's no rational function that gives only prime numbers?

4:19 AM

Let $f: R \to R$ be a continuous function and $f(x) = f(2x)$ is true $\forall x \in \mathbb R$. If $f(1) = 3$, then the value of $\displaystyle \int_{-1}^1 f(f(f(x)))\mathrm{dx} $ is ?
A) $3f(0)$
B) $0$
C) $3f(3)$
D) $6$

can anybody help me on this?

@robjohn hi sir

4:48 AM

hey chat.

I'm trying to construct an artificial extension to a field and, long story made short, I want to prove that $f \in F \implies f \not \in \{ F, \{s, F\}\}$ for any $s$

that's just plain set theory, and foundation should suffice, but I can't see how foundation implies what I want.

oops, $f \ne \{F, \{s, F\}\}$

a friend told me that $F \in f \in F$ is a contradiction.

5:24 AM

you know, things like this is why I like Lean's type theory much more

you don't ever need to worry if two types accidentally share some elements

(types are the analogue to sets)

@LucasHenrique if $f \in \{F, \{s, F\}\}$ then either $f = F$ or $f = \{s, F\}$, and in the first case we have $F \in F$ (contradiction) and in the second case we have $F \in f \in F$ (contradiction)

$F \in f \in F$ is a contradiction because of the axiom of foundation

why is $F \in f \in F$ a contradiction?

@LeakyNun I've tried that but...

consider the set $\{f, F\}$

by AF, it contains an element that intersects trivially with itself

i.e. either $f \cap \{f, F\}$ is trivial or $F \cap \{f, F\}$ is trivial

the axiom of foundation would be a lot nicer if you could say that every element of your set is disjoint from it

but the former contains $F$ and the latter contains $f$

oh, that's right

thanks, @Leaky :)

5:28 AM

@LucasHenrique that is false: consider $\{\varnothing, \{\varnothing\}\}$

its element $\{\varnothing\}$ is not disjoint from it

uhh... yeah

lol...

my assumption was stupid, jeez. any von Newmann natural $n$ has exactly $n-1$ counterexamples :p

5:41 AM

if a manifold is locally euclidean, what object is locally semi-riemannian?

I think it's a Finsler space

5:56 AM

\o @Lelouch

6:27 AM

Let $f: R \to R$ be a continuous function and $f(x) = f(2x)$ is true $\forall x \in \mathbb R$. If $f(1) = 3$, then the value of $\displaystyle \int_{-1}^1 f(f(f(x)))\mathrm{dx} $ is ?
A) $3f(0)$
B) $0$
C) $3f(3)$
D) $6$

Can any body help on this ?

@BalarkaSen yo Balarka

What's new

Also good morning

Bott and Tu exercises are harder than I thought

I am trying to compute the compact supported cohomology of R^2 - {0,1} bare hand

yeah good morning

Hm I don't remember the exercises in B-T, good book though

basicallly I'm stuck on showing that $g dx dy$ is exact iff $\int_{\mathbb{R}^2} g = 0$

one direction is trivial, and other direction is probably trivial as well, basically I need to find functions $h_1, h_2$ such that $h_1 - h_2 = g$ and $\int_{x} h_1(x,y) = 0$, $\int_{y} h_2(x,y) = 0$

I was trying random stuff idk but couldn't find suitable $h_1, h_2$

@BalarkaSen are you attending MM's hyperbolic geo lectures ?

6:39 AM

Yeah how did you know

I mean I'm attending, that's why I was asking

We get a mail for zoom seminars

Ah OK great

any idea about the prereqs that's expected ?

Nope, no clue.

Hopefully he'll begin with something basic, like classical geometry on $\Bbb H^2$.

WeavingBird1917

“the limit of f(x), as x approaches a, equals L” means we can make the values of f(x) arbitrarily close to L by restricting x to be sufficiently close to a but not equal to a.

What if sufficiently close is like |x - a| = 10?

6:47 AM

@WeavingBird1917 yeah, it can be, depending on what you mean by "sufficiently close"

WeavingBird1917

It's the formal definition of a limit I found in a few texts, but it seems to work on things that aren't actually limits, since it depends on what you define as sufficiently close.

I don't get you ?

wdym by "but it seems to work on things that aren't actually limits"

WeavingBird1917

For example, there might be a graph of f(x) = {2 x<=0, 4 0<x<=4, (x-4) x>4} and and f(x) will get arbitrarily close to 3 if you define sufficiently close as 10.

ok, suppose I want f(x) to be less than .0000000001 distance from 3

what's your counter distance

WeavingBird1917

Then x would be close to 7, and |x - 0| = |7 - 0| < 10, which might be sufficiently close?

6:52 AM

Well yeah, because f is continious at 7 and f(7) = 3

I don't understand your objection

WeavingBird1917

Yea, but then based on that definition, we can say a limit exists at 0, when it doesn't.

Since I am saying a = 0.

do you mean something like this: "Let f(x) = 0, if x < 0, and 1 if x >= 0. Then if we define "arbitrarily close" by less than or equal to 1.5, then the limit exists at 0" ?

If that's the case, then well the limit don't exists because 1.5 isn't arbitrary -- you can't replace 1.5 by, say, 0.0001

WeavingBird1917

I'm trying to understand the formal definition of the limit, but it seems that if you can define "sufficiently close" as whatever value, then you could say that a limit exists when it actually doesn't.

@Lelouch Hm, try this: $\Bbb R^2 - \{0, 1\}$ can be covered by two charts, call them $U$ and $V$; choose a partition of unity $\rho_U, \rho_V$ subordinate to this cover. Pick a compactly supported function $h$ on $U \cap V$ such that $\int_{\Bbb R^2} h = \int_{\Bbb R^2} \rho_V g$, then $h - \rho_V g$ integrates to $0$ over $V$, so $(h - \rho_V g)dxdy$ is exact by Poincare lemma.

WeavingBird1917

As in the function above, it can get arbitrarily close to 3 since we can to go from the left and right side of x = 7. As in x = 6.999 and 7.0001, which might be sufficiently close to a = 0 depending on what is defined as arbitrarily close.

6:57 AM

@BalarkaSen uhh I know, I can do it using Poincare Lemma. The entire point was doing it bare hand using just the definition

How do you show that $H^2_{comp}(\mathbb{R}^2) = \mathbb{R}$ ?

I have reduced the prpoblem to this

By Poincare lemma my dude.

are fuck it's in section 1

poincare lemma is in section 34

*section 4

I don't know, just use whatever you know.

Also my argument actually generalizes to show $H_c^n(M; \Bbb R) = \Bbb R$ for any noncompact orientable $n$-manifold $M$

Basically how do you show that, if $\int_{\mathbb{R}^2} g(x,y) = 0$, then you can find $h_1, h_2$ such that $h_1 - h_2 = g$ and $\int_{x} h_1(x,y) = 0$, $\int_{y} h_2(x,y) = 0$ for all $x,y$

I don't know but good luck.

I am not going to do calculus

7:00 AM

lol ok.

i'm probably going to leave it for now, spend way too much time on this

@Lelouch My point is this is exactly the statement of Poincare lemma for $\Bbb R^2$.

A compactly supported $n$-form which integrates to $0$ over $\Bbb R^n$ is exact

This is doable by hands of course

Yes, but that's essentially doingn the proof for the $n = 2$ case. Probably that's the intended way, I can't think of any other

I mean the proof is simple, just realize that this gives you an $n$-form on $S^n$ which integrates to $0$ on $S^n$

Therefore is exact on $S^n$ (Stokes'). It's obviously $0$ in a neighborhood of infinity

So you can adjust the antiderivative to be $0$ in a neighborhood of infinity as well -- this is the trick part.

Bott-Tu's proof is coming down by a factor of $\times \Bbb R$ every time, which is an integration over fibers trick -- but just $\Bbb R^n$ is much simpler.

@Lelouch Do you know how to get a compactly supported antiderivative

@BalarkaSen Probably smooth using a partition of unity supported in a neighbour of infinity ?

Which antiderivative are we talking about here ?

"smooth"?

I mean let $\omega$ be the compactly supported form on $\Bbb R^n$, that gives a form on $S^n$ by one-point compactification and $\int_{S^n} \omega = 0$ as $\int_{\Bbb R^n} \omega = 0$. Find a compactly supported $(n-1)$-form $\alpha$ on $\Bbb R^n$ such that $\omega = d\alpha$.

7:11 AM

sorry, I mean if the original function is $f$, consider $f - \rho g$ where $g = f$ in a neighbourhood of infinity and $\rho$ is supported in a neighbourhood of inifinity, and the integral is zero

Huh?

What function man

nevermind, I thought you were asking something else (about the integral formulation I stated before)

Forget that formulation, that's some nasty symbolic garbage

@BalarkaSen Yes, this I know, I have seen it in Donaldson

OK.

So that's pretty bare hands to me. Bare hands doesn't mean mess with a weird linear PDE!

7:15 AM

basically you let $f(x) = \int_{t} g(x,t) dx$, and then rerplace $g(x,y)$ with $g(x,y) - \rho(y) f(x)$

and iterate once and you're done

Yeah you had the right idea but why do you write in terms of functions, that's horrible.

So how do you think of the exactness condition without thinking the differential equation it satisfies ?

Here's how I would write it: $\int_{S^n} \omega = 0$ means $\omega = d\beta$ for some $(n-1)$-form $\beta$ on $S^n$. On a neighborhood of infinity, $\beta = d\gamma$ for some $(n-2)$-form on that neighborhood, bump it up to $\rho \gamma$ where support of $\gamma$ contains that nbhd at infinity and look at $\beta - d(\rho \gamma)$. Call this $\alpha$; this is zero on a closed nbhd at infinity so is compactly supported as an $(n-1)$-form on $\Bbb R^n$ and $\omega = d\alpha$

Isn't exactness basically satisfying a bunch of linear differential equations?

@BalarkaSen Thanks. I should probably practice thinking in terms of differential operators instead of the differential equations explicitly

Actually I don't think in terms of forms so I am not the right person to ask for forms intuition

7:21 AM

@BalarkaSen Maybe I'm mistaken, but how do you show $\int_{M} \omega = 0$ implies $\omega$ is exact without assuminig Poincare's lemma ?

The only proof of Poincare for $n = 2$ that I know is using the integrals explicitly as I mentioned

We're not using compactly supported Poincare's lemma here, only Poincare's lemma.

That $\int_{M} : \omega \mapsto \mathbb{R}$ is an isomorphism ?

ok nevermind

sorry

Yeah just for $S^n$ even, so you can prove that explicitly.

Bott-Tu does deRham cohomology and compactly supported de Rham cohomology separately right?

I don't remember

@Lelouch I guess you get something super explicit; write the form $\omega = (\int_{x_0}^{x} g(t, y) dt ) \, dy$ so $d\omega = gdx \wedge dy$; this gives the normal Poincare lemma. Then do your trick to get the compact supported version, assuming $\int_{\Bbb R^2} g = 0$.

7:40 AM

@BalarkaSen Yes, they do it separately. But they put in the exercise after defining the two deRham cohomologies.

Hm OK

@BalarkaSen That's the sort of anwer I was looking for. Thanks again !

Hi @TedShifrin

@Lelouch I keep asking people to learn Borel-Moore homology but they ignore me instead.

We should sit down and learn it someday; all of this compactly supported junk gets clearer in that setup I think

what do you call the parts of $\Bbb R^n$. For example, $\Bbb R^2$ has "quadrants"

for R^3 they are called octants, and by extension in higher dimension as well

$2^n$ants

lol

7:53 AM

okay $2^n-$ants

oops there goes another 2^n ants

lol

cuz hes got high dimensions hes got hiiiigh dimensions

Today it's from 3 PM right @BalarkaSen

3 to 5 right?

yeah

7:56 AM

Wait, I thought it's of 1 hour

yeah nvm, 3 to 5

i'll be happy if I can follow the first half hour

yeah i dunno lets see

i thought this was a tifr course, how come cmi gets notified for this

user image

@BalarkaSen ^^

Ah

Pity that ISI doesn't get any of this shit

is the ISI mail pretty much defunkt ?

I mailed a bunch of ISI profs for summer internship back in Feb, still didn't get any reply

lmfao

No we use it, that's bizarre

whom did you email

8:00 AM

are you getting any spam mails lately ?

@BalarkaSen Sury

Not lately but a couple weeks back yeah

so ended up doing with an IISc prof

There were multiple people from Large International Company asking if we wanted to do interns

@BalarkaSen no it's not like that

Ah ok Sury is great

I thought he checks emails

8:01 AM

so basically, you get a mail saying subject as Prof X

and the mail will say that he needs money

you reply you fucked up

LOL no

we didn't get that

probably because it's easier to stalk CMI people than ISI people

everyone got a personalized spam mail, on the positive side

we have a stupid squirrelmail that sucks

well you don't forward it to gmail ?

we use roundcube

not by default; you have to ask them for linking it to your gmail account

i dont care enough to

8:04 AM

them as in whom ?

as in, the tech dept

huh lol

CMI is probably overall run way better than ISI Bang

i hear the campus sucks

maybe, but who cares about admin. stuff

@Lelouch is it just a wild rumor that CMI server goes down whenever the Cooum river floods lol

8:08 AM

@BalarkaSen I don't know, I was not here when the last flood happened.

hmk gotcha

WeavingBird1917

8:42 AM

f(x) approaches the limit L at a, if we can make f(x) arbitrarily close to L by requiring x be sufficiently close to, but unequal to a.

I keep reading the definition of the limit like this: For every ε > 0, there exists x and δ > 0 such that if 0 < |x - a| < δ then |f(x) - L| < ε

The actual definition is: For every ε > 0, there exists δ > 0 such that for all x, if 0 < |x - a| < δ then |f(x) - L| < ε, but I don't get how the formal worded definition means this. Been staring at it for probably 2 hours now.

9:06 AM

To parse that second definition slightly, "for every ε > 0, there exists δ > 0 such that anything within δ of a (not including a) gets taken by f to within ε of L"

The reason the first is invalid is that it says "there exists x"---that's not nearly enough. You need everything in some radius around a to be mapped within some tolerance around L.

WeavingBird1917

Thanks, I think I understand it better after looking at the definition for limits as x approaches infinity: f(x) gets arbitrarily close to L whenever x is sufficiently large. I was thinking of x as a specific value, rather than a range.

-1

Q: Each "octant" of $\Bbb S^3$ is diffeomorphic to $\Bbb R^3.$

I'm thinking of a space $\Bbb S^2$ in which each "quadrant" of $\Bbb S^2$ is diffeomorphic to $\Bbb R^2,$ and a space $\Bbb S^3$ in which each "octant" of $\Bbb S^3$ is diffeomorphic to $\Bbb R^3.$ To clarify, the "quadrants" and "octants," are disjoint submanifolds whose union's closure is the w...

geometry reference-request

can anyone speak to my extensive edit?

WeavingBird1917

"whenever" made more sense to me than "by requiring", the latter still confuses me.

Epsilon-delta is a subtle and weird thing. I think pretty much everyone has a slightly different internal language for it.

Fun little aside: the open sets containing all x where |x| > M (for some M) are called "neighborhoods of infinity", for the same reason that an interval (a - ε, a + ε) is called a neighborhood of a.

9:23 AM

Why are complex numbers so important in electronics

AC circuits alternate their currents and voltage in a sine-wave fashion. This means that you can represent these things as 2D vectors spinning around in a circular fashion (because a sine wave is just a coordinate of a circle). Complex numbers are great for this because they're naturally good at handling circles and rotation, because complex multiplication is just rotation and scaling.

To put it another way, using complex numbers means you don't constantly have to remember trigonometric sum and difference formulas for sines and cosines when doing calculations; the complex numbers handle all of that without you having to do any thinking.

10:01 AM

in Calvin's Chat Room, 55 mins ago, by Knight

If $f$ is a continuous function on $[a,b]$ such that $f(a) \lt f(b)$ then by Intermediate Value Theorem can I say “ $f$ assumes all the value between $f(a)$ and $f(b)$” ?

I know I’m not very rigorous in the above conclusion, but hopefully you know what I’m trying to say.

Yes. That's basically the exact content of IVT.

Thanks Fargle.

10:28 AM

@BalarkaSen What was the name of hte book he told ?

in the beginning

Brideson-Haefliger

and section 3.8 or chapters 3 to 8 ?

chapter III.H

oh ok

Hi @Alessandro

Alessandro Codenotti

10:29 AM

Hi

can you give an example of any group theoretic question which is made easier using the machinery of hyperbolic geometry ?

i dunno lol theres too many applications

give me a basic one

heres an easy one: any hyperbolic group has F_2 as a subgroup

what is F_2 ?

10:32 AM

free group on 2 generators, sorry

Alessandro Codenotti

Free group on two generators

@Alessandro are you still in vacation with shitty internet

Alessandro Codenotti

Except for $\Bbb Z$ which is the weird group in between hyperbolic and amenable

Yep

yeah i should have said nonelementary hyperbolic groups i.e., not virtually Z

or virtually trivial lmao

(virtually X means X upto finite index btw @Lelouch)

ok thanks, this looks interesting

10:34 AM

i say easy but i cant write proofs, it just seems kind of clear once you get used to the idea of hyperbolicity

i'll read B-H to learn to write proofs lol

What's is B-H lol ?

Brideson-Haefliger

You didn't read it yet ?

nah

seems he is back

10:36 AM

see you on the other side

Konformist Liberal

might be obvious, however couldn't prove: is every topological space coproduct of connected spaces (it's connected components)?

ok find a reference

found*

Not the coproduct in Top, no---$\Bbb Q \subset \Bbb R$ under subspace topology is totally disconnected, but this topology isn't discrete

Konformist Liberal

@Fargle thanks, best counterexample to keep in mind

11:15 AM

@BalarkaSen What was the theorem regarding fixed point of isometries ?

Fixed point set of Riemannian isometric involutions are totally geodesic submanifolds

it's a good exercise

you can bunk involution as well

@Konformist for what it's worth, this is true if the space is locally path-connected

every topological space is the disjoint union (as sets) of its connected components and the connected components are always closed subsets

but for the space to be the disjoint union (as topological spaces) of its connected components, you need the components to be open subsets as well

@BalarkaSen ok, I can see it's a submanifold but ig proving it's totally geodesic requires some theory

not really

draw the picture for an involution

cases in which this holds include when the space is locally path-connected or when there are only finitely many connected components (neither of which holds for $\mathbb{Q}$)

11:22 AM

if you have geodesics joining two points on your fixed point set which escapes the fixed point set involuting gives a distinct geodesic, roughly speaking -- but then suppose your endpoints were very close then this would be impossible because locally geodesics are unique

OK, I think I can prove it if I assume that for two points $P,Q$, and a vector $v \in T_p(P)$ there can be atmax one geodesic $\gamma$ such that $\gamma'(t)$ lies in the span of $v$. But is this true ?

you're assuming too much there. there is a unique geodesic through any point and any initial vector

thats just because of uniqueness of ODEs

@BalarkaSen there is always one, or there can be atmax one ?

existence and uniqueness of ODEs -- existence gives one such geodesic (arc)

uh, I have fixed both endpoints

11:26 AM

geodesics satisfy an ODE, those always have local solns

how is this true even in $\mathbb{R}^2$

@Lelouch But why?

this doesnt seem to be useful

sorry, nevermind. Isn't this actually quite simple: So basically let $\Psi$ be the involutive isometry, and let $P$ be a fixed point. For any $v \in T_p(P)$ (tangent space as considered wrt the fixed point submanifold), consider the geodesic at $P$ starting with initial direction $v$. Then, by uniqueness of geodesics wrt the submanifold, this is unique. So any geodesic $\gamma$ in the original manifold and same inital direction must coincide with this

i dont follow, why is the geod at P starting in the direction v contained in the submanifold?

$v$ lies in the tangent space to $P$ at that submanifold

11:35 AM

geodesics in the submanifold with the induced Riemannian metric can be wildly different from the ambient geodesics so you must produce some argument

@Lelouch Ok, and then?

So, let $\gamma$ be the geodesic in with the same direction wrt the original manifold. Then $(\Psi \circ \gamma)'(0) = \gamma'(0) = v$, so $\Psi \circ \gamma = \gamma$ by uniquness applied to the original manifold

and thus $\gamma$ must lie in the fixed point submanifold

that's correct, so $T_p P$ is $d\Psi_p$ invariant is what you're using

yeah

17 mins ago, by Balarka Sen

if you have geodesics joining two points on your fixed point set which escapes the fixed point set involuting gives a distinct geodesic, roughly speaking -- but then suppose your endpoints were very close then this would be impossible because locally geodesics are unique

your argument is this written in symbols :p

Yes, I basically made it explicit because I was not sure I had interpreted what you said correctly or not

especially the "your endpoints were very close" part

11:41 AM

always a good thing to do

very close means inside the injectivity radius of $p$

@robjohn morning sir

How are you?

man i have to read this tome of a book

and i have to start at chapter 3.H

goddamnit

all the books on this hyperbolic shit are too thick

lol. I also need to prepare for endsems

any idea how long this is going to be ?

HAH i have no endsems

suck it

this as in the course?

and I have the first exam on 13th :( probably gonna clash

@BalarkaSen yeah. 10 weeks ?

11:47 AM

something like this, i don't know

12:07 PM

1

Q: What are some best practices when trying to design a reward function?

Generally speaking, is there a best-practice procedure to follow when trying to define a reward function for a reinforcement-learning agent? What common pitfalls are there when defining the reward function, and how should you avoid them? What information from your problem should you take into con...

reinforcement-learning markov-decision-process rewards reward-shaping inverse-rl

mic check

@Lelouch yo do you want to ram through B-H

i feel like i should just pick this stuff up in an hour or two but have no real motivation to lol

12:28 PM

one hour or two ?????

only ??? I can't read story books in this speed. wtf this is insane.

@BalarkaSen Yes, I am interested of course

I mean how hard can this be

its like 20 pages

u want to start now

oh. I thought you meant entire book

that's quite doable ig

No lol I dont care about CAT(k) stuff

its good but theres too much

@BalarkaSen sure. ig he only did III.H.1 today ?

yeah

im gonna read prop 1.6

so basically $d(x, \gamma)$ is of the order of $\log \ell(\gamma)$; why does this make sense?

Travelling $O(n)$ far from the geodesic between $p$ and $q$ requires you to walk $O(2^n)$ much -- of course, right?

if u gonna walk in a way that avoids geodesics in hyperbolic spaces you have exponentially high cost to pay

Nice Fig H.2 is the proof lol

12:43 PM

hey what is this CAT(k) stuff

bunk it

its not worth the investment

cool, yeah doing that

so if I'm interpreting prop 1.6 correctly, if you thicken any geodesic $\gamma$ by $2^{l(\gamma)}$, you'll cover the entire space ?

Ehhh $\gamma$ is the random path at hand

endpoints $p, q$

sorry

so basically, the distance between any curve $\gamma$ joinint two points, and the geodesic joining them can't be greater than $2^{l(\gamma)}$

you mean (of the order of) $\log_2 \ell(\gamma)$

not $2^{\ell(\gamma)}$ I mean

12:48 PM

wait shit, sorry. yeah

12 mins ago, by Balarka Sen

Travelling $O(n)$ far from the geodesic between $p$ and $q$ requires you to walk $O(2^n)$ much -- of course, right?

this is a better reformulation I think

being $O(n)$ far from the geodesic means there is some $x \in [p, q]$ such that $d(x,\im(\gamma)) = n$, say. Then $\ell(\gamma)$ must be of the order of $2^n$

linear deviation from geodesic forces exponential time spent on walking your goddamn path

1:13 PM

@BalarkaSen Why $y_n$ lies in an interval of length exactly $\frac{l(C)}{2^n}$ ? Isn't $y_n$ on the geodesic joining two points of $c$ which are $\frac{l(C)}{2^n}$ away from each other in $c$ ? (so we can say that it's upper bounded by this)

sorry I was afk for a while

that geodesic has length $\ell(c)/2^n$

the geodesic joining $c(\frac{m}{2^n})$ and $c(\frac{m+1}{2^n})$ will have length $\frac{\ell(c)}{2^n}$ ?

Of course we have $d(c(\frac{m}{2^n}), c(\frac{m+1}{2^n}) \leq \frac{\ell(c)}{2^n}$ (because $c$ parametrizes image proportional to arc length), but why it's equality ?

sorry if this is bovious

We arclength parametrized $c$ in the beginning, right?

yes

so the distance between $c(\frac{m}{2^n})$ and $c(\frac{m+1}{2^n})$ along $c$ is $\frac{\ell(c)}{2^n}$. The geodesic joining them can have smaller lenght, right ?

oh sure

i mean the stuff between $c(m/2^n)$ and $c((m+1)/2^n)$ is length $\ell(c)/2^n$

inside $c$, not on the geodesic -- misspoke

here's the idea, you dont really have to read the proof: just triangulate the region bounded by $c$ and $[p, q]$ by $\delta$-thin triangles, then observe that for any $x \in [p, q]$ and $y$ constructed as before, $d(x, y) \leq d(x, y_1) + d(y_1, y_2) + \cdots + d(y_N, y)$ which is less than $\delta N + \ell(c)/2^{N+1}$ or whatever

so at most $\delta N + 1$ aka at most $\delta \log_2 \ell(c) + 1$

the estimation for the geodesic arc is needed for the $d(y_N, y)$ term

the length of the whole geodesic is at most $\ell(c)/2^N$ like you said, but $y$ is the closest endpoint, so that divided by $2$ aka $\ell(c)/2^{N+1}$

@Lelouch but correct he wrote this iffily, he meant at most in at least one place (pun pun)

BTW I didn't know Andre Haefliger was into geometric group theory

1:28 PM

Thanks, I understood the proof but was not sure of the portion I mentioned. So ig a high level summary would be we would search for a point on the curve with the distance from $x$ bounded by $\log_2(\ell(c))$ by some analogue of "binary searching", using the $\delta$-hyperbolicity condition. It don't quite work, but we get an approximation $y_n$ so we can finish by triangle inequality

yeah it is exactly binary searching

for your portion note that $d(y_N, y) \leq (\text{length of geodesic between c(m/2^n) & c(m+1/2^n)}/2 \leq (\ell(c)/2^N))/2 = \ell(c)/2^{N+1}$ as you yourself pointed out

because the geodesic has less length than the arc on $c$ which is $\ell(c)/2^N$

Theorem 1.7 is Morse lemma, very useful

how do you think about quasi-geodesic ?

it's geodesic but upto bounded scale and translation; on average you walk minimal length but sometimes you go haywire and flare up and down, but only a bounded amount of distortion

how much you go up and down is controlled by $\varepsilon$, how fast you go up and down is controlled by $\lambda$ (eg if $\varepsilon = 0$ it's just a $\lambda$-Lipschitz embedding, a geodesic but with constant velocity $\lambda$)

Morse lemma is saying that all $(\lambda, \varepsilon)$-quasigeods uniformly shadow goedesics (uniformity constant = $R$)

sorry for this noob-ish questions, this is the first time I'm seeing quasi-isometry

yeah no probs

1:38 PM

a quasi-isometry needn't be continious/bijective, right ?

i mean its a good question

its certainly continuous

sorry, yes of course

is a quasi isometry necessarily bijective ?

need not be bijection; you can collapse bounded diameter subsets

any bounded diameter space is quasi isometric to a point!!

wha are some nontrivial examples of quasi isometry ?

nontrivial as in, the image is not a point/countable set

except the normal isometries

How to calculate stock returns integrally? Correct me I am wrong, but I think if I invest 100 INR, after 1% gain, it will be 101. And after another 1% gain, it will be 102.01. So how do I compound it every instant instead of daily/weekly?

1:41 PM

Cayley graph of $\Bbb Z$ with generating set $1, -1$ v Cayley graph of $\Bbb Z$ wth generating set $1, -1, 2, -2$

@VarunAgw that's how euler's number got invented

construct a quasi-isometry between them

@Lelouch calculatorsoup.com/calculators/financial/…

According to it, 300% gain = 20x returns compounded. Seems too good to be true. Maybe I am doing something wrong.

@BalarkaSen so we map $e_{m, m+1}$ to itself, and break $e_{m, m+2}$ into two equal parts and map the first one to $e_{m,m+1}

and second half to $e_{m+1,m+2}$

oh nice, thanks

i cant read symbols but im sure youre doing this right

basically two things are quasi isometric if you have bad eyesight and you think they are isometric

and you're looking at them from afar

all those 3 things

1:45 PM

I suppose the integers and the real line would satisfy that defintion

yes

but I doiin't think there's a nonconstant map from the latter to the former

*nonconstant continious

for obvious reasons

and isn't quasi-isometries continious ?

ah youre right no theyre not continuous

you just have that distortion condition on f

yeah

amount of discontinuity is just controlled by that

1:48 PM

ok, now I get it. cool, nice to see this "zooming out" formalized

hm, im a little scared what quasi geodesics really look like now then

i forgot qitries will not be continuous in general

@Lelouch I guess then a quasi geodesic can also do odd shit like in a graph it might not be a path at all but a collection of edges emanating from a bunch of vertices

star-like things I mean

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