Mathematics - 2024-02-29 (page 2 of 2)

EE18

21:01

OK, derived the contradictions for $\alpha_{p-1} < 0$ or $\geq q$ so all good. Thanks again @Thorgott!

psie

I will continue ruminating about this :)

Thorgott

@EE18 no problem, glad you figured it out

Jakobian

21:17

@Thorgott Is ChadK saying that we should use $\text{Top}$ with continuous open maps here?

Balarka Sen

@Ted, @Thorgott: Consider a paraboloid $c^2 z = x^2 + y^2$ ($c > 0$ a constant) with the induced metric from $\Bbb R^3$. Delete the origin $(0, 0, 0)$. It seems plausible at a glance that this is still geodesically convex.

The problematic points are of course $(\pm c \alpha, 0, \alpha)$ for some $\alpha > 0$, but are we sure that the arclength of the parabola $c^2 z = x^2$ joining $(\pm c \alpha, 0)$ does not get beaten by the arclength of the arc of the circle $x^2 + y^2 = c^2 \alpha$ joining $(\pm c\alpha, 0)$ (which is, of course, $\pi c \sqrt{\alpha}$).

I'd have to compute arclengths, I guess.

Thorgott

@Jakobian he just seems to be wondering cause you haven't actually defined a functor. perhaps the question is not well-posed. idk if restricting to open maps will still give a reasonable subcategory.

@BalarkaSen yeah, would have to run a computation

Jakobian

@Thorgott But why do we need to define a functor for the question to be well-posed?

Balarka Sen

More generally, are there very steep curves $y = f(x)$ with $f(0) = 0$, $f(x) = f(-x)$, and $f \in C^1$ such that the surface of revolution minus the origin is geodesically convex?

Thorgott

well, we don't, but the idea seems to be that what was written down on objects should be part of a reflection functor, which is obviously dubious if we can't produce such a functor in the first place

Jakobian

21:26

From what I read, reflective subcategory only demands objects

oh okay. So he's saying that its implausible to follow with the idea

I'm more concerned with the objects anyway so it won't bother me if maps get changed

Balarka Sen

@BalarkaSen We need $2\int_0^t \sqrt{1+f'(x)^2}dx > \pi f(t)$, correct?

Near $t = 0$, especially.

$f'(0) = 0$ is forced

Let's try something, $f(x) = x^2$. Then the left side is $2 \int_0^t \sqrt{1 + 4 x^2} = t \sqrt{1 + 4t^2} + \mathrm{sinh}^{-1}(2t)/2$, and the right side is $\pi t^2$.

Works, no?

Near $t = 0$, the arclength of the meridians of the surface of revolution beats the lights out of the arclength of the parabolic arcs going through origin.

So a little cap of the surface of revolution of $y = x^2$, with height $\leq \varepsilon$, minus the origin, should be geodesically convex?

Ted Shifrin

21:45

@BalarkaSen The arc of the circle is not a geodesic, of course.

I’ll think about this with paper later.

Balarka Sen

Yep, but since it has smaller length than the parabolic arcs going through $0$, the latter can never be minimal geodesics.

Ted Shifrin

I think I disagree.

Balarka Sen

If $\gamma$ is a minimal geodesic between a pair of points $x, y$, it should minimize length over the set of all paths joining $x$ and $y$. That is to say, there should not be another path $\sigma$ between $x$ and $y$ with $\ell(\sigma) < \ell(\gamma)$, regardless of if $\sigma$ is a geodesic path or not.

The variational equation for arclength (or energy, rather) has minima at exactly the minimal geodesics, or not?

Ted Shifrin

Of course I know that. But what if you break symmetry? Go to a point just across the origin, not one at the same height.

Clairaut describes the geodesics explicitly.

Balarka Sen

I remember the picture for the ones you're speaking of but I think those geodesics never cross origin.

Call the antipodal points on the meridian $x, y$. Then the minimal geodesics should dip down from $x$ to some height not 0, and climb up to $y$

Ted Shifrin

21:55

Yes, uniqueness tells you that.

Balarka Sen

So then we should agree that $0$ is always missed by all the minimal geodesics :D

Ted Shifrin

Unless it’s one of the parabolas.

Balarka Sen

Granted, in which case you'd have to trust me regarding my calculation but that is the very thing I am untrustworthy at

Ted Shifrin

I’ll think about this later. I need to go now.

Balarka Sen

Cya.

Jakobian

21:59

well I felt bad for downvoting FShrikes answer for being wrong, which I had to do so someone doesn't get the wrong idea that it might be correct, but someone decided to upvote it anyway...

so annoying

okay its fine now

Jakobian

22:18

I've panicked a little

sorry

amWhy

22:32

Howdy to @TedShifrin, and @robjohn: I'm looking forward to a bit of green in your identicon ;)

psie

22:55

@Thorgott how would you define completion? I think my problem would be solved if I can find an isometry $f$ from $X=C_c(\Omega)$ (the space of continuous, compactly supported functions) to $Y=L^p(\Omega)$. Also, $\Omega$ in my original question was a subset of $\mathbb R^n$. I need to show too that the image space $f(X)$ is dense in $Y$. That $Y$ is complete is also a requirement in the definition of completion I have in front of me, but that is already given.

leslie townes

the isometry is just the inclusion there (or whatever passes for the inclusion should you define L^p(Omega) in some galaxy brain way)

that's X with the L^p norm, btw, psie

you keep leaving out express mention of what the norm is and i don't know if this is an expository choice or something that is maybe slipping through the cracks here

Xander Henderson

@leslietownes transcend the galaxy brain. You need UNIVERSE brain.

2

Jakobian

@psie A completion of $X$ is any metric space $Y$ such that $d_Y\restriction_{X\times X} = d_X$, $X$ is dense in $Y$, and $(Y, d_Y)$ is complete

that's the definition

Thorgott

no

leslie townes

"C_c(Omega)" can sit inside a lot of function spaces and have lots of norms and other things put on it, but in context, you're thinking of C_c(Omega) as being normed with norm using integration that you might more commonly associate with L^p spaces

there's nothing stopping you from thinking of that integral norm as a norm only on an incomplete space such as C_c(Omega)

which is what the author is implicitly doing way up above

Jakobian

23:00

@Thorgott corrected myself

Thorgott

ok, then i agree

the category theorist in me wants to say that the completion is defined by a universal property that expresses it as a reflection of a metric space to the subcategory of complete metric spaces (morphisms in either category being isometries for this purpose)

leslie townes

i used to bother people more than i do now about using notation for function spaces without being express about how they are defining it (in particular, if it's a normed space, what the norm actually is) but this is a good example of a time where it helps to be express about it

Thorgott

another option is to offer an explicit construction of the completion, which can e.g. be done by taking the space of all Cauchy sequences in $X$ and defining an appropriate equivalence relation on it, similar to how $\mathbb{R}$ is constructed from $\mathbb{Q}$

leslie townes

because you can define what something like "C_c(A)" is as a set of functions without ever putting a norm on it

and what norm you put on it actually matters here

Thorgott

i also agree with leslie's point

Jakobian

23:04

Kuratowski embedding

In mathematics, the Kuratowski embedding allows one to view any metric space as a subset of some Banach space. It is named after Kazimierz Kuratowski. The statement obviously holds for the empty space. If (X,d) is a metric space, x0 is a point in X, and Cb(X) denotes the Banach space of all bounded continuous real-valued functions on X with the supremum norm, then the map Φ : X → C b ( X ) {\displaystyle \Phi :X\rightarrow C_{b}(X)} defined by...

this is way better

Thorgott

i remember proving this when i studied ANRs

Jakobian

you can't use it to construct $\mathbb{R}$, but it has more appeal. Using Cauchy sequences is ugly to me

and for $\mathbb{R}$ you can just use Dedekind cuts

and yes, Dedekind cuts are beautiful in my book

oh I guess this goes back to our discussion of "Cauchy sequences vs Dedekind cuts"

one of the examples was that you can use Cauchy sequences to complete a metric space, but not Dedekind cuts

but as above shows, why use Cauchy sequences in the first place?

leslie townes

[anti piracy video font] you wouldn't USE DEDEKIND CUTS to COMPLETE A METRIC SPACE ...

Jakobian

so Cauchy sequences lose with Dedekind cuts

not only is the construction applicable to arbitrary total orders, the only reason for why one might prefer Cauchy sequences was eliminated

psie

@leslietownes I'm a slow thinker here. Sorry for leaving out the norm, it's been slipping through the cracks I must admit. The norm, once and for all, is the $L^p$ norm. When you say inclusion, do you mean the set inclusion in $C_c(\Omega)\subset L^p(\Omega)$?

Jakobian

23:20

what Leslie was alluding to is that elements of $L^p(\Omega)$ aren't functions but equivalence classes, so you need to interpret it as sending a function $f\in C_c(\Omega)$ to its equivalence class $[f]\in L^p(\Omega) $

leslie townes

psie: yes, whatever you use to think of a function on Omega as an element of L^p(Omega)

Jakobian

I've hit 10k today :D

@Thorgott and you hit 11k today :D

Ted Shifrin

@amWhy Don't look forward to mine. It never changes.

leslie townes

an analogy maybe worth making is that (C_c(Omega), L^p norm) is to (L^p(Omega), L^p norm) as (rationals, usual norm) is to (reals, usual norm)

the static around whether functions in C_c(Omega) are literal elements of L^p being very similar to the static around whether a rational is also a literal real number

Thorgott

the best way to construct the reals is clearly using coarse endomorphisms of the integers modulo bounded ones

@Jakobian steady dripping wears away the stone, or however they say it

Jakobian

23:26

I could have made it easy and just answer a bunch of calculus questions

@Thorgott what does that mean?

Ted Shifrin

@Thor That's erosion.

leslie townes

if you fling enough mud at the MSE walls, some of it will stick and turn into points

Ted Shifrin

@leslie Are you calling my answers mud?

Jakobian

@leslietownes you wish it was just mud...

leslie townes

washes out my eyes

ted: your answers are great i was describing my own posting strategy

Jakobian

23:34

am I interested in patterns?

leslie townes

if jakobian were to adopt it, he too could be closing in on 8700 rep after 12 years

but alas, the path i walk is not for everybody

Jakobian

what does it mean to be interested in patterns

Ted Shifrin

Could be geometric patterns; could be arithmetic patterns. Shrug.

leslie townes

very young babies are often interested in visual patterns

Jakobian

I don't think I ever saw a pattern worth contemplating but I guess I would if there was one?

> I am interested in the patterns or correlations of events.

Ted Shifrin

23:40

With actual events, people confuse causation and correlation.

EE18

I am in the very concerning position of having followed a proof of something but not understood how it can be true

The claim is the very last line of the above. My reading of what we prove (not pictured, it's the next page) is that no positive real is represented by any sequence of digits which terminates with infinitely many digits $q-1$ when we are working in base $q$

Oh shoot, maybe that does make sense

I was going to say doesn't 2/3 =: 0.66666666... when in base 10

but that's not a counterexample

leslie townes

you don't include "the algorithm" that this screenshot keeps referring to in the above (and i'm not saying paste it), but perhaps this is a situation where e.g. in base 10, when a number has multiple decimal representations, like the terminating 1.0000 and the non terminating 0.9999...., "the algorithm" would select the terminating one

Thorgott

@Jakobian it's a literal translation of a German idiom, not sure if it actually exists in English. it's meant to describe how doing something continuously will eventually yield results (modeled on erosion, as Ted says)

Jakobian

@Thorgott the construction of reals

leslie townes

or maybe they're otherwise doing a construction of R where the non-terminating representation doesn't arise

Thorgott

23:52

oh lol i got the reply wrong

EE18

I think you're right Leslie, thank you. yes, in the construction of a correspondence between these symbols and R (introduced axiomatically) it's shown that the subset of symbols which terminate with (q-1)(q-1).... do not arise

Thorgott

it's a reference to the obscure Eudoxus construction, don't ask me about the details

leslie townes

EE18 so yeah i would think of this more as a statement about how that construction works, than how one might make sense of decimals in general

EE18

What would you mean by the latter?

leslie townes

the slight non-uniqueness of decimal representation is sufficiently annoying that i could see someone wanting to define it away

EE18

23:53

I had thought that the point of this construction was precisely to make contact with the decimal (and other base) systems?

I see. i guess in this construction we are simply not allowing for symbols like 0.999999999....

by definition of the space of allowable symbols

Ted Shifrin

So the algorithm would already have given the decimal 1.0 for that.

leslie townes

EE18: please don't make me read that entire book haha but it's perfectly possible to make sense of "0.9999...." as a "real number," and outside the world of some construction people often do, but whenever you do, you are forced to allow it to be the same "real number" as "1.0000...."

and i could see someone attempting to construct the reals via digit sequences avoiding that problem by defining it away

EE18

Touche that I too often rely on you being able to read my mind :)

leslie townes

so that reals can be literal digit sequences, instead of things that merely have digit sequences (sometimes more than one) associated to them

EE18

but ya agreed with you and Ted, i think i follow now

Ted Shifrin

23:56

The guy who taught my Rudin real analysis course my freshman year at MIT spent the first week of the course trying to define arithmetic rigorously using decimals. After we were in agony from that, he spent the next week doing Dedekind cuts. Yes, we did eventually get to "actual" analysis. And, yes, he was a probabilist.

EE18

Rudin IIRC leaves the development of decimals to the reader

Your faves (the Hubbards) also go through the decimal route IIRC

Ted Shifrin

Interestingly, when he lectured the multivariable calculus class for 300+ students, when he got to polar coordinates, he did all his integrals $\iint r\,d\theta\,dr$ because never in his life had he encountered a region that was not symmetric about the origin. We had to teach his students the right way, since every one of the self-paced exams for that unit had regions with $r=r(\theta)$ ...

Maybe that's in some appendix of Hubbard^2 now, EE18. I certainly have no recollection of its being in the first edition (which is what I used for a few months). I would not have taught it, regardless.

leslie townes

my rudin analysis instructor was similarly hopeless about foundations, but also (thankfully) avoided them

Ted Shifrin

It is no fun to define addition of decimals, and defining multiplication is beyond torture.

leslie townes

i didn't realize how hopeless until i asked about an unassigned problem and learned that the instructor wasn't entirely sure if a [particular, non-obviously convergent] absolutely convergent series even defined a real number

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