Mathematics - 2025-01-09

handan_toddler

01:43

copper.hat

03:49

I want that book.

hopefully the fires are not impacting @leslietownes...

handan_toddler

04:07

@robjohn ...and robjohn

and Ted

Soham Saha

05:11

@NirbhayKrishnan The key step is observing that $T_i=a^ib+a^{i-1}\left(\frac{b^i-1}{b-1}-1\right)$, after that it should be easy enough.

Homesick Iguana

05:35

I spent months re-inventing the wheel

Let $\Delta: \mathcal{I} \to \mathcal{I}$ denote a $\Delta$-action, and define the slicing plane $\Sigma_{x_1} := \{ (x_1, x_2, x_3) \in \mathcal{I} \mid x_1 = 1/2 \}$ as the plane through which we partition $\mathcal{I}$. This plane intersects $\mathcal{I}_{\Gamma}$ and defines two subregions:

1. $\mathrm{Right~Half~Surface}: \mathcal I ^+:=\{ p \in \mathcal{I} \mid x_1 > 1/2 \}$
2. $\mathrm{Left~Half~Surface}: \mathcal I^{-}:=\{p\in \mathcal I \mid x_1<1/2\}.$

We define the twist map $\tau: \mathcal{I}^+ \to \mathcal{I}^+$ as a 90-degree rotation about the axis normal to $\Sigma_{x_1}…

because this is basically a Dehn twist

copper.hat

05:48

And @robjohn of course. I think Ted is in a different state.

Homesick Iguana

I'm okay everyone.

I'm surprised there are this many folks in the chat at this hour

reminds me of a time when we had a topology pow wow and then all of us returned 7 hours later to discuss more

after sleeping

does anyone know what "oud" is?

I have a candle that says "sandalwood & oud"

Sine of the Time

06:19

@HomesickIguana it's resin

Homesick Iguana

08:00

wow that is cool

never knew what that was

@LukasHeger I tried asking the same question on mathoverflow but I guess it didn't work there either. Honestly I'm about to throw in the towel on math.

They honestly think I'm trolling - i know that I am not. LOL

that is the disconnect

🤷

handan_toddler

08:52

@copper.hat did he move from San Diego

Sine of the Time

10:15

en.wikipedia.org/wiki/…. @HomesickIguana

nickbros123

10:39

just realised this course I am taking is actually a galois theory course

the title of the course, though, is fields, modules and algebras

im in for a nice ass beating semester

Jakobian

11:03

@nickbros123 field theory is basically Galois theory

at least in the sense that when you study one, you study both

Nirbhay Krishnan

@SohamSaha yep i have got the same general term great :)

Jakobian

link.springer.com/book/10.1007/978-1-4612-4040-2

I like this book for field theory, I studied from it the bits that interested me

it was for real-closed fields, so things related to transcendental degree

nickbros123

11:26

@Jakobian nice. this book is also in the recommendations for the course (primary one being Dummit and Foote)

some people, I have noticed (in my class), have already figured out what they want to do research in and narrow down themselves to it already in undergrad

like, I know a guy obsessed with number theory, algebraic geometry, so he only really cares about that.

Jakobian

@nickbros123 that's good that they did

the sooner the better

nickbros123

interesting you advocate this. I am not sure for myself though

I really like analysis, my department is also analysis flavoured

but I dont think i have seen enough of stuff

to make a choice like that

Jakobian

11:42

it's like with choosing a university or what are you going to study there

the sooner the better

you haven't seen enough, most people didn't

you'll waste your time trying to, no?

it's not a time to be indecisive

Jakobian

12:07

@nickbros123 I'd suggest something you could try to learn that I found pretty interesting, but at the same time deep

take a look at Stochastic differential equations and diffusion processes by Ikeda and Watanabe

nickbros123

@Jakobian what is that something

Jakobian

probability theory

Homesick Iguana

I used to think statistics was synonymous with probability theory

Jakobian

and you were wrong

Homesick Iguana

and that is why I succeed :P

nickbros123

12:13

@Jakobian thank you very much. I actually will be doing probability theory the upcoming summer (come what may, rain or shine, but in india its mostly shine). Doing measure theory (little by little) in this semester to prepare myself a bit

Jakobian

@nickbros123 why not download the book I just mentioned? You can learn the gaps in your knowledge as you go

I feel like that's a good motivation

Homesick Iguana

Is math sometimes never not well motivated? For example, the triangle. Wasn't that just a curiosity?

Jakobian

not what I was saying

Homesick Iguana

yeah i was just asking a question independent of your statement lol

Jakobian

when you ask a question like this right after I have said something, it can be put into context that you are trying to ask something in relation to what I said

even if you didn't mean that, this is how it will appear to most people

so it's natural that, regardless of what you mean, I would clarify this

isn't that obvious

nickbros123

12:25

@Jakobian sure.

Homesick Iguana

@Jakobian I see what you are saying yeah

Jakobian

how can I make this to not show up in my chat list

hm. It disappeared now

Thorgott

12:54

@Jakobian you know anything about this by chance? math.stackexchange.com/questions/5021145/…

Jakobian

@Thorgott homogeneous spaces?

If $X$ is homogeneous and $x, y\in X$ then there is a homeomorphism $f:X\to X$ such that $f(x) = y$, so that $X\setminus \{x\}$ and $X\setminus\{y\}$ are homeomorphic

Thorgott

yeah, it's sufficient, but not always necessary

Jakobian

sure, but isn't homogenity the reason of OP's question

Sine of the Time

13:11

@Jakobian the banner at the top of the page with upcoming events is annoying; plus the name of the events seems made by a 3 years old. Did you manage to remove it?

Thorgott

all the examples they know are probably homogeneous, but it's still an interesting question IMO

I think the sufficient condition for the converse in my comment can be improved by a lot

Jakobian

@SineoftheTime no

@Thorgott David Gao provided example of $\omega_1$ which is not homogeneous

and I believe any ordinal $\omega_\alpha$ should work

in this question you want to look at what, the group of homeomorphisms of course

homogenity says that for any $x, y\in X$ there is $f\in H(X)$ such that $f(x) = y$

Thorgott

sure, but ordinal spaces are still decently ugly depending on how you look at them

Jakobian

this questions asks about slightly distinct property, that if $x, y\in X$ then $X\setminus \{x\}$ and $X\setminus\{y\}$ are homeomorphic

Thorgott

for a concrete question, I wonder if there is any counter-example that is non-compact, LCH and connected

Jakobian

13:17

@Thorgott what? Ordinal spaces are very nice

Thorgott

as I said, matter of perspective

Jakobian

you never saw an ugly space

as for this, maybe $[0, 1)\times \omega_2\setminus \{(0, 0)\}$ with lexicographic order works, the longer line

this doesn't work

Thorgott

isn't it homogeneous?

Jakobian

maybe something like long circle

@Thorgott long line would be, but I used $\omega_2$ and not $\omega_1$

but this doesn't work because it doesn't have the homeomorphism property

so I think you want to wrap it around like a circle

Thorgott

@Jakobian and why is that not homogeneous?

Jakobian

13:25

@Thorgott because $(0, \omega_1)$ is a point where this space is not first countable

Thorgott

urgh, ok

perhaps it's not so easy to get a sufficient converse in the non-compact case

Jakobian

I have an idea like this

take an open shape homeomorphic to the circle, and puncture it in countably many places

add closed intervals, countably many

maybe I should draw this

the green dots are holes, the red is where there's no boundary, blue is where there is boundary (but not on the edge)

oh maybe the gaps being so large won't make it homeomorphic when we remove the points from blue ones

so perhaps just puncture the x-axis by countably many points

no matter which point you delete from this, it should still be homeomorphic to itself

in $\mathbb{R}^2$, locally compact, path-connected

Explicitly by a formula, say, $\mathbb{R}\times [0, \infty)\setminus (\mathbb{Z}\times \mathbb{N}_0)$

(where $\mathbb{N}_0$ is natural numbers with zero)

Thorgott

is it clear that satisfies the property?

it's clearly a manifold with non-empty boundary and thus not homogeneous

Jakobian

which property

Thorgott

13:40

that the point complements are all homeomorphic

but I suppose you can just wiggle everything and it checks out

Jakobian

well intuitively it does, since if you remove a point from above the x-axis then you can move it upwards

and the rearange those points

and if you remove one from the x-axis then you mess with that area of your space

formal proof - that would be a chore

Thorgott

yeah, I mean, away from the $x$-axis we are homogeneous anyway

removing a point from the $x$-axis is the critical case, but you should be able to just rearrange the nearby punctures above the $x$-axis

@Jakobian yeah I agree

Jakobian

@Thorgott yeah I think you can just move the x-axis back in place, and then rearange the points above so they get in their original place

I think if you have the half-plane, and remove discrete closed infinite countable set from both the upper plane and the x-axis, then that's always this space

Thorgott

yeah, that has to be true

Jakobian

I believe you can have some kind of dendrite without a point as well

ah no how would that work

I can't think of anything 1-dimensional

For subsets of the real line, of course nothing works if we want it to be connected

but I was thinking about 1-dimensional subsets of the plane

आर्यभट्ट

15:37

Who is cleo?

Jakobian

@आर्यभट्ट an user that existed on math.se famous for solutions to questions about integrals without explaining them

I think math.se banned cleo, and it made a lot of people disappear from the site, or something like that

Homesick Iguana

16:13

@XanderHenderson Will you sponsor my paper for ArXiv?

Lucas Henrique

hi chat. so why do we define the geometric genus only for smooth (complete) varieties over $k$? as long as $X$ is a proper $k$-scheme, we have that $\Omega_{X/k}$ is coherent, using the description of finitely presented $A$-algebras for the module of differentials $\Omega_{B/A}$. so $H^0(\Omega_{X/k}, X)$ is a finitely generated $k$-vector space and $g(X):=h^0(\Omega_{X/k}, X)$ still makes sense.

Homesick Iguana

16:31

you're right. but iirc the focus on smooth complete varieties is because $\Omega_{X/k}$ behaves more predictably as a locally free sheaf. Also $g(X)$ aligns better with geometric intuition and other invariants like Hodge numbers and the canonical divisor. And I think that singularities make $g(X)$ harder to interpret geometrically

You might use alternative invariants like arithmetic genus for non complete varieties.

Xander Henderson

16:47

@HomesickIguana No. I don't know you, and I don't know your work. I'm also pretty sure that I don't know the general area of your work, and I don't have the time to get caught up.

Lucas Henrique

@HomesickIguana properness is ok for me. but i'm wondering why the smoothness hypothesis is necessary, as the geometric genus would still be well defined. well, guess i just have to accept

Jakobian

17:03

@XanderHenderson hmm... why are they asking you this all the time

Xander Henderson

@Jakobian No idea. I'm not even sure that I could sponsor someone.

Jakobian

what does it mean to sponsor someone on ArXiv anyway? I thought the service is free?

Xander Henderson

@Jakobian arXiv doesn't allow just anyone to upload a paper. In order to upload your first paper, you need to be endorsed by someone who is "trusted" by arXiv.

I am not exactly sure how they establish "trust". I think that anyone who has already submitted a paper to arXiv can endorse a new author, but I am not certain of that. In any event, I don't think that I have an arXiv account, so despite the fact that I have a couple of papers on arXiv, I am not sure that I could endorse anyone else.

Jakobian

ah, I see. Interesting, since people still go there to post their proofs of RH and Collatz conjecture

Almanzoris

Greetings.

@SineoftheTime would you like to play another chess match one of these days?

copper.hat

17:19

my h-index must be in the pico range

Vladimir Lysikov

@XanderHenderson You need to be registered as an author of some amount of papers in a category (like math.RT) in the last five years to be able to endorse a paper in which this category is primary.

Xander Henderson

@VladimirLysikov Well, then, I'm useless. I am not sure that I ever created an arXiv account, and my papers are all more than five years old.

Thorgott

and even if they weren't, I'm pretty sure they're listed under a different category

Xander Henderson

@Thorgott almost certainly

psie

17:59

In Folland's text, specifically on product measures, he defines "the" product measure $\mu\times\nu$ to be the restriction of the outer measure induced by the premeasure $\pi(E)=\sum_1^n\mu(A_j)\nu(B_j)$ on the algebra $\mathcal{A}$ of finite disjoint union of rectangles (indeed, he uses the word 'the' when saying "We call this measure the product of $\mu$ and $\nu$ and denote it by $\mu\times\nu$.").

Someone told me once that by $\mu\times\nu$, Folland denotes a unique measure (the restriction of the outer measure from Carathéodory's extension theorem).Maybe unique is not the right word here, but if I remember correctly, this is what I was told once. Folland goes on to say if $\mu,\nu$ are $\sigma$-finite, $\mu\times\nu$ is the unique measure with the property $\mu\times\nu(A\times B)=\mu(A)\nu(B)$. Has someone read this part and has the same doubts concerning the notation $\mu\times\nu$?

I guess by his last remark, $\mu\times\nu$ is not always a unique measure on the product $\sigma$-algebra.

Thorgott

I don't understand the concern, $\mu\times\nu$ is an explicitly defined thing.

It has a certain property and that property may or may not uniquely determine it, but that's neither here nor there.

Jakobian

@psie I think we already went over this and even had some example of non-uniqueness

psie

yes, that's how I'd phrase it too; $\mu\times\nu$ is a specific measure. I believe one can't restrict the outer measure induced by the premeasure and obtain (again) another measure. So it's a distinct object.

Jakobian

what's of concern here is the definition of product measures

Folland defines them that way, while someone else might define it as any measure satisfying the given property

psie

@Jakobian yes, possibly. Currently I'm reading a text where the product measure is not constructed through Carathéodory's extension theorem, but through a $\pi$-$\lambda$ argument.

Jakobian

18:14

huh. Interesting

$\pi$-$\lambda$ is a tool to check if something is a $\sigma$-algebra, so not sure how that would be applied

psie

@Jakobian it has a corollary by which you can check uniqueness of measures

ILikeMathematics

18:28

I want to prove that $f:[a, b) \to \mathbb R$ is uniformly continuous on $\mathbb R$ $\implies$ $\lim_{x \to b} f(x)$ exists.

I have the following lemma available:

Let $f : [a,b) \to \mathbb R$ be a function. If for every monotonically increasing $(x_n)_{n \in \mathbb N}$ with limit $b$ the sequence $f(x_n)$ in $\mathbb R$ converges, then the limits match and $\lim_{x\to b} f(x)$ exists.

So if we can show the premise, this should immediately follow. Any ideas?

Soumik Mukherjee

A proof of the Riemann Hypothesis??

►

Somewhat related, he also talked about what it requires to post on arXiv.

Thorgott

18:45

@ILikeMathematics hint: a uniformly continuous function takes a Cauchy sequence to a Cauchy sequence

Jakobian

19:12

@ILikeMathematics This follows from theorems about completions. Interval $[a, b]$ is the completion of $[a, b)$, and so any uniformly continuous function $f:[a, b)\to \mathbb{R}$ extends to $\tilde{f}:[a, b]\to\mathbb{R}$

Jakobian

19:54

@Thorgott I must say, the discussion about when $X$ is Alexandroff extension of $X\setminus\{x\}$ was pretty interesting

"Alexandroff extension is functorial with respect to homeomorphisms (more generally proper maps) so homeomorphism $X\setminus \{x\}\cong X\setminus \{y\}$ extends to $X\cong X$"

what does functorial mean in this case?

If you mean that it's a functor then isn't this not really saying much

the latter seems to be part of checking if it is a functor (or er, that it makes sense to define it)

but yeah, it should be pretty clear that a homeomorphism extends to homeomorphism of Alexandroff extensions in a pretty obvious way

Thorgott

21:17

@Jakobian a proper map $f\colon X\rightarrow Y$ induces a continuous map $f^+\colon X^+\rightarrow Y^+$ that maps $x\mapsto f(x)$ and $\infty\mapsto\infty$, and this construction is compatible with identities and composition

Jakobian

21:55

@leslietownes something for you. I had some ideas but I pass

Sine of the Time

22:05

@Almanzoris ok I can next week

leslie townes

23:01

jakobian: good idea. similar to what robert israel is doing here: math.stackexchange.com/questions/1227253/a-dense-subspace-of-l2 (it doesn't seem to matter that n is an integer, only that it is a sequence going to +oo)

the other solution in that exercise shows another idea, which is to somehow go through holomorphic functions. i don't immediately see how to get it to uniform approximation, but the 1/(z-a)'s are clearly dense in H^2(the disc) by the argument given there (the 1/(z-a) represents evaluation at 1/a, or -1/a, or some scaled version of that)

there are a lot of pretty baroque reuslts in approximation in C[0,1], patterned on simple classical results for polynomials like the muntz-szasz theorem. this exercise doesn't seem to invoke anything near that level of substance because it doesn't care about how the sequence goes to infinity, and the lot of the conditions you see in polynomial or rational approximation absolutely do care about stuff like that. that is where they shoot off into outer space.

Claudio

23:22

I was wondering: at first sight, does this surface look regular to you guys?

spoiler it is not: I couldn't draw it so I checked all the requirements and found out that the normal vector $\mathbf{\phi}_u \times \mathbf{\phi}_v$ is not non-zero in the interior of $K$

lmao, if only I'd been able to draw it from the getgo I wouldn't have wasted all this time checking all the requirements (fun fact it satisfies 2 out of 3 of the latter)

leslie townes

this more general form may help you recognize what kind of surface it is a part of, and why the equations have the form they do math.stackexchange.com/questions/1578756/…

these folks even drew a picture web.maths.unsw.edu.au/~rsw/Torus/index.php

Claudio

well that explains it, on my part, I must admit I've never encountered the parametrization of a torus :)

now it makes sense

oh wait

now that I think about it it also looks familiar to the parametrization of the moebius strip I've seen in my book

@leslietownes as a person of my age would say: this is absolute fire

the moebius strip parametrization components are in fact similar, but $K = [0,2\pi] \times [-1,1]$ so the second parameter is not an angle

leslie townes

23:37

the picture in the second link suggests thinking of the formula for g(u,w) as a vector sum of two things, a formula that sweeps out a circle in the xy plane of radius R_1 (here 2) that sweeps out the center of the donut that the torus is the surface of, and then a formula that sweeps out a circle in a plane containing the z-axis

if you sort of think that through symbolically you do get that gibberish

Claudio

@leslietownes that's a neat way of constructing it indeed

leslie townes

and yeah if you let things twist around the origin you get a moebius strip type formula, all of this mysteriously paired cos sin stuff is reflecting rotational symmetry or near symmetry about the origin

Claudio

I see

leslie townes

someone who thinks in pictures can say this all in like 20 seconds while drawing it perfectly at the same time, and make it look easy

and then you go home and stare at your notes and think ".. what"

Claudio

yeah I wish I could do that :P

maybe a differential geometer, or whatever they are called

leslie townes

23:41

it should be a regular surface though. at any given point the "x_u" and "x_v" will be a little tangent vectors pointing along their respective directions and their cross product should be a nonzero thing that points normal to the surface of the donut

lots of opportunities to make sign errors or errors with trig identities here, that is probably what is going wrong if something is going wrong

Claudio

the normal is $[(2+\cos u) \cos u \cos w, -(2+\cos u)\cos u \sin w, -(2+\cos u)\sin u]$ which clearly is zero at some point in $(0,2\pi) \times (0,\pi/2)$

leslie townes

that expression isn't "clearly" anything :)

but if you compute its length there's a lot of squaring and a lot of opportunities for sin^2 + cos^2 = 1 to make it look different from how it looks now

Claudio

oh wait maybe no :)

cos and sin are never both 0

no ok it is regular then, but the normal vector clearly can't be continuous from the picture

wait, I must check in the interior of $K$, so maybe the picture is misleading

leslie townes

why can't the normal vector be continuous from the picture? you could probably draw it by hand anywhere, except maybe on the absolute edge where there isn't enough "surface" for the normal to make sense

but that's just a result of the parameter restrictions they made in the problem (and is not present on the interior of the parameter space)

Claudio

no yeah I thought the edges were the problem

leslie townes

23:52

if you let that 0,pi/2 parameter run over some larger a,b you just get a bigger or smaller slice of donut, or if you let it run over all R you get the whole thing

Claudio

but then again one can probably extended even to the edges so that it attaches smoothly

@leslietownes Yeah you're correct, it is indeed continuous

leslie townes

yes, if you let it run over all of R you no maybe no longer have what your book would describe as a regular coordinate patch for a different reason - different pairs of parameter values can map to the same point in R^3

but the normal vectors you'd compute (using the same formula as above, for different angles u, v) would actually coincide for different choices of parameter mapping to the same point of R^3, so even then you have a well defined normal :)

that maybe doesn't happen in the moebius strip case

Claudio

what you said above is that the normal vector does not depend upon the choice of pairs of parameters?

leslie townes

for choices of parameters that represent the same point

it obviously depends on the choice of parameter, but if you allowed yourself negative angles or whatever you'd find that the normal you'd get corresponding to the parameters (u,v) = (0, 0) would be the same as the normal you'd get corresponding to the parameters (u,v) = (-6 pi, 0)

so in this case it doesn't matter which parameter values you choose to represent a point of R^3, the normal you get using those values will be the same

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