Mathematics - 2024-03-24 (page 1 of 2)

mick

00:29

0

Q: The sequence of algebraic numbers $C_n$ in $\int_0^{2 \pi} \ln(\sin(x)^{2n+1} + C_n) dx = 0$

Consider the following integrals $\int_0^{2 \pi} \ln(\sin(x)^{2n+1} + C_n) = 0$ All of the $C_n$ are algebraic numbers. In fact all these $C_n$ can be given as zero's of some integer polynomial with degree $v$ where $0<v<2n+2$. Conjecture 1 : All these $C_n$ are algebraic numbers of degree exactl...

for the fans of calculus sine and polynomial sequences

Jakobian

@Thorgott well I'm not sure why its true, but apparently it is in the ring I'm considering

which is of course, $C(X)$

Thorgott

01:01

@BalarkaSen yeah, I like this perspective, probably the least unintuitive

@BalarkaSen idk, spheres, tori (the donuts, not the flat tori), I guess anything bounding a convex solid, projective spaces too iirc

@Jakobian I have no clue what prime ideals in this ring generally look like

surely this is in the Rings of Continuous Functions monograph

Jakobian

what they look like? Not really

I have some strategy for it, but I wasn't successful so far

My idea is to prove that the ideal $(P+Q)/P$ is prime

Because ideals of $C(X)/P$ are really nice, they form a chain, and there is a description of so called "upper ideals"

so what I want to do is show that $(P+Q)/P$ is a union of upper ideals

upper ideals do have some sort of description

the only issue I'm experiencing so far is that I need to use that $P+Q$ is a convex ideal somewhere in my argument i.e. if $0\leq a\leq b$ and $b\in P+Q$ then $a\in P+Q$

not sure how to show that, maybe its obvious

there is no real way to handle cases like this from what I know

Thorgott

that much is definitely proven in the monograph

Jakobian

@Thorgott that $P+Q$ is convex?

I'm doing exercises from that monograph and I don't recall there being anything that I didn't mention already that would be of help

Thorgott

01:18

perhaps not

you know this stuff better than me

Jakobian

might have overlooked something

Thorgott

01:35

one observation you may have already made is that every prime ideal is contained in a unique maximal ideal, hence if $P+Q$ is not the entire ring, the unique maximal ideals containing $P$ resp. $Q$ are the same ideal

but this does not seem to necessarily that $P\subseteq Q$ or vice versa

Jakobian

yeah I think I've got the case of $P, Q$ contained in the same maximal ideal case down actually

I just had an idea

See theorem 14.9

if they're not contained in each other, one can just use that to show that $P+Q$ is even a prime $z$-ideal

If we consider two maximal chains of prime ideals $\mathcal{A}$ and $\mathcal{B}$ extending that of prime ideals containing $P$, $Q$ respectively, then by the theorem we should have $P+Q\subseteq J = \bigcap \mathcal{A} + \bigcap \mathcal{B} \subseteq P+Q$

so $J = P+Q$ is a prime $z$-ideal (again, the inequality $P+Q\subseteq J$ is due to the assumption that $P, Q$ don't contain each other)

and I think that if $P, Q$ are not contained in the same maximal ideal then $P+Q = C(X)$

Thorgott

@Jakobian yeah, that's what I meant

not sure about how you're applying 14.9 though, it doesn't say anything about sums

Jakobian

$(I, J)$ means $I+J$

I guess they're using this notation to be more closer to "ideal generated by elements of I and elements of J"

thanks by the way, you did help me, I was stuck thinking that 14.9 is not useful for this problem

Thorgott

01:56

ah ok

glad it worked out

user10478

@Jakobian Okay, I was thinking about this more for some examples in $A = I, \mathbb{R^n} = \mathbb{R}$. I believe LPs have a property I'm tentatively calling "vague linearity." If I'm correct, then $(extr_{\vec x}\ (k\ f(\vec x))$, $A\vec x \leq \vec b$, $\vec x \geq \vec 0) = (k\ extr_{\vec x}\ f(\vec x)$, $A\vec x \leq \vec b$, $\vec x \geq \vec 0)$

and $(extr_{\vec x}\ g(\vec x)$, $A\vec x \leq \vec b$, $\vec x \geq \vec 0) + (extr_{\vec x}\ h(\vec x)$, $A\vec x \leq \vec b$, $\vec x \geq \vec 0) = (extr_{\vec x}\ (g(\vec x) + h(\vec x))$, $A\vec x \leq \vec b$, $\vec x \geq \vec 0)$, where $extr()$ picks out whichever of $min()$ or $max()$ is needed to make the linearity conditions true.

Put another way, $(min_{\vec x}\ (-3\ f(\vec x))$, $A\vec x \leq \vec b$, $\vec x \geq \vec 0) = (-3\ max_{\vec x}\ f(\vec x)$, $A\vec x \leq \vec b$, $\vec x \geq \vec 0)$; you have to toggle between mins and maxes to pull out negative constants. Similarity for additivity with negatives.

I think this might even be fine for nonlinear programs (aside from the matrix vector notation, I don't really know how to notate nonlinear programs but the linearity here appears incidental to the form of $f$, $g$, $h$, and the constraints).

Soumik Mukherjee

05:08

Hi Chat

Dannyu NDos

How would you compare Richard F. Bass' Real Analysis to Gerald B. Folland's Real Analysis?

My univ has two campuses and they have different textbook choices.

leslie townes

05:35

well, course offerings and textbooks are entirely different things, you can't really compare one by comparing the other. folland is a pretty common textbook choice. i had not heard of bass but maybe because it is newer (i think the first edition of folland was in the 1980s, the first edition of bass seems to be ~ 2010 ish)

looking at its index it looks like approximately similar coverage

this type of book is usually used for a graduate level class, or at places that might not have a grad school or organize their classes that way, in a second class in analysis as opposed to a first exposure to it

for that kind of use, i personally feel like the choice of text really doesn't matter very much. i took a class out of folland, and thought it was fine. like other books of that type, it can be used to cover a lot of related material in a unified way, and that is its main value, and maybe not so much that it is particularly good at any one thing

nickbros123

can the proof for "every linearly independent set of vectors in subspace $W$ of finite dim vector space $V$ is part of a basis for $W$", be extended to countable steps?

Dannyu NDos

@leslietownes Thanks. I'm looking at Bass, and I find it more intuitive.

leslie townes

nick for infinite dimensional vector spaces people often end up assuming some form of the axiom of choice, which implies that any linearly independent subset of any vector space can be extended to a basis of that space, irrespective of any limit or lack of limit on the cardinality of anything involved

nickbros123

i see

leslie townes

05:51

if you are the kind of person who is interested in things that slightly weaker than AC, you might have to be more specific about what you are willing to assume about W and/or V (i.e., are they assumed to have bases, what cardinalities are involved, and what kinds of choices are you allowed to make from sets that might not be finite sets)

nickbros123

i havent thought about it that much :P

leslie townes

which seems to be something that interests set theorists more than people are just interested in linear algebra, who generally seem to be comfortable with assuming "every vector space has a basis"

nickbros123

im ok with full fledged AC

though, i havent studied the ramafications

leslie townes

yeah then its just a zorns lemma type argument on a collection of linearly independent subsets of W

35

Q: Vector Spaces and AC

I know that the proof that every vector space has a basis uses the Axiom of Choice, or Zorn's Lemma. If we consider an axiom system without the Axiom of Choice, are there vector spaces that provably have no basis?

linear-algebra vector-spaces axiom-of-choice hamel-basis

apparently "every vector space has a basis" is not just implied by AC, but in some sense equivalent to it

nickbros123

thats cool

i was just revising some of my old LA notes and was wondering which theorems hold for infinite dim case and which ones dont

leslie townes

05:58

@DannyuNDos i took a class out of folland and while i thought it was fine in context i remember thinking it might not be a very good book for self study

not a lot of express signposts in there about what is routine and what is subtle, and it's pretty hard to make inferences about that from how much detail he includes (seems to vary more based on his interest in the material than its subtlety)

@nickbros123 a lot of LA stuff remains true or at least can be recovered in spirit in important and useful cases even when it is not generally true. the bare concept of a basis in the raw linear algebra sense is not as useful in the infinite dimensional setting

once you begin to think about constructions that might go on forever and only converge to something that you want, instead of terminate in something that you want in a finite number of steps, whether you realize it or not you are wanting to put a topology on your vector space, and finite sums become less important than convergent series do

nickbros123

06:14

I see

Sanjana

07:30

Is adjoint irrep (and its conjugate) the unique irrep for a non-Abelian algebra whose dimension match the dimension of the algebra itself?

nickbros123

09:27

Suppose, $S$ is a subset of vector space $V$ which is countably infinite; $S=\{ \beta_1, \beta_2, \beta_3 \dots \}$ and that $\text{span}(S)=V$.
Is it true that for every linearly independent set $\tau$ in $V$, a subset of $S$ is surjective to $\tau$?

leslie townes

"is surjective to" is not quite how anybody would phrase it, but yes

3

Q: Is the cardinality of linearly independent set always less than or equal to of a spanning set?

I am trying to prove (or disprove) this statement: Let $V$ be a vector space, and let $L,S\subseteq V$. If $L$ is linearly independent and $\text{span}(S)=V$, then $|L|\leq |S|$. Is this statement true? I have consulted many linear algebra books, but they only provide the proof under the ad...

linear-algebra

nickbros123

damn, thanks for the link

leslie townes

again, maybe it gets subtle without choice, but with choice (which makes cardinal arithmetic relatively predictable) its an unqualified yes, if S spans V and L is a linearly dependent subset of V then there is a surjection from S onto L

nickbros123

isnt choice beautiful (idk anything about choice)

leslie townes

it makes a lot of things behave very much the way a lot of people would expect them to behave

i'm not sure its particularly 'useful' in the way that finite dimensional linear algebra is useful, but its nice

i'm for it

Jakobian

09:39

axiom of choice is an extremely intuitive axiom

leslie townes

i meant to write "L is a linearly independent subset . . . " up above, hopefully context makes that clear

Jakobian

so intuitive, that you forget when you're using it

and that's why I'll never be a set theory person...

there's just too many things that "depend on axiom of choice" to keep track of

that I don't feel like I want to consider

and I think most people are the same. Refuting axiom of choice is even weirder than accepting it, too

and those things to keep track of, they are in your calculus course already

so you have to relearn those results to check if you didn't assume axiom of choice there somewhere, for example

leslie townes

yeah, there's basically no reason for anyone who isn't a set theorist to care about when or exactly how one is using AC, except in keeping track of a vague and holistic sense of those situations where you can't expect to be able to "get your hands on" something that theory tells you exists

and even then maybe who cares

Soumik Mukherjee

Is the term square free popular? Yesterday I wrote an example where there was a question : If a finite group G has square free order and H is a subgroup that intersects every subgroup non trivially then prove that G=H

Jakobian

When I can't obtain any concrete version of object, I will know from the proof anyway

or at least, I will know that the proof doesn't give me one

leslie townes

09:51

i think there's value to the nonspecialist of knowing "this example sucks but people use it because it's really easy to see that it is an example" vs. "this example sucks because the existence of anything like this is equivalent to some version of choice"

Jakobian

@SoumikMukherjee square-free is a pretty popular term

leslie townes

soumik: specifically for positive integers (as there) it is so common that i wouldn't think twice or even notice if someone used it without defining it

Soumik Mukherjee

I wrote an exam* not example

leslie townes

if someone transferred "square free" over to some other kind of object (like a finite group) i'd definitely immediately have questions (like "are you saying G is square free to mean that the order of G is square free")

Jakobian

you wrote an example on an exam, or an exam of an example?

Soumik Mukherjee

09:54

This was a question in the exam

Yesterday I wrote an exam* where there was a question:

Jakobian

@leslietownes yeah idk what a square-free group would mean. I think it'd be fine to use the term in a UFD

Soumik Mukherjee

Idk why I thought square free means the order is not a perfect square:"

leslie townes

all misunderstandings of that sort are understandable in an exam setting

even if an exam expressly included the definition, i could see someone blazing past it or misreading it

Jakobian

Erdos square-free conjecture: $\binom{2n}{n}$ is not square-free for $n > 4$

leslie townes

ask me what 2+3 is on an exam and there's a good chance i'll say 6

Jakobian

09:59

@leslietownes what's 2+3 in Z/5Z ?

"6? Correct!"

wait no

leslie townes

we are all one in Z/1Z

Soumik Mukherjee

I actually solved the question, but kept thinking that I have partially solved it. I started with let |G|=p_1^r_1×p_2^r_2...p_i^r_i and showed that the assertion is true when r_i's are at most 1

Jakobian

yeah it'd have to be Z/1Z, lol

Soumik Mukherjee

I actually solved the question, but kept thinking that I have partially solved it. I started with let |G|=p_1^r_1×p_2^r_2...p_i^r_i and showed that the assertion is true when r_i's are at most 1

Jakobian

@SoumikMukherjee You probably did a good job, question like this induces fear in me even right now

Soumik Mukherjee

10:03

Well this was an admission test, which I don't think that I will be selected.

Jakobian

even though I did so many exercises from group theory in the past

Soumik Mukherjee

There were some questions I could not understand where to start

nickbros123

@Jakobian ive used axiom of choice without knowing what it is many times, for proofs that dont even need it rofl

Soumik Mukherjee

@Jakobian Same here. The main problem is that the questions seems to be easy and anyone can understand the solutions once they see it but writing the solution on their own in exam hall seems to be the hardest part

nickbros123

i thought you were a grad student @SoumikMukherjee, do they have exams?

Jakobian

10:09

@SoumikMukherjee Here's an exercise: Show that if $|\{(x, y) : xy = yx\}| > \frac{5}{8}|G|$ then $G$ is abelian

I don't remember how to solve it so, good luck :D

nickbros123

I suppose the assertion that set of all finite subsets of $S$ cardinality is $\vert S \vert$ is the part that uses choice?

X4J

Hey all, I'm trying to determine wether the improper integral $\int_{e^{-1}}^1 (ln|ln(t)|)^7 dt$ converges. I noticed that $|ln(t)| = -lnt(t)$ at a sufficient small left neighborhood of the point 1. Furthermore, I know that the improper integral $\int_{0}^{1} ln(t) dt$ diverges. However I am struggling to combine this information :/

Soumik Mukherjee

@Jakobian I think I have seen this one before, but I don't know how to solve it, I will try

leslie townes

nick: yes, that assertion (for infinite S) is usually shown using choice and is maybe equivalent to some version of it

Soumik Mukherjee

@nickbros123 I am currently applying at some places for PhD. This was the tifr exam(2nd stage)

This question was also on the exam. I did solve it. math.stackexchange.com/q/118536/1093844

nickbros123

10:13

oh, good luck! TIFR is amazing

i hope u get it

Soumik Mukherjee

Well I don't think I will, but thanks:)

Jakobian

@X4J yeah, $(\ln |\ln(t)|)^7 = (\ln(-\ln(t)))^7$ on this interval but I don't know if thats helpful at all

You might try substituting $u = -\ln(t)$ first

X4J

@Jakobian I have tried but it just made it more complicated for me to solve and I am not sure if it was meant to calculate the integral directly

Jakobian

$dt = -\exp(-u)du$ so you end up with $-\int_1^\infty \exp(-u)(\ln u)^7 du$

this looks like at $u\to \infty$ it converges, definitely

and near $u = 1$ there's no issue anyway

@X4J checking convergence of an integral is far more easier than determining its value (by which I mean, also proving the integral exist... and not just giving an unjustified formula)

Soumik Mukherjee

@nickbros123 You are in iiser tvm no?

nickbros123

10:20

@SoumikMukherjee yep

Soumik Mukherjee

How is the math department there?

@Jakobian Can you occasionally give me some topology questions to solve?

nickbros123

as a 2nd year student i dont know if i can gauge it. a lot of applied math / computational math people here, PDE people also are high in ratio. they are hiring new profs though, the newest one was C* algebra guy. overall algebra there are only like 3-4 people, analysis perhaps a little more. one guy I know was diff geo and math physics guy

Soumik Mukherjee

@Jakobian Can you occasionally give me some topology questions to solve?

Sorry I am having network issues

nickbros123

The Head of math department is a rly great guy though, works in PDE. Some of them suck at teaching though, if thats something

Soumik Mukherjee

Sent the same message twice

nickbros123

10:25

id say IMSc is my favourite in all of india (partly perhaps because thats where my home is :) )

Soumik Mukherjee

The network connection is even slower than my brain

Jakobian

@SoumikMukherjee handpick it?

I could, I have something in mind you could try already

Soumik Mukherjee

@Jakobian Means?

@nickbros123 Ah ok

Jakobian

> select carefully with a particular purpose in mind.

Soumik Mukherjee

Ohh ok

Jakobian

10:27

hit me up when you want me to give you one

Soumik Mukherjee

@nickbros123 I got an interview call from there but didn't go for the interview cause the department seems to be too much algebra centred

@Jakobian Okay, for starting, can you give me one now?

nickbros123

@SoumikMukherjee yeah definitely, every other person is a lie algebra / rep theory guy

but theyre very friendly, i love them

i was there once in 10th grade

Soumik Mukherjee

I heard that too, a friend is in Cmi, he told that all the professors of Cmi and Imsc are very friendly and kind

nickbros123

IMSc is also a beautifully constructed, small, technical institute, the environment there will be most conducive for productivity. they really had research as a primary thing in mind when they constructed it, and it shows. its a proper think tank

@SoumikMukherjee yeah another bonus point is the linkup btwn imsc and cmi, u can visit one as the student of another, attend classes there etc

Soumik Mukherjee

@nickbros123 Ah nice

nickbros123

10:34

only down side ofc is, Chennai sucks as a city

Soumik Mukherjee

That's actually a great thing, like a buy 1 get 1 offer

@nickbros123 why? Because of the weather?

Jakobian

@SoumikMukherjee I'll have to remember the precise statement

nickbros123

@SoumikMukherjee the weather, roads suck, theyre always building the metro, they were building it when i was an infant, theyre still building it 18 yrs later, theyll keep building it after im gone. but i suppose the areas near IMSc, and IIT madras will be well kept, there are governement official buildings around that place, so theyll maintain that decently I suppose

Jakobian

well for know you can satisfy yourself with this one I guess

nickbros123

have u applied for ICTS TIFR?

Soumik Mukherjee

10:39

No, too much physics oriented

@nickbros123 Oh, I never knew this things about chennai

Jakobian

@SoumikMukherjee Prove that if $X, Y$ are metric spaces and $f:X\to Y$ is a continuous map then the following are equivalent: 1) $f$ is proper, 2) $f$ is perfect, 3) $f$ is surjective and if $(x_n)\subseteq X$ and $(f(x_n))$ is convergent, then $(x_n)$ has a convergent subsequence.

Soumik Mukherjee

Thanks, I will try

Jakobian

Definition of proper: surjective continuous map such that $f^{-1}(K)$ is compact for compact $K$. Definition of perfect: surjective continuous closed map with compact fibers

Sanjana

There must surely be some positive integers which can never be the dimension of an irrep of a given simple Lie algebra. Do the numbers which indeed do form the dimensions form a nice sequence or something?

X4J

@Jakobian I dont under why the improper integral boundaries aren't $0$ to $1$ rather than $1$ to $\infty$ and I think that's why I got stuck

Jakobian

10:46

@X4J mistake on my part

Alessandro Codenotti

@leslietownes it looks equivalent to full AC to me

Jakobian

@X4J then its divergent because $\exp(-u) = $ const. $\neq 0$ near $0$ and $\int_0^1 (\ln(u))^7 du$ should be divergent

X4J

@Jakobian Yeah I get that intuition I am trying to reason it

Jakobian

comparison test justifies dropping of $\exp$ part

and for the logarithm part I guess one way would be just to directly see that $(\ln(u))^7 \leq \ln(u)$ for $u$ small

(note that here those are negative so we end up with what we want...)

and $\int_0^1 \ln(u) du$ is divergent by some argument you already have so I don't even have to think about why its divergent

(seems like direct calculation though)

X4J

@Jakobian So it seems like from this part it is relatively easy and the main thing was to replace the absolute value and use the correct substitution

Jakobian

10:59

yeah. Good bounds for integrals often emerge after substitutions

this was basically my Bachelor thesis

X4J

Thing is after the substitution I tried to compute the integral and I didn't notice that the "new" integral (the one after the substitution) is easy to understand

Jakobian

its easy to understand in terms of convergence issues at least

X4J

yeah sorry

X4J

11:24

@Jakobian Actually I think this converges

Jakobian

does it

Xander Henderson

@leslietownes People who didn't make this assumption are such perverts, and should be locked up (both for the good of society, and their own safety).

Sanjana

Does anyone have any idea what David Hill meant when he said one can read off the roots from Dynkin diagrams of classical algebras here?

nickbros123

12:21

@XanderHenderson I saw a ice berg meme once where top level was analysis geometry algebra etc, 2nd level was naive set theory, 3rd was ZFC etc, 5 th ish level was set theory without choice, and set theory with negation of choice

The last level was "170! Is the last natural number "

Balarka Sen

12:37

@Thorgott Spheres, yes. Convex solid is precisely right. $\Bbb{CP}^n$ is correct.

The donut is not good: think of an upright donut like in Morse theory. There are two saddle singularities; at each of them the donut is tangent to... well, a saddle. $z = xy$. However, this has Gaussian curvature $-1$ (it curves "positively" along the smile of the saddle, and curves "negatively" along the frown of the saddle. These are the principal curvatures. Product of them is Gaussian curvature, which is negative)

This is happening because the donut is not convex. Arbitrarily near the saddles, there are always pairs of points contained within the donut such that the joining line segment gets out of the donut. However, there are points on the donut near which it is genuinely a convex body, where curvature is indeed positive.

Some corollaries of your examples are quotients by groups acting freely on $S^n$. But beyond this it is hard to come up with examples.

For instance, one can prove that the fundamental group of an everywhere $\geq \varepsilon > 0$ sectional curvature manifold is necessarily finite.

Shaun

12:59

in Helpful Commentary, 23 secs ago, by Shaun

Please may I have some feedback on the following?

0

Q: For $g\in\operatorname{SL}(2,q)$, do we have $\operatorname{tr}(g)=\operatorname{tr}(g^{-1})?$

The Question: For $g\in\operatorname{SL}(2,q)$ and $q=p^r$ for a prime $p$ with $r\in\Bbb N$, do we have $\operatorname{tr}(g)=\operatorname{tr}(g^{-1})?$ Here $\operatorname{tr}(h)$ is the trace of the matrix $h$. Thoughts: I think so. According to a preprint, it holds for $\operatorname{SL}(2...

linear-algebra matrices finite-fields trace

Thorgott

13:17

@BalarkaSen oh duh, I goofed

@BalarkaSen was this Myers? I think I've seen this before

I think you can use this Milnor-style to get that a semi-simple Lie group has finite $\pi_1$

though IIRC there also is a de Rham-cohomological argument for that

Jakobian

13:39

@Thorgott You've done goofed

Peter

13:53

To everyone having the right to delete-vote questions : We need helpers in CURED !

X4J

14:21

How can one determine the convergence of the improper integral $\int_{0}^{1} \ln^{k}(x) dx$ for a general $k \in \mathbb{N}$ without using lHopital for $\infinity$?

Jakobian

comparison test with $\int_0^1 \ln(x)dx$

Koro

Why is the last integral 0?

@Thorgott

nvm

I think it all boils down to tangent spaces.

dw is an n+1 form on M hence 0.

Thorgott

14:38

indeed

Koro

Does it make sense to talk about degree of a map $f:\mathbb R^2\to \mathbb R^2$?

Shaun

I'm after feedback, thank you very much.

Xander Henderson

14:55

@Shaun What I would suggest is that, in the future, you ask for feedback before you post a question.

Koro

I have seen the definition only for compact oriented domain and codomains.

Xander Henderson

When a user links to a lot of their own questions in here, it looks very much like they are advertising those questions, and looking for upvotes. That may not be your intention, but it is very much how it comes across to other users.

Shaun

I thought this was a solid question, @XanderHenderson. It's only downvoted, I think, because people here don't like me asking for feedback and the answer turned out to he trivial.

Xander Henderson

@Shaun I don't care about the question. I am not looking at the question.

I am describing your your behaviour is perceived by many people in this room.

Shaun

I got that impression long ago. I guess I'm naive, thinking others would be more reasonable.

Not enough people frequent "Helpful Commentary".

Xander Henderson

14:58

As you point out, I think that people get tired of seeing the same people posting links to their questions on the main site over and over again. The act of advertising a question, rather than just asking for help with the mathematical content of that question, gives the appearance of one who is trying to get upvotes.

@Shaun Other people are not behaving "unreasonably", and I would suggest that you remain civil and not impugn the motives of others.

Shaun

I'm trying to get feedback. Please don't be so cynical.

Xander Henderson

@Shaun I never said that you weren't trying to get feedback.

Shaun

@XanderHenderson Pot. Kettle. Black.

Xander Henderson

What I said is that your behaviour gives others the impression that you are trying to game the system.

Shaun

I see.

Xander Henderson

15:00

And I am trying to give you ways to get feedback without creating that negative reaction in others, i.e. ask the question here, first, before posting in on the main site.

@Shaun Remain. Civil.

Shaun

What do I do, then, when I want feedback?

Xander Henderson

@Shaun If you have a genuine mathematical question, and all you are interested in is getting an answer to that question, you can ask here.

If nothing else, the people here might be able to help you better craft a question which will be well received on the main site.

Koro

Posts can be downvoted without ANY REASON.

Shaun

@XanderHenderson That's what I'm aiming for.

Xander Henderson

@Shaun You are missing the order in which I am suggesting that you do things.

Shaun

15:02

@Koro Yes, but four times?

Koro

yes, even 5 times.

Xander Henderson

Ask the mathematical question here, first. If you don't get an answer, then post a question on the main site.

Koro

The best is to just move on with that. Who cares about downvotes?

Shaun

I put a lot of effort into my content here.

Koro

As long as I get the answer to my question, I'm fine with it.

Xander Henderson

15:03

@Koro Or, less cynically, posts can be downvoted for ANY REASON.

Koro

@Shaun Then perhaps someone interested in the subject will answer it.

Shaun

Perhaps you could change the description of what a downvote means, since, often, it's not serving its intended purpose.

Xander Henderson

@Shaun I'm not sure how you expect others to respond to that. In life, you are not rewarded for "effort". You are rewarded for what you do or produce. I put a lot of effort into my masters thesis, only to have the main result scooped two months before I was to defend. I put a lot of effort into producing a new result in those two months. But at the end of the day, all anyone cared about was the final document.

No one cares that I had to throw away a chapter because someone else managed to do it first (and better).

Koro

@Shaun That's it. I hope someone will see that if the question is not deleted by you.

There are some posts of mine wherein I put lot of effort. They got downvoted.

Xander Henderson

Koro

15:06

But they were upvoted after some time (say 2 months later).

Xander Henderson

I don't see a problem with that text.

A downvote can mean "I don't think that this question is useful."

Koro

Perhaps, someone saw my effort in that post and decided to interact with the post.

Xander Henderson

What is meant by "useful" can vary a lot from person to person. However, I am sympathetic to the point of view that if the answer to the question evokes the response "Oh, I was just being careless" or "Oh, I overthought it" or "Oh, that was really simple... d'oh", then the question probably isn't all that useful.

Koro

The point I'm trying to make is: in general, there is nothing one can do to guarantee an answer within a stipulated time.

"one day, it will be answered" is the only thing that one can hope for.

Shaun

@XanderHenderson That's fair.

Xander Henderson

15:11

I will also not that when you advertise a question and ask for "feedback", votes (including downvotes) are a form of feedback. You should not be surprised when you get feedback after you ask for feedback.

Shaun

Okay, well, I'll calm down a little. I might not respond for an hour or so.

Xander Henderson

(It looks to me like those downvotes came today, likely in response to the request for feedback.)

Koro

is one allowed to downvote just for fun?

Xander Henderson

@Koro There is nothing in the system which prevents it. It is not in the spirit of the site, but it is not something which can, in any way, be enforced.

The only kind of vote which is expressly forbidden is a vote which contributes to a pattern of voting which targets a person, rather than content.

Koro

@XanderHenderson yeah, leading to voting irregularities.

Koro

15:26

a) what to do?

b) seems wrong as I said earlier.

Peter

To everyone having the right to delete-vote questions : We need helpers in CURED !

Xander Henderson

15:49

@Peter 1) please keep in mind that CURED is a content moderation room. The "U" stands for "undelete", and the "R" stands for "reopen". It is not just about deleting content.

2) I think that those people who are interested in content moderation are aware of the existence of the room. Please try to limit your advertisement of the room.

Maybe, as a general note to everyone in this room: people get annoyed when they see the same kind of content over and over again from the same person (whether it be requests for feedback on their posts, requests for input in other chatrooms, or advertisements for their own personal projects or websites). Please try to limit these kinds of comments.

3

This is a chatroom. We are supposed to be chatting. :/

Soumik Mukherjee

@XanderHenderson Who came up with that name? Quite a thoughtful name.

Jakobian

@XanderHenderson huh. That's very subtle, but it did sound kinda cynical

Thorgott

@Koro they probably mean $\mathbb{R}^2\setminus\{0\}\rightarrow\mathbb{R}^2\setminus\{0\}$

Koro

no idea, it's perhaps not the degree as I understand it.

3

Q: Finding the degree of a map

I am having trouble computing the degree of a certain map using the fact that $f: N \rightarrow M$ where $M$ and $N$ are both $n$-dimensional manifolds induces a homomorphism between the nth cohomology groups. Also if there is a more algebraic topology approach as opposed to differential manifol...

differential-geometry algebraic-topology differential-topology

Lucky Chouhan

@koro

Xander Henderson

16:00

@SoumikMukherjee The room used to be called "CRUDE". I think I might have been the one to suggest the name change, but I don't recall.

Jakobian

what does E stand for

Soumik Mukherjee

Edit I think

Xander Henderson

@Jakobian "edit".

Soumik Mukherjee

@XanderHenderson Somehow both names make sense.

Balarka Sen

@Thorgott Yes indeed.

@Thorgott Indeed correct. For simplicity, assume your semi-simple Lie group $G$ is compact, by passing to the maximal compact subgroups (as this is a homotopy equivalence, $\pi_1$ is unchanged). Now, $G$ admits a bi-invariant metric, and one can compute the sectional curvature for this metric to be $Sec(X, Y) = \frac{1}{4} \|[X, Y]\|^2$, if I am not wrong. Here, by $Sec(X, Y)$ I mean the sectional curvature of the plane spanned by $X$ and $Y$, upto a normalization, possibly by $\|X\|^2 \|Y\|^2$.

Thorgott

16:19

think there's a sign somewhere

Balarka Sen

The problem with this is that $G$ contains many tori, especially if it is higher rank. So if $X, Y$ are in the Cartan subalgebra, $[X, Y] = 0$. So the sectional curvature is only non-negative, not quite positive.

Thorgott

the Killing form should be negative of the sectional curvature, I think?

Balarka Sen

No-- that's the fix. If you compute Ricci curvature, that's negative of the Killing form.

And Myers' theorem goes through with Ricci curvature bounded below by a positive constant.

The Ricci is a sort of in-between of sectional curvature and scalar curvature. It is a partial average; for example, $Ric_p(X, X)$ is average of $Sec : \mathrm{Gr}_2(T_pM) \to \Bbb R$ over all the planes containing $X$.

And then $Ric_p(X, Y)$ is defined bi-linearly from just knowing $Ric_p(X, X)$.

Koro

Killing is the name of a person :)

Killing form as in "Killing form of U(2) etc."

0

Q: How to compute degree of $f: S^1\to S^1$ defined as $e^{ix}\mapsto e^{iax}$?

How to compute degree of $f: S^1\to S^1$ defined as $e^{ix}\mapsto e^{iax}$? It is known that $f$ induces a vector space isomorphism $H^1(f): \mathbb R\to \mathbb R$ which is multiplication by a number that we call degree $(f)$. The following theorem is also known to me: If $M,N$ are oriented, co...

multivariable-calculus differential-geometry manifolds differential-topology

Thorgott

ah yeah, it was Ricci

I remember working out all these formulae based on Milnor's paper, but hell if I remember them

the bottom line is that semi-simple Lie groups become Einstein manifolds or something like that

Balarka Sen

16:34

@Thorgott That sounds kind of correct.

Thorgott

the constant is $1/4$ or something

if you take the bi-invariant metric obtained from the negative of the Killing form

Balarka Sen

But OK so the underlying point is this stuff seems to fall under the rubric of rigidity that you spoke of earlier. After all the examples of spaces with positive scalar (or Ricci!) curvature we can think of are spheres, quotients of spheres (elliptic manifolds), certain homogeneous spaces of semisimple Lie groups (eg, CP^n), ... and then we are out of ideas

Thorgott

cause semi-simple <=> negative-definite Killing form <=> finite pi_1 etc.

Xander Henderson

@Koro How much did that blow your mind when you first learned it?

(I think that Chari thought I was an idiot, and I think that part of that is that I expressed surprise when I learned that fact in her class.)

Thorgott

how much did it blow your mind when you first learned that Levi-Civita is just one person

(I had a physics prof who never learned this)

Xander Henderson

16:39

@Thorgott It didn't. You can tell from the hyphenation.

And I've never heard the name said out loud. :)

Balarka Sen

@Thorgott Now, here's the shocker, and why I had ventured to tell you about this in the first place. Relax to positive scalar curvature. Then you suddenly have a zillion trillion examples.

They look nothing like convex bodies

Thorgott

you always hyphenate when things are named after multiple people too

Xander Henderson

Not with an n-dash.

n-dash for one person, m-dash for several.

Balarka Sen

For instance, any finitely presented group is fundamental group of a positive scalar curvature 4-manifold.

Thorgott

@XanderHenderson guess I'm the one whose mind gets blown

@BalarkaSen crazy

what are topological obstructions for the existence of positive scalar curvature metrics?

Jakobian

16:43

I think you can also tell because if it were two authors then it Civita would be first

Balarka Sen

Why are you hell-bent on obstructions? Let me tell you about the vastness of the class of these spaces first!

Here's the proof, which is missing some details but gets the idea across. Take your favorite finite 2-complex $X$. Embed it in $\Bbb R^5$ by Whitney embedding theorem. Take a $\varepsilon$-neighborhood $N_\varepsilon(X)$. The boundary $\partial N_\varepsilon(X)$ is a closed $4$-manifold. More or less, this looks like a fiber bundle over $X$ with fibers $S^2$. Decrease $\varepsilon$ as much as you like so that this spherical fiber is very tiny, which means it has high Gaussian curvature.

It's so high that it dominates all the other sectional curvatures of all the other 2-planes

Thorgott

@Jakobian don't think this is always true

Balarka Sen

Average: you get positive scalar curvature.

In general: the property of admitting a positive scalar curvature metric is preserved by surgery in codimension $\geq 3$. This was observed by Gromov and Lawson. The reason again is that you can fiddle around with an $S^2$, make it very short so highly curved.

Koro

@XanderHenderson I learned it about 3 days before!!

I was shocked and I guessed that to be a German name.

and it is indeed a German name.

Thorgott

interesting, you're taking advantage of the topological flexibility to dominate some parts by others

Balarka Sen

16:50

Exactly. It's so simple. And suddenly you have flexibility in Riemannian geometry.

@Thorgott The only systematic obstruction comes from two avenues: Atiyah-Singer index theory and minimal surfaces. I only understand the first of these two: If $(M, g)$ is a psc manifold which is also spin, then consider the Dirac operator $D$ on the spin bundle.

It satisfies an identity $D^2 = \Delta + \frac{1}{4} scal$, you don't question this identity. If $scal > 0$, then $D\psi = 0$ has no solution (feed $\psi$ to the equation above and take inner product with $\psi$ and self-adjointness of $D$).

Which means index of the Dirac operator is $0$. But Atiyah-Singer computes the index for you, it's a topological quantity.

It was open for a long time if $T^n$ admits a psc metric. This is known as the positive mass conjecture. Solved by Schoen Yau for $n \leq 7$ using minimal surface theory and then by Gromov-Lawson using Dirac operators for all dimensions.

Around 1980's.

Jakobian

It would be funny to name something after yourself just for the sake of annoyance of others

Thorgott

analysis is like black magic man

Balarka Sen

Gromov has some beautiful lectures on this topic scattered across the IHES youtube channel. Recommend; that's where I learnt whatever little I know.

I never realized psc is such a flexible condition.

Koro

is there any approach to find the degree of a map?

not the theoretical set up that is used to prove only theorems.

a concrete method to find the degree of a map f: M--->N

Xander Henderson

17:20

@Thorgott Hey, Killing blew my mind.

Heaviside, too.

@Jakobian Not necessarily. m-dashes are used to distinguish between the names of people on named theorems. Often, the "most famous" or earliest name goes first.

Sorry---hyphens are used in names, n-dashes for multiple names. I am embarrassed to have messed that up. Though, to be fair, I haven't completely memorized the AMS style guide.

Thorgott

Krull-Akizuki theorem comes to my mind

Xander Henderson

17:36

Radon-Nikodym, too.

Thorgott

oh yeah, nice one

i've also recently encountered that there seems to be no coherent choice in the literature as to whether a certain set of invariants ought to be called "James-Hopf invariants" or "Hopf-James invariants"

Jakobian

those are just one person

you can't convince me otherwise, its what I chose to believe

Koro

17:54

I'll join you in believing so.

no one, absolutely no one cares about it though.

robjohn

@XanderHenderson this page seems to have a different opinion. -–—o—–-

Xander Henderson

@robjohn Does it? It doesn't seem to.

Hyphens for compound words (such as hyphenated names), but it doesn't talk about the names of theorems.

robjohn

@XanderHenderson it seems to say that en-dashes are only for date or quantity ranges.

Xander Henderson

@robjohn Yeah, but it isn't considering mathematical theorem-naming conventions.

robjohn

I see—

Xander Henderson

18:06

See page 93 of the AMS Style Guide.

Thorgott

you're telling me mathematics has an equivalent to PEP-8??

Xander Henderson

PEP-8?

robjohn

I don't like the fourth bullet point of 12.7.8

Xander Henderson

@robjohn *shrugs*

I mean, you don't have to like it. But if you want to publish in a journal that uses the AMS style guide, and if the editor is on their game, you kind of have to tolerate it.

robjohn

just my point of view

Lucky Chouhan

18:21

Do you like Markdown(.md)?

Koro

So here's what I did:
Suppose a to be an integer. Then note that if $f(e^{it})= (e^{it})^a=1$, then $e^{it}\in $ {|a|th roots of unity } and hence $f^{-1}(0,1)$ is a set of $a$ elements.
Taking a chart $\phi: (-1,1)\to S^1$ at (0,1) such that $\phi(t)= e^{it}$ and $\psi: f\circ \phi(-1,1)\to (-1,1)$ defined as $\phi(e^{iat})= t$, $\phi\circ f\circ \psi$ is an identity map and hence orientation preserving.

Therefore by the theorem I wrote in my post, deg (f)= a.

robjohn

Is "|a|th" supposed to be $k^\text{th}$?

oh the $k$ changed to $a$ nvm

Koro

@robjohn :-)

robjohn

18:37

@Koro should "a set of k elements" be "a set of $a$ elements"?

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