Mathematics - 2024-03-27 (page 1 of 2)

12:41 AM

In Gamelin and Greene's Introduction to Topology, there is the following definition of local compactness:

> Definition 1 $X$ is locally compact if every $x \in X$ has a neighborhood whose closure is compact.

Contrast this to another common definition;

> Definition 2 $X$ is locally compact if every $x \in X$ has a compact neighborhood.

Now, Gamelin and Greene use the following definition of neighborhood; for $x \in X$, $V \subset X$ is a neighborhood of $x$ if $x \in U \subset V$ for some open set $U \subset X$.

I'm totally lost on how the two definitions are equivalent, if they are. It seems like the Definition 2 is more common, but maybe this follows from another definition of neighborhoods...

Jakobian

They are equivalent

psie

are you sure? I think they depend on the definition of neighborhoods one uses.

Xander Henderson

Definition 1 says that $\overline{V}$ should be compact. Definition 2 says we just choose $\overline{V}$ in the first place (that is, we just choose $V$ to be closed from the start).

Jakobian

If $x\in X$ has a compact neighbourhood $U$, then $V = \text{int}(U)$ is a neighbourhood with compact closure

conversely, if $x\in X$ has a neighbourhood $V$ with compact closure, then $U = \overline{V}$ is a compact neighbourhood

@psie Sure, here its understood that neighbourhood is a set $U$ with $x\in\text{int}(U)$, as is most common

otherwise you wouldn't really talk about compact neighbourhoods

@psie Actually under the definition in Gamelin and Greene, both are equivalent, the other definition of a neighbourhood demands that its open

if they were open then it wouldn't be equivalent, but also you wouldn't use definition 2

psie

ah ok, interesting, if neighborhoods would be open then they wouldn't be equivalent, good to know

Thorgott

12:51 AM

@Jakobian I'm not sure if this is true

Jakobian

@Thorgott then get sure

closed subspace of compact space is compact

Xander Henderson

@Thorgott A neighborhood of $x$ is a set $V$ which contains an open set $U$ which contains $x$. So, if $V$ is a neighborhood of $x$, it has an open subset $U$, and $U \subseteq \operatorname{int}(V)$. So the statement holds water.

(I don't like Jakobian's use of $U$ and $V$, since $U$ is always an open set in my world, but the statement is otherwise correct.)

Thorgott

@Jakobian obviously, but nobody says the closure of $V$ has to be contained inside $U$

in a Hausdorff space, sure

in general, this should fail

@XanderHenderson yeah, that part I have no issues with

Xander Henderson

@Thorgott Yes, I am seeing your objection, now. I agree that $\overline{\operatorname{int}(V)}$ needn't be compact.

Thorgott

consider the space $\mathbb{N}\cup\{\eta\}$ with the topology whose open sets are the empty set and sets containing $\eta$. the closed sets are the entire space and sets not containing $\eta$. in particular, $\{\eta\}$ is a compact neighborhood of $\eta$, but the only closed neighborhood of $\eta$ is the entire space, which is not compact, because it contains the infinite closed discrete subspace $\mathbb{N}$.

so I say this is one of the dozen inequivalent ways of defining "locally compact", but they all happen to agree if we additionally assume Hausdorff

Jakobian

12:59 AM

@psie I made a mistake. Definitions 1 and 2 are not equivalent. Definition 1 is stronger, and implies definition 2, but definition 2 doesn't imply definition 1.

psie

ok

Jakobian

@Thorgott Yeah I was making unconscious assumption that my compact set is closed. Rookie mistake

Xander Henderson

EVERY SPACE IS HAUSDORFF! GET OVER IT!

In fact, every space has a topology which is induced by a metric. And the metric is life. And the metric is love. All hail the metric!

Thorgott

yeah, assuming everything is Hausdorff makes the world go round

non-metric spaces I insist on, however, I need CW-complexes that aren't necessarily locally finite

Xander Henderson

@Thorgott Those are dumb. Abandon such evil things, and walk in the light of The Metric.

Jakobian

1:08 AM

law of stupidity: The more confident I am, the more stupid I feel after being proven wrong

one potato two potato

2:31 AM

@Thorgott then shouldn't the result be $\langle a.b.c.d\mid ab =c ,a = d^m, b = d^k, c = d^n\rangle$?

one potato two potato

2:44 AM

Wait then $d^{m+k} = d^n$...strange

one potato two potato

3:15 AM

I need to be care of the orientation I have chosen. $a = cb$ here and the resulting is cyclic group

$\langle a,b,c,d\mid a = cb, a = d^m, b = d^k, c = d^n\rangle$ so $\langle d\mid d^{n+k-m}\rangle$.

I think the 2-cell comes from the 2-cell in a pair of pants

Wait hmm...

Dannyu NDos

3:44 AM

Alright, the new edition of Fraleigh.

leslie townes

for extra credit, what does the image on the cover have to do with the content of the textbook?

Dannyu NDos

Dunno; ask the authors or the editors.

leslie townes

4:06 AM

someone likes triangles is my guess. my edition of fraleigh also had something triangular on the cover

one potato two potato

7:25 AM

@Thorgott I think I almost follow this, but if that relation is from $c = ab$, then shouldn't it be $\gamma \alpha^n \gamma^{-1} = \alpha^m \beta \alpha^k \beta^{-1}$? because $\beta$ is a loop from $a$ to $b$ and $\gamma$ is a loop from $a$ to $c$ to $b$ is represented by $\beta b \beta^{-1}$ and $c$ is by $\gamma c \gamma^{-1}$

Mad

11:18 AM

giving some inequality. $f(x) \geq g(x)$ does taking the infimum left or the supremum right change this inequality? i thought of examples where taking the infimum or supremum would result in equality, but not that the inequality sign flips.

$ 1/n \geq 0$ taking the infimum would give us equality.

which would me think if given f>g then infimum or supremum would give $ \geq $?

Soumik Mukherjee

12:40 PM

@Mad Yes

@Mad At best you can get equality but the signs won't flip ever.

Mad

thanks

Consider the shift operator, we want to show its isometry. I am stuck:
$ T: \ell^\infty (\mathbb{N}) \rightarrow \ell^\infty (\mathbb{N}), \quad (x_1, \dots) \rightarrow (0, x_1, \dots) $
Then apply:
$ \sup_{s \neq 0} \frac{\| TS \|}{\| S \|} = \sup \frac{\| S' \|}{\| S \|} = \sup_{S} \left( \sup_n \frac{\| S'n \|}{n} \right) / \left( \sup_n \frac{\| S_n \|}{n} \right) = 0, \quad n=1, \frac{x_{n-1}}{x_n}, n>1 $
We should get the norm to be that of $ \Vert S \Vert=sup_n x_n/n $ ?right

Maybe i am using the definitions wrong..

Thorgott

1:01 PM

@onepotatotwopotato this is almost equivalent to what I gave, except a) I realize now your labels and mine differ by a permutation. my $a$ winds around $n$, $b$ winds around $m$ and $c$ winds around $k$ times, b) I worked off of $a=bc$ instead of $c=ab$. these relations are not consistent. which one you want depends on the choices of orientations.

one potato two potato

1:25 PM

oh I see. it's more or less straightforward now. It's just attaching one 2-cell by a word you wrote.

one potato two potato

2:24 PM

Just to be sure: If I change a circle $\alpha$ as a boundary of a torus with a disk removed, then $\pi_1$ of it is just adding one more relation of the form $aba^{-1}b^{-1}$ right @Thorgott ?

Ryder Rude

what do u think about platonism ?

also, is platonism necessary for one to talk about "the standard model of natural numbers"?

becuz all theories have non standard models, we can only claim that the theories can only approach the platonic natural numbers. is this correct

Jakobian

@RyderRude its stupid

but everyone has their own choice in beliefs

Xander Henderson

@RyderRude "Standard" just means "the one with which the largest number of people are familiar". This has nothing to do with Platonism (nor any other philosophical outlook).

Jakobian

@RyderRude no, if anything you want to talk about formalism and not platonism

If you were trying to talk about the "standard" part, see Xander's message

Ryder Rude

2:43 PM

@XanderHenderson @Jakobian yeah i think "standard" usually means what ZFC can prove about natural numbers. but what about the statements that are undecided here too?

Jakobian

I think platonism is an outdated concept from times where people were still trying to connect symbolism to geometry

not that some people are still not trying to do so, but it was pretty much what Platon was talking about if I'm not wrong. He was giving meaning to platonic solids for some odd reason

Xander Henderson

@RyderRude That isn't actually a model for the natural numbers. ZFC is an axiom system. A model for the natural numbers is some set which satisfies those axioms, e.g. the von Neumann model.

Ryder Rude

@Jakobian i personally prefer formalism over platonism

@Jakobian yeah the new platonism is probably different from what plato was talking about

Jakobian

If anything, I think axiomatic systems show that our understanding of mathematical objects as existing out there is invalid in some sense

Xander Henderson

@Jakobian No, this is not what is meant by "Platonism" as a description of a philosophical standpoint. Platonism is probably best understood in terms of Plato's parable of the cave. He suggested that the world that we perceive is only a shadow of some true, fixed (perhaps "perfect") world.

A Platonist would argue that the objects studied by mathematicians actually exist in some meaningful sense, in a fixed and perfect realm, and we only see vestiges of those perfect objects through our limited senses.

Peter

2:49 PM

@RyderRude I would like to participate the discussion. What exatly do "platonism's" not accept in math ? If you give an example , I will tell you my opinion.

Ryder Rude

@XanderHenderson yeah.. but the model itself is sort of just another theory... so there must be undecidable statements still left?

Jakobian

@XanderHenderson I said why I think its a thing right now

Xander Henderson

@RyderRude No, a model is an object which satisfies the axioms of some theory, thereby demonstrating that the theory is not totally vacuous.

Ryder Rude

@Peter did u mean "what do platonists not accept in math?" in either case, i dont think the practice of mathematics differs all that much between platonists and formalists

@XanderHenderson oh

but there r some peopl like finitists who somewhat want to re-write all math without using infinities. these people differ in their practice

Peter

@RyderRude That's what I mean. Where are the differences in the points of views ? Or must I google "platonism" to participate ?

Sahaj

2:53 PM

why the channel for moderation activities named CURED?

Jakobian

@RyderRude even if you are a finitist you can still accept, say, set theory, and study it.

Ryder Rude

the differences are mainly metaphysical. platonists believe mathematical structures exist in a metaphysical universe @Peter

@Jakobian correct. only some finitists want to re-write math

maybe because they think infinite math is inconsistent

Xander Henderson

@Sahaj It is an acronym: "Close Reopen Undelete Delete Edit".

Ryder Rude

i came across one such finitist

Peter

I have a clear attitude against finitism , in particular ultrafinitism. I do not consider this as math.

Jakobian

2:55 PM

are you talking about Wildberger

Ryder Rude

let me check

@Jakobian yes

i think finitists are somewhat platonist. they reject infinite sets perhaps because these sets do not "exist" according to them

and they may think peano arithmetic has inconsistencies

Xander Henderson

@RyderRude I think that most mathematicians are somewhat Platonist, even if they won't admit it to others.

Ryder Rude

@XanderHenderson makes sense

Peter

Hm, I tend to disagree , although I still do not exactly know what platonists believe. But denying infinite series is bullshit.

Ryder Rude

i think the rules of logic and numbers are borrowed from.the real world. so these structures are probably platonic

to me, logic and numbers almost qualify as physics as much as they qualify as mathematics

Jakobian

3:01 PM

I think the part that you feel like is platonistic is our imagination. Okay, yeah, but does that really make us platonists? That we can imagine those objects in our heads?

Peter

math and the (physical) reality are utter different things !

Ryder Rude

yeah... but i can never believe in "mathematical universes which exist in some other realm" :P

the closest i can get to platonism is the belief that math exists in our own universe

Jakobian

Wildberger is a curious case, because he acts in ways that a leader of a cult would. Calling things revolutionary all the time, and rejecting infinity and standard mathematics like we are an opposition to him

Straight up calling axiom of infinity wrong

Ryder Rude

math with infinite sets is useful and convenient. so unless there is any hint of inconsistency, there is no reason to re-invent the wheel, irrespective of one's philosophical stance

this is why finitists, formalists, platonists can work together easily. they only differ in their interpretation of math. no one denies the usefulness.

Peter

3:29 PM

What about , say , Graham's number ? It clearly has no physical meaning , its magnitude is utterly incomprehensible. But no serious mathematician denies its existence.It exists in the mathematical universe.

Ryder Rude

i believe only our universe exists. i dont believe in the existence of any other realm. my stance on such large numbers is that they only exist as ideas (in the sense that u can use symbols to refer to them and manipulate those symbols) @Peter

and these ideas can also be useful, as large numbers have been used in proofs

or if space is unbounded, then such numbers physically exist too. time at least goes infinitely far into the future according to our current models of physics @Peter

Mad

Any tips on that norm i asked about?

Xander Henderson

4:01 PM

@Peter I deny the existence of any non-natural number larger than 3.

Such things don't actually exist.

I'm a bit doubtful about 1, 2, and 3, as well. But I generally accept their existence.

Which is not to say that I have problems working with larger numbers, and negative numbers, and real numbers---I just don't believe that they exist.

Peter

Well , then I am out. I cannot agree this point of view.

Xander Henderson

Jun 23, 2018 at 20:31, by Xander Henderson

Personally, I am three-ist. I don't believe that there is any number larger than 3.

Soumik Mukherjee

in In praise of Math.SE site and its users, Aug 4, 2023 at 4:44, by Martin Sleziak

Xander Henderson is a moderator already for three years.

Oh no! you can't be a Mod for any longer

Xander Henderson

@Peter The point that I am making is that I don't accept Platonism. I don't believe that any of the objects studied by mathematicians actually exist (where the notion of "existence" I am referring to is one more closely aligned with what natural scientists study---I have no problem with the existence of stars, or water, or mooses; I can see, touch, or otherwise interact with those objects---I cannot see "three").

Lucky Chouhan

@SoumikMukherjee means?

Astyx

4:06 PM

the plural of "moose" is actually "meese"

(not)

Xander Henderson

@Astyx Pretty sure it is actually "mice".

Lucky Chouhan

dictionary.cambridge.org/dictionary/english/moose

Xander Henderson

(I almost wrote mooses and gooses, just to make the joke more gooder, but jakobian gets angry at me when I make bad jokes and/or explain my bad jokes, so I stuck to one intentional screw up.)

Soumik Mukherjee

@LuckyChouhan Jk

Lucky Chouhan

@XanderHenderson what are perks of being a moderator on MathSE?

Xander Henderson

4:09 PM

@LuckyChouhan Suspending users who ask me what the perks of being a moderator on MathSE are?

:P

Lucky Chouhan

Oh my goodness! I won't ask this question anymore ;) haha

Xander Henderson

I am not sure what you mean by "perks". Mostly, it is just work: handling flags, responding to meta questions, getting flamed for doing the above, etc.

I guess I have an SE T-shirt... is that a perk?

Astyx

making bad jokes and not getting banned

Jakobian

thats the whole charm

Xander Henderson

@Astyx My jokes, sir, are EXCELLENT!

(Only I get to call my mother fat, and only I get to call my jokes bad.) :P

Koro

4:43 PM

If A doesn't like any political party, why should he vote?

Xander Henderson

@Koro No one should vote. Voting doesn't matter. Arrow showed this. :P

Koro

if no. one votes, then there will be re-elections!!

Xander Henderson

More seriously, when I vote, I typically leave large portions of my ballot blank---in some races, there is no candidate that I like; I also refuse to vote for judges, because I do not believe that judges should be elected. So A is more than welcome to cast a blank ballot.

Or to vote only in those races that they care about.

Koro

advertisement costs and all is taxpayers' money. 😑

Xander Henderson

Or to write in candidates.

@Koro Not in the US, but I wish it were so.

Koro

4:46 PM

@XanderHenderson ahh blank vote!

or just don't show up for the voting.

Xander Henderson

...which is the same as not voting.

Generally speaking, if a person doesn't want to vote, I see no reason to encourage them to do so.

Koro

exactly!!

Xander Henderson

@Koro So what was the point of the question chat.stackexchange.com/transcript/message/65417023#65417023 ?

Or are you A?

and you are just looking for people who agree with you?

Soumik Mukherjee

There should be an option for 'I hate this candidate'.

Koro

But not voting is frowned upon. I don't know why. The argument is: You're wasting your vote by not voting. Even if you don't like any of the parties, there must be some that you least dislike. You should've voted that.
But A never thought about which one they least disliked. 😅

@XanderHenderson Just a hypothetical question. Elections are around the corner here (and in your country too) so probably news and all made me think about it.

Xander Henderson

4:53 PM

@SoumikMukherjee "None of the above" is an option on Nevada ballots.

Koro

@SoumikMukherjee if that candidate wins, he'll take note of that and give you blessings.

NOTA is an option here as well.

Astyx

probably A should start thinking

Koro

It would be amazing if NOTA wins :-)

Astyx

not voting is one thing, abstaining completely from political matters is another

Koro

but that'll lead to re-elections with some candidates replaced by other ones.

Xander Henderson

4:55 PM

@Koro There is an argument to be made that in a democratic society, you have a responsibility to choose your leadership, and that failing to vote is an abdication of that responsibility. The argument goes that if you don't vote, you have no right to complain about how society is run.

Koro

yes fair point.

Jakobian

I don't have such responsibility

and I think that even if I don't vote, I have full rights to complain

Soumik Mukherjee

@XanderHenderson Nota is available here too but that is not the same as expressing hatred. I don't like any of the candidates don't mean I hate them.

Koro

But suppose that A's issues are not addressed by any of the parties, then A has nobody to complain.

Soumik Mukherjee

@Koro Anonymous voting ofc

Xander Henderson

4:58 PM

Indeed, at the local level, I am sympathetic to that point of view---I live in a town of 5000 people. Voter turnout is typically around 1000, and elections are typically decided by 10s of votes. And local government has a much bigger impact on my life than state or national government.

(Though the state is really f*ck*ng with our budget right now, so there's that...)

Jakobian

my town has 100,000 people

Xander Henderson

That being said, I think that there are civic duties that are much more important than voting, e.g. jury duty.

Soumik Mukherjee

My city has 393,000 people

Koro

@XanderHenderson that's very low %.

Xander Henderson

@Koro Not really, considering that something like 1500 people are not old enough to vote, and we have a somewhat significant population of non-citizens.

Jakobian

5:04 PM

thats a huge amount of young people

Xander Henderson

@Jakobian I think that it is rather that we don't have a ton of old people.

This is a hard place to retire, as there are not a lot of services.

It may also be worth noting that Holbrook is behind the Zion Curtain, and has a large LDS population. LDS families tend to have a lot of children.

So even if you ignore the lack of olds, I would expect a larger-than-typical number of youngs.

Soumik Mukherjee

What is LDS family?

Xander Henderson

@SoumikMukherjee LDS = Latter Day Saints (Mormons)

Soumik Mukherjee

Okay, I have to google, I don't have any knowledge regarding this

John Zimmerman

6:28 PM

example of a mathematical object exhibiting both a non abelian symmetry group as well as a continuous symmetry group?

John Zimmerman

7:01 PM

oh it was simple all along lol 😂

Jakobian

I searched for my gloves everywhere, under the bed. Turns out they were under some clothes in the bathroom. Sigh

John Zimmerman

dam!

Obliv

How does one count the number of finite groups of order $n$? I think it was called $C(n)$ in class

was an excursion i think so I don't really have to learn it

Jakobian

@Obliv I don't think there is a formula, but I think you use $p$-Sylow theory

Thorgott

this question is terribly difficult

Obliv

7:05 PM

wait yeah maybe it was number of subgroups of a finite group with order $n$

er wait that's the same question

Jakobian

its a different question

I remember using Sylow's theorems to find all groups of orders $\leq 60$ before

Xander Henderson

@JohnZimmerman Do you need a beaver?

John Zimmerman

@XanderHenderson hahahaha

Xander Henderson

@Obliv On your fingers.

Use your toes if you run out.

Or get a friend to help.

John Zimmerman

@XanderHenderson I miss your jokes.

Xander Henderson

7:08 PM

hash tag dad jokes

John Zimmerman

miss(ed) plural

past

How do i get my foot in the door to illustrate a math book?

I'm trying to make it big as a mathematical illustrator

Jakobian

I don't think that pays for much

John Zimmerman

last year i learned the kosher way to draw a torus

Jakobian

or is authors choice, probably

to illustrate math books you'd probably need to work for a publishing company

Obliv

are permutation groups the same thing as symmetric groups?

Jakobian

7:12 PM

@Obliv Sometimes people call any subgroup of symmetric group a permutation group

Obliv

in this book a permutation group is a bijection on a set equipped with function composition. Idk why they're called symmetric groups

Xander Henderson

@JohnZimmerman My mother was / is a freelance technical illustrator. She mostly works in archaeology, but did book on cosmology in the late 70s. It is not a great way to earn money, particularly now that computers make it a lot easier for any schmuck to generate their own images.

Jakobian

Permutation group

In mathematics, a permutation group is a group G whose elements are permutations of a given set M and whose group operation is the composition of permutations in G (which are thought of as bijective functions from the set M to itself). The group of all permutations of a set M is the symmetric group of M, often written as Sym(M). The term permutation group thus means a subgroup of the symmetric group. If M = {1, 2, ..., n} then Sym(M) is usually denoted by Sn, and may be called the symmetric group on n letters. By Cayley's theorem, every group is isomorphic to some permutation group. The way in...

or maybe always according to wikipedia

John Zimmerman

@XanderHenderson good info to know thanks. I will try something else

Jakobian

@Obliv read what you just wrote

Xander Henderson

7:14 PM

yep

Bam.

that's perfect!

It is both kosher AND halal.

Obliv

i mean what if it's non-abelian?@Jakobian

Jakobian

7:15 PM

@Obliv what?

Imagine me staring at you weird

Obliv

I get it now, $S_n$ is symmetric after $n$ iterations

Jakobian

what?

Obliv

they're called symmetric groups because they are finite & cyclical. For example,
standard reflection is $S_2 \cong (\mathbb{Z}_2,+)$

or something whatever

Jakobian

they're not cyclic

Obliv

HUH..

Jakobian

7:20 PM

what?

Obliv

oh I guess I remember a problem stating that a group with finite order doesn't necessarily require elements to have finite order

Jakobian

no, if a group has finite order then every element has a finite order

@Obliv the symmetry is that of a set $\{1, ..., n\}$ of $n$ elements

its not a geometric type of symmetry, but literally symmetry of an $n$ element set

so any bijection

psie

@Thorgott of the two definitions I gave yesterday of locally compact, which one is more appropriate when the space is not Hausdorff? I have a hard time accepting inequivalent definitions...

Obliv

jk it was that an infinite group can consist of only elements of finite order

$\mathbb{Q}/\mathbb{Z},+$

Jakobian

yes, such groups can exist, for example if you take the complex circle and only consider $n$th roots of unity

Obliv

7:26 PM

@Jakobian cyclic means every element is generated by one element, which is not the case in $S_n$ is what you were getting at?

Jakobian

so called Prufer group

Thorgott

@psie the appropriate thing to do is work with Hausdorff spaces, but if I have to make a choice on how to define local compactness in a non-Hausdorff setting, it would be neither of the two options you gave yesterday

Jakobian

@Obliv no this is not what cyclic means

cyclic means the group is generated by one element

$S_n$ is not cyclic unless $n\leq 2$

@psie different results work in different generality. I recall a couple working with the first and a couple working with the second definition

but this is something people who like to classify spaces care about, rather than a typical learner of topology

People usually work with locally compact Hausdorff spaces

psie

ok 👍

Obliv

"Identity element in S_n is even, but not odd" why not just say it's even.. or not odd

is there a reason to state both?

oh theres even a lemma that states no permutation in S_n is both even and odd. talk about being thorough

Jakobian

7:33 PM

@Obliv I don't know, parity is a certain homomorphism $S_n\to \{-1, 1\}$

Thorgott

I care about exponentiation, so the definition of locally compact I choose is the one that makes it work

i.e. every point has a nbhd base of compact nbhds

Balarka Sen

@Thorgott check washington dc

Jakobian

@psie on pi-base you have weakly locally compact space, locally compact space and locally relatively compact space

the other common definitions are by either adding assumption of regular, or of Hausdorff, so they're not important

either way you can see on pi-base what properties are being implied by each one

psie

nice, thanks, I will check it out

Jakobian

weakly loc. compact is the most weak of the assumptions

Obliv

7:38 PM

I find it so weird seeing abstract algebra theorems being named after 18th century mathematicians. It reminds me of the chinese remainder theory section

this 20th century formalization of algebra just does not look very "isomorphic" to what they probably had in the first couple centuries

Jakobian

the assumptions of loc. compact and loc. rel. compact don't imply each other

according to the properties on pi-base, not much is being implied by loc. rel. compact, and indeed its not something I see very often

most people assume weakly loc. compact, and if that isn't enough then they assume loc. compact

or, you know, just add Hausdorff assumption

definition 1 would be loc. rel. compact, and definition 2 would be weakly loc. compact

so among those two, definition 2 is the more common one from what I've seen

which is justified by being weaker than definition 1. I actually don't see definition 1 in practice at all, its either definition 2, or assumption of basis of compact neighbourhoods

(well the most common is to assume spaces are Hausdorff...)

John Zimmerman

sorry but what notion of transversality for 3 manfiodls is the right one?

Sanjana

Does anyone know a method to compute the explicit matrix form of a non-commuting generators of a simple Lie algebra in the fixed basis in which the Cartan generators are diagonal and are given by the root vectors? E.g. This is equivalent to deriving the Pauli matrices for $A_1$ and the Gell Mann matrices for $A_2$

John Zimmerman

I don't know actually

Koro

@Thorgott: Given $f:S^1\to S^1: z\mapsto z^n$. I'm trying to show that Df_p is orientation preserving: Let's try to show it at p= (1,0). Tangent space at p is one dimensional and is spanned by basis vector $\partial/\partial y$. I assume S^1 to be oriented anticlockwise. I'll show that $Df_p (\partial/\partial y)$ is positive multiple of $\partial/\partial y$

Take any C^oo(p) function $g$. Then $Df_p(\partial/\partial y) g=\partial/\partial y|_p(g\circ f)$

how to go from here?

Jakobian

7:59 PM

@psie Note there is no example of weakly locally compact but not locally compact nor locally relatively compact space on pi-base. I've already noticed developers of pi-base about this, so there might be an example of this coming up. Or not, who knows

John Zimmerman

I want to be able to compute the MCG of $M_k=\overline{\Bbb H^2 \times \Bbb R}_g$ intersect them for $k=1,2,3,...$ with Gromov boundary being homeomorphic to a sphere. Does that object exist in mathematics yet? Moreso interested in classifying certain sub-spaces of this infinite beast.

Thorgott

@Koro think in terms of curves

or note that $S^1$ is oriented as the boundary of $D^2$ and $f$ extends to a function on $D^2\rightarrow D^2$ that is holomorphic, hence orientation-preserving

PM 2Ring

A few weeks ago, in an Astronomy.se question, the OP posted a sphere diagram from an old astronomy textbook. I think the diagrams are from the original edition, published in the 1970s. Back then, technical illustrators used various tricks to draw ellipses. They look a bit quaint these days.

Koro

@Thorgott I'm not sure.

0

Q: Derivative of the map $f:S^1\to S^1: z\mapsto z^n$ is orientation preserving or not?

Given $f:S^1\to S^1: z\mapsto z^n$. I'm trying to show that $Df_p$ is orientation preserving: Let's try to show it at $p= (1,0)$. Tangent space at p is one dimensional and is spanned by basis vector $\partial/\partial y$. I assume $S^1$ to be oriented anticlockwise. I want to show that $Df_p (\pa...

I posted it just in case.

psie

@Jakobian indeed. Would be nice to have an example.

Thorgott

8:14 PM

third suggestion is that $f$ is just given by multiplication with $n$ in the obvious chart

but if you want to insist on inconveniencing yourself, I won't stand in your way any further

Jakobian

@psie an example of such space seems to be disjoint union of two spaces $X$, $Y$, both which are weakly locally compact, first is not locally compact, and second is not locally relatively compact

such spaces $X$ and $Y$ exist so we just need to show that this example works

I think its pretty clear that $X\sqcup Y$ is has those properties, actually

For example, if $x\in X$ and $V$ is a compact neighbourhood of $x$ in $X\sqcup Y$, its clear that $V\cap X$ is a compact neighbourhood - given a covering, just add the open set $Y$ to it to extend it to covering of $V$.

and this should work with compact closures too, with no issue with "closed implies compact" like I made a mistake of earlier

and now I'm tempted to say, what about connected examples? But I really don't want to consider this any longer

psie

8:32 PM

I understand :) let's take a break, but thanks

Alessandro Codenotti

8:42 PM

@Jakobian what kind of space do you want?

Jakobian

@AlessandroCodenotti weakly locally compact, not locally compact, not locally relatively compact, connected

Alessandro Codenotti

I'll regret asking but what are weakly locally compact and locally relatively compact?

Jakobian

weakly locally compact: compact nbds, locally relatively compact: closed compact nbds

Alessandro Codenotti

Oh non Hausdorff spaces. I'm out bye

(what is locally compact if that's weakly locally compact? Because to me compact nbhds = locally compact)

Jakobian

neighbourhood basis of compact sets

Alessandro Codenotti

8:49 PM

Ok so to get a connected example can you not take $X$ and $Y$ as above but look at their wedge? Take the wedge at an $x\in X$ and $y\in Y$ that are not the witnesses for the failure of local compactness and local relative compactness to be safe

Xander Henderson

I really dislike this question:

0

Q: Help Antie evaluate Gauss curvature of a smooth surface using ruler and a protractor

Antie, a smart ant living on a smooth surface $S$ of $\mathbb{R}^3$, would like to evaluate the Gauss curvature $K$ at a certain point $P\in S$. Antie is aware of Gauss Theorema Egregium, according to which Gauss curvature may be evaluated using the first fundamental form of the surface. So, Anti...

differential-geometry surfaces curvature

As a mathematical question, I really like it.

But all the extra fluff is just... annoying and obfuscatory.

Koro

@XanderHenderson why would an ant want to evaluate Gauss Curvature?

What's the purpose?

Xander Henderson

@Koro It is a smart ant, according to the question.

Honestly, I think that the question has most of the required context, but it is buried under so much silliness about an ant that the actual question is hard to see.

I hates it.

Koro

Removing an ant from the picture would not change a thing.

@Thorgott I think I can show that. But why does it show that Df is orientation preserving?

Balarka Sen

@XanderHenderson Disagree.

Upvoted.

Thorgott

9:04 PM

@Koro what's the determinant of multiplying with $n$

Koro

You probably mean determinant of derivative of multiplying with n?

PM 2Ring

That ant probably should be female. OTOH, maybe the boy-ants occupy their time with differential geometry while they're waiting around to mate and die. :)

Koro

if yes, then that's n. But why does it show that Df is orientation preserving?

What result are you using to conclude Df is orientation preserving?

Thorgott

technically true, but the derivative of multiplying with $n$ is the same thing as multiplying with $n$, it's a linear map after all

the result I'm using is the definition

Koro

9:21 PM

no, not linear technically. charts are on open intervals...

Df_p is from T_p S^1 to T_{p^n} S^1

so what definition did you use?

Xander Henderson

@BalarkaSen Sure. We don't have to agree on anything, and I've done nothing more than express my opinion. I hate all the fluff.

Balarka Sen

@Xander How do you explain to someone what an "intrinsic property of the manifold is" without explaining them exactly what a very tiny ant feels while walking on the surface?

I do not know of a single better way of putting it than the ant analogy

My opinion is that your opinion is bad, and your hate is misdirected annoyance at informal language many mathematicians use in (sometimes private, but also public) communication, which is nevertheless quite effective.

Thorgott

9:53 PM

@Koro doesn't make any difference to what the derivative sees

@Koro orientation-preserving :<=> positive determinant

we're at $p=(1,0)$, so this is a linear map from a vector space to itself

Koro

Say $\phi,\psi$ are charts such that $\phi\circ f\circ \psi^{-1}$ is multiplication by n. Its Jacobian determinant is n. Fine.

But why is Df orientation preserving?

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