Hmm, I did not know that.

Replace $\rho^*$ with $\rho$.

And this book does not seem to prove it.

Well, if you go back to yesterday's transcript, you'll see I tried to say that to you once or twice.

I know!

And the whole time I've been asking for why that's true!

I don't have Jost's book here, so I can't cite it.

@TedShifrin This is literally what he says in the proof text.

But he does not prove it, I want to know how to prove it.

It's proved in Chapter 3 of DoCarmo's graduate book.

Geodesic convex neighborhoods ...

Thank you, that's all I wanted.

Even though I looked there already...

I don't like it personally, @Jasper, but it's a standard reference. I have always done a mish-mash of Spivak, Chern, and other sources for my grad courses.

I prefer to base everything on differential forms, and doCarmo pretty much avoids them.

Also, @0celo, look at Theorem 14 in Chapter 9 of Spivak Volume I.

@TedShifrin Is Theorem 3.7 in do Carmo what I need?

page 72

His notation is a bit different, I'm trying to translate it

You need Proposition 4.2 on p. 46-47.

Oh, 3.7 should do what I said, yes.

@TedShifrin In?

Proposition 4.2 makes a stronger statement about distance-minimizing geodesics staying inside the normal neighborhood.

Yes, Proposition 3.7 in doCarmo is what I was saying to you, yes.

46-47 is a bunch of exerices...

are you talking about Spivak?

oh sorry, I mistyped ... I mean 76-77. Apologies.

Ok, thanks.

I'm not sure how Jost expected the reader to come up with this though.

It's really just inverse function theorem stuff on the tangent bundle.

@Paradox101 The fraction part is $\frac{1+\frac ax}{1+\frac bx}\to1$. The sine part is $\sin^2\left(\frac1x\right)\to0$

Your manifolds background may or may not be everything you need for Riemannian geometry.

@TedShifrin Maybe...

hi

hi @TedShifrin

@TedShifrin !!

So here's a question. What are the most important "simple" definitions in math? For example, limits, groups, and compactness are all pretty simple and very useful. What would you add to this list?

@TedShifrin what does that mean?

@Twink You're in over your head mathematically.

@0celo7 I already read so

But English is not my first language and I don't understand the expression

Ok I already found it dictionary.cambridge.org/es/diccionario/ingles/…

Thank you @TedShifrin but I don's see math as a diffucult situation that I cannot get out of, I see it like a challenge and I like to learn everyday to become better, but thank you for your encouragement, I bet your students were always very enthusiastic having a so motivating teacher

Can anyone explain why my question was closed as a duplicate?

I made it exceptionally clear how it differentiated from the chosen duplicate.

The concepts within the answers there are far too complicated for most 8 year olds to swallow, nevertheless someone's who just started learning about multiplication.

The question was aimed at how to explain to kids something that would take a teenager/adult to understand fully.

Perhaps, you should try MathematicsEducators.SE

^

@Paradox101 No, it doesn't, because the graph always approaches 1 on the right side. Pay attention to the graphs!

Hey @anon

heya

Just a question

If I consider the following paths in $X = R^2 - {0}$ and consider $f(s) = (cos(\pi * s),sin(\pi s))$ and $h(s) = (cos(\pis),-sin(\pi *s))$, so here we have two circles in opposite directions

stupid latex

consider $f(s) = (\cos(\pi *s),\sin(\pi * s))$ and $h(s) = (\cos(\pi * s),-\sin(\pi * s))$

@KarimMansour latex cannot be stupid

so something/someone else here has to be...

:)

:P

anyway

the reason we can't deform f(s) to h(s)

protip: \sin, \cos make the operators into, well, operators

$\sin x$ vs. $sinx$

cool

that comment is offensive

@Twink I'm sure it is.

What? It's never just nice to see me?

I'm just here so people can get in and out of the room

the reason we can't deform f(s) to h(s) is because we have to first deform f(s) to a point and then go back to h(s) since they are in opposite directions, but 0 isn't in the domain and so we would have a discontinuity so f(s) and h(s) can't be homotopic

Guess there's no need to share these cookies then.

correct ?

@anon?

intuitively speaking

I don't even know who that is

@Doorknob I tend to shoot doors open with my 12ga.

How do you close them then?

@KitZ.Fox Carefully, of course.

@0celo7 door murderer! >:O

@KarimMansour intuitively I suppose

How was I lured into the midst of a topology discussion.

@Doorknob I generally tend to shoot the knob/lock area.

remind me to stay away from @0celo7...

@KitZ.Fox Impressive that you know topology when you see it

Okay, come on, stop flagging things. Flags are for serious issues only.

@0celo7 Really? Why is that?

Stuff is still being flagged?

@AlexClark prove it

I haven't trolled today

Well, OK, but if someone has questions about topology, that's really a first order kind of discussion.

@KitZ.Fox My dad has an MS in applied physics and had never heard it until I mentioned it to him.

And I am loathe to interrupt it with my usual pleasantries.

And judging by your profile you're not a STEM guy

Seriously?

@0celo7 You shouldn't judge me by my profile.

Why are all the mods in here now

Because.

Because flags.

Somebody's been bad

Multiple flags?

Seriously?

Also, I mentioned that I brought cookies.

@0celo7 Flags summon every 10k+ user and moderator on the network (who's on chat at the time).

I know

That, and I was interested in seeing if the math folks were talking math again.

@0celo7 all on you, yes

But how are there flags

@anon For Pete's sake

I'm going to get banned now

They don't even belong here

infer what you will

Whatever

Mods should be mods only in their rooms

@HDE226868 They were! About topology! And now somebody is having trouble with a difficult subject and it's buried.

they shouldn't have power in other rooms

@Twink Are you suggesting that I get a room?

@KitZ.Fox Ooh, this bit's over my head.

@Twink Take it to Meta. Oh wait. It's already there.

you already have one

Um?

This is our room.

How many flags are there?

We're allowed to come here out of free will.

And cookies.

I didn't say you're not

Look, the important question here is whether @Karim got his question answered.

but you shouldn't have power here

@RobertHarvey Ooh! Chocolate Chip!

here you're none

@RobertHarvey Oh, yeah. Here. Share them around.

nom nom nom

@Twink You said your other account got a 12 hour ban less than 12 hours after it got banned

So you're either lying about being banned or lying about being another user

Or you evaded

@0celo7 I'm sure the math mods are aware of any situation.

Psh. Nothing to see here.

And terrifically well-equipped to deal with it.

In other news, your site confuses me. All those HINT answers, and boy you'd better know the mathspeak.

@KarimMansour ^^ Yes or no? I feel bad that we kind of interrupted.

yes

I got it answered :D

OK good. Is this your first semester of topology?

@RobertHarvey the majority of the answers that begin with "hint" are full answers

Great.

I blame Dubuque

There are a lot of mods in here...

Seems as though Twink landed himself a 24 hour suspension.

@anon Like I said. Confusing.

@robjohn Hey! Have a cookie.

@0celo7 Probably for disrupting the learning environment.

@KitZ.Fox what's the buzz?

@robjohn Just now noticed?

@0celo7 Yes. Now it would be ideal if the conversation returned to math-related topics...

@AlexClark That's crazy

@Doorknob ok

someone explain to me what a topos is

@RobertHarvey just now got here.

looking at some of the flags that seem needless that have been cast in the last while

@0celo7 You mean, like the toothpaste?

@KitZ.Fox sure

what's a toothpaste

Wait, that's Topol.

<--- needs teeth brushed

A topos is a three dimensional projection of a surface.

IIRC.

<--- tries to kill teeth brushers

I can't remember how to define the sets for one though.

@0celo7 don't bite the tooth that brushes.

@KitZ.Fox yes

I am taking next semester algebraic topology

this semester point set topology

something something something field equation something something something dark side something something profit?

@KarimMansour That's a lot of topo. What's your field of study?

but this thing I am doing right now is for my project which will proving Ulam Borsuk theorem in dimension 2.

I am undergrad

last year

Oh. I'm quite impressed.

@robjohn that makes no sense

do the teeth brush each other?

or do my teeth brush his teeth?

@0celo7 teeth can brush against things...

wise words of wisdom

also holy cow there's a lot of gum under this desk

Don't chew it.

I have my own gum

not pre-chewed

I wonder if there's a market for pre-chewed gum.

hmm

for y'all's info Costco has really good prices on bulk gum

@user You weren't banned by mods of another chat room.

Post about it on your Meta.

$10 for a 10 pack of 5gum

Don't use a sockpuppet to come back in and immediately complain.

is @user = @Twink ?

yes

Are we just allowed to switch accounts whenever we get suspended??

brb making a second account

@0celo7 $$\huge\text{No}$$

@0celo7 Wait, I have to suspend you first.

@KitZ.Fox no

What have I done to you

What?

so @anon I am looking how they defined something called groupoid on certain elements of equivalence classes of the homotopy classes

For the record, use of a sockpuppet account to take actions that would normally be prevented with a single account is in clear violation of network-wide policy and can cause the second account to be nuked

Oh. You mean "no!!1!"

@HDE226868 $$\mathfrak{Nay,\,Canadian,\,I\,shan't\,obey}$$

so what is the difference between group and groupoid ?

it seems very similiar

a groupoid only has a partially defined binary operation. if a group is a category with one object and only isomorphisms, then a groupoid is a category with only isomorphisms but any number of objects. the fundamental groupoid is I think the best illustration of what it is supposed to be.

@KitZ.Fox I meant "no11!1"

get it right

Dude. pssht.

i see

I'm outta here. Let you guys get your maths on.

@HDE226868 $$\Huge\mathfrak{Yes}\huge\mathfrak{Yes}$$

is there a bigger thing than \huge?

oh I see

liar

so I guess it is defined on some elements not all elements

while a group is defined on all elements

\Huge works

Seriously

@AlexClark Will they, though?

If this has been going on for 2 years...

How many socks do you have, anyway? Let's take inventory.

2 on me

^^

@AlexClark anti-gay, don't be lame

@anon dude, I got banned for "big b**ty"

I'm taking NO chances here

this place is crazy strict

:25290671 Please stop using sockpuppet accounts to circumvent system restrictions. Your accounts will continue to be suspended for as long as your main account, and possibly removed.

I think his account is gone

The post won't load properly

... okay, yeah, someone nuked it

@AlexClark the sylow q-subgroups of a p-group are trivial (if p=\=q) or the whole group (if p=q). proper subgroups of a p-group are not sylow p-subgroups.

I'm scared now

@0celo7 Harsh room.

You should try EL&U. Pretty much anything goes there.

well, that's what I don't understand

sometimes anything goes

other times someone flags something and people get suspended

it's crazy

I will never understand stack exchange moderation

It's timing.

And most of your flags look pretty bad in isolation.

That's up to the site mods.

@KitZ.Fox Wait, what?

Are you talking about the ones today?

@0celo7 Nothing.

Or am I infamous?

No. Definitely not.

Well, can you please tell me what got flagged?

If I did something that "looks pretty bad"

Should I not be made aware of it?

You know what got flagged. You mentioned it just now.

The BDSM allusion?

I'm just yanking your chain, trying to be friendly fredley. Sorry. Let's talk maths!

I genuinely don't know if you were yanking my chain or just avoiding this now.

I was just goofing. I'm writing right now and it makes me a bit...unconscious of what I'm saying to other people that might then be misconstrued.

Especially with online chat, I tend to forget that I'm not writing dialogue, so I'm not actually in control of both ends of the conversation.

So the lesson here is that I should close this chat window and go back to writing.

In an effort to maybe steer the conversation back to something math-related, I'll rehash an earlier chat question of mine. Maybe someone new can answer it, although I feel like it's really simple and I'm just being a bit thick:

@HDE226868 $c\in\Bbb R^3$?

I have a question! :0

@anon Whoops. Edited. I was used to the $^3$ too much.

omg, so much better with LaTeX rendering, thanks @robjohn.

anyway, yes, $\Bbb R^3\setminus\{(0,0,c)\in\Bbb R^3:c\in\Bbb R\}$ is all points except the line $z=c$

@anon Okay, thanks. Interpreting it separately from the definitions of related objects made it make more sense.

@PerplexedGuest sounds like stirling numbers

Haha yes thank you @anon! :D

Hm, there is an explicit formula for Stirling numbers of the second kind, which is fantastic, but they count the number of ways to place $n$ labeled objects and I need unlabeled ... I could probably do that on my own. P:<

are the places to put them distinguishable or indistinguishable?

Indistinguishable. P:

distinguishable: integer partitions. indistinguishable: integer compositions.

also look up: stars and bars (combinatorics)

Oh my god it's just ${n-1 \choose k-1}$ ...

That's going to help out my research! xD

Thanks so much!!!

np

But wouldn't that also be the $J(g)$ for $g$ acting on $M_n(\Bbb R)$ by right multiplication? Which can't be right, because $G={\rm Aff}(1,\Bbb R)$ has disagreeing left and right invariant measures.

@TedShifrin I'm contradicting myself over Haar measures. If $G\subseteq{\rm GL}(n,\Bbb R)$, the left-invariant measure should be $\int_S J(g)^{-1}\prod dx_{ij}$ where $J(g)$ is the Jacobian determinant of $g$ acting on $M_n(\Bbb R)\cong \Bbb R^{n^2}$ by left multiplication. Since $g$ acts on columns independently, and the Jacobian determinant of it acting on a single column vector is $|\det g|$, we should have $J(g)=|\det g|^n$.

err, $dg_{ij}$, whatever

hmm, if G has lower dimension I don't think $\prod dg_{ij}$ makes sense, since that's $n^2$ differentials

@anon then if I have to find lims sup of $\frac{x+a}{x+b} sin^2(\frac{1}{x})$ I simply check the limit of the whole thing?

Hey@TedShifrin

@anon: The left-invariant one-forms on $M_n$ are the entries of $g^{-1}dg$. When you wedge them together, there's some serious non-commutativity going on.

hi @Remember

@anon: For subgroups, you have to restrict (pull back) to the subgroup. So it's a nontrivial computation.

And, anon, you have to use the linearly independent ones once pulled back to the subgroup. Consider $O(n)\subset GL(n)$, for example.

@Paradox101 I guess you have to first find the sup and then take the limit. For example $\lim_{x\to 0} sup\sin(\frac{1}{x})=1$

Self avoiding walks are so ... freaking ... complicated ...

I've been gone for a while, but I was hoping someone here could explain what's happening in a circular convolution in this step $\mathcal F_n(c)\mathcal F_n(x)$

What kind of multiplication is this? Is it dot-product?

what?

@anon so I am currently reading on the product and box topologies

their is something I don't understand in the definition of

infinite cartesian product

Let Let J be an index set $\{A_{\alpha \in \mathbb{J}}\}$, so the infinite cartesian product is defined to be the set of all functions $x : \mathbb{J} \rightarrow \bigcup_{\alpha in \mathbb{J}}$ such that $x(\alpha) \in A_{\alpha}$ for each $\alpha \in \mathbb{J}$

\{\}

so it is all set of functions, so we know given a function it is a subset $J \times U_{\alpha \in J}A_{\alpha}$

but what I don't understand why this definition reduce to the one for finite cartesian product

a tuple $(x_1,\cdots,x_n)\in A_1\times\cdots A_n$ is a function $I\to{\rm blah}$ where $I=\{1,\cdots,n\}$ and $x_i\in A_i$ for each $i$

I see but this after we apply a mapping while for example $x \in A_1\times ... A_n$ in normal cartesian product means that x is either one of them.

so isn't there some distinction ?

huh?

I mean for regular definition of finite cartesian product we have that for $x \in A_1 \times ...A_n$ means that $x \in A_1\ or \ x\in A_2\ or \ ... \ or \ x \in A_i$

no

like one is defined in terms of maps and the other is defined in terms of just element

this is wrong

you're confusing cartesian product with union

very different

nvm

yeah my mind confused union with cartesian product for some reason !!!