Mathematics - 2015-04-29 (page 1 of 5)

Mike Miller

12:01 AM

@samuel still working on your foray into rings and stuff?

Samuel Yusim

well I took basically all the undergraduate algebra courses this year so I dunno what I'll do now in that regard

I did groups and rings, fields and galois theory, and commutative algebra

Mike Miller

if you still like it, you can surely work by yourself on something (Karl here read Matsumura's ring theory book), or find a professor to work with you

Samuel Yusim

that's true

Mike Miller

after all, undergrad courses usually only cover math from about a century ago

half-century if you're lucky

Karl Kronenfeld

lol not cover to cover by any means.

also hi @MikeMiller

Samuel Yusim

12:04 AM

I want to try something new, and I have to take differential geometry in the fall, so I'm learning topology

Mike Miller

so you couldn't teach me the theory of derivations of crings, @Karl?

Samuel Yusim

I'd like to see a different flavour of math for a while anyway, I think

Mike Miller

well, I certainly think topology is a good choice

Karl Kronenfeld

@MikeMiller it would be a very poor job. somewhere between rtfm and uh lemme look up the definition

Mike Miller

we covered a little under half in my commutative algebra course last fall

not strictly sequentially

Pedro Tamaroff

12:33 AM

Hello.

How's it going?

Ted Shifrin

12:54 AM

hi mr @Pedro

hi @Karl

Mike Miller

@TedShifrin: If I have a codim 1 foliation, do the leaves all have to be diffeomorphic?

Ted Shifrin

no

Mike Miller

Are we on bad terms again? I apologize for my Beowulf comment.

Ted Shifrin

LOL ... no, we're not on bad terms

Incurrence

@Ted!

Hi

Ted Shifrin

12:56 AM

you don't like it when I give too much away, @Mike. So I figured two letters would do it.

hi @Incurrence

Mike Miller

Fair enough.

Can you tell me the dimension of your example?

Ted Shifrin

1-dimensional leaves

you can make them 2 if you like

Mike Miller

Sure, if you can do it in a surface, just multiply by $S^1$.

Ted Shifrin

Well, that wasn't quite my example ... but close

Incurrence

Is showing that the 3 by 3 Heisenburg group over the integers can be written as a central extension and showing that it doesn't split a classical exercise?

I worry that I will work on this for awhile and the second person doing a presentation on central extensions will also do it

(and if he goes first, I will have nothing good)

Ted Shifrin

1:01 AM

I don't know that particular stuff well enough to comment, @Incurrence.

Disciple of Barney

Why don't you ask what the other person is doing ( I don't think it is a classical exercise) @Incurrence

Incurrence

@DiscipleofBarney I worry I will mention it, and then he will start working on it...

I mean I could ask him and pretend that I have nothing, but then he will judge me for being unproductive

And I will potentionally be working with him on assignments next semester since he is taking algebraic topology, advanced algebra and algebraic methods of physics with me

Mike Miller

Ok, @Ted, I have one on the annulus where the boundary circles are fibers and the other fibers are $\Bbb R$.

Ted Shifrin

right, Reeb foliations ... :)

Disciple of Barney

@Incurrence It sounds like you are worrying about nothing. If it actually matters, maybe you can have the prof mediate, so the prof knows what both of you are doing and can tell you both whether it is the same thing.

Ted Shifrin

1:06 AM

I'm always amazed at how much time people will spend here bothering those of us who don't know what their professors know ... rather than just talking for a minute or two with their professors.

Mike Miller

why are you working on the heisenberg group if you're doing tensor products?

Ted Shifrin

we talked him into switching, @Mike

Mike Miller

well now who's going to tell me what the point of the tensor product of groups is?

Disciple of Barney

He decided tensor products is to hard ( I didn't talk him out of it its all @TedShifrin fault)

Ted Shifrin

most things here are my fault ...

Mike Miller

1:07 AM

@Ted: I bet if you restrict to compact leaves it becomes true.

Disciple of Barney

@MikeMiller Maybe I will look into for you :)

Ted Shifrin

Ehresmann theorem somehow, @Mike?

Disciple of Barney

It sounds interesting

Incurrence

@MikeMiller Hahahaha not much :P

Ted Shifrin

It must depend on the leaf space

Mike Miller

1:08 AM

Your leaf space is going to be nicer when the leaves are compact

You can't get dense leaves eg

Incurrence

@MikeMiller Tensor product of non-abelian groups is extremely localised, and for abelian groups it is uninteresting[from what little I understood :P]

Mike Miller

@Incurrence Yeah but I care about the extremely localized one

Ted Shifrin

you'd need the projection to be proper, and the leaf space to be a manifold ... then ...

Disciple of Barney

Uninteresting?

Mike Miller

for abelian groups it's not terribly difficult, but the tensor product of modules is far from uninteresting

it's the underpinning of huge swaths of mathematics

Incurrence

1:09 AM

@TedShifrin I can't just talk to my lecturer about trivial matters for him...

Mike Miller

yes you can

that's what office hours are for

Ted Shifrin

I talk to my students all the time, @Incurrence ... I don't know why you guys think you have to be geniuses to talk to your professors.

4

Disciple of Barney

It looks. judging from the authors who seemed to have invented it it has to do with higher dimensional van Kampen theorems (Ronald Brown) @MikeMiller

Incurrence

I have anxiety with approaching people for some reason

Mike Miller

I have thought about flagging his posts as spam, @DiscipleofBarney, because that's what they are

Incurrence

1:11 AM

@MikeMiller Mine?

Mike Miller

No lol

Disciple of Barney

Haha, he does seem sort of obsessed and maybe (a little?) self promoting

Incurrence

Oh okay haha I was hurt, and I do try to talk about math

Cbjork

@Incurrence Can confirm @Ted's statement.

I mean I can confirm

Mike Miller

He has never posted anything that's not about his own work or his book @Disciple

It annoys me

Let me find where I read before about the tensor product of groups - it was in topology, but not related to his stuff

Incurrence

1:13 AM

I can't find where you guys are referring to him, I can't find his account

Mike Miller

Ronnie

Incurrence

I see

Mike Miller

@DiscipleofBarney It showed up in this book on rational homotopy theory. In dimensions above 1 the rational homotopy groups are $\pi_i(X) \otimes \Bbb Q$ - so one needs to define this in dimension 1...

Pedro Tamaroff

@MikeMiller What posts?

Mike Miller

All of them.

Pedro Tamaroff

1:17 AM

What user?

Disciple of Barney

@MikeMiller Okay, he is a homotopy theorist so maybe it does have to do with his work.

Ted Shifrin

hi mr @Pedro

Mike Miller

@DiscipleofBarney: I know what he does; it's not. It's showing up for different reasons. He cares about them because something something crossed modules. These guys care because one wants to know what the rational fundamental group is.

If I actually thought they should be flagged as spam, I'd have flagged them, @Pedro.

Cbjork

Hey @KajHansen

Kaj Hansen

@TedShifrin, I feel rather insecure talking to professors about things I think should be trivial. For one, I feel like asking them is neglecting my responsibility as a student to understand as much as I can on my own first. So sometimes I'll sit stuck for hours, but when I finally arrive at the answer I feel like I've learned more than what I would've had I "given up".

Mike Miller

1:20 AM

In any case, I looked at the book just now @Disciple and it's not very satisfying from a group-theoretic POV. It's not interested in the group theory.

Incurrence

So if $G\cong H\times N$, $N\cap H = \{1\}$ and $N,H$ both normal we can take a short exact sequence

$$1\hookrightarrow N \hookrightarrow G \twoheadrightarrow H \twoheadrightarrow 1$$
Or, since they are both normal, we can take
$$1\hookrightarrow H \hookrightarrow G \twoheadrightarrow N \twoheadrightarrow 1$$
Since for both we can take the cannonical projection: $\phi: G\to G/N$ or $\phi_2: G\to G/H$.

Whereas if only $N$ was normal in $G$, we would have the semidirect product rather than the direct product - so $N\cap H=\{1\}$ and $N,H$ subgroups where only $N$ is necessarily normal, we w…

Mike Miller

That's fair, @Kaj. I rarely ask my advisor anything outside scheduled meetings, especially because I can just ask other students first...

Ted Shifrin

sometimes we'll tell you to go think yourself, @Kaj, but often not

hi @AlexW

Alex Wertheim

Hello @Ted :) Busy in here now, I see.

Kaj Hansen

Of course, nowadays, I will always make sure to get myself unstuck in the end, whereas back in the day I'd let my insecurity get the best of me. And I have more mathematical maturity for things so it's easier for me to get unstuck by reading or Googling.

Ted Shifrin

1:23 AM

Grad students have a built-in community, @Mike

Incurrence

Quick verification from anyone on my short exact sequences above?

Mike Miller

My point exactly @Ted

Undergrads build one too, usually

Ted Shifrin

not the ones around here ... except the UGA students really do

Mike Miller

The main issue is when there aren't people to build a community with

Ted Shifrin

or when they're too anxious to talk to them ... which really needs to be fixed

Incurrence

1:28 AM

I talk to my fellow students :)

Ted Shifrin

ok, @Incurrence, that's good

Incurrence

But I don't have many good fellow students

Mike Miller

I think we should have students hold no-holds-barred cage matches, to establish a proper pecking order

Incurrence

@MikeMiller That I would win

Disciple of Barney

@Incurrence Just because you have a normal subgroup, doesn't mean you have it is a semidirect product

Incurrence

1:29 AM

@DiscipleofBarney I meant with the other conditions

@DiscipleofBarney I have $G=NH$ where $N,H$ are subgroups and $N$ is normal in $G$ and $N\cap H = \{1\}$

Disciple of Barney

@Incurrence What other condition?

Pedro Tamaroff

@MikeMiller OK, Mike. Just trying to address your concern. =)

Disciple of Barney

Oh okay

Incurrence

But if I stengthen that to $N,H$ both normal

I can put $N$ or $H$ in the 2nd or 4th position of that short exact sequence

E.g. a direct product

Disciple of Barney

Yes

Ted Shifrin

1:32 AM

@Pedro: Did I miss a hello?

Pedro Tamaroff

@TedShifrin Did you?

Ted Shifrin

I helloed you twice ...

Pedro Tamaroff

How's it going for you, Ted?

Ted Shifrin

finished my last class ever ... bittersweet ... house goes on the market tomorrow.

Pedro Tamaroff

@TedShifrin Did they bake you a cake?

Incurrence

1:33 AM

@TedShifrin And then you move to Straya next door to me, and help me become a mathematical genius

Ted Shifrin

LOL, no ... @AlexW promised me one, though :)

Alex Wertheim

LOL. :)

Pedro Tamaroff

@TedShifrin Good!

Ted Shifrin

I gave you a cake, @Pedro, so you can't complain.

Alex Wertheim

@Ted was afraid I would poison it! I assured him I would never do such a thing.

Ted Shifrin

1:34 AM

@Incurrence: I don't think geniuses need help. :D

Pedro Tamaroff

@TedShifrin Well your house is very nice Ted. It should sell fast. =)

Disciple of Barney

It seems good, was there something in particular about what you said that you don't think works? @Incurrence

Incurrence

@TedShifrin Help me become, and then I will shed you like a butterfly from a cocoon

Ted Shifrin

We'll see ...

I've been shed by pros, @Incurrence.

Incurrence

@DiscipleofBarney No, I was just not comfortable so I wanted to display everything, to make sure there were no chinks

Disciple of Barney

1:35 AM

Okay

Stan Shunpike

@TedShifrin what do the ++ in $\Bbb{R}_{++}^n$ mean?

Ted Shifrin

I have no earthly idea, @Stan.

Look in your book for notation.

Stan Shunpike

Can't. found it on a set of lecture slides.

I would have looked. That was my first instinct. But it doesn't seem to be labeled anywhere. Weird

Ted Shifrin

what's the context?

Mike Miller

@AlexWertheim: But what if you're a bad cook, and you poison him by mistake?

Stan Shunpike

1:40 AM

He's talking about a function that inputs prices and a utility level and outputs a demand vector. Prices are defined over R_++

Alex Wertheim

Well that would bring me great shame, @MikeMiller. But if my cooking is so bad that I can't bake a cake without poisoning someone, perhaps there are bigger issues...

Mike Miller

Mine probably is, @AlexW

Alex Wertheim

I'll remember that for next year, @MikeMiller. I'll know to be suspicious if you ever offer to cook.

Stan Shunpike

I'll go back through earlier slides. Maybe it's labeled somewhere. My class doesn't even use vectors. I kinda wish there was like a math track econ within the honors econ track. You can major in math and specialize in econ, but I don't think that really changes how the econ is taught. The econ just doesn't use enough basic math tools imo. But other people seem content so I'm probably in the minority

Disciple of Barney

Well $ ^+ +^+ _+ \Bbb R ^+ {++}_{+_+++}$ means finite subsets of $\Bbb C$, so you can figure out what that notation means from there :) @StanShunpike Do you have a link to the notes?

Stan Shunpike

1:43 AM

Yeah i do hang on.

Mike Miller

I can cook a delicious pasta sauce, @Alex, and popcorn. Be wary of any brownies I cooked unsupervised.

Stan Shunpike

pitt.edu/~luca/ECON2100/lecture_06.pdf

Page 2

Alex Wertheim

How do I know you're not just trying to winnow the number theory crowd to plot the ascent of the topologists, @MikeMiller?

Mike Miller

A new topologist is coming next year and the year after, @AlexW. I don't need to plot our ascent; it's already happening.

Alex Wertheim

Hmm, good point. Ok, I guess I can believe that then. :)

Mike Miller

1:52 AM

What are you working on today, @AlexW?

Alex Wertheim

Unfortunately, @MikeM, the thing I've most been concerned with has been keeping fever down. I've got the stomach flu, sadly.

Mike Miller

Oh dear. I wish you the best.

Alex Wertheim

Thanks! I'm already feeling much better, thankfully. I hate sitting around doing nothing.

To give a more mathematically meaningful answer, I've moved onto chapter 2 of Atiyah-Macdonald. I've just come to the section on tensor products of modules.

Mike Miller

Good progress. Be sure you really have those first 2 or 3 sections down, as they're essential for A-M, the qual, and life in general.

You wouldn't believe how many days are ruined because the local butcher forgot that localization is flat.

Alex Wertheim

LOL. I'm certainly trying. I'm taking pretty copious notes, and have done almost every problem from the first chapter, so I hope I'll have learned something at the end of all this. :)

I'm particularly excited for chapter 2 because I've been particularly ignorant about both modules and tensor products for far too long.

Mike Miller

1:57 AM

Remember to think about them in terms of what they do (the universal property), not how you construct them.

Alex Wertheim

Will do! A&M actually make a funny (at least I found it funny) remark to that effect: "We shall never again need to use the construction of the tensor product given in (2.12), and the reader may safely forget it if he prefers."

Mike Miller

For good reason.

Pedro Tamaroff

@MikeMiller I have to print flyers with that.

And throw them in every Algebra II class.

Alex Wertheim

I really digging commutative algebra. I was enjoying topology as well, until I came to a problem that asked me to prove a particular map is continuous. I hate proving that explicit maps are continuous. =P

Pedro Tamaroff

@AlexWertheim Are you taking a commalg course?

Alex Wertheim

2:05 AM

No @Pedro, just trying to study some before school starts in the fall.

Pedro Tamaroff

Ah. A&M is definitely a good source. You can also peek at Eisenbud's book.

It has some good stuff.

Alex Wertheim

Indeed, it's great. I love the problems. I've heard good things about Eisenbud and Matsumura, with the latter probably being a bit advanced for me right now.

pjs36

You're making me want to pick up A&M... I've been meaning to do some algebra, but got caught looking at Berger's geometry books.

Pedro Tamaroff

@AlexWertheim Hehe, Matusumura has some serious mathematics in his book. I've peeked at them, but never really read the book. I think Pete Clark has some notes on commutative algebra, too, since he gave a course.

Alex Wertheim

@pjs36: do it! It's a fabulous book, in my opinion. Of course, take that with a grain of salt, since I'm not very far in it. :)

Mike Miller

2:14 AM

Nobody should be saying good things about Matsumura. There's no motivation or geometric insight.

pjs36

@AlexW Haha I just might; we can be study buddies then. I'll badger you, since you'll be at least a chapter ahead :P

Alex Wertheim

@Pedro: indeed, I'm hoping to get acquainted with it at some point. Dr. Clark's notes are great, as always. I'm a big fan of both his and Keith Conrad's expository materials.

Mike Miller

It's a great reference, not a great source to learn.

Pedro Tamaroff

@MikeMiller Who did? =)

Alex Wertheim

@pjs36: please do! It'll help with keeping on track. I'll be much more productive in a week or so, once I've left my job.

Mike Miller

2:17 AM

@PedroTamaroff: Just being safe.

Alex Wertheim

@MikeMiller: interesting, that's a shame. Do you know of any better books? Matsumura was described to me as the gold standard, so that's rather disappointing.

Mike Miller

It's the gold standard of references. Everything is in there.

It's just that it's hard to learn from.

pjs36

Anyone have an opinion on Isaacs's Algebra? I imagine it's heavily geared towards character theory, but I've never managed to make it very far.

Mike Miller

I don't

Anthony

2:35 AM

@MikeMiller Ayo.

Mike Miller

hey

Anthony

2:56 AM

@MikeMiller Sorry to be dumb, but I'm actually still confused about what you said last night. Do you have a minute?

Pedro Tamaroff

@Anthony Hey there.

What is it?

Anthony

Ayyyy.

It was a question about quadratic residues... Certainly not suppose to be confusing lol.

Pedro Tamaroff

OK?

Anthony

I'm just supposed to find the quadratic residues in $\mathbb Z_{77}$ that only have two roots.

My professor said part of it would be computational, but I'm just confused on the general form of quadratic residues mod $\mathbb Z_{pq}$ with only two roots.

I don't know the relevant theorems about any of this, but I got the feeling they would be something like the QR in $\mathbb Z_p$ times q, and the QR in $\mathbb Z_q$ times p. But that's not right...

Pedro Tamaroff

@Anthony Do you know about the Legendre symbol/character?

Anthony

3:04 AM

No.

I can look it up.

Pedro Tamaroff

Sorry, too much Real Analysis for me.

I meant Legendre.

Anthony

lol

Still no.

Pedro Tamaroff

Oh. It's a useful thingie.

But you do know CRT, yes?

Anthony

I know of it- for some reason I've never had to use it in my classes.

I'll look at the both of them, this stuff is kind of tangential to what we've been doing in class, so I don't think he was expecting us to prove anything, but I'm certainly thoroughly confused now.

Thanks!

Mike Miller

I don't have time to explain, but I'll repeat what I said to @Pedro so he can explain. :)

The point was to find what elements of $\mathbb Z_{77}$, not coprime with 77, are quadratic residues.

I was claiming they're quadratic residues iff they are quadratic residues mod 7 and mod 11.

The forward direction is thus. If $a$ is a residue, $b^2 = a$ mod 77. Take this mod 7 and mod 11 to see that $a$ is a residue mod 7 and 11.

I did not give details in the other direction.

Anthony

3:14 AM

Thanks @MikeMiller.

Does the other direction use the CRT?

Mike Miller

Yes.

Anthony

Ugh.

Thanks.

2mkgz

3:35 AM

Does anyone have a conjecture as to why this question has 15 upvotes and 11 favorites?

pjs36

I could see the favoriting (if you're into Diophantus), but the upvotes are a little weird. I guess people just love textbook problems?

Pedro Tamaroff

@2mkgz People are enticed by what they don't understand, but they know some words about.

2mkgz

Hm. If I post "Solve x^2+3xy+y^2 = ((x+y)/5 + 2)^4" will it also get 16 upvotes?

Disciple of Barney

@2mkgz Try and see :)

Maybe I will post "Solve $x^2+3xy+y^2 = ((x+y)/7 + 2)^5$" after

Incurrence

4:07 AM

@2mkgz Coz it has two squares and the middle one is like both of the variabeelz

Trufa

Either something was wrong with math or something was wrong with my brain, guess which one it was.

Incurrence

On a serious note, because it looks simple enough to be interesting to non-math folks. Does someone have a link to that post that explained downvotes via a story about someone, I believe it was, losing their keys, I haven't been able to find it

@Disc It seems that the definition of having a split extension is strongly connected to semi direct products?

E.g. if $G$ is a split extension of $H$ by $N$, then we equivalently have $G\cong N \rtimes H$

and perhaps if $G$ does not split, and we have $1\hookrightarrow N \hookrightarrow G \twoheadrightarrow H \twoheadrightarrow 1$, then $G\cong N\times H$? E.g. they are the direct product?

Also @Disc, why did you never respond(or maybe you deleted it) to the answer to your question on math overflow?

Disciple of Barney

Direct product is a special case of semidirect product. @Incurrence (maybe you should prove that)

Incurrence

@DiscipleofBarney indirect product = semidirect product?

Disciple of Barney

lol

Incurrence

4:15 AM

Haha

Disciple of Barney

What do you mean about the MO questions?

Incurrence

Special case when you take away the normality of one of the subgroups

@DiscipleofBarney Yoy never responded to the post that answered your question

Disciple of Barney

What should I have said?

Incurrence

I dunno, normally I thank people, and more normally I have follow up questions

I want to do the following exercise, so please don't give me any hints, I just want to clarify one thing:

Wait one sec, how do I put something on top of something else in latex?

Like putting $\phi$ over $\to$?

$\overset{\phi}{\to}$

Oh wow lol

Must have used that once, since I just tried it seemingly randomly

Okay

$$0\longrightarrow \Bbb Z \overset{.2}{\longrightarrow} \Bbb Z \longrightarrow\Bbb Z / 2 \Bbb Z \longrightarrow 0$$

Disciple of Barney

Yah, I could have thanked YCor I guess, maybe should have, although this disagrees with that practice.

Incurrence

4:19 AM

What the heck does $.2$ mean there

Disciple of Barney

Its times two, \cdot, so $\cdot 2$

Incurrence

Hmmm

Oh that was silly of me

Mike Miller

I think one shouldn't thank. Having elaborative questions - you don't understand something in the answer - is fine. Asking more questions is, imo, usually not.

Incurrence

@DiscipleofBarney Wow, my bad

Look at any answered and accepted question of mine...

Oh god, there are two comments on the star wall about me having bad study habits...

Disciple of Barney

It is not the worst practice, and I am pretty sure different communities have different preferences on that rule.

Haha

Mike Miller

4:23 AM

Three, if you count the rabbit hole comment.

Incurrence

@MikeMiller True :S

Well I dropped tensor product of finite groups for these reasons I guess

I really need to learn module theory first I think, and that won't be for awhile I imagine

Disciple of Barney

Did you read the answer to my MO question?

Mike Miller

great answer

admittedly I thought the answer was obviously yes when I first read it

Incurrence

Yep

I read it three times I believe

Disciple of Barney

Yah, totally agree, plus more than my analysis classes I had, it got me thinking about how useful that idea is

@Incurrence And you didn't upvote it? That is the way you say thank you in SE-speak

Incurrence

4:30 AM

A natural embedding is a monomorphism?

What exactly is an epimorphism? A cannonical projection?

@DiscipleofBarney I didn't because I wasn't automatically logged in

Happy? I linked my account just to upvote

Disciple of Barney

@Incurrence It depends on your category you are working with. A natural embedding will just be the "obvious" injection (the identity basically) and the canonical projection, depends what you are talking about but I am guessing in your context it is $a \mapsto [a]$ or $a \mapsto aN$, or whatever

Incurrence

Indeed

So $1\hookrightarrow N \hookrightarrow G \twoheadrightarrow H \twoheadrightarrow 1$

Is two monomorphisms, and then two epimorphisms

Mike Miller

I don't understand why canonical is showing up here

Incurrence

where $G=NH$ $N,H$ are subgroups

Mike Miller

Epimorphism has a technical definition; canonical does not

Incurrence

4:34 AM

and $N$ is normal in $G$. and $N\cap H=\{1\}$

Mike Miller

In Grp, monomorphisms are just injective homomorphisms, and epimorphisms are just surjective ones

Incurrence

Ok

I suppose the first case of these monomorphisms and epimorphisms(when proving the semi-direct products equivalent definitions) they were referred to as a natural embedding and a natural projection, and some definition of the canonical projection seemed equivalent to the definition of natural projection

Pedro Tamaroff

You should try to come up with an epi in $\rm Top$ that is not surjective.

Incurrence

So I guess I lost marks on that part

Pedro Tamaroff

Think about Hausdorff spaces and dense subspaces.

@Incurrence Don't torture yourself. =P

@MikeMiller I cannot count.

Mike Miller

4:36 AM

@PedroTamaroff: No. Epimorphisms in Top are surjections.

Epimorphisms in HausTop can fail to be surjective.

Incurrence

@PedroTamaroff Okay, give me 6 months for when I am doing Topology :P

Is that not Category: HComp

Mike Miller

No.

Disciple of Barney

What is HComp?

Pedro Tamaroff

@MikeMiller Yes, yes.

Mike Miller

presumably compact hausdorff spaces

Incurrence

4:38 AM

Indeed

Disciple of Barney

Okay

Mike Miller

@PedroTamaroff I think this is an important point.

Incurrence

(I just so happened to be on the wiki page for epimorphisms when they started talking about these things, and HComp was the category listed directly under Top)

Remember me

Hi@Incurrence

Incurrence

@MikeMiller This is true in Grp and FinGrp?

@Rememberme Hello

Mike Miller

4:41 AM

Yes, and in AbGrp.

Incurrence

That is abelian groups?

Mike Miller

It's not true in CRing or HausTop.

yes

Remember me

I need a small help from you @Incurrence if you are free?

Incurrence

@Rememberme Hmm I am free if it is short and sweet :)

Remember me

It is about linear independence

Disciple of Barney

4:42 AM

Its true in Grp (epi iff surj, monic iff inj)

Incurrence

Monomorphic = Monic here?

Disciple of Barney

Although the epi implies surj in Grp is nontrivial

Incurrence

Oh that is a standard shortening

@Rememberme Sure

Disciple of Barney

@Incurrence Yes

I guess to be consistent I should have said epic

Incurrence

I prefer the full word :-)

user143442

4:46 AM

How to prove that if $X$ is compact and Hausdorff, then the cone of $X$ can be visualized as the collection of lines joining every point of $X$ to a single point?

Pedro Tamaroff

@user That's not really something you want to prove.

user143442

I found something similar here math.stackexchange.com/questions/204684/… but it's not exactly the same problem

user143442

do you think I could define a similar homeomorphism?

user143442

@PedroTamaroff why not?

Pedro Tamaroff

That statement is just saying "you can draw intuition from the drawing."

user143442

4:50 AM

unless I prove that the set of those lines is homeomprphic to the one pont compactificacion of $X \times [0,1)$, then I can use that answer

Pedro Tamaroff

But it is not something you "prove".

The point is that the drawing might not faithfully represent what goes on with the cone -- say for the cone of $\Bbb Z$.

user143442

no, but it's actually a problem I need to prove

user143442

but $\Bbb Z$ is not compact

Pedro Tamaroff

The cone has a topology, and this topology might not coincide with the topology "you think" a drawing of your spaces and lines joined to a point has.

user143442

if $X$ and compact and Hausdorff, I take a point $p \notin X$ and then the set of all lines from all $x \in X$ to $p$

user143442

4:52 AM

we can suppose that $X \subset \Bbb R^n$

user143442

and give this set of lines the usual topology

Disciple of Barney

I am still curious as to whether the case for countably infinite generated groups could be strictly between $\aleph_0$ and $2^ {\aleph_0}$.... I think there must be... By the comment of Emil Jeřábek we would have to have $\aleph_1$ normal subgroup which is really interesting.

user143442

of $\Bbb R^{n+1}$

user143442

then it's homeomphic to the cone

user143442

$X \subset \Bbb R^n$ and $p=(0,0,...,0,1) \in \Bbb R^{n+1}$

user143442

4:58 AM

$K(X)\subset \Bbb R^{n+1}$, $K(X):=\{l_{px} : x \in X\}$, where $l_{px}$ is the segment of line going from $p$ to $x$.

user143442

then $K(X) \cong Cone(X)$.

user143442

if $X$ is compact

user143442

where $Cone(X)=(X \times [0,1]) / (X \times \{1\})$

Balarka Sen

5:35 AM

@user joining lines only makes sense for cones in $\Bbb R^n$. indeed, in $\Bbb R^n$, the geometric cone and the topological cone coincides, upto homeomorphism.

@JC574 yes. in fact, CW complexes in general are locally path-connected.

Remember me

5:48 AM

@BalarkaSen I don't get the question I asked you yesterdat I wrote them using Euler's formula and it gave me that $f_2(x)=e^{ix}=cosx+isinx , f_3(x)=e^{-ix}=cosx-isinx$

Balarka Sen

what is $x$ here, in particular? does the question say anything about it? is it a real number, a complex number, an integer, or what?

Remember me

Its says V is a vector space over the complex numbers of all function from R into C

Balarka Sen

ahh.

so that's an important information.

@Rememberme ok, so why can't you prove that $f_i$s are independent? just use the definition.

$f_1, f_2, f_3$ are all considered to be functions from $\Bbb R$ to $\Bbb C$, note that.

Remember me

Ok I have show that there exist scalars a,b,c such that for all a,b,c a+bf_2+cf_3=0 and a=b=c=0

Balarka Sen

no, wrong quantifications.

you have to show that for any scalars a, b, c, if af_1 + bf_2 + cf_3 = 0, then a = b = c = 0.

@Rememberme still false.

Remember me

5:58 AM

Ok fine then

Balarka Sen

no, absolutely not fine. you've got to get used to using correct quantifications. haven't you read about them in Hammack?

Remember me

I read them but I take some time to get them right

Balarka Sen

do the exercises in Hammack.

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$Mathematics$ Mathematics