morning

morning, just got back from grandma's

@AlexClark are you there?

can someone explain how free groups are "particular examples of coproducts"?

the free group on $n$ letters is the coproduct of $n$ copies of $\Bbb Z$ in Grp

okay, I'll try to make sense of that

thanks

try it for $n=2$ first

okay

that'll give me the free group on two generators, right?

yes

Goodnight @Mike

hm

morningg @MikeMiller

hello

why is it true that a quotient map $q: X \to Y$ must send saturated open sets of $X$ to open sets of $Y$?

what's a saturated open set

oh, I have no idea what definitions are well-known, sorry

a subset $U$ of $X$ is saturated w.r.t. $q$ if $U = q^{-1}(V)$ for some subset $V$ of $Y$.

oh, that saturated open sets are sent to open sets should be the definition of the quotient map. what's your definition?

...yeah, that's all it is here too

I really need to get my act together with quotients

it takes a while for me to internalize new concepts usually

well like, subspaces, products, and disjoint unions were all alright for me

i agree that quotient maps are much harder to intuit

Question. So suppose $\lambda \neq 0$ is an eigenvalue of $A$. This book claims that the set of eigenvalues corresponding to $\lambda$ along with the vector $0$ form a subspace of the column space of $A$. Is that even right?

you should really try to think of $X \to X/A$ as the canonical example of a quotient map. sometimes important quotient maps can be a little harder than that, but not usually

and again quotient maps can be mad screwed up, $[0,1]^2$ is a quotient of $[0,1]$

yeah, how's that possible?

pick a space-filling curve $f: [0,1] \to [0,1]^2$, like the Peano curve. this induces a continuous map $[0,1]/\sim \to [0,1]^2$, where $x \sim y$ if $f(x)=f(y)$, essentially by the definition of quotient space. because $[0,1]/\sim$ is compact and $[0,1]^2$ is Hausdorff this is a homeomorphism (again, you'll learn this later; it's not a deep fact, follows with a little thinking from the definitions)

so if you can write down a continuous surjection $[0,1] \to X$, $X$ hausdorff, then $X$ is a quotient space of $[0,1]$. this includes all compact connected manifolds... so, for instance, spheres of every dimension

Is this topology?

yeah

Oh.

@SamuelYusim: somewhere on mse there's a question about visualizing this quotient map. the answer, IIRC, was only a little better than 'lol good luck'

Anyone have any thoughts on my question? I don't think that's possible, since that would HAVE to mean that the column space of $A$ is a vector space, which obviously isn't always true... I think?

wow, sucks to be me then

@Clarinetist: the column space is defined to be the span of the columns of $A$. the span of some set of vectors is automatically a vector space

@SamuelYusim: don't worry about quotient maps that aren't nice.

not really any need to.

I'm slowly being convinced that topology is the most messed up human endeavor in history

meh

"Point-set topology is a disease from which the human race will soon recover." - can't remember who

I thought that quote was about set theory

Nope. Poincare, it seems.

link.springer.com/article/10.1007%2FBF03024067

Never said, according to Wikiquote.

no, the actual quote is about set theory, and also probably was never said

oh, samuel beat me to it

Okay, utterly stupid question. Why is it that the eigenspace mentioned above is a subset of the column space of $A$?

looks like the point-set topology one is also listed as never having been said

there's no need to call a question stupid. remember that $v$ is an eigenvector if $Av = \lambda v$ for some $\lambda$. note that $Av$ is in the column space; and hence $\lambda v$ is in the column space; and hence $v$ is in the column space provided that $\lambda \neq 0$.

@SamuelYusim: if it's not clear to you, by the way, the things topologists are interested in are not these sort of pathological ideas. the modern-day geometric topologist works with manifolds and maps between them; the modern-day algebraic topologist works with CW complexes (aka 'nice spaces') and maps between them. occasionally they work with related spaces. but it is very rare that someone thinks much about, say, quotient maps $[0,1] \to [0,1]^2$ except in a sort of morbid curiosity

it'd be a shame if point-set pathologies turned you off of topology; if you're to be turned off, i'd prefer if it were for other reasons :)

yeah, I know, and I'm certainly not turned off. Just because it's messed up doesn't mean I can't like it.

@SamuelYusim One thing, that you probably already notice, is that topological spaces, and their maps, are absurdly general objects, so not much can be said in general, so most people(all?) end up working with only nice spaces. So if you have a statement about all topological spaces it is probably trivial or wrong (sort of stealing that from Gromov)

I look forward to moving out of the pathology zone though

i see, that's good to hear. i've misunderstood your occasional comments, i think

The messiness can be fun though...

i don't really like to think about pathological spaces and objects myself. some people do

I think I can tend to sound like I'm complaining even when I'm not, so yeah

I'm doing this on my own so if I didn't like it there'd be no harm in just stopping

i like to know about them so that i don't risk trying to use some horribly general statement that's just not true

@Clarinetist, if $\lambda \neq 0$ is an eigenvalue and $x$ is an eigenvector then $x = \frac1{\lambda}Ax$ is in the column space of $A.$

Okay, so I had to prove that any group in which the square of every element is the identity is cyclic is a commutative group. Take any two elements $g, h$ and observe that $e = g^2 = geg = gh^2g = (gh)(hg)$ and $e = (gh)^2$, so by cancellation $gh = hg$. Correct?

that looks good to me

why's the word cyclic there?

yeah

that looks good

aside from having the word being cyclic

it can be described by the following

oh, dunno, must've been a brainfart

skip the "is cyclic"

sorry

you could also prove it as follows hg = ehg = $(gh)^2)*hg$ = ghghhg = ghgeg = ghgg = ghe = gh

yeah, that's sort of shorter

cool

can someone explain what it means to say that $F_2$ is the fundamental group (what's that?) of the figure-eight shape?

I know no algebraic topology, mind you.

Or topology, really.

then probably not, no

oh

i'll wait, then

in any space you can look at the set of maps from the unit interval into that space such that the endpoints both go to some basepoint. these are called loops. there is a notion of being able to deform one such loop into another; this is called homotopy. the set of homotopy classes of loops forms a group with group operations (essentially) composition. this is called the fundamental group.

that's probably all the details you can get without learning what these things mean

It basically means no loops you trace on the figure-eight can not be tightened to a point, unless your loop backtracks

simplistically

@MikeMiller yeah, I think I understood most of what you said

Is $F_n$ ever isomorphic to any "famous" groups?

Apart from $F_1 \cong \mathbb{Z}$.

$F_n$ is pretty famous

@AlexClark: I just noticed how asymmetric your gravatar is and I hate it

Did you modify it yourself

aaaaaa

Oh my god, @AlexClark what have you done?

@MikeMiller Any other famous groups, I mean?

define 'famous group'

yeah, it's awful

it's a holiday here, i hate holidays

i got no food at home and everything is shut down

lmao

christmas is alright i guess

but whit monday? what the fuck?

i wanna go to lectures and buy groceries

@AlexClark Now you got out of just trolling yourself, and got MIke

"NSA" and "Arschersetzer" are also good choices

@SohamChowdhury Lots of "famous" groups have $F_2$ as subgroups, look up the Tits alternative

All hyperbolic groups too

For example most surface groups

Actually that all hyperbolic groups thing is false, but all torsion free hyperbolic groups, that have two elements that don't commute, have $F_2$ as a subgroup

@AlexClark I used to, then I quit sleeping in once and for all. Haven't slept in once in the last year. $\checkmark \left(\color{green}{\frac{10}{10}}\right)$

Damn, I failed hard there

No, that's just a byproduct of having a made-in-China respiratory system

I'm all clogged up all year. :(

And asthma is fun too lol

@AlexClark what are you working on?

^ Yeah, that's my nose

Not anymore

I did do a bit of Aluffi at grandma's. I did my final pass of the first four sections of the groups chapter.

@AlexClark I remember some guy on HN who had 2400 tabs open

I used to have ~30 until last Sunday. Now I restrict myself to five.

As in, not yesterday

My mom was quizzically looking at my browser right now.

Apparently "tits alternative" looks suspicious

@AlexClark Just decide to, like Feynman did (he said to himself one day, "I'm only eating chocolate ice cream from now on" because choosing takes time)

Haha

Bookmark them. And then forget about them. I have >1000 I'm sure I'll never see again.

Yes.

The >1000 is from pre-exams though. I doubt I've made 10 bookmarks in the past month. (Apart from really relevant things like those "learning roadmaps" on M.SE)

Just getting disciplined.

Universal properties are cool, whatever Balarka says.

I have 1 bookmark: Start ChatJax.

Well pretty interesting ... did scuba diving for the first time !!@AlexC

Hello@soham

@alexC I just did Heine Borel... And now i am feeling i have weapons....Attack!!

I've never been scuba diving. It looks like tons of fun

@Rememberme hey

What book are you doing?

Rudin and Munkres@soham

What have you done before this?

What do you mean?

Books.

What other books have you done?

I think i gave you list

Oh, yes.

Set theory, linear algebra and everything?

Including Hoffman-Kunze.

I remember.

Not everything.. Basically nothing

See the sidebar, haha

Star panel?

@AlexClark Pat yourself on the back

$\checkmark$

@Rememberme yes

Ha lets go and answer some questions...

On Heine-Borel?

Lets see what i encounter

@MikeMiller @PaulPlummer @AlexClark this looks suspiciously like a vector space basis to me.

dramatic pause

I know.

@AlexClark your gravatar contains more information than everyone else's (I mean those with similar "default" ones)

yeah, in addition to the standard metadata, it tells us that Alex is a dick

^ Why does this remind me of Russell's paradox? ;)

@MikeMiller is there any connection between free abelian groups and vector spaces?

eh, you guys talk too much.

hello@balarka!!

@SohamChowdhury no, you don't.

proving that the fundamental group is really a group is non-trivial

"understood" = "have a sort of faint idea of what you're talking about" in that context, not "can prove everything you said"

I just wanted something better than total cluelessness, really.

But thats not what you mean when you say i understood..........@Soham

@Balarka I want to prove that a k-cell is compact How to go about it?

by k-cell you mean disk D^k?

define what you mean by it.

I didnt get you

I don't know what you mean by a k-cell.

Let a ∈ R and b ∈ R. If a_i < b_i for all i = 1,...,k, the set of all points x = (x_1,...,x_k) in R^k whose coordinates satisfy the inequalities a_i ≤ x ≤ b_i is a k-cell...@Balarka

Thats the definition in Rudin....@Balarka

@Rememberme huh? that's a weird terminology.

Second chapter?

well, try to draw a picture.

@AlexClark btw, you don't need both $t$ and $u$ to exist for your sequence to split, as you wrote here. Having a $t$ automatically gives you a $u$, and vice versa (splitting lemma).

Did you talk about central extensions, then?

Make sure you talk about inverse limit of rings, though :D

sup

that'd be nice. did you do much ring theory, then?

hi @iwriteonbananas

balarka, have u studied complex analysis?

@iwriteonbananas nope, not a lot

ok, shame

what do you want to know

@AlexClark ideal of which ring?

if f is continuous if $D\subset \Bbb{C}$ and holomorphic on $D-L$ where $L$ is a straight line in $D$, then $f$ is holomorphic in $D$.

can we do this:

ideal generated by $a$ and $b$

Have you thought about maximal ideals of $\Bbb Z[x]$, then?

hmm

probably we should use that $f$ is holomorphic in $D$ iff it is locally integrable in $D$

Nice, because it's a good question.

@iwriteonbananas: pick a point $l$ in the line. pick a point $x$ near $l$ so that the taylor series of $f$ around $x$ converges at $l$ ($f$ is continuous there, so there is some such taylor series). then the taylor series agrees with $f$ where it converges in $D-L$; and because $f$ (and the taylor series) are both continuous in a small neighborhood of $l$, and they agree near $l$, they actually need to be equal in that neighborhood of $l$. so $f$ is holomorphic in that neighborhood.

repeat for all points in $L \cap D$.

ok i thought about something like that too

@AlexClark it's a good question.

there are plenty of details in the sketch above to be filled in; particularly sketchy is the bit where I say they're the same function in a neighborhood of $l$

i think i see how to make your argument work too

can we make an argument with integrals over triangles being 0?

seems like it, but you'd need to work locally and the best way I see to do this is by doing a limiting argument by decomposing your triangles into arbitrarily small ones

@MikeMiller in that bit, did you use some identity theorem?

yeah

(triangle idea: the issue is that there's no reason to believe a triangle that crosses over $L$ should integrate to zero; so cut your triangle into sub-polygons or whatever with $L$, and then decompose into smaller triangles so that the ones that contain part of $L$ are arbitrarily small; invoke continuity to bound the integrals, and take a limit so that this bound goes to zero)

ok

@MikeMiller fair enough, the taylor series argument seems a bit cleaner than that

this argument is at its core what you'd have to do for your local integrability argument

(well, locally integrable and path-independent)

yeah

ugh : my rep is at 2899. it's annoying to see something so close to 2900 but not quite it.

:P

i can fix that

✓

glares hard

that didn't help, as it's still not 2900.

but I appreciate you for downvoting that answer

but it's farther away

so no worries about how close you are anymore

my highly upvoted answer on summing a series, which is ok.

right

Which comment?

she'll be back

I didn't see it

Its not

not on my pc

ok, to anyone with some understanding of pde. Let c = {x in R^n | x_n < |x|} (like a sphere without a cylinder I would say). Now u is a solution to the dirichlet problem, u = |x| on the boundary of C, and u harmonic inside C. Why is u(x) >= |x| for x in C?

@Balarka I have to think of a k-cell So a k-cell in any dimension is fine?

@fubal: i don't think i understand your domain. isn't that {x in R^n} \ {(0,0,0,...,x_n) : x_n \geq 0} by the triangle inequality?

if so, i don't really get the sphere without a cylinder comment

@MikeMiller you are right. Correction: C = {x in R^n, |x| < 1, x_n < |x|*a} for some a >0

hmm, @Mike. did Hatcher really asked for a \Delta-complex structure or did he ask for orientation on the resulting manifold? this questions says that it's the former, but I don't recall.

if it's a \Delta-complex structure, it should not really be that hard

it is more like Ball without a cone

i see, so we're interested in the case a<1

gotcha

yes exactly :D

@BalarkaSen: ctrl+f tells me it's on p109, so you probably want to look there

no, apparently it's a problem of 2.1. i just checked and it really asks about a \Delta-complex structure. that's odd

@BalarkaSen i remember discussing that problem w/ you

no, i think we discussed 2.1.10(a)

yes

that's what i meant

it's actually 2.1.10(b) that's bothering OP.

i never solved (b), curious to see if someones gonna answer that question

i have never wrote it down explicitly, but i don't think it should be hard either.

pick a simplex from you collection, orient so as to make it $\Delta^2$.

write a response then!

now you're pasting other simplices from your collection by gluing the edges pairwise

i don't get it. well, why can't any orientation work?

(i am probably being silly)

@Balarka Why cant i use Heine Borel to prove that every k-cell is compact

i don't get your mucked-up definition of a k-cell, and i am not bothering to think about it right now. ask someone else, i guess.

sorry, not thinking about the problem right now. gotta finish complex analysis

Fine thanks...

@fubal: i can't get this to work :( one standard trick is to take a soln f, replace it with some related function v that's, say, bounded on the boundary; show that v has to be constant by the maximum principle; use this to get the desired bound on f. but i haven't been able to fiddle to find the right v.

there's probably a more natural approach

you're probably better off asking on main or asking @DanielFischer

@Mike

upps sorry

@MikeMiller what is a soln f and what is main? thanks anyway! if |x| would be subharmonic I could use the maximum principle to show the desired result but for n >= 2 it is superharmonic :/

i just mean a solution to your equation with given boundary solutions; main meaning the main site, math.stackexchange.com

yeah, i thought about that too

for shame

@MikeMiller ok then thanks for the help. I also asked some collueges but if they can't figure it out I will probably post it on main

good luck! interesting problem

yeah it is part of the proof that the outer cone condition (is it called like that) is enough to show regularity of a point

so it should be somewhat obvious but I can't see it

@MikeMiller.... are you free?

mesmerizing!

Beautiful pics arent they?

I think being beautiful is not the point :P

The maths is the point for you,currently for me pics are the point :p@Balarka

yes

What?

it's essentially a consequence of Bolanzo-Weierstrass.

Bolanzo-Weierstrass whats that

google it.

I am pretty sure it's in the next chapter of Rudin though.

each bounded sequence in R^n has a convergent subsequence.. Ahh

What you say is not a field.

Set up an isomorphism with something that is not a field.

Well, recall how you proved that $\Bbb R[x]/(x^2 + 1) \cong \Bbb R[i]$

That's not an explicit isomorphism.

To prove two things are isomorphic, you need to set up an isomorphism between them in the first place.

Yes, but even before that you need a homomorphism.

Ack, I gotta run right now, but hint : do similar things with $\Bbb R[x]/(x^3 + 1)$ and use the fact that a field has no nontrivial proper ideals.

The Cantor set is a percfect set right?

Which set is more important $\Bbb{Q}$ or $\Bbb{R}$

Important?

As in more considerable

@BalarkaSen Bolzano*

AS in more specifically why do we care about real numbers because aren't they from the set of all rational number@AlexC

$\sqrt{2}, \pi$ etc. for another

Not yet

Which chapter in rudin

Hey @AlexClark, are free abelian groups some kind of vector space?