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12:00 AM
why ?
 
Prove it!
 
Proving stuff is hard, though. xD
 
oh I see it is an earlier result before using the quotient map
just a sec
 
12:17 AM
@BalarkaSen Are you there?
 
1:12 AM
I found now the following theorem:

Let $f,g\in \mathbb{C}[e^{\lambda}x]$. Suppose that in the ring of analytic (over $\mathbb{C}$) functions $f \mid g$, i.e., there is an analytic (over $ \mathbb{C}$) function $h$ such that $g=fh$, or equivalently each root of $f$ (in $ \mathbb{C}$) with multiplicity $m$ is a root of $g$ with multiplicity at least $m$. Then $f \mid g$ in the ring $\mathbb{C}[e^{\lambda}x]$, i.e., there is a $h\in \mathbb{C}[e^{\lambda}x]$ such that $g=fh$.
What exactly is an analytic function? @robjohn
 
1:27 AM
Are there examples of finite length modules with an infinite number of composition series?
or an infinite number of simple submodules?
 
@MaryStar anything about the wikipedia page on "analytic function" you don't understand?
@dREaM consider vector spaces
 
Lol
$\mathbb R^2 $
 
any line through the origin followed by the whole space right?
 
mr @Pedro !! I thought of you today as I drove by San Pedro (near LA) :)
hi @anon
 
1:31 AM
@TedShifrin Hope you're pronouncing that right. =)
 
$0\leq L \leq \mathbb R^2$
 
You still owe me Spanish lessons.
 
@anon right?
 
hi @Pedro @Ted, yes @dREaM
 
Balarka thinks his puns have improved when he wonders about a cobel prize? Oy.
No mas.
 
1:33 AM
thanks anon
 
better: "comathematicians are a device for turning cotheorems into ffee."
 
That one's old, @anon.
 
mmhmm
and makes sense
 
Maybe not as old as I, but close.
 
1:36 AM
I think somehow it's a coabel joke with one fewer a than it should have
 
@Pedro: Don't worry, I can roll r's. I speak French and German. :)
still wonders if @MikeM food poisoned family
 
No.
 
@Pedro: They're all horrrrrrrible.
Well, @MikeM, I made it to Santa Barbara and back in about 27 hours.
Did I miss any good math?
 
No.
 
@anon: Sense only if co$^2$ is trivial.
 
1:42 AM
thatsthejoke.tex, yes
 
Oh, it's a joke? :D
It was in the co-category, too.
@anon: So your Lie groups course is almost over?
 
Your jokes are in close competition with Balarka's.
 
I wasn't joking.
I have no sense of humo(u)r. Just ask chris'ssis.
Oh wait.
 
yeah, we might continue to meet and talk next semester tho
 
Are you guys Schur you want to keep trying?
 
1:44 AM
jfc
 
You're afraid we'll decompose, @Pedro?
What else do you have planned for next semester, @anon?
 
I was gonna ask for some help but @TedShifrin is here and he's gonna tell me not to "command"
I'll come back later
 
nothing really. trying to read about clifford algebras and octonions and stuff in order to understand accidental spin groups.
 
pondering whether to bother commenting
awesome stuff, @anon ... about time to learn some diff geo :P
 
@TedShifrin That's dark.
 
1:47 AM
Only if you're irreductive, @Pedro.
Maybe I should quit coming here. After all, there are at least two people who think I'm a czar who ruins it for everyone else.
 
No need to be so dramatic.
 
2:02 AM
Love the use of "czar".
 
2:37 AM
hi @TedShifrin
@TedShifrin ?
 
3:01 AM
@TedShifrin? @L33ter Why are we pinging @TedShifrin?
@L33ter I think he left @L33ter
 
:D
I was just kidding with him
 
Alex I have a question
how come if have $h : S^1 \rightarrow S^1$ and if $h_{*} : \pi_1(S^1,b_0) \rightarrow \pi_1(S^1,b_0) $
if h* is non-trivial this will imply h is not null-homotopic ?
?
 
Are you talking to me?
 
yes
 
3:09 AM
Why am I Alex?
I don't know anything about what you just wrote :D
Is that homological algebra?
I'm only up to commuative algebra
 
algebraic topology
 
You paged me, Karim?
 
I haven't got to do any of that yet, but I think I will some time.
 
yes
I have few question about some theorem
 
Yes?
 
3:12 AM
I thought I could ask you instead of posting it on main
 
Or you could wait a few hours and ask Balarka.
 
if $h : S^{1} \rightarrow S^{1}$ is continuous and antipode-preserving then h is not nulhomotopic.
I understand mainly the proof in munkres
 
That's a correct statement.
 
but why is enough to show that $h_{*}$ is not trivial
 
What does it mean for $h$ to be nullhomotopic?
 
3:13 AM
that is not homotopic to the constant function
 
And homotopic maps induce what sorts of maps on $\pi_1$?
Oh, typo in your answer just above. Nullhomotopic = is homotopic to a constant map.
 
I don't understand this question the * maps are just homomorphisms of $\pi_1$
yeah *
 
But if $f$ is homotopic to $g$, how are $f_*$ and $g_*$ related?
 
they are isomorphic
 
No. Be precise.
Maps aren't isomorphic.
 
3:17 AM
bijective
 
I asked how two homomorphisms are related.
You need to understand your earlier reading.
 
I think you are working through your book too fast @L33ter
 
1 sec thinking
so any continous function that has the property that $f(x_0) = y_0$ induces a homomorphism from $\pi_1(X,x_0)$ to $\pi_1(X,y_0)$
 
Yes.
(Well, really $(Y,y_0)$, but in your case $X=Y$.)
 
denoted as $f_{*} : \pi_1(X,x_0) \rightarrow \pi_1(Y,y_0)$ where $[g] \mapsto [f g]$
yeah I meant to write Y
 
3:23 AM
OK ... and if you have two homotopic maps $f$ and $g$ [this $g$ has nothing to do with the one you just wrote], what do you know about $f_*$ and $g_*$? What is the meaning of $\pi_1$, after all?
 
Can somebody help me with this? (it's not a command) math.stackexchange.com/questions/1543985/…
 
@user29232 Why did you change your name?
 
I've been working on the same problem for so long I've started to dream about it.
 
@I'mmostlyjustanidiot why not?
 
@user29232 It's confusing sometimes.
 
3:25 AM
You should probably edit the title so that it is correct, @Sajindia. You need $E$ to be simply connected.
 
I know but they can read it in the problem
 
@Perplexed: Sometimes that goes on for months.
 
I can't write everything in a short title
 
oh oh I see they will both agree on every element of $\pi_1$ as we are considering the equivalence classes and here the equivalences classes are actually homotopy classes
 
@Ted!!!
 
3:26 AM
Why did Twink put a bounty on it? Do you know that user? @user29232
 
I understand all the proof of the theorem, I just don't get why they omit to prove that $\phi$ is bijective
 
@AlexW !!!!
 
@AlexWertheim !!!!
 
Because he's my friend @I'mmostlyjustanidiot
 
@user29232 OK.
 
3:26 AM
:D How's it going? :)
@L33ter: hello! I'm sorry, I don't recognize the name :(
 
So you agree that $\phi$ is continuous, @Sajindia. Can you determine $\phi^{-1}$?
 
@I'mmostlyjustanidiot you're asking too many questions
 
(I'm guessing you're someone I do know though)
 
@AlexW: It's Karim.
How're you doing, @AlexW? I've thought about you several times.
 
Ah. Well that makes sense!
 
3:27 AM
@user29232 Sorry.
 
I drove past UCLA yesterday and today (well, past on the freeway).
 
lol
oh I see
 
Mathematically, not so bad. Algebra is going well - differential topology, maybe less so.
 
I understand now @TedShifrin
 
Cool, Karim.
 
3:29 AM
Aw, shame you didn't stop by, would've loved to say hi. How are you, @Ted?
 
because here if we do the same thing for the identity in place of g
I am sorry
 
Not identity, Karim.
 
if the constant function in place of g
 
Well, @AlexW, I didn't get off the freeway. But we can arrange something in the future.
 
I am writting stuff that I don't mean lately which sucks.
 
3:30 AM
@TedShifrin It was a big deal to prove that $\phi$ is continuous. I wish it's not that difficult to prove that its inverse is also continuous.
 
that is because of not sleeping too much
 
Well, if you want to talk about manifolds, etc., sometime, @AlexW, let me know.
It's not, @Sajindia. What do you know about $G$?
 
it's simply connected and locally pathwise connected
oh
about $G$
hmm
 
nods
 
hey @TedShifrin I was talking to balarka about some prof guy who doesn't believe that infinite uncountable sets exists.
 
3:30 AM
it's a group
 
Bingo; it's a group.
 
I would love to, @Ted. It's slow going learning about them, I'm afraid, and it's a fast paced course. I feel like I only have fuzzy conceptions of things right now, which doesn't jive well with how I learn math.
 
Yes, @AlexW, but working through stuff with classmates and talking about it is helpful — I've made this recommendation before :)
 
so if $\phi \in G$ then $\phi^{-1} \in G$
 
Unfortunately, grad school is generally faster paced than anything you've done.
 
3:31 AM
I hate fast paced classes
 
Well, by definition, grad school is fast-paced. :)
Right @Sajindia. Are we done?
 
like I am taking this graduate course in algebra the prof rushes through the material.
but I am taking 5 classes as well so I don't really give it the time that I should give it
 
just let me see what's the inverse and I'll ask you if it's right
 
Grad school (as opposed to undergraduate) expects students to make studying a full-time job (40-80 hrs a week).
 
my poor friend @twink put a bounty on this question and it was so easy
 
3:32 AM
yeah
 
LOL, uh huh @Sajindia.
 
Haha, I do try. It's been hard for us to all meet, though. I tend to badger Mike a lot, but that's not quite the same, since we're in very different positions w/r/t the material.
 
@TedShifrin this guy youtube.com/watch?v=h26w-MyHEW8
 
Yes, @AlexW, I'm sure Mike is helpful, but there's stuff you need to talk through yourself and having someone else say it fast won't do. It doesn't need to be all. Find one or two.
 
Right, @Ted. Working on it, hopefully I'll have better plans as far as this goes next semester :)
 
3:34 AM
what I will do for graduate school is the following which I didn't have time to do as undergrad
is to read everything before attending class
 
Anyhow, @AlexW, if there are particular things you want to chat about, I'm happy to help.
 
Hopefully I'll have better approaches to the material in general.
 
and do assignments on on the day I receive it
 
People often don't follow texts slavishly in grad school, either, Karim.
 
14 hours ago, by I'm mostly just an idiot
A $K$-homomorphism $\sigma : L\to M$, that also is an isomorphism, is called a $K$-isomorphism. A $K$-isomorphism $\sigma:L\to L$ is called a $K$-automorphism.
 
3:35 AM
I'm not going to watch that lecture, Karim. How can $\Bbb Z$ not exist?
 
Admirable, @Karim, but optimistic. Life doesn't stop getting in the way once you start graduate school, I assure you :)
 
Question 5. v2: So a $K$-isomorphism has that $L$ is isomorphic to $M$, and a $K$-automorphism has that $L=M$?
 
But it's stronger than $L\cong M$, @I'mmostly
 
yeah he is a weird guy
yeah your right @AlexWertheim
 
The isomorphism is required to preserve elements of the base field $K$.
 
3:36 AM
@Ted: I'll probably ping you in the next week (not here). Unfortunately, I think I need to just work through a lot of examples with some of these things, which takes time.
 
@TedShifrin And that $\sigma|_K$ is the identity
 
Right, @I'mmostly.
 
What is this life people speak of?
 
OK, @Alex. I'm happy to talk on Skype if that helps.
Because any two finite fields of the same order are isomorphic, you mean, @I'mmostly?
 
Nvm that was my second question when I misunderstood the first thing
 
3:37 AM
I have no idea, @PVAL.
 
Might, depending on the nature of the question. Trying to use TeX elsewhere can be a massive pain :)
 
Indeed, @AlexW ;P
 
@TedShifrin - I've been working on it for months, haha.
 
@Perplexed: In the days I did research, I often spent years on single questions. Kept falling in traps and had to untangle myself.
Even with coauthors ...
 
May I ask what you've researched? :0
 
3:39 AM
Mostly overlaps of differential geometry and complex algebraic geometry.
 
Someone in my department is apparently handing out some smallish computational topology funded projects as an alternative to teaching. I'm contemplating if I should apply.
 
If it is of interest to you to spend time on it, sure, @PVAL. You need teaching experience when you apply for jobs, but not every single term.
 
@PVAL what is a question in computational topology that I might understand?
 
I'd make the obvious joke, @Ted, but I think I've said it more now than you've actually used it in book titles =P
 
@TedShifrin But if I construct $\alpha$ such that $\alpha \circ \phi = Id_E = \phi \circ \alpha$ that implies $\alpha$ is the inverse of $\phi$ in $G$? but what if I can construct a non continouos $\alpha$ which has that property? it cannot be in $G$
 
3:41 AM
Sorry, @AlexW, it is undeniable that geometry has been a large part of my life :P
 
Geometry ... now that's a complex and varied beast. xD
 
Geometry is great! :D
 
But the text has proved any $\phi\in G$ corresponds to a continuous mapping, right, @Sajindia?
 
Ok, well, what little I know about geometry is great.
 
is topology part of geometry?
 
3:42 AM
Not necessarily.
 
no, that's the deffinition of $G$. the homeomorphisms of $E$ which preserve the fibers
 
Some people might say topology is closer to set theory than geometry.
Depends on what kind you're doing, I suppose.
 
@TedShifrin I have only taught once under a professor and that happened to be a role with very little work for me (its mostly lecturers here). I do not know if things are going to improve in that sense. I think I will need to count on research for my first application for jobs.
 
the only thing that I know of geometry is euclidean geometry but I am sure that is like thousand years old
 
they constructed a $\phi$ and want to prove $\phi \in G$
 
3:43 AM
But if $G$ is a group and $\phi\in G$, then $\phi^{-1}\in G$ as well, so the proof shows that $\phi^{-1}$ is also continuous, @Sajindia.
 
I don't know anything of modern geometry
 
I think tomorrow I shall hijack one of my university's computers and just run it through some truly terrifying Mathematica code. I need the computing power. xD
 
Most postdocs still want to hear about your teaching, @PVAL. If all you do is recitations, so be it, but you need some faculty to observe you and write a teaching letter.
 
if $\alpha \circ \phi = Id_E = \phi \circ \alpha$ then $\alpha=\phi^{-1}$?
is that because the inverse is unique?
 
Yes.
 
3:45 AM
@TedShifrin I have applied in mcmaster as one of my options to work with a sympletic geometer
 
it's the only function with that property, hmm
 
We're using inverse in the group to deduce inverse as functions on the topological space, however.
 
You need to know a lot about manifolds and differential forms to get started with symplectic geometry, Karim.
 
I see
 
3:52 AM
The over-arching goal of my research right now is to find an explicit formula for this: oeis.org/A007764
 
@TedShifrin is the inverse just $\alpha(e)=e \cdot [\tau^{-1} \sigma \tau]^{-1}?$
 
There's quite an infamous paper in symplectic geometry (that was published in the annals around 2005) that turned out to be completely wrong (somewhat recently). Apparently a lot of people's work (which depended on it) became completely invalidated as well.
 
so all results are now not valid ?
or I guess they would have to make it work using other theorem
or something
 
4:15 AM
@TedShifrin But to prove $\phi^{-1} \in G$ I need to prove first that $\phi \in G$
and that's what I want to prove
that $\phi \in G$
I just have that $\phi$ is continuous, but that's not enough to say $\phi \in G$
 
@TedShifrin here?
 
no
 
4:57 AM
Can anyone please help me with weak star limits
 
5:08 AM
@anon maybe you could help me understand one final thing
 
k
@PVAL oh?
 
consider the following map q from $S^1$ to $S^1$ given by $q(cos\theta,sin\theta) = (cos(2\theta),sin(2\theta))$
why is $q_{*}$ injective?
 
intuitively, any loop around n times becomes a loop around 2n times, which isn't trivial. if you want you could assume the image of something nontrivial were trivial and lift the nullhomotopy (because it's a covering map) to get a contradiction if you want some rigor.
 
oh I see
so it is just multiplication by 2
I guess it is 2Z
oh I guess that also answer my second question of why we needed even that q to be a covering map
we need it for this final result.
thank you @anon
 
5:59 AM
@anon consider $q : S^1 \rightarrow S^1$ where I defined it as before it is a covering map. Let we can see that if f~ is any path in $S^1$ from $b_0$ to $-b_0$ then the loop $f = qf~$ represent a nontrivial element of $\pi_1(S^1,b_0)$ because it will just go around the circle twice start at $b_0$ and end at $b_0$ correct?
so it is not the constant path
 
6:46 AM
@MikeMiller I sent you the paper. If you read the abstract of the dissenting paper, you will get an idea of what some people think about the paper.
 
Quick verification: A finitely generated ideal is an ideal that can be described in the form: $\langle x_1,x_2,\cdots,x_n\rangle$ for some $n\in \Bbb N$, is that correct?
Question 6: What is an example of a non-finitely generated ideal(of some ring of your choice)?
 
7:03 AM
try $\mathbb{C}[x_1, x_2, x_3, \dots]$
 
@TedShifrin No, the relevant puns are Ado and Luri(e)d.
 
@BalarkaSen
YAAAAAYYYY
:D
ok I finished understanding the proof
but I would like to discuss something with you
some stuff
I was gonna go to sleep now
 
Sure.
 
but good thing that you came online
 
Did you finish proving that $S^2 \to S^1$ thing?
 
7:11 AM
yes
we note that $\tilde{f}$ is any path in $S^{1}$ from $b_0$ to $-b_0$ then the loop $f = q \tilde{f}$ repesent a non-trivial element.
ok I see this
if we draw it
but they I guess give it more rigour by saying that $\tilde{f}$ is a lifting of f to $S^1$ that begins at $b_0$ and does not end at $b_0$
I have looked before in the book
they have the following lemma
Let $p : E --> B$ be a covering map and let p(e0) = b0. any path f : [0,1] --> B begging at b_0 has a unique lifting to a path begging at e_0
 
First you have to understand why $f$ is a nontrivial element. Drawing it is fine, not at all nonrigorous.
 
yes I understand
I understand the intuition
now I need to understand the rigour
 
What rigor?
You don't need to know that $\tilde{f}$ is called the lifting of $f$.
That's just a name.
 
why do they need it here then ?
what does this information help with ?
 
Why do we need what?
 
7:16 AM
do we need that f~ is lifting
 
I don't know where we need $\tilde{f}$. You're the one reading this proof.
 
you see they mentioned
For $\tilde{f}$ is a lifting of f to $S^1$ that begins at $b_0$ and does not end at $b_0$.
 
Seems merely like a remark to me.
It's just a restatement of $f$ being homotopic to the identity loop.
 
I see
where do we even use that q is a covering map
in the proof ?
 
step 2, it seems.
 
7:20 AM
but why we need it ? I mean they show that k* is non-trivial
already
 
"$p^{-1}(U)$ consists of..."
 
I know they showed it in step 2 I understand that part but why do we need the fact that q is a covering map ?
where in the proof do we use properties of covering maps
 
@L33ter I just directed you to the statement.
Also, it seems Munkres as a typo. He meant $q^{-1}$ instead of $p^{-1}$ up there.
 
well that part that you directed me to shows that q is a covering map
but I don't see anywhere where we use property of covering map to prove something..
 
@L33ter Ah, fair enough.
Well, I guess you don't need to spell out the fact that $q$ is a covering map explicitly.
 
7:27 AM
I guess it is used in the part to show that $q_{*}$ is injective map
?
 
You can do that without spelling that $q$ is a cover out. But, if you want.
 
yeah
yeah all I have left now to do is to prove rigorously that $q_{*}$ is injective..
because I see it intuitively
but I want to prove it..
I guess this is where we use that q is a covering map...
because we can use the lifting property..
 
You can do that without spelling out q is a covering map still :P
 
really why ?
 
Just prove the homotopy lifting property holds for $q$ in particular.
 
7:29 AM
oh I see
 
No need to say that it's a covering map :)
 
I see
I will do it tomorrow I am tired atm
or maybe I will ask it on a question on main I already did alot of proofs today
 
You can do it easily by htpy lifting if you think about it carefully.
I recommend against asking on MSE.
 
yeah I will not take it on MSE its good to think about such stuff.
 
One does not learn that way.
 
7:31 AM
yeah
I agree
I will think about it tomorrow however
 
Prove in general that if $p : (E, e_0) \to (B, b_0)$ is a cover, then $p_*$ is injective.
 
I am tired atm
 
Sure.
 
1 last question
I want to make sure I have completely understand that part.
the reason it is sufficient to show that $h_{*}$ isn't trivial to show that h is not nullhomotopic
is that
if indeed that h is homotopic to the constant function $e_{x_0}$ lets call the homotopy H
 
If $f : X \to Y$ is null, then $f_* = 0$. Prove this.
 
7:34 AM
if $f : X \rightarrow Y$ is null, so what does it mean it means that $f(x) = c$ for all x in X.
so
 
No, it doesn't mean that
 
I am sorry
 
$f$ is nullhomotopic. It's just homotopic to the constant map, not necessarily the constant map itself.
 
yes yes sorry making stupid mistakes
f null homotopic means that is is homotopic to constant map
so what does it mean
it means that there exists a homotopy H that deforms f to the constant map
 
homotopy, not homotopic function.
 
7:36 AM
yes
let $[g] \in \pi_1(X,x_0)$
so $[g] \mapsto [f \circ g]$
but you see $f_{*}(H)$ will be a homotopy between $[f \circ g]$ and $[e_{x_0} \circ m]$
 
It doesn't make sense to say $f_*(H)$ is a homotopy. It's a group-level map.
 
I guess $f_{*} \circ H$
 
Huh? $H$ has codomain $Y$, $f_*$ has domain $\pi_1(X \times I)$.
 
nvm
 
What does it mean to compose them?
 
7:41 AM
yeah I just realized that
I will just think about it tomorrow when I am not tired
because I am making now stupid mistakes
 
ok.
 
its embrassing
good night
 
8:20 AM
1 hour ago, by I'm mostly just an idiot
Quick verification: A finitely generated ideal is an ideal that can be described in the form: $\langle x_1,x_2,\cdots,x_n\rangle$ for some $n\in \Bbb N$, is that correct?
1 hour ago, by I'm mostly just an idiot
Question 6: What is an example of a non-finitely generated ideal(of some ring of your choice)?
 
@I'mmostlyjustanidiot Samuel Yusim just gave you an example. $(x_1, x_2, x_3, \cdots)$ of $\Bbb C[x_1, x_2, x_3, \cdots]$.
@I'mmostlyjustanidiot Yes.
 
@BalarkaSen Thanks
Thanks @SamuelYusim
 

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