Onto? There is no such cover.

sure there is. R \to S^1

@MikeMiller Trivial ones.

I knew someone would say that, @DanielFischer

@Balarka Last I knew, there was no onto map $0 \to \Bbb Z$.

by onto i don't mean surjection

oh and i mean normal too hehe

p : E \to B is a galois cover if \pi_1(p)(\pi_1(E)) is normal in \pi_1(B)

@Kaj: Well, lots of faculty don't get their grades done until literally the last minute, anyhow, so under the old system you still wouldn't have known until the night of December 22. Some of us entered them earlier and then were, typically, bombarded by emails — mostly of a whining or begging nature, but occasionally of a grateful or happy nature :P

Oooh, Balarka is inventing an onto homomorphism from $0$ to $\Bbb Z$. Yummy.

Which is very different than the first thing you said. OK. Thanks.

@TedShifrin That was mike interpreting me differently

grumph just unignore me will you

say please nicely again, as your lawyer I advice you to do this :D

already said that yesterday and also apologized

ok. please have mercy on this kid and unignore me @Ted.

I would like to put two conditions on the terms in a sum.. so $\sum_{v\in\calT \,\mid\, \ell_v\geq \sqrt{n}} \!\!\!\!\!\!\!\ell_v$

@Ted By the way, if you were curious, here's the gauge theory course description.

$\sum_{v\in T \,\mid\, \ell_v\geq \sqrt{n}} \!\!\!\!\!\!\!\ell_v$

What in the blazes is that, @Lembik?

LEL

hmm. not working. let me try again

$\sum_{v \in T | \ell_v \geq \sqrt{n}} \ell_v$

@TedShifrin fixed now

@Mike: We ran a year-long seminar going through Lawson's CBMS lectures (the first thing on Ciprian's bibliography), and we also went through some of Donaldson/Kronheimer's book. It's still on my shelf :P

so I want to sum over all v's such that they are in T and this property $\ell_v \geq \sqrt{n}$ holds

how should I write that?

Are you asking how to LaTeX that with the conditions in two lines?

@TedShifrin well I am really asking if I just put them in two lines, would people read that as two conditions that have to hold or that one of the two have to hold?

both

what do you mean>

?

I've forgotten how to do it in LaTeX, but I've done it before, so I'll find it if you make me :)

I can do the latex.. it's what I should latex that I don't understand

I want both conditions to have to hold

not just one of them

@Ted It's pretty exciting. Donaldson's theorem in particular... this is still stuff from decades ago but I feel like I'm learning modern things :)

Right. If you put the two lines under the sum sign, it will be understood that you mean both conditions.

oh that's great!

@Mike: Given how old mathematics is, it's very modern.

thanks

@TedShifrin well. ..would it then be clear that it was over the v's?

I want to learn homology theory properly... sigh.

Munkres gives a very unsatisfying definition.

@TedShifrin and not over $\ell_v$'s

Sure, if the $v$'s are on the first line. I think so. I'll ponder.

thanks!

Stop whining and do Hatcher, @Balarka.

@Lembik $$\sum_{\substack{v\in T\\ \ell_v \geqslant \sqrt{n}}} \ell_v$$

Pedro understands CW complexes and he started after you!

@DanielFischer thanks.. so my question is, is it clear we are summing over v's?

I will, but first I will finish Munkres first @Mike

Well, it's not my fault if Munkres doesn't start with CW complexes!

@DanielFischer is that how people will read it.. not that you are summing over v's and $\ell_v$s

@Lembik Pretty much, yes. By convention, the first line gives the summation variable, and the later additional conditions.

Well, one doesn't need CW complexes to do point-set topology, @Balarka - one needs them to do algebraic topology.

@DanielFischer thanks!

@MikeMiller I am not going point set topology.

Munkres has a book on algebraic topology.

Do you seriously think I am studying this stuff from nowhere, @Mike? Like Sanath?

What are you talking about, @Balarka? I assumed you were reading the last chapters of Munkres' intro book.

Yes, I am.

@Studentmath I didn't understand how to use a simple graph with 5 vertices to find how many ways to make a connected graph with 4, 5, 6 edges.

Munkres's topology has a part II which is on alg topo entirely

My point was that he doesn't introduce CW complexes because you don't need them for what he does in his book.

@MikeMiller You said "one doesn't need CW complexes for point-set topology"

Which is mostly what he does, yes. You also don't need them for the (rather small) amount of algebraic topology he does.

Ah OK. But I think I should strengthen up my basics with Munkres first. I'll do Hatcher when I finish Munkres.

Sure.

can someone tell me the name of the integral looking symbol in the final comment to math.stackexchange.com/questions/1059379/… ?

Hatcher's book is very advanced and I think I need some background first. My mathematical potential is very low.

I can't grasp stuff that fast.

or is not really a recognised symbol?

Any advice on this? "Naoki and I are playing a game with an unfair coin that is rigged to come up heads with probability $\frac35$ and tails with probability $\frac25$. Naoki goes first, we take turns, and the first player to flip a tail wins. What is Naoki's probability of winning?"

what does $\oint$ represent mathematically?

what is it called? I don't recognise it

Contour integration @Lembik.

I'm glad you finally want to go slow on something @Balarka

@BalarkaSen oh thanks!

@MikeMiller That's because algebraic topology is far from the stuff I am comfortable with.

I can't handwave out 80% of the stuff in there.

It's such a pain, checking all the details, going through all the calculations.

Partially it's also the reason I am trying to develop my understanding on the connection of galois theory with covering spaces, so that I can handwave more :P

@BalarkaSen can you parse the final comment of math.stackexchange.com/questions/1059379/… ?

mmhmm

That is an absolutely awful motivation, @Balarka, and the next time I hear a handwavey argument from you that you don't intend to fix it's curtains

@Lembik by "parse" you mean if I understand it?

@MikeMiller what sort of curtains?! :)

@BalarkaSen yes!

then nah.

The bad kind, @Lembik

@BalarkaSen me neither :(

OK, OK, @Mike. I don't intend to give a handwavy argument to you.

It's just so I get the intuition. I can't possibly hope to do math without knowing what's really going on out there.

:P

I knew you'll get mad reading it.

Of course. The point of math is to understand what's going on, and the point of intuitive understanding is to be able to write down an actual proof starting from these ideas (and get a more fundamental understanding)

Yeah.

Hi, tanned @Jasper

@TedShifrin Hi.

LEL "tanned"

Can anyone persuade @Ted to unignore me, please?

No

I didn't even ping you @Mike

But the answer to your question is no.

But why not.

I don't want @Ted to ignore me cries

Note that your question was "can anyone persuade", not "can anyone try to persuade". The answer to the latter is likely yes. The answer to the former seems, based on all evidence available to me, to be no.

OK, can anyone try to persuade him to unignore me?

Given the answer to the first question, asking the second one seems like a waste of time!

It was your answer, @Mike, not Ted's.

OK, @Mike, I have proved the existence of a short exact sequence.

Congrats!

Note, by the way, that for a regular cover $\text{Aut}(X,Y) \cong \pi_1(Y)/p_*\pi_1(X)$, from which the existence of your sequence should immediately follow.

You're right, but I guess I wanted to do it the plain old way, thinking about the Deck transformation group as Galois groups.

So there is a SES $1 \to Aut(X, Y) \to Aut(X, Z) \to Aut(Y, Z) \to 1$

I'm glad you did, which is why I didn't mention that fact until now. Didn't know if you knew it.

I know it.

It comes from combining monodromy representation and the covering correspondence.

There is an iso $\pi_1(Y, y_0)/\pi_1(p)(\pi_1(X, x_0)) \to p^{-1}(x_0)$

And there is a homom $Aut(X, Y) \to p^{-1}(x_0)$ by lifting loops and thinking of the endpoint.

Now combine these two.

OK, back to what I was saying. There is a SES $1 \to Aut(X, Y) \to Aut(X, Z) \to Aut(Y, Z) \to 1$

Set $X = \tilde{X}$, universal cover, and $Y = X/G$ and $Z = X$.

$1 \to \pi_1(X) \to \pi_1(X/G) \to G \to 1$ follows as a consequence @Mike.

that is awful notation

I know LEL

how do you propose that there's a covering space $X/G \to X$...?

Yikes.

I mean $\tilde{X} \to X \to X/G$

Very nice.

I wonder what's the dual $\mathbf{Fix}$ functor involved.

Dual is the wrong word here. Analagous would be the right word.

hehe now i am going to develop galois theory for covering spaces

@MikeMiller I mean $Aut$ is a functor. what's the cofunctor?

Cofunctor is not, like, a thing.

?

It's not a word, at least not in the context you want to use it in.

OK, then, I won't spell names I am not familiar with.

(I know what you mean, though, which is really all mathematical terms are for... I'm being such a hypocrite here)

Question : What's the corresponding Fix functor involved?

:D

:P

HEY WAIT

I wonder if there is a Galois theory involved for $\pi_n$s too.

For covering spaces?

not covering spaces, but, say, fibrations.

in that case the SES is no longer a SES but only right exact

we know of a left-exact galois theory (galois theory in groups)

we know of a short exact galois theory (fields, covering spaces)

what about a right-exact galois theory?

another interesting project would be to recover hilbert's theorem 90 in covering spaces.

that the Fix functor in galois groups is exact is precisely hilbert's theorem 90

what about the Fix functor in covering spaces?

Oh dear, Thomas withdrew from the elections.

So I think I will vote for Daniel, Jyrki and Pedro.

I wonder why people decide to compete and then withdraw.

They should at least give it a shot till the very end.

@BalarkaSen What undergrad institution will you be studying in?

no idea.

i'll concentrate on mathematics than institutions for a few (say, 4) years from now.

@BalarkaSen Are you sure Ted is ignoring you?

yes

@BalarkaSen Do you want me to tell him to stop ignoring you?

OK, sure.

I'd appreciate that.

@TedShifrin, @BalarkaSen wishes to apologise and hope you will unignore him, for your consideration.

Thanks.

@JasperLoy are you the real one?

@user130018 Yes. I changed my blue square to brown square. You can tell it's me by looking at my classic short answers.

@JasperLoy It doesn't look brown

@JasperLoy More like a burgundy or crimson on my monitor

@user130018 It says brown in GIMP, lol.

@JasperLoy What is the color code in GIMP for that colour?

@user130018 #A52A2A

It's maroon, I think, @user130018

But in any case I wouldn't label that color in the family of brown

I always find that Did is very sarcastic man, but seriously, he is very funny with his intelligent jokes :D

Brown is a weird color. in HSB, it is yellow with low brightness. But it obviously has red looking stuff in it.

@MikeMiller There's one drawback of the analogy though.

$\mathbf{Gal}(\overline{\Bbb Q}/\Bbb Q)$ is a profinite group while $\pi_1(X)$ isn't.

Oh wait I see why Grothendieck wanted the algebraic fundamental group then.

I don't know what you mean. $\pi_1(X)$ can be anything you want.

Well it's not necessarily a profinite group is it?

Neither is $\mathbf{Gal}(L/K)$ for arbitrary $L,K$.

Gal(L/K) for arbitrary L, K is not analogous to \pi_1(X), but rather analogous to it's subgroups.

Fine, neither is $\mathbf{Gal}(\overline K, K)$ for arbitrary $K$.

The right analogy is Gal(\bar k/k) with \pi_1(X), for some field k.

@MikeMiller it's profinite for all nice fields k.

Right, I also forgot ARtin-SChreier.

Whoever just joined for the hat, please don't star random stuff. Come on.

what's the problem?

There's plenty of stuff already starred. It'd be nice if you unstarred all the garbage you just did and starred stuff that was already up.

@JorgeFernández Star the already starred messages

and in any case just starring messages of two users won't get you the hat. hehe.

you have to star messages of 8 different users

I just got the "Interesting" hat. I wonder what it is for.

There's a warm wind blowing the stars around.

@skullpatrol is the bad muzak day?

@robjohn Upvoting a highly voted answer, I think.

@MikeMiller ?

@MikeMiller what about Fascinating Ma'am?

Question.

You got it @robjohn

ah

just did. let's see what happens.

alright guys

@MikeMiller I don't understand what you mean by Artin-Schreier though.

You mean the abs gal group over A-S field is not profinite?

No, I meant that you couldn't realize all groups as absolute galois groups.

sure, but i never claimed it

I know you didn't.

Z is an easy example of a group that can't even be realized as galois group of any field ext

it's not residually finite.

let $(q_n)$ be a sequence with $q_n \to \infty$. how does it follow from $$ \| u \|_{p+q_n} \leq \|u\|_{\infty}^{q_n/(p+q)} \cdot \|u\|_p^{p/(p+q_n)}$$ that $\limsup_{p\to \infty} \|u\|_p \leq \|u\|_{\infty}$ ???

i know it looks ugly

oh yeah, also $\|u\|_{\infty}$ is finite

@MikeMiller so you agree with me about the drawback of the analogy?

I don't really care about the analogy :)

well at least alg fundamental groups makes sense now

I feel that I am being imprisoned by my dark thoughts. I need to break out of this prison. AAAAAAAA!

But a long time ago you said you wanted to realize that one group geometrically; I mean, you can still realize it as a fundamental group of something.

@JasperLoy take some mushrooms

@iwriteonbananas Drugs are illegal here and may give you the death sentence.

@JasperLoy mushrooms are a vegetable

@MikeMiller maybe, maybe not

it's a problem i keep at the back of my head

It's so sad Thomas withdrew. He could have been elected.

@iwriteonbananas Your inequality gives you $$\limsup_{n\to\infty} \lVert u\rVert_{p+q_n} \leqslant \lim_{n\to\infty} \lVert u\rVert_\infty^{q_n/(p+q_n)}\cdot \lVert u\rVert_p^{p/(p+q_n)} = \lVert u\rVert_\infty.$$ Now you use e.g. Hölder's inequality to go from $\limsup\limits_{n\to\infty} \lVert u\rVert_{p+q_n}$ to $\limsup\limits_{p\to\infty} \lVert u\rVert_p$.

@DanielFischer amazing, thank u

now i just need to prove the original inequality

@JasperLoy?

@DanielFischer You could aim to get 100k this year.

@JasperLoy It will happen or not.

@skullpatrol Yes?

You ok?

@DanielFischer After you get 100k, you can focus on moderating and not answering, lol.

@skullpatrol I am not having an emergency, if that's what you mean.

icic

@skullpatrol Thank you for your words. I hope you find a miracle too.

No miracle can change what has already happened.

@skullpatrol If you need someone to talk to, you can always email me.

@DanielFischer actually i'm not sure how we go from $\limsup\limits_{n\to\infty} \lVert u\rVert_{p+q_n}$ to $\limsup\limits_{p\to\infty} \lVert u\rVert_p$ using Holders inequality. can u elaborate on that pls?

@iwriteonbananas If $p < q < r$, you can use Hölder's inequality to bound $\lVert u\rVert_q$ in terms of $\lVert u\rVert_p$ and $\lVert u\rVert_r$.

@DanielFischer right i've seen that before

Voting begins in about 1 hour from now.

Ignatz repeatedly rolls a fair 6-sided die. What is the probability that he rolls his first 5 before he rolls his second (not necessarily distinct) even number? The probability that he rolls a 5 is 1/6, the probability that he rolls an even number is 1/2. What do I do next?

Anybody?

Get a die and try it :-)

@skullpatrol I'd prefer not...xD

@skullpatrol Got any hints?

My new hat is not showing up yet.

Anybody? Still no?

@DanielFischer ok, any idea on how to prove that $$\| u \|_{p+q_n} \leq \|u\|_{\infty}^{q_n/(p+q)} \cdot \|u\|_p^{p/(p+q_n)}$$?

@iwriteonbananas Write down $$\int \lvert u\rvert^{p+q_n}\,d\mu$$ and look at the right hand side of your inequality.

@DanielFischer oh yeah that's easy

Do you maybe have an idea for this exercise: math.stackexchange.com/questions/1074884/… ??