A demic is called epi if it is surjective

Guys, could someone give a hint how to construct the iso? I tried working with the exponential, by my problem is that the points $m+n\tau$ don't scale linearly

@EdwardEvans rip

I think exp is the right call

But...

@MikeMiller Nothing's stopping me. I was worried the entire time about whether you get trivial second homology with this construction (the standard construction of the torus lurking in the back of my mind the entire time), but I just realized there is no issue when we pick an abelian presentation, because then the degree formula implies that being in the kernel means attaching along the trivial loop, which we don't do.

I should tweak it right

You want the kernel of that map, for which you need to specify an antecedent of q

Well, I was thinking, if $q=r e^{i\theta}$

I'm thinking $q = e^{\tau}$

but then

$\tau$ is indeed sent to $q$

but

what was the question

e^{n+m\tau}=e^n q^m

So I was thinking, including $2\pi i$?

So that the effect of $n$ gets annihilated

but that felt ugly

but maybe that just is it

Yes, you can take $q=e^{2\pi i\tau}$

oh welps ok

that one I had, but I felt too insecure about that map

thanks

It's just an homothety

(if that's the spelling)

show it's a compact Riemann surface, use covering space theory to calculate the fundamental group, deduce its genus and then quote classification of compact Riemann surfaces

lol

what was the trivial second homology question retart

we just wanted to construct a CW complex with arbitrary abelian group as first homology and trivial homology in other degrees

reduced*

Take a Moore space

Which is a product of M(Cyclic, n)'s, which is just the coMobius

well, constructing a Moore space was the point

we didn't assume f.g.

Every group is direct limit of fg ones right

Homology commutes with direct limit

You just take a homotopy direct limit of the Moore spaces

is the category of CW-complexes cocomplete?

I don't know what words mean but you have these Moore spaces $M_n$ I guess you just take the mapping cylinder construction on $M_n$'s

Moore space is not functorial so you might have to do work

those words mean "is the direct limit of some Moore spaces a CW complex?"

First of all you have to tell me what the maps between Moore spaces are

also true

but you were the one who suggested to take a homotopy direct limit

It's a good suggestion

Might work

not saying it won't

@Thorgott Yeah, I recognized the concern. I was just phrasing it in a way to hopefully make the point clear

The fact that the relations form a free abelian group is crucial

@Thorgott That's not a category that's a collection of objects

why doesn't bouquet of circles work?

For what?

@Thorgott s question

is there any ambiguity in talking about the category of CW-complexes or what's your point?

worried whether I just want continuous or cellular maps?

yeah

that's a huge concern

lol

we're passing to the homotopy category, so doesn't matter

are you kidding me

yes it does

every map is homotopic to a cellular map but that is not unique up to homotopy!

a cantor set lives in the homotopy category

you can make objects by upto homotopy cellular shit which is not cellular

@Thorgott however I appreciate the bluntness here, we've trained you well

Now that I have said this i need to remind myself of examples lol

Ah OK

$S^1$ and $T^2$

Hmmm no I think I need something more complicated

Bah I've been a fool, the point is that the right notion of homotopy now is cellular homotopy, and yes every homotopy of cellular maps is itself homotopic to a cellular homotopy between those maps

Fine!

But you definitely do not pass to the homotopy category when you say "cocomplete!"

I botched my point at the beginning but this is indeed a subtlety and ought to matter on the point-set details of your choice of category of spaces

Messy

Fun fact $M(\Bbb Q, n) = K(\Bbb Q, n)$ for odd $n$

@TedShifrin Sorry you had to see this. One never wants to start off their day with a homotopy category.

I think that's the only K(G, n) I know how to write down other than n = 1 and K(Z, 2)

lol

Well I also know how to write down K(Z/2, 3) but not why I can write it the way I can

Well, I sort of do but whatever

I blame Balarka for bringing up the homotopy category

@MikeMiller Is there a good notion of localization/completion of nilpotent groups at a prime?

Look in May, Concise course 2

Good notion = preserves exactness

Yeah pi_1(K(G, 1)_p) does not do that

You get pi_2's in the Postnikov tower

But that's all you get

Then the answer must be no

Localilzation tho I believe

Hm

I trust you

Don't trust me look at that beast of a book

Lol yeah

It's actually More Concise I think

random aside question, how do I pronounce Palais (as in Richard Palais)

Gotta love J P M

s silent

pa-la-ee

I think

Pa lay

Hm

@BalarkaSen It certainly is not more concise

But it is quite encyclopedic

I look there for any basic question about localization and completion

And also Hopf algebras those are in there too

Guys , please do check this .

alright, thanks

But I think tommaszo tammo bombadillo is bigger

Tammo Tom Bombadil is very scary both in content and size

@BalarkaSen youtube.com/watch?v=t989-ukRYTY

Balarka is not very french

Hey Thorgott did you prove isotopy extension yet

Moser's paper with his trick in it is interesting. He proves his theorem a different way first, and the trick is seemingly just a "bonus" off to the side.

His local argument is important in generalizations to lower regularity IIRC. I think in low regularity the flow argument will fall apart.

what the hell happened with my perfectly simple question and why are you guys now at homotopy limits

and argueing about categories

Your question is still simple

yes

@MikeMiller what do you mean by "low regularity"?

You don't have many derivatives.

Oh, so in like C^k for a finite k

Yeah, though I think you ought to be fine for k>1.

Maybe even k=1. But for C^0 volume forms you will have to do something else, because the flow argument wants to differentiate the volume form.

I don't remember though, spitballing from a talk I heard once.

Heh.

Well thanks, that's interesting.

Just be careful, I have surely made an error somewhere here. Don't trust any of the above without seeing references :)

Indeed, I will just aim my spitballs lower if I ever try. ;)

Why is regularity an issue you can just nudge a volume form to be smooth

not yet, but I'm gonna return to thinking about it tomorrow

classic Balarka

currently writing down the last details for my seminar talk tomorrow, so that takes precedence for now

Its a nonzero section of the last exterior power if you nudge it its still nonzero

Because the set $$\{(\omega, \psi) \mid \exists \varphi \;\;\; \varphi^* \omega = \psi\} \subset \Omega^n \times \Omega^n$$ is not obviously open, is it?

Seems like you would conclude it's fiberwise open.

But this is what your argument is trying to use.

Ah I see I didn't remember what Moser's trick tried to prove

Of course nudging should be open

Actually I think you want this set closed.

Right, you want to say "nudge $\omega, \psi$ to smooth forms, construct $\varphi$, now take a limit of the $\varphi$'s as these guys move to the C^0 forms"

And this seems like a mess

yes thats the proof

Yeah but they need not have limits, you can always take your sequence $\varphi_n$ and adjust it by precomposing with volume-preserving diffeos and you have a new sequence.

This should be a working argument for things involving G-actions of compact groups but not by Diff

Subsequential limits brah

Gromov trick

plus Diff is open in C^infty

I don't buy any of this

I gotta write linear algebra

Limits always exist

- Gromov

Did you know thats what he said when people objected to his proof of polynomial growth theorem

In Moser's argument, he uses the Hodge decomposition. I don't recall this being an issue in the symplectic case. Is this fixed by closedness of the symplectic forms and by the choice of cohomology class?

He said some sequence of spaces has a Gromov Hausdorff limit when it obviously doesnt

You have to use ultrafilters

He just responded by saying of course it does, everything has limits

mans doesnt care

this is what I call the Gromov trick

Take limits of things which do not have limits by saying obvious

he realized that there is no such things as non-existent limits, only working in the wrong category

yes

hes big on categories nowadays

kyategworikal

(in his accent)

@anakhro This is just how he thinks about cohomology. He's explaining why, if two closed forms lie in the same cohomology class, there's a smooth path of closed forms between them in the same coho class.

lol he uses Hodge decomp instead of w + t*da

?

@MikeMiller ah, I see!

@BalarkaSen Yeah it's a little unclear why.

Analyst brain

Massive legend Moser

is there any clear example of a semigroup that "becomes" a monoid whenever you remove/relate all but finitely many elements from/in it, but is otherwise just a semigroup?

is there any clear example of a semigroup

other than the naturals that's a good question

is there any clear example of algebra

is there any clear example

I might allow murky example for our convenience

It's not very clear to me what kind of examples are you looking for

1920 algebra brain

lololol

My favourite magma is Rock-Paper-Scissors.

cringe

2020 algebra brain is the same but with "object" appended after every structure in that picture

lol

@AlessandroCodenotti and the arrows are actually meaningful in some obscure category

@AlessandroCodenotti think of big presentations and you kind of add relators

That's not an answer to his question lol

If you have a question it comes from somewhere

where this isnt just logical implication

@MikeMiller who's question

yours, but I'm done with this game I think

I don't understand

@anakhro and I'm done with this game

Can you formulate your question in a precise manner? You say the naturals are an example, then talk about relators when the naturals are a free semigoup, so I have no idea what kind of examples you're asking for

I'm done with this game... time to hop on this train, I've got way too much fame, everybody knows my name, I can't be so tame, friends call me by my name, I say what I say, then hide my face, cause I'm a disgrace, I pay for my grace, I go at my own pace

don't know how I can be worse than a random generator in a game

@AlessandroCodenotti sorry that part was a joke wrt Balarka's question about an example of a semigroup

@BigSocks to get it straight, you are looking for a semigroup S without an identity, a monoid M and a surjective semi-group hom S-->M?

@BalarkaSen Was this a serious question btw? Because I have a good serious answer in that case

No dude

@AlessandroCodenotti What, $e^{-t\Delta}$?

Yeah

I guessed thats what he had in mind

No those are ugly semigroups

Well, any dynamics

Isn't that a monoid though

I was thinking about the Ellis semigroup in dynamical systems

@MikeMiller yes but people call it a semigroup for some reason

@MikeMiller Massive brain

@anakhro probably, but I would like the semi-group hom to be really constructible like not just proving the existence of one type of thing

Heat flow go forward in time but what if you stop time like in the movie Clockstoppers

ideally both algebraic objects are at most countable and not too messy

@MikeMiller Ew

why would you watch that

A long time ago an algebro geometer gave me a puzzle which was like

I watched it as a kid and again about a year ago. I was pretty disappointed with the rewatch

Can you make an inverse system of nonempty objects and surjective morphisms with 0 inverse limit

There's like three good scenes

What I told him immediately was take a ODE which blows up backward in finite time

This doesn't work but I'm proud of the idea

He was like wtf

@BalarkaSen makes you double check at least

@BalarkaSen there's examples even in Set right? Or am I misremembering?

yeah

you cannot make a sane geometric example

@BigSocks there are "trivial" ones which are immediate, I think. Just take one of the semi-groups of order 2 which are not monoids, and then map it under the constant morphism.

the indexing set is in 2^(2^R)

or something like this

Something like look at order embeddings from countable ordinals into $\Bbb R$ with "extending" as order, then the limit if it existed would embed $\omega_1$ into $\Bbb R$

yeah

actually your weird trees work

you have $\omega_1$-trees with empty ends right

@anakhro yeah ok maybe I should emphasize the semigroup should be countable and not finite to begin with, but it was a good example

Aronzjain trees

or whatever

and trivial is usually good imo, but I allowed it to be too trivial

yeah but that's very similar

The usual construction of an $\omega_1$-Aronszajn tree is by looking at particular functions from countable ordinals into $\Bbb Q$

ok ok no one asked for the construction

You need to take some care because you also want to keep every level countable, but the reason there's no $\omega_1$-branch is exactly the same

also it doesn't have to be a semigroup, like you could in some way mess with associativity to get left and right 1 not the same, but I haven't thought of it too much

You need to take some care

I shant

@BigSocks take the ring of continuous functions with compact support in R, and then the ring of continuous functions [0,1]-->R. The former is a proper semigroup (with multiplication), and the latter is a monoid. The map is just restriction to [0,1].

@AlessandroCodenotti that actually doesnt sound too bad

It's not countable, but at least it's useful.

pretty cool example, but yeah, a little big. still, I had not thought about this kind of example so thank you

Imagine trying to derive the Lorentz boost in 4D using $e^x - e^y - e^z - e^t$ coordinates

straightforward but very computationally involved

-John Lennon, ca. 1970

it's literally a problem from non-heaven

He imagines that there's no heaven so that fits in

@BigSocks here is a countable one. Let M be your countable monoid. Then make S as the set of finite words in M (it's countable). Then the semigroup operation is letter-wise multiplication in M, with a truncation to the shortest word length. Mapping to the first letter gives you a surjective hom into M.

Voila.

as in, you take any word in S and you map it back to M as the first letter of the word, so an element of M?

Yes.

(nevermind for this remark)

hmm that is a good one, but I just realized how doesn't fit the original criteria because the map doesn't remove all but finitely many of the semigroup's elements

That would make M finite, wouldn't it? And that's not desirable?

so I suppose, in light of this, it would be favorable to consider $M$ to be finite

maybe it still works

M finite works, but I thought oyu wanted everything to be |N|.

yeah, I thought I did, but I think I don't actually. good that it still works with M finite though

this is cool, thank you for thinking so much about this and giving me such a good solution

(hint: what would the identity have to be in order for it to be a monoid)

empty word?

empty word doesn't multiply with anything in M. :P

right empty word is not something in M

in particular, not in S.

What's a good arithmetic geometry textbook

@EdwardEvans @LeakyNun @anyotheralgebraist

I am reading Lorenzini "An invitation to arithmetic geometry" and it seems good

Ok ill check it out

@BalarkaSen do you have anything more specific

not really

i dont know anything more to be specific

jmilne.org/math/CourseNotes/LEC.pdf

J.S. Milne Etale Cohomology

@anakhro still kinda stuck on this- you are referring to S and why it's not a monoid as a collection of words?

@BigSocks Yes.

right and it's not bc you don't have an empty word in S because there would be no analogue in M I guess

Not quite. The empty word wouldn't function as an identity either.

let M just be {1,0} for example.

S is a bunch of strings of just 1

ok

I changed it to binary.

S is $\{0,1\}^*$

So finite binary strings, and the operation is that you truncate the longer of the two strings, and then letter-wise addition in M.

oh ok then I didn't initially understand the operation

It's not concatenation.

so it's $S \times S \to S : (r,s) \mapsto (r_i * s_i) \forall i \leq min(len(r), len(s))$

sorta

and $*$ is $*^M$

min of the length

But yeah

of course

Hi all iv'e got a general PDE and wondering if anyone knows of any particular examples of it which are interesting. The form of the PDE (solved for $u : \mathbb{R}^+ \times \mathbb{R}^d $) is $$ \partial_t u = \div ( u A ( \frac{\nabla p (u) }{u} + \nabla f ) ) $$

@anakhro yeah I was blinded by thinking this

where A is a singular matrix, $p:mathbb{R}\to \mathbb{R}$ is "the pressure" (take that to mean whatever you want) and $ f : mathbb{R}^d \to \mathbb{R}$ is an "confining force" (take that to mean whatever you want)

ok so an identity in $S$ would have to be $(1^M, 1^M, ...)$, but there is no such thing because we only have finite strings

@LeakyNun thx

@BalarkaSen that is on my to-read list buried somewhere deep...

we can read

@BigSocks $A+$

if my mood isnt too bad

also in other news $\bigoplus_{i=0}^\infty \Bbb F_2$ has no multiplicative identity

right

deep

never noticed it

@anakhro eyyy, very neat! thank you for also making sure I got it lol. you a logician?

No. I am a crappy math student

hahaha not if you ask me

@LeakyNun nice connection

@anakhro these are not mutually exclusive properties

lol

@BalarkaSen there seems to be a discrepancy between the courses you (have to) take and your actual knowledge

slight discrepancy

@BalarkaSen Seems like you should just prove this in set and invoke universal property

But maybe they want you to prove naturality of the associator

:D

Commutative heptahedra

yoneda moment

what do you call functors between natural isomorphisms