my layout is just really bland

@LeakyNun fun fact: I'm showing an equivalence of three statements where every single direction uses Zorn's lemma

@Daminark for each open $U \subseteq \Bbb R$, declare $\mathscr F(U)$ to be the functions $s : U \to F$ such that $\pi \circ s = \operatorname{id}|_U$, i.e. a sort of "partial" sections

@MatheinBoulomenos gets triggered

explodes

ugh @Mathein

that's just to get a hold of infinite-dimensional reps

@Daminark and then this will give us our original sheaf. If we started with a presheaf, we'll still end up with a sheaf, called the sheafification (you might have heard of this)

you can ignore it for finite-dimensional ones and I said that

@Daminark and this is why elements of $\mathscr F(\Bbb R)$ are called "global sections"

the space of global sections, Leaky

to be fair, the only Zorn's lemma you need for the direction that is actualy applied to representations is the existence of maximal ideals

@TedShifrin corrected

@Daminark it is in this sense that it is the space of global sections to the projection from the espace etale

from l'espace etale

de l'espace etale

if you check a commutative algebra book like A-M and look how many lemmas use the existence of prime ideals, then you'll find that a lot do

c'est dans ce sens qu'il est l'espace des sections globaux a la projection de l'espace etale

Hmm, I'll have to write this on a board to be sure I have details but I do see the idea for sure, thanks!

Problem: Let $M$ be a module over the integral domain $R$. If $x \in Tor(M)$ is nonzero, show that $x$ and $0$ are "linearly independent"...Question: what does linearly independent mean in this context?

Nothing with $0$ is ever linearly independent. Say what?

@user193319 the same as it would mean in vector spaces, I'd imagine

but 0??

There's a quotation mark and it's not honestly linearly independent, also it's torsion, so let's see for real

That's what I was thinking, but what Ted brought up is precisely what is confusing me.

I think you're supposed to show that if you have a torsion element $x$, then $\{x\}$ is linearly dependent

Sup nerds

Yo

the statement as you wrote doesn't make sense

That sounds better, since $R$ is an integral domain, @Mathein.

heya nerd Eric

How goes

but that's what I would guess as the intended problem

Hey @Eric

Ah, so a singleton isn't always linearly independent in a module? I know it's true for vector spaces.

see what I mean about the importance of good (not to mention correct) exercises, @Mathein? :P

@TedShifrin true

correct proofs are also important

Most people would say that. :P

@user193319: Linear independence with modules turns out to be highly subtle, as I recall.

@user193319 try to do the exercise as I wrote it

@Daminark espace etale is so amazing

It seems interesting. I don't know enough about AG to really tell what you can do with it but hopefully soon enough this stuff will become clear

@daminark do u know what ur doing now

@Leaky, Demonark: Next, understand how it's different from a covering space, topologically.

we were working through Zagier's proof of the Eichler-Selberg trace formula in the appendix of Lang in our modular forms lecture, the prof is sick, so a post doc (and friend of mine) has to the proofs, there are some really nasty errors and stuff like "that's easy with the substitution bla", but substitution bla just leads to a really nasty integral

hmm

what exactly do you mean by different?

lol i guess this is what vakil meant from "Some people strongly prefer the espace etale interpretation"

@loch of what?

In particular, what does it spit out for the sheaf of C^0 functions?

subtle errors are really nasty when the proof is just multiple lectures of calculations (that aren't even carried out in the text)

@loch I think it gives good intuition

UGH... WHY IS IT SO BRIGHT OUTSIDE!?

@MikeMiller C^0 is continuous right

Projection from the espace étalé to the base is a local homeomorphism, just like with a covering space. But what else is needed for a covering space, and can you see where that goes wrong?

It's called summer, Xander.

Summer can suck it. :(

LOL

Where is my protective layer of smog blown in from LA?

Oh.

Maybe Mike's holding on to it.

if the étale space is good/useful for topology, that's a proof that non-Hausdorff spaces do matter

iirc that's non-Hausdorff even for totally reasonable sheaves

@MikeMiller it does - for me it's kind of just something that's good to know behind the scenes rather than thinking too much about it though - but as balarka mentioned earlier maybe they're important in topology

well, not in the analytic/algebraic category, @Mathein :P

Idk about them being important in topology

@MatheinBoulomenos Why would you ever want to consider a space that doesn't possess a complete metric?

I just think it's good to have the picture at the back of your head

I have taught sheaves as topological spaces every time in R.S. and complex manifolds, but ultimately it's the presheaf of sections we care about.

@TedShifrin right, the fact that distinct functions can restrict down to the same function is the culprit

that makes the preimage not a disjoint union

That's not the issue, @Leaky. It's a local question, not a global one.

@XanderHenderson I like $\Bbb Q_p$, too, no doubt about that

Sheaves are local to global information; the escape etale is just the clearest way of saying this immediately

@MikeMiller i agree

Especially because it's clear that sections form a sheaf!

But $\mathbb{Q}_p$ is complete! >:(

He of course meant Q with the p-metric

@XanderHenderson I was just saying there are some spaces I consider that do have a complete metric

Ah, okay.

but most don't

@TedShifrin if we have open $U \ni x$, we can't partition $\pi^{-1}(U)$ into disjoint sets, each homeomorphic to $U$, since what I said makes the obvious partition not disjoint

Have of my Ph.D. thesis consists of examples in $\mathbb{Q}_p$

adele rings are also local to global things (with local to global in a different sense), so maybe I should think of etale spaces as the topological analog of adele rings

I'm getting really vague here

Fun fact: there is a reasonable definition of fractal such that singleton points in $\mathbb{Q}_p$ are fractals. :\

@TedShifrin in particular, the fibres of a point, i.e. the stalks, do not have the discrete topology

but you can do adele rings for Riemann surfaces, too iirc

no, Leaky, as the espace étalé is germs ... so near the point $p$ you have only a unique germ, so on small open sets, only one function.

No, they do have the discrete topology.

:o

You're missing what goes wrong.

Even in the holomorphic category.

isn't that like all of the criteria already

according to wiki

Yes, even covering is the main thing to look for in a covering space.

isn't it satisfied

You were claiming it wasn't, but for the wrong reason(s).

if it has discrete fibers, the only thing that can fail is that it's not a fibration, so it doesn't satisfy the homopy lifting property

oh, joy! my people are home! later, all!

That's too hard for my brain, Mathein.

I want to look for even covering.

Later @XanderHenderson

bye Xander.

@Eric so the options which came up were group cohomology, modular forms, and just raw ANT (class groups, all that) First two are more likely at the moment

See you @Xander!

perhaps try running through some examples, like a skyscraper sheaf

Nah.

I'll suggest the sheaf of germs of holomorphic functions on $\Bbb C$.

But continuous/smooth will have more issues. :P

Hmm, so I'm still not well-versed enough in this but are germs of holomorphic functions somehow easier by virtue of analytic continuation?

No there are just fewer of them

the same goes through my mind

ok let's say we have open $U \ni x$. Then, $\pi^{-1}(U) = \{ (z,U,f) \mid z \in U, f:U \to \Bbb C \}$

I always like telling people to understand the espace etale of the trivial sheaf with fiber R, and that of the sheaf of sections of the trivial vector bundle R

"What is the difference between R and R?"

one has discrete topology, the other one hasn't?

on the fibers, I mean

no, both do, @Mathein

Is Mathein his first name now?

but vector bundles don't have the discrete topology on fibers @Ted

why don't we just call him Learn then

I've always called him that. I think I have his actual German name somewhere ... but ....

but the espace étalé for the sheaf of sections does, @Mathein

you can call me Lukas if you prefer

ah I see

Doxxed

Why are u guys spoiling

I'll call you Learn Want

Dark Amin

@LeakyNun I prefer that

We still haven't hit the main point, @MikeM.

@Daminark lol, some people here are under their clear name

isn't it true that two things there are equal iff the functions are equal?

because of analytic continuation?

:P

@Ted so it is indeed a disjoint union?

There's indeed a disjoint union, yes. But ...

There is something wrong there.

What is a germ of a function at $x$?

each $f:U \to \Bbb C$ defines an open set $\{(z,U,f)\}$

@TedShifrin $(x,U,f)$

@Daminark cool

Can you use the same $U$ for every $f$?

I see

this is BS

I see

interesting

@Ted have I solved it?

that equality isn't true

No, it isn't. Can you be more precise? :)

hmm

I can define a function with domain smaller than $U$

that cannot be extended to the whole of $U$

and that function will define an open set in the espace etale

that will not be in any subset of $\pi^{-1}(U)$ homeomorphic to $U$

OK ... so in complex analysis land the term "radius of convergence" might be relevant

right ...

yeah that's kinda what i mean by domain

so it's a covering space iff the sheaf is flasque?

cannot be analytically continued to $U$

ok so the culprit is the non-surjectivity of $\operatorname{res}_{V,U}$

Um, I've forgotten flasque, @Mathein. I always dealt with fine.

flasque means the thing is surjective

res

flasque means all restriction maps are surjective

:)

I imagine it's fine if you don't want to worry about flasque sheaves

chuckles

Oh, then I guess so. ignores Demonark

Is flasque flabby

yes

@EricSilva yes, flasque is French, and flabby is the translation I think

So even the fine sheaves (things with partitions of unity) won't work, of course, because we can still blow up ...

Lol French

You're flabby

@Daminark do you know how to define a sheaf by only specifying the sections on a basis?

I sure is.

@MikeMiller how rude

I am quite skeptical that covering space implies flabby. It should be much weaker

Hi everyone

No, the other way.

@MikeMiller does flabby imply covering?

Oh sure that's clear enough, but he said iff

Hi @Jalapeño, whoever you are.

Oh, I missed the extra f.

Hi Professor Shifrin :)

his full name is Jalapeno Nachos ... o.O

Do we know him under another non-de-plume, Leaky?

it was a question, I'm not sure

In German we say "welke Garben" for flabby/flasque sheaves, so one theorem is "welke Garben sind schlaff", "flabby sheaves are acyclic" which sounds really funny in German

I think most things do

so things asleep = acyclic?

when studying a rigid body in some paper that gives model equations for its motion, why do the authors only care about the center-of-mass velocity? what about the instantaneous velocity at the other parts of the body? the papers make a "quasi-steady" assumption - does this mean that the velocity is assumed to not change throughout the body for a time, t_0?

hi @LeakyNun :)

@MatheinBoulomenos everything anyone ever says sounds funny if you listen to it really hard

center of mass velocity gives "net" translation, and other than that you do instantaneous rotations about the center of mass, @Jalapeño

schlaff is more like slack, limp etc. @Ted

limb??!!

@TedShifrin ah, I see

I mean limp

oops

Limp?

@TedShifrin oh and now we have an alternative definition of the support of a function :P

LOL

Auslautverhärtung strikes again

We do, @Leaky?

This chat is warzone where everyone uses sniper rifles

You supply most of the ammunition to everyone, Demonark.

@TedShifrin if we have an abelian group-valued sheaf $\mathscr F$ on topological space $X$, then the support of a global section $s \in \mathscr F(X)$ is the points $p \in X$ for which the germ $s_p$ is nonzero

@TedShifrin then what does the model being "quasi-steady" mean? I don't see what's quasi-steady about it; I had guessed that it meant that the velocity does not change throughout the body for some time, t_0

OK, @Leaky, the same definition.

alright

I don't know the definition, @Jalapeño. Is "quasi-steady" that the center-of-mass velocity is constant?

@TedShifrin hmm the CoM-velocity is definitely not constant - the rigid body is falling

so what's the definition?

lmao

> What I know is $\left(\dfrac pq\right)\left(\dfrac qp\right)=1$

I've never liked that notation.

@LeakyNun it's just cancellation of fractions, duh, don't know why Gauss wrote so long proofs of that

Well, the OP thinks it's a regular fraction in parentheses.

and even for the wrong statement!

Why not a better Legendre symbol notation?

none of those things is a ratio

this is just wrong

huh?

why do people keep mixing up ratio and fractions

oh

I don't get upset about confusing a ratio and a fraction. They're isomorphic.

there was a time a friend of mine in high school asked me why 3:4:5 = 6:8:10

if 3/4/5 is not 6/8/10

Ugh.

I've never in my life written a "multi-ratio" like that.

But I know that some people do.

because he thinks a ratio is the same as division

those brackets in a lot of variations are well-established in number theory (and partially algebra), you use them for Legendre symbols, Jacobi symbols, Kronecker symbols, Artin symbols, Hilbert symbols and for generalized quaternion algebras as elements in the Brauer group

@TedShifrin he means (3/4)/5 and (6/8)/10

It's bad enough that $(a,b)$ can be a vector or an interval.

Yeah, yeah, @Leaky. You never give me credit for figuring out stuff.

sorry

@TedShifrin Quasi-steady-state-approximation: A method to reduce the number of variables of a system that includes processes on different time scales which can be separated into slow and fast. One assumes that the fast processes are always in a steady state, which changes on the slow time scale.

That's too complicated for me.

I'm TAing intro number theory and so far there was no confusion with the notation for Legendre symbols

Sure, @Mathein. That doesn't mean we couldn't have a better notation.

I agree but I really don't think it's a big issue

Sure, we all recognize it in context if we're advanced enough.

But there are mathies all over who aren't.

It's like the difference between $D$ and $d$ in multivariable calculus. In particular, $D^2$ and $d^2$. :P

@TedShifrin yeah, the system is described by a force model - so, not a great deal of variables - position, velocity, acceleration, torque, so, I'm not really sure where the quasi-steady assumption was applied to

notation for differential things in diff geo is a mess. It doesn't help that everyone has his own notation

I was talking calculus/analysis, not diff geo, @Mathein.

Stop it with your prejudices.

Differential geometry notation is ok if you stay away from the Riemannian geometers

And remember that sometimes [v, v] really needs a 1/2 out front

Riemannian geometry was what we did in diff geo

@Jalapeño: Maybe the net change in rotational stuff is of a size comparable to the translational velocity. I dunno.

Rip

and I wasn't the only one in that course confused by the notation

I did tell you to stay away from them