point-set-topology

[sleek thicket]

Meaning that every set is open

[quasi forum]

Why does that mean every set is open?

[marsh forge]

every subset of the space

is an open set

[sleek thicket]

yes, sorry

[gritty widget]

(by definition)

[sleek thicket]

So unions of open sets are open right?

[marsh forge]

The proof is a cute exercise

[quasi forum]

Oh, is that what discrete means in this scenario?

[marsh forge]

dont tell them

[sleek thicket]

Yup

[quasi forum]

Yes, fully agree

[marsh forge]

discrete topology = every subset is open

sham gave you one of the two things you need to notice to prove the claim

the other is the important one tho

[quasi forum]

So since a metric space is a topological space, there must be a collection of open sets contained in X.

But all possible subsets are finite, so the only option is that the finite subsets are open. 🤔

[shut moat]

probably better for your mental health

[gritty widget]

oh no

i assume this person is an unpleasant fellow

[uncut surge]

i don't even think he was that bad

i don't get why he has that reputation

[gritty widget]

my sympgeo prof is a very nice person so i think i'l just ask her

[quasi forum]

So wait, does that mean a finite metric space is just a collection of isolated points?

[uncut surge]

Yeah

If you just have finitely many points, there's always going to be some ball around every point that only contains the point itself

[quasi forum]

Ohhh, that's all there is too it. 😆

I was worried I couldn't jump to that conclusion, but it makes perfect sense

[uncut surge]

Yup yup; if you have a well-defined notion of "distance", you can always just choose a ball around a point which is smaller than all the distances between all the finitely many points

[marsh forge]

Hank is the worst lol

truly an annoying person

[uncut surge]

It's pretty intuitive I feel

[marsh forge]

one tier below negi

[quasi forum]

Since their are only a finite number of distances for a given metric, so just choose the radius of the ball to be the minimum of the distances.

[marsh forge]

also a meme

because he got so mad he left hopf once

because he said the 0 map was nonlinear

and got roasted to hell

[uncut surge]

Maybe it was just the rest of the server which made hank bearable to me back then

[shut moat]

he did??!

[uncut surge]

But all cringe and toxic

[marsh forge]

yeah hahaha

[shut moat]

omg lol

[quasi forum]

So how would this claim break in a general topological space?

Wait, not that one.

[shut moat]

I feel like you're approaching this question in the wrong direction

[sleek thicket]

didn't hank used to be the name for

[shut moat]

you start with a set (be it finite or infinite), and then you give it a topology

[quasi forum]

This right here, Eshamrock said this is generally untrue.

[sleek thicket]

or no it had like an egg thingy up top

[marsh forge]

yes

[uncut surge]

You could equip the two-point space {a,b} with the topology { {}, {a}, {a,b} }

That's a pretty standard and boring one

[quasi forum]

Is the left most entry the empty set?

[marsh forge]

it is generally untrue yes

if you space has closed points it is always true

i.e. if the set X-{x} is open always

[cloud owl]

better yet, on a set X you can just have {{}, {X}} i think

[uncut surge]

And yeah, the leftmost set is the empty set, that must be in every topology

[cloud owl]

best topology

[marsh forge]

Thats the trivial topology

[uncut surge]

oh true, the indiscrete boy

[marsh forge]

yes

Anyway I think your statement is equivalent to a space having closed points

it is

[quasi forum]

So are you saying this isn't closed?

Sorry, still trying to process

[uncut surge]

The set {a} in this example is open, but not closed, since its complement {b} is not open

[quasi forum]

Ohhhhhh!

[uncut surge]

And you can do the same with every set X that contains more than one element, if you equip it with the indiscrete topology {{}, X}, since then no single point is closed, because the complement of no single point is in this topology

That was the other example that was mentioned earlier

(i like calling this one the "indiscrete" one rather than the "trivial" one because i feel like the discrete topology is also pretty trivial so a priori i wouldn't know which one of the two you mean)

[quasi forum]

Oh, i think I get it. Because the only sets you are considering in that topology is the empty set and X itself. So $X\setminus {x}$ is not in the topology.

[gentle osprey]

dackid

[marsh forge]

(assuming X\neq {x})

[uncut surge]

tru

[quasi forum]

That's a good thing to point out

[marsh forge]

but anyway

most nice spaces

have closed points

meaning that X-{x} is always an open set

in this case any finite subset is closed

(and its an if and only if)

[quasi forum]

So any finite subset is closed iff X{x} is an open set?

Is that the proposition?

[marsh forge]

any finite subset is closed if and only if X-{x} is open for all x in X

[quasi forum]

Okay cool. So I had the right idea, just needed to be more specific

[marsh forge]

(this is another good exercise that is just about the axioms)

[quasi forum]

So really, it's not that there is a collection of open sets in X that really matters, but it's how the collection is defined.

That makes it far more tangible to understand

[marsh forge]

Im not sure what you mean exactly

a set doesn't have a topology until you give it one

[quasi forum]

Oh for sure

[marsh forge]

and in general the same set can have many many different topologies

and they can be nothing alike

[quasi forum]

But the way the topologies are different is based on the open sets you consider in the collection

[sleek thicket]

yup, that's the definition of a topology

can you have a strictly finer topology X be homeomorphic to a strictly coarser one?

I don't think about gen top stuff often

(this isn't a question directed at you dackid)

[quasi forum]

I was really struggling to see how this generalizes. Thank you so much for the help. I feel like I am getting a grasp of general topological spaces now

[shut moat]

I feel like this shouldn't be true

[quasi forum]

We've only been dealing with metric spaces, so it has been difficult seeing how to generalize into topological spaces. This channel has been really helpful for grasping that intuition.

[marsh forge]

i feel like it should not be true too

and because of that

i am certain it is possible

point set is always bad

[shut moat]

[gritty widget]

metric spaces have nice things like boundedness, and sequences actually work in them

and then you have non metrizable spaces

non-hausdorff spaces

[shut moat]

agh fuck

that is so weird

[marsh forge]

oh man

thats so cursed

i knew it

[sleek thicket]

lol

Love it

I was actually thinking about this kind of theorem

Like a big recognition principle for topological spaces

Which would generate an example

But I couldn't remember any concretely

The real example is cuter but less wtf

[flint cove]

That's a really interesting example. Let's call Xi=:(ℝ,τi).
I think the thing that makes this work is that the continuous map c: X1→X2 induced by the identity (c for coarsening) takes the odd integers and turns them into generic points.
If we look at the T0 quotient, i.e. identify all topologically indistinguishable points, for X1 we get cofinite ℕ plus a point * whose closure is everything in the first case, and in the second case the same with 2ℕ. the map c now descends to *+ℕ → *+2ℕ, but not by multiplication with 2, but by mapping all odd numbers to *, since every odd number, is after coarsening also just a point whose closure is all of X1.
I'm wondering whether that still works when the coarsening induces a bijection on the T0 quotients, i.e. it doesn't introduce any „new“ topologically indistinguishable points.
(If taking T0 quotients is not a functor or I made any other mistake please shoot me)

Wasn't a point whose closure is everything called “generic point” or something?

[tough imp]

Yes

[rotund thicket]

somebody drew this before my topology lecture lol

idk if its topology tho

[gritty widget]

found an answer

[sleek thicket]

Yayyy!

wait I think we're doing this in my bundles class rn

lmfao

I haven't been in a week

[gritty widget]

[sleek thicket]

dω = K dA apparently

[gritty widget]

[sleek thicket]

Proof

[flint cove]

What ist $\mathscr J(M, \omega)$? Isn't that complex structure stuff, and is $(M, \omega)$ a symplectic MF?

[gentle osprey]

lux

[gritty widget]

$(M, \omega)$ is a symplectic manifold and $\mathcal{J}(M, \omega)$ is the set of $\omega$-compatible almost-complex structures on $(M, \omega)$

[gentle osprey]

(T*Terra, dqⁱ ∧ dpᵢ)

[gritty widget]

so ya

[flint cove]

Ah, almost-complex structure was the word

We just had one lecture about that in our diffgeo course, and it was confusing

[gritty widget]

it's neat stuff

[sleek thicket]

I didn't realize my class was doing thus

[flint cove]

(or at least we were missing context about why that should be of interest. It didn't feel complicated per se)

[sleek thicket]

Would've gone if I knew

[gritty widget]

the lecture i had earlier was a bit confusing so im reading around now

i guess it's of interest because it's kind of an intersection between symplectic and kahler complex geometry?

i dont want to make any definite statements since i litrally got introduced to this a few days ago

[flint cove]

from my incompetent pov almost complex structures, symplectic, and kähler geometry are all in the same corner, lol

[sinful pecan]

kahler is symplectic

[gritty widget]

hm maybe i should have said complex geometry

[sinful pecan]

https://en.wikipedia.org/wiki/Nearly_Kähler_manifold

[gritty widget]

not "almost kahler"

[sinful pecan]

oh yea i had them backwards

https://en.wikipedia.org/wiki/Hermitian_manifold#K.C3.A4hler_manifolds is the right one

[gritty widget]

[quartz edge]

yall really be one semester ahead of me talkin about all this crap

[gritty widget]

just skip a semester

[quartz edge]

i want to be a true master.

[gritty widget]

that's what i did

/s

[flint cove]

idk, to me a lot of topics I don't know about feel like „here are arbitrary definitions A, B, C, and D is interesting because it translates between the three“, and I'm sitting here asking myself why I should care about any of these definitions
Usually takes a long time for me to answer that satisfyingly.

[quartz edge]

im just speedrunning uni

[flint cove]

took me years to accept that I should care about char > 0

[flint cove]

Are they still optimizing the route, or is it just grinding at that point?

[quartz edge]

i dont know anymore

but in any case the goal is sufficiently monolithic to keep me focused on actually learning math so

that monolithism is useful in keeping on the right route with such a high workload

sometimes i forget english and just say words too, and i dont realize it until a day or two later

this may be one of those times

[shut moat]

I'm low key tempted to take the grad manifolds class at my uni instead of complex analysis next semester but that feels like an unhealthy decision

[gritty widget]

why not both?

both are good mathematics imo

[shut moat]

that'd be like 6 classes total

[gritty widget]

what does the complex analysis course cover

[sinful pecan]

ngl it'd be choice to take manifolds before complex analysis

[shut moat]

uuh lemme check

[gritty widget]

looks like a fun course

what about manifolds

[quartz edge]

brb conformal metrics

on manifolds

[gritty widget]

[shut moat]

seems super fun

[gritty widget]

i'd go for that one tbh

the complex analysis course seems like pretty standard material

[fast minnow]

why not both

[gritty widget]

^

[shut moat]

is the manifolds class not standard material?

[fast minnow]

yolo

the first few topics look standard

[gritty widget]

the complex analysis course looks like things you can find in any complex analysis course

[fast minnow]

but it looks like the class is gonna be a topics class? idk

[gritty widget]

topics classes are comfy 😌

[shut moat]

yeah topics classes seem fun

with all that exotic shit

[fast minnow]

topics classes are almost always free A's

and i can't learn without fear

[shut moat]

how hard are grad math courses compared to the ug ones?

[fast minnow]

depends on a lot of things really

[gritty widget]

depends on if it's a core/qual or not

[sinful pecan]

same difficulty if you actually meet the prereqs

[fast minnow]

topics are harder but in terms of grades generally easier unless it's a core sequence

[shut moat]

but in terms of grades generally easier
oh wtf

[fast minnow]

yeah cause they're designed for grad students who are genuinely interested in the stuff

and also professors don't want to grade

[marsh forge]

grad students taking classes post-quals don't have like

[fast minnow]

so it's a win-win for everyone

[marsh forge]

real grades

lol

[shut moat]

ok that's a win win then lmao

[gritty widget]

sympgeo topics course is a very pleasant break from what i'm used to

weekly homework in ug courses

[fast minnow]

if you wanna get into grad school though i'd recommend just taking core classes since good grades in those reflect better

[shut moat]

hmmm

[fast minnow]

idk how it works at your school but at least at my school complex analysis is a hard sequence and diffgeo is easy

[shut moat]

I'm still a freshman so idk the reputations of these classes, ig I should ask around first

although like

I've peeked at the higher level complex analysis course

stuff looks fucking insane

[fast minnow]

in that case i'd prob recommend the complex analysis class more since diffgeo is usually for more "advanced" students

[gritty widget]

higher complex analysis

[fast minnow]

yeah once you get past the basics complex analysis gets hard really fast

[gritty widget]

i am aware

t. got fucked by complex 2

[flint cove]

I know half of the basics. What keywords should I be afraid of?

[fast minnow]

i think subharmonic functions is where things get messy

at that point it's like real analysis but with C instead of R

[gritty widget]

ew

[shut moat]

this horrifying arrow shit is on the front page of the MIT OCW grad cpx anal page

[sinful pecan]

horrifying??

[flint cove]

Just because it has a lot of arrows doesn't mean it's horrifying

rather the opposite, lol

[fast minnow]

pretty tame compared to all the horrible estimates you do with entire functions

[gritty widget]

diagrams like this are nice

[shut moat]

Cat theory stuff looks like black magic to me idk lol (if that's what this is)

[sinful pecan]

what's that multipage diagram proof...

[fast minnow]

everything's category theory if you squint hard enough

[shut moat]

interesting

[tight agate]

ye this gives you the hodge decomposition

[cloud owl]

how is set theory category theory

[fast minnow]

just consider the category of sets

[flint cove]

Everything is determined by morphisms of the terminal object into other objects

[cloud owl]

is that a thing? damn

[fast minnow]

yeah you can view like everything as a category

usually not in a deep way xd

[shut moat]

ok for some reason I assumed anything with like 100 arrows and a bunch of words like Hom is cat theory

[cloud owl]

no, that sounds about right

commutative algebra vs capitalist algebra

www

[shut moat]

ok cool, ty guys

I'll email my advisor and see how he feels about it. He used to teach the course

[flint cove]

Anyway… what's that saying? I don't know the context for that picture

are the Ωs Kähler differentials?

[shut moat]

https://ocw.mit.edu/courses/mathematics/18-117-topics-in-several-complex-variables-spring-2005/

was from this site

[gritty widget]

i've seen this guy's name pop up a bunch in reference lists for sympgeo pages

[flint cove]

Hm, that should be interesting. I only know the Kähler complex from the abstract derivations pov.
I should look into how these two definitions relate.

[shut moat]

Guillemin or Campbell?

[sinful pecan]

guillemin

[gritty widget]

lmao in the notes theres a page on symplectic reduction

this looks fun

bookmarked

ty

[shut moat]

np uwu

[obtuse meteor]

we defined a cell complex and the homology groups today in algtop

who is ready for "prove meyer-vietoris" on homework

[marsh forge]

im so confused

what do you actually do in class

if things like mayer are relegated to hw

[obtuse meteor]

if you want you can see my notes from class to get a feel

we sketched a proof of van kampen in class

but like

classification of based covering spaces--homework

[marsh forge]

weird lol

[obtuse meteor]

we did all of the "certain actions induce covering spaces and they relate to fundamental groups nicely" on homework

We sketched the proof of effect of gluing in disks on pi1 from Hatcher in homework

tons of stuff like this

[marsh forge]

I'm torn on this

[obtuse meteor]

Here's the notes btw

http://www-personal.umich.edu/~alephnil/notes/MATH-592-notes.pdf

[marsh forge]

on the one hand it sounds like good stuff to learn details of I guess

[obtuse meteor]

and all the homeworks can be found here

http://www.math.lsa.umich.edu/~jchw/2021Math592.html

[marsh forge]

on the other hand I feel like it would make me reluctant to assign too much practice actually using these theorems

[obtuse meteor]

(as well as quizzes and a midterm)

we also have a good bit of practice using them

we got a lot of practice using the theorems in reviewing for the midterm

[marsh forge]

Interesting

[obtuse meteor]

the midterm was fully like

true/false and computational

and there are a good few like

using things

on the homeworks too

Like this homework has these problems:

1. Prove Scheier index formula from group theory with alg top
2. Give / prove the Galois correspondence (we've already done the other parts of the classification, so this was pretty easy)
3. [from a qualifying exam] give an example of a 3-sheeted covering space which is not normal/regular
4. Show that genus g surfaces have regular covers with any number of sheets
5. Covering spaces using quotients by nice actions, like all of that stuff
6. Lens space [apply 5 to construct a cool 3-manifold and compute it's fundamental group]
7. Try and discover for yourself a relationship between the monodromy action and the action of the normalizer(pi1(covering space)) via deck transfromations on a fiber

Which I think is a good balance

of apply and prove big result

[marsh forge]

It is a good balance

seems like a long pset

how long do you have for it?

[obtuse meteor]

the psets are pretty long on average

a week

[marsh forge]

I suppose its fine

[obtuse meteor]

my solutions were 18 pages long typed?

so not too bad

[marsh forge]

idk thats longer than most of my psetst

[obtuse meteor]

gotta remember we've done the classification theorem so like

[marsh forge]

by a long shot

[sleek thicket]

That's on the upper end

I mean I consistently do like 10-15

[marsh forge]

im also very concise tho

[obtuse meteor]

I'm anti-concise

[sleek thicket]

but 18 would make me raise my eyebrows

[obtuse meteor]

like I'm used to 30-40 page psets from my intro analysis class

[sleek thicket]

a dude in my analysis class told me his first homework in our class was 75 pages long today

[marsh forge]

jesus

[obtuse meteor]

granted I had bigger margins then

[marsh forge]

thats cancer

[obtuse meteor]

but like

[marsh forge]

so another Q

are you expected to like

[sleek thicket]

The way he described how he did stuff sounded the same as what mine was

[marsh forge]

pretend hatcher doesn't exist

[sleek thicket]

But mine was 15 pages

[obtuse meteor]

no

[marsh forge]

for these "prove result found in any textbook" type problems

[sleek thicket]

and we got the same grade

[obtuse meteor]

we can read it for some things, and in fact are expected to read it for some of them

(and instructed to if we need it)

[marsh forge]

interesting

well if you seem fine w it im sure its fine

[obtuse meteor]

on this problem we had like

[marsh forge]

yeah my psets are always like

half the length

[obtuse meteor]

https://estrogen.fun/i/qna3.png

[marsh forge]

of my other friends

and if you made my formatting denser

probably even less

[obtuse meteor]

for this one question

[marsh forge]

because i do a lot of inline stuff

bc i hate long paragraphs

[obtuse meteor]

I see

[marsh forge]

lmao my rep theory final which took a fuckton of hours

ended up being like

4 pages

idk what people write on their psets to make them so long

[fast minnow]

20pt font and 2 inch margins

[marsh forge]

i think i have like

one inch margins

or so

[obtuse meteor]

If you want I can share one of my psets with you in dms

[marsh forge]

im eyeballing it

[obtuse meteor]

to show you

[marsh forge]

no im not that interested i wont make you lol

just bored

[obtuse meteor]

lol

[marsh forge]

i think what happens is like

i spend a lot of time understanding why results are true

but once i understand it i have no problem under-explaining it

lol

so i just am very concise as long as I can back it up

normally graders seem okay with this

[flint cove]

That doesn't mean that was the intended solution tho

[marsh forge]

if im writing a pset longer than say

20 pages

im gonna go ahead and assume im doing something wrong

i would at least reach out to a friend or something

lol

[obtuse meteor]

lol

[marsh forge]

75 pages is mind boggling

[obtuse meteor]

yeah that is a huge amount

[marsh forge]

i can't even write 75 pages for like

[obtuse meteor]

Mine are like

[marsh forge]

an reu

lol

[obtuse meteor]

I'm just really detailed about things

[flint cove]

what's a reu

[marsh forge]

math summer camp

[obtuse meteor]

research experience for undergrads

do math

for a while

at a uni

[marsh forge]

except normally u get paid

[obtuse meteor]

over summer

[marsh forge]

instead of paying

[flint cove]

wat

[obtuse meteor]

based reu

pays you

[marsh forge]

paid = small stipend

[flint cove]

a) how is an undergrad in any way able to do research
b) who would pay money for that

[obtuse meteor]

yeah it works out to like

10 an hour of work

[marsh forge]

1. they really aren't most of the time

2. NSF + university

[obtuse meteor]

so it's pretty good tbh

the idea is it pays off in the long run

[marsh forge]

does it

[sleek thicket]

Yeah I have no idea how that dude squeezed out 75 pages

When I did 15

And I'm usually too verbose

[marsh forge]

i dont think any school actually makes money off their reus

[obtuse meteor]

by training those who go on to become researchers, and letting people know ahead of time if they don't like it

[sleek thicket]

Pays off in the sense of paying back into the community

[marsh forge]

how does that pay off

[fast minnow]

same + i make diagrams and my max is like 15

[obtuse meteor]

I think like "pays off" as in "for the mathematical community at large"

[marsh forge]

oh okay

sure

[sleek thicket]

training the next generation or whatever

[obtuse meteor]

ye

[marsh forge]

i was gonna say

[flint cove]

I think it's actually an interesting challenge to do assignments as short as possible
I tried to do Hatcher's LieAlg exercises with the constraint to not use more text than the assignment

worked at least for the first chapter, lol

[obtuse meteor]

Hatcher's LieAlg?

[flint cove]

Umrh, ofc not hatcher
…whatsthename…

[marsh forge]

i think the optimal writing style is like

[sleek thicket]

Humphreys?

[flint cove]

humphreys

[marsh forge]

verbose enough to explain anything actually hard

[flint cove]

ah, you were faster

[marsh forge]

but as concise as possible shy of that

[sleek thicket]

I have a copy in line of sight 😉

[marsh forge]

i like books that leave out details for me to check

or work out

it keeps my attn

[obtuse meteor]

hmmm

I guess?

I like being detailed

but I think like

the aim of a pset versus a paper to explain something to people feels differently to me

[flint cove]

But I completely mislike books who don't commit to either being complete or being concise
When stuff is so verbose it takes quite some time to navigate through the chapters, I become quite angry when the author leaves out nontrivial details

[obtuse meteor]

if I need to explain something to people I want to convince them that it's possible to check the details

if I'm writing a pset I want to make sure I can handle all the details and ensure the grader of this

[flint cove]

otoh, when things are concise and most proofs are sketches anyway, I don't mind

[gritty widget]

gromov

[ivory dragon]

he's not wrong

[summer jolt]

What does it mean for a manifold to be conformally embedded?

So if the manifold is a Riemannian then we can just the metric to check this?

[marsh forge]

what are angles

[sleek thicket]

angles are the inverse cosine of g(X, Y)/|X| |Y|

[gritty widget]

Is there a definition of "straight line" in riemannian manifolds via angles? (I would guess not, right?) As opposed to "a straight line between points is a path that minimizes length"

[sleek thicket]

Are you looking for a characterization of geodesics in terms of angles or a new concept that could reasonably called a straight line?

[gritty widget]

I guess the former (maybe the latter could be interesting too if the resulting "straight lines" are not too weird)

[gritty widget]

here's a necessary condition

[gentle osprey]

(T*Terra, dqⁱ ∧ dpᵢ)

[gritty widget]

the second equality comes from the fact that if you know parallel transport then you can recover the connection via that derivative (proof: choose a parallel orthonormal frame along the curve and expand)

[tough imp]

bro tterra quit being a nerd

no one knows anything about parallel transport

[gritty widget]

Ah, and parallel transport is an additional structure on top of the riemannian metric?

parallel transport comes from the riemannian metric

(well from its levi civita connection, which is uniquely determined by the metric)

Oh, somehow my intuition is that the riemannian metric can't relate things between points, ie the metric at two points are sort of independent

that's what parallel transport does for you

lets you "transport" data between two tangent spaces

so although the metric itself isn't relating data between two different points, its connection is

[chrome dew]

pulls out formula for christoffel symbols in terms of derivatives of the metric tensor

[gritty widget]

lovely

I'm having a hard time intuitively seeing how a metric tensor gives rise to a parallel transport

I guess what makes sense to me so far is if I have a geodesic, then the derivative of the geodesic at 2 points should be parallel to each other, despite that they are in different tangent spaces. But I'm not sure about the other n-1 other directions besides the derivative

i googled a bit and someone on MO says to take a peek at the appendix to arnold's mathematical methods in classical mechanics book

[flint cove]

I legit didn't know that there was a formal definition to „parallel transport“ that wasn't „a connection“

Are these concepts 1:1?

Or are there restrictions like parallel transport always inducing torsionless connections

Or however this works

[gritty widget]

i think there is indeed a one-to-one correspondence between connections and (suitable notions) of parallel transport

googling time

[fiery remnant]

Anyone working on mobile sensor networks/evader detection? (Just curious)

[marsh forge]

this is not topology or geometry really

[fiery remnant]

What do you mean? These problems were essentially the birth of applied topology to my knowledge

It's a mix of topology, dynamical systems, and combinatorial structures. But topology is the foundation

[gritty widget]

✋ does network topology count as topology?

I guess also stuff like graph theory

[fiery remnant]

It should, tbh. Maybe not in this chat though, I guess. I honestly don't know how network topology isn't topology though

I guess if that question is too specific, then is anyone working on applied topology?

[quasi forum]

So when considering an ultrametric space we normally have this inequality $|a+b|\leq \max{|a|,|b|}$.
However, I am dealing with p-adic numbers, and I am pretty sure this is actually an equality for p-adic distance.

[gentle osprey]

dackid

[quasi forum]

If I am wrong, can you please help me break out of that intuition?

[obtuse meteor]

I know this is true in alot of cases

[quasi forum]

Oh really? So I guess a better question is when wouldn't this be true in ultra metric spaces?

[obtuse meteor]

https://encyclopediaofmath.org/wiki/P-adic_valuation

one example where it's not true

take a = b = 1/2

in 2-adic valuation

then |1/2 + 1/2| = |1| = 1

but |1/2| = 2

[quasi forum]

Hmm, okay, so it really boils down to when the p-values are the same.

It also seems this is unique to 2-adic valuation

[chrome dew]

it's an equality when $|a| \ne |b|$

[gentle osprey]

Merosity

[chrome dew]

it's not hard to show either, it's kind of a fun trick

wlog assume $|a|>|b|$ then $|a| = |a+b-b| \le \max (|a+b|, |b|)$. Since $|a|>|b|$ we know $|a|\le |a+b|$. And of course $|a+b| \le \max(|a|,|b|) = |a|$ so we have shown inequality both ways and $|a+b|=|a|$.

[gentle osprey]

Merosity

[chrome dew]

Just a little more discussion in the p-adic case we can additionally look when |a|=|b| by normalizing to |a|=|b|=1 and then we can reduce a,b mod p. Then it's just looking at if they're negatives of each other mod p. When they're negatives of each other you get |a+b|<max(|a|,|b|) and when they're not negatives you get |a+b| = max(|a|,|b|)

[quasi forum]

Thank you for that explanation Merosity.

[chrome dew]

you're welcome 👍

[gritty widget]

can anyone link me a way to see that compact manifolds have finite dimensional singular cohomology without using de rham + hodge + elliptic operators have finite dim ker and coker

tyty

[uncut surge]

Every compact manifold is homotopy equivalent to a finite CW complex, and cellular homology is equal to singular homology? That might be a nice way

Of course that first statement is quite heavy but eh

[obtuse meteor]

hmm

how heavy is that first statement?

You just have to prove that every manifold is homotopy equivalent to a CW complex. And manifolds by definition are essentially built by gluing together patches of euclidean space

so if this is hard to prove I'm going to be angery

[median glade]

idk, the gap between "intuitively clear" and "easy to prove" seems to be pretty large in topology

[obtuse meteor]

we're just going to believe

that it's easy

[sleek thicket]

Hmm

I'm thinking of something silly like

Choose a metric

Cover by convex open sets

(so that all intersections of cover elements are contractible)

Try to inductively do mayer vietoris?

So the idea is something like "if a space admits a finite good cover then it has finitely generated cohomology"

[obtuse meteor]

choose a metric

bad! bad geometer!

[sleek thicket]

The metric is just used to produce a am open cover where all finite intersections are empty or contractible

I don't know any other way to do this

[shut moat]

possibly dumb question- how do you motivate riemannian geometry? Like, I get that riemannian metrics define a notion of size and distance on manifolds, but suppose you've never heard of the things and wanted to add more structure to a smooth manifold. Wouldn't adding a metric (in the metric space sense) that's compatible with the manifold topology be the more obvious choice?

[viral atlas]

@ terra

[obtuse meteor]

that doesn't give you any smooth data @shut moat is probably one intuitive reason

[shut moat]

smooth data?

[obtuse meteor]

like requiring the metric agree with the topology

doesn't respect any of the structure a smooth manifold has

[shut moat]

ah true

[gritty widget]

how do you measure lengths with a topology-metric? area? angles?

an inner product on each tangent space lets you do so infinitesimally (i.e. on each tangent space), and then you'd like to do that everywhere; putting inner products together in a smooth fashion gives you a riemannian metric

we already know from linear algebra how to measure things like angles, and a lot of things on manifolds are done infinitesimally in the linear algebra setting and then put together smoothly. from this point of view a riemannian metric is a reasonable thing to consider (imo)

[shut moat]

i suppose something along the lines of the limiting case of $\sum_i d(\gamma(t_{i+1}), \gamma(t_i))$

[gentle osprey]

~S^1

[gritty widget]

hmm

that could work, yeah

it'd be very hard to use in practice though, i feel

(not like the integral formula for arc length is any easier to use, but it's certainly simpler)

also what faye said

shamrock fields still hasn't gotten back to me

[sleek thicket]

Hmm, I'm pretty sure the riemannian metric is determined by its metric space metric. I wonder whether you could figure out a condition for when metric space metrics come from a riemannian metric?

[shut moat]

coincidentally I was just looking at a stackexchange page talking about that

https://math.stackexchange.com/questions/106508/existence-of-a-riemannian-metric-inducing-a-given-distance

[sleek thicket]

hahaha

[shut moat]

I think you could resolve this by requiring that d: MxM--> R is smooth?

[gritty widget]

that's not even true for the usual metric

on R^n

[shut moat]

oh shit

[gritty widget]

the graph of d(0, x) will be pointy near the origin

[sleek thicket]

r i p

[gritty widget]

but it's a good thing to think of

[shut moat]

hm

is the distance function associated with a metric unique?

[gritty widget]

wym by "associated?"

i'd say it would be just by definition

[shut moat]

so in principle, if I have a distance function that can be induced by a riemannian metric, then there is only one metric that does so?

[gritty widget]

on a connected manifold

this is an exercise in lee irm lol

i was just looking at it

[shut moat]

NICE

[gritty widget]

let me show

part b: if two riemannian metrics determine the same distance function then they must be equal

[gritty widget]

compare part a to this

[shut moat]

wtf is that weird limit notation

[gritty widget]

t decreases to 0 (i.e. approaches from the right)

[shut moat]

also hell yeah \o/

hipsters smh

oh nvm

[obtuse meteor]

what about requiring (d(x, y))^2 to be smooth

[shut moat]

intuitively that seems like it should work

but also it was intuitive that a coarser topology can't be homeomorphic to a finer one so I guess that can't be trusted lmao

[obtuse meteor]

lol

[pliant marsh]

I have the following exercise.

and I started off a proof like this,

I want to show that U' and V' are non-empty and in particular they contain an element that is in the class of x and y respectively, it seems easy but I can't think of a way to show this.

[marsh forge]

It's not true

Let X be R and G=Z/2. Let the action be given by x\mapsto -x. Then this is a continuous group action.

Take a small neighborhood of 5

the intersection of this neighborhood and the respective one around -5

is empty

You are correct that you want to use finiteness in a creative way

[pliant marsh]

Oh hm, I was mainly looking at this answer and the comments https://math.stackexchange.com/questions/1508271/prove-that-x-g-is-hausdorff and was trying to fill in the gaps.

[marsh forge]

Would you like a hint

[pliant marsh]

Sure yeah, I'm a bit lost after what you said

[marsh forge]

Finiteness is as important for playing games with hausdorffness as it is for playing games with open set intersections

Like

given a finite number of points you can find neighborhoods separating all of them

I can give more of a hint but my suggestion is you think about what it means

to have neighborhoods separating [x] and [y] in X/G

and what this lifts to in X

Maybe even draw it for a nice example like the one i gave above

[pliant marsh]

Yeah I think that is my main problem, I find it very hard to visualize what X/G actually is in terms of what I think in topologies I'm used to

[marsh forge]

So

what would my example look like do you think

R/(Z/2)

[pliant marsh]

R^+?

[marsh forge]

Don't forget 0

but yes

[pliant marsh]

oh right yeah

[marsh forge]

so

let's do this in reverse

we know we can separate 4 and 5 in R^+

what do those neighborhoods lift to in R?

[pliant marsh]

like (3.5,4.5) and (4.5,5.5) for example?

[marsh forge]

Sure this works

what is their preimage in R

under the Z/2 map

(hint: every neighborhood should become |G| neighborhoods in the original space)

[pliant marsh]

(-4.5, -3.5) and (3.5,4.5) same with the other one?

[marsh forge]

Yep!

So

a separating neighborhood in X/G

comes from

a bunch of separating neighborhoods in X

[pliant marsh]

Oh so, if I look at the class of x and y for example and look at all the points that map to it, I want a neighbourhood of all the points that are equivalent to x and a neighbourhood of all the points equivalent to y such that they are disjoint?

[marsh forge]

Yeah kinda

so these preimages of x and y are both finite

so you can use hausdorffness to make sure you get "small enough" neighborhoods around all the preimages

and then you can make sense of a "smallest" one in some sense

idk how much hint you want i think you're very close

[pliant marsh]

sorry just tryna think how I can incorporate all this into a proof

[marsh forge]

my suggestion is to draw more pictures

i am a big fan of generic looking ovals being sent to other generic looking ovals

[pliant marsh]

I get the example you gave, but I'm not really sure how to draw this in a general setting

[marsh forge]

I'm trying to think of a good example that includes the technicalities I didn't give you

Basically you want to take small nbhds around all the points

make sure none of these intersect

but thats not quite enough

because like

(-3.9,-4.1) is a small enough neighborhoods around -4

and (3.5,4.5) is likewise for positive 4

but the two don't give you a well defined neighborhood in the quotient

does that make sense?

[pliant marsh]

I can't really see that why that's the case no

[marsh forge]

So

basically (-3.9,-4.1) gets sent to (3.9,4.1)

which is not the same as

(3.5,4.5)

so (-3.9,-4.1) is not g(3.5,4.5) for any g

[pliant marsh]

oh so they're not equivalent in a sense?

[marsh forge]

Yeah so the idea is like

you want to simultaneously choose a neighborhood of gx for all g

and likewise for y

so that they give a well defined neighborhood of the equivalence classes

in X/G

[pliant marsh]

and I assume this U' does exactly that

[marsh forge]

the correctly chosen U' would yes

[pliant marsh]

hm I'm trying to see why that is the case

[marsh forge]

Sorry why what is the case

[pliant marsh]

U' does this

[marsh forge]

We haven't come up with a suitable U' yet

[pliant marsh]

oh the intersection of all the gU's

[marsh forge]

thats empty

in general

that approach isn't going to work

it's almost the right idea though

[pliant marsh]

but this was the one suggested in the stack exchange post

[marsh forge]

You mean the answer?

like by Rob Arthan?

[pliant marsh]

the very last comment

[marsh forge]

Yeah that comment is wrong

[pliant marsh]

oh I see

hm then lemme try to think up such a neighbourhood

[marsh forge]

fwiw i imagine the commentor had what I have in mind

but wrote it incorrectly

[pliant marsh]

So gU, etc are open nbhd's of gx for each g and similarly for y and I can force all the gU's to be disjoint from the gV's

[marsh forge]

well

a priori you can choose $U_{g}$ and $V_{g}$ for all $g$ so that $gx\in U_{g}$ and $gy\in V_y$ and they are all pairwise disjoint

there is no guarantee that like

U_g should be gU

if that makes sense

[pliant marsh]

oh and the intersections of these work?

[gentle osprey]

sweetgreen simp

[marsh forge]

No

at least, not a naive intersection

they don't overlap in X

Can you think of a place where they do overlap?

[pliant marsh]

in X/G?

[marsh forge]

Yes

So

let q be the quotient

all the q(U_g) are neighborhoods of q(x)

[pliant marsh]

oh so if I map all these Ug then that intersection works

I think

[marsh forge]

Yes, you take the intersection in X/G

prove that these are open sets and disjoint

and contain [x] and [y]

[pliant marsh]

alright will have a go

is it alright if I give a ping if I'm stuck?

[marsh forge]

Yeah go for it

ill be studying french all day

so id much rather do this

lmao

[pliant marsh]

haha

[pliant marsh]

@marsh forge alright so I've gotten everything except them being disjoint, I was thinking of showing p(U_g) \cap p(V_g) is empty but this isn't necessarily true I think.

[marsh forge]

my suggestion is to proceed by contradiction

let U' be what happens when you intersect all the p(U_g) and likewise for V'

Suppose z is an element of both

[pliant marsh]

hm alright let me see where this gets me

[marsh forge]

basically you can derive a few different contradictions i think

[sleek thicket]

I think this works

Do induction on the number of elements in the cover

You MV on U1 cup ... cup Un and U_{n+1} to get this big LES

2 out of 3 terms are finitely generated by induction so the last one is too (this is in answer to your question @gritty widget )

[pliant marsh]

@marsh forge I think this works, but I'm not 100% sure

[marsh forge]

I think the argument is more or less correct but I am not sure the set manipulations in the last sentence work

You need to be care about like

how a priori

maybe U_h intersects V_g

[pliant marsh]

yeah that part I wasn't sure about

[marsh forge]

I would avoid taking unions at all and say something like, okay we have some lift of z in every single U_g and V_g. Prove that these guys are all disjoint. Then you can conclude after making an observation

[pliant marsh]

lift of z means preimage?

[marsh forge]

yes sorry

Im thinking of X sitting "over" X/G

[tight agate]

I'm working on this bundles problem

[gentle osprey]

Brofibration

Brofibration

Brofibration

Let $L$ be a line bundle on the torus defined by $\mathbb{C}\times \mathbb{C}/ (z, v) ~ (z+1, v), (z, v) ~ (z+ \tau, \phi(z)\cdot v)$

Brofibration

[tight agate]

and the problem asks us to show that the line bundle has degree 1

and I think you can do it by computing the divisor associated to the corresponding theta function

[tight agate]

Let $L$ be a line bundle on the torus defined by $\mathbb{C}\times \mathbb{C}/ (z, v) \sim (z+1, v), (z, v) \sim (z+ \tau, \phi(z)\cdot v)$

[gentle osprey]

Brofibration

[tight agate]

or another way might be to compute it using curvature and stuff

[pliant marsh]

@marsh forge sorry I'm not able to think up a contradiction, I have that the preimage is contained in the union over g in G of all gU_h same for V_h for each h in G, but I can't think up an argument to show that the intersection of these are empty

[marsh forge]

Again my advice would be to not take unions

Note first that all the U_gs and V_hs are disjoint from eachother

note that there must be some lift of z in each

now the contradiction I have in mind follows from counting how many z's that means you have to have

and comparing it to how many z's you know you have

[pliant marsh]

in order to argue with these U_g and V_g wouldn't I have to find the inverse image?

[marsh forge]

So you constructed U'\subset X/G

by starting with a bunch of disjoint stuff

and then taking intersections

you can prove from there that they have to lift to a bunch of disjoint open sets

[coral gale]

When a smooth $m$-manifold $M$ is compact, choose ${(B_i,\varphi_i)}{i=1}^n$ a covering by coordinate balls. Use bump functions $\chi_i$ so that ${\chi_i}|{B_i}=1$. Define
\begin{equation}
F(p)=(\chi_1 \varphi_1(p),\hdots, \chi_n \varphi_n(p))\in\mathbb{R}^{nm}.
\end{equation}
We can show this is a smooth injective immersion $M\to\mathbb{R}^{nm}$, and hence an embedding since $X$ is compact. My question is, how does one show this is an immersion?

[gentle osprey]

yetiyeti

[obtuse meteor]

question about simplicial homology computation

So I have this Delta-complex structure on a mobius band

https://estrogen.fun/i/0nlf.png

and I've computed that

https://estrogen.fun/i/3nys.png

[pliant marsh]

@marsh forge is it possible to demand that each U_g is disjoint from eachother or would that mess things up, then I could argue by placing different g_iz in each U_g, but then V_g cant be disjoint from U_g as one of these gz's are inside V_g? Reason I'm asking is because I think it's possible to place gz for some g in each U_g, then g'z in each V_g for another g' then everything would work out still but z would be in each p(U_g) and p(V_g). (also sorry this is probably getting pretty annoying by now)

[obtuse meteor]

unfortunately, this doesn't mean that im(d2) = <a, c - b1 - b2> because you run into parity issues

is there a way to deal with the parity issues?

(I'm sure there is, more asking how do I think about computing this image)

maybe instead as generators I can take <a - b1 + c, a + b2 - c>

but that is grody

[gritty widget]

that's nice tyty n.n

[obtuse meteor]

Ah I think I see how to do this, neat

[obtuse meteor]

it turns out it's not grody and everything is fine

one must just compoot

[obtuse meteor]

Chad professor assigns tai danae bradley reading

https://estrogen.fun/i/pj2k.png

[quasi forum]

So I am having a bit of trouble. I need to show that p-adic valuations satisfy the ultrametric inequality. I am trying something, but it seems like overkill/ I don't think it is getting anywhere.

I could use some assistance on getting a good starting point

Nevermind, I went about it case by case and it worked out well

[viral atlas]

[violet sun]

To show lim S is a closed set, Can't I just say if (p_n) converges to p,then p is a limit of S,implying p is in lim S?

[quasi forum]

But does that tell you it's closed? The argument seems a bit circular 🤔

[violet sun]

If all limits are in S,S is closed

That's how my book defines closed sets

[quasi forum]

Wow, really?

I mean, there is definitely truth to it

[tawdry valve]

just make sure to pay attention that p_n is a sequence of limit points, not points in S

[violet sun]

So,p_n need not be in S?

[tawdry valve]

yup

[quasi forum]

The sequence specifically needs to be in lim S

[quasi forum]

This does make the problem more interesting

[long hornet]

You can use the facts cl(S) = S' U S and that cl commutes with union to do it.

cl(S) = S u S' = cl(cl(S)) = cl(S') u cl(S) = S'' u S' u S, and thus S'' is in S u S'

So S'' is contained in S'

Hence cl(S') = S'' u S' = S', i.e., S' is closed.

[violet sun]

Nice proof

[long hornet]

This step makes sense to me but I didn't really justify it.

[gritty widget]

closure is sus

[long hornet]

cl(A u B) = cl(A) u cl(B)?

I know it's true because it is given in Munkres.

[flint cove]

To see it directly observe that x is not in the closure of something precisely if it can be separated from this something with an open nbhd

[marsh forge]

what defn of open do you have

where M_rp isn't trivially open

[flint cove]

Oh, I was missing the backlog. I thought this was abstract talk about the Kuratowski closure axioms.

[gritty widget]

A projective curve gives us an affine curve in each chart. Do you have any good geometric intuition about these different curves?

Say we are working in P^2, a projective curve X is smooth if each of the curves three curves X_i:=X\cap U_i are smooth

I want to find a projective curve that is not smooth, but one of the X_i is smooth

[sleek thicket]

Take zy^2 = x^3. This isn't smooth because it's not smooth in the chart z ≠ 0. But in the chart y ≠ 0 it is smooth since it's the curve z = x^3

does that seem right @gritty widget ?

[tardy meadow]

how could one deduce that all real intervals are connected just from [a,b] being connected? (a < b) My slightly silly idea was to use it to show that R is connected from that (by the intervals [-n, n], each connected with connected union because their intersection is non-empty) and then construct continuous maps from R to each type of interval, but surely there's a nicer way?

[ivory dragon]

suppose [a, b] is connected but (a, b) is disconnected - i.e. there is a pair of disjoint open subsets U, V such that their union is all of (a, b)

derive a contradiction

[long hornet]

Path connectedness? Just construct a continuous bijection.

[marsh forge]

oh i had a different proof in mind but ig you need to know path connected=>connected

[ivory dragon]

that works too

[tardy meadow]

yeah ideally it'd be appealing straight to connectivity

[marsh forge]

namis proof works

[tardy meadow]

had this idea but for some reason I thought it wouldn't work

[long hornet]

Alternatively, (a, b) is homeomorphic to R

[tardy meadow]

yeah but I'd have to use explicit constructions I'd imagine, not sure if I can just quote that

I'll give that direct method another think

wait I'm completely stupid

one sec

[ivory dragon]

note here that "open" means "open wrt the subspace topology"

since [a, b] and (a, b) and whatnot are being considered as subspaces of R

[tardy meadow]

I'm not really sure how to "add in" a and b since that's what I assume we're trying to do (main difficulty is that {a}, {b} and {a,b} are not open in [a,b])

[ivory dragon]

no, but singletons are closed, hence their complement is open, hence the intersection of their complement with an open set is open

if you cant justify why these are closed, consider (a, b+epsilon) intersect [a, b]

or alternatively (-infty, a) U (a, infty)

[long hornet]

So {a} is clopen? But that disproves connectivity, doesn't it?

[ivory dragon]

huh?

[tardy meadow]

yeah I get that singletons are closed

[tardy meadow]

nah not in [a,b]

[ivory dragon]

okay so we can consider (a, b) as a subspace of [a, b] with the subspace topology

U and V are open in (a, b), which means there are open sets U', V' in [a, b] with U = U' intersect (a, b)

[long hornet]

Oh my bad

[ivory dragon]

you can do a casewise argument on whether these contain a, b or not

but ultimately thats a small technical detail

continue the argument and you get that [a, b] must be disconnected

a contradiction

[gritty widget]

Yes thank you!

[ivory dragon]

aside: it may be logistically easier to go from (a, b) disconnected -> (a, b] disconnected -> [a, b] disconnected

or something like that

just because that lets you consider a single endpoint

and then the argument is basically the same to include the other one

[tardy meadow]

nah it was supposed to be [a,b] connected -> any interval connected

think I see it now

and then consider countable unions for when one end is infinite? (idk if you're supposed to consider that case but doesn't seem to be too much of a big deal)

[ivory dragon]

well if you have (a, b) connected for all (a, b)

then yea

you can just countably union

assuming youve proven that countable unions preserve connectedness at least

er wait

(a, b) U (a, b+1) U (a, b+2) U ... is simpler

[tardy meadow]

ye that sort of thing

so U' \cap (a,b) and V' \cap (a,b) being disjoint means that they could only share a, b. But I don't see where to go from there - in my head you're going to have to pull out sups/infs again?

like if either U', V' have a, b you can get separations for some half-open or closed interval but I can't see how you can guarantee which one, may just be being dense

wait one sec

i see why you say it's better to go the other way around, cos you can suppose a separation [a, b] = A \cup B then A \cap (a,b) and B \cap (a,b) is a separation of (a,b). Seems a bit clunky this way round

is it not better to instead of consider [a,b] and (a,b), to consider (a - epsilon, b + epsilon) instead? Like if you have a separation (a - epsilon, b + epsilon) = A \cup B, then A \cap [a,b] and B \cap B [a,b] would separate [a,b], contradiction. Seems to work better

think I've got it that way

only adding/taking epsilon works main point being [a,b] \subset [that]

[pliant marsh]

Quick question, if I have a subset A in X equipped with the subspace topology, and I know that A\cap B is open in A can I immediately conclude that B is open in X?

Ah I see alright

Is there any condition that I can impose on B such that is does hold?

[sleek thicket]

That B is contained in the interior of A

[shut moat]

why're you trying to do that Athurus?

[pliant marsh]

I'm trying to prove something is a quotient map and I have that g^{-1}(U) is an intersection of the subset with something else (which I want this something else to be open)

[sleek thicket]

it was a serious suggestion!

[pliant marsh]

but I definitely don't have this interior thing so I can't use that

[sleek thicket]

lol

[pale sky]

so these are my notes from class:

why does f(1)=/=f(-1) imply there does not exist a lift?

[sinful pecan]

you need p(f(1))=p(f(-1))

[pale sky]

why is that?

[sinful pecan]

because you're lifting p

[pale sky]

sorry could u explain alittle more, i suck at this alot 😦

i thought we are lifting f

[sinful pecan]

oh oosp

anyhow though, imagine if you didn't have that condition - if f(1) != f(-1) then 1 and -1 arent "above" the same point, but we know that they must be since p makes them equal

you've seen more visual examples on e.g. S^1 already right

[pale sky]

im confused to as why p makes them equal since we have $p(\tilde{f}(z))$ does $p(\tilde{f}(1))$ have to be $p(\tilde{f}(-1))$?

[gentle osprey]

Not A Horse

[pale sky]

the helix on S^1?

they call the covering p, -1 and 1 are both mapped to 1, so they are identified as the same, the lift should not distinguish between these? Also im confused about this word lift and covering, $p^{-1}$ goes to the covering space right? and $f$ takes it to the space as if we are lift something to the covering space?

[gentle osprey]

Not A Horse

[sinful pecan]

the covering space here is C/0 and p takes the covering space to the base space. a lift of f:C/0 to C/0 is a map from C/0 to the covering space so that when projected back down it is just f, i.e. p(\tilde{f}(-))=f(-).

[pale sky]

so in this case our f(-1)=/=f(1) is originally the case, but if we suppose there is a lift, we can never project back down to satisify this?

[sinful pecan]

yes

[pale sky]

$p(\tilde{f}(-1))=f(1)$ will never be the casE?

[gentle osprey]

Not A Horse

[pale sky]

why is this true?

[sinful pecan]

no its not

[pale sky]

i mean the fact that it will never be the case

[sinful pecan]

to be accurate, you need to pick a value for \tilde{f}(-1), either f(-1) or f(1)

[pale sky]

the induced map $f_$ seems to tell us this since $f_ (n)=2n$ but i dont see why this implies the result

[gentle osprey]

Not A Horse

[sinful pecan]

you don't need the induced map, let me think for a sec

wait so you accept the definition of the lift right? that f=p\circ \tilde{f}?

[pale sky]

yes

if tilde{f} exists

[sinful pecan]

of course

[pale sky]

what yuou explained i understand now, thank you

that first part

[pale sky]

this part

[sinful pecan]

ohhhh oops

[pale sky]

i think the problem is $\tilde{f}(1)=1,-1$ but based on the drawing with the circle, it looks like $\tilde{f}(1)=1$, how can we see that $ \tilde{f}(1)=-1,$ is also valid ?

[gentle osprey]

Not A Horse

[pale sky]

i think i get how the drawing works

if u set like n=1, it takes 1 to -1, if u set n=2, it takes 1

[quasi forum]

So I kind of hit a road block. I am trying to show with the ultrametric inequality that if $d(x_n,x_{n+1})\to 0$, then the sequence $(x_n)$ is Cauchy.

[gentle osprey]

dackid

[sinful pecan]

@pale sky if you haven't figured it yet, i'm pretty sure the answer is just that \tilde{f} must be continuous.

[pale sky]

oh i figured it out, its that tilde{f} is not well defined

no matter how u try to define a tilde f, it will send 1 to -1 and 1 based on that circle picture

thanks though!

in this picture, it seem like if you decide to not move, you get tildef(1)=1, if you decide to move you get tildef(1)=-1

[trail ibex]

@quasi forum To go from xn to xm, for m>=n big, you can make small steps ; but the ultrametric inequality tells you travel (at most) as far taking several small steps and taking just one small step. Thus if m is big enough you win since you can have that one step makes you travel an arbitrarily small distance

[quasi forum]

I'm so sorry, I'm not sure I am following

I kinda have the idea, but I think clarity will be helpful

[trail ibex]

you go from xm to xn by going xm, xm+1, xm+2, ... and you apply the ultrametric inequality at each step

For instance d(xm,xm+2)<=max(d(xm,xm+1),d(xm+1,xm+2))

[quasi forum]

Thank you! My friend got to the explanation basically right before you did.

But I definitely see it now. I was trying to apply the ultrametric inequality in a not very useful way earlier.

[chrome dew]

nice now you can easily prove $\lim_{n \to \infty} a_n = 0$ iff $\sum_{n=0}^\infty a_n$ converges

[gentle osprey]

Merosity

[quasi forum]

But it's not true.

Or do you mean in ultrametric spaces?

[chrome dew]

yeah

[quasi forum]

Taking a picture so I can ponder it later 😁

[chrome dew]

it should be very easy using what you just proved about cauchy sequences

if you ponder you're thinking too hard

[quasi forum]

Because it is Cauchy, there is a point where you are adding terms arbitrarily close to one another

You are right I am thinking too hard

[chrome dew]

welllll it should be a bit more rigorous than that

what does it mean for an infinite sum to converge?

its partial sums $s_n = \sum_{k=0}^n a_k$ converges

[gentle osprey]

Merosity

[quasi forum]

If the sum is S, then $d(S, S_n)\to 0$

[gentle osprey]

dackid

[chrome dew]

so write the condition of s_n being a cauchy sequence

maybe I'm assuming translation invariant metric

[quasi forum]

I'm sorry. It's 3 am. This may seem easy, but my brain is not having it at this hour

[chrome dew]

no it's my bad, I was assuming it was a translation invariant metric

[quasi forum]

Oh okay. So it is not generally true?

[chrome dew]

I don't think I've spent any time thinking about ultrametric spaces that weren't normed vector spaces so I don't know lol

[trail ibex]

It's true if you have a complete ultrametric space

because sn+1-sn=an tends to zero

But for instance $\sum_n p^{n}$ doesn't converge in Z for the p-adic valuation

[gentle osprey]

Othenor

[trail ibex]

Um wait I'm not so sure about that one I should double check

But basically take any x in the p-adic integers Zp not in Z, take its p-adic expansion x=sum an p^n and voila

[chrome dew]

1+p+p^2+p^3+... = 1/(1-p) so if p=2 then 1/(1-2) = -1 actually does converge in Z

[chrome dew]

but can you prove that for a metric that isn't translation invariant? The critical step: $d(s_n, s_{n-1}) = d(a_n,0)$

[gentle osprey]

Merosity

[gritty widget]

is there a name for the map, from a space X to its class group sending a point of that class

[cursive glade]

is the directional derivative of vector field along another just the dot product of the vectors at a point?

[gritty widget]

the directional derivative of a vector field with respect to another will be another vector field, but the dot product is a scalar

[cursive glade]

yeah, but i was thinking you just get the gradient of that one then. I'll have to go back over definitions, i dunno what im missing

[gritty widget]

Second Countable Topological Spaces implies First Countable Topological Spaces
but
First Countable Topological Spaces doesn't imply Second Countable Topological Spaces

right?

Finite Topology?
also I definitely am

My book uses:

The space (X,T) is said to be first countable iff there is a countable local base at each point of X

C is a local base at p iff each neighborhood of p contains a member of C

Have C be the collection of neighborhoods of p?

hmmm

i see where youre getting at

lemme think

A T1 space?

Do you have a link I can use to find out?

since I'm way off, im misunderstanding something.

a limit point?

i should

Discrete because it's all sets of X, including the neighborhoods of p?

What do you mean by naming it?

by "base for x" you mean any point right?

A local base C1 in X , U in C1, and x exists in C1, U, and X

the local base C exists in X and each neighborhood of x contains a member of C

can i get a hint?

Does anyone have a hint for this? We are working on a compact Riemann surface X.

If [p]-[q] is a principal divisor, does this mean we must have some function that has a pole over order 1 at q and and a zero of order 1 at p?

would it be {x}?

[gritty widget]

de Rham cohomoloy

[quasi forum]

I was wondering if this is an efficient proof for this problem or if I need to go in to more detail. If so, I am having a little trouble figuring out how to elaborate and can use some pointers

[shut moat]

you could do it directly- d(x, A) = 0 implies that every open neighborhood of x intersects A. So x is a limit point and therefore an element of \bar{A}

[median glade]

I believe you're missing a direction too

you've shown => for now

[quasi forum]

The other way is pretty straightforward

[median glade]

is it?

I think the other direction slightly harder

still a one liner, but the justification is a bit heavier

[quasi forum]

If x is in the closure A, then every ball of radius r intersects A

And so the infinum is 0

[median glade]

right yeah

but concluding that the infimum is zero from this case is a bit heavier

[quasi forum]

Well in the case for the part I am doing, I am taking the contrapositive

There we go. I like this better

And then the other direction is since x is in the closure of A, there is a sequence (a_n) that converges to x. Therefore, the infinum is 0

[tidal cedar]

d² = 0, ez

[gritty widget]

d=0

[shut moat]

bespoke

[gritty widget]

i found this https://www.mat.univie.ac.at/~michor/dgbook.pdf

it's got a lot

[viral atlas]

[shut moat]

god I've downloaded so many diff geo pdfs that I won't read

[gritty widget]

do all the exercises in all of them

[shut moat]

if you can't literally latex the books from memory you haven't learned anything

[gritty widget]

🥴 physics

symplectic geometry is abstract physics

[shut moat]

the only reason it's interesting

[viral atlas]

Isn't a lot of diff. geo. motivated by physics?

[gritty widget]

owo