point-set-topology

[obtuse meteor]

Hmmm

[empty grove]

We do, because we need finite non empty sets to not be open

[obtuse meteor]

Follows from these being a basis I think no? Bc these are all infinite sets

[empty grove]

Yep

[obtuse meteor]

Yeah

[empty grove]

The way it's written suggests that any topology containing them would work

[honest terrace]

I mean I don't think the author was trying to give details of how the proof goes

well yeah, which are the two things that a "technical" in the proof

[empty grove]

Finally I get to sully ultra

Wait that doesn't sound right

[pseudo crane]

[obtuse meteor]

[marble socket]

[honest terrace]

also it's not necessary to prove this, is it ?

viewing it as a subbasis should suffice, no ? 🤔

like just take the coarsest topology that contains all the S(a, b) as open sets

but that makes proving the fact that no open set is finite a bit longer, ig ?

[empty grove]

You'd want to prove then that finite intersections of those sets can't be finite

And that's pretty much proving that they form a basis

[honest terrace]

yeah right, good point

[thin jewel]

What's an easy way to compute a (Zarisky) orbit closure of some group acting on a matrix?

[gentle osprey]

bdobba

bdobba

[thin jewel]

I'm trying to figure out a way to do it in Macaulay2

I only need the ideal I should say

[quasi forum]

My study group and I were trying to do this problem (section: connected spaces), but we were completely stumped. Does anyone have any pointers so we can figure this out?

My thought is to focus on the function |f(x)-x|. And I've already shown such a function is continuous. Since [a, b] is connected, so is the range, and so the range must be an interval.
However, the issue I am having is that I am not sure how to show 0 is in the interval 🤔

Hmmm. I see where you're trying to go here. I'm just not sure how to sure there are positive and negative values of f(x)-x

Son of a bitch

Since a is the endpoint, f(a) must either be equal to or bigger than a

The opposite holds for b

Thanks for the help mate!

[gritty widget]

ty ultra

[sick smelt]

https://mathworld.wolfram.com/FixedPointTheorem.html

[quasi forum]

Yea I know. Tbh, I am not so sure why I struggled with that one: I have done this kinda problem before.

But it's okay. I got it now

[sick smelt]

[neat mango]

okay how do these guys work

like is it just you imagining it as an infinitely stretchy square

and then rotating stuff so that it matches

But ive been having trouble imagining RP^2

and when i googled it ppl were cutting them and stuff so idk

how can that be justified

[pseudo crane]

wdym "cutting"

it doesn't actually intersect itself as far as I'm aware

oh cutting as in physically constructing

[tawdry valve]

idk if it really helps at all, but you could also pull the corner in a lil bit to make it a disk, and then the gluing is just identifying antipodal points

[neat mango]

hmm

[tawdry valve]

maybe it's more helpful to just look at the disk before gluing, and then view the quotient as having portals?

[neat mango]

hmm

hMM

visualization hurts my head smh

smh

[pseudo crane]

ultra can visualize in 4 dimensions

(so can everyone else)

[neat mango]

chads

In the beginning i had trouble

imagining a plane

not joking

[pseudo crane]

https://tenor.com/view/airbus-plane-airplane-aviation-take-off-gif-14487641

pretty easy if you ask me

[neat mango]

lmao

[pseudo crane]

sorry im turning this into #chill lol

ill go now

[neat mango]

shoo

[neat mango]

okay so im thinking about this stuff and boy

this self intersection shit

is kinda cringe

like ngl

Doesn't this mean that wrt the fundemental polygons

we are gluing more than one thing together

like more things than we want to glue

so what's the deal with this

or am i missing sth and this is not a problem

bcs with this same idea i can come up with a bunch of different representations of "the" torus that are not homeomorphic, starting with the defining fundemental polygon

[pseudo crane]

you mean like

well

there are no embeddings of RP^2 in R^3

[neat mango]

yeah that immersion shit

So my main question is how do these fundemental polygons even define

[pseudo crane]

so I believe it is a similar situation as one that you might have heard with a Klein bottle

[neat mango]

these topological spaces

[pseudo crane]

the self intersections where we don't want them aren't actually self intersections

they just appear that way because you've put it in R^3

[neat mango]

hmm

So the space formed from the fundemental polygon is simply the object in the smallest? dimension such that we have no self intersections?

[pseudo crane]

uh

im not sure about smallest

but the point is that we are doing nothing to the square besides identifying the edges in a particular arrangment/orientation

self-intersection not at these points is explicitly not allowed

idk if that answers ur question

[neat mango]

hmmm

I guess that's fair

i shud not be thinking about this it's very late rn

[pseudo crane]

the reason that is the case is because it's defined that way

lol

same

[neat mango]

well yea

i mean these were given w/o context in the book

so im trying to clue out the defn

[gritty widget]

a fundamental polygon defines a cw-complex structure

[pseudo crane]

this helps show the explicit quotientiation, at least

[neat mango]

i cee i cee

i still don't intuitively get what a CW complex is

[gritty widget]

lol

[pseudo crane]

google

[neat mango]

tried

[flint cove]

My go-to mental picture is that of a 1-skeleton (i.e., a graph) where all of a sudden a wild 2-disk appears and gets welded to said skeleton on its boundary

Another mental animation I have is taking a circle as your 1-skeleton, and then imagining the 2-disk slightly different: You make an incision , take the midpoint in your hands and place it above the circle, and then wrap it around a circle to form something like an ice-cream cone. Ofc the incision has to align perfectly once you're done wrapping (so we actually get a continuous map
And then I imagine the same ice cream cone procedure just „wrapped around twice“, i.e. every point on the circle meets two layers of 2-disk-waffle and the incision still ligns up perfectly, and I have my example of an attachment with „double boundary“ (i.e. ∂(that 2-cell) = 2*(Σall the 1-cells))

not sure if that helps you in any way tho.

[neat mango]

hmm i see

i am just going to bed rn tho so

will look at this later

thank u

tho

[empty grove]

donut

[honest terrace]

🍎

[gentle osprey]

squirtlespoof

[gritty widget]

probably means to draw a picture

or describe one accurately

[cedar pebble]

I figured out the trivialization thing it's really cute

Here's a fun illustration of this

Suppose you have a unipotent vector bundle with flat connection (E,\nabla) on an elliptic curve X=C/(Z+\tau Z) where the monodromy representation π_1(X,0)->GL_2(C) sends \alpha (the straight path from 0 to 1) to the unipotent matrix (1,a;0,1) and sends \beta (the straight path from 0 to \tau) to the unipotent matrix (1,b;0,1).

Show that (E,\nabla) is holomorphically trivial precisely if a\tau-b=0

[pearl holly]

So this might be an odd question but I've just read the definition of a topology and I am already confused. I've heard that topology studies "rubbery geometry", so how can the definition of a topological space have any connection to the "rubbery geometry"?

Oh okay. So, in simple terms, what does topology study then?

So what are homotopies?

[ivory dragon]

particular pseudoalgebras of sets

a more serious answer is that it studies "qualitative" properties of maps and spaces

[cloud owl]

well just look at the latin root of the word, homo-topy; same-place. homotopies are just places in topologies that are all the same as each other. yep.

[ivory dragon]

if that sounds really vague

and like it could apply to like

half of math

then yeah

topology is very broad

the most invoked topological property is that of continuity

which youre probably familiar with from analysis.

the topological definition of continuity, however, is divorced from fiddly epsilon delta constructions and whatnot

"a function such that preimages of open sets are open"

[pearl holly]

So like the classical example of a mug "transforming" into a donut is a homotopy?

[ivory dragon]

thats a homeomorphism.

[pearl holly]

oh

[ivory dragon]

homotopy talks about functions

homeomorphisms about spaces

you can view a homeomorphism as a special case of a homotopy for that reason

[pearl holly]

Man, it really feels like I am getting into topology without even knowing what it is... I hope this will be fun

Well I am trying to self study it without even knowing the reason behind my choice

[cloud owl]

many of the more abstract fields in maths are just like. in order to work out what you can do with it you have to understand it

topology is very useful tho

[summer jolt]

Hey, I'm trying to understand the solution to this question about finding lines on a cubic surface.

The question given is the following

I don't understand the last line in the picture above

Why must every line intersect the line $L_1$?

[gentle osprey]

snypehype

[summer jolt]

You mean something like $\mathbb{P}^1 \rightarrow \mathbb{P}^3 : [u:v] \mapsto [0:0:au+bv:cu+dv:eu+fv]$?

I don't see how that helps

[gentle osprey]

snypehype

[summer jolt]

I meant $\mathbb{P}^1 \rightarrow \mathbb{P}^3 : [u:v] \mapsto [0:0:au+bv:cu+dv:eu+fv]$

[gentle osprey]

snypehype

[neat mango]

Question why isn't every CW complex contractable

say i have this guy

I can contract the disk guy into a point

hmm

it's paint

oh word?

i guess that's why

Yeah you're right

I need to get used to spheres being the boundaries

and stuff

why so

I thought that

you could contract the disk down to a point

then you'd have S^1

And i guess i erroneously thought that was contractible too

hrm

but regardless when we contract the disk to a point doesn't that bring the edges of the lines closer together? Like necessarily?

yeah

Like the peanut looking thing is supposed to be a disk

up to bad drawing

it's supposed to be D^3 glued to D^1 at two points

hrm

one sec lemme look back at notes

oh word

yeah that's right

sorry just learning about this stuff

so sorta new

Yeah these look much less general than what i thought they were

so for instance like

I mean in the sense of

simple things

I thought these would just be a generalization of strings and balls etc just attached

And they don't generalize that entirely

could you elaborate on this

yesterday i was having trouble visualizing RP^n too

Ah i see

*RP^2

Hrm

what about that boy's surface thingy

[pseudo crane]

immersion not embedding

[neat mango]

Hmm i see i see

Ahh

okay that makes this clearer

basd

*based

[tight agate]

you can imagine RP^2 as a circle bundle on the circle

or wait can you

no you cant

sorry

[neat mango]

I've heard the name, what are they?

yeah

Hmm

So it's just a generalization then

So G(n+1,1) is RP^n

i see i see

G(2,0) would be normal plane?

err no

[pseudo crane]

what are 0-dim subspaces

[neat mango]

oh yeah sorry , empty set has dim 0 i think, no?

trueeee

damn

yeah

[gritty widget]

grASSmannian

[sweet wing]

lmao

agree

[thin jewel]

Grassmainians have a nice embedding in the form of the Plucker embedding too

In the alternating product

They essentially live as points on a quadratic in higher dimensional spaces

[tough imp]

bdobba

Also the Plücker embedding, I did an exercise about it and it was painful.

[thin jewel]

The coordinate free version is hard but I think it's not too bad just to think of the Plucker coordinates as maximal minors of some Matrix

[gentle osprey]

bdobba

[thin jewel]

Wow typing latex on mobile sucks

[sharp yoke]

Im trying to apply Lee theorem 6.35 to the function $F: X$x$R^n \rightarrow R^n$ given by $F(x,v)=x+v$

[gentle osprey]

『Urahairywizard』

[sharp yoke]

Im having trouble showing this function is transverse to Y

[tough imp]

Do we know the automorphism group of affine space? This is equivalent to knowing what Aut(k[x1,...,xn]) is, and I think this might be the Jacobian conjecture lol

[snow shadow]

Could someone help me with the following problem:

[gentle osprey]

bim

[sweet wing]

this should be in #groups-rings-fields but try to construct the valuation directly (hint consider Z_p)

[snow shadow]

I took a guess since this is a problem from a course in algebraic geometry

[thin jewel]

Just talk about Spec(R) then you're good ;)

[pearl holly]

So a subbasis S for a topology X is a collection of subsets of X whose union equals X. So a subbasis is a set that contains sets and those sets contain sets?

[empty grove]

Should be 1 level less

Set of subsets of a space

[pearl holly]

So what does the union refer to? Is it that the union of S should be X?

Or is it that the union of sets in S should be X?

[empty grove]

This

Like collection of all open intervals in R for example

Or for a topological space, the topology itself is a subbasis for itself

[pearl holly]

So the union refers to ALL the elements of S? Or just some arbitrary sets of S?

[empty grove]

All

It means the union of all the sets that S contains

[pearl holly]

Ohhh okay!

So Us = X where U is the union and s is the subbase?

[empty grove]

Yeah

[pearl holly]

Okay thank you! One more thing, if B is the collection of all finite intersections of elements of S then given x in X it belongs to an element S. But how do you know that x is in some element of B?

[empty grove]

Finite intersections includes the intersections of just 1 set

Which is doing nothing, you just get that set itself

[pearl holly]

Okay so every element in S is also in B?

[empty grove]

Yep

[pearl holly]

But what does it even mean to take the intersection of just 1 set?

[empty grove]

You can view it as intersecting that set with itself if you want

But another way to view it is just $\bigcap_{i=1}^1 X_i$

[gentle osprey]

Moldilocks

[pearl holly]

Oh okay I get it. So if A is in S then A intersect A should be in B, right?

[empty grove]

Yep

[pearl holly]

Okay well now I get it! Thank you so much! This is just so confusing. It will be fun they said... But they all lied

[empty grove]

Lol np

It becomes fun when you get to actual theorems instead of just set theoretic definitions

[pearl holly]

Yeah I hope so. I just have a hard time grasping and remember the definitions and without a teacher guiding me it's becoming harder and harder

[verbal wraith]

"In topology, a subbase (or subbasis) for a topological space X with topology T is a subcollection B of T that generates T, in the sense that T is the smallest topology containing B. "

[empty grove]

Yeah it will be a bit abstract initially, but you'll see examples and it will become clearer

[verbal wraith]

what's the problem with this definition provided by wikipedia

[empty grove]

I don't see any

[verbal wraith]

Maybe you're having trouble with how some operations on a set can generate a larger set which is closed under those operations?

[pearl holly]

Well this definition is a lot easier to understand, but how can this definition be equivalent to the one I presented?

Well I guess that this is somehow connected to prime fields? In the different definitions

[verbal wraith]

"Sometimes, a slightly different definition of subbase is given which requires that the subbase ℬ cover X.[1] In this case, X is the union of all sets contained in ℬ. This means that there can be no confusion regarding the use of nullary intersections in the definition.

However, this definition is not always equivalent to the two definitions above. "

also from wikipedia

[empty grove]

Yep this

[verbal wraith]

looks like your prof is using a non-standard definition

[empty grove]

Not really non standard both are pretty standard

[pearl holly]

Unfortunately, I have no professor. I am still in high school reading Munkres

[empty grove]

The difference is that if you allow nullary intersections, you get X automatically, as not intersecting any sets just gives you the whole space

[verbal wraith]

oh true, actually, you're right

[empty grove]

So in that case, any collection of subsets forms a subbasis as finite intersections would always give you a basis

[verbal wraith]

who even thinks about not allowing nullary intersections

In general finite intersection and union operations are a hack

If you notice in topology nothing ever depends on the finite operation

[empty grove]

Apart from this difference, it's essentially the same. The B you described earlier is clearly the smallest basis for a topology that contains a given subbasis S, and let's say that B generates a topology T by unions. Then T is the smallest topology containing B and hence also the smallest containing S.

[verbal wraith]

It's just there cause we want finite intersections and no other reason

[empty grove]

The difference being that you're defining a subbasis before the existence of a topology, then using the subbasis to generate the topology

While in the other case you'd be given a topology and checking whether a subcollection is a subbasis

[pearl holly]

Ohh okay I think that I get it!

[verbal wraith]

and also if you're just learning topology I would add that the definitions won't make any conceptual sense from the POV of open sets

open sets have some nice properties related to this: https://en.wikipedia.org/wiki/Invariance_of_domain

But closed sets are the ones that actually make sense

[pearl holly]

Okay but are there any other way of thinking about open sets instead of some fancy math definition?

[verbal wraith]

they are the complements of closed sets

[empty grove]

I just like to think in terms of metric spaces

That's what motivates topology in the first place

[verbal wraith]

with a topology the main idea is that you have some closed sets and an intersection operation

and then you also have a vestigial finite union operation

[empty grove]

There are many equivalent ways of looking at a topology (you can also look at kuratowski's definition in which a topology is just a unary operator on P(X)) but what you find best will depend on your background

[gritty widget]

I think it's just something you get used to lol. I remember being confused for quite some time before it started being to make sense.

[empty grove]

Almost all books use open sets

[verbal wraith]

I mean that's a terrible reason

[empty grove]

Because most students are comfortable with working with those in metric spaces, and you don't have to think of any algebraic closure operators etc

[pearl holly]

So in a metric space, what would an open set represent?

[empty grove]

The way you described topology is way too abstract given the apparent background here

[gritty widget]

In standard euclidean metric, an open ball.

Or a union of open balls.

[empty grove]

To think in terms of operators abstractly like that

[verbal wraith]

No it's just the direct explanation for why you would consider a topology

[empty grove]

I would really suggest you see metric spaces before topology

[pearl holly]

Yeah I have looked at it a bit before but I don't know how much I actually need

[empty grove]

Or at least see a bit of theory with them, because topological spaces generalize things like continuity for when you don't have a distance function but only a notion of relative closeness

Like continuity is defined using the distance by using epsilons and deltas

[gritty widget]

I don't know, I think both approaches have their pros and cons. By learning metric stuff first you kinda get used to having a metric and don't really get the 'general' case (at least thats how I felt)

[verbal wraith]

This is historical precedent but it's largely unhelpful for getting the big picture

[empty grove]

But you don't actually need exact distances, intuitively you can say that things that are very close stay very close

Without metric spaces I cannot see how you would motivate something like continuity

The topological definition of continuity seems very weird compared to the one in metric spaces, but it's easy to prove that they are equivalent for metric spaces

[verbal wraith]

It's because pointset topology is poorly understood

[gritty widget]

Maybe you're right, although I think Munkres explained point set pretty well. (response to Moldilocks)

[verbal wraith]

the definition in pointset topology is convoluted

there's another definition of continuity in general topology which is intuitive

[pearl holly]

Okay so I have two book that I borrowed. One is Munkres which is the one I am reading now. The second one is "Introduction to topology" by Theodore W. Gamelin which starts out with metric spaces instead of topology directly. With one would "better" to start with for an "intuitive"/better understanding?

[gritty widget]

Honestly, try both and see what you like more (just read first chapter or sth).

[empty grove]

I feel metric spaces provide the visualisation, without which it might seem like just a bunch of formal notions

[verbal wraith]

In pointset topology you have two semi-lattices. P(X) and T

A morphism of topologies is therefore two maps

one on X and one on T

so a continuous function is a function on X that induces a function on T which has some property

which is why that definition is hard to understand

for any function f on X the function that you get on T is cl o image_f

if that makes sense

[empty grove]

No idea

[pearl holly]

Sorry but I honestly don't understand anything

[empty grove]

Are you mapping the lattice T_X to the lattice T_Y or what

[verbal wraith]

yeah

[empty grove]

And is T_X the collection of closed sets

[verbal wraith]

yeah

[gritty widget]

@pearl holly Ignore them lmao.

[pearl holly]

Well I am going to read a bit of both. Thank you all so much!

[empty grove]

But closed sets may not go to closed sets

[verbal wraith]

that's why u use the closure

[empty grove]

oh ok

I don't see how this helps with intuition

[verbal wraith]

But the real property of continuity is that the induced map f_T is a left adjoint

[empty grove]

Bruh

[verbal wraith]

which may sound complicated

[empty grove]

You wanna do cat thy before metric spaces

[verbal wraith]

but actually means that you have a first isomorphism theorem type result

you don't need to do any category theory for this

[empty grove]

So you just specialise to the specific case?

[honest terrace]

still not sure if Nikita is trolling or not

[verbal wraith]

uhhuh

So the first isomorphism theorem

it states that the image map and the pre-image map are left and right adjoints

of P(X) and P(Y)

[empty grove]

Are you trying to see how far you can go before I realise you're trolling or are you actually saying this legit

[verbal wraith]

but basically this result only requires the fact that P(X) and P(Y) are semi-lattices

lool

I should just type it up so that people have a good resource for it

[empty grove]

Yeah because as it is I do not see how this could help when seeing topology for the first time when people already find it too abstract with the usual definition

[verbal wraith]

I mean how is the first isomorphism theorem not intuitive

[nimble pebble]

i havent worked my way thru a full semester of point set topo yet and i think its because i just dont understand cat theory yet

[empty grove]

Idk bruh I find the epsilon delta thing way more concrete and from there it's very easy to see equivalence to topological continuity

[verbal wraith]

A morphism should always be a thing such that the first isomorphism theorem holds

[honest terrace]

I wouldn't say it's easy but atleast it's not abstract nonsense moldilocks

[empty grove]

I've never seen a first isomorphism kinda thing for continuous functions so idk

Wait are you talking about quotient topologies

Or is this a first isomorphism theorem on lattices

[nimble pebble]

i always found myself feeling numb when looking at topo continuity bc it doesnt feel real

[verbal wraith]

yeah so a quotient of T_X is isomorphic to a subobject of T_Y

that's what I mean here

[empty grove]

I see

Yeah well I don't immediately see the link so will probably have to think about it

[empty grove]

But I like how you said this

[verbal wraith]

XD

[honest terrace]

[pearl holly]

For 1, does'nt A=U?

[honest terrace]

no

for example take ]0, 1[

for any x € (0,1)

there's some epsilon_x

[nimble pebble]

A could just be any neighbourhood contained in U

[honest terrace]

such that x € (x-epsilon_x, x-epsilon_x) C (0,1)

[bleak helm]

Be gone frenchman!

[honest terrace]

so it'd be open

[gritty widget]

https://tenor.com/view/funny-as-hell-french-cat-what-are-you-french-kitty-gif-17067952

[honest terrace]

I changed my notation

[nimble pebble]

epic embed fail

[honest terrace]

while writing

lol

[gritty widget]

How to embed

[nimble pebble]

usually being inconsistent is the worst sin

but its better than consistent ][

[honest terrace]

anyway did you understand @pearl holly ?

wait

Ok I'm dumb

I understood your question wrong

[gritty widget]

🤔

[honest terrace]

[gritty widget]

[honest terrace]

so a better answer:
take (0, 1)

again

and you have 1/2 \in (1/3, 2/3) C (0, 1)

but (0,1) != (1/3, 2/3)

[pearl holly]

Yeah now I understand. I read the question wrong lmao

[honest terrace]

so I'm not the only one who can't read

[pearl holly]

So I want to show that if ${\tau_{\alpha}}$ is a family of topologies on X then $\cap \tau_{\alpha}$ is a topology on X. So let $A_i \in \cap \tau_{\alpha}$ and let $x_i \in \cap_{i = 1}^n A_i$. This means that $x_i$ is in every single of of the $A_i$'s. Since $A_i$ per definition is in $\cap \tau_{\alpha}$ I have shown that $\cap \tau_{\alpha}$ is closed under finite intersections, right?

[gentle osprey]

older sister

[pearl holly]

And the same method can be used for finite unions and we also know that X and the empty set is in the set so it is a topology. Is this right?

[summer jolt]

Does anyone know how to find all points of order 2 in a curve?

[verbal wraith]

Maybe you mean the intersection of A_i is in each tau_alpha and thus in the intersection of the tau_alphas?

[pearl holly]

Well I don't really know what I meant, but is my solution correct?

Because I have shown that each element of the intersections of A_i is indeed in a set of the topology, right?

[viral atlas]

You don't really need to consider elements, you just have to check that finite intersections of A_i's are in intersection of tau_alpha

Same for arbitrary unions

[pearl holly]

Yeah but don't I need the elements for that?

Oh wait. The intersection of every A_i (where A_i is in the intersection of every tau) is in the intersection of every tau since every A_i must be in each tau and tau is per definition closed under intersections?

[verbal wraith]

sounds right

[pearl holly]

Okay great! Thank you! But just out of curiosity, is my earlier solution correct?

[verbal wraith]

no

[pearl holly]

Why not? I really feels like I have shown that every element in the intersection of A_i is in some set of the topology, which means that the intersection is in the topology, no?

[verbal wraith]

Oh actually, I think I see what you're saying

well yeah it's right, too. I guess its just a very simple problem that you should try to highlight key things in a proof

so you just didnt state the conclusion of the part of your argument that uses x_i's

[pearl holly]

Yeah that's true. I will keep this in mind. Thank you!

[tardy meadow]

sanity check - I consider cts maps f:\cup A_i -> {0,1} and use induction to conclude that f(A_i) must be equal to the same singular point for each i. So f(\cup A_i) is equal to that singular point so f is constant so \cup A_i is connected. Is that all correct?

had something in the back of my mind saying that didn't work for some reason

[frosty sundial]

seems good to me

[strong heron]

Can a non-orientable surface be a cover for an orientable surface?
Specifically, I have to find 3-fold connected covers of X= torus-{disk}. I used the property of Euler characteristic to conclude that the Euler characteristic of a cover of X must be -3.
But I'm not sure how to rule out all the options that I'm getting based on this information alone.

[light plinth]

hello guys, im going to try to give a good explanation of my question !
I have a theory on a possible solution to my master thesis that is about genetic algorithms (no need to understand what that is to understand my question, but geometry and topology is one of the main advanced components that are deeply rooted into this topic)...
in order to achieve and test my hypothesis I need to be able to, in any n-dimensional sphere, calculate the cartesian coordinates of each point in the line that unites the center of that sphere with its boundary with a certain angle... my objective is create a spherical scan (assuming im not doing only this line im exemplifying but all possible lines in the 360 degrees possible) that possesses every point in that line
think of it like you're finding all the points that are inside that red line... if I want to check all the space that is in that line I need to get all of the points of that line (assume r = 100, I have to get the cartesian coordinates of every unit "1,2,3...till 100" --> assuming the smaller distance is one unit)
then I will change the angle (assume the smaller unit is 1degree) and if I do that for all the 45 lines i will get each and every point on that quadrant cartesian coordinates)

[light plinth]

😦

[verbal wraith]

Not sure what you're asking man....

[light plinth]

basically im asking how can I get cartesian coordinates of a point in 32d for example

n-d

[verbal wraith]

Are you trying to get points on a sphere that are spread out approximately at a uniform distance from one another?

[light plinth]

yes!

but not on a sphere, but on a n-d sphere..

[verbal wraith]

why don't you just generate some random points and delete ones that are too close to each other until you get what you want

[light plinth]

hmm not sure i can do that in my problem

by close enought

enough, i mean in the same direction

like a straight line but in nd

cartesian coordinates of a 2d line within a nd space

[gritty widget]

I think you should also draw the situation in 3d

[frosty sundial]

a + bt

where b is a vector

and a is a point on the line

[light plinth]

this is probably a tensor problem not sure

[frosty sundial]

it sounds like youre just trying to generate points uniformly on the sphere

you can just generate points uniformly in the cube [-1,1]^n and then scale them to be on the sphere

[verbal wraith]

If you just want a great circle in n-d space that's not too hard. you just pick two perpendicular directions and then it's the same as the 2D case.

[light plinth]

yeah, but that would be only in the direction of each axis... for example if I have something 5D and i make 3 points lets say [1, 0,0,0,0] ,[2, 0,0,0,0],[3, 0,0,0,0] this would be in the same direction

but if I only scale 1 of the dimensions I would be only searching in each axis direction

and we know there are more... ofc

[frosty sundial]

what do you mean?

I'm saying to generate points randomly in [-1,1]^n

all coordinates are between -1 and 1

and then scale them so they lie on the sphere

[light plinth]

i want to search a space, point by point, in a single direction (a straight 2d line through the space), then if i dont find anything, ill look on another direction that is like changing an angle in a 2d circle radius (radius being an example of a straigh line through a 2d space)

[frosty sundial]

what is a 2d line

a line is a line

[light plinth]

yeah mb

a line *

[frosty sundial]

pick two vectors on that circle, call them u and v

then cos(theta)*u + sin(theta)*v will parametrize the circle

[chrome dew]

are you wanting to know how to go through spherical coordinates in higher dimensions or something? can you articulate what is preventing you from doing what you want to accomplish

[frosty sundial]

multiply by r and you get lines

[light plinth]

this yeah

like it would be easy to show you context

im sure this isnt a hard problem

[frosty sundial]

https://en.wikipedia.org/wiki/N-sphere#Spherical_coordinates

[light plinth]

ok that might help!

because I'm basically wanting to do that in order to analyse points (data points) through a landscape

but not in a random way

they literally have to be on the same line since the center till the border of that n-d sphere

and then, if i cant find anything within a certain threshold (a rule) I must change that line to one that is point towards a close to the previous space

[chrome dew]

sounds like he wants to systematically go through the angular components, and for each angle, go from 0 to R along the radial component

[light plinth]

yeah, what you're saying is really making sense to me

[chrome dew]

and do this for discrete points in a certain order

[light plinth]

but I cant imagine how i do that in a 32D for example

[chrome dew]

so you need a way to detect what the next point is

as you rotate through an angle, what's the next point

or set of points

well, I suppose you could just convert from rectangular to spherical coordinates

then sort by the angles in the way you want to move through the list

for instance in 3D your points will be like (r, theta, phi) you could sort them like:
(.2, 0, 0)
(.42, 0,0)
(1, pi/3, pi/7)
(.3, pi/5, pi/2)
...

so that you are looking at going through theta first, then phi, but for those fixed angles the same you go from smaller r to larger r

once you've converted to spherical coordinates in n-d then it's just about sorting a list how you like to go through the angles

good luck

[light plinth]

ok, that last sentence was one of my problems and I can see how u are solving it but how should I change the angles, but if we have like a 15d space, we will have how many angles?
im obligated to keep them as rectangular (cartesian) coordinates...

14?

so I would have (r, theta, phi, x,y,z,p,j...., 14th angle) right?

and I would need to maintain the choice of angles throughout the dynamic R

that makes sense for me

but is there a way in which i can do that but for rectangular coordinates?

[chrome dew]

use the wikipedia link from earlier

it shows how to convert from rectangular to spherical

if you want to sort by spherical coordinates, I would just convert it

maybe there's some trick to doing it in rectangular coordinates that sorts it as if it's spherical coordinates directly but idk

[light plinth]

yeah, I was hoping that trick would be maybe tensor related?

[chrome dew]

nah no tensors

[light plinth]

something like a matrix operation of those starting coordinates that would simulate what we are trying

(a direction through space)

[chrome dew]

all I'm describing is a change of coordinates

and then sorting the coordinates

there's no magic tensor stuff happening

[light plinth]

ok

thx anyway for this elucidation, it certainly helps in the journey

[chrome dew]

you're not the first person in the world to want to do something like this I'm sure

[light plinth]

once im able to do that in my code, i will be able to test it

[chrome dew]

try googling for something like 'sorting points with rectangular coordinates by their angles' or something

[thin jewel]

How much diffgeo do I need to make it though a module on Lie algebras

I’m thinking of either that or a module on reflection groups (my advisor has mentioned that either one of these two things would help for a potential project)

I feel like the Lie Algebras course is more general but it’s going to end up way more work

[strong heron]

Anyone?

[verbal wraith]

No because the orientation on the orientable surface would induce an orientation on the surface covering it.

[strong heron]

Can you elaborate just a little bit more?

[verbal wraith]

Just take any loop on the cover surface, it's image is a loop on the orientable surface.

you can assign an orientation to every loop on the cover surface based on the orientation of its image.

[strong heron]

Ah, I see!

Thanks

[dusk heron]

Is it correct to define a principal bundle as a map $\pi:P\to X$ together with a fiber-preserving and free right $G$-action on $P$ which acts transitively on each fiber? More precisely, does this automatically imply that $\pi$ is a quotient map, so that $X$ is homeomorphic to the quotient space $P/G$?

[gentle osprey]

gustavn64

[strong wren]

I have a strange topological challenge of sorts. I think I need help with it.

Let me know if you're interested and I'd be happy to discuss my problem. I'll reply in a bit, bout to drive.

[gentle osprey]

slimvesus

[empty grove]

I can see the independence from A_0, because if a loop goes around the origin some zeta times in R2, then an orientation preserving invertible linear map is just like distorting the axes slightly. A proper way to see it will be as a composition of 3 maps. If e1 and e2 are standard basis of R2, one would be sending e2 somewhere in the upper half plane, another is rotation by some angle and another is rescaling, and all 3 preserve zeta.

I'm also confused by the zeta + 2deg alpha thing. Visually it seems to me that there should be an eta instead of a 2.

Ohhh

I was just thinking of RP1 as S1 itself because they're homeomorphic lol but I guess the homeomorphism has degree 2

Still don't see why there's no factor of eta in that term tho

So do you see why a loop in RP1 won't change degree by application of A_0

Think of R2

Rotation won't change the number of times a loop goes around the origin

The whole loop just gets rotated

Loops in RP1 starting at [1] are just paths in R2 minus 0 that start at 1 and end at ±1

Well if alpha is the identity map, the longitudinal loop on the torus gets one meridional twist it seems

I'm thinking of first coordinate as longitude

Oh wait no that actually agrees with the example

Take the constant loop

It's image is the constant loop regardless of deg alpha

So 2 deg alpha seems wrong

Because it corresponds to (0,0) in the fundamental group

Should be 2 eta deg alpha I think but I only have visualisation not a proof yet

Yeah so what I'm thinking is that F twists each meridian in the torus by alpha(x), so if alpha has deg 1, the loop going around the torus once along the longitude will get 1 meridional twist

And if you go around the longitude n times, you end up with n meridional twists

That's why I say 2 eta deg alpha

n meridional twists = actually 2n because RP1

Because a meridional twist is unaffected

Like if x is fixed

Yeah I'm being very reliant on my visualisation lol

Assume for now that A_0 is identity. So if you have the loop (1, e^2ti pi), thinking of S1 x S1 for convenience, then F maps this to (f(1), e^(2ti pi + alpha (1)))

Since alpha(1) is a constant, you're just rotating this loop, that doesn't change its degree

Ah right

t in [0,1]

Yeah so I think I can actually prove it formally now

Just give me some time

[empty grove]

rotation by 2pi alpha(x) you mean?

[gentle osprey]

slimvesus

[empty grove]

I didnt get it

Let $a(t) = (1,[e^{i\pi t}])$ and $b(t) = (t, [1])$ be loops $[0,1]\to S^1\times \mathbb{RP}^1$ based at $(1,[1])$ ($t$ is a loop in $S^1 = \mathbb R/\mathbb Z$, not putting []). Then their equivalence classes generate $\pi_1(S^1\times\mathbb{RP}^1,(1,[1]))$, so it is enough to see the effect of $F^*$ on these.\\
$(F\circ a)(t) = (f(1), [A_0 e^{i\pi t + 2\pi\alpha(1)}])$. $A_0$ doesn't affect the equivalence class because it doesn't affect the equivalence class of the standard generator of $\pi_1(\mathbb{RP}^1,[1])$, which is $[e^{i\pi t}]$, because this loop remains monotonous (because of orientation preservation) and only visits $[1]$ at its end points. So $(F\circ a)(t) \sim (f(1), [e^{i\pi t + \alpha(1)}])$, which corresponds to $(0,1)\in \mathbb Z\oplus \mathbb Z$.\\
$(F\circ b)(t)\= (f(t), [A_0e^{2\pi\alpha(t)}])$\$\sim (f(t),[A_0e^{2\pi\alpha(0)}]) \cdot (f(1), [A_0 e^{2\pi\alpha(t)}]) $\$\sim a'^{\deg f} \cdot b'^{2\deg\alpha(t)}$\
where $a'$ and $b'$ are $a$ and $b$ but rotated/translated so that the basepoint becomes $F(1,[1])$.

[gentle osprey]

Moldilocks

[empty grove]

good thing I didnt type this in discord directly

Decompose any loop as a product of a and b

And apply F*, you'll get the result

You will get 2 eta deg alpha

Np nice problem

[pearl holly]

So let's say that X is a metric space with a subset Y. Then x_1 is not adherent to Y (because the space is too small to contain a ball that reaches Y) but x_2 is. Have I understood the concept correctly?

[rugged swan]

To be sure, is the canonical line bundle on the projective space P_k^n the same as O(1) ?

[empty grove]

Nope. A point x is not called adherent to Y if there is some open set of x intersecting Y, but if every open set containing x intersects Y. Another way to think about it is that if I give you a positive real number r, you can find a point in Y which is within r of x (so that any open ball around x of radius r contains at least one point of r, which is exactly the earlier definition)

[pearl holly]

Ohhh okay. So x_1 is adherent to Y because I don't need to draw a circle but instead some curved path consisting of circles?

[empty grove]

No, as you've drawn it, neither is adherent to Y

Because for both, you can find an open set not intersection Y

[pearl holly]

Oh okay, I got it

[empty grove]

Whether it has to be a circle or a general open set turns out to not matter

[marble socket]

another way to think about adherent point is distance of x from the set Y\{x} should be 0

[empty grove]

Because if every ball around x intersects Y, then every open set around x (which must contain a ball around x) also intersects Y

[pearl holly]

So the closure of Y is literally Y with it's boundary?

[empty grove]

Yep

[pearl holly]

Ohhhh

So if Y is closed then it only contains the boundary?

[marble socket]

(it also contains its interior...)

[empty grove]

Not only, it could also contain the stuff in the interior

[pearl holly]

Oh yeah, that's right

Okay okay I think I get it now. Thank you all so much!

[empty grove]

@marble socket where's the eevee emoji wtf

[marble socket]

(lol i didn't help at all)

(i use the emoji to say "thanks for thanking me", "the current problem is over" and "look eevee is so cute")

[summer jolt]

Hey, I'm trying to understand the solution to this question here (part b)

First of all what is w' ? Is just a loop or is the composition of two loops?

The solution is the following

I don't quite how they worked out part b

[stray bane]

I asked my question in the question channel earlier, but it was probably the wrong place to go. I am studying the topology book off munkres, and i am trying to find an example of two sets X and Y, aswell as a subset $Z\subseteq X\times Y$ such that the projections of Z onto X and Y are both open, but Z itself is closed wrt. the product topology. I really can't wrap my head around how this could be possible, could anyone give me hints to work out an example?

[gentle osprey]

Frozaken

[stray bane]

Sorry this is correct, it should say "not open" and not closed

I am fairly sure an open disc minus a smaller open disc would satisfy this right? that is my best guess atleast

However i find it really really hard to prove that this is correct

it would not be open since it has a boundary towards the interior, however this is intuition that comes from analysis, and i am not sure how i would ever show this in the context of the product topology

[pseudo crane]

would Z=X×Y not work?

sort of trivially

and closed

oh did they say not open

mb didn't see

[stray bane]

The projections would be open since they are just the projections of the larger open disc onto both subsets i think

both sets*

X and Y

Perhaps this example is simpler?

Let $X$ and $Y$ both be $\mathbb{R}$. and let $Z$ be the union of the x and y axis. $\pi_X(Z)=\pi_Y(Z)=\mathbb{R}$, thus both are open.

[gentle osprey]

Frozaken

[stray bane]

however the union of both axes is clearly not open

I really struggle to argue this from the product topology though, i guess i have been working with the analysis way of looking at openness for too long

I guess i first use that it is written as a union of two axes(we can because it is a topology) and the use that either axis cannot be written as a union of cartesian products of two open sets (since the singleton {0}) is not open in the standard topology on R

[empty grove]

that is an incorrect proof, openness of the union of a bunch of sets does not imply openness of each set in the bunch

[empty grove]

Take any point in the union, eg 0, and show that theres no neighbourhood of this in the union of the 2 axes. If there were, then some point of the form (x,x) would be there for a non zero x

[pseudo crane]

they're proving closure

er

[empty grove]

closure of the union? why is that needed?

wasnt the condition just "not open"

[pseudo crane]

dammit I thought it was "closed and not open"

[empty grove]

yeah they changed it later

[pseudo crane]

im really bad at reading apparently

lol

[empty grove]

ono

ah I see

lol smh

[empty grove]

Those 2 are the same thing, a composition of loops is also a loop

For b part, instead of RP^n, think of S^n. Conveniently enough, w' lifts to a loop in S^n, based at 1 (do you see this?). On S^n, for n>1, you might know that all loops are contractible, so you can extend the map to a map from the disk to S^n. Take the composition of this map with the quotient map to get the desired map from D^2. What that solution is doing is just writing down the extension to D^2 unnecessarily explicitly

[summer jolt]

w' lifts to a loop in S^n
what do you mean by lift?

On S^n, for n>1, you might know that all loops are contractible, so you can extend the map to a map from the disk to S^n.

I'm aware that all loop are contractible but why does this imply that we can do such extension?

[empty grove]

are you familiar with quotients?

one way to think about it in this specific case is to say that the loop is an equatorial loop in S^n and so you can map the disk to one of the hemispheres, but this isnt the most obvious for n>2

[summer jolt]

I always struggled with them, it took me a week to understand the universal property of the quotient😂

[empty grove]

oof

Well a homotopy between the loop and the constant loop is a map from the cylinder S^1 x [0,1] to S^n such that S^1 x {0} is the given loop and S^1 x [0,1] is the constant loop. You can then quotient by the equivalence relation that identifies all points of S^1 x {1} to a single point (making the cylinder into a cone) and by the universal property, the homotopy factors through the cone so you get a map from the cone to S^n which maps the base of the cone to the given map, so this is the extension of the loop to the cone. But the cone is homeomorphic to the disk with homeomorphisms that dont affect the circle, so you get an extension to the disk

this can be done without quotients, lmk if you cant visualise the quotient

[summer jolt]

Ah yes I see so the cone is homeomorphic to the disk! I forgot about that.

[stray bane]

Huh interesting, how about this case though where i know from the get go that my set is the union of two sets, if i show that none of them are open then Z cant be open though? Or am i misunderstanding

[empty grove]

2 not open sets can union to an open set

think of indiscrete topology on 2 elements

[stray bane]

Wow okay i did not think of this. Im not sure i can just look at the neighbourhood, as this is mostly the analysis way of showing it i believe, but i should rather use the basis of the product topology

[empty grove]

You can view it as a product topology proof too. Any neighbourhood of 0 will contain a basis element (a,b) x (c,d)

then if r = min{-a,b,-c,d}, then (r/2, r/2) is there in that basic open set

ultimately any proof would rely on the metric of R, because thats how the topology on R is defined

well not necessarily, you could view it as an order topology and the above as a proof based only on the order and archimedean property of R rather than on the metric of R, if you want to look at it as less analytic

[stray bane]

Alright thanks, i only recently started looking into topology, so this is very helpful, thanks

[obtuse meteor]

A more analysis example. Union (-1, 0] and [0, 1)

[stray bane]

Could i not just use that for $U$ to be open in $X\times Y$ we must have for each $x\in U$ a basis element $B$ such that $x\in B$ and $B\subseteq U$, however from the definition of the order topology, on $X\times Y$, all of the basis elements are on the form $A\times B$ where $A$ and $B$ are open wrt the standard topology on $\mathbb{R}$, we can now simply pick an arbitrary $x$ in our set consisting of the union of two axes, and show that it is in fact imossible to find a basis element where it also holds that $B\subseteq U$, since all of the basis elements are "too thick"

[gentle osprey]

Frozaken

[empty grove]

yep thats essentially what ive done, just a bit more explicitly

[stray bane]

I guess just using your intuition of picking out a single point and showing that its neighbourhood is contained is very useful, ill work the details out myself i think 👍

[bright acorn]

How can I prove that the ring of germs of holomorphic functions $\mathcal{O}_{X}$ on $x$ defined on an open subset $X \subset \mathbb{C}^{n}$ and $x \in X$ is a local ring?

[gentle osprey]

MisterSystem

[stone cipher]

,tex Is it true that the connected components of an open set $U\subset \mathbb{R}^n$ are open sets of $\mathbb{R}^n$, thus path-connected?

[gentle osprey]

RaD0N

[bright acorn]

This is true since $\mathbb{R}^{n}$ is locally connected

yeah

[gentle osprey]

MisterSystem

[empty grove]

In a local ring, the maximal ideal is exactly the set of non units. Can you characterize units?

[bright acorn]

Oh yeah

The units of $\mathcal{O}_{X}$ in this case are germs of holomorphic functions of the form $\frac{1}{f}$ where $f : X \subset \mathbb{C}^{n} \rightarrow \mathbb{C}$ is holomorphic and non zero at $x$.

[gentle osprey]

MisterSystem

[empty grove]

yeah, and whenever a germ is non zero at x, it will be a unit

[bright acorn]

But how do I prove this maximal ideal is indeed unique?

[empty grove]

If the set of non units is an ideal, then any proper ideal (which cannot contain units) must be a subset of the ideal of non units

[bright acorn]

hm

Oh I see it now

Thanks

[stone cipher]

Thx

[stone cipher]

,tex I thought that it is true because $\mathbb{R}^n$ is locally path-connected. It is sufficient to be locally connected?

[gentle osprey]

RaD0N

[bright acorn]

You only need the fact that $\mathbb{R}^{n}$ is locally connected to prove that given any open subset $U \subset \mathbb{R}^{n}$ and a connected component $G$ of $U$, then $G$ is open.

[gentle osprey]

MisterSystem

[bright acorn]

But in order to prove that, indeed this component $G$ is also path connected, then indeed you need to use the fact that $\mathbb{R}^{n}$ is locally path-connected, because in a locally path-connected space open and connected implies path connected.

Get it?

[gentle osprey]

MisterSystem

[bright acorn]

Proving this is also not that hard btw

[stone cipher]

Got it. Thanks!

[empty grove]

How are A and B mobius bands when m=n=2? Seems to me that A and B are bands with 2 edges, just that one of the edges is twisted twice over and overlapped with itself (or being turned into RP^1) I dont see the klein bottle thing either. I tried thinking in terms of CW decompositions but they seemed very different

[cedar pebble]

hint: for m=n=2 you're getting this folding

[empty grove]

I dont see it, isnt the top blue edge going to paste onto itself rather than onto the bottom blue edge?

because points with second coordinate 0 are being identified with points with second coordinate 0 and same with 1

[cedar pebble]

no

you're gluing the red edges together and the blue edges together

this gives you a Klein bottle

[empty grove]

I meant in the given description

[empty grove]

I dont see how that matches this picture

[cedar pebble]

gluing the red edges gives you the S^1 part of this

gluing the blue edges gives you this remaining quotient

[empty grove]

This is what I am getting from the description of the equivalence relation

ignore the blue arrowhead on the red arrow

[cedar pebble]

right

what you have drawn is like

two copies of this fundamental polygon glued together

so as to obtain the fundamental polygon of a torus

[empty grove]

wait im not drawing this

the top edge and bottom edge are different colours

blue and green

[cedar pebble]

hmm

[empty grove]

because we are identifying (x,1) with (-x,1)

and (x,0) with (-x,0)

[cedar pebble]

wait this has got to be a typo then

[empty grove]

so it should change the second coordinate in both cases?

[cedar pebble]

you see how you get the picture I'm posting if you identify (x,1) with (-x,0) and (x,0) with (-x,1)?

[empty grove]

yeah

[cedar pebble]

yea okay

[empty grove]

I see, cool

thanks

[cedar pebble]

hah okay I see why this was confusing now

I've literally done this with a paper where I didn't realize the author made a typo and confused myself for like 2 weeks straight

[empty grove]

RIP

[empty grove]

What are N and n?

[empty grove]

No idea what the principal normal is

[uncut surge]

Sooo it's well known that if $G$ is a Lie group with Lie algebra $\mathfrak{g}$, then the exponential $\exp : \mathfrak{g} \to G$ is a diffeomorphism onto its image if restricted to a small enough neighbourhood $U \subset \mathfrak{g}$ around the identity $0 \in \mathfrak{g}$. Can one choose this neighbourhood to be Ad-invariant, so $\text{Ad}_g U = U$ for all $g \in G$?

[gentle osprey]

Lartomato

[uncut surge]

Intuitively this feels like a very stronk requirement and if it holds, probably only in some really special cases like compact or semisimple groups, but idk

[sleek thicket]

hmm

So I want to think about the case of a compact group

Because in that case, you can choose a bi invariant metric and turn this into a RG thing

And RG has a good description of like, the amount of stuff you need to get an injective/surjective exponential

I also wonder whether like exp(U) is conjugation invariant

Like I think C_g(exp(X)) = exp(Ad_g(X)) in exp(U) if X in U

[uncut surge]

Ye that should be right if you think about matrix lie groups

[sleek thicket]

Yup

The exponent map is natural

okay so that's something

If V = exp(U) then V is conjugation invariant

The subgroup generated by V will be normal

well that's not super interesting, it's going to be the connected component of the identity I think?

[uncut surge]

Oh that sounds like a good point though

[sleek thicket]

Yeah, I feel like V being conjugation invariant is key

Since we're basically assuming the infinitesimal version of that

[uncut surge]

If the subgroup is normal and your Lie group is, say simple and connected, then the subgroup has to be the whole group, and then you're basically demanding that your exponential map is a global diffeomorphism between the Lie algebra and G

which I guess can't be

[sleek thicket]

No but at some point I took the subgroup generated by something

The exponential won't be surjective

[uncut surge]

Hmmm, ah yeah, i guess exp(U) isn't automatically a group

Awright, but this is still a good perspective, I'll think about that for a bit

[sleek thicket]

Yeah, it will rarely be such

Sure, but unfortunately exp(U) generating the connected component of the identity is just always true

The subgroup generated by an open subset is open and open subgroups are actually clopen

[uncut surge]

Haha true

[sleek thicket]

Oh I guess I'm assuming U is connected

Which is probably bad

Since your assumption is about U

But still

It's going to be a union of components maybe

Instead of a single component

Anyways yeah my first instinct is to look at the compact case by trying to fit into the RG framework

Choose a bi invariant metric and look at the injectivity radius

[uncut surge]

Yup, alright, will do! Thanks

[pearl holly]

[1, 2) is not in the K-topology on R because [1, 2) can't be written as an open interval, right?

Isn't the proof that the K-topology and the lower limit topology (both on R) are not comparable kind of "trivial", at least not too difficult?

[marble socket]

well kind of...

[marble socket]

yea its pretty simple... you just need to find a set which is open in one and not in other and vice versa.

[pearl holly]

But do I need to prove that [1, 2) is not in the K-topology or is it "trivial"?

[marble socket]

just prove it with definitions?

(its trivial in the sense that proof isn't hard at all... but you still need to give a proof)

[pearl holly]

Oh okay, let me think about it

Well [1, 2) can't be written as (a, b) U ((a, b) - K)

[marble socket]

well i can just say (1, 2) u (2, 3) u (3, 4) can't be written as those... does that mean its not open?

(also (a, b) u (a, b) \ K = (a, b))

[pearl holly]

So I want to show that [1, 2) intersect something in the K-topology can't be in the K-topology?

[marble socket]

just show [1,2) is not in K-topology?

you just made the question one step longer?

if you can show that for every something, then you can show it for something = (-5, 5)

[pearl holly]

Yeah sorry I got carried away. But there's no interval in the K-topology that is equal to [1, 2) so what is there to prove?

[marble socket]

well like i said there is no interval in K-topology that is (1,2)u(2,3)... then why is the second thing still open?

[pearl holly]

Well that is a union of open intervals that also is the empty set. But the K topology can't contain half open intervals, right? It only contains open intervals, right?

[marble socket]

that's not the empty set

[pearl holly]

oh sorry, I thought that it was intersections

[marble socket]

well you need to prove that... the definition of topology is by a basis. so it tells you a few open sets... it doesn't tell you what are not open sets

we want to show [1,2) isn't open

and you can intuitively see that the problem is at that "closed-ness" at 1

how do you define a set to be open given the basis?

[pearl holly]

Okay so each basis element is of the form (a, b) U (a', b') - K. Then the open sets are unions of these. But the unions of these can't be half opened sets, right?

[pearl holly]

The union of some basis elements, right?

[marble socket]

yep

so if [1, 2) was open then you can find a basis element containing 1!

[marble socket]

i can see why but not completely happy. since you didn't really explicitly use anything about K. if K = [0, 1) then this statement is false... (0,2) \ K = [1, 2)

[pearl holly]

But what does this give me?

[marble socket]

how do the basis elements that contain 1 look like?

[pearl holly]

Like this: ?

[marble socket]

lol what

[pearl holly]

lmao sorry but I have no idea

[marble socket]

there are two types of basis elements right

(a,b) and (a, b) \ K

[pearl holly]

yeah so (a, b) must contain 1

[marble socket]

but then it also contains stuff less than 1!!!

[marble socket]

if this was true, then we would find a basis element contained in [1, 2) which also contains 1

do you see the problem?

[pearl holly]

no

oh okay I see now

So for example if one basis element contained 1 then the basis element should look something like (-1, 2) U (a, b) \ K.

[marble socket]

(okay maybe we've read different texts... my book defined the basis as {(a,b) | a < b} u {(a, b)\K | a<b})

that's a minor problem though...

[pearl holly]

my book does it as well

[marble socket]

(you kept saying this... so i kinda got confused)

this is not a basis element... its union of 2 basis elements.
(a, b) U (a', b') - K

I'll summarise the argument:

[a, b) is not open because if it was then there would be a basic open set L containing a and contained in [a, b). But then L = (c, d) or L = (c, d)\K. in either case, one can find an element x in L smaller than a. But this contradicts that L was contained in [a, b)

[pearl holly]

"L containing a and contained in [a, b)" how do you know this?

[marble socket]

we defined open-ness as union of basis elements

if [a, b) = union of {L_alpha | alpha in some indexing set} where each L_alpha is a basis element

[pearl holly]

Yeah I think that I know why L must contain a, but how do you know that L must be contained in [a, b)?

[marble socket]

now a must be in some L_alpha on right right

well union of those L_alpha is all of [a, b)

how can they not be a subset of [a, b)

[pearl holly]

ohhhhhhhhhhhhhhh

[marble socket]

you probably saw this type of open-ness definition while doing metric spaces

a set is open if each of its point is contained in a small enough open ball which is entirely in the set

[pearl holly]

Yeah I read that somewhere before

[marble socket]

here the open balls serves as a basis

[pearl holly]

Okay okay I get it now..... I really feel like a moron lmao. Thank you for helping me, seriously!

[marble socket]

it's Daijobou

[fading vale]

kinda confused here

i thought the mapping cylinder as a pushout was like

[gentle osprey]

Moth In Shambles

[fading vale]

is this somehow equivalent?

or i guess maybe its like

a(i_0(X)) = j(id(X)) so in the first your shooting the first X into X x {0} and identifying that with the top of the cylinder

and a(i_1(X)) = J(f(X)) so you identify the second X with f(X) in Y

yea i think that makes sense

gamer moment part 2: they mean qj = id and qJ = f right??

because J sends f(X) in X + Y into Z(f) while j sends X into Z(f)

[empty grove]

Yes

[empty grove]

No, it seems correct. j embeds X into the cylinder by x \mapsto (x,0), which then maps to f(x) under q, and J is y \mapsto y on which q acts as identity

[fading vale]

yeah i got confused because it typoes as f decomposing into q and J instead of q and j

but other than that it makes sense

[empty grove]

Ahh

Right

[sweet wing]

yea it's equiv you'll see this diagram in ch5 on cofibrations xd

[pearl holly]

Is it true that the upper limit topology (has all (a, b] as basis) is not comparable with the K-topology on R?

I have used this lemma but I am not sure if I got anything wrong

[empty grove]

Yes they aren't comparable

[pearl holly]

Okay thank you!

[empty grove]

Wait no

They are I think

[pearl holly]

Oh

[empty grove]

Upper limit should be finer

Lower limit is non comparable

[pearl holly]

Yeah I proved that with some help of det

[empty grove]

0 will be the only potentially problematic point

See if you can write basic nbhds of 0 as unions of basis elements of the K topology

There are only 2 kinds of basic nbhds of 0

[flint cove]

what is the K-Topology?

[empty grove]

K = {1/n}

For n natural

[pearl holly]

I did it like this using the lemma. Let x be a real number. All bases in the K-topology that contain x are in the form (a, b) where a<x<b or (a, b)-K. But then there doesn't exist a basis B in the upper limit topology such that x is in B and B is a subet of (a, b) or B is a subset of (a, b) -K

[empty grove]

And K topology has the basis {(a,b)} u {(a,b) \ K}

[flint cove]

ah, thanks

[pearl holly]

This would work if (a, b) - K would "dissapear"

[empty grove]

There is a B in both cases

In the (a,b) case, (a, (b-x)/2] works

Or rather even (a,x] lol

[flint cove]

So closed sets in the K-Top are the usual closed sets as well as K (and intersections ofc). Interesting. 🤔

[pearl holly]

yeah I thought about that, but what about the (a, b) -K part?

[empty grove]

Yep, it's just introducing a new closed set essentially

[empty grove]

If x is less than or equal to 0, same thing works. If x is greater than 0, it will either lie between two 1/n's or be greater than 1 so you can probably see how you can handle that

[pearl holly]

Oh yeah, I could somehow just chose a small enough interval that does contain x but not the 1/n parts

[empty grove]

Yep

[pearl holly]

Okay thank you so much! I will try to formalise this in some way

[summer jolt]

Hey, I'm trying to understand rel {0,1} homotopies and came across this:

Given paths $\alpha, \beta :I \rightarrow G$ and $e$ the constant closed path given by $e: I \rightarrow G; t \mapsto 1$, construct the rel {0,1} homotopy
$$m(\alpha) : \alpha \bullet e \simeq \alpha : [0,1] \rightarrow G$$
where $\bullet$ denotes the concatenation path.

Now, the answer turns out to be:
$$m(\alpha)(s,t) =
\begin{cases}
\alpha(\frac{2s}{1+t},t)&\text{if}, 0\leq s \leq \frac{1+t}{2}, 0\leq t\leq1\
1&\text{if}, \frac{1+t}{2}\leq s \leq 1, 0\leq t\leq1\
\end{cases}$$

There is a few points about this construction that are confusing to me:

1. How was this constructed?
2. $\alpha(t)$ seems two require two inputs rather than one, which I don't get because the domain of $\alpha(t)$ is the unit interval so it should only require one input variable.

[gentle osprey]

snypehype

[summer jolt]

This is my guess on how alpha was constructed

but I still don't get why it needs two inputs

[empty grove]

Yeah that seems like a typo

It should be $\alpha((2-t)s)$ I think

[gentle osprey]

Moldilocks

[empty grove]

Or at least that's the one that you've drawn

And bounds might need to be changed if they're going for something else entirely

But there shouldn't be 2 inputs definitely

Actually bounds stay the same

[empty grove]

You just put this expression and you get exactly what you've drawn

[summer jolt]

but shouldn't $m(s,t) = \alpha(t)$ for s =1 at all times?

[gentle osprey]

snypehype

[empty grove]

For s = 1 it should be constant e

For t = 1 it should be alpha(s)

So we're just working with exchanged coordinates

[empty grove]

But I think this is the convention, the second coordinate t is thought of as the time coordinate, when you think of the homotopy as a deformation over time

[summer jolt]

Ok so $m(1,t) = \alpha(1) = e$ right?

[gentle osprey]

snypehype

[empty grove]

Yeah

[summer jolt]

and we have $m(0,t) = \alpha(0) = e$

[gentle osprey]

snypehype

[empty grove]

Yes

Wait alpha(0) is not given to be e

Alpha could start at any point we just know it ends at identity

[summer jolt]

Hmm, this was the definition I had

Shouldn't this imply $\alpha(0) = e$?

[gentle osprey]

snypehype

[empty grove]

Oh right yeah

Wait no

[empty grove]

That is just saying in this diagram, the whole left edge and the right edge are constant

But they could be different constants

alpha • e and alpha both have the same starting point, but it doesn't have to be the identity

[summer jolt]

Ok but with your formula $\alpha((2-t)s$ we get $m(1,t) = \alpha(2-t)$ right?

[gentle osprey]

snypehype

[empty grove]

Only when t is 1

[empty grove]

See the bounds

Because 2-t is greater than 1 whenever t is less than 1, so it won't be in domain alpha for t less than 1

[gentle osprey]

Moldilocks

[summer jolt]

ok so what is $m(1,t)$? It's not clear to me that it is a constant

[gentle osprey]

snypehype

[empty grove]

If t<1, then it fall into case 2, so m(1,t) = 1

If t = 1 then m(1,1) = alpha((2-1)1) = alpha(1) = 1

[summer jolt]

Ah I see

[strong heron]

Is this channel free now?

[empty grove]

You can post

[strong heron]

I have to calculate connected 3-fold covering spaces of RP^2 \vee RP^2 up to covering space isomorphism.

I did the analysis with the action of the fundamental group on {1,2,3} and I got one diagram

But I'm kinda confused with the "covering space iso" part

[empty grove]

Isn't RP^2 simply connected?

Wait nvm I'm probably wrong

[strong heron]

Hmm, no. Identifying antipodal points on the 2-sphere. It's fundamental group is Z/2Z.

[empty grove]

Ah right

[fading vale]

just making sure im not confused: this is what the homotopy pushout looks like right

you have f: A -> B, g: A -> C, and then glue the bottom of A x I along f(A) and the top along g(A)

[fading vale]

u know honestly i think im just gonna accept the church of blackboxing ugly homotopies

[maiden oracle]

why is it so hard to find a topological space that fulfills T3 and T4 but not T2

[cedar pebble]

@fading vale yes, while the usual pushout is a gluing construction, the homotopy pushout is a gluing up to homotopy; see example 2.2 in Dugger's primer:

(https://pages.uoregon.edu/ddugger/hocolim.pdf)

[honest terrace]

check pibase @maiden oracle

[maiden oracle]

pibase ?

[fading vale]

yea i saw this i think

like we define it this way because you can have A htpic to A', B to B', C to C' but the pushout of A->B, A->C might not be homotopic to the pushout of A'->B', A'->C'

so the homotopy pushout fixes that basically

[honest terrace]

well

so

T_3 implies T_2 @maiden oracle

[cedar pebble]

yes exactly

[maiden oracle]

no

[maiden oracle]

only if T1

[honest terrace]

https://topology.jdabbs.com/

pi base disagrees here

[fading vale]

does literally anyone care about T3 but not hausdorff

[cedar pebble]

it depends on how you define it

pretty much any reasonable source defines T_3 as T_1 + other conditions

[fading vale]

having Tn not imply Tm for m < n is stupid

[honest terrace]

[maiden oracle]

well i in the lecture notes i have T3 is defined as "for every point x and every closed set with x not in A there are disjoint open subsets that contain A and x respectively

and it is explicitly stated that T4 => T3 => T2 when T1 is fulfilled or equivalently if singleton sets are closed

[maiden oracle]

that's a pretty cool resource tho, thanks for sharing

[honest terrace]

[maiden oracle]

bruh, im a little salty for not coming up with the first one but hey i think i tried long enough

i tried a lot longer already before i wrote into this discord

[gritty widget]

sometimes it's the super trivial examples that we all miss

general topology bad

find a T_475749959 space that isn't T_4040000010010

[maiden oracle]

i think this is finally the assignment where i include the famous quote from poincare

xd

[gritty widget]

.pin

[gritty widget]

is it just me or are the other pins gone

[empty grove]

Ye I only see this

[maiden oracle]

i dont think so, the last pin i see before that was from may 1st

or well not i dont think so but not for me

[gritty widget]

[gritty widget]

hmm

mobile discord bad

[cedar pebble]

condensed math will soon take over

[sleek thicket]

Condensed math

oat math

Bagged math

[shut moat]

@sleek thicket WELCOME BACK

[gritty widget]

[sleek thicket]

👋