point-set-topology

[pulsar thunder]

[marsh forge]

never let reason be the enemy of dunking

[pulsar thunder]

looks like im gonna be dunked a lot

[marsh forge]

rule 2 is never dunk on someone trying to learn

[hollow harbor]

i don't like rule 2

can't you dunk and teach simultaneously

[marsh forge]

dunking while teaching is allowed iff you are punching up

[pulsar thunder]

so im safe

[pulsar thunder]

[marsh forge]

like if @frosty sundial was trying to learn something i could tease him but thats only bc hes clearly ahead of me in knowledge and career

[frosty sundial]

max teach me about spectra

[marsh forge]

add preimages and equivalences until -\smash S is invertible

[frosty sundial]

idk if i told you this before but a couple years ago I asked emerton about this like

so there are some mysterious connections between the sphere spectrum and number theory

for example, bernoulli numbers are a big deal in number theory and are apparently related to the sphere spectrum

also the stable homotopy groups of the sphere spectrum are conjecturally the K-groups of the field of one element

and I asked matt what the deal was

and he said something like "in 20 years, algebraic topology will be a prereq for number theory and you won't be able to do NT without knowing stuff about spectra"

[marsh forge]

max nt arc

[obtuse meteor]

homotopy theory + algebraic geometry eating all fields of mathematics

[marsh forge]

tbh i think spectra are a very natural thing

[frosty sundial]

hahaha

[marsh forge]

and i think Brown Peterson is like, a clear proof that thats so

sorry

brown rep

[obtuse meteor]

frick I had a like nice introduction to spectra + related stuff

that I was reading

[marsh forge]

i always confuse the naming

[obtuse meteor]

and

it disappeared

[marsh forge]

i think every intro to spectra ive read was bad for a specific reason

Like, when spectra were being invented

the people more or less knew what they wanted and why

and all the various models for spectra were not randomly made, but there were guiding principles

[obtuse meteor]

makes sense (tm)

[marsh forge]

but every fucking intro to spectra steamrolls through the motivations

[obtuse meteor]

this is common in mathematical writing / pedagogy

[marsh forge]

ironically the nLab intro might be the best one

just because it at least explains why we want this category

so i think if i just spew a formal defn of a model cat of spectra at you

its like, understandable, but theres a lot of "so what"

and it takes awhile to get to punchlines

i wonder if lurie's intro is any better

you need to know infinity categories though

[obtuse meteor]

simplicial sets time

[cerulean oriole]

owo moth pedagogy is like maths pedagogy?

[fading vale]

@marsh forge whats the motivation for spectra

go go go

[marsh forge]

Two things

First you want to force the loops-suspension adjunction to be an equivalence

Then one reason you might want to do this is that the suspension iso on homology and space-wise brown representability suggested that the individual homology groups induced by the representing spaces ought to compile together in a category with naturally graded hom groups, i.e. you want to see each representing space as a suspension of the previous one

Sorry thats poorly phrased and is two seprate things

the first is that homology is suspension stable, so forcing suspension to be invertible should allow us to still read off homology

and second is that the suspension loops adjunction relates that functors $[-,K(G,n)]$ and $[-,K(G,n+1)]$

[gentle osprey]

MaxJ (Cali Surfer Arc)

[marsh forge]

so that maybe entire cohomology theories should really be suspension-loops stable

things like this

also

you want a category where homotopy groups are the stable homotopy groups

[fading vale]

that kind of makes sense

probably just need to read more

[obtuse meteor]

is there a general construction of this sort where you have a pair of adjoint functors you want to force to be an equivalence?

seems like not just a loop space / suspension kind of thing people might want to do sometime

[marsh forge]

I am not aware of such a construction

Do you mean some kind of triangulation axioms or?

[tight agate]

Is there a proof of the BBD recollement theorem in english?

[pearl holly]

So there's an example in my book that shows that the unit sphere in $\mathbb{R}^n$ for $n > 1$ is path connected. He first defines $g: \mathbb{R}^n - {\mathbf{0}} \rightarrow S^{n-1}$ by $g(\mathbf{X}) = \mathbf{X}/||\mathbf{X}||$. Then he says that it is continuous and surjective and "and it is easy to show that the continuous image of a path-connected space is path connected". What does he mean by the continuous image of a path-connected space? Don't you need to find a continuous function from a closed interval to the sphere?

[gentle osprey]

Tokidoki

[gritty widget]

"the continuous image of a path connected space" means the continuous image of a path connected space

[hollow harbor]

The author is showing that "if R^n - {0} is path connected, then S^(n-1) is path connected"

the author isn't showing path connectedness of S^(n-1) directly by contracting paths to points

[pearl holly]

hmm okay. But what is the continuous image of a space?

[hollow harbor]

Just the image of the space under some continuous function

Like

this statement says "if X is path connected, and f : X to Y is a continuous function, then f(X) is path connected as a subspace of Y"

where Y is any topological space you like

[pearl holly]

ohhh okay now I see

[hollow harbor]

you could similarly say "continuous images of compact spaces are compact"

[pearl holly]

okay I don't know what compact spaces are but I will learn that soon. Thank you!

[gritty widget]

the compact image of a continuous space is compact

[empty grove]

[gritty widget]

[frigid patrol]

https://www.youtube.com/watch?v=fyaWMEqfomw

[kind sluice]

Hey there, im stuck with a problem, wonder if i can get some help

Let $X$ be the left-cosets of $B$ in $GL(n,R)$, where $B$ is the set of all inferior triangular matrices. We give $X$ the quotient topology of $R^{n^{2}}$. I want to show that $X$ is compact

[gentle osprey]

Maikel

[kind sluice]

So, i was hinted all equivalence classes contained a matrix of determinant 1, and that i could use that.

I proved that, and also i noticed that the projection from GLn to X could be restricted to the special linear subgroup SL(n,R), but thats as far as i have come

[shy moss]

hi

why simplicial n-homology group trivial implies no n-dimensional holes?

[marsh forge]

That is almost better seen as a definition

if you are wondering why its a definition, consider the 2d case first

if I have a triangle of 1-simplicies

but that triangle isn't a boundary

that means its not filled in

i.e. its a triangle-shaped hole

this basic idea generalizes

if I have a bunch of filled in triangles that form a pyramid but aren't filled in, that means I have a 3D hole

and so on

(the reason I say this is more of a definition than a theorem is that I think for n>2 or 3 it's hard to give a proper definition of a hole that doesn't amount to saying what I just said)

@shy moss

[shy moss]

thanks

[kind sluice]

I'll bring my question back so it can be seen ._.

[honest terrace]

hint: show that SLn(R) is compact @kind sluice

[kind sluice]

Oh

But i thought it wasnt

Thats why i wasnt trying to do that

[honest terrace]

wait I may be brainlagging

[kind sluice]

Its ok 😂

I think its not compact, but maybe i could restrict the projection even more and it will still be surjective?

[honest terrace]

yeah you're right, it's not compact sorry

it's not bounded

[kind sluice]

Indeed

Ive been hours in this...

[silver umbra]

what are some examples of functions which are continuous and invertible

but its inverse is not continuous?

[river granite]

think of the usual parametrization of the unit circle

[gentle osprey]

derivada.schwarziana

[ivory dragon]

this is a topologically important example btw

like super duper important

[silver umbra]

so its inverse would be f(x,y) = arcsin(x) ?

[ivory dragon]

not quite

(for multiple reasons)

(note the domain)

[silver umbra]

ah

its [0,2pi)

im reading on MSE that phi cant be a homeomorphism because the circle is compact

and the half open interval isn't

which makes sense

what does the inverse actually look like explicitly tho

[river granite]

some piecewise function involving arctan

it's not too important, one way to see it isn't continuous is by looking at points near (1,0)

[ivory dragon]

^^

[river granite]

or another way using stronger results is this

[ivory dragon]

you can derive it but its neither nice nor insightful

the point is to observe it cant be continuous

do you see why?

[silver umbra]

what goes on at points near (1,0)

?

oh

it jumps

from the beginning of the interval

[ivory dragon]

right

[silver umbra]

to the end

ahhh

[ivory dragon]

you can imagine phi as "wrapping" the interval into a circle

and if you try and "unwrap" it

and follow along the circle as you do so

you have to jump from one end of the interval to the other

near (1, 0)

so its not continuous

that is to say, your "unwrapping" necessarily introduces a "break" in the circle

if this phrasing sounds very informal, it is.

you can do the rigorous details yourself if you want

but hopefully that gets across the idea

[silver umbra]

yeah the intuition definitely makes sense ty

to show that its not actually continuous

do i need to consider the topology of the unit circle

[ivory dragon]

well, your argument that it cant be a homeomorphism because compactness is a homeomorphism invariant is sufficient

since we know it has continuous inverse, so the only other possible way it cant be a homeomorphism is if the function itself is not continuous

(which it aint)

but if you prefer a more "grounded" approach

yes, youd have to look at the topology of S^1

in particular, neighbourhoods of (1, 0)

[everywhere else is fine, that single jump is the problem]

[silver umbra]

right

ty for the explanation

def cleared things up for me

[gritty widget]

gauss bonnet

$$\int_S K,dA = 2\pi \chi(S)$$

[gentle osprey]

R2T2

[gritty widget]

lovely

[sleek thicket]

😳

[gritty widget]

it is very nice

it says quite a bit about the topology of surfaces

all of those big results i saw back in my RG class are just generalizations of this thing's topological implications

[hollow harbor]

😌

[gritty widget]

i'm going historically backwards

but this does make everything seem a bit more coherent

[gritty widget]

Is this true or false ?

How we prove ?

what do you think

what have you tried?

I only know the definition of homeomorphism

what would it mean for the distance between h(A) and h(B) to be zero? how would you show that?

the distance between two sets is the infimum of the distances of all pairs of points from the sets

d(X, Y) = inf{d(p, q): p in X, q in Y}

[gritty widget]

Yes yes l know that

[gritty widget]

Yeah

[chrome dew]

is the intersection of A and B empty?

[gritty widget]

Might be. Might not be

General condition

[chrome dew]

give an example of both scenarios for me please

[gritty widget]

Wait

I had an example

A= {N} and B={n + 1/n such that n in n}

A and B closed.

A Intersection B is empty.

Distance [A,B]=0

[gritty widget]

These two sets are closed

But intersection non empty

And the example of two close set . Intersection non empty is obvious.

Then it will be easy to prove this statement....

That is what I thought in exam.

But found the example later

what about the other scenario

[gritty widget]

Two closed sets intersection non empty ?

wait i might have misread

A and B be closed subsets of R^2

Intersection of A and B be non empty.

Implies that
There exist "x" such that

x is a element of A intersection B

Implies that
h(x) belongs to h(A) intersection h(B)

Implies that
distance [ h(A),h(B) ] = 0

[gritty widget]

Proved for non empty intersection

[gritty widget]

Now we need to prove for empty intersection.

[empty grove]

Empty intersection cant occur, if 2 closed sets have distance 0 then they must intersect

wait nvm thats for compact

[gritty widget]

[empty grove]

🤡

[gritty widget]

x axis and graph of 1/x are currently coping and malding

[empty grove]

[strong heron]

Does the following work?

d(A,B)=0 implies there exist sequences such that d(a_n,b_n) --> 0. By homeomorphism, we get d(h(a_n),h(b_n)) --> 0 which gives d(h(A),h(B))=0.

[gritty widget]

please expand on the "by homeomorphism" step

[dim meadow]

The sequences a_n and b_n here are not convergent sequences

So the argument you are using does not work

@strong heron

[dim meadow]

let your first closed set be the y axis and let your second set be the elements in the first quadrant determined by the equation x=1/y. Let your function be f(x,y)=(x+yx, y) in the first quadrant and f(x,y)=(x,y) else

The y axis is fixed by this function, while the elements of the form (1/y,y) get mapped to (1/y+1,y)

That the function is invertible is pretty trivial (it is increasing for each fixed y), and that the function has continuous inverse follows from the inverse function theorem except for on the boundary of the first quadrant, where you have to piece together the agreement of the inverses of the piecewise functions this guy is made of, but this isn't too hard

(unless I'm making some silly mistake because it's 4 am, but if so someone please correct me)

[gritty widget]

One thing is for sure

[gritty widget]

If this function given is a "homeomorphism". Then the given statement is false

[gritty widget]

.

But I am not able to understand how to prove this function as homeomorphism

i.e.

f(x,y)=(x+yx, y) in the first quadrant and f(x,y)=(x,y) else

[dim meadow]

The inverse function is g(x,y)=(x/(y+1),y) in the first quadrant, g(x,y)=(x,y) else

The proof I wrote before was very haphazard but it's pretty easy to see that this is an inverse function which is continuous

The reason I'm using a piecewise function is to avoid the bad behavior of the inverse at y=-1

[gritty widget]

Ok then.
I want to say something.
I am not very good at these problems.

Is there some easy or understandable way for me to prove that the function and it's inverses are continuous functions.....

......

I can in no way see how to prove them continuous

[dim meadow]

You show that the two functions which you are combining agree on their intersections

[gritty widget]

On the axises .....?

[dim meadow]

Yep

[gritty widget]

The inverses function.

How is it continuous at the points

(x,-1)

[dim meadow]

That's why I defined it piecewise

Read the definition of the function

[gritty widget]

The inverse function.

At the point

(0,-1)

The two "pieces" meet ?

i.e
The limiting value and the functional values are the same ?

[dim meadow]

The boundary of the first quadrant is the set of pairs (x,y) where x and y are >=0 and one of them is 0

That is where the two pieces meet

It's not hard to show that if one of x or y is 0, then (x(y+1),y)=(x,y) and (x/(y+1),y)=(x,y)

Does this make sense

I am only making these functions not the identity in the 1st quadrant

Else they are the identity

So when y is -1 we are fine

[gritty widget]

Ok ok ok ok

(0,-1) is not a point of the first quadrant !!!!!!!!!!

[dim meadow]

Yes

[gritty widget]

Okay. I will think for some time.....

[gritty widget]

The conclusion:

We have constructed a homeomorphism
f: R^2 --> R^2

f(x,y)=(x+yx, y) in the first quadrant and f(x,y)=(x,y) else

A={ y axis} B= { 1/y: in the first quadrant} are two closed sets.

Here,
distance [A,B]=0
distance [ f(A),f(B)] >0

Thus the given statement is false.

@dim meadow thank you for your help

[strong heron]

I also thought about one counterexample. So you let h(x,y)=(xe^{-y},y) when x>=0 and h(x,y)=(x,y) when x<0. This is a homeomorphism which "straightens" the y=logx graph. Now you can take A={(0,y):y is real} and B={(x,logx):x>0}. Then h(A)=A and h(B)={(1,logx):x>0}={(1,y):y is real}. So, d(A,B)=0 while d(h(A),h(B))=1.

[gritty widget]

Ok . I understand this (counter)example. Thank you for your help ✌️

[gritty widget]

I have one more problems in this plz someone help.

The two functions we constructed

I know if they meet at the "joints". Then they are continuous

But what about inside the quadrants....

Do we use some theorem to prove that...

.

f(x,y)=(x+yx, y) in the first quadrant

let h(x,y)=(xe^{-y},y) when x>=0

How do we prove these two functions continuous......

How do we apply the definition of continuity in them.....

Or is there some theorem we need to apply......:pensive:

Yo, sup. Someone have any pdf about Algebraic Geometry?

rising sea

by vakil

Thanks @gritty widget

[strong heron]

Aren't these component-wise continuous? I don't understand your question.

[gritty widget]

If the function is "components waise continuous" then the "function" is continuous ?

[strong heron]

Oh, yes. When you apply the epsilon-delta definition of continuity, if each component is continuous, you can choose appropriate delta for any given epsilon and prove that the whole function is continuous.

Good exercise to try writing that, if you're unsure.

[gritty widget]

Another way might be.

Limiting value = functional value

Iff

The function is continuous

[gritty widget]

I think to use this will be easier here

@strong heron
Anyway.
Thank you for your help.... ✌️

[vast estuary]

I'm trying to prove the above inequality. Let me explain the notation.

[gentle osprey]

Hausdorff

Hausdorff

[vast estuary]

What do I do next? Please help

[vast estuary]

(Sorry I missed the \epsilon here multiplied with B^n_2 - it's a typo)

[tight folio]

When does bijecting two topologies uniquely define a homeomorphism?

[dim meadow]

Well first off you need the property that your bijection respects inclusions

And then if the points are closed in your spaces you are good

Cause the complements of points get mapped to the complements of points

[verbal wraith]

A topology consists of two collections X and T, so a morphism of topologies: (X, T) -> (Y, S) should consist of two functions (f_X, f_T). But we have cl o f_X = f_T o cl, and T includes into P(X). This means that f_T can be obtained from f_X as f_T = cl o Image_f_X. So what we say is that we have a continuous map f_X and an induced map f_T. f_X is a homeomorphism if (f_X, f_T) is an isomorphism i.e. if both f_X and f_T are bijective.

[marsh forge]

When you have a bijection of the topologies that can induce a homeomorphism (respects the structure) then the map on the underlying set should be fully determined so long as no two points share all open nbhds

This is basically saying that you need the topology to see all of the points distinctly

[tight folio]

Ok that’s what I thought

That’s a weaker condition than being hausdorff?

[dim meadow]

Yeah having closed points (which I'm not sure is what max is talking about) is weaker than hausdorff

[tight folio]

Well closed singletons would definitely imply my thing

Because then you just see what complements of singletons map to

[dim meadow]

Yes

I think you can show the converse as well (if there is some point which is not closed so that its closure is a set then you can define multiple homeomorphisms from a single (suitable) bijection on the topology)

[tight folio]

I feel like it should be true but nothing obvious jumps out at me

[dim meadow]

A nicer way of saying this is there is an automorphism which fixes the topology

[tight folio]

Oh, intersect every open set containing a point

Either the result is a singleton or it’s not

There’s your result

[marsh forge]

i didnt specify hausdorff bc i was worried hausdorff is too strong

[tight folio]

Nvm, that’s not enough?

[marsh forge]

you can probably get away with like

sober spaces

or something

[tight folio]

I feel like ‘no two points have the exact same inclusion index’ deserves a name

[tight folio]

Huh, cool

Feels like that should be the first listed definition, it’s so simple

[gritty widget]

let $M$ be a two-dimensional riemannian manifold, and fix $p \in M$. if $A(r)$ denotes the area of the geodesic ball of radius $r$ at $p$, then the gaussian curvature at $p$ is given by $$K(p) = \frac{12}{\pi}\lim_{r \to 0^+}\frac{\pi r^2 - A(r)}{r^4}$$

[gentle osprey]

R2T2

[gritty widget]

neat fact

[hollow harbor]

ok, after staring at it for a bit i'm not COMPLETELY confused. but i'm still wondering

1. why can't A(r) have a cubic term?
2. WHY 12?!?!

[honest terrace]

2. Don't ask this kind of questions, just accept it

[tight agate]

because 12 is the 69th fibonacci number

12/pi is the quantum golden ratio

lmao

[hollow harbor]

I guess I just don't know how to compute the area of a pringle off the top of my head

[gentle osprey]

R2T2

[hollow harbor]

I see

It's integral(4K(p)rho^3)/r^4

And that's just K(rho)

The question has successfully been transferred to "where does the 1/3 come from in that taylor expansion?"

[gritty widget]

that one is a long computation

very long

$$g_{ij}(x) = \delta_{ij} - \frac{1}{3} R_{iklj}(p)x^kx^l + O(|x|^3)$$ is the general case

[gentle osprey]

R2T2

[hollow harbor]

Something something symmetries of the riemann curvature tensor

I buy it

[gritty widget]

it's so fucking squished

[river granite]

https://tenor.com/view/community-ken-jeong-for-ants-gif-11383549

[honest terrace]

Someone: "I'm not bad at maths, I'm not bad at mental arithmetic"
every mathematician ever: "noo, mathematics isn't just computing stuff "
TTerra: "yes "

[hollow harbor]

Please don't tell me you just wrote that tterra

Good lord

Sometimes I'm like "you know geometry is really cool" and then I see proofs like this and it bring me back to grad diff geo psets

And it's all over

[gritty widget]

i already had it written

[hollow harbor]

Sensible

What if someone could tell me in one english sentence "the 1/3 is here because blah blah blah this symmetry blah blah"

That would be awesome

It's probably just not possible and that's sad

[gritty widget]

doing so would require actual brainpower

not something i'm capable of

[hollow harbor]

I don't think it's possible looking more closely at the computation

It's just product rule combinatorics

Nothing pretty

[crisp helm]

Can you used algebra to get an expression for the 14, then try to interpret it that way

It's not that dumb

[gritty widget]

🤔

[crisp helm]

O

[cerulean oriole]

If A is a subset of X × Y and the intersection of A with any "horizontal line" or "vertical line" is open (in the "line") then is A open in X × Y?

[honest terrace]

what is an horizontal or vertical line, here, @cerulean oriole ?

X x {y} for y € Y and {x} x Y for x € X ?

[cerulean oriole]

yes

[gritty widget]

||X = [0, 1] with discrete topology, Y = {1, 2} with indiscrete, A = X x {1}||

i hope

[cerulean oriole]

Pretty sure if X is discrete it's true
Because X × Y is the disjoint union of the vertical lines {x} × Y.

[cerulean oriole]

Isn't A open in X × Y since X is discrete?

Oh wait

[gritty widget]

have i forgotten point-set again

[river granite]

Y is slightly larger than {1} though, and has the indiscrete topology

so if X x Y has the product topology, A is indeed not open in X x Y

[cerulean oriole]

Intersection with vertical lines is not open, because (x,1),(x,2) are indistinguishable

[gritty widget]

oops

you said the intersection should be open in the line

mine's open in A

sadness

[cerulean oriole]

I got a counterexample from someone else

[gritty widget]

i am interested

[cerulean oriole]

A = open unit disc \ {(t,t) | 0 < t}

[gritty widget]

[cerulean oriole]

[gritty widget]

point-set topology moment

[true robin]

I just saw the question, but can we do R^2 minus a point?

[cerulean oriole]

It's open anyway.

[true robin]

Oh, i see

[gritty widget]

.

[true robin]

R^2 minus positive reals should work? I hope it does atleast

[obtuse meteor]

.pin

[gritty widget]

the line R x {0} intersects A in (-\infty, 0] x {0} which isn't open in R x {0}

[cerulean oriole]

Cut the ray diagonally and it works

[cerulean oriole]

.

[true robin]

Oh you already had a solution, sorry

[true robin]

Damn, feeling horrible now

[gritty widget]

[honest terrace]

(wow my first pinned message )

[obtuse meteor]

that message was A+ <3

[vast estuary]

did anyone get this

[tight folio]

@cerulean oriole in a similar vein: there is a non-measurable subset of the plane which intersects any line at exactly two points

So intersections with straight lines definitely don’t tell the whole story abt the regularity of a set

[cerulean oriole]

Wow

[pearl holly]

So my book defines path connectives with the “normal definition” but the path function should have the domain [a,b]. In Wikipedia, they use the same definition but with the domain [0,1]. But these definitions are equivalent because [a,b] is homeomorphic to [0,1], right? I think that I have proved this formally but I’m on mobile and too lazy to type it out

[vocal wharf]

if a < b yes

[pearl holly]

Okay great!

[vocal wharf]

take like x \mapsto (b-a)x + a

for [0, 1] to [a, b]

and then you can compose those to switch between definitions

[pearl holly]

Just to clarify, this is the definition in my book but with [a,b] instead of [0,1]

[pearl holly]

Okay that’s great! That’s exactly what I’ve done on paper, I’m just lazy to type it out lol

But thank you so much!

[strong heron]

It is common in mathematics to use the unit interval when an interval is to be considered. It is somewhat like a 'canonical' interval choice. People sometimes also denote it with a capital I.

[pearl holly]

And this implies that a product of two path connected spaces are connected because of the fact that a function is continuous if it’s continuous in each coordinate, right?

[pearl holly]

Okay thanks for the info!

[reef shore]

how are you using that exactly?

[reef shore]

ye

[pearl holly]

By this maybe?

[reef shore]

i dont see how its immediate

[pearl holly]

Hmm well I know that if X is path connected and if Y is connected then there are function from [0,1] to X and Y respectively. But then the function from [0,1] to X x Y is also continuous

I hate being on mobile

[reef shore]

right so you are connecting some 2 pooints

[pearl holly]

Yes exactly and since the function to the product “behaves” like the path function in each coordinate then we can connect two points

I think

[reef shore]

yeah

another way to do this is to just move along the axes

and travel in an L shape

[vocal wharf]

how did you define product topology?

[pearl holly]

Ahh I see

[reef shore]

open rectangles as bases

[pearl holly]

By the product topology, not the box topology

But those are the same when we have finite products

[pearl holly]

Yes that’s it

[reef shore]

I think loch was asking if it was defined by a basis, or a universal property or whatever

[pearl holly]

Oh okay sorry

[vocal wharf]

ye

because if you have the universal property, this is indeed immediate

otherwise do the L shape thing

[pearl holly]

Yeah but Munkres proved the universal property for the product topology so it’s fine I think

Who are you malti? Moldis brother?

Or like sister?

[reef shore]

Im moldi but with more power

because I can use

[pearl holly]

WHAT

How

[reef shore]

hax

[pearl holly]

Blocked

Banned

[frigid river]

@mods

[pearl holly]

Malti fake

There’s no moldi 2

[reef shore]

[pearl holly]

There’s only one real moldi

[empty grove]

tru

[pearl holly]

And if A is path connected then the closure of A doesn’t need to be path connected, right? A counter example is the topologist sine curve or whatever it’s called

[reef shore]

ye

[pearl holly]

Okay great ez, thank you!

[reef shore]

[frigid river]

If I understood correctly, a homotopy is just a continuous function $H: X \times [0;1] \to Y$ such that $H(X,0)=f$ and $H(X,1)=g$, where f and g are maps from X to Y.
If that's correct, I don't see how that's the same as the definition given in Hatcher's.

[gentle osprey]

verydisturbedperson

[marsh forge]

Is that not the definition in hatcher

oh does he only every use f_t?

[empty grove]

Hatcher does a family of maps thing I think

[marsh forge]

Ah

This comes from something called currying

[frigid river]

That's the stuff in Hatcher's, and I don't get that :/

[marsh forge]

Where a map X x [0,1] ->Y is the same thing as a family of maps f_t:X -> Y and the condition in Hatcher that the latter vary continuously also is equivalent to the condition that the former be continuous

yeah the condition (2) there is what I am talking about

[empty grove]

That's the specific case of path homotopy

X = I

And endpoints remain fixed throughout

[marsh forge]

Oh

also

this is the specific case of a based homotopy

[empty grove]

Other than this extra condition it's the same thing

[marsh forge]

Where the definition you posted is a free homotopy

or just a homotopy

[empty grove]

Based homotopy

[marsh forge]

You can fix the definition you sent to be based homotopy as well

(tldr the reason you can't see the equivalence is that the two notions aren't equivalent only conceptually the same)

(and small modifications can make them actually equivalent)

[frigid river]

And what does he mean by the endpoints being independent? They are fixed anyway, aren't they?

[marsh forge]

it means that it is true for all t

and the same

so its just saying for all t the endpoints are the same (indepedent of the choice of t)

[frigid river]

And the second line is just saying that all the functions of that family must be continuous, right?

[marsh forge]

No not just that

It's saying that the functions vary continuously as well

[frigid river]

I can't even pretend to know what that means 0.0

[marsh forge]

for example consider the following family of maps $f_t:\bR\to \bR$ given by $f_t(x)=1$ for all $0\leq t<1$ and $f_1(x)=0$. Then all the $f_t$ are continuous but they don't vary continuously

[gentle osprey]

MaxJ (Cali Surfer Arc)

[marsh forge]

the picture in your head should be that like

you have some slider for t that you control

and as you slowly change t

the functions f_t should change slowly

in my example the functions "jump" from 1 to 0

[frigid river]

Oh, I think I get that.

Okay, thank you. I will see if I get the examples now.

But them varying continuously isn't a condition for the regular homotopy, is it?

Nevermind, I think it is -.-

So what's so based about path homotopy? Is it just that the end points are fixed?

[marsh forge]

oh

based is an adjective meaning like cool and good

but also

based means something mathematically

[frigid river]

0.0

[marsh forge]

in the sense of like "based at [location]"

so he was making a pun

so a based homotopy is just a homotopy with fixed endpoints

[frigid river]

Okay, but is based homotopy based?

[marsh forge]

Based homotopies are generally more important to topolkogists

topologists*

In fact based spaces are generally more important to topologists

[frigid river]

that's based

[marsh forge]

(it turns out if your space is good enough the base-point restriction isn't actually that big of a deal because a free homotopy can often be altered into a basepoint-preserving one

[pliant parcel]

Is this channel the right one for fairly open ended diff geometry questions? (Sorry am new)

[viral atlas]

Sure, go ahead.

[pliant parcel]

https://projecteuclid.org/journals/michigan-mathematical-journal/volume-20/issue-4/The-volume-of-a-small-geodesic-ball-of-a-Riemannian/10.1307/mmj/1029001150.full

From this article (Volume of small geo. balls on a riemannian manifold) they derive a series expansion formula for the volume of spheres on a manifold. It might take a bit of a read, but my question is how one might find the variational derivative (in terms of the metric) of equation 9 or 10 in section 3. The expression is:

$S(r) = \int_{exp(S^{n-1}(r))} i(\omega)$

where S is the volume, i is the inclusion of the n-1 volume and $\omega$ is the volume element.

[gentle osprey]

Anteater

[frigid river]

I hope Anteater isn't mad for me posting this, but which two (continuous) functions actually aren't homotopic?

[marsh forge]

you were first anyway

the easiest example is on the circle to itself

you can have the identity map

or you can wrap the circle around itself twice

these are not homotopic

any two functions R->R are homotopic

continuous*

[frigid river]

And a path always "has a domain" of [0,1], right? Is there a reason for this, or is that just a definition? 0.0

[empty grove]

That's the definition of a path, and it's defined that way because it matches our intuition of paths

[frigid river]

so only functions with a domain of [0,1] could be path homotopic?

[marsh forge]

uh

[empty grove]

Yes path homotopy is specifically a homotopy between paths

[marsh forge]

path homotopy means homotopy of paths

this is like asking whether a bear can be an African elephant

[frigid river]

but my intuition of a path tells me that not all paths are of "equal domain"

[empty grove]

Continuous functions don't preserve length

[marsh forge]

(there is some nuance here i feel obligated to say. in other fields people refer to anything of the form [a,b]->Y as a path rather than just a=0 and b=1 but this makes no difference because up to homotopy we can reparameterize without a problem)

[empty grove]

And [a,b] is just homeomorphic to [0, 1]

[marsh forge]

(for a<b)

[frigid river]

so it doesn't matter?

[pliant parcel]

I'm sorry I was busy for a second. Oh this is a different question, continue on 😆

[empty grove]

Ye if you do all your theory from that perspective you still get the same results

[frigid river]

Okay, thanks.

[marsh forge]

if all you want to do is consider which paths in a space are homotopic you dont lose any information just using [0,1]

i.e. if you only care about the paths and how they act in the codomain

but there are places where it matters outside of topology

[pastel thistle]

Need Help. I have I=[-1,1] and the following relations given by

~_1 : {1/2, 1}, the rest related to itself

~_2 : {1/2, 1}, {-1/2, -1}, the rest related to itself

I need to who that [-1,1]/~_1 and [-1, 1] /~_2 are not homemorphic any ideas ???

[swift fjord]

maybe a connectedness argument?

[pastel thistle]

I mean the first relation is just every number in [-1,1] related to itself and 1 related to 1/2

the second relation is the same but we add -1 related to -1/2

it should be something really basic because Im starting to study quotients spaces

I understand that one quotient is homemorphic to something like a straight line with a curl
the otherone is homeo to a straight like with two curls

[swift fjord]

Could be something like suppose they were homeomorphic by a homeomorphism f, and remove point x from one space, then this decomposes it into some number of connected components, but this would be homeomorphic to the other space minus the point f(x), but the other space has at most some other number of connected components when removing a point

you might need to remove 2 points for this argument

but I think that should work

And you know the number of connected components is preserved under homeomorphism

[pastel thistle]

okeeey

I get it

[swift fjord]

Again, this is just at a glace but it should work since one space is a line with 1 curl and 1 space is a line with 2 curls

[pastel thistle]

I understand that the line with 1 curl has two connected components maybe ?

[swift fjord]

I'm not sure if the spaces as they are are connected

[strong heron]

I don't think a connectedness argument will go through

Do you know fundamental groups?

[pastel thistle]

I know the first fundamental group but I should try to use something more 'fundamental' than that lets say

[swift fjord]

smth like this?

sorry for my awful handwriting

[pastel thistle]

could not take a point in the straight like

line

[swift fjord]

if you remove 2 points, no matter where, in Y you will disconnect the space

[pastel thistle]

if you take a point in the straight line you disconnect the space with just one point

[strong heron]

This is nice. I was unable to see it immediately.

[swift fjord]

Right, but that's true for both spaces. You want to find a property that happens only in 1 space but not the other

[gritty widget]

i like the fundamental groups proof too

[pastel thistle]

maybe Im getting it wrong Im kinda new with this disconeccted arguments

[strong heron]

Actually one can do better and just remove one point from the central line in X

You cannot disconnect Y by removing just a one point.

Ah, you can. Sorry

[swift fjord]

Yea I tried doing it with 1 point but it doesn't work

not that I can see

[strong heron]

Yeah, I was thinking of wedge of two circles and one circle. But those are homotopy equivalent spaces to these ones. My bad.

Your proof is good, ShiN

[swift fjord]

The point is, if X is homeomorphic to Y by some f, then X-x will be homeomorphic to Y-f(x), and you can keep subtracting points like this (In general, if $X \cong Y$ and $A \subseteq X$ then $A \cong f(A)$). With connectedness arguments you usually want to find some property of X that is preserved after removing some number of points (Say, connectedness or the number of connected components), such that no matter which points you omit from Y (Meaning, it's independent on the map you choose), that property won't hold

[gentle osprey]

ShiN

[swift fjord]

so they must not be homeomorphic

[swift fjord]

If only topology proofs were as simple as drawing them

Not a fan of working with quotient spaces

Like this would be simple to write but not the cleanest

[strong heron]

Most topologists would be perfectly happy with this picture proof.

[marsh forge]

yeah picture proof is fine here

[pastel thistle]

allright !

I see the argument know

[swift fjord]

Great!

[gritty widget]

picture proofs are always fine, imagine actually writing down details formally

[swift fjord]

lmao

Wish my profs would agree

[pastel thistle]

oh was going to ask for that

[gritty widget]

on my point-set topology final i drew a picture with a few words for a problem and got full marks

[pastel thistle]

like picture proofs are enough formal right ?

[swift fjord]

fr, that would never fly in my class

[pastel thistle]

okey that gives me confidence

[gritty widget]

if your prof is an analyst, no. if your prof is a topologist, yes

[swift fjord]

You might just want to see you understand how the proof is actually done

[pastel thistle]

hahah nah I heard my professor say that picture proofs are okey a couple of times

[swift fjord]

lucky fucks

For one of my HW assignments I had to calculate a stereographic projection from the punctured sphere to the plane

Like, it's not that bad, but it's just too analytical for me

[strong heron]

Stereographic projection is pretty standard. Those formulae are actually important.

[swift fjord]

I mean yea but the point of the exercise was to understand why these spaces are homeomorphic

the rest of the details are a bit superfluous

[gritty widget]

computation is good

[swift fjord]

Gross

[gritty widget]

(see pins)

[marsh forge]

terrible

[marsh forge]

wrong

[swift fjord]

I just wanna draw commutative diagrams all day

is that too much to ask

[marsh forge]

yes

[gritty widget]

[strong heron]

Oh, okay. To just see they are homemomorphic, you don't need explicit formulae.

[marsh forge]

this is an anti ugct channel

[gritty widget]

shamrock in shambles

[marsh forge]

shamrocks not a ugct

[swift fjord]

what's ugct

[gritty widget]

undergrad category theorist

[marsh forge]

he has a very fair relationship w categories that is only cringe bc he seems weirdly confused and torn about it online lol

[swift fjord]

nono

[gritty widget]

let me find the meme

[swift fjord]

I meant in the sense of algebra

not pure category theory

[gritty widget]

[swift fjord]

Like, i'm taking algtop next semester and i'm super excited

and the semester after that i'm taking a commutative algebra class

[marsh forge]

commutative diagrams are like a scalpel

very useful in the right context

but its a bad thing if they are all you know how to use

[swift fjord]

I mostly just wanna get really good with tikzcd to flex on the grad students that will be taking the course with me

[marsh forge]

lmao i promise the grad students are probably better

[swift fjord]

The ones i've spoken to aren't

and there's like

10 in the whole department

ok maybe more like 30 but still

[marsh forge]

no one even uses tikzcd anymore

[swift fjord]

why not?

[marsh forge]

they use https://q.uiver.app

writing tikzcd by hand is bad

[swift fjord]

Oh yea that's what the TA for one of my classes uses

ok this is pretty dope

didn't know that was a thing, def gonna use it

[marsh forge]

my opinion is that tikzcd is subsumed by quiver and tikz proper should be replaced by illustrator

it makes commutative diagrams trivial

and easy

and therefore unexciting which is good

[strong heron]

All the arrow diagrams in Hatcher were done using Illustrator, so yeah, there are much tastier alternatives to tikz.

[obtuse meteor]

I pray that I am welcome 😌

[tight agate]

copresheaf

[frigid patrol]

Copresheaf.

[swift fjord]

Brofibration

Maybe the real gluing axiom was the bros we made along the way

[river granite]

cope sheaf

[marsh forge]

we believe in rehabilitation

[obtuse meteor]

Fffffff

[honest terrace]

@mirza

wait she's not here

[gritty widget]

[cerulean oriole]

I am scared

[empty grove]

Mirppa wasn't even ug goddamn HSCTs

[little dew]

I'm kind of confused about the notation here, this is the def of the equivalence relation for constructing a compact n-dim torus, but the "[n]" is kind of confusing me.
Intuitively I feel like it should mean some set of n indices, but I've never seen this notation used in this context, given that n is a dimension. Is it some sort of equivalence class?

[reef shore]

[n] = {1,2,...,n}

This is saying that x ~ y iff x_i and y_i are 0 or 1, or both are neither but equal for all i in [n]

[little dew]

ty!

[little dew]

is this common convention?

[chrome dew]

yeah, semi common

[little dew]

tyty

[marsh forge]

the only potential thing to be careful about is that some people use $[n]={0,1,...,n-1}$ but this is obviously a small difference in 99% of situations

$|[n]|=n$ is basically always true

[gentle osprey]

MaxJ (Cali Surfer Arc)

[little dew]

good to know tyty

[lofty stratus]

I read some algebraic topology on how to construct a cw complex space such that its fundamental group is isomorphic to a given group with presentation $G=<S| R>$. First, each 0-cell corresponds to each element of the group, then the $1$-cell from $x$ to $y$ if $y=ax$ for some generator $a$.

Then we glue a $2$-disk to each loop corresponding to a relator r in $R$. I worrry about the set $R$ when it includes some "useless" relation. For example, considering $\mathbb{Z}\times\mathbb{Z}$ with generators a, b. R must include $aba^{-1}b^{-1}$, but it may also include $a^2ba^{-2}b^{-1}$.When it does include such relators, do we have problems?

[gentle osprey]

mqcase1004

[coral pivot]

i mean not really, you arent modding out any other loops right so redundant relations wouldnt be a problem i would think

[empty grove]

Think of it like the sphere, which is 1 0 cell and 1 1 cell and 2 2 cells. Putting 1 2 cell is enough to make the group 0 but the other 2 cell doesn't allow any new homotopies. More generally if you attach a redundant 2 cell, see from van kampen that it doesn't affect the fundamental group, and therefore attaching finitely many 2 cells won't affect the fundament group, and then attaching infinitely many won't because else by compactness of the closed interval you can reduce to the case where finitely many redundant cells reduce the fundamental group which isn't possible

[gritty widget]

https://realniranjankrishna.github.io/math/Munkres-Chapter2-Solutions-Part1/

It would be of great help if somebody could verify these solutions. Thanks in advance

[pastel thistle]

I have X =$\mathbb{R}^2$-{(0,0)} and xRy iff $\norm{x} = \norm{y}$
I need to show that $p: X \to X/R$ is open and closed any help ?

[gentle osprey]

Tomás Sentagne

[pastel thistle]

any idea ?

[empty grove]

For open, see that the image of an open ball is open, and then the rest is unions

[pastel thistle]

thats actually not true

you could have an open map on the basis thats not open at all

[empty grove]

Isn't f(union of S_i) = union of f(S_i)

[pastel thistle]

I think is a subset

[pearl holly]

hmm isn't it a subset if it's intersections?

[empty grove]

f(S) = {f(x) | x ∈ S}

If you put union inside

The right side splits as a union

[pastel thistle]

Im trying to find where I read that it was false

im maybe wrong

[empty grove]

hmm

[pastel thistle]

oh right I read it for sub basis

so stupid

[reef shore]

ah right there you have intersections

so for closed, I am thinking that you can show both spaces first countable, hence sequentially closed is equivalent to closed

[pastel thistle]

any idea using saturated sets or something like that

[reef shore]

hmm let me think

[pastel thistle]

y know that "p" open iff p^(-1)(p(U)) is open

[reef shore]

ye ye

right if you saturate any given closed set C of R^2, then it remains closed. Since you only saturate it, the image doesn't change, and the images of saturated closed sets are closed

so it remains to show the first thing

Suppose you have a sequence x_n in C which converges to some x in R^2. We need to show that x is in C. Since x_n converges in R^2 it is a bounded set, and so must have some subsequential limit in R^2 which must be in C because C is closed. This subsequential must be equal to x

[pastel thistle]

😫

okeeey

[reef shore]

is that not the kind of proof you wanted

There might be a way to do it without sequences

[pastel thistle]

is the second exercise in my notes about quotient spaces

I thought it should be easy

[reef shore]

oof

[honest terrace]

Just sharing a nice topology question a friend asked me:

[hollow harbor]

wait

[honest terrace]

What are the integers n such that there is some norm N on R² whose unit ball is a regular n-polygon

[hollow harbor]

why is there another moldilocks

[reef shore]

one of me wasnt enough

[hollow harbor]

maybe for the 3 free months nitro

you already had it in the past

[reef shore]

yes

[hollow harbor]

HA!!!!!

[reef shore]

caught

[honest terrace]

[hollow harbor]

you can call me "private y"

because i am the best sleuth around here

anyway this n-gon thing is so weird

[obtuse meteor]

This feels illegal

[hollow harbor]

it has to be symmetric under x -> -x

so definitely even

[honest terrace]

(I must say that my answer is just a bazooka with a more general theorem, but you definitely can do without it )

[obtuse meteor]

This being topological scares me

You said it was a topology question

But all n-gons are the same….

[hollow harbor]

ok

this makes me think about gauges from functional analysis

or minkowski functionals i guess

[honest terrace]

I said topology just because it involves norms

[reef shore]

rice why does it need to be symmetric under negation?

[obtuse meteor]

[obtuse meteor]

Norm laws

[honest terrace]

absolute homogeneity moldi

[hollow harbor]

if you take the minkowski functional for a given set, as long as it's convex, balanced, and absorbing, it's a seminorm

[reef shore]

oh right lol

I was thinking of metric

[hollow harbor]

this is some theorem i forget how to prove

[supple locust]

I have written a detailed solution for this question. Can someone take a look? I am not very sure about the last line in my solution. \
We will prove that $X $ is homeomorphic to $S^1 \times [0,1]$. Consider the function $f:\mathbb R \times [0,1]\rightarrow S^1 \times [0,1]$ given by $(x,y)\mapsto (exp(2\pi i x),y)$. Note that $f$ is surjective and continuous and so the map induced $X\rightarrow S^1\times [0,1]$ is surjective and continous we also note that it is injective. So it remains to prove that $f^{-1}$ is continuous. Open set in $X$ is such that its preimage in $\mathbb R \times [0,1]$ is open, but f is open so the induced map is open.

[gentle osprey]

bert

[hollow harbor]

but an absorbing set is a set $U$ for which $\bigcup_{r > 0} rU$ contains the whole space, and the minkowski functional for $U$ is the function on the space $\rho_U(v) = \inf{r : v \in rU}$

[gentle osprey]

RYC

[cerulean oriole]

For a given n, there's also only one possibility for the norm, right?
Upto rotoations, dilations, etc.

[hollow harbor]

so the idea is that these functionals are like "norms" whose unit ball is the set U

[cerulean oriole]

And the only thing that may or may not hold is the triangel inequality

[honest terrace]

I don't know for sure but yes I think so

[hollow harbor]

it is a fact from functional analysis that a minkowski functional is a seminorm if U is absorbing, balanced (x in U implies -x in U) and convex (this is for triangle)

[honest terrace]

as ryc pointed out, absolute homogeneity holds only for even n

[obtuse meteor]

This solution is not detailed precisely bc you should justify continuity, subjectivity, and most importantly openness of the map

[supple locust]

I thought they are pretty obvious

i.e. I think they are obvious ( but I am doubtful about openness in the end)

[obtuse meteor]

I think in a detailed solution you should give a quick explanation of each of these things

If you are afraid of writing details then you should

[reef shore]

is there an even n>2 that doesnt work? it seems that convexity of the n-gon gives the triangle inequality

[obtuse meteor]

t. Max

[cerulean oriole]

Yes so convexity ensures the triangel inequality

[obtuse meteor]

As it is you are essentially just asserting that this map is a homeomorphsim

Which is not very convincing

[hollow harbor]

so note that in finite dimensions, these minkowski functionals of absorbing balanced convex compact sets are actually norms

not just seminorms

[supple locust]

so f is continuous because it is product of continuous maps. similarly it is product of open maps so it is open

[hollow harbor]

actually

i don't see how it can ever be zero even in infinite dimensions

[reef shore]

Rice in his own world

[cerulean oriole]

You should also mention that the map "induced" on X is actually well-defined, I think.

[obtuse meteor]

You should see that the complex exponential is open

In my opinion this is something that should be verified at least once if you haven’t shown it before

[supple locust]

right. pretty obvious tho

[hollow harbor]

oh i get it

if the set we have is unbounded

then it can be a seminorm

then it's zero because every r works

ok

anyway, my point is that all 2n-gons for 2n = 4 and up can be done

[reef shore]

[hollow harbor]

the norm is literally "inf{r : v is in a 2n-gon of radius r}"

[reef shore]

whatever rice said, am I correct shika?

[hollow harbor]

proving subadditivity is easy because it's a convex set

proving homogeneity is only possible because the number of sides is even

[honest terrace]

RYC is correct

[cerulean oriole]

Yes I just thought of U = whole space :kek:

[reef shore]

[hollow harbor]

and by compactness, it's positive definite. there's always some r which makes the 2n-gon smaller than v.

[honest terrace]

also

I think you meant sup

not inf

[hollow harbor]

i did not...

[honest terrace]

wut

[cerulean oriole]

If in an overly generalised way

[honest terrace]

oh wait lol

I'm tired sry

just woke up

yes not sup, sorry

[hollow harbor]

v is in any very very large n-gon

[honest terrace]

yeah lol sry

[cerulean oriole]

Problem solved here lol

[honest terrace]

yes but RYC gave the details

[hollow harbor]

it would be something like sup if i did rv in the unit 2n-gon

but then i would need to take 1/ at the end

i think?

idk

[supple locust]

its open because image of an open interval is $exp(2\pi i t), \ t \in (a,b)$. So for any point in this image there is $x\in (a,b)$ such that the point is $exp(2\pi i x)$. So the intersection of $B( exp(2\pi i x), r)$ with $S^1$ where $r= \min (|a-x|,|b-x|)/2$ is contained in the image.

[hollow harbor]

very nice problem! i've never thought about how norms can have pretty much any unit ball like that before

[honest terrace]

There's a more general theorem actually

[hollow harbor]

as long as it's convex, bounded, and balanced i guess?

[gentle osprey]

bert

[honest terrace]

what's balanced ?

[cerulean oriole]

defined above

[honest terrace]

I didn't read everything ryc said lol

there was a lot of minkowski nonsense

[supple locust]

too many conversations simultaneously

[cerulean oriole]

[honest terrace]

mmh I don't know then, the characterization I know is any unit closed ball is a convex, neighborhood of 0, symmetric around 0 and compact set (and for any each set, we can find a norm whose unit ball is this set, defining it in a similar way than what you did here)

[hollow harbor]

Yes

Wait

Compact though?

[honest terrace]

I'm not sure about what the balanced part implies or does not imply here, maybe its equivalent to what I'm saying

[hollow harbor]

Only in finite dimensions

[honest terrace]

yeah in finite dim

in R^n

idk for greater dims

[hollow harbor]

Surely you can just say bounded and the whole thing generalizes to arbitrary vector spaces I think

Maybe not

[honest terrace]

mmh maybe

[hollow harbor]

I guess it needs to be a topological vector space to say what "neighborhood of 0" means

Ah

That's funny

Actually I really don't like that

Hmm

Oh, something just clicked that I learned like 2 years ago

Wow

You need local convexity in order to guarantee the neighborhood of zero wont be wacko

[honest terrace]

in the R^n case, since every norm is equivalent there's no problem not giving more details on what "neighborhood of 0" means, but ig it becomes more annoying in infinite dim ?

idk

[hollow harbor]

And this is why a locally convex topological vector space is exactly a space with a family of seminorms!

Wow

Because you just take minkowski functionals on every convex compact balanced neighborhood of 0

:catGlad:

[honest terrace]

those "ooh wait now I get it" moments are so nice

[cerulean oriole]

It has to be compact?

[cerulean oriole]

balanced = symmetric about 0

[cerulean oriole]

How are you defining bounded when you're yet to give the nor?

norm*

[honest terrace]

The closed ball must be closed, and obviously must be bounded

oh are you asking if the condition is redundant ?

[cerulean oriole]

It can only bounded after you define a norm with it

At which point it is already bounded

[honest terrace]

you don't need to already have a norm defined

[cerulean oriole]

[honest terrace]

every norm is equivalent in R^n

[cerulean oriole]

...

Oh

R^n

That is still
Weird

Can you define bounded without a norm in R^n?

[honest terrace]

with a metric yeah

[cerulean oriole]

😑

[honest terrace]

Joke aside, you probably can

but it'll probably feel like cheating

[cerulean oriole]

Oh I think

Image of U under any linear functional is obounded

[honest terrace]

bounded for what norm ?

[cerulean oriole]

Can confirm
My first idea was "pick a basis and make it bounded in some norm, say infty-norm"

[honest terrace]

[cerulean oriole]

Real numbers norm

[honest terrace]

well I mean, yes then, but definitely feels like cheating too

like a linear functional is something of the form $(x_1, \cdots, x_n) \mapsto a_1x_1 + \cdots + a_nx_n$

[gentle osprey]

Shika

[cerulean oriole]

After you pick a basis

[honest terrace]

there's a canonical basis on R^n so who cares

[cerulean oriole]

At least this is visibly

basis-independent
norm-independent
doesn't need a norm in the first place

[cerulean oriole]

Oh
When you said R^n I mentally made that finite-dimensional vector space lol

If it's R^n just assume Euclidean norm

Also canonical

lol

[honest terrace]

yeah I mean, what I said obviously also holds for any finite dim R-vector space, but working in R^n is just a bit easier because everything's canonical

[cerulean oriole]

We could try this in the infinite-dimensional case

As well

[hollow harbor]

Great question, something like "convex but contains no vector subspaces"

Besides zero of course

[cerulean oriole]

Wait
Aren't all topological vector spaces given by a family of seminorms?

[hollow harbor]

No

[honest terrace]

but yeah, image through any linear functional is bounded doesn't sound bad

[hollow harbor]

Seminorms only induce locally convex topologies

[cerulean oriole]

IIRC any uniform space is given by families of seminorms

[hollow harbor]

[cerulean oriole]

And TVSes are uniform spaces

[honest terrace]

yeah I recall seeing this too

[marsh forge]

topological vector space

[hollow harbor]

Uniform spaces are defined by families of pseudometrics

Hmmm

[honest terrace]

oh wait

yeah

[cerulean oriole]

Oh right

OK but

[honest terrace]

Also Raghuram

[hollow harbor]

"a homogeneous, translation invariant pseudometric admits a seminorm"

Ok

So something is wrong there

[cerulean oriole]

For commutative groups
Pretty sure you can make the pseudometrics translation invariant

[cerulean oriole]

What is homogenous?

[hollow harbor]

Must be that non locally convex tvs's are defined by inhomogeneous pseudometrics, that kind of makes sense

[honest terrace]

If we could define bounded without referencing any norm, but while conserving the fact that for any norm, bounded for the norm <=> bounded for the independent def, we'd get that every, wouldn't we get that any norm is equivalent on any vector space ?

[cerulean oriole]

Ooooh

[hollow harbor]

It would be like d(rx, ry) = rd(x, y) where r > 0

[honest terrace]

or maybe not

[cerulean oriole]

The part about scalar mltiplication

That is not there for uniform spaces

Right

[hollow harbor]

Yeah

Somehow this is the difference between having convexity and not

[cerulean oriole]

Yes

[hollow harbor]

Which is weird

[cerulean oriole]

[hollow harbor]

Since usually that's triangle

[honest terrace]

Actually not necessarily

nvm

what I said was wrong

[cerulean oriole]

How?

Above ==> bounded equivalent for every norm
==> for all norms p1, p2; unit p1-ball bounded in p2 ==> p1 finer than P2
=> all norms equivalent

[honest terrace]

ok yeah I was right, I just brainlagged nvm

I mean you're right, and what I was initially saying was right, not what I just said when I said I was wrong

[cerulean oriole]

[honest terrace]

wow that sentence is so bad

[cerulean oriole]

Very weird
If it's just one pseudometric, then it's locally convex, right?
Since balls are convex

Was the result you mentioned

all locally convex spaces are given by a seminorm
?

[feral copper]

Hello ! In a lemma in an article, I don't understand what the authors mean by “... framings induced by F”, would someone be kind enough to explain please ? (might be dumb)

That is, point #3 in the lemma

Are they talking about the framing for the link ? (like the extension of the embedding) In that case, what could that mean to have a framing induced by F ?

[marsh forge]

i assume they mean a framing of the knot

@feral copper https://math.stackexchange.com/questions/2138942/going-between-heegaard-diagrams-and-framed-link-diagrams/2141225

hopefully this helps? i dont know this stuff

okay @frigid river go ahead

[frigid river]

Quick dumb question: f(x)=x(x-1) is considered a loop, right?

(from [0;1] to R)

[gritty widget]

sure

[frigid river]

okay, thanks.

It's just not how I imagined a loop 0.0

[gritty widget]

well do you have a definition for "loop?"

[feral copper]

Yeah that's what I had guessed, but I don't know what they mean by a framing induced by the boundary... Oh and here, the 2-handles are 4-dim too

[gritty widget]

4-manifolds

[frigid river]

yes, but I couldn't really connect that to the adjacent pictures

[feral copper]

A loop in R is something that would go over itself, you don't have room for interesting stuff as it's 1-dim (don't confuse the loop in the space and the graph of the function)

Imagine a loop an arc in R as the path an ant would take on a spaghetti

[marsh forge]

path is probably the term verydisturbed knows

a loop is just a path that starts and ends at the same point

you can do that by walking around your house

or by walking in a straight line and then walking right back to where you started

or by just standing still

[gritty widget]

what if walking around my house doesn't bring me back in front of it

[feral copper]

Then you live in RP² ?

No, actually no, on its cover xD

And you end up in a copy of your house

With a copy of your wife/husband XD

[gritty widget]

[frigid river]

This image still bother me though. How are you going to describe the loop, a, as a function?

[feral copper]

Take your favourite parametrization of the torus

Then your curve has an equation f(t)=(x(t),y(t),z(t))

[frigid river]

[feral copper]

(because in our parametrization we embed everything in R³)

[frigid river]

thanks!

[feral copper]

So here you need to choose a curve as g(t)=(theta(t),phi(t)) and compose with the parametrization

[marsh forge]

maybe a more topologist-y answer

a function here is just like

a choice of point on the torus for each time t

and if you look at that picture

its certainly doing that