point-set-topology

[coarse kestrel]

just feels weird to me

[west spindle]

you get that 5 and -5 are zero units apart...

[coarse kestrel]

Yep

[worldly spindle]

Lol

[limpid mural]

Let $a, b, c \in \mathbb{R}^{+}$ such that $ a \geq b \geq c$ and let $S$ be an ellipsoid given by the equation $ (\frac{x}{a})^{2} + (\frac{y}{b})^{2} + (\frac{z}{c})^{2} = 1$. Let $ \gamma : \mathbb{R} \to \mathbb{R}^3$ a smooth curve such that $Im(\gamma) \subset S$ and $\gamma '(t) \neq 0 \forall t \in \mathbb{R}$. Find the minimum value of the curvature of $\gamma$ at a point, on varying $\gamma$.

[gentle osprey]

emme:

[limpid mural]

So we fix a point and we find the curve that has the minimum value of the curvature at the point fixed, among all the smooth curves on the ellipsoid.

So I think that the curves that minimize the curvature are the geodetics, am I right?

[honest narwhal]

Okay gonna think out loud about topological groups a bit

Join me on this adventure through wonderland

So if H is a subgroup of a topological group, so is its closure

[gentle osprey]

Dami:

Dami:

[gritty widget]

@honest narwhal yap that's true

[honest narwhal]

Okay perfect

Ugh while I'm here it's impossible to just focus on things

Okay so now if H is normal

[gentle osprey]

Dami:

[coarse kestrel]

If we define the metric $d(q,p)=1-\frac{1}{1+|q-p|}$ on $\bbR$ then is $\bbR$ bounded?

[gentle osprey]

Whoever:

[umbral surge]

is every point contained in B_1(0)

i woulda just done arctan(normal dist) tho

[dim meadow]

@coarse kestrel boundedness is not a Topological property

It is a metric property

You can just take the metric min(d, 1)

And that makes any metric space into a bounded metric space

[coarse kestrel]

I mean, I didn’t define a topology

I gave a metric

[dim meadow]

The metric defines a Topology

[coarse kestrel]

Ok

.

Every time I see this phrase I die a little bit

[dim meadow]

What phrase?

[coarse kestrel]

Metric defines a topology

And I have no idea what that means

[dim meadow]

Take all the epsilon balls

That gives you a basis

[coarse kestrel]

Although I’ve hears this phrase from multiple ppl

[dim meadow]

Look at the unions of metric balls

Aren't you studying Topology?

[coarse kestrel]

Not really

[dim meadow]

Or am I thinking of someone else

Oh ok

[marsh forge]

A topology is a way of studying when points are “close” to each other

A metric is another notion of such a thing

[coarse kestrel]

Oh

[marsh forge]

Quite naturally, a metric can define a topology as a result

[dim meadow]

You know what an open set is right?

A Topology on a space is the set of all the open sets

We want that to have certain properties

For continuity reasons

Sorry @coarse kestrel

[coarse kestrel]

Yes

Well

I don’t know about continuity

The preimage of open set is open?

[marsh forge]

Yes

[honest narwhal]

Since no one has talked for 7 minutes I'm hijacking this

Okay so I didn't commentate here but I went through the proof that quotients of topological groups by closed subgroups are topological groups

(Where we require topological groups to be Hausdorff since we're decent human beings with a moral compass)

Now time to learn a bit about some "classical" Lie groups except for now just the topology

So M_n(R) is just R^{n^2} and it's a topological group because limits

a_n + b_n -> a + b and -a_n -> -a

That should do it I think since continuity is controlled in metric spaces by limits of sequences

determinant is continuous since it's a polynomial

So GL_n is open

Matrix multiplication and inversion is continuous because polynomials and Cramer's rule and all

SL_n is a closed subgroup since det^{-1}(1)

Works for R, C, and apparently GL_n(H) is okay but that takes work since no determinants... should I care?

So the map A -> AA^T - I is continuous

Since it's also a polynomial

(The fact that shit's a polynomial will eventually imply that we're Lie groups because regular values but let's not be impatient guys)

O(n) is closed since it's preimage of a point

And it's bounded in operator norm

So it's actually compact :0

U(n) too

Though I guess my "eventually it'll be a Lie group" may not carry over to say it's a complex manifold since you need to conjugate

Actually U(n) is a compact subset of C^{n^2} so it's def not a complex manifold because max modulus

[marsh forge]

I’m gonna start hijacking category theory

[honest narwhal]

And apparently there's a quaternionic spin group

Fuck do the quaternions actually matter?

[marsh forge]

Give me a dm once you’re giving lectures dami

I’ll come out to see one as a meme

I feel like they will be hilarious

[honest narwhal]

It depends on how much I prepare and how ambitious I am

I have had some ambitious lectures which were epsilon suboptimally prepared

And those were a fucking time

Actually did I?

I had one lecture where I tried to cram 2 hours of stuff into 1

And that didn't go well

I've been disorganized in various rants, also I at one point gave an impromptu math club lecture which was disorganized but it wasn't too fast at the beginning so... 🤷

[gritty widget]

👀

[real notch]

however bad those lectures may have gone, i've done worse I feel

[honest narwhal]

Okay so

We have a map f:O(n)->S^{n-1} by f(A) = A(0,...,0,1)^T

We can think of O(n-1) in O(n) by block matrices

Just stick a 1x1 block with the entry equal to 1

Obv works

And then O(n-1) fixes (0,...,0,1)^T

In particular f(AB) = f(A) if A \in O(n-1)

So this descends to a map O(n)/O(n-1) -> S^{n-1}

And now we wanna say this is a bijection

I guess visually you can sorta see that O(n) acts transitively on S^{n-1}

I'll make a note to think about doing this rigorously

But okay so if you fix (0,...,0,1)^T, then yeah you have to be in O(n-1) because just look at the matrix entries, bottom right thing has to be 1 and since we're in O(n) yeah

So this is a bijection, O(n) is compact so so is O(n)/O(n-1), and then S^{n-1} is Hausdorff so this is a homeomorphism

And next up is continuous group actions so before proceeding I should think about the transitivity of the action

OHHH

Okay so

We want to say that if v is a vector of norm 1 then we can find an orthogonal matrix whose last column is v

But that's just finding an ONB of R^n including v

Which we can do

Gram-Schmidt or just say restrict to orthogonal complement

Either way lel

Okay time for actions of topological groups

So G(x) is the orbit and G_x is the stabilizer, for reference

Also called isotropy

Also we call the action effective if it has no kernel

(Also continuity is G\times X -> X, not G->homeo(X). At least not a priori)

So if X is Hausdorff and G is a compact topological group, then the map G/G_x -> G(x) is a homeomorphism

Orbit-stabilizer gives that it's a bijection and then G(x) is Hausdorff since it's a subspace of X, G/G_x is a quotient space of G so it's compact

Oh wait this makes my earlier shit easier

Obviously the stabilizer of (0,...,0,1)^T is O(n-1) because look at the matrix

Okay and yeah same for U(n), the inclusion works since you're sticking 1 so no conjugation

U(n)/U(n-1) is S^{2n-1}

And yeah same for quaternions

[coarse kestrel]

O my goodness

[honest narwhal]

Okay so time for some Stiefel manifold thing

So k-frame means k orthonormal vectors

And V_{n,k} is the set of k-frames in R^n

So the set of nxk matrices A such that A^T A is the kxk identity matrix

So that's its topology

It's pretty Hausdorff

And O(n) acts on it, I guess by matrix multiplication on the left

Since that's what makes sense

I guess I should confirm that it spits out the right thing

So let's say M \in V_{n,k} and A\in O(n)

nxk matrices is R^{nk}

And this is a subspace

So (AM)^T AM = M^T A^T A M = I

So this is an action because multiplication and it's continuous because polynomials

We'll see what I'm looking at since I'm not sure yet

I'm just going along with Bredon

So O(n) acts and it should act transitively because... Hmm...

Okay so given two sets of k orthonormal vectors I guess you can complete each to orthonormal bases of R^n

And mapping one over to the other should be orthogonal?

Let me just be sure in my mind one sec

Oh lol yeah I was doing it out but that's easier thanks

Okay so yeah action is transitive and the stabilizer should be a copy of O(n-k)

Yeah I guess if you play with the entries in your mind this works

I guess you don't need matrices, just say it's a map which acts on the orthogonal complement however it wants

But it has to fix these k guys

So up to a base change it's a copy of O(n-k)

The story ends there pretty much, now O(n)/O(n-k) is V_{n,k}

Since compact and Hausdorff etc

And yeah there are analogies in the unitary and symplectic cases

Also I totally thought for a hot minute that the analog of orthogonal for quaternions was spin since it's written Sp(n)

But no it's symplectic

Also turns out what I said works for special orthogonal/unitary groups, that's pretty easy to see

But H doesn't have determinants so no special symplectic group

Also with Stiefel

You can flip signs so transitivity should still be okay

And then including 1 will keep special things special

Okay time for some problems

Oh wow, component of the identity in a topological group is a closed normal subgroup

Okay closed duh

[coarse kestrel]

Um i don’t want to interrupt the conversation (so ill delete this message afterward) but I have a question about constructing a compact set with countable limit point in R, where should I ask?

[honest narwhal]

Like, countably infinite?

[coarse kestrel]

Countably many limit points in R

[honest narwhal]

Well the thing is some say countable = countable or finite

But I guess in this context that would make the problem stupid

[coarse kestrel]

Yeah, countably infinite

[honest narwhal]

Hmm, I guess you want a bunch of convergent sequences

[coarse kestrel]

$E_n:={\frac{1}{k}\cdot \frac{1}{(n+1)n}+\frac 1n|k\in\bbZ^+}$

$\bigcup_{n=1}^\infty E_n$

No

[honest narwhal]

That's not compact

You forgot { before infinity

[gentle osprey]

Whoever:

[coarse kestrel]

Yeah

[honest narwhal]

Okay now you're good

[coarse kestrel]

Yep, now is this infinite union a possible candidate?

[honest narwhal]

Yeah I can see that working, like if every 1/n is a limit point kinda thing

[coarse kestrel]

Yeah

And it’s closed, because it contains every limit point

And it’s bounded

So I think it’s compact in R

Wait

[honest narwhal]

I'm not sure about the details here

[gentle osprey]

Whoever:

[honest narwhal]

Oh that makes more sense now

[coarse kestrel]

Lol ok

[honest narwhal]

Or does it actually?

[coarse kestrel]

Idk that’s why I asked

[honest narwhal]

Is 1/n in E_n?

I guess you can stick it on by proxy

[coarse kestrel]

Yeah, but 1/n is included in every previous E_n

Lol

I just realized

[honest narwhal]

Proving that sounds like work

Just add things by proxy if you need really

[coarse kestrel]

Lol is there a better set?

Ok

[honest narwhal]

I mean I'm adding 1/n to E_n and then adding 0 so that shouldn't be an issue

[coarse kestrel]

Ok

I feel like this should work

But I also feel like there might be other problems

[honest narwhal]

Just adding one convergent sequence and its unique limit point shouldn't add any other limit points

The basic issue is proving that the only limit points are 1/n

[coarse kestrel]

Oh

[honest narwhal]

Yeah that's what I was thinking originally

[coarse kestrel]

That’s so much better

How is it compact tho

I can see that it has countable infinite many limit points

[honest narwhal]

Sniped me

[dim meadow]

So something shitty you can do is make a set with countably many limit points that's closed in (-infty, 0), and then take e^ that set

[honest narwhal]

That's clever

[dim meadow]

And then add in 0 and 1 or something

[coarse kestrel]

very clever actually

[honest narwhal]

I really like that actually

[night pivot]

e^(-n+1/m) haha

[honest narwhal]

I'm declaring it the official answer to this question

[coarse kestrel]

Lol

That was what I was thinking with e^ stuff

[honest narwhal]

Anyone caught using other answers will be shot

[coarse kestrel]

e^(-n+1/m)

[honest narwhal]

Survivors will be shot again

[night pivot]

with a camera

[coarse kestrel]

Because e^ (-infty,0) is way too clever

[honest narwhal]

@gritty widget because that requires no thought, I had to think for a sec why that had only countably many limit points

[coarse kestrel]

Um

[honest narwhal]

But even my mom had the epiphany immediately once liquid told me the answer

[coarse kestrel]

Well ik what he said

I don’t think i need to type out all the details

Lol

[honest narwhal]

@gritty widget no the point is you pick countably many convergent sequences in (-infinity, 0)

Then you push it across

Oh oh

[coarse kestrel]

He meant that e^x where x in (-infty,0] is not countably infinity

[dim meadow]

Thanks

[coarse kestrel]

@honest narwhal sorry for interrupting your long lecture tho I understood nothing you said

[honest narwhal]

It wasn't really a lecture

It was me teaching myself topological groups out loud

[coarse kestrel]

Oh ok

[honest narwhal]

But yeah back to that

So why is the connected component of the identity a normal subgroup?

Don't spill it yet if you know already but if you wanna think with me do join

It's gonna be something cheesy I can feel it in my soul

So let's say g is in the connected component of the identity, g^{-1} is because continuous and connected etc. If g and h are

Then oh yeah product of connected sets I guess

And yeah

Okay

[dim meadow]

Yeah if you multiply the connected component by the inverse of g, g will get mapped to the identity, so the connected component will get mapped inwards

And congugation fixes the identity

So I guess the same thing will happen there

[honest narwhal]

Next problem, surjective hom G->H, kernel is closed normal subgroup, and if G is compact then G/K is homeomorphic to H

Which is just tilt your head for a microsecond

Closure of {g^n} is a subgroup if you're compact, what about if you're not?

Hmm

Oh I guess like

Oh wait hmm

[dim meadow]

Obvious counterexample is R I guess

To being a subgroup

Is that not the case

Hmm

Addition let's say

[honest narwhal]

Addition that's just discrete

So it's already its own closure

[dim meadow]

Oh yeah you're right

Idk what I was thinking

Goodnight

Oh btw I decided not to take the Algebra quals @honest narwhal

[honest narwhal]

Goodnight

[dim meadow]

Gonna focus on Topology

[honest narwhal]

And yeah prob not a bad idea

Once I get through this I'll prob focus on algebra and then study for topology in the day between the two tests

I'm not really really under pressure here

[dim meadow]

I'm semi under pressure because I need permission to take an algebraic Geometry course

[honest narwhal]

Which is probably bad I should feel more pressure but it feels far since I've got some shots

Oh rip I can just take whatever I want

[dim meadow]

And I need to show them I know some Algebra and Topology for that

And that I won't get stuck on the quals

Sometimes even later

And when the current chair took over he purged a bunch of students

So I don't have as much freedom as I would want

Have you looked at Pontryagin's Topological groups book btw?

[honest narwhal]

Wait purged meaning kicked them out?

[dim meadow]

Yeah

[honest narwhal]

And no I'm reading Bredon

Wait so was it like, a technical time limit to passing quals but they let it go?

[dim meadow]

I think so

They were very lax about it for a while

[honest narwhal]

Or was he like okay effective immediately anyone who doesn't pass by the end of second year is out

Oh okay that makes more sense

[dim meadow]

And it hurt the grad program a lot

[honest narwhal]

I thought they originally didn't have a real time limit and then they changed their minds and knocked people off

Which would've been a bit dickish

Oh really? Wait how?

Oh I guess grad program makes sense

I was like wait you judge departments by their profs

But I guess if the students are slow and not getting postdocs that looks bad

[dim meadow]

I think the quality of your cohort is a big thing as well

A lot of people are the type who get their PhD so they can get a cushy job in a small college teaching

[honest narwhal]

I see

I guess you'll stand out if that's any help

[dim meadow]

They've been trying to change that recently though

[honest narwhal]

Go to conferences and all

Oh that's a bit of an uphill climb

Since you gotta either attract the good ones who were previously like nah or you gotta beat ambition into your people

[dim meadow]

So they can attract good people

A big problem is also money

Their funding is shit

And it doesn't look like that's going to change anytime soon

[honest narwhal]

I mean yeah my point is that whatever the circumstances are a change has to be made if you're not currently getting who you want

[dim meadow]

That's fair, hopefully they're trying to implement changes

It seemed like they were from talking to the grad director

[honest narwhal]

Tell them to give you a ton of money to go to conferences and represent them

[dim meadow]

Gotta go to sleep now, good night

[honest narwhal]

See you

[coarse kestrel]

Let $E$ be the set of all $x\in[0,1]$ whose decimal expansion contains only the digits $4$ and $7$. Is $E$ countable? Is $E$ dense in $[0,1]$? Is $E$ compact? Is $E$ perfect?

[gentle osprey]

Whoever:

[coarse kestrel]

I can’t answer whether E is compact or perfect

Any hint is appreciated

[vocal wharf]

did you figure out if E is closed yet

then you can use Heine Borel

and whether it's perfect should be easy as well

[coarse kestrel]

Oh ok

But I haven’t figure out that it’s closed

[vale panther]

yes?

[bitter yoke]

Do you have a question

[coarse kestrel]

How is x a limit point

Lol

[vale panther]

For any neighborhood of x, you can always find an x(n)#x in the cantor set that is in that neighborhood. That is what they explained in the paragraph "To show that P is perfect, "

[coarse kestrel]

Yes that is the definition, every point is a limit point

But I don’t understand how x is a limit point

[vale panther]

Because every neighborhood of x has a point different than itself which is in P.

Or do you mean that you didn't understand why that is true?

So you didn't understand the paragraph "To show that P is perfect, "

?

[coarse kestrel]

Every does it say every neighborhood?

[vale panther]

"Let S be any segment containing x."

[coarse kestrel]

Oh...

I see

Thank you!

Makes sense now

[small obsidian]

@coarse kestrel
What's the book?

[vale panther]

Rudin

[coarse kestrel]

Yea

@small obsidian

[small obsidian]

Fair! I should have known lol

I'll take another look at it. I don't remember the definition for perfect

[coarse kestrel]

Every point in the set is a limit point

[coarse kestrel]

I guess I can use this to prove that stupid set with digits in decimal only containing 4 and 7 is perfect

[dim meadow]

You know that the Cantor set is perfect right?

You can construct a homeomorphism between this and the cantor set

[coarse kestrel]

😐

What exactly is homeomorphism

@dim meadow

How can I prove this set is perfect

[bitter yoke]

homeomorphisms are isomorphisms for topological spaces

[coarse kestrel]

Ok I got another proof

So I can just do something like

Let $x\in E$ and $\varepsilon>0$ be arbitrary. Select $n\in\bbZ^+$ such that $\varepsilon>1/10^n$. Then just change the $n+1$ digit of $x$ from 4 to 7 (or 7 to 4) and let this number be $x’$. Then $|x-x’|<1/10^n<\varepsilon$

Pretty trivial I would say

[gentle osprey]

Whoever:

[coarse kestrel]

Ugh

[bitter yoke]

yeah that works

[coarse kestrel]

I’ve proved (a), (b), and (c)

How do I prove (d)

[night pivot]

Maybe we can map it to the real numbers?

With an onto function

Or at least [0, 1]

[bitter yoke]

If your two points are x,y then show that for every k in (0,1) there must be a point z such that d(x,z) = k d(x,y)

[night pivot]

that would work.

let's suppose no such point existed for some k

Then consider part c, the ball with radius kd(x,y) around x. x would be in the set A, and y would be in set B, so they aren't connected

[coarse kestrel]

Oh ok

I see

Thanks

I’ve asked way too many questions

[night pivot]

asking a question is an effective way to learn

[coarse kestrel]

Tru

[dim meadow]

Solving problems yourself is important also

[coarse kestrel]

Well

I tried every problems the best I could

I’m not like asking every question in rudin ch.2

[coarse kestrel]

Nope I’m going to stop at 23 I think I had enough of this

[lucid turret]

OK this is kinda stupid, but does topological space ∅ have 2 distinct covers? one cover is the empty cover ∅ (our space is equal to union of all elements of cover ∅), the other cover is {∅}

[small obsidian]

If ∅ is the entire space, then you can only use subsets of ∅

[lucid turret]

um... all elements of the empty cover are subsets of ∅

[small obsidian]

Exactly. {∅} isn't one of them

[lucid turret]

by {∅} I mean open cover which consists of 1 subset, the whole ∅

See I'm getting 2 different covers, one with 0 elements and the other with 1 element

[small obsidian]

There can't be a cover with one element, because there's no such thing as an element

Not in the empty space

[lucid turret]

No, I mean there is cover which consists of exactly one subset

Every topological space X has cover which consists of just 1 subspace, the whole X

[small obsidian]

Hmm, wait I think I see what you're saying

Interesting find. Why would that be a problem if it has two covers?

You're asking if it can

It doesn't seem to contradict the definition

Empty sets are weird

[lucid turret]

*set

[small obsidian]

Someone admits to two covers here
https://www.google.com/amp/s/amp.reddit.com/r/math/comments/9zzbmc/what_is_the_strangest_property_of_the_empty_set/

[lucid turret]

yes

good

[versed pivot]

is it correct to say that the pullback of a (1,1)-form (e.g. a hermitian metric) by a holomorphic map is zero? because of the (0,1) part? or am I making a naive mistake?

[fleet rapids]

this might be a basic question for this channel sorry if it is, but I've seen Euclidian space refer to R^n, n in N, but do we also assume Euclidian geometery when refering to a Euclidian space, does it vary?

ok yeah nevermind I forgot about that, so its just R^n for some n in N right, like we just have that and maybe field structure?

(I can't recall if R^n is a field, I guess no particular structure is assumed with just the term "Euclidian Space" though?)

with scalar multiplication and vector addition if anything, idk maybe thats nonstandard

ah ok thanks

[versed pivot]

well it's a field for n=2...

but not n>2

[vale panther]

You need multiplication

To have a field

[fleet rapids]

wait C is a field though

so shouldnt R^2 be?

[vale panther]

Yes it is, but R^2 isn't the same

[versed pivot]

of course I meant with some multiplication, since none was giving in the first place

[fleet rapids]

ah bc its scalar instead of complex multiplication

ok I see

whoops

[versed pivot]

to answer the original question, I would say it depends on the context

sometimes people mean euclidean in the geometric sense

[fleet rapids]

Thankyou seoin

[versed pivot]

other times it's in a topological sense, as in the term "locally euclidean" being used to define a manifold in only the topological sense

[fleet rapids]

hmm?

whats our multiplication for Rn?

[versed pivot]

Well all R^n can be given addition and multiplication that make it a field
that's famously known to be false for n>2

[fleet rapids]

please be direct

[versed pivot]

what is the multiplication on R^3?

[fleet rapids]

mhmm

[versed pivot]

R^n with the usual addition

[fleet rapids]

ok so we just dont have multiplication with Rn, n > 1?

Like we have scalar mult in linear algebra I know that at least

[vale panther]

Bijection doesn't mean isomorphism though

[fleet rapids]

just not in field theory I guess

[versed pivot]

this kind of confusion is exactly the kind of problem @fleet rapids asked about initially. Jan is considering R^n as nothing more special than a set of continuum cardinality, I was implicitly considering it at least as an additive group

[vale panther]

oh got you

[fleet rapids]

I mean all I'm saying is that R^n is given scalar mult in linear algebra but I guess if we only care about isomorph in abstract algebra then I guess thats why it wouldnt matter

I just wanted a more direct answer to my questions is all

ah because then we're taking from 2 different sets I see

so I'm thinking of vector space not abstract algebra ones like group/ring/field

hmm fair enough

I guess the connections between abstract and linear come up a lot when you get to field extension theory right?

e.g. minimum polynomials and max number of eigenvalues, etc.

[versed pivot]

well you shouldn't forget about the scalar multiplication... ideally you would like your prospective field multiplication to be compatible with it

[fleet rapids]

ok now im confused again

please elaborate seoin

what seoin said

[versed pivot]

because Stephen was mixing up the scalar multiplication with the product

I said the scalar multiplication isn't irrelevant even in a field

you can have all these structures existing together

for example in the nice fields you're familiar with like R and C, they have both a field multiplication and a scalar multiplication

I'm not sure what Q has to do with anything, they are acted on by all real scalars

[fleet rapids]

umm I guess I was just noting that you need 2 different sets, so the scalars would form a field if we have the vector space Rn, but Rn cant itself be a field

at least not with scalar mult

[versed pivot]

@fleet rapids right, with the usual addition except in the case of R or C

[fleet rapids]

So you do need 2 distinct sets to have a vector space right? Or can they be the same, b/c if they need to be distinct then fields and vector spaces are disjoint

[umbral surge]

whats happening in here

[versed pivot]

you need a scalar field and an additive group for the scalars to act on, yes, but they can be the same sure

because any field is a 1-dimensional vector space over itself

[fleet rapids]

ah ok that makes sense

but wait what you were saying before seoin

with except in the case of R or C, is that where R or C is the vector space or the field with scalars?

[versed pivot]

as vector spaces, over R

your question was about making vector spaces into fields, so those are being considered as vector spaces here

[fleet rapids]

ah ok R and C are fields and vector spaces over themselves, but R^2 and higher Rn arent fields with the ususal addition and scalar mult

[umbral surge]

field multiplication in R^n
would be multiplying 2 vectors together, and getting a vector

[fleet rapids]

Sorry about the algebra overload in the topology chat by the way

[umbral surge]

scalar multi is totally different

[fleet rapids]

was my fault

[umbral surge]

its not a big deal

[fleet rapids]

is the field multiplication in R^n something like dot/cross product?

[dim meadow]

There isn't any field multiplication

[fleet rapids]

(not binomial multiplication right?)

[dim meadow]

Except for a couple of cases

[fleet rapids]

I'm trying to figure out what Sigma meant here

[umbral surge]

dot product returns a scalar

[fleet rapids]

right ok my bad

so how do we get field multiplication in Rn @umbral surge ?

[umbral surge]

i doubt cross in R3 would have an identity

uhh

[dim meadow]

We don't

[umbral surge]

i was testing some but couldnt get one

[fleet rapids]

hmm

[dim meadow]

This is a very famous fact

[fleet rapids]

maybe try prove it doesnt exist generically?

so were saying theres no vector multiplication that gives a field for Rn n > 2 right?

[dim meadow]

^

[fleet rapids]

ok thanks

I should prove some of these things to myslef more, just about time though ususally

[dim meadow]

I think this sort of thing is hard to prove

[fleet rapids]

so what kind of math area would that be in roughly if we can pick a category?

[dim meadow]

I think there's a K-theory proof

But yeah, Algebra

[fleet rapids]

hmm ok thanks

[versed pivot]

well I learned something new from this conversation:
R^n actually does admit a field structure with the usual addition, for any n
because they have the same dimension over Q, they are all isomorphic as additive groups
so using any bijection with R or C you really can make (R^n, +) isomorphic as a field to R or C
it's just that the scalar multiplication isn't compatible (and hence no contradiction to the result we were all thinking of)

[coarse kestrel]

i don't think R^n has a field structure

it's more like a module

[vale panther]

I think he means group structure.

Or something?

Because he wrote (R^n, +)

[coarse kestrel]

i guess

[vale panther]

So I'm pretty sure he means group with addition

[versed pivot]

I mean a field multiplication compatible with the additive group structure

[vale panther]

But the fact that you can't add multiplication means it can't be a field.

compatible with the usual additive group structure

[versed pivot]

we had been talking about this an hour ago, but when we all said it was impossible to find a field structure compatible with the additive structure we weren't careful enough

of course the real result references normed division algebras, not just fields

it turns out they can be made into fields without changing the addition

I was even the first one to make that misconception

[vale panther]

Oh. Ok then xD.

[weak plaza]

Hey guys, I asked a question in beta regarding a circle that circumscribes three other circles. Could anyone here give me a hand over there? :)

[dim meadow]

This doesn't belong here

#geometry-and-trigonometry

[soft cloak]

figured, sorry

[gritty widget]

So much quieter and less junky

[honest narwhal]

So one thing I can do is to say that two paths f,g:[0,1]->X are equivalent if there's a homotopy H such that H(0,t) = f(0) and H(1,t) = f(1) for all t

The picture you should have in mind is that I'm basically sliding one curve to another smoothly

[gritty widget]

That's actually what I had in mind

[honest narwhal]

Nice

But yeah given two spaces X and Y, we say they're homotopy equivalent if there are maps f:X->Y and g:Y->X such that f\circ g is homotopic to the identity on Y, and g\circ f is homotopic to the identity on X

And we kinda say two spaces are the same if they're homotopy equivalent

[gritty widget]

Makes sense

[honest narwhal]

For example, the punctured plane is homotopy equivalent to the circle

The map from the circle to the punctured plane is just the inclusion, and the map back is the map x/||x||

The idea being that you can kinda "slide" the identity map on the punctured plane to the x/||x|| map

[gritty widget]

Yeah I actually had to do a problem very similar to that

[honest narwhal]

So here's something cool. Let X be a space and let x be a fixed point. I'm gonna take the set of maps f:[0,1]->X such that f(0) = f(1) = x

So basically these are loops at x

Given two loops, I can concatenate them. If you think of [0,1] as time somehow, them a loop is tracing the path of a point. So I just go around one at double the speed, and then the other

If we identify homotopic loops, then this actually becomes a group operation

So we call this the fundamental group of X at x

Denoted π_1(X,x)

If X is path connected, it turns out that this doesn't depend on X, so you just get π_1(X)

This is the start of algebraic topology

Turns out homotopy equivalent spaces have the same fundamental group, so one way to distinguish spaces is to compute these and show they're different

For example, the sphere has π_1 = 0

Which means any loop in the sphere can be contracted to a point

[gritty widget]

Sphere as in the standard one in Euclidean space? Or something else?

[honest narwhal]

You can see this visually for nice loops

Yeah standard one

But if you think of the torus, you can see that a "vertical" loop, sorta ring shaped, can't be contracted

So it has a non-trivial π_1

Hence the two aren't homeomorphic

Does this make sense at least visually?

[gritty widget]

Yeah!

Sorry, I briefly lagged out

[honest narwhal]

It's all good

But yeah so, this kind of thing ends up going very far

So if you give me a loop, that turns out to be the same thing as a map from the circle into your space

You can use quotient spaces to make that formal, since the map e^{2πix} going from [0,1] to S^1 is a quotient map identifying 0 and 1

[gritty widget]

That actually clarifies a lot

[honest narwhal]

So I can define π_n to be the same construction but I instead consider maps from S^n into X, again a priori I send a fixed point in S^n, say the North Pole, to x\in X

But it turns out this is very hard to compute, and a lot of algebraic topology is using heavy algebra tools to learn about these

[gritty widget]

Ah cool

[honest narwhal]

There are also other objects you can associate to a space, each with their own associated geometric notions

Here is an interesting theorem: if X is a topological group, meaning it's a group such that the group operation X\times X -> X is continuous as well as the map sending a point to its inverse, then its fundamental group is abelian

[gritty widget]

I don't learn about abelian groups until next week 😂

(Well formally, I have seen this word so many times that I have looked it up.)

[honest narwhal]

If your professor says to give examples of abelian groups, say "the fundamental group of a topological group!"

Also turns out π_n is abelian for n>2

[gritty widget]

He was my topology teacher, so I will ;)

[honest narwhal]

But yeah there are other things you can associate which are much easier to compute (computing all the π_n for even the 2-sphere is still an open problem)

[gritty widget]

:O

[honest narwhal]

π_1 is pretty easy though

But yeah you have these things called homology and cohomology groups (for each natural number n), and it turns out these have a very similar structure to other types groups you can associate

And one thing people care about is the structural relation between these theories

Which geometrically seem like they have nothing to do with each other

The differential topology side basically uses the smooth structure that comes with a manifold

For example, if you give me two compact n-manifolds that don't have an "edge" (think of the boundary circle as the edge of the disk, but that the sphere doesn't have an edge), I can define the degree of a smooth map between them

[gritty widget]

With manifolds?

[honest narwhal]

Yeah

Oh actually you need to have a notion of orientation (so the Mobius strip doesn't work)

But yeah so, given two maps from such a manifold to the n-sphere, they're homotopic if and only if they have the same degree

This is not an easy theorem

(Basically a degree k map is one which is usually k to one, except maybe at bad points

But yeah in a nutshell when you introduce these elements into topology you get weird connections between things that you don't expect

Differential forms and triangulations and pairs of pants and vector bundles and these π_n groups all have to do with each other

And that's something I like

[gritty widget]

:)

One last question

[honest narwhal]

Go for it. We'll see if I can stretch the answer out to become unreasonably long :P

[gritty widget]

"So that stuff is mostly setting up toward other things. A lot of current topology is along different lines" (referring to point-set topology)
So what is point-set topology setting up for?

[honest narwhal]

A lot of it is really a backdrop for other areas of math, especially analysis

It's sort of convenient to be able to phrase things in terms of topological spaces and continuity

For instance, uniform convergence can be thought of as convergence in the metric space C[0,1]

There are some other things, you can define some very weird topologies which are useful but don't have a nice geometric notion

So I feel a lot of the importance of point-set topology is providing a language for other areas of math

But if you talk to a topologist nowadays they really don't care about bullshit like [0,1]\times [0,1] with dictionary order

[gritty widget]

😂

Maybe one day I'll take another topology class

It's a shame that class is the only topology class that they offer. I don't think most nearby schools offer anything more advanced

[honest narwhal]

Probably edit/delete that message because information

But yeah one book I like is Rotman "Introduction to Algebraic Topology"

[gritty widget]

Hmmmm... I'll do it just to be safe

[honest narwhal]

If I end up doing topology in grad school it'll be with this one guy who tries to adapt stuff you can do for smooth manifolds with solution spaces to polynomials that aren't necessarily smooth

For example, the solutions to x^2 + y^2 - 1 form a circle

[gritty widget]

Wait, do you start grad school soon or are you already in grad school?

[honest narwhal]

But if you look at the equation y^2 - x^2(x+1) = 0, it has a self-intersection

So, my first semester of grad school is starting in a couple weeks

Just that I don't yet know who my adviser is gonna be and all that

[midnight jewel]

So one thing I can do is to say that two paths f,g:[0,1]->X are equivalent if there's a homotopy H such that H(0,t) = f(0) and H(1,t) = f(1) for all t
well for path-homotopies we also required that the endpoints stay fixed under the homotopy

[midnight jewel]

as opposed to any old homotopy which could move all points about

[digital peak]

we just impose the coherence conditions that the left and rights loops are contractible

[honest narwhal]

Pretty sure I mentioned that either earlier (this continued a convo in chill) or later

[midnight jewel]

I didn’t see it at a cursory glance

which is why I mentioned it

[gritty widget]

Two dimensional beings: "In three dimensional universe the mobius strip wouldn't have to intersect"

[coarse kestrel]

How does a mobius strip in 2D look like 🤔

[coarse kestrel]

[umbral surge]

try to flatten a 3D mobius in between 2 sheets

thats it

[coarse kestrel]

Oh so it’s just a simple projection

Idk if that’s useful or not but ok

[marsh forge]

Uhh worth noting that this isn’t “a Möbius strip in 2D” in any topological sense

Like it’s not homeomorphic to a Möbius strip

[dim meadow]

Very sad

A Mobius strip in 2D is S^1

Hot take

[coarse kestrel]

How do you state the property of mobius strip in terms of topological properties?

like how do you use topology terms to say mobius strip only has 1 surface

[dim meadow]

It is the only nontrivial line bundle over S^1

Is how I would say it

Oh wait

That's not what you meant

Non orientable is a good word, although it doesn't describe what you are saying

In fact non orientable is kind of THE Mobius band property

I am talking about the Mobius bundle, not the band actually

[marsh forge]

Nonorientability is exactly the property you’re talking about

[rancid shard]

Is every continuous curve is a closed set in R^n with respect to the standard topology?

[marsh forge]

Not a space filling curve

Actually, I’m not sure if you could actually fill Rn w space filling let me double check

Wait when you say curve

[rancid shard]

Can we even define a curve that fills R^n though

[marsh forge]

What’s your domain

[rancid shard]

R

[marsh forge]

I want to say yes but a proof isn’t immediately clear to me

[rancid shard]

I'm not sure how to exactly phrase my statement. I'm just doing an exercise in topology where I have to prove that a line is closed in R^2, as well to the circle, ...

So my intuition says that if we have a reasonable curve ( like continuous)

then it is closed

[marsh forge]

Ok 1) I was wrong you can’t fill Rn with a “space filling curve”

2. the statement is that img f for f: R\rightarrow Rn is closed

For f cont

[rancid shard]

I see, thank you vm !

[marsh forge]

Still, I think this might only be true for curves parameterized by closed domain

But I can’t realllly think of why

I’m gonna sit on it

[rancid shard]

hmmm, makes sense

The curve doesnt have a boundary when the domain is not closed

at end point

Let f(t)=(t,t) for t in (0,1) be our curve. Then we have (0,0) and (1,1) as our limit points ( every open ball should cut the curve). The curve clear doesnt contain these points, so it is not closed. It is not open either, as every open ball centered at a point on the curve will contain a point outside the curve ( I think this fact can be generalized to any other curve as every curve is 1-dimensional while space has >=2 dimension). For curve R->R this degenerates to open sets.

But what funny is, the precise definition of a curve is a continuous function whose domain [0,1] as per Wikipedia https://en.wikipedia.org/wiki/Space-filling_curve

So if we accept this definition then I guess it is true

[marsh forge]

Yeah

I was just coming to point out that it wasn’t true

Yeah haha this is obviously not true now that I think about it

But yes curves normally come from closed intervals (which are compact)

Which is why I asked what your domain was

[dim meadow]

@marsh forge you can fill R^n with R very easily

Just tile R^n with I^n's

And use space filling curves I -> I^n

In fact there was a interesting thing I found a while ago

[marsh forge]

Huh?

[dim meadow]

A function g:R^n -> R^m is continuous iff for every continuous map f:R -> R^n we have g°f is continuous

[marsh forge]

You can’t use I as the domain tho

[dim meadow]

?

[marsh forge]

Wikipedia requires a space filling curve have domain I

Not R

That’s why I said that

[dim meadow]

Oh lol

[marsh forge]

Yeah I know you can do it w R

[dim meadow]

I was interested in this sort of thing like 6 months ago

So I got excited

[gritty widget]

I need some assistance here please

[tight folio]

#geometry-and-trigonometry perhaps is better

[coarse kestrel]

Right

This channel is not for this kind of geometry

[covert oyster]

I made up a problem

[gentle osprey]

nowhere dense:

[coarse kestrel]

Interesting

Can't really see it even when n=2

🤔

[covert oyster]

but there are always infinitely many, you can't chose n, n is indexing only

the problem is that they may be really small balls

[bitter yoke]

I think he's talking about R^n

[covert oyster]

ooh right

[dim meadow]

For part 1 you can just take a countable dense subset of S^n-1 and attach one ball at each point. Then shove each ball in a tiny bit

Part 2 looks like it would involve some actual (basic) Geometry so I'm not going to do it

@covert oyster is part 1 right?

Oh I guess you would project inwards from that set of dense points and put the ball on the corresponding point on the sphere of radius min(1, 1/r) where r is the radius of the ball

[covert oyster]

It's right!

Part 2 is not so geometric

Actually I would say part 3 is more geometric

Or maybe there is a more geometric solution for part 2 as well, we will see

[dim meadow]

Was the first thing I said right btw?

I made the change because I wasn't so sure

Oh it definitely wasn't right

[covert oyster]

I think it wasn't right yeah

[dim meadow]

Just take balls whose circumference sums to a small number

Or something

And you can prove that wouldn't work

[covert oyster]

The problem is similar to the construction in which you put over each rational number an interval centered on it of length 2^-n (n here comes from an enumeration of the rationals)

This doesn't cover all the reals

[dim meadow]

Yeah

I guess I believe part 2 actually

[covert oyster]

Mm so I think you didn't solve part 1 correctly now

[dim meadow]

Why?

Maybe you don't use 1/r but you do something similar

[covert oyster]

Yes, you need to explain that choose

[dim meadow]

I think I meant r

Not 1/r

Oof

Yeah I definitely did

[covert oyster]

r sounds better

But you need to choose something bigger than r

Otherwise 0 is in the ball

[dim meadow]

Nah I don't think you do

You shove it in r/2

And when you pull it back to the sphere I guess it's lower bounded by some constant

Is what I want

Yeah that is what I want

I don't want to figure out the constant rn lmao

[covert oyster]

Haha

Ok, I will go to sleep now its getting late here

[dim meadow]

Alright have a gn

[covert oyster]

Bye

[dim meadow]

Was this your solution btw?

[covert oyster]

You got the important part

That is ||the radious of the ball doesn't matter because the amount of rays it cover doesnt change once you apply a homothety, hence you just need to find a solution with balls of any size||

[dim meadow]

Oh cool

What is a homothety?

Is that a homotopy in a metric sense or something?

[covert oyster]

Just multiplying by some scalar factor

[dim meadow]

Oh lol

So fancy

[covert oyster]

Haha

[rancid shard]

How do I prove that R is regular ? ( Euclidean topology)

[coarse kestrel]

By definition. For R the closed sets contain all limit points

[sweet wing]

i think it shouldnt be too hard to find a basis that encloses a point and is disjoint from a set, disjoint from the point directly from definition of reguar
T3 include T2 proof

[rancid shard]

@coarse kestrel "closed sets contain all limit point"s is a statement that is true for all topological spaces. Hence it can't be used to infer about regularity.

[coarse kestrel]

Well for R^n in this case

Oh just R

If x is not in a closed set E then it’s not a limit point and thus there is a neighborhood of x which doesn’t intersect E

[rancid shard]

I mean

could you be more specific

which neighborhood ?

[coarse kestrel]

It’s the euclidean (standard) topology right?

[rancid shard]

yeah

[coarse kestrel]

So there is an open ball, which is an open interval, or open neighborhood, of x

[rancid shard]

Can you construct it exactly in set builder notation

I would like a constructive proof

[coarse kestrel]

Ok but there really isn’t anything I feel like

Suppose $E$ is closed and $x\notin E$, then $x$ is not a limit point of $E$ and there exists a neighborhood of $x$ that doesn’t intersect $E$, or there exists a positive real number $r$ such that ${p\in\bbR:|p-x|<r}\cap E=\varnothing$

[gentle osprey]

Whoever:

[coarse kestrel]

Lol maybe I misunderstood

Actually that might quite possibly be the case

[vocal wharf]

that doesn't quite prove regular yet

just let C be a closed subset, x a point. then there exists an open ball around x of radius r that does not intersect C (just take the distance between x and C)

now B_{r/2}(x) is an open ball around x

and the union of all B_{r/2}(c) for c in C is an open neighbourhood of C

as union of open sets

the intersection is empty because triangle inequality

@rancid shard

[rancid shard]

@vocal wharf The proof u showed is quite general. As for abitrary set which does not contain x we can do the same: take the union of the open balls which are less than half of the distance between any two points in the set and X. So I suppose this is true not only for closed sets?

[dim meadow]

In full generality you can do this between any 2 closed sets

That are disjoint

You can't do this without the closed condition because of limit points

@rancid shard

[vocal wharf]

this is also true for open sets, but that's more trivial

open sets are their own open neighbourhood

[rancid shard]

Nice, I understand now, thank you everyone !

[rancid shard]

Is regularity preserved by subspace ?

It is, i just found out

[midnight jewel]

yep (unlike normality!)

(though counterexamples for that are annoyingly difficult)

is there an easier counterexample to normality not being preserved than (0,1)^ℝ as a subspace of [0,1]^ℝ?

(which I know is true but can’t prove myself)

(in particular the step that ℝ^ℝ is not normal I don’t really know how to show)

[dim meadow]

So normality is true for metric spaces, and subspaces of metric spaces are metric spaces

Because of that easy examples aren't so easy to find

[midnight jewel]

yea, even just an example of a hausdorff but not normal space is hard enough, though I do know a nice one

upper half plane with boundary, basis for the topology are:
•all open balls not touching the boundary
•open balls with center on the boundary but excluding all non-center points on the boundary

this is hausdorff, but it’s not regular: every subset of the boundary δ is closed, but there are no open sets separating (0,0) and δ \ (0,0)

[west spindle]

doesn't that have a name of some sort

the niemytzki plane or something

[midnight jewel]

it probably does have a name

[vocal wharf]

yeah or moore plane

[dim meadow]

Can you do the classification of compact surfaces using cellular homology?

By viewing a compact surface as a cw complex with 1 0 cell, a finite collection of 1-cells and a single 2 cell

And then using degree theory to get the homology

[dim meadow]

@honest narwhal

So let's talk about CW complexes

Well to start off

Given an inclusion A -> X, we have a short exact sequence of chain complexes 0 -> H_n(A) -> H_n(X) - > H_n(X, A) -> 0

(I'm not going to do exact sequences in latex btw)

This induces a long exact sequence

Which we are going to use for cellular homology

So given a CW complex X, let's look at it's k skeleton X^k

[honest narwhal]

Wait you mean H_n?

[dim meadow]

Yeah

Oof

[honest narwhal]

Or actually C_n

[dim meadow]

Oh yeah

And that gives you the long exact sequence

In homology

But okay

[honest narwhal]

And just to remember, C_n(X,A) is defined how?

I know it'll be C_n(X)/C_n(A) or something

[dim meadow]

Yeah

That

[honest narwhal]

But is that the definition or a theorem or what?

[dim meadow]

Definition

[honest narwhal]

Okay perfect so exactness is lmfao

I'm with you now, sorry about that, go on

[dim meadow]

Okay, so we know something about H_n(X^k/X^k-1)

So this is the homology of the CW complex which is attaching a bunch of k-cells at a point

[honest narwhal]

Ah, D_n mod boundary is S^n

[dim meadow]

Yeah

[honest narwhal]

So this is a wedge of spheres?

[dim meadow]

It is

And we know the homology for wedges

[honest narwhal]

Okay so its homology is gonna be Z^{number of k-cells}

[dim meadow]

Yes

In top dimension

And Z in bottom

And 0 else

[honest narwhal]

Okay good I still "know" homology

[dim meadow]

Note also that we have a isomorphism between H_n(X^n/X^n-1) and the reduced homology of H_n(X^n, X^n-1)

This is a previous result which I'm not going to prove

So the deal with cellular homology is that we can make a chain complex whose entries are H_n(X^n/X^n-1)

In other words just Z^{number of n-cells}

And we make the boundary maps by stapling together a bunch of exact sequences

We use the long exact sequence I talked about earlier here

[honest narwhal]

Oh no

[dim meadow]

This isn't too bad

Let me see if I can reconstruct the diagram chase needed to show this is a chain complex

[honest narwhal]

That's not too bad

[dim meadow]

Yeah it really isn't

[honest narwhal]

Since it contains \delta_n j_n

[dim meadow]

Okay right

The next part is showing that this gives you an isomorphism with singular homology

Which is just purely Algebraic

[honest narwhal]

I'll take it on faith for the moment unless there's an idea that's worth knowing

[dim meadow]

Not really

It's purely follow your nose

It's worth it to do on your own tbh

The really important thing

[honest narwhal]

Like I'm also pretty much black boxing the proof of singular = simplicial until after quals as well, in my mind it's just "the axioms hold"

[dim meadow]

And the thing that makes this worth it

Is that you can describe the maps in terms of degree theory

[honest narwhal]

:0

[dim meadow]

This is the sort of thing which is worth going through

So basically the way we do this

Is you take a n cell

Look at the map from S^n-1 to X^n-1

[honest narwhal]

Okay kinda had to steal it but I can thwart the cops

[dim meadow]

Then collapse all of X^n-1 - a (n-1 cell) to a point

Note that that collapsing is the same as identifying the boundary of D^n-1

So you get a quotient map from X^n-1 to S^n-1

By composing the attaching map from the boundary of the chosen n-cell with this quotient map, you get a map S^n-1 -> S^n-1

Which induces a map on H_n

And the degree of that map is the coefficient for the n-1 cell in d_n(n-cell)

Here's a shitty example of this

Classification of surfaces

[honest narwhal]

Wait something's not type checking in my mind

[dim meadow]

You can make a compact surface have a cw complex with 1 0-cell, n 1-cells, and 1 2-cells

Sorry go on

[honest narwhal]

So we have an attaching map S^{n-1} -> X^{n-1} along which we glue a D^n

[dim meadow]

Yes

Btw I'll try to present the proof of this later tonight

Cause I would have to prepare

So you have an n-1 cell e_\alpha^{n-1}

So for clarification the n-1 cell is the image of the interior of the ball

In case that's what was bothering you

Not the image of the closed ball

[honest narwhal]

Oh

[dim meadow]

So the collapsing of $X^{n-1} - e_\alpha^{n-1}$

[gentle osprey]

Liquid:
Compile Error! Click the reaction for details. (You may edit your message)

[dim meadow]

Gives you a disc with it's boundary identified

So S^n-1

So here's a example from classification of surfaces

Consider the surface with 1 0-cell, 4 1-cells a,b,c,d and 1 2-cells

The surface word corresponds to the attaching map from the 2 cell to the 1-skeleton

Let's say the surface word is aba^-1b^-1cdc^-1d

So we can compute the map from H_2(X^2/X^1)

Which is just Z here

The coefficient for the 1-cell a

Would be aa^-1

=0

They are all zero except for d, which has degree 2

So the map H_2(X^2/X^1) -> H_1(X^1/X^0)

Just sends the 2-cell to 2d

So H_1 in this case will just be Z^4/(2Z) = Z^3 \oplus Z/2Z

Because the map from H_1(X^1/X^0) is trivial

[honest narwhal]

Ah so then that works with Euler char

[dim meadow]

Yeah I think you can get the classical definition of Euler characteristic using cellular homology very easily

Assuming the homology definition

Tbh there's nothing really deep going on in the proofs for cellular homology

They're purely Algebraic

And that's my shitty crash course in cellular homology

@honest narwhal any questions?

Oh so a little detail

I've been saying H_n(X^n/X^n-1) instead of H_n(X^n, X^n-1)

This is fine except in bottom dimension

But for bottom dimension it's just connected component stuff

So it shouldn't really matter

But it actually uses H_n(X^n,X^n-1)

[honest narwhal]

Yeah that's all fine, main question is how do compute degrees IRL in general. Does it match with smooth degree from AT?

n->1 maps at regular values up to orientation etc

[dim meadow]

Yeah it does I think

I've read that it does in the past

In this case it's much easier than in that case

Cause your only dealing with S^n

And looking at the induced map Z->Z on homology

I need to brush up on my degree theory tbh

But I've been doing it by eye and it's not too bad

[honest narwhal]

Hmm, I guess I'm less familiar with how to compute those degrees as well. If it aligns with difftop degree I'm somewhat happier

[dim meadow]

If you have something explicit, you can actually compute the map on homology

Oh there are some results for degree theory

You should read

One sec

This sort of thing

Also there's another big result

On 135-136 of Hatcher

[rancid shard]

I'm not sure how to relate the definition of continuity as preimage of an open set is open with the normal definition in analysis as : for all x and y, if |x-y|< delta => |f(x)-f(y)|<epsilon . Mainly because the forward direction of inference makes it a bit awkward for me to prove equivalence

nevermind i got it

[west spindle]

for all x and y, if |x-y|< delta => |f(x)-f(y)|<epsilon
that's not the definition of continuity at all

[coarse kestrel]

I think that’s uniform

[rancid shard]

that's just me abbreviating the definition

[west spindle]

no

[gritty widget]

lol

[west spindle]

it's you mangling the definition to the point that it is incorrect

[rancid shard]

It should be something like for all x and y, for all postivie epsiolon, there is a delta st .... and the rest, right ? And for the uniform continuity we will have the delta qualifier right after epsilon

Could u cite me the correct definition then ?

[marsh forge]

Also, I’d recommend proving the equivalence of the two definitions

You can ask for help ofc but the best way to understand it is to give it a shot

It’s not too hard tbh depending on how much topology you know

[rancid shard]

i wrote "nevermind i got it". Also I was just being informal but people here seem to be keen on getting everything correct

[marsh forge]

Tbh your choice of informality was closer to uniform cont than it was to normal cont

So it’s not surprising people pointed that out

As your question is false using uniform continuity

[rancid shard]

oh i forgot that it might not be true for uniform continuity

thanks for pointing that out

[marsh forge]

np

[fleet rapids]

kindof a basic question, but Ive seen S^1 used to mean the unit circle, and S^2 for the unit sphere

is there a discrete dynamical system that relates S^n to S^m for n and m in N?

and what gets us from S^1 to S^2, I assume theres no homeomorphism between them right?

[west spindle]

wdym by "what gets us from S^1 to S^2"

ofc the two aren't homeo

[fleet rapids]

maps

b/c we cant continuously and inveribly deform 1 to the other right?

[west spindle]

well no of course not

i mean

there do exist continuous maps from S^1 to S^2 and from S^2 to S^1

but of course there exist continuous maps between any two topological spaces

so i'm really not sure what exactly you're looking for here

[fleet rapids]

basically just looking for the strongest topological relation between them to explain why they are named almost the same things but with slightly different superscripts

and I thought a homeo is just a continuous inverible map of top. spaces, no?

[west spindle]

uh

okay so first off

the same could be said of R and R^2

or R^2 and R^3

[fleet rapids]

yeah bc Rm is bijective to Rn

[west spindle]

no

NO

[fleet rapids]

huh?

I thought they had same cardianlity

[west spindle]

yes but that doesn't matter

S^1 and S^2 have the same cardinality as R as well

[fleet rapids]

but that directly implies theres a bijection between them

[west spindle]

yeah but none that is continuous and whose inverse is also continuous in their standard topologies

[fleet rapids]

ah ok I see my bad

key word continuous

am I correct that homeo is just continuous inverible mapping though?

[west spindle]

no

the inverse also needs to be continuous

[fleet rapids]

ok so continuous, invertible, and inverse map is continuous, missing anything still?

so I guess the third requirement is whats preventing S^1 from being homeo to S^2 right?

[west spindle]

no that's the definition of a homeomorphism

and no

if anything, the last requirement stops S^1 and [0, 1) from being homeomorphic

S^1 and S^2 can be shown to be non-homeomorphic by other means

[dim meadow]

A nice proof is removing 2 points from both

[fleet rapids]

ok

my question was that is the inverse map not being continuous a sufficient condition that holds between S^1 and S^2 though?