point-set-topology

[gritty widget]

I am having trouble dealing on how to write that "the only things that count is the n-th skeleton of Y"

[gritty widget]

Up?

[dark dew]

Down?

Left?

Right?

[gritty widget]

Lol

[reef eagle]

I'm having trouble understanding this bit of a proof of a theorem on finite total curvature.

If anyone has any insight it would be greatly appreciated

[small obsidian]

Reading a different book on manifolds, but got to this:

That example ii)
The function isn't even C^0 is it?

Oh pfft I read the wrong part of the definition

[vestal furnace]

radius r, and center P. Show that the image of S under F is a circle.
[Hint: Let S' be the circle of center F(P) and radius r. Show that F(S)
is contained in S' and that every point of S' is the image under F of
a point in S.]```

Do you really need to prove every point in S' is image under F of S?

I don't get the reason to have it mentioned

[vital sail]

yeah, you're essentially proving S' = F(S)

so you prove S' ⊆ F(S) AND S' ⊇ F(S)

[vestal furnace]

I get the subset thing

But how will F(S) even have any point in it other than the image of S?

Or it's just the usual formal thing?

[vital sail]

F(S) is defined as the image of S

S' is defined differently

[vestal furnace]

Oh wait

Thanks!

I missed that

[vital sail]

np

[floral gust]

Is z to ź a homeomorphism?

where ź is complex conjugate

[bitter yoke]

on what domain/codomain

[floral gust]

a region I guess? (Why would it matter ?)

Oh yes I see the reason why it matters

Like C to C

[bitter yoke]

Then yes

[floral gust]

I suppose it should be a homeo since it is its own inverse.

Okay thnkx

and obviously cont.

@bitter yoke It is a homeomorph also in the case right where it is a mapping from the domain to the image of the function?

[bitter yoke]

yes

[floral gust]

thnx

[midnight jewel]

AlgTopo_irl

[covert cove]

Hey can anyone help me understand the outline of problem 1 section 52 in munkres?

[hexed prawn]

Hey guys... I wasn't sure if I could post a question here so I asked a topology question in #help-0 if any of you want to take a gander.

[gritty widget]

You can ask questions here

[fleet rapids]

So is all that's in Homeo(S^1) (where S^1 is thought of as [0,1) mod 1) rotations and reflections (Since expansions and contractions aren't bijective)? I was trying to think of S^1 as lim n-> infty (D_2n) but not sure how that would be formalized or helpful in this case.

[dire warren]

You can have many other maps in homeo S^1

Though they aren’t easy to write down explicitly

[fleet rapids]

oof

[ripe thunder]

In particular, think about contracting part of a circle and dilating outside of it to account for the contraction

[fleet rapids]

ah ok, I can see why this is a homeomorphism, ty

[sleek thicket]

My manifolds book just proved that taking global sections of a vector bundle over M is full and faithful

And defined subbundles to be embedded submanifolds whose fibers are subspaces

With the restriction of the projection map and all

[honest narwhal]

Full and faithful functor to what?

@sleek thicket

[sleek thicket]

Oh sorry C^infty(M) modules

Got distracted helping someone in algebra, I'll finish this question in a sec

So define an inclusion of vector bundles (over M) to be a bundle homomorphism ι : D -> E which is a smooth embedding

This has to be injective, so it's injective on fibers, and gives us a consistent way to identity D_p with a subspace of E_p

Is there an algebraic condition on a C^infty(M)-module homomorphism φ : Γ(D) -> Γ(E) that tell us that the induced map D -> E is an inclusion?

[honest narwhal]

I'm skeptical

Or hmm so in the algebraic setting the problem is basically that you don't always have that many global sections but I guess in smooth manifolds you can play bump function games

So maybe there is something here

[sleek thicket]

You saw nothing

[honest narwhal]

:0

[sleek thicket]

So like the best possible thing would be to just require φ to be injective

And I haven't been able to find a counterexample

But that's mainly because I don't know very many bundles or bundle homomorphisms

Also, if we consider vector fields on the sphere as maps S^2 -> R^3, crossing with the unit normal gives a C^infty(M)-module endomorphism of the set of vector fields

What does this look like?

This is like calc 3 or earlier geometry

But I'm having trouble visualizing what crossing a vector field with the unit normal does

[gritty widget]

this is not the kind of geometry that goes in this channel

#precalculus

[vestal furnace]

Okay!

[gritty widget]

Suppose M is a set and d, d' are two different metrics on M. Prove that d and d' generate the same topology on M if and only if for each x ∈ M, there are some r₁, r₂ > 0 with Bᵈ'(r₁, x) ⊆ Bᵈ(r, x) and Bᵈ(r₂, x) ⊆ Bᵈ'(r, x).

i'm not sure how to start the backward direction

could anyone give a hint

[fading vale]

whomst ping

oh my god

i remember

ok fine :|

imma throw away this ice cream

*ice cream cup

[gritty widget]

okok how did you do the backward direction

[fading vale]

@gritty widget what have you tried so far?

[gritty widget]

idk i'm kinda stuck

for the forwards direction i used open iff union of open balls

[fading vale]

okay well, what does it mean exactly for two sets to generate the same topology

[gritty widget]

they generate the same open sets

[fading vale]

correct

so what do we want to show?

[gritty widget]

that the existence of those r₁, r₂ show every open set in (M, d) is open in (M, d') and vice versa?

[fading vale]

i mean, more concretely, we want to say that an open set O under d is open in d'

and vice versa, right?

[gritty widget]

ya

[fading vale]

so, start be letting O be open wrt d

how can we show that it is also open wrt d'?

(start by taking an x in O. What can we say about that x, since O is open?)

[gritty widget]

every x is an interior point of O

[fading vale]

think about it in terms of open balls and metrics

[gritty widget]

for some s>0 you can find an s-nbhd of x contained in O?

[fading vale]

i mean yes, so for every $x \in O$, we have $x \in B^d (r, x) \subset O$

[gentle osprey]

Sloth:

[fading vale]

what relationship do we have between the $B^d (r, x)$ and balls around x under the metric d'

[gentle osprey]

Sloth:

[fading vale]

(think about your hypothesis. reread the statement of your proof)

[gritty widget]

they're contained in B^d (r,x)?

wait

[fading vale]

be a little more specific

[gritty widget]

idk

[fading vale]

well, its literally stated in the hypothesis right

[gritty widget]

yes

[fading vale]

for each $x \in M$ and thus each $x \in O$, we have $B^{d'} (r_1, x) \subseteq B^d(r, x)$

[gentle osprey]

Sloth:

[gritty widget]

right so from 'for each x ∈ M, there are some r₁, r₂ > 0 with Bᵈ'(r₁, x) ⊆ Bᵈ(r, x) and Bᵈ(r₂, x) ⊆ Bᵈ'(r, x)' we're trying to prove that d' and d generate the same topology

oh

[fading vale]

do you see now?

[gritty widget]

no

sorry

[fading vale]

Well, we've shown that if O is open in the d-topology, each $x \in O$ has an r-ball around it (with the metric d)

[gentle osprey]

Sloth:

[fading vale]

and our hypothesis states that any r-ball with the metric d has an r1-ball with the metric d', such that $B^{d'}(r_1, x) \subset B^d(r, x)$

texit u there

[gentle osprey]

Sloth:

[fading vale]

@gritty widget do you know what it means for a set to be open in terms of balls?

[gritty widget]

$\forall x \in G: \exists r>0: B(r, x)\subset G$?

[gentle osprey]

mátt:

[fading vale]

right

so we just showed that for every $x \in O$, we have an $x \in B^{d'}(r_1, x) \subseteq B^d (r, x) \subseteq O$

[gentle osprey]

Sloth:

[fading vale]

right?

[gritty widget]

yes

oh

oh

right

[fading vale]

hi zoph

[gritty widget]

cool ty

[fading vale]

yeah

and you can see that its like

the exact same thing for showing sets open in the d'-topology are open in the d-topology

[gritty widget]

yes

ty

sorry dumb q

[fading vale]

nah its fine

[gritty widget]

Let $(M,d)$ be metric space, let $c$ be a positive real number, and define a new metric $d'$ on $M$ by $d'(x,y)=c\cdot d(x,y)$. Prove $d$ and $d'$ are generate the same topology.

[gentle osprey]

mátt:

[gritty widget]

So I chose $r_1 = r_2=r/c$. Then $B^{d'}(r_1,x) \subset B^{d}(r,x)$ and $B^{d}(r_2, x) \subset B^{d'}(r,x)$, so $d'$ and $d$ generate the same topology

[gentle osprey]

mátt:

[gritty widget]

but the solutions say r_2 = rc

am i going crazy or is this wrong

[gritty widget]

scratch that r_2=r/c and r_1=c/r

[sick jewel]

difference between strong continuity and just regular continuity? I couldn't find anything useful online

In particular, for p: X -> Y, U open in Y iff p-1(U) open in X is what I find for both definitions

[bitter yoke]

no, the definition of continuity is U open in Y implies p^{-1}(U) open in X

it only goes one way

[sick jewel]

ohhhh

so strong continuity says p^{-1}(U) open in X implies U open in Y?

and vice versa?

[bitter yoke]

in addition to continuity yes

[sick jewel]

kk thx

[soft aurora]

can u help me figure out what this transform into? (all the points labeled with c go into the same point and the and segments labeled with a and b go to the segment with the same letter, in the direction indicated by the arrow)

[soft aurora]

i have an easier question if someone wants to help me out

I have to proof the following statement: In a finite topological space, the arbitrary intersection of open sets is an open set.
If the space is finite, there must be a finite number of open sets, and by the definition of topology the intersection of a finite number of open sets is an open set.
but i'm not entirely sure if that proves it

[vocal wharf]

what makes you unsure

[soft aurora]

idk haha

"If the space is finite, there must be a finite number of open sets" this makes me unsure

[vocal wharf]

what is the finest topology on a set

i.e. with the most open sets

[soft aurora]

the discrete topology

[vocal wharf]

how many elements has it

[soft aurora]

2^n with n the number of elements?

[vocal wharf]

yes, in particular finite

all subsets of finite sets are finite

[soft aurora]

hmm ok

ty ^^

and do u knwo about the other question? idk if i explained it correctly xD

[vocal wharf]

well

by joining segments/point i assume you mean passing to the quotient space

if you identify the sides a and b with each other you get T^2, i.e. the standard torus

not sure what exactly you mean by joining the c's

can you formalize it?

[soft aurora]

yeah i meant passing to the quotient space
i know that identifying the sides u get the torus
the segments labeled with c go to the same point

if there was only one segment with c i know its the torus with one point "collapsed"

and im afraid its going to be hard to formalize and im supposed to be able to do it "graphically"

[vocal wharf]

what exactly are you supposed to do?

draw a picture or find a "better" space it is homeomorphic to?

[soft aurora]

find a better space it is homeomorphic to, but its most likely a space that can be drawn xD

or at least a well known space

[rugged swan]

it is

[soft aurora]

ive tried this too

thats like a sphere but the c line is annoying me xD

[rugged swan]

but that's the sphere

you've just sketched a drawing

[soft aurora]

but the sphere shouldnt have the c line in the middle

maybe 2 spheres with 1 point in common?

[rugged swan]

with you torus pov you can also glue the c line

[soft aurora]

yeah but i cant figure out how to glue the c xD

so u think its a sphere??

[rugged swan]

it is clearly a sphere

why can't you glue c ?

you can

[soft aurora]

in my last picture, i can see clearly that without the c line it would be a sphere

but with the c line i cant visualize it xD

[rugged swan]

it is as you were pinching the line

that makes the sphere

[soft aurora]

well i have to keep thinking on it a bit more, but thank u very much Zak and Lochverstärker

[rugged swan]

I think it's clearer with the torus drawing

[grim coyote]

what is a characteristic map (relating to CW complexes)

[sleek thicket]

@grim coyote what do you know about CW complexes?

A CW complex structure on a space X consists of a partition of X into "cells" {e_α}. Each cell comes with a map φ_α from the closed n-disk (n dependent on α) into X which restricts to a homeomorphism of the open n-disk onto e_α

You need other axioms too

But φ_α is the characteristic map of e_α

[red atlas]

I'm trying to work through Hartshorne right now. Does anyone know what is meant by the zero set of x_i here? https://imgur.com/a/T1RC1Ji

[honest narwhal]

There's the polynomial f(x_0,...,x_n) = x_i

Just find the zeroes of that guy

[red atlas]

Hmm, okay

Thank you

[sleek thicket]

@red atlas I don't think Hartshorne chapter 1 is very good

I started with it and switched to Gathmann's notes

Sorry for the unsolicited advice

[red atlas]

Haha, that’s good to know

[gritty widget]

how can a submanifold be a set?

it's in the definition

[gentle osprey]

mátt:

[gritty widget]

so this is one direction of the proof

idk if it's the simplest

but does it hold up?

this is for me?

oh no

ok sorry

but look at the definition of submanifold

interesting

[little hemlock]

@gritty widget $x \notin \partial A$ gives you that any open neighborhood $N$ containing $x$ does \emph{not} intersect points from both $A$ and $X \setminus A$. In other words, $x$ belongs to either the interior or exterior of $A$. Well, since $x \notin A$, it must belong to the exterior.

[gentle osprey]

kxrider:

[little hemlock]

I think everything else after that is alright tho

wait I misread what you said. I think read "or" as "and" somewhere

[gentle osprey]

mátt:

[gritty widget]

@little hemlock ya ty

[gentle osprey]

mátt:

[gritty widget]

Anyone here with knowledge in algebraic topology?

[honest narwhal]

I'm iffy but in principle yes

[sick elm]

Do we ever deal with tensors of non square matrix representation in Diff Geom. ?

[sweet wing]

probably not i have never seen such a thing

[versed pivot]

something like a vector valued 1-form? one index ranging up to the dimension of the base, the other index ranging up to the rank of the bundle

[pseudo roost]

anyone know where I can get some questions like this ? preferably with answers

[bitter yoke]

The knot book by Adams has questions like this

There are no answers but

[pseudo roost]

Thanks for the suggestion looks similar to my course, thanks

[floral gust]

Can somebody tell why is Dg(x_0) nonsingular

Or more particularly why is Dpi(x_0) non-singular

[honest narwhal]

D\pi isn't non-singular, it has a kernel, but it's got maximal rank

Since pi is linear, Dpi = pi

Now beta is what we'd need to know about to infer non-singularity of Dg

[little hemlock]

hm im kinda stuck on this one

[sleek thicket]

@little hemlock did you know this is true when A is open or closed?

[little hemlock]

nope

[sleek thicket]

Well that's a good place to start

It tells you that a broad class of examples won't work

What have you tried already?

[little hemlock]

I started by trying to do it with the mobius strip, but then I figured it would probably work with a simpler example. Like a space where you take an interval [0,1] and identify its end points.

[sleek thicket]

Yeah that's what I was thinking

Do you know what we call the space where you identify the endpoints of [0, 1]?

[little hemlock]

nah, but I imagine its something homeomorphic to S1, right?

[sleek thicket]

Yeah I mean I call it the circle

So how can we sort of screw this up?

What A might we want to choose?

[little hemlock]

maybe [0, 1)?

[sleek thicket]

Yeah, that's a good idea (since it isn't open or closed)

So what's the image of A?

[little hemlock]

it has the same image as [0,1] under the identification map where you send points to the subset they belong to.

[sleek thicket]

Right

So let's think about why this can't be an identification map

Is there any geometrical sort of reason you see?

Draw a picture of a half open interval and a circle maybe

[little hemlock]

yea this is kinda where I got stuck. hmmmmmm

[sleek thicket]

hmmm

So we start with p : [0,1] -> S^1 given by p(t) = e^(2 π i t)

then restrict this to q : [0,1) -> S^1 given by the same formula

What does it mean by definition that q is an identification map?

Also are you using Brown's book? I haven't seen the terminology identification map anywhere else

[little hemlock]

q is surjective and the topology on S^1 is the largest for which q is continuous. i.e. U is open in S^1 iff q^-1(U) is open in [0,1)

nah, this is armstrong

[sleek thicket]

Ah okay

Well then I'm not sure if it's in there but there might be another definition of identification map. Do you know what a saturated set is?

[little hemlock]

nope, what is it?

[sleek thicket]

Don't worry about it

So let's try to think about sets U which aren't open but whose preimage is open

In most places on the circle, this map looks like p, which we know is an identification map, right?

So U had better involve the weird point

[little hemlock]

yep, i agree.

can it really just be some set of this form?

[sleek thicket]

what happens if you take the preimage of such a set?

[little hemlock]

something like [0, a]U[b, 1) which is closed so hmm

[sleek thicket]

It's definitely closed

Oh sorry no

I think you chose a slightly wrong set

I mean, would that set even have open preimage under p?

How could we modify it so that the preimage under q is open?

[little hemlock]

well if we just removed one of the endpoints, we would get something like
[0, a]U(b, 1)

which is hmmm, looks like is not open nor closed

[sleek thicket]

That's true, but we want the preimage to be open, right?

However if we remove the other endpoint, then the set itself becomes open

So maybe we can't just take any set containing the bad point. Any other ideas?

[little hemlock]

something like

so its preimage would be [0, a), right?

[sleek thicket]

Right, and what does that tell us?

[little hemlock]

welp, since [0,a) is open in [0, 1) that would be it. no identification map

[sleek thicket]

Yeah, that's exactly right

Here's maybe a shorter proof

q is injective, right?

well an injective quotient map is a homeomorphism

But the circle is not homeomorphic to an interval

(proof: an interval has a point you can remove to make it disconnected, while the circle has no such point)

[little hemlock]

ah yea thats slick. Thank you for guiding me through! appreciate it

[sick elm]

hmm, I am learning about connections on a manifold. I understand something like $\nabla_{X} Y $ where Y being a vector field produces a vector field. But I don't understand what $\nabla_{[X, Y]} Z $ is supposed to be

[gentle osprey]

bibek22:

[sick elm]

where on the lower slot it somehow involves product of two vector fields? How is that defined and how does it relate to the definition of connection

[honest narwhal]

@sick elm it's called a Lie bracket, takes in two vector fields and spits out a third, so that still type checks

It's a kind of commutator basically. Tbh I don't know diffgeo well enough to explain further but at least that's the term to Google

[vital sail]

x,y = xyf - yxf

vector field is defined by its action on scalar fields

[sleek thicket]

you can also think of [V, W] as a derivative of W as it goes along the flow V

If T and Y are vector fields, and φ is the flow of Y then [Y, T] equals

Where we're taking the derivative in the tangent space at p

[vital sail]

you don't need a connection to define [x,y]

[sleek thicket]

was that in response to me?

Because I don't even know what a connection is

[vital sail]

no, it's a response to "how does it relate to the definition of connection"

[gritty widget]

@sick elm try computing the components of the riemann tensor

or even the torsion tensor

in that example the lie bracket just makes in the torsion tensor tensorial

[midnight jewel]

isn’t the right way to think about [X,Y] to be a measure of how hard schwarz’s theorem (interchanging of derivative) fails?

if one thinks of X(f) as the derivative of f in direction of X

in ℝⁿ, [X,Y]=0 for all vector fields by schwarz’s theorem

in arbitrary manifolds it can tends to be nonzero

[gritty widget]

that's exactly what the curvature tensor does

$R(\omega, Z,X,Y) = \omega (\nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z)$

[gentle osprey]

mátt:

[sleek thicket]

@midnight jewel uhh [X, Y] is not always zero in R^n

Your thing only applies to coordinate vector fields

Edit: the lie bracket is R-bilinear but not C^infty(R^n)-bilinear, so you can't just pull the coefficients of the coordinate vector fields out. The lie bracket satisfies a sort of product rule, which you can derive from the lie derivative formulation I posted above

[midnight jewel]

hmm

thanks for pointing out my misunderstanding

I’m struggling a lot with diffgeo and have definitely too many half- (or un-)truths memorized

but yea, makes sense

I didn’t think about it carefully enough

it does hold for “constant” vector fields though, right? like, ones that always point in the same direction

[sleek thicket]

I'm not sure what you mean by that

Oh yes I am

Yeah so if the coefficients of the coordianticr vector fields are constant, then you can use R-bilinearity

And the fact that the bracket of two coordinate vector fields is 0

[sick elm]

is any of the option here incorrect ? all of it looks right to me

[rugged swan]

only one here is correct

[midnight jewel]

am I dumb? I can only see one that I find obviously incorrect
||the first one is definitely wrong as it’s supposed to be the union (modulo details with retaining the basepoints)||. what are the errors in the other ones?

||the second one is correct iirc. the third one one could argue about the word “inherited” which probably should be “induced” but that’s nitpicky. the fourth one is correct as far as I can tell. the last one depends on your definition of tangent vectors, but “element of the tangent bundle” doesn’t seem to me like the worst definition||

[sick elm]

yeah sorry. first is incorrect, i see that now. what's wrong with the last three

[rugged swan]

sry, what does mean "the target space of a smooth vector field" ? For me it means that it is a vector field, it can't be true

but yes, 2nd, 3rd and last are true

[sick elm]

A smooth vector field is a smooth section $\func{\sigma}{M}{ TM} $ such that $\pi \circ \sigma = id_M $ where $\pi $ M and TM constitute a tangent bundle

isn't it so ? and then wouldn't total space be the target space ?

that latex got messed up sorry

[gentle osprey]

bibek22:

[rugged swan]

oh ok it's the definition

[little hemlock]

if you take something like a sphere and identify the points along its equator, what quotient space do you get? I feel like it would be a union of two spheres sharing a point, but idk

[rugged swan]

it's the wedge sum of two spheres

[sleek thicket]

(the wedge sum is the union sharing a point you mentioned)

This is actually really important when constructing higher homotopy groups

well for spheres the 2nd homotopy groups

Also note that this works for circles too (but then the equator is just two antipodal points)

[little hemlock]

i see...

[sleek thicket]

Sorry, that's probably not meaningful to you right now

If you have two maps f, g : S^2 -> X such that the value of f on the south pole is the same as the value of g on the north pole, you can "concatenate" them to get a map S^2 -> X which first squashes down the equator, then evaluates f on the top hemisphere (which is now a sphere) and g on the bottom

Does that kind of make sense visually?

[little hemlock]

ah, is this function glueing stuff?

[gritty widget]

@sick elm are you watching schuller's lectures

[sick elm]

@gritty widget yessir! these are great

[gritty widget]

idk anything about diffgeo but it makes my Ricci tense 😫

[floral gust]

Ricci tenser

[little radish]

Is there a general procedure for determining a T that maps a R^n space to an R^n space such that all the regions are relatively compact and are nearly of equal size like in the following

sorry for bad ms paint picture

[floral gust]

Doesnt the fact that a is a coordinate patch on a Manifold M of class C^r implies that even a^-1 is of C^ r as well

Where U is in R^k and V is in R^n

[floral gust]

<@&286206848099549185>

[honest narwhal]

I'm guessing that in a few lines he'll cite something like the inverse function theorem

Or implicit maybe in this setting

But it's possible for a smooth map to be bijective but not have an everywhere smooth inverse, e.g. x^3

But if you know the derivative is everywhere full rank then yeah inverse/implicit games will give it to you

[floral gust]

Yes that he does

But I dont need to do all that for the purpose of concluding that the composition is C^r right?

[honest narwhal]

Well see the example of x^3

Smooth bijections need not have differentiable inverses

[floral gust]

But, so my argument is this. If "a" is a coordinate patch on manifold of class C^r, then, by definition, a from U to V and a^{-1} from V to U are C^r functions.

[honest narwhal]

a o a^{-1} is the identity

The problem here is "a^{-1} from V to U is C^r" isn't obvious

You need to prove this fact

[floral gust]

Wait

Oh right right... the definition of coordinate patch just has that the inverse is continuous not C^r. Soz soz

[honest narwhal]

Uh, tbh I have no idea what you're trying to prove or what your definitions are

[floral gust]

Regardless you showed me my error xD. I was trying to prove that if a1 and a0 are coordinate patches then the map a1^{-1} o a0 is C^r.

[honest narwhal]

I see

[floral gust]

In part B they claim that U ⋂ H^k+ is open in R^k. How can they claim this? For this to be open in R^k, both sets should be open in R^k. H^k+ is open in R^k by definition. However U is only open in H^k which doesn't mean that it is open in R^k.
The way I tried to resolve this was by taking U = A ⋂ H^k where A is open in R^k and then consider A ⋂ H^k_+.

[vocal wharf]

U is open in H^k iff there is V open in R^k, s.t V \cap H^k = U. then U \cap H^k+ = V \cap (H^k \cap H^k+) = V \cap H^k+ which is open in R^k

oh nvm that is exactly what you did

so yeah, you are right

you can also argue that the boundary of U (in R^k) does intersect H^k+ only trivially, so U \cap H^k+ (in R^k) does not include its boundary as a subset

[gentle osprey]

mátt:

[gritty widget]

that is on the 4th line of the proof

[honest narwhal]

Cut down from r_1 to r_1/2

The closed ball of radius r_1/2 lives inside the open ball of radius r_1

[gritty widget]

oh cool thank makes sense

tyty

[gritty widget]

also since #advanced-analysis is taken, how did rudin get (c) in this proof

[gritty widget]

actually dw figured it out

[swift flare]

Does anyone have a smug idea for figuring out the following problem? I have N points distributed in space (known). I want to pick a percentage of those points and figure out what set of points is the most homogeneously distributed.

Using % * N amount of points

[small obsidian]

How do you measure how "homogeneously distributed" a set of points are? Is this space in R²? @swift flare

[swift flare]

No R3. I am not sure how to define the homogeneousnous of a set of points. Basically I have a crystal lattice with hydrogens attached to it (it's a semi sphere, deformed because the crystal is very small). I want to be able to pick a % and then I want to calculate the set of hydrogens that maximizes the "spread"

So maybe I should maximize the st.dev between the points?

@small obsidian

[small obsidian]

That's a strategy. You could go with a least squares. But there's probably some intense physics that is used to actually predict what the system should look like

[soft radish]

Hey please help me clear out a very stupid doubt that I have

I just studied about path homotopies, covering spaces and the fundamental group of a circle which i understanding the proof is isomorphic to Z

So intuitively i am thinking the generator path for the homotopy classes to be the single loop over the circle

Clockwise being negative and anti clockwise positive

And similarly x² is the double loop then triple and etc

So i tried to really see if I can't find a homotopy between f(x) = e^(i2πx) and g(x) = e^(i4πx) being the single and double loop respectively

So F(x,y) = e^(i2πx(y+1)) has F(x,0) = f(x) and F(x,1) = g(x)

And it is continuous

So i am thinking maybe my intuition is wrong? 🤔

[tepid totem]

For a path homotopy you would also need that for all y, F(0,y) = 1 = F(1,y)

[vocal wharf]

it doesn't keep endpoints fixed, so no path homotopy

damn, too slow

[tepid totem]

sniped

[soft radish]

Ah yes right

[grim coyote]

What makes a CW complex regular?

[gentle osprey]

Token:

Token:

[vocal wharf]

i assume closure in R^2 with standard topology?

in that case, no

[gentle osprey]

Token:

[vocal wharf]

do you think that $(0, \frac{3}{4})$ is in the closure?

[gentle osprey]

Lochverstärker:

Token:

Token:

[vocal wharf]

what

[gentle osprey]

Token:

Token:

[vocal wharf]

well ye

and (0, 0)

you should sketch that subspace

and try a formal proof

[gentle osprey]

Token:

[vocal wharf]

essentially

[gentle osprey]

Token:

[vocal wharf]

you don't need density

just metric stuff, take any point not in the closure and find an open ball around it that still fits

[vocal wharf]

sure, but that just follows from regular metric stuff

first consider a point on any axis, that is not in the set and not the origin, then it's between $\frac{1}{n}$ and $\frac{1}{n+1}$ for some $n$

[gentle osprey]

Lochverstärker:

[vocal wharf]

then find some open ball and show that it also doesnt intersect that set in 2 dimensions

if your point is not on the axes, just project on both of them and do the same

then take the min of whatever epsilon balls you found

[sick elm]

suppose the set of g-null vectors is a straight line through origin. How could i infer that the signature of g is one of either (++0) or (--0)?

[loud scarab]

is there any intuition behind a 'compact' set? Like I just went through the definition and theorems and I still don't know what to think when a 'compact' set is mentioned

[bitter yoke]

Well, your intuition for euclidean space should be that a set is compact if and only if its closed and bounded

[modern holly]

in a course on analysis in the euclidean setting, i dont think it makes sense to even introduce compactness because it doesnt differ from being closed and bounded and thus it just confuses people to think that the concepts are the same
some notes

1. in the general topological setting the compact sets are somewhat small with respect to the topology
2. in the metric setting compactness is equivalent to sequential compactness (i.e. every bounded sequence has a convergent subsequence)
3. in the euclidean setting compactness reduces to just being closed and bounded
   we want a property that is invariant under continuous maps, and here we notice that neither closedness or boundedness is preserved by continuous maps in general, im not sure of the motivation to distinguish compact sets in and sequentially compact sets in the general setting for i never outside of a topology class saw spaces which are just one but not the other
   also one last thing i like to think is that compact sets are sort of a continuous case generalization of finite sets

[loud scarab]

that last line is interesting

[honest narwhal]

That's a theme that's kind of a thing in math re compact = almost finite

But analysis is often taken by people who have reason to learn other math later where the distinction matters, so I'd rather not give people an impression that they have to unlearn later.

[gritty widget]

what if R was compact

[sweet wing]

you can have a 1(or)2 point compactification of R

[honest narwhal]

I think presenting the open cover definition of compactness as the definition and proving that you have equivalent criteria shouldn't be inordinately difficult to anyone, so they're free to focus on sequential compactness and closed/bounded as they're comfortable with and just keep covers in the back of their mind until they reach a setting where things don't coincide

[midnight jewel]

in my analysis course we did about the following:
-first semester, analysis over ℝ: only defined “compact interval”, as one of the form [a,b] with a,b∈ℝ
-second semester, we defined compact metric spaces as “a metric space X which satisfies any of the following equivalent conditions” (and proved equivalence); among these conditions were countable cover compactness, finite intersection property, sequential compactness and “every continuous function X→ℝ attains a max and a min”; we also showed that for X a subset of ℝⁿ, compactness of X as a metric space with the induced metric was equivalent to it being closed and bounded in ℝⁿ

fourth semester, in topology, we then defined compact topological spaces in generality

[jaunty nebula]

I've got a set O = [1,2] union {2} and I'm supposed to prove that {2} is open and closed in the relative topology of set O.

My first instinct would be to say that a point is closed and so that's easy to satisfy.

But being open is a bit more difficult. I read the example on the book Topology without tears and it didn't help me much.

[dim meadow]

A set A is open in the subspace topology if there is an open set U in the larger space with $A = O \cap U$

[gentle osprey]

Liquid:

[dim meadow]

Your set O doesn't seem right

@jaunty nebula

[jaunty nebula]

Let O be the following. Show that {2} is open and closed in the realtive topology of O. That's the wording from my exercise

[dim meadow]

Okay that's better

Yeah so showing 2 is open should be trivial now

You just need to find an open set which contains 2 and does not intersect [0, 1]

Open set here just means open in the standard topology on R

[jaunty nebula]

Hmm. If I'm choosing U, should everything within U be inside O or am I free to choose U = (1.5, 3) for example?

[dim meadow]

That would work

U is an open set in the bigger space

Not in the subspace

[jaunty nebula]

Ah thank you, this helps a lot

[dim meadow]

🐴

[jaunty nebula]

Does this rule work for examining if something is closed as well? For example, U = [1,3] to check if {1}U{2} is closed?

[dim meadow]

Yeah

But you have to prove it

[midnight jewel]

(the proof is like 1–2 lines that shoulnd’t be hard to find if you understand the definitions involved well)

it’s literally just repreated applications of definitions

[modern holly]

precisely why its important to prove it instead of remembering the result as a "rule"

since it requires you to internalize the definitions

[fading vale]

hmm

im trying to prove that a metric space is 2nd countable iff it is separable

ive shown that 2nd countable implies separable already, and ive made some decent process on separable implies 2nd countable, but im not quite sure where to go from here

[gentle osprey]

Sloth:

Sloth:

Sloth:

Sloth:

[fading vale]

im guessing i have to use the triangle inequality somehow but tbh im just not sure what to do

(pls ping if you have something)

[bitter yoke]

You want to take q < r/2 essentially, this will imply that x \in B_{r/2}(y) and that B_{r/2}(y) \subseteq U by triangle inequality things

[fading vale]

the triangle inequality is quickly becoming my least favorite thing ever

[bitter yoke]

Take y such that y \in B_{r/2}(x)

Which of course implies that x \in B_{r/2}(y)

so all you have to do is show that B_{r/2}(y) \subseteq U

but you can show that B_{r/2}(y) \subseteq B_r(x)

[fading vale]

ohhh

yeah i think i see

cuz B_{r/2} (x) is also open

[bitter yoke]

That's where you use triangle inequality, if z \in B_{r/2}(y), then d(x,z) \leq d(x,y) + d(y,z) < r/2 + r/2 = r

[fading vale]

right

yep

thanks zoph

[gritty widget]

So i'm trying to show that $0$ is the only limit point of $$E=\left{\frac1n: n\in\mathbb{N}^*\right}.$$ Suppose $x\in\mathbb{R}\setminus{0}$, and choose $\epsilon < \min{d(x,y): y\in E}$. Is it enough to say that $x$ is not a limit point of $E$ because $B_\epsilon (x)$ contains at most one point of $E$ (that is $x$ if $x\in E$)?

[gentle osprey]

sexy minion:

[little hemlock]

@gritty widget no. You don't address 1. why the min{...} exists (i.e. u should take inf, and 2. why it should be >0.

I think the correct strategy is something like this:

1. when x < 0. no limit points here obviously.
2. x = 0 (is a limit point)
3. when x > 0, you can find a positive integer n such that 1/(n+1) < x < 1/n... then carry on

[gritty widget]

right so then if $x\not\in E$, $$\left(\frac1{n+1}, \frac1n \right) \cap E = \varnothing$$

[gentle osprey]

sexy minion:

[gritty widget]

and if $x\in E$, then $x=1/n$ for some $n\in \mathbb{N}^*$, $$\left(\frac1{n+1}, \frac1{n-1} \right)\cap E = \varnothing $$

[gentle osprey]

sexy minion:

[gritty widget]

@little hemlock sorry for the late response

[little hemlock]

yep.

[gritty widget]

cool stuff ty

[little hemlock]

npnp

[gritty widget]

wait $$\left(\frac1{n+1}, \frac1{n-1} \right)\cap E \neq \varnothing$$ but it contains no other point of $E$ then $x$ so it's still fine

[gentle osprey]

mart:

[little hemlock]

right.

not sure if its worth including for you, but existence of such an n depends on the archimedean property.

for any x>0 there is a positive integer n such that xn > 1

[gritty widget]

yup got it

rudin so ya probably have to include that

[little hemlock]

haha yep :p

[gritty widget]

ty for all the help n.n

[little hemlock]

np

[gritty widget]

https://www.youtube.com/watch?v=Z7rd04KzLcg

did anyone watch the vid above?
any big-brained individuals in here? (grad+ level) that can talk about Weinstein's theory (I'm a noob myself and just wanted to know if it's good or not) (ping me)

[ivory dragon]

weinstein simply hasn't given enough information about his model for it to critically be evaluated by anybody

his model hasn't made any predictions, it hasn't presented any falsifiable claims

it hasn't concretely explained what new novel statements it brings up

and just in general, weinstein hasn't done the intellectual burden it takes for scientists or mathematicians to respond to his proposal

we don't have a paper, we don't have predictions, we don't have suggestions for experiments, we don't have a concrete list of claims or of derivations or of proofs

all we have is a lecture series from a few years ago

what this translates to, in plain english, is: "don't bother worrying about weinstein's stuff"

for what it's worth, he's not a total nutcase or whatever - he's moderately good at math and physics

but his model is, as far as we can tell, empty sentiments and vague air

@gritty widget

[gritty widget]

his model hasn't made any predictions, it hasn't presented any falsifiable claims
@ivory dragon that's most of high end theoretical physics

[ivory dragon]

most?

[gritty widget]

like string theory, loop quantum gravity, etc

[ivory dragon]

i mean ok string theory but thats mostly because

string theory is so fucking broad

as a term it covers like 5000 different proposals

you cant falsify them all, but individual formulations can be (and have been) falsified

e.g. various string theories have made predictions about heavy ion collisions

which were later falsified

not that this matters to a mathematician

[gritty widget]

yap

which is why I asked my question in the first place

if there were mistakes in his theory

[ivory dragon]

anyway the thing with string theory is

there's mountains of mathematics behind it

like thousands of papers

weinstein has published 0 papers

and 0 mathematics

and it's not like he's been rejected or covered up or whatever

as far as we can tell, he hasnt even tried

he presented a lecture series in oxford (iirc?) which he probably got via media connections

[gritty widget]

he does have a PhD in mathematical physics from Harvard

[ivory dragon]

and that's the lump sum of what we know

yeah, as i said, he knows actual math

and he did, of course, publish papers during his thesis

he just hasnt published anything concrete from his "geometric unity" model

the string theory literature is full of back-and-forths

someone makes a paper proposing x addition to y model to resolve z problem

some other paper is published bringing up w issue and how simplifying x to x' can help eliminate that

[gritty widget]

sure

[ivory dragon]

another response paper argues that this is overcomplicating it

etcetc

[gritty widget]

I'm guessing his intention was to give ideas to other people to work on because he himself doesn't have enough time to work on it further since he runs a company

[ivory dragon]

anyway there are "fundamental" claims made by string theory

that WOULD falsify it

if disproven

for example, lorentz invariance

or negative cosmological curvature

disprove either of those and you've falsified all of string theory

admittedly, disproving those things is very very hard, but they are concrete claims backed up by mathematics

magnetic monopoles and coupling constant unification also come to mind

as for loop quantum gravity, that also makes falsifiable claims - for example, it predicts different hawking spectra than what we currently believe to be the case

which technically falls within the margin of error but

is very much on the edge

LQG also makes similar predictions to string theory, like lorentz invariance and quantum mechanical universe, etcetc

anyway, my honest advice is to not worry about weinstein's stuff, because there is nothing there to worry about

if you wanted to "study" weinstein's model

you'd be watching like, 8 talks on youtube

and then you'd run out of stuff to do

there's nothing else

on the bright side, i guess this reduces the workload significantly

the reason string theory and similar theories are attractive is because the mathematics works out very, very nicely (this has even provoked accusations of "they're just adding dimensions until the math works", which honestly, kinda has some truth to it). "geometric unity" makes sweeping claims about the mathematics, and it's not even unrealistic to see mathematical connections there

in fact, there are inklings of good ideas present IIRC

but it hasnt been fleshed out enough to really study or respond to

anyway, perhaps you're right that he was just "hoping to encourage others to look into it", but in that case the best thing a hobbyist interested in this stuff could do would be to learn alg geo, diff geo, and modern physics

since those provide the backdrops to weinstein's model

[gritty widget]

apparently Weinstein is on this server

@ivory dragon

also if you listen to his podcasts

he is pretty angry at the academic system in general

and he sees string theory as a completely failure

because it's been around for how long

40-50 years?

and so far it's still that same useless theory

also Weinstein's own theory is falsifiable

but he says he doesn't know how to build a system to prove it

so it's like the string theory situation

[gritty widget]

https://youtu.be/2xiEEtoa-_4

https://www.math.toronto.edu/drorbn/Gallery/KnottedObjects/PlanetHopf/

" What's shown? The pullback to S3 of a (partial) map of (spinning) planet Earth, via the Hopf fibration S3→S2, projected down to R3 by the stereographic projection. Much like the cheese wheel on the left, a wedge-shape of openning 80o is cut out of R3. The whole map of Earth can be seen on both sides of the cut (each in itself, a half-plane), with an 80o rotation between the sides. For improved visibility, the full pullbacks of only the G8 countries are actually shown, and only the pullbacks of Canada, Japan, and Italy are continued through the cut, in a translucent form."

[fading vale]

someone pls cure my monkey brain:

show that a subset A of a topological space X is nowhere dense in X iff any non-empty open U of X has a non-empty open V subset U, with V cap A = null

[wanton marsh]

nowhere dense means what ?

[fading vale]

ah sry, here its defined as int(closure(A)) = null

wait actually

[wanton marsh]

I feel like this is just a case of unfolding the definitions

[fading vale]

i think i may have something

yah i got it

@wanton marsh sorry about that lol

[forest rover]

i dont have a topology question per se

but how does one pronounce the X/A as in the quotient topology

like in abstract algebra that would be X mod A

oh

thanks

so the quotient space D^2/the boundary of D^2 is pronounced D^2 mod the boundary of D^2

[pseudo roost]

how's this not a triangulation ?

[rugged swan]

it is ?

[chilly locust]

can some help me on a geometry test

[pseudo roost]

It isn't my textbook / online notes say its not but not the reason

[rugged swan]

that's weird, I can orient it

[honest narwhal]

Might be one of those things which is a "delta complex" but not a simplicial complex

There are two edges between the same two vertices in the hole.

Officially in a simplicial complex an edge is determined uniquely by its vertices

[pseudo roost]

it's because they meet on two different lines and the union of those lines is not a simplex

[forest rover]

No it's pronounced S^3 lol
@gritty widget actually its S^2

[gentle osprey]

Kogasa:

[unique wyvern]

Is the complement of a relatively open set relatively closed?

[ivory dragon]

by definition.

[unique wyvern]

Really?

But using demorgan's laws you get the union of a closed set with the complement of the base set you're working with?

[ivory dragon]

yes, and

?

ok, maybe its better to ask

how are you defining open sets and closed sets?

[unique wyvern]

set is open iff complement is closed

I'm aware of topological definitions but this is in my MVC course

So im looking at that proof of f is continuous on the domain iff all preimages of open sets are rel open iff all preimages of closed sets are rel closed

and working on equivalence between the second two statements but from what you're saying it's a triviality?

[dim meadow]

Yeah it's a triviality

[unique wyvern]

Ok I think I was assuming that because I had a union involved it couldn't be an intersection of an open set and the domain

which is obvs false

[dim meadow]

Preimage of all opens is open implies preimage of a complement of an open is the complement of an open

And vice versa

[unique wyvern]

Yeah I knew this

Thing is I was going by our definition of relative open sets / relative closed sets

A in U is relatively open if there exists an open set O s.t. A is the intersection of O and U

analagous for relative closed

[dim meadow]

This is just the definition of continuity though

But sure, I think it's obvious that the complement of a relative open set is relative closed and vice versa

I don't see why you're having a problem with this

@unique wyvern

[unique wyvern]

Because by demorgan's laws it's not immediately obvious?

[dim meadow]

Lol

[unique wyvern]

IDK I like to prove stuff in ways that I'll remember

[dim meadow]

Take the open set O in the bigger space, take the complement

[unique wyvern]

Yh

[dim meadow]

You get a closed set in the bigger space which intersects U to give you the complement

(here U is the subset whose topology we are looking at)

[unique wyvern]

Ohhhhh

It's the intersection of the complement with U?

Complement is closed

[dim meadow]

Well you can really have 2 definitions of rel closed sets

One definition is the complement of a rel open set

The other is the intersection of a closed set with your subspace

Both definitions are kind of trivially equivalent

I'm not sure what your definition of rel closed was

[unique wyvern]

The latter

[dim meadow]

Oh ok

[unique wyvern]

That's why I didnt find it immediately obvious

[dim meadow]

In that case what I said shows that the complement of a rel open set is rel closed

But if you know a bit about topology it should have been obvious tbh

[unique wyvern]

I haven't done as much topology as I should have

[dim meadow]

Cause the rel open sets form a topology

Called the subspace topology

Oh ok

[unique wyvern]

Yh I've only recently caught up on definitions of topology

ie indiscrete and discrete topology

So its a rather new concept to me

[gentle osprey]

khaled014z:

[fading vale]

hmmm i'm not entirely sure if this is true but it feels like it should be correct:

let X be a topological space, A a subset of X. then any open set which intersects the closure of A also intersects A

wait nvm im literally dumb lol this is easy to show

its definitional almost

@sweet wing dont me ur mean :(

[warm mirage]

How to approach this:

Show that the space of Weirstrass functions from R to itself is dense in C(R)

Yeah

Thanks

[little radish]

Anyone have a good computational geometry book

possibly focusing on higher dimensional spaces

[gritty widget]

I'm reading Vassiliev's Intro to Topology. In a couple of the problems/examples, Vassiliev refers to a topology on some set of matrices (e.g., $End(\mathbb{R}^2), GL_n(\mathbb{R}), O(n, \mathbb{R})$), but doesn't actually say what the topology is. My first expectation would be that they have the subspace topology inherited from the Euclidean metric on $Mat_n(\mathbb{R}) \cong \mathbb{R}^{n^2}$. Is this correct/reasonable, or is there some other natural topology that I'm missing?

[gentle osprey]

refutations:

[bitter yoke]

That's probably right, but there are a lot of other norms you can put on matrices

Like the operator norm that comes from them being linear operators, or the frobenius norm

But since this is a finite dimensional vector space, all metrics are equivalent so it's not a huge deal

[modern holly]

you mean all norms are equivalent, all metrics are not
but you are right its not a huge deal because equivalent norms generate the same topology, you can prove this rather easily

[hazy token]

Would someone be able to talk with me about the surface classification theorem? I kinda understand a simple proof of it but I wanted to be able to discuss it

[bitter yoke]

[hazy token]

Open in what topology? @bold merlin

Also was that in response to me? Like could I get help from you? @bitter yoke

[bitter yoke]

just ask

my memory is fuzzy but I'll see what I remember and maybe someone else can help too

[hazy token]

Do you know what the definition is of an open set is in the Euclidean topology?

Okk thank you!

So let F be a closed surface (2-manifold). The phrasing of the theorem I’ll try to get to is “F is either the connected sum a sphere with a finite number of tori or with a finite number of real projective planes” @bitter yoke

[hazy token]

Ok so let F be a closed surface (2-manifold). The phrasing of the theorem I’ll try to get to is “F is either the connected sum a sphere with a finite number of tori or with a finite number of real projective planes”

Surgery can be performed along both orientable and non-orientable closed curves:

Let 𝛾 be a non-separating, orientable, closed curve on a closed surface F.

Construct the regular neighborhood N(𝛾 )= 𝛾 x[0,1]. N(𝛾 ) is the belt region of a 2 handle H in E3

Perform surgery on F by cutting out N(𝛾) and gluing in the attaching region of H into F so each of the two disks fill the two holes made in F. Two faces were added, so the Euler number of F increases by 2 after surgery along an orientable curve.

Let Ψ be a non-separating, non-orientable, closed curve on a surface F.

Construct the regular neighborhood N(Ψ )= 𝛾 x[0,1] with the antipodal identification. N(Ψ ) is a möbius strip.

Perform surgery on F by cutting out N(Ψ) and gluing a disk in F to fill the hole made in F. One face was added. So the Euler number of F increases by 1 after surgery along a non-orientable curve.

Consider the following algorithm:

Begin by asking if the surface contains a non-separating closed curve

If yes, perform surgery and then loop back to the first question

If no, then F is S2 by the Jordan curve theorem

The algorithm must terminate in a finite number of steps as performing surgery only increases the Euler number (by 1 or by 2) and S2 has a [finite] Euler number of 2. Therefore every surface is the connected sum of a sphere with a finite number 2 handles and disks replaced by mobius strips.

A disk replaced by a mobius strip is a real projective plane and adding a 2 handle to a sphere creates a torus, so every surface is the connected sum of a sphere and a finite number of tori and real projective planes. The connected sum of a torus and a real projective plane is the connected sum of three real projective planes, so every surface is either the connected sum of a finite number of tori or real projective planes.

Could someone critique my understanding of the proof?

[bitter yoke]

It looks okay to me, but someone better at this stuff should probably take a look

<@&681259184582688842>

[hazy token]

I think it’s correct in principle, but I’m worried about detail stuff

[loud scarab]

whats the difference between a map and its image? for example the differentiable curve $ \alpha : R \mapsto R^{3} $ , say the map alpha is smooth, the curve can have corners, such as the 'cusp' at the origin, and that is not smooth overall, so im confused

[gentle osprey]

khaled014z:

[modern holly]

strictly speaking a curve is the map but often one calls the image a curve
the curve you linked is not smooth because the map is not smooth at the origin

[gritty widget]

curve you linked is not smooth

it can be parametrized by a smooth curve

usually you require the curve to be regular, that is, alpha'(t) != 0 for all t

then you can parametrize it by arclength and then the image does indeed look smooth

[modern holly]

ah yeah my bad it is actually smooth you are right its the regularity

[loud scarab]

i dont get it

can a curve with the derivative being equal to zero be smooth?

[gritty widget]

yes

smooth = has derivatives of all orders

the constant curve is smooth

[loud scarab]

how can I know if a curve is not smooth at a point where the derivative is zero?

like the constant curve has derivative zero everywhere and intuitively it is smooth but according to this paragraph, all the points t are singular

[modern holly]

smoothness doesnt require existence of the tangent line, as sonja said its the regularity (i.e. nowhere singular) that makes the curve "appear" smooth because (in this case the tangent vector varies continuously)
smoothness of the map i dont think you can determine based on the image, you have to check the coordinates are smooth functions, i.e. the map has continuous derivatives of all orders

[loud scarab]

oh my god

so say i have a map that has continuous derivatives of all orders

i conclude that it is smooth

but the image does not 'appear' smooth

the curve has to be regular so that the image has to 'appear' smooth

is that right?

if thats right then why attach a meaningless 'smoothness' criterion for the map itself and not the image

since the image determines the actual geometric structure?

[modern holly]

the smoothness criterion is more important its just unfortunate for intuition that an image of a smooth map can appear not smooth
we cant really make a complete theory out of "smooth images" but out of smooth maps we can
also one point to make is that when you compare to maps from R to R, you have this intuition that the graph appears smooth, this intuition carries over
if you graph on R^3 the graph of the map t -> (t^2, t^3) that you linked the image of earlier you get a smooth looking curve even if the image has a cusp at the origin

compare to your earlier image

[loud scarab]

how do you graph that on R3?

[modern holly]

graph is defined as the set of points (t, f(t))

so in this case you graph (t, t^2, t^3)

[loud scarab]

oh so just treat f(t) as 2 separate numbers

okay

i will try that out

thank you for your help

[modern holly]

no problem

[thin trellis]

I have two 3-sphere hyperspherical coordinates as in (r,ψ, θ, φ) and I want to find the great circle distance between them. I understand I have to derive this from the law of cosines, but I'm kind of stumped. Anyone want to help?

In my case r is always 1, unit length.

[wanton marsh]

I would just convert to cartesian coordinates and measure the angle between the two points from the origin

[gritty widget]

so i'm trying to prove that every point of an open set $E$ in $\mathbb{R}^2$ is a limit point of $E$. Since $E$ is open, every $x\in E$ has some $\varepsilon > 0$ so that $B^\varepsilon (x)\subset E$. Take $\delta>\varepsilon$. Then $B\varepsilon^ (x)\subset B^\delta(x)$, so $B\delta^ (x)\cap E \neq 0$. Take $0>\delta>\varepsilon$. Then $B_\delta^* (x)\subset B_\varepsilon^* (x)$, so $B_\delta^* (x)\cap E \neq 0$.

[gentle osprey]

mart:

[gritty widget]

is this enough?

[chrome dew]

0 > delta > epsilon looks suspicious

what's your definition of limit point

like open ball of arbitrary size around the point always contains another point

[gritty widget]

$x$ is a limit point of $E$ if $\forall \varepsilon > 0: B_\varepsilon^* (x)\cap E\neq 0$.

[gentle osprey]

mart:

[chrome dew]

what's * mean, open ball of radius epsilon I assume

[gritty widget]

punctured epsilon ball around x

[chrome dew]

oh ok

I don't get what you're doing, seems like you do too much

Once you start taking delta > epsilon is where I get lost, why not just stop there and use B*_epsilon

what's the extra larger punctured ball for?

@gritty widget you still there?

[gritty widget]

sorry omg

so i know that each $x\in E$ is an interior point of $E$, that means there exists some $\varepsilon>0$ so that $B^\varepsilon (x)\subset E$. Then I have to show that $x$ is a limit point of $E$, that is \textbf{for every} $\delta > 0, B\delta^ (x)\cap E\neq 0$.

[gentle osprey]

mart:

[gritty widget]

@chrome dew

[chrome dew]

yeah set epsilon = delta

[gritty widget]

what about when delta > epsilon

[chrome dew]

well the intersection of those two would be the smaller one

so that's why I ignored it

[gritty widget]

ok ty n.n

[chrome dew]

I guess I'm taking the "there exists some epsilon" to mean

there's one fixed one, and we get all smaller epsilon for free

but you're making delta do that separately

maybe I'm just sloppy about how I'm doing it lol

do you also have to prove or explain that the punctured disks are nonempty

or is that just obvious in R^2 that it doesn't matter

I guess literally average the punctured point and a point on the radius if you need to show the specific existence of a point

[sonic totem]

why not just go epsilon/2 in any direction (say, up, or to the right) from x

oh, this is a really old discussion

[snow shadow]

Hi, im wondering if someone could help me understand and solve some problems in differential topology?

I have a 2-form defined on R^2-{origo} and a two dimensional manifold in R^3 that is an oriented, im suppose to show that the restriction of this form to my manifold is equal to a given expression. However we are given a hint and it feels like the hint is the entire solution, so i dont get it.

[pine heath]

Can someone explain how donut equal ONE COFFIS CUP.

[ivory dragon]

question of our times

[bitter yoke]

Almost as good as "Where will Estonia go with category theory?"

[ionic locust]

I think he truly do be wanting to know how that donut be a coffis cup though.

[bitter yoke]

I forgot where namington found this

[gritty widget]

lol

it was some cranky medium article

about how functional programming "got it wrong"

[sweet wing]

[sleek thicket]

I found it on Twitter

Posted it in general

[warm mirage]

Any good books on synthetic geometry?

[steel trellis]

try Elementary Synthetic Geometry (Classic Reprint)?

[loud scarab]

what is this?

does he mean that e1 is tangent to the line v=v0?

so e1 has the same direction of the line v=v0 , so it is tangent

or is it something more

[warm mirage]

NotWeird thanks

[pine heath]

No one explained how donut=one coffee please so I just am confused about.

[warm mirage]

Hint: imagine they're made of rubber

[limpid pine]

Ahoy. Let [A \in \mathbb{R}^{3\times3}] be nonsingular and [E = {y | Ax = y , x^Tx = 1}]. Im trying to prove E is an ellipsoid and I'd like a hint. I know I can do it myself I just seem to be lacking good ideas.

[gentle osprey]

RamJam413:

[limpid pine]

Basically A(S^2) = E is an ellipsoid. I wanna use x^Tx = 1 and get the formula for an ellipse out of it but it isnt clicking

[small obsidian]

x^Tx
Is the dot product of x with itself and can instead the condition can be expressed as:
|x|² = 1

x is a vector on the unit sphere, Ax is a linear deformation of the unit sphere @limpid pine

If instead A were a 2×2, that would instead be a linear deformation on the unit circle

[limpid pine]

Right. I believe I am on board with everything you've just described. I even intuitively understand why the ellipsoid's principal axis are scaled by the singular values and in the direction of i believe the left singular vectors. I just cant seem to get the formula for an ellipse out of this information thus far

[chrome dew]

can you get the equation of a circle

[hot holly]

is it true that a closed curve with rotational index of -1 has to be simple?

[gritty widget]

@hot holly

[hot holly]

ahhh haha thank you, i just got to the conclusion that its not true

appreciate it ❤️

[knotty pasture]

Hellooo

I need a help for an exercise of differential topoligy

I have to prove that every n-smoth manifold is contained in a n+1 oriented smooth manifold

Help me please xD

[pine heath]

Look I am friendly I swear

Im just want to know why coffee cup = one doughnis

I ned one explain form topologist

[knotty pasture]

Im just want to know why coffee cup = one doughnis
@pine heath Because this 2 object as the topological point of view is the same thing (have the same homology group)

[pine heath]

Thank you.

[loud scarab]

Hello, in showing that a sphere is a regular surface, we divided it into 6 parts to cover the sphere, so we showed that each part is a regular surface, but how do we know that by covering the sphere with each individual map that the sphere is a whole is a regular surface? Is the union of regular surfaces a regular surface?

[dim meadow]

@knotty pasture no

[loud scarab]

An example of the map x and the set U

[knotty pasture]

@knotty pasture no
@dim meadow sorry, what no?

[dim meadow]

what you said is bad

[knotty pasture]

what you said is bad
@dim meadow ?

[dim meadow]

having the same homology groups is not the same as the same from a topological pov

[loud scarab]

@pine heath please ask a more detailed question or watch an example on youtube, i am friendly too i swear.

[dim meadow]

@loud scarab the definition of a regular surface is that around each point there is an open nbd with certain properties

so in order to show a surface is regular you have to cover it with regular patches

[loud scarab]

open in the standard topology of R3 am I correct?

[dim meadow]

no

open in your manifold

yeah in this case open in the subspace topology

[loud scarab]

hmm I still dont know what a manifold is unfortunately

but what does 'cover' it with regular patches mean

where is this piece in the definiton

[dim meadow]

that the union of the regular patches is the whole surface

[loud scarab]

im trying to show that the union of the regular patches == regular surface

[dim meadow]

no

the definition of regular surface

is it is covered by regular patches

[loud scarab]

but then what is 'regular' patches

[dim meadow]

oh

[loud scarab]

cuz im using something not defined to define something that needs that something

[dim meadow]

you're trying to show that the union of the patches you have is S^2?

[loud scarab]

yes

well

the union is obviously S2

im trying to show that S2 is regular

by combining the patches

each patch is regular

by the definition

what gives me the right to combine the patches and claim that the union is regular

am I making sense?

I used this to prove that each patch I have is regular

[dim meadow]

lmao

[loud scarab]

x given its properties and the differential ofcourse

[dim meadow]

so you are basically done then

[loud scarab]

oh okay im retarded

[dim meadow]

yes

[loud scarab]

i just noticed where i was confused

well its not cool to call me retarded

[dim meadow]

lmao

[loud scarab]

in my opinion, i am retarded

[knotty pasture]

having the same homology groups is not the same as the same from a topological pov
@dim meadow yes, you're right, it was a mistake born from my wrong idea. You have an example of a pair of topological space with the same H_n sequences that aren't isomorphic? And in the category of smoth manifolds?

I have to prove that every n-smoth manifold is contained in a n+1 oriented smooth manifold
@knotty pasture i up my previous question

[knotty pasture]

eheheh yesss

BD

[knotty pasture]

and this? "I have to prove that every n-smoth manifold is contained in a n+1 oriented smooth manifold" Nobody have idea?

[jaunty nebula]

I'm supposed to prove that this is continuous. The example I'm being given is of a proog that f(x) = sqrt(x) is continuous and it's proven with the use of I_a and I_b. I fail to see the connection here

[jaunty nebula]

Ah I think I got it.

[limpid pine]

Hello. Still struggling with the damn ellipsoid. Here is my thinking thus far.

Let (A \in \mathbb{R}^{3\times 3} ) be nonsingular and (E = {y| y = Ax , x^Tx = 1}). So E is the ellipsoid of A, the image of the unit circle under A.

The SVD of A, (A = P\Sigma Q^T\ = p_1\sigma_1q_1^T + ... + p_3\sigma_3q_3^T) gives me (Ax = p_1\sigma_1q_1^Tx_1 + ... + p_3\sigma_3q_3^Tx_3)

and thus ((Ax)^TAx = q_1p_1^Tp_1\sigma_1^2q_1^Tx_1^2 + ... + q_3p_3^Tp_3\sigma_3^2q_3^Tx_3^2 = \sigma_1^2x_1^2 + \sigma_2^2x_2^2 + \sigma_3^2x_3^2)

Which I can feel is really close, but I need to scale it equal to 1, and I probably made some mistake that pulls the bottom out somewhere, and idk. Does someone know where to go from here?

[gentle osprey]

RamJam413:

[limpid pine]

@small obsidian
I know
(x_1^2 + x_2^2 + x_3^2 = x^Tx = 1) and (\sigma_1^2x_1^2 + \sigma_2^2x_2^2 + \sigma_3^2x_3^2 = (Ax)^TAx). I wanna say ((Ax)^TAx = |A|_2^2) because x is unit length but from there I am just not able to click it into the ellipsoid formula. I must be missing something obvious

[gentle osprey]

RamJam413:

[loud scarab]

so in the definition of a regular surface, there's a condition that the parametrization be a homeomorphism between an open set in R2 and an open set of R3 intersected with the surface

and in order for x to be a homeomorphism, not only does it have to be one-to-one, but also onto, but how is it in this case x: U --> R3 is onto? if I pick an arbitrary point of R3, it certainly need not be an image of x, right?

I highlighted the part where he mentions an inverse without first checking that x has an inverse, how does it follow here that x has an inverse?

The only way I can see this resolved is that if x is an imbedding of U in R3

not sure if that's the case though.

[midnight jewel]

it needs to be surjective not onto that open set, but onto the open set intersected with the surface

the inverse is also only defined on that surface

[loud scarab]

oh well then I guess the notation should be $x: U \mapsto V \cap S$ where V is open in $\mathbb{R^{3}}$ and $S$ is the surface so now surjectivity makes sense

[gentle osprey]

khaled014z:

[digital anchor]

Hello all i have a qeustion

I need to find 24 points evenly spaced around a circle where the ceneter of the circle is 0,0 and the radius is 33

I have no idea how to do this im no math man

[sleek thicket]

@digital anchor this isn't an appropriate place to post this, go to #geometry-and-trigonometry

[digital anchor]

ah

[unreal mist]

Is anybody familiar with Geometric Algebra (clifford) ? I've read literally every single google/duckduckgo result on Geometric Algebra and Clifford Algebra and I still don't get it

[bitter yoke]

Just ask about what you're confused about

[unreal mist]

Well some basic things. Can you divide by non scalers/pseudoscalers, Ect can you divide by Vectors/Bivectors in say (3,0) space (or pick a space that would be easier)?

No I think you can now that I think about it. It's just the inverse of multiplication. I got some better questions

Can you take the square root of Vector/Bivector in say (3,0) space. Even if it won't be unique . Can you even take non integer powers? Or raise to a non scaler?

[white wren]

dude i trully want to help you but i dont understand i am not used to do mathematics in english sorry

does what you want to know have matrices?

[unreal mist]

No. Geometric Algebra is funny in that Matrices are totally optional.

I've read literally every single resource on this. And I can't find anything confirming or denying the square roots thing.

It's supposed to be an alternative framework.

[white wren]

sorry i dont see. maybe i wil after few days

i tryed

[unreal mist]

I'm not expecting you to know but I'm hoping someone would be familiar with the subject.

[bitter yoke]

<@&681259184582688842>

[unreal mist]

I have no idea what this even has to do with Topology besides being requested to post here

[white wren]

@bitter yoke i study early topology and i dont see anything related to it

all i can seeis the space is maybe a NORME (it's what we call it in my country)

[bitter yoke]

This is the topology and geometry channel, we don't have a geometry tag, but most people who have the topology tag know geometry too

[white wren]

i should study geometry now but with coronavirus...

[unreal mist]

Will a problem ever call for taking a square root of something I can't take?

If so, what do I do then?

[ivory dragon]

[if you are, you can explain why it's "impossible" i guess]

[i.e. "noninteger powers are not defined on this algebra because suchandsuchreason" or whatever]

[but such questions will usually be phrased as "If possible, compute ...." or whatever]

[white wren]

i dont think at this lvl you wll have this kind of a situation

[unreal mist]

But wait isn't using Euler's formula a 'non-integer' power. And isn't its use necessary for Geometric (clifford) Algebra?

[ivory dragon]

what?

i mean, yes, but that'll be in contexts where it makes sense

[unreal mist]

Where exactly would it make sense?

[ivory dragon]

on algebras where it's defined?

that is, C

[unreal mist]

Anywhere I can find more resources on clifford algebras that isn't just a google search away (Since I have everything there)

where's all the so called 'math material' anyway all I find is 'physics materials'

[ivory dragon]

well, clifford algebras arent a particularly popular subject amongst mathematicians - they're often considered "full of unnecessary structure"

[chrome dew]

you originally said you were stuck at learning line integrals and started trying to learn clifford algebra to get a second perspective

[ivory dragon]

that said, there are some math-focused texts

[chrome dew]

I suggest not doing that

[ivory dragon]

wait what the fuck

oh god dont

[chrome dew]

lol

[ivory dragon]

i was about to recommend Porteous

lmao

[chrome dew]

just focus on learning multivariable calculus and ask questions on that in #multivariable-calculus instead

I think that's more the level you're at right now

[unreal mist]

Porteous? What level is Clifford Algebra?

[ivory dragon]

graduate level

or at least upper undergrad

[chrome dew]

it's like, not very mainstream version of differential geometry and tensor calculus sorta stuff that you're not ready for yet

[ivory dragon]

most mathematical resources i'm aware of assume you know basic diff geo

and a couple semesters of group theory

universal algebra is helpful too

[white wren]

you mean major maths lvl ?

[unreal mist]

But Hestenes says he wants to teach it to people right away.

[ivory dragon]

hestenes' viewpoint is very fringe

it's not necessarily wrong

but it doesnt have widespread adoption

in any case, he proposes it as a replacement for linear algebra

but in particular, the linear algebraic parts are

bsasically unchanged

maybe some words are swapped but thats about it

so it wouldnt really present an "alternate perspective"

if youre just working with calculus over R^n and C^n lmao

[chrome dew]

like using quaternions is cute, but it's kind of a waste of time when you have linear algebra

it's just more cumbersome in my experience trying to use any clifford algebra

no real reason to use it as far as I'm aware, but I'm pretty ignorant

[ivory dragon]

anyway, instead of trying to learn a 2-semester course in pure mathematics to eventually develop enough theory to maybe gain an "alternate perspective" that may or may not be insightful and doesn't really have many resources available (nor, in all likelihood, acceptance by your prof or TAs)

why not just practice more line integrals?

[chrome dew]

I will say I think there is an appeal to thinking of the whole wedge product/bivectors thing instead of cross products intuitively speaking

[unreal mist]

That too.