point-set-topology

[gritty widget]

discuss how geodesics

[lean marten]

shortest

[sleek thicket]

locally shortest

[pearl holly]

So I am trying to show that B = {(a, b) | a<b, a and b are rational} is a basis for a topology that generates the standard topology on R. The exercise wants me show this using the following lemma

[gritty widget]

NOOO DONT WRITE ON THE BOOK

[bleak helm]

Yesss write on the book

[fathom cave]

just use a property of rational numbers 🤔

[pearl holly]

So I want to show that for each real x there exists (a, b), both rational, such that x in (a, b) subset (a', b') where a' and b' are real. Can I argue like this: There are uncountable number of irrational points in this interval but there are countable number of rational points. This means that I could always find this (a, b) interval for both rational a, b

[pearl holly]

It's my math teachers book, I didn't write that I just borrowed the book

[fathom cave]

u wana show for each open set U and each x in U, there is a set of the form (a,b) where a,b /in Q

[pearl holly]

yeah so can I argue like I did?

Because in this case the open sets are (a', b'), both real. Right? Since we are looking at the standard topology

[fathom cave]

i mean the main point is, ur always going to find a rationl pair a,b but ur reasoning doesnot sound convincing

[pearl holly]

yeah I know, that's why I came here

[fathom cave]

given two numbers, there exists a rational number between two,.. should work

[pearl holly]

Between two irrational points?

[fathom cave]

any

[pearl holly]

Is that it? Don't I need to prove it or something?

[fathom cave]

ur using a property
i dont think u will need to prove why this is true
u do this in real analysis

[strong heron]

But you should prove it to yourself at least once

[pearl holly]

Would this work as a proof?

or is it not rigorous enought?

Actually, I think that it doesn't prove anything

[fathom cave]

how does a set being countable guarantte ur always going to find a rational number?

right

set of integer is countable aswell

[strong heron]

Yeah, infinity arguments are always subtle

Have you heard of something called Archimedean property?

[pearl holly]

No. What is that?

[strong heron]

For every positive x, there is a natural number n such that nx>1

[pearl holly]

Well that sound trivial... But it probably took mathematicians 1000 years to prove lmao

[fathom cave]

ya it does sound trvial but the proof is a bit involved

[strong heron]

Yes, it will however help you prove that rationals are dense in reals

[pearl holly]

Okay I will try to use it in some way, but what is "dense"?

[strong heron]

It just means that between every two reals, you have a rational

[pearl holly]

Oh okay! Thank you so much!

[strong heron]

Did you understand how should you start?

[pearl holly]

Well I don't know how to start but I know that I should use the property

[strong heron]

Okay. Let x and y be two arbitrary reals.

Without loss of generality, you could say x<y, right?

[pearl holly]

yes

[strong heron]

Then you have y-x>0

Now use Archimedean

So you get a natural n s.t. n(y-x)>1. Fine?

[pearl holly]

yes

[strong heron]

Open brackets to see this means ny-nx>1

When two real numbers are more than a unit distance apart, what can you say from this info?

[pearl holly]

That there's some point between them, right?

[strong heron]

Some point is always there since they're not equal. What kind of point?

[pearl holly]

Oh, (ny-nx)/2 must be in between, right?

[strong heron]

That's true, but not really helpful. Do you see there will be an integer between them?

[pearl holly]

yes

[strong heron]

Okay, great. So we have there exists m integer s.t. nx<m<ny

Right?

[pearl holly]

yes

[strong heron]

Divide by n throughout. Notice n is positive, so inequality remains preserved.

[pearl holly]

ohhh

[strong heron]

What do you get?

[pearl holly]

I get that x<m/n<y and m/n is rational

[strong heron]

Precisely!

[pearl holly]

Ohhh okay I get it now. Thank you so much!

[strong heron]

No problem! Yeah, this proof is a nice consequence of Archimedean. I remember being quite delighted when I first understood it.

[pearl holly]

Yeah the proof is quite elegant!

[empty grove]

Did you end up figuring it out?

I just read some theory on this so I might be wrong, but the fundamental group of that is Z/2Z * Z/2Z which seems to have exactly 1 index 3 subgroup, and 3 sheeted coverings should correspond to index 3 subgroups of the fundamental group

[strong heron]

Yes, I got just one cover also. But if you take covers with basepoints, perhaps you get 3. That's the part where I'm not too sure.

Upto homeomorphism, there's just one cover. No doubt about that.

[empty grove]

I think my approach gives covers with basepoints

Because if I'm not wrong, index 3 subgroups correspond to 3 sheeted pointed coverings and conjugacy classes of index 3 subgroups correspond to 3 sheeted coverings

If we have 1 index 3 subgroup then both of those things are 1

[lament anvil]

Suppose im working with a tangent surface, and when calculating its principle curves on the tangent surface, and i calculated the principle directions as (1,-1) and (0,1) or Xt-Xs and Xs what does that then make the principle curves?

[strong heron]

True, but the index 3-subgroup that you got, is it normal?

What is the subgroup that you found actually?

[empty grove]

Writing the group as <a,b | a²=1, b²=1>, it is the subgroup <a, ababab>

I think it's normal but I didn't try proving that is index not preserved under conjugation for infinite groups?

[empty grove]

Or if you view it as the infinite dihedral group <r,s | r²=1, rsrs = 1> via isomorphism r → a, s → ab, the subgroup is <r,s³> which I found easier to work with

Wait <rs, s³> and <rs², s³> should also be index 3

And are distinct I think

And I think these 2 are conjugates but <r,s³> is normal?

Guess I need to work this out again lol

Because that seems wrong

Oohh all three are conjugates

[empty grove]

So to summarise, 3 index 3 subgroups, given by
< a, ababab >
< b, ababab >
< bab, ababab>
and all 3 are conjugates (and have no other conjugates)

So there should be 3 pointed coverings and 1 covering upto isomorphism

[strong heron]

Precisely!

Great, so this confirms my analysis as well. Thanks!

[empty grove]

[strong heron]

[gritty widget]

A lie algebra is a vector space with a lie bracket, pretty much that means we require
[x,y]+[y,x]=0
[x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0
Do we get any interesting structure or use out of extending this to higher cycles?
[x,[y,[z,w]]] + ... =0 for example

[summer jolt]

Hey, in this question am I correct in deducing that the fibre is just S^1?

Since any neighbourhood of inverse q(C{0}) will be homeomorphic to C{0} x S^1 ?

[wanton marsh]

no

[summer jolt]

Why?

[wanton marsh]

Y isn't C\ {0} x S1

the fiber of a point is just the preimage of that point

I'm not sure why you're talking about neighbourhoods

[summer jolt]

Hmm ok I see, I'm having trouble visualising Y. I don't quite get what y/|y| = z^2 intuitively means

[wanton marsh]

I assume that here S1 = {z in C | |z|=1 }

[gritty widget]

Draw the complex plane

Mark a point

And try and see what y/|y| means

[summer jolt]

Ok my guess would be that is two lines in the complex plane

The line z = +1 and z= -1

[wanton marsh]

pick your favourite complex number y, then solve for z in the equation y/|y| = z²

z should be in S1

that means |z| = 1

[summer jolt]

oh ok so the first z is not the same as the last z in z^2?

[wanton marsh]

the first z ?

they are the same z

[gritty widget]

I think you are getting confused from trying to do too much at once

[summer jolt]

because I though y and z were real here. The question mentions $(y,z) \in C{0}$

[gentle osprey]

snypehype
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[wanton marsh]

no y and z are complex numbers

[gritty widget]

Try and understand what Y is saying before you do the question

I don’t mean this in a cheeky way

[wanton marsh]

y is in C\ {0} and z is in S1

[gritty widget]

But sometimes it’s easy to over look it

[wanton marsh]

(y,z) is in the cartesian product (C\ {0}) x S1

[summer jolt]

oh ok that's what I misinterpreted I thought that they meant (y,z) in complex plane

Ok so $\frac{y}{|y|}$ is just the unit circle in the complex plane right?

[gentle osprey]

snypehype

[wanton marsh]

I'm not sure what you mean

[summer jolt]

I mean that $\frac{y}{|y|}$ just is the set of points intersecting the unit circle and lines through the origin

[gentle osprey]

snypehype

[wanton marsh]

y/|y| is a complex number

[summer jolt]

but isn't $\frac{r \exp(i \theta)}{r} = \exp(i \theta)$?

[gentle osprey]

snypehype

[wanton marsh]

yes ?

[strong heron]

Consider $z=e^{i\theta}$. Then your $y$ is a complex number such that $\frac{y}{|y|}=e^{2i\theta}$.

[gentle osprey]

whysee

whysee

[wanton marsh]

no ?

to find q-1(y) you have to solve for z in the equation y/|y| = z²

[strong heron]

Oops, you're right @wanton marsh, I didn't see the map is from Y.

[tall gale]

Hey

I have set of points in complex plane (sequence) $ (Z_1, Z_2, Z_3, \ldots )$
Therefore sequence to converge in Z I have condition:
$$|Z_n - Z| < \epsilon$$.
I am reading Isham - Modern differential geometry for physicists , it states that:
$$\text{tails } T_n := { Z_k | K>n}$$
for all $\epsilon > 0$ there exist $n_0$ such that $T_{n_o} \subset B_{\epsilon}(z)$ . where $B_{\epsilon}(z) = {Z' \in C | |Z - Z'| < \epsilon}$

My problem : let n is 7 and let ${Z_7, Z_{10}, Z_{11}, \ldots}$ inside the disk $B_{\epsilon}(z)$ and $Z_8, Z_9$ outside the disk , then $T_7 := { Z_k | K>7}$. Hence it looks like $T_{n_o} \not\subset B_{\epsilon}(z)$. But is not it contradictory to the definition of tails.

[gentle osprey]

Aster Kleel

[strong heron]

$q^{-1}(y)={(y,z): \frac{y}{|y|}=z^2}$

[gentle osprey]

whysee

[summer jolt]

Perhaps I'm being obtuse here, but my reasoning was y/|y| represented every complex number rescaled so that they have modulus 1

[summer jolt]

Ok so to solve this eqn I can set $\exp(i\theta) = \exp(i2\psi)$

[gentle osprey]

snypehype

[digital peak]

how to show that a killing form of a nilpotent lie algebra is zero

[sleek thicket]

The trace of a nilpotent matrix is zero

Bc all of the eigenvalues are 0

So you just need to know ad(x) ° ad(y) is nilpotent

But this follows from the lie algebra being nilpotent

[empty grove]

Definition says that there is some tail T_(n0) in B_ε(x). Here you proved that T_7 is not in that ball, but there could be some other tail that is, for example T_10 will work

[bright acorn]

How hard is it to prove in general that the tangent bundle of a smooth manifold (in this case I mean C^infty) is trivial?

I get the example of why TS^1 = S^1 × R

It I would like to see more

And if there's an equivalence between saying the tangent bundle of a manifold is trivial and something else that makes it easier to prove

[sleek thicket]

Yes

There are easy to compute things which tell you if the bundle isn't trivial

Eg the stiefel whitney classes

But they don't give you an affirmative "yes this is trivial", it just gives a potential way to show nontriviality

[bright acorn]

Hmm

That's cool

I will take a look into those

[sleek thicket]

This post is good: https://math.stackexchange.com/a/3114539/408287

for a smooth manifold embedded in R^n, there's an explicit way to write down a classifying map for the tangent bundle

and so it might be possible to show the bundle is trivial by thinking about whether that map is null homotopic

but whether a map is null homotopic or not is still hard

[bright acorn]

Hmmmm

So like

This is somehow the idea behind Borsuk-Ulam and the reason why S^n for n≥2 doesn't have a trivial tangent bundle, right?

[sleek thicket]

That's not true, S^3 and S^7 have trivial tangent bundle

[bright acorn]

Oh

That's weird then

Is it even known

All spheres that have trivial tangent bundle then?

[sleek thicket]

yup, it's only in dim 0,1,3,7

this is connected to the fact that the real line (dim 0+1), complex plane (dim 1+1), quaternions (dim 3+1), and octonions (dim 7+1) are the only "normed division algebras" over R

I don't actually know the proof that odd-dim spheres outside of dimension 1,3,7 are not parallelizable, I think it relies on k theory. I've been meaning to learn it for a while. that even dim spheres aren't parallelizable follows from the hedgehog theorem (there is no nonvanishing vector field on an even dim sphere)

[bright acorn]

Yeah

Hm

That's interesting

So

I actually am quite interested about knowing exactly what K-theory is about

All I know is that it somehow

Is linked to the study of fiber bundles and generalizes some homology theories

[sleek thicket]

Yup, more specifically vector bundles

[bright acorn]

Yeah, but I don't know exactly how it is done

Hatcher has a book on it

I intend on reading it after learning better algebraic topology

[sleek thicket]

sure, so the easiest case is the 0th K group of a space

space here means like, CW complex or something

paracompact hausdorff space

so you want to think about either real or complex vector bundles

you take all of these bundles together

there's a way to "sum" two of them together

the whitney sum

have you seen this?

[bright acorn]

Yeah

If you have two vector bundles on a topological space X

You construct a new vector bundle

Whose fibers

Are given by the direct sum

Of the fibers of those other two vector bundles

[sleek thicket]

right

so this sort of makes the set of vector bundles on X into a monoid

(monoid = group but you might not have inverses)

but there's two problems with this statement

the first is that there isn't a set of vector bundles, there are too many of them. the second is that the operation isn't literally associative

if $A, B, C$ are vector bundles on $X$ then $A \oplus (B \oplus C) \neq (A \oplus B) \oplus C$, but they are isomorphic

[gentle osprey]

shamrock

[sleek thicket]

similar two how the direct sum of vector spaces or cartesian product of sets isn't literally associative, it's just associative up to isomorphism

[bright acorn]

So like

We could try to define the sum on the space of all vector bundled on X in the same spirit as we define the product on a loop space of X at a fixed point x_0, in the later case it would be associative up to homotopy. But in this case we are trying to study, it would be nice if we considered this operation up to an isomorphism.

Is that the idea?

[sleek thicket]

yeah!!!!

I had already written out: we can solve both of these problems by considering the set of *isomorphism classes* of vector bundles. it might not be clear that this is a set, but it is. the idea is that every vector bundle is determined up to isomorphism by (1) a cover of the space X on which it is trivial and (2) the transition functions from one trivialization to another

[bright acorn]

Lmao

Hmmm

[sleek thicket]

but I'm glad you saw the idea too

[bright acorn]

So I am familiar with this idea

Nice

[sleek thicket]

(another thought would be to look at it as a monoidal category, we're not doing that)

but yeah so we can associate to every space X a monoid Bun(X) of isomorphism classes of vector bundles with the whitney sum

[bright acorn]

I have studied a bit of that

But like

I always forget the formal definition lmao. But the idea I have in mind is that monoidal categories are categories equipped with a tensor product that is associative up to an isomorphism.

[sleek thicket]

yeah, same idea

[bright acorn]

I have studied that a bit in commutative algebra

[sleek thicket]

actually there is another monoidal structure on the category of vector bundles using the tensor product

but this one would be the whitney sum instead

but anyways

not going to talk about it rn

[bright acorn]

Hmmm

[sleek thicket]

so Bun(X) is not the K group

[bright acorn]

So the 0-th...

Oh ok lmao

[sleek thicket]

sorry yeah I should clarify

it's not the 0th K group

:^)

but yeah why is that?

well it's not (usually) a group!

if you think about just a point

What's Bun(pt)?

[bright acorn]

I think it should be just all products {p} × X where X is a topological space in this case, right?

[sleek thicket]

Not quite, we're only considering vector bundles

[bright acorn]

Yeah

[sleek thicket]

but it seems like you're thinking "all bundles are trivial"

right?

[bright acorn]

Yeah

[sleek thicket]

so it would be (up to isomorphism) all the trivial vector bundles R^n

or C^n

[bright acorn]

We are considering only finite dimensional vector spaces, right?

[sleek thicket]

Yup, good point

so what monoid is this? like, up to isomorphism?

You've seen it before, I promise

[bright acorn]

The 0 monoid

It has only one equivalence class

[sleek thicket]

You're saying all trivial vector bundles are isomorphic?

[bright acorn]

I guess it should be

Oh shit

No

R^n

[sleek thicket]

Yeah

[bright acorn]

It should be isomorphic to R^n or C^n

Depending if we are talking about R finite dimensional vector spaces over R or C

[sleek thicket]

Yeah so just to be clear, we only ever consider either R^n or C^n

not both at the same time

[bright acorn]

Because the equivalence classes

[sleek thicket]

so what's the monoid?

you're saying it's R^n or C^n?

[bright acorn]

Would be just the vector spaces of a fixed dimension

[sleek thicket]

for what n?

so the isomorphism classes of bundles over a point are like R^0, R^1, R^2, ...

right?

(or C^0, C^1, C^2, ...)

[bright acorn]

It would be R^infty yeah lmao

What I meant really is

[sleek thicket]

But there's only countably many elements

[bright acorn]

The equivalence classes can be identified to R^n for every n

[sleek thicket]

Right

[bright acorn]

And this monoid

Could be identified

[sleek thicket]

but R^infty has way more points than that

[bright acorn]

To the direct sum

Of R^n for n≥1

Is that it?

[sleek thicket]

Nope

[honest terrace]

as shamrock already said you only have countably many elements

the equivalence classes are (R^0, R^1, R^2, ...) and that keeps going like that

[bright acorn]

Oh my god

N?

[sleek thicket]

yup!

[bright acorn]

The natural numbers

[honest terrace]

Yeah

[sleek thicket]

R^n (+) R^m = R^(n+m)

[bright acorn]

Lmao

Idk how I didn't see that coming

[sleek thicket]

so the point of this example was

Bun(pt) = N is not a group

[bright acorn]

So you consider the Grothendieck group of N

Right?

[sleek thicket]

oh yeah

where have you seen the grothendieck group before?

Anyways yeah

K0(X) is the grothendieck group of the category of vector bundles on X

[honest terrace]

grothendieck group of N is just a fancy word for Z

[bright acorn]

Yeah

[sleek thicket]

Ie the groupification of the monoid of iso classes of bundles on X

[bright acorn]

I would say that nut my cellphone just turned off rip

Anyway

I have seen the Grothendieck group

Reading a but of hatcher's book

On K-theory

[sleek thicket]

nice!

So this is the easy part of defining K theory

[bright acorn]

But I didn't really get much...

[honest terrace]

oh btw welcome back shamrock 😄

[bright acorn]

Btw

This is weird because

I still have a lot to learn about algebraic topology

Idk why I would be reading about K-theory in the first place

For instance

[honest terrace]

(I'd love to be able to say that, that would mean I atleast know some algebraic topology )

[bright acorn]

I still don't really have a good intuition of what the singular (co)homology groups with integers coefficients of a topological space mean.

I know how to define them

Etc

Even how to compute some

[sleek thicket]

Yeah

[bright acorn]

But I still don't have a good intuition of what they mean really

[sleek thicket]

it's tricky

[bright acorn]

And they are

Theoretically

The simplest (co)homology theory

There is

That and simplical (co)homology for CW complexes

So I still need to get an intuition of these things

I am still at that point where I am getting confident with the definitions and so on

[bright acorn]

Btw

[honest terrace]

I'm still at the point where I'm trying to dive into some AT book and not give up a day later because it's too hard for me

[bright acorn]

Would it be much if you tried to give me a more visual intuition of what the singular (co)homology groups are?

For instance

[sleek thicket]

¯_(ツ)_/¯

I think of homology in terms of simplicial homology

singular homology is a generalization where you use as much data as possible

and miraculously it coincides with simplicial homology

when both are well defined

[bright acorn]

I know that the n-th Homotopy groups of a topological space at a base point are just homotopy classes of loops φ : S^n -> X where the operation is given in a similar way as the product of paths

[sleek thicket]

sure

[bright acorn]

And this is kind of intuitive for me

I still have no ideia of what the operation on the n-th homology group should mean tbh lmao

[sleek thicket]

ah well

that's sort of because the operation is completely formal

like

you start out with a set of chains and just formally turn it into a group

the interesting thing is that when you pass to homology, weird algebraic relations show up

hahaha

but the operation is still kind of just formal

it's that you've imposed a bunch of weird relations on top of that

idk

lmao

I guess I think of it as like, the boundary of a simplex is sort of like a union of lower dim simplices with marked orientation

this is sort of like an alternating sum

so you impose enough algebraic structure to write down this alternating sum

[empty grove]

If you have 2 simplices a and b, then you can think of a+b as the map you get from disjoint union of the standard simplex with itself, to the space

really just viewing it as 2 simplices but together lol

[bright acorn]

I think that I just gotta get used to it lmao

And turns out even more abstract (co)homology theories are ahead of me

Like

Why

[empty grove]

[sleek thicket]

at least we have de rham cohomology

beautifully geometric

[bright acorn]

This one is really nice

Good thing

This was the first one

I learned

[empty grove]

I think you can get pretty good intuition from just thinking about why the n-th homology for R^n-{0} is not trivial

I hope I have my indices right

[bright acorn]

https://youtu.be/uEidnW0gN44

[sleek thicket]

you do not

R^n \ {0} is the n-1 sphere

[bright acorn]

These lectures are really good

[empty grove]

oof

[bright acorn]

The intuition he gives for the de Rham Cohomology is trying to study things like orientation

And also

Solutions to certain differential equations

On Smooth manifolds

Which really is nice

Anyway

Sham

[sleek thicket]

☘️

[bright acorn]

So you define the 0th k-group of a space as follows, you consider a topological spaces, look up at all isomorphism classes of vector bundles over X with the Whitney Sum and the take the Grothendieck-group of this monoid.

Right?

[sleek thicket]

right

well you're going to want to restrict your category of spaces in a sec

but yes

[bright acorn]

So I suppose that for the n-th k-group you prolly abstract this in terms of something else

[sleek thicket]

you would think so, right?

:^)

[bright acorn]

I just have no idea what it should be lmao

[sleek thicket]

you would be wrong!

it's so much weirder

so if you think about singular cohomology

you can prove the relationship $H^p(X) \cong H^{p+1}(\Sigma X)$

[gentle osprey]

shamrock

[sleek thicket]

where $\Sigma X$ is the suspension

[gentle osprey]

shamrock

[sleek thicket]

you can see this from the mayer vietoris LES

\Sigma X splits up into two contractible pieces and then a thing homotopy equivalent to X

yeah?

[bright acorn]

I know Mayer-Vietoris sequence

But let me think about this

I haven't seen this result yet

Yeah

I might want to read on that later

[sleek thicket]

sure

so if you've ever seen the Eilenberg-Steenrod axioms

the axioms for a cohomology theory

you can actually formulate them in terms of this suspension axiom instead of the mayer vietoris axiom

[bright acorn]

Oh

[sleek thicket]

so the thing is

[bright acorn]

So you mean the Eilenberg-Steenrod axioms?

[sleek thicket]

this is great for us, if we want K-theory to satisfy those axioms

oh

yes

lmao

sorry haha

topologist's have too many names

imo

okay yes anyways

this helps us a lot

because it tells us how to define half of all of the K groups

namely, the negative ones!

we define $K^{-n}(X) = K^0(\Sigma^n X)$

[gentle osprey]

shamrock

[sleek thicket]

this is uh, sort of silly

it still doesn't tell us how to define the positive K groups

which are what we need for a cohomology theory

but here's the absurd thing

these groups K^{-n}(X) are actually periodic in n

with period 2 (in the complex case) or 8 (in the real case)

this is called bott periodicity

that's insane!

magical!

So I'm claiming $K^0(X) \cong K^0(\Sigma\Sigma X)$ in the complex case and $K^0(X) \cong K^0(\Sigma\Sigma\Sigma\Sigma\Sigma\Sigma\Sigma\Sigma X)$ in the real case

[gentle osprey]

shamrock

[sleek thicket]

hello tterra

[gritty widget]

[sleek thicket]

anyways @bright acorn, to be consistent with this we just define K theory to be periodic in positive degrees too

so e.g. $K^6(X) = K^{6-8}(X) = K^{-2}(X) = K^0(\Sigma \Sigma X)$

[gentle osprey]

shamrock

[bright acorn]

Hmmm

On the chapter I am in

Hatcher's only mentioned Bott's periodicity

But didn't really say much more about it

Rather than

[sleek thicket]

ah there's a couple different forms of it

[bright acorn]

Atiyah's proved it

Using K-theory

And was one of the motivations

To study the subject

At least at the beginning

Ok so

K-theory tries to study a topological space

By studying its finite dimensional vector bundles

[sleek thicket]

Yes

[bright acorn]

In such a way that

[sleek thicket]

and magically you can extend this into a cohomology theory

[bright acorn]

For each natural number n

You assign a functor from the category of topological spaces Top to Grp. (You haven't talked anything about any of these constructors actually being a functor but I will just assume this because we want our constructions to be nice behaved anyway haha)

[sleek thicket]

yup

(the groups are abelian and the functor is contravariant)

(K^0 being a functor follows from the fact that you can pull back vector bundles and f^*(X (+) Y) = f^*X (+) f^*Y, other K groups are a composition of this and suspension, which is a functor)

[bright acorn]

And the other's follow from Bott's periodicity, right?

Which in this case I think we are taking as an axiom.

[sleek thicket]

wym?

[bright acorn]

For the theory

To construct the other positive K groups

[sleek thicket]

yeah so the other ones are compositions of K^0 and suspension

some number of times

and suspension is a functor

[bright acorn]

I see

Man that's weird

One thing I find weird is that

I can see how this could be used to study manifolds

Because I have seen explicit non-trivial examples of vector bundles

Like the tangent or the cotangent bundle

Or tensor products of these

Which have nice interpretations

But for a general topological space

I really haven't seen much about even fiber bundles being useful to study them

[sleek thicket]

well, who cares about the homology of a general topological space? why does anyone care about general topological spaces at all?

[bright acorn]

Lmao

You are right

But I think that it will take me some time

Until I see some other examples

But I think historically all this started by study manifolds, right?

[sleek thicket]

Yup, that's my understanding

[bright acorn]

And somehow generalizing the tangent bundle to more general topological spaces

Because the tangent bundle is really useful

One thing that I am interested in studying

Is a bit more of history

Of topology

Because I think there's some things I might understand better

If I take a look at what they were thinking at the time

Anyway uh

I think that I have to go now

Thanks for the good time

I just asked here a question at first to get some insights on a problem

And then I got a full explanation on K-theory and (co)homology that's so nice

Thanks

[gritty widget]

[digital peak]

how?

seems nontrivial to show

[sleek thicket]

What's your defintion of nilpotence of the lie algebra?

[digital peak]

the lower(?) derived series converges to 0

[sleek thicket]

So there's an n such that for any x1,...,xn, y we have [x1, [..., [xn, y],...]] = 0

Right?

[digital peak]

lower central series i the name

[sleek thicket]

yup

[digital peak]

yes

[sleek thicket]

Right

Well

[digital peak]

ad(x_1) . ad(x_2) ... ad(x_n) = 0

[sleek thicket]

(ad(x)°ad(y))^n(z) = 2n iterated lie brackets applied to z

= 0

[digital peak]

oh

@sleek thicket is the converse true?

[sleek thicket]

That if the killing form vanishes the lie algebra is nilpotent?

That feels wrong to me

Google says it's false

[tardy meadow]

sanity check: this proof makes sense right?

making sure i'm not missing anything obvious

oh wait I should really say that since y \in f(X) the pre-images are non-empty

[empty grove]

is good

[summer jolt]

How do you calculate this homorphism

?

Do I just compute f(1:0:0)?

Sorry I'm a bit unfamiliar with the language, what do you mean by generator?

@gritty widget Would that be a curve in the equator of Rp2?

[summer jolt]

So if I compose f with this loop I should get some other loop in RP2 right?

Hmm I see, I just can't think of any explicit path that would make this work

[summer jolt]

Would (cos(pi x),0,0) work?

hmm I'm not sure but I guessed another one: cos pi t, sin pit ,sin pi t?

Ok so only one of the coordinates is changing right?

sorry two of them

let's say z is constant

Then we can use something like cos 2pi t, sin 2pit,0?

Ah yes

The only problem is that f is given in terms of x,y,z

Ok so we would get f(cos pi t, sin pi t, 0) = (cos^2 pi t - sin^2 pi t : 0 : 2 cos pi t * sin pi t) correct?

I'm still a bit puzzled about how to compute the homomorphism though😅

Yes, it should be that f(gy) = f(g) f(y) right?

[gentle osprey]

slimvesus

[summer jolt]

This is the one that was given to me

So the map alpha --> falpha would be it?

We have alpha right?

I think the identity will be just sent to identity (constant path??)

I suppose I could just one representative and see right?

Just the circle on the sphere?

we can just take to be [x:y:z] --> [1:0:0] I guess since it was in the question

You're right of course, I'm just being obtuse

so just I --> [1:0:0]

Hmm I don't get quite what you mean

[gentle osprey]

slimvesus

[summer jolt]

So what would be gamma in my case?

@gritty widget ok thanks a lot for your help.

I might try tonight

[pearl holly]

How is {1/2} X (1/2, 3/2) open in the dictionary order topology on RxR? Is it because {1/2}X(1/2, 3/2) = (1/2x1/2, 1/2x3/2) where 1/2x1/2 means the ordered pair consisting of 1/2 and 1/2?

[bright acorn]

The dictonary order topology is generated by the open intervals of the order topology of R×R

So yup

This one is, in particular, an open intervals of the dictonary order topology

Do it would be open

[pearl holly]

Okay so {1/2}X(1/2, 3/2) = (1/2x1/2, 1/2x3/2) is correct, right?

and since 1/2 < 3/2, the above is a basis element for the dictionary topology on RxR?

[bright acorn]

What does (1/2×1/2, 1/2×3/2) means in the order topology?

This is the set of all points such that 1/2 × 1/2 < x < 1/2 × 3/2 sure

But what is the order relation here?

Well

In order for 1/2 × 1/2 < x < 1/2 × 3/2 we need to check two things

Writing x = (x_0,y_0)

We need to check two things, either 1/2 < x_0 < 1/2, or, if x_0 = 1/2, we would have 1/2 < y_0 < 3/2

But the first option is impossible

So we have that x_0 = 1/2 and 1/2 < y_0 < 3/2

The implies that any x in (1/2 × 1/2, 1/2 × 3/2) is in {1/2} × (1/2, 3/2)

And the other inclusion is trivial

So indeed both are equal

Just use the definition, there's not much to do

Try to prove this in general tho

[pearl holly]

Okay, very good explanation! Thank you!

[strong heron]

The given map was nullhomotopic, right?

[obtuse meteor]

thinking about this, this is because it has a lift RP^2 -> S^2

we can be conservative and say it's not nullhomotopic just 0 on pi_1 via this though

(the map lifts because swapping signs does not change the signs in the output)

you can also lift to R^3\{0} if you want since scaling the input by lambda scales the output by lambda^2

before functoriality, man lived in caves

yes

That's a Brian Conrad quote

hm

to me the explicit computation is just an instance of this realization

like you compute in pi_1(RP^2) by lifting loops to S^2

the way you lift the loop in this case is by lifting the function

and the realization that the generator goes to zero is that this makes sure it remains a loop in S^2

[rugged swan]

if X is a scheme and x a closed point, what does it means to say "k(x) is the structure sheaf of x" ? What is k(x) ? for me it's the field OX,x/mx

[gentle osprey]

the-last-knight

[sleek thicket]

They're probably thinking about how you have a scheme map Spec κ(x) -> X whose image is just the point x. Idk though

[fading vale]

@cedar pebble if you have time can you explain what exactly the codiagonal in h-Top is to me

dieck uses it to define the operation on [Sigma(X), Y]

[cedar pebble]

what, the map (id,id):X⨿X->X?

[fading vale]

yeah

[cedar pebble]

what is confusing about this

[fading vale]

is there like an explicit description of it i guess? like the diagonal sends x to (x, x), what would the codiagonal send (x, y) to?

[cedar pebble]

I mean I already wrote an explicit description but if you want more words

X⨿X is a disjoint union of two copies of X

so the codiagonal maps the first copy of X to X by the identity

and maps the second copy of X to X by the identity

that's all it is

[fading vale]

so just like this or?

[cedar pebble]

yes pretty much

[fading vale]

oh wait im literally dumb lol its disjoint union not cartesian product

nvm

[cedar pebble]

yea

[fading vale]

brain bad

[cedar pebble]

A point in X⨿Y is a point in X or a point in Y

so yea now it should be clear

[fading vale]

thanks!

[cedar pebble]

I can see why you would be confused if you thought it was the Cartesian product

but yea you use the codiagonal (id,id):X⨿X->X to define the cylinder object Cyl(X)

[fading vale]

right

and for the suspension u basically just take the bottom half and suspend that and the top half and suspend that and wedge them together

[cedar pebble]

mhm

[fading vale]

ok so that makes sense i think

[cedar pebble]

right the universal constructions to keep in mind are like

Cyl(X) is the factorization of the codiagonal by a cofibration followed by a weak equivalence

while Path(X) is the factorization of the diagonal by a weak equivalence followed by a filtration

[fading vale]

cofibrations and fibrations

next chapter

this one is on elementary htpy theory and its mostly group cogroup loop space stuff and it ends with the fiber and cofiber sequences

which seem very cool

[cedar pebble]

nice

yea once you get into like fiber and cofiber sequences or model structure you can start doing some abstract homotopy theory

also if you haven't seen it before there's a really nice way to think of a lot of homological algebra in terms of model structures

where projective/injective resolutions can be stated in terms of fibrant/cofibrant replacements

[fading vale]

that seems very very cool i hope dieck gets into it

he does all the homotopy up to basically intro stable htpy stuff and then spends a while on singular homology and homological algebra

[cedar pebble]

ah hmmm

yea I forget if Dieck covers this

[fading vale]

then he does bundles and manifolds for a while before cohomology and duality

and then the last few chapters are like

characteristic classes, homology and htyp together, and bordism

[cedar pebble]

mhm

[pearl holly]

Wait wait... Is (-1, 1) - K open in R in the standard topology? Here K = {1/n | n is a positive integer}

It's not, right?

[ivory dragon]

not saying youre right/wrong, but why do you think it isnt open?

[pearl holly]

I think that it is not because if it were then 0 should be in a interval (a, b). This can not happen because I could always find a 1/n in (a, b) which is a contradiction

[ivory dragon]

right

what about (-1, 1) - (K U {0})?

i.e. what if we deleted 0

would that make it open?

Solution: ||Yes, because we could then write it as a union of the open interval (-1, 0) with every open interval between every 1/n, 1/(n+1)||

||this is an infinite union of open sets, hence open||

[pearl holly]

Oh, I see

[ivory dragon]

the "moral" is that, when working with metric spaces, you should understand openness by looking at the boundary behaviour of sets

in this case, the problematic part was the 0 on the boundary

everything else was fine

but 0 ruined openness

[pearl holly]

yeah we should kill it

Okay but seriously now, then (-1, 1) -K should also not be open in (-1, 1) intersect U where U is a open set in R, right?

[ivory dragon]

what if U = (-1, 0)

[pearl holly]

good point

[fading vale]

if this channel is free

[gentle osprey]

Moth In Shambles

Moth In Shambles

Moth In Shambles

Moth In Shambles

[fading vale]

But idk how I'd show that this is well defined at all

I'm not even sure that it is, like we only have stuff up to homotopy

But we need to use it to define a function

[gritty widget]

Just a sanity check

The pull back of a lie bracket by f

Is just f*[ -, - ] = [f(-), f(-)]

i would imagine so

[silver umbra]

if C is a compact subset of R2

how can i define the orthogonal projection of C onto a line

[gritty widget]

are you asking how one would express "orthogonal projection onto a given line" as a function R^2 -> R^2?

because you can find an expression for that and then just take f(C)

[fading vale]

how would i show that pushouts in Top (say f: A -> B, j: A -> X with j an embedding) induce homeomorphisms on the quotient spaces?

i think i got surjectivity

i just have no idea how to do injectivity

[sleek thicket]

Let $X$ be a scheme and consider the canonical map $X\to \mathrm{Spec} \Gamma(X, \mathscr{O}_X)$

[gentle osprey]

shamrock

[sleek thicket]

This may fail to be surjective, eg in the case of a an infinite disjoint union of spectra of fields

But it is dominant in that case (I think), it's dominant in the case of an integral scheme X (i think), and it's obviously bijective if X is affine

Is this map always dominant? When is it surjective?

[tough imp]

Wait what? The infinite disjoint Union of spectra of fields is affine

How can it be obviously bijective in that case?

[sleek thicket]

No it isn't??

[tough imp]

The disjoint union of affines corresponds to Spec of the prodict of the rings

[sleek thicket]

For finite disjoint unions

[tough imp]

I don’t think this fails for infinite things

Aren’t disjoint unions exactly the coproduct?

[sleek thicket]

If you believe Sándor's claim I can prove it's not affine to you

[tough imp]

Nvm

I looked it up

Wtf

[sleek thicket]

this is like my second example of a non affine scheme

After P^n

[tough imp]

Shouldn’t Spec respect even infinite stuff being an equivalence of categories??

[sleek thicket]

The colimit in Aff is different than in Sch

Limits are computed the same but not colimits

[tough imp]

Oh fuq

Tfw

[sleek thicket]

What you can say is that the global functions on a colimit of affines are the limit of the global functions

Since Γ : Sch^op -> Ring is a right adjoint

[tough imp]

I see

[sleek thicket]

Here's a simpler example of this phenomenon

P^1 is the pushout of Spec k[x] and Spec k[y] over Spec k[x, y]/(xy-1)

In fact, any scheme is the colimit of its affine open subschemes

Reposting my question: is the map X -> Spec Γ(X, O_X) always dominant? When is it surjective?

So if $x \in X$, this map sends $x$ to $\mathfrak{p}_x = \ker(\Gamma(X, \mathscr{O}_X) \to \kappa(x))$

[gentle osprey]

shamrock

[sleek thicket]

Then I think the image of this being dense says that $\bigcap_{x \in X} \mathfrak{p}_x$ equals the nilradical of $\Gamma(X, \mathscr{O}_X)$

[gentle osprey]

shamrock

[sleek thicket]

Which doesn't seem automatically correct

[honest terrace]

Any of you know some good doc that sums up how to construct stereographic projections in dim n properly ?

[cosmic wedge]

you know circles have points, (cos(x),sin(x)) is there a generlization of that for any dim?
Im tryna get it for 4 dim

[pearl holly]

So this exercise wants me to show that the projections $\pi_1: X \times Y \rightarrow X$ and $\pi_2:X \times Y\rightarrow Y$ are open maps. So if $U$ is in the product topology of $X \times Y$ then it I can write $U = \bigcup_{i \in I} U_i \times Y_i$ where $U_i$ are open in $X$ and $Y_i$ are open in $Y$. Then $\pi_1(U) = \bigcup_{i \in I}U_i$ and so $\pi_1(U)$ is open in $X$? Is that it or am I missing something?

[gentle osprey]

older sister

[pearl holly]

(where $\pi_1(x, y) = x$ and $\pi_2(x, y) = y$)

[gentle osprey]

older sister

[cosmic wedge]

@pearl holly I believe that that's it

[pearl holly]

Okay thank you so much!

[thin bramble]

Does anyone know Covariant derivative?

[gritty widget]

do you have a question or are you just curious to know if people in this server know what a covariant derivative is?

[naive hatch]

We know that (X,d) is a metric space, but we need to show it is complete. To show this, I was thinking of taking a general cauchy sequence in X and then finding a convergent sub-sequence which would imply the whole sequence converges in X. However I'm not really sure how to construct that convergent sub-sequence, anyone have any ideas?

[empty grove]

Put into words, the distance between x and y is 2^-n where n is the first position when x and y disagree. If you have a Cauchy sequence, then the first position where terms of the sequence disagree tends to infinity as you go down the sequence. You can then define the limit sequence from there pretty easily (pointwise limit)

[naive hatch]

thats another thing i was thinking, and intuitively it makes sense why it would converge then. I'm having trouble trying to put that into "proof language" though. it wouldn't be sufficient to explain it like you did then say "then there exists a limit point that the sequence approaches", would it?

[empty grove]

Yeah you have to prove it

Try proving that each coordinate is eventually constant

[pearl holly]

Just to be sure, the basis for a product topology X x Y consists of EVERY element of the form u x y where u is open in X and y is open in Y, right? Not just SOME elements?

[lunar yoke]

That is a basis. In general you can have many different bases. For example it would suffice to choose a basis B for X and a basis B' for Y and then consider all sets U x V where U iterates over B and V over B', as basis for the product topology of X x Y

[naive hatch]

proving each index is eventually constant is pretty straight forward. could you then say that as the limit of each term exists, the limit of the whole sequence of terms exists? like saying lim(x1,x2,...xn) goes to (lim x1, lim x2, ..., lim xn) - n would be infinite but maybe this is a way of understanding it

[pearl holly]

Okay so to be sure: if X x Y is a subset of X' x Y' then every single open set in X must also be open in X' because every base element of X xY must be in X x Y, right?

[lunar yoke]

no, i don't think that works. If X is a subset of X' with the subspace topology, then open sets in X need not be open in X'. Example X' = real line, X = [0,1]. Then (0,1] is open in X, but not in X'

But maybe this helps: If U is open in X and V is open in Y, then U x V is open in X x Y

[empty grove]

Yeah the pointwise limit will be turn out to be the actual limit

[pearl holly]

Okay but does this argument work: Every basis element in X x Y is u x y where u is open in X and y is open in Y. This basis element must be in X x Y. If X x Y is a subset of X' x Y' then u x y must be in X' x Y'. Since X' x Y' has a basis consisting of u' x y' where u' is open in X' and y' is open in Y', then u must be open in X'.

Because since u x y is in X' x Y', then u must be open in X' by definition. Am I missing something?

[lunar yoke]

I'm not sure I understand you correctly, but as I see it, you are taking some arbitrary u open in X and deducing (via some argument involving the product topology) that it is open in X'. This can't work, since I just gave a counterexample above

[pearl holly]

Yeah I am taking an arbitrary basis element of X x Y which I call uxy, the Cartesian product between an open set in X and Y. Then, since X x Y is a subset of X' x Y', uxy must be in X' x Y'. This means that $u \times y = \bigcup_{i \in I} u' \times y'$ where $u'$ is open in $X'$ and $y'$ is open in $Y'$. Since $\bigcup_{i \in I} u' \times y'$ is open in $X' \times Y'$, then so must $u \times y$. But then $u$ must be open in $X'$

[gentle osprey]

older sister

[lunar yoke]

but just because u x y is contained in X' x Y', this does not imply that it is open in X' x Y', hence you may not be able to write it as this union of basis elements of X' x Y'

[pearl holly]

Oh yeah that's true. I don't know that it is open in X' x Y'

[lunar yoke]

anyway the argument cannot work

[pearl holly]

Yeah okay thank you so much!

[honest narwhal]

Interesting

[sleek thicket]

it's dominant in the quasicompact case but not in the general case, e.g. if X is the disjoint union of Un = Spec k[x]/(x^n) over each n and f is the function which is x on each Un then f is not nilpotent but does vanish at each point of X, which implies that the vanishing locus of f is a proper closed subset of Spec Γ(X, O_X) containing the image of X -> Spec Γ(X, O_X), so that map isn't dominant

thanks to daniel litt on twitter for wrongly stating the map was always dominant

it made me want to find a counterexample more

[wanton timber]

Nice example. I wonder if there are irreducible examples.

[sleek thicket]

Good question. I imagine so but don't have an example

There's no examples in the reduced case

[gentle osprey]

squirtlespoof

[versed pivot]

@vague zodiacStill stuck?

Ok, so you know that the binormal is constant B(s) = B_0

You can try differentiating alpha(s) dot B_0

Since alpha(s) dot B_0 = c is the equation of a plane

(As a side note, you should also assume that the curvature is non-zero everywhere so that the binormal vector makes sense at all.)

[silver umbra]

really simple question:

is the interior of [-1,1] x R

(-1,1) x R?

[empty grove]

yez

[silver umbra]

ty

[bright acorn]

Is proving that a closed subgroup of a lie group is a lie subgroup harder than it looks?

I was thinking about maybe doing it as an exercise

before looking at the proof

given in the book I am reading

Any ideas?

[sleek thicket]

Yes

How were you thinking of proving it?

[bright acorn]

Yiikes

[bright acorn]

I stared at it for like

5 minutes

And then I noticed

Yeah

It may be harder than it looks lmao

So I didn't really think about anything useful

[gritty widget]

all i can say is uh

something about the exponential map

[sleek thicket]

lmao

Something something foliation something something frobenius

[gritty widget]

the council is judging you

[versed pivot]

Using the orthogonality you should find (alpha(s) dot B_0)' = 0

Right exactly

And that's the equation of a plane, so alpha(s) lies in a plane

[sharp yoke]

idk if manifold stuff goes in here but voila

[versed pivot]

If you want to be more specific the equation is really (alpha(s) - alpha(0)) dot B(0) = 0 so the statement is that alpha(s) - alpha(0) lies in the plane orthogonal to B_0

[gritty widget]

this is the place for manifold stuff

[sharp yoke]

ok

[vast estuary]

Hello

[gritty widget]

what's h_{(1-t)x+ty}?

[vast estuary]

check the definition of f_z

it is that, just with z replaced by (1-t)x + ty

[gritty widget]

ah, okay

[vast estuary]

a small typo, let me repost

[gentle osprey]

the-last-knight

[gritty widget]

i could prove it if there was a reverse holder inequality 😛

[vast estuary]

lol i think we don't have to use any inequalities besides the first one that's given

just gotta get that to R^n-1 somehow and in integral form

[vast estuary]

prekopa leindler is sometimes famously called reverse holder inequality, which is what i wanna prove with this XD

is there any way to relate the following integrals? if yes maybe we are done

$$\int_{\mathbb R^{n-1}} f_z \text{ and } \int_{\mathbb R^n} f(\cdot, z)$$

[gentle osprey]

the-last-knight

[vast estuary]

are they the same?

[gritty widget]

one is the intergral of the other, right?

[vast estuary]

how

[gritty widget]

$\int_{R^n} f = \int_{\mathbb{R}} \int_{\mathbb{R}^{n-1}} f_z(x) dx dz$

[gentle osprey]

8da 💕

[vast estuary]

Right but in the LHS isn't z fixed?

[vast estuary]

in the 2nd integral here z is fixed right

[gritty widget]

if z is fixed, then how can we integrate over R^n?

[vast estuary]

oh good point then it would diverge or something?

[gritty widget]

diverge?

[vast estuary]

umm not sure

[vast estuary]

okay, got it!

i actually had to use the induction hypothesis of prekopa inequality here

let me know if you want more details, and thanks!

[gentle osprey]

mebi_bugkata

[uncut surge]

If you're including 0 in the interval, then the log is not defined there (tending to -infinity), so that's not a well-defined function into the real numbers

[sick elm]

oh of course, thanks!

[gentle osprey]

squirtlespoof

[pearl holly]

So somethings wrong here I think. Let $U$ be an open set of $X$ and $y$ be an open set in $Y$. Let's say that $U \times y = \bigcup_{i \in I}(U_i' \times y_i') $ where $U_i'$ is open in $X$ and $y_i'$ is open in $Y$. Define $\pi:X \times Y \rightarrow X$ as $\pi(x, y) = x$. If we act with $\pi$ on both sides of the equation then we get $U = \bigcup_{i \in I} U_i'$. It feels wrong doing this. Anything wrong with this argument?

[gentle osprey]

older sister

[empty grove]

No it's good you just need to make sure that no y_i' is empty

Because you can always add in X x {} into the union without changing it

But that will change the union of U_i to be X itself

[pearl holly]

Oh okay, thank you so much! It just felt like a weird argument for some reason

[thin jewel]

how strong do I need to be on Differential Equations for a module on Lie Groups?

it's the only prerequisite listed on the course website that I don't have

(apart from Calc 2 )

[gritty widget]

Systems of linear ODE with constant coefficients would be enough for some basic examples of lie theory

(in my point of view you don't need to know anything about DE to understand Lie Groups)

[thin jewel]

ok, thanks

[cosmic wedge]

coughs

$\int_{yournan} dV > \infty$

[gentle osprey]

NotKitten

[cosmic wedge]

coughs

[cedar pebble]

see #advanced-analysis for a dreadful Hodge theory question which I posted in analysis since it's about harmonic functions/Laplacians

[pearl holly]

So this is the exercise that I am working with: Let X be an ordered set. If Y is a proper subset of X that is convex in X, does it follow that Y is an interval or a ray in X? I think that it does. Let a and b be the smallest and largest element in Y respectively. Then (a, b) is a subset of Y since it's convex. If c is an element in Y then it must be between (a, b) so we get that Y = [a, b]. If it has no largest or smallest element then it is a ray. Is this right?

[pearl holly]

I mean the first part is okay I think, but does it really hold when Y has no smallest or largest element?

[fading vale]

Is X totally ordered here?

[pearl holly]

I honestly don't know. It just says that X is an ordered set

[bleak helm]

What's the definition of ordered set in your book?

[pearl holly]

I don’t think the book defines what an ordered set is, he just defines an order relation

[pastel linden]

wrong channel

so I started getting back into smooth manifold stuff today but just wanted to review some topology

[gritty widget]

#geometry-and-trigonometry

[pastel linden]

currently trying to prove RP^n is a topological manifold

[solemn coyote]

my bad sorry i read geometry

[pastel linden]

So I did the following but I'm not sure how to proceed

I belive thats true that it suffice to show the quotient map is open right? But not quite sure the best way of doing that

[thin bramble]

Does anyone know how to do 16.7?

[gritty widget]

can you post your pictures not sideways

[thin bramble]

Yea, give me a second

[empty grove]

Ordered set usually means totally ordered, partial order is usually stated explicitly

[empty grove]

Yeah there may not be a smallest and largest element so that doesn't work

Hint: || Try to prove that Y has a maximal interval subset ||

|| ie subset that is an interval/ray ||

[empty grove]

You can factor the quotient map through $S^{n+1}$, so you have
$\bR^{n+1} \setminus {0} \to S^{n+1} \to \mathbb{RP}^n$
where the first map can be view as a projection and hence is open. For the other map, show that if $U$ is open in $S^{n+1}$ then so is $V(U)$ where $V(U)$ is $U$ along with all the points which have antipodies in $U$. Then $V(U)$ is a saturated open set with the same image as $U$

[gentle osprey]

Moldilocks

[thin bramble]

@gritty widget check

[jagged ocean]

Do the exterior derivative and exterior covariant derivative agree more than they do in general on a principle bundle with Abelian structure group?

[cerulean oriole]

I think this is false when X is not complete (or bounded-complete), e.g. X = Q, Y = ( (-sqrt(2), sqrt(2)) intersect Q ).

But if X is bounded-complete (bounded sets have GLB/LUB) and Y is bounded, something similar to your claim will be true, using b = sup(Y) and a = inf(Y) instead of maximum and minimum when Y doesn't have maximum/minimum elements. In that case, you also can't guarantee that a, b are in Y i.e. the interval could be open at either end.

[sweet wing]

dedekind cut time

[pearl holly]

Okay thank you so much!

[pearl holly]

Let I = [0, 1]. The dictionary order topology on I x I is not a subset of the product topology on I x I, right? A basis element in the dictionary order topology can be seen as "lines" in a plane with "half lines" in the end points. The lines in the middle can be represented as a rectangle (open set in the product topology), but the "half lines" can't so therefore the dictionary order topology is not a subset of the product topology on I x I. Is this right?

[gritty widget]

How can a line be a rectangle? (Neither full lines or half lines can be rectangles)

[pearl holly]

yeah but I mean that I could look at all those lines as a rectangle

[empty grove]

not as a rectangle without boundary

[pearl holly]

Man every single solution online say different things... One guy first claims that one is strictly finer than the other and then in the end he claims from nowhere that they are not comparable. Why can't Munkres just have the solutions in the back of the text???

[empty grove]

the inclusion is the other way

[pearl holly]

You mean this inclusion?

[empty grove]

wait no

non comparable

[pearl holly]

That's what I got as well!

But since every solution is different I got confused and I obviously can't learn anything if I got something wrong without knowing it

[stray bane]

I am looking to prove that if $X$ is hausdorff, then for a continuous function $f$, the fixed points of $f$ is closed in $X$. We already know the diagonal $A$ is closed in $X\times X$. Thus we now define $h(x)=(\mathrm{id}_X(x), f(x))$. Again $h^{-1}(A)$ is closed, which is the set ${x\in X\mid f(x)=x}$. This proof seems a bit too easy? However i feel like this proves the statement, wouldn't it?

[gentle osprey]

Frozaken

[empty grove]

It is correct

In fact you can prove using the same technique that the set of points where 2 continuous functions agree is closed

By replacing id with the second function

[stray bane]

Yes! my thoughts exactly, furthermore we wouldn't even need to require those functions to be maps between two hausdorff spaces, only that they map into one

[empty grove]

Yep

[gentle osprey]

bdobba

[empty grove]

Why not?

Self intersection doesn't imply existence of a point with 0 derivative

[cerulean oriole]

{ 1/2 } x (0,1) is open in the first but not the second, so yes.
For the opposite, (1/4, 3/4) x (1/4, 3/4) is open in the product topology but not the dictionary order topology <-- wrong example 😅.
So neither one is finer.

[pearl holly]

Yes okay good! I also found that [0, 1] X (1/2, 1] is not open in the dictionary topology because {x} X (1/2, 1] can't be open in the dictionary topology so I can't "fill in the rectangle". Thank you so much!

Or you could just show that [0, 1] X (1/2, 1] can't be open in the dictionary topology directly, but I like to think of it as lines because that's intuitive.

[frigid river]

Are those statements the same?
$$ \mathscr{S}={\pi_1^{-1}(U) \ | \ U \ open \ in \ X } \cup {\pi_2^{-1}(V) \ | \ V \ open \ in \ Y}$$
$$\mathscr{S}= (U \times Y) \cup (X \times V)?$$

[gentle osprey]

𝔙eryhappyperson

[frigid river]

Let me just ping myself: <@&286206848099549185>

[pearl holly]

Well according to Munkres, if $\pi_1: X \times Y \rightarrow X$ is a projection then $\pi_1^{-1}(U) = U \times Y$ of $U$ is open in $X$ so I believe that they are the same. Don't trust me tho, I am a noob

[gentle osprey]

older sister

[cerulean oriole]

$pi_1^{-1} (U)$ is indeed $U \times Y$ for any subset $U \subseteq X$ (no topology needed) if that's what you wanted to know.

[gentle osprey]

Raghuram

[cerulean oriole]

Exactly.

[frigid river]

So $U$ doesn't have to be open in $X$?

[gentle osprey]

𝔙eryhappyperson

[cerulean oriole]

Not for the equation I mentioned to hold; that's just set theory.

However, the product topology is formed using only the sets of the form you gave for U open in X/V open in Y respectively.

[gentle osprey]

Raghuram

[cerulean oriole]

Given a bounded measure-0 set E within a compact set U (in Euclidean space), can E be covered with countable collections of balls of arbitrary small total volume, such that each ball is a subset of U?

(In case there are others, the definition of measure-0 I'm using is the property above, except that the balls need not be subsets of U.)

I'm not sure, but it might be necessary to assume that U is a neighbourhood of E rather than just a superset.

Right.

[fading vale]

whats the intuitive meaning behind a space being well-pointed/the base point being nondegenerate

like when the inclusion of the point into X is a cofibration

[sweet wing]

it's more of a technical tool for later on :p

[fading vale]

the closest thing i can think of is maps f: X -> Y and I -> Y define homotopies of X into Y

interesting

oh i think i kinda see it

like you have a path in Y and then a map f: X -> Y which attaches the basepoint x to the start of the path in Y and then you can sorta slide the image along the path basically

neat i guess?

[sweet wing]

something something like that

you end up seeing it behaves very well under smash products and homotopy gruops

[fading vale]

[sweet wing]

oh yea one thing tom dieck didnt mention but i found pretty nice

In TOP if we have i: A->X, f:X->Y such that fi=*, then f factors through X/A

If i is a cofibration then if fi is nullhomotopic it factors as well

you can kinda think of chapter 5 as like

wow h-Top sucks!

lets fix it

i feel like a lot of topologyfancy categorical constructions appearing is also like

wow Top sucks! got to fix it

[empty grove]

No, the first set doesn't depend on the value of U and V (those 2 are quantified) and the second one does

[gentle osprey]

Moldilocks

[fading vale]

even ch 4 is kinda this like htpy pushouts and stuff

[empty grove]

Is that different from the universal property of quotients? Seems like just a specific case

Talking about the first part

Idk what a cofibration is

[fading vale]

no but the idea is we are working in h-Top (everything is up to homotopy) not Top

so in Top we have fi = * -> X/A, we are saying that for cofibrations f, fi = * in hTop (meaning fi is nullhomotopic) implies that f factors through the quotient

its what ari means by "we want to make hTop nice"

there r a lot of categorical things that dont quite work up to homotopy

e.g pushouts

if f and g are two maps from A into X, Y with pushout Z then we can have A', X', Y', f': A' -> X', g': A' -> Y' with everything homotopic to its corresponding thingy (A' simeq A, X simeq X' etc) but the pushout Z' is not necessarily homotopic to Z

so htop kinda ugly

but its ok we can fix it

[pseudo crane]

brushie brushie

[fading vale]

[empty grove]

Ah I got confused because the message said Top

Didn't see in context

[fading vale]

cofibrations are cute

[empty grove]

[fading vale]

theyre not scary!

[empty grove]

I'm scared because that's how det talks

"_____ is cute!"

[gentle osprey]

Moth In Shambles

[empty grove]

Oh ye ik what fibrations are

[fading vale]

oh

[empty grove]

But idk how you would dual that

[fading vale]

basically you extend homotopies instead of lift them

[empty grove]

Oh wait homotopy extension property

[fading vale]

so we have an $H \colon X \times I \to Y$ with $H(x, 0) = f(x)$ and $H(i(a), t) = h(a, t)$

[gentle osprey]

Moth In Shambles

[fading vale]

the whole diagram ends up looking something like

[empty grove]

Wouldn't you want to say HEP for A?

That's what we do with HLP

[fading vale]

no like given a map i: A -> X we are extending a homotopy of A into Y along f:X -> Y

its relative to the map i

not the map f

[sleek thicket]

moldilocks, are you thinking of how we might say "HLP for cubes" or something?

[empty grove]

wait what's the completion of this sentence?

[fading vale]

this

[empty grove]

I'm just thinking of defn of HLP

[empty grove]

Got it

[fading vale]

given a map f: X -> Y and a homotopy h of A into Y we get a homotopy H of X into Y that extends h

[empty grove]

So cofibration would be if you have HEP for all Y?

[fading vale]

also hi shamrock!!

right

[empty grove]

I see

[fading vale]

but it turns out its actually not so hard to verify because you only need to think about it for the mapping cylinder of i

(the pushout in h-Top basically)