point-set-topology

[flint sinew]

the "other direction" in this case referring to transforming the image back to preimage?

[woeful oasis]

$f: U \to V\subset\mathbb{R}^d$ is a homeomorphism, then $f^{-1}:V\to U$ must also be continuous.

[gentle osprey]

Apopheniac

[woeful oasis]

By "the other direction", i mean the continuity of f^{-1}

[flint sinew]

I see

[woeful oasis]

Then the map f also maps open subsets of U bijectively to open subsets of V. So as far as topological data goes, these two open sets are equivalent via this homeo f.

[pearl holly]

You here?

[frigid river]

yep

[pearl holly]

[frigid river]

Toki and I have been doing trying some exercise again.
We argued if there existed a retraction, then the homomorphisms between their fundamental groups should be surjective, which it isn't, because it is of the form f(x)=ax and it is from Z to Z.

Is that correct?

[pearl holly]

this is the proposition that we used just to clarify

We also know that the fundamental group of the Möbius strip is Z and that the circle is a deformation retract of the Möbius strip since Hatcher said so in his book

[frigid river]

(we know the proof is wrong by now, but is this the right direction?)

[pearl holly]

Yeah because what if a = 1? Then f(x) = x is surjective

[frigid river]

then crap

[pearl holly]

We need to show that a can't be 1. Don't know how to do that tho

[woeful oasis]

What does the boundary circle get mapped to via the induced map?

[pearl holly]

To the Möbius band, right?

[woeful oasis]

The induced map would be on the level of groups, so homotopy classes of loops

[pearl holly]

So I'm like "projectring" a circle that goes around itself n times to the Möbius strip by the i_*[w_n] = [i o w_n]

Hmm but wait

[woeful oasis]

Not just any n

[pearl holly]

f(x) = ax in "the language of fundamental groups", this would mean that I'm taking a loop that goes around itself x times and getting a loop that goes around itself ax times in the Möbius strip. But if I do a "whole rotation" in the Möbius band, I actually get "two rotations" in the circle. So a can't be one because a loop that goes around itself x times can't go around itself x times in the Möbious strip

[woeful oasis]

I think that's the idea yes.

[pearl holly]

Because it takes two full rotations to get around the möbius strip

[woeful oasis]

If you're only allowed to traverse a path along the boundary circle, then you end up going around twice

[pearl holly]

so in fact, the homomorphism should be f(x) = 2x

Okay so just like Max said to me yesterday, I should just relax and think of the intuitive geometry instead of working with symbols all the way lmao

Okay anyway, thank you so much Apopheniac!

[woeful oasis]

🙂

[pearl holly]

happyperson, you got that?

[woeful oasis]

I think a helpful tool in proving stuff like this is that you already have a special chosen deformation retraction of the Mobius strip onto the equator (the circle along the middle). So this gives you a specific way to tell what homotopy class a loop in the Mobius strip has.

[frigid river]

Now I did.

Thank you.

[fading vale]

So for context here p: Y -> X is a map over X, and H is isomorphic to a subgroup of the automorphism group of Y over X

I dont really see how the projections give us a space over X?

like whats the map project onto the second factor and then compose with p?

so only the second projection is being used or

[marsh forge]

Can you send the G-action

or just the whole page

[delicate hollow]

Just learned what locally finite means, so a collection of sets A, of a topology X is locally finite if:

$\forall x, \exists U \ni x$ such that $U \cap A = {x_{i}}_{i=1}^n$

The example the book gave was $A = {(n,n+2) \mid n\in\mathbb{N}}$ and they say it is a locally finite set of topology on R

[gentle osprey]

Abdo Kiza

[delicate hollow]

I dont see how

[gentle osprey]

Abdo Kiza

[delicate hollow]

maybe i misread the definition

yea i did, nvm

[reef shore]

This is not correct. The last part should be "U intersects finitely many sets in A". U intersection A doesn't make sense as U is an subset of the space and A is a collection of subsets of the space.

[gentle osprey]

Moldilocks

[delicate hollow]

ty i realized my fault when i reread definition

original definition i gave in symbols would work for some topologies not a lot

im wondering working with the example i gave and the false definition i gave

what topologies this applies to on R

im thinking no topologies

[reef shore]

Well U \cap A will always be empty (assuming no set theoretic pathologies where points of the space themselves are subsets)

[delicate hollow]

maybe discrete topology

and thats it

[reef shore]

It will be empty because U is a set of points, A is a set of subsets

[delicate hollow]

oh

i forgot A is a set of subsets

[reef shore]

So all topologies on R would make all collections locally finite with that definition

[delicate hollow]

i think i like the correct definition in words because it is descriptive and easy for me to understand

[reef shore]

Yeah it's easier to read

[delicate hollow]

also the name locally finite is intuitive

[reef shore]

yep

[sullen ether]

Guys, one fast question. I have to show that product of topological groups is topological group. That is easy if there is finite number of topological groups involved but what if there is infinitly many of them?

[empty grove]

Product is defined componentwise, and the product function is continuous iff it is continuous in each coordinate (universal property of product topology) and the latter is known to be the case

Same for inverses

[cedar pebble]

right you can just use the fact that maps into a product are continuous iff all compositions with the canonical projections from the product are continuous

what you do is you take your infinite family of topological groups {X_a} and consider the product \prod_a X_a with the product topology, then the main claim is that the product and inverse and so on are continuous with respect to this topology

but this is immediate from this remark about projections: it means you can check continuity pointwise

[pearl holly]

Okay so I started to think about the deformation retract in this question late at night yesterday and I became really confused lmao. I didn't want to ask it yesterday because it was late and my mind didn't work back then. So I will ask it now.

Like I said yesterday, the circle is a deformation retract of the Möbius strip because Hatcher said so. This means that there's a homotopy F: X x I -> S^1 such that for all x in X and a in S^1

1. F(x, 0) = x,
2. F(x, 1) in S^1 (i. e F(X, 1) = S^1),
3. F(a, 1) = a.

Now we can define a continuous map f: X -> S^1 by f(x) = F(x, 1) and this will be a retract since f(a) = a for all a in S^1 by 3. Don't we now have a retraction of the Möbius band into its boundary circle, S^1? There's something that I misunderstand since the exercise clearly stated that there's so such retraction. So what goes wrong here?

So there's something that I misunderstand about deformation retractions and retractions. Isn't every deformation retraction a retraction? So if the circle is a deformation retraction of the Möbius strip, that should also define a retraction from the Möbius strip to the circle, right? But this is wrong and I don't understand why!

[pearl holly]

https://math.stackexchange.com/questions/1373143/showing-that-mathbb-s1-is-a-deformation-retract-of-the-mobius-strip-rigorou

This guy apparently shows that S^1 is a retract of the Möbius strip. Doesn't this contradict the exercise?

[reef shore]

the exercise clearly stated that there's so such retraction.

are you sure about this?

It might have said something about retractions to specific embeddings of the circle

Like the boundary, which would make sense because that goes around the strip twice

[ivory dragon]

the exercise.

[pearl holly]

"Show that there is no retraction of the Möbius band onto its boundary circle" doesn't this mean that S^1 can't be a retraction of the Möbius strip?

[reef shore]

the embedding matters

[pearl holly]

and since the boundary circle is a deformation retraction of the Möbius band, the circle should be a retraction of the Möbius band

Hmm what do you mean by embedding? Like how many times S^1 loops around itself?

[reef shore]

like how the circle you asre trying to retract to sits in your space

take S^1 disjoint union D^2 for example

[regal sonnet]

what is S^1

[reef shore]

here you can embed S^1 in 2 different obvious ways

take the wedge sum not the disjoint union

[pearl holly]

Well I'm not that comfortable with wedge sums and disjoint unions

[regal sonnet]

will it be like Coordinate Transformation?

[reef shore]

its just a disk and a circle, attached at their respective (1,0) point

[honest terrace]

unit circle

[regal sonnet]

ok, in which Topological Space?

[reef shore]

There are 2 natural embeddings of the circle in this, one to the boundary of the disk and the other to the copy of S^1 itself

and only one of these has a retract as its image

[reef shore]

or take the figure 8, and fill in one of those circles

[honest terrace]

Since it is usually considered as a topological space (atleast it is in this case), there's two distinct definitions but the two spaces are (in a fairly natural way) homeomorphic @regal sonnet:
{z € C | |z| = 1} and {(x, y) € R² | x² + y² = 1}

(I'm assuming you constructed C as a civilized person, i.e not by defining some laws on R², else the two definitions are exactly the same 🐒)

[reef shore]

is this ring theorist agenda

[honest terrace]

Algebra invading the topology channel

[reef shore]

[honest terrace]

oh btw moldi remember that isometry group S_n of n+1 points in R^n problem ? The existence of the said points is actually ridiculously simple, but it's kinda hard to think of it

[reef shore]

yeah its just a tetrahedron no?

[honest terrace]

In R^3, yes, but how do you generalize that in R^n ?

[pearl holly]

I still am convinced that the "counter argument: the core circle (as Hatcher describes it) is a deformation retract of the Möbius strip -> the core circle (which I assume is the boundary circle) is a retract of the Möbius strip" works tho

Is there maybe a difference between the core circle and the boundary circle?

[honest terrace]

I had the tetrahedron in mind, but I struggled to generalize it to higher dims

but it's actually really simple

the solution is 4 words long

||canonical basis of R^(n+1)||

with a bit more details ||consider the affine hyperplane of R^(n+1) generated by its canonical basis. That gives you a space of dim n containing n + 1 equidistant points||

[reef shore]

yeah, "it is obvious bro"

[honest terrace]

it's not obvious, it's simple*

[reef shore]

lol

[honest terrace]

it's kinda different, the solution is super simple but I honestly don't think I could've thought about it even with more time

(not sure if you clicked on the spoilers)

[regal sonnet]

yeah, I clicked

[reef shore]

I didnt lol im in the middle of a game I havent tried the problem yet

[honest terrace]

ok ok

hf

[gritty widget]

lol u lost

[reef shore]

[regal sonnet]

😂

[reef shore]

Yes

The core circle is the one in the middle of the band

The boundary one is the edges of the band

[pearl holly]

well frrriiiiccck

[reef shore]

and the boundary circle goes around the band twice while the core goes around once

[pearl holly]

Then my solution is wrong god daimmit

[reef shore]

F

[pearl holly]

Because then I can't use this proposition right

To show that the induced homomorphism is a isomorphism

[reef shore]

yeah you cant

[pearl holly]

Yeah because the boundary circle is not a deformation retract of the Möbius strip, its the core circle

and those two are different

right? Have I understood that correctly?

[reef shore]

yep

its not even a subspace

[pearl holly]

Okay great thank you so much! So now I'm not confused but angry because happyperson and I have to redo the exercise lmao

[regal sonnet]

it took So much Time!

[delicate hollow]

boundary circle of mobius strip isnt a subspace of mobius strip?

[reef shore]

Boundary isnt subspace of core

I thought that's what toki said

oh I just misread

mb

[hazy nexus]

Lmao, I was also confused by that

[delicate hollow]

core circle is S1 in middle of mobius strip right?

boundary circle is the one that intersects

[marsh forge]

the boundary is the edge

[delicate hollow]

the edge intersects itself right?

also its included in the mobius strip

so it is a subspace

[marsh forge]

are you thinking of a klein bottle

self-intersection is not a topological property

[delicate hollow]

maybe its just the optical illusion?

[marsh forge]

the klein bottle doesn't self intersect in R^4

[reef shore]

only when you embed in R^2

[delicate hollow]

it has one side but it changes orientation

oh ok

i have r2 brain

[marsh forge]

i honestly dont even know what embedding ur talking about

[delicate hollow]

its r2 embedding

[marsh forge]

yes

i got that part

[delicate hollow]

i only seen mobius strip in 2d afterall

[marsh forge]

what

You can project a mobius strip onto R^2 and I guess it self intersects

yeah i mean the projection won't be injective so sure

but this clearly doesn't need to self intersect in R^3

[delicate hollow]

yea my r2 brain sees the twist as an intersection

[marsh forge]

in fact you can build one yourself with some tape and paper

[delicate hollow]

i know it doesnt was a mental fuck up on my part

i have to always use my hands to visualize it sadly

[pearl holly]

Okay I realised (I think) that I don't even need any propositions to show that there's no retraction between the boundary circle of the Möbius strip to the Möbius strip. It will involve a very similar argument that I presented but I won't be using any propositions.

So first lets assume that there is a retraction $r: X \rightarrow S^1$ where $X$ is the Möbius strip. We know that the inclusion map $i$ is a right inverse to $r$, meaning that $r \circ i = e$ where $e$ is the identity. But this means that $r_* \circ i_* [f] = [r \circ i\circ w_n] = [w_n]$ so $r_$ is surjective since it has a right inverse (can you also use the pigeon hole principle to show that $r_$ is surjective?). So $r_*[f_n] = [r \circ f_n]$ where $f_n$ is a loop in the Möbius strip around the boundary circle. We also know that $[r \circ f_n]$ is a element of $\pi_1(S^1)$ so it must we of the form $w_m$ where $m$ is the winding number. But $[r \circ f_n] = [f_{2n}]$ because one loop around the Möbius strip results in two loops around the boundary circle. This would correspond to a function $g: Z \rightarrow Z$ defined as $g(n) = 2n$, but this is not surjective

[gentle osprey]

Tokidoki ✓

[pearl holly]

It's basically the same argument that happyperson and I presented but this time we work with r instead of the inclusion map

[reef shore]

yep works

[pearl holly]

ayyy

Thank you so much!

[reef shore]

idk about the pigeonhole principle thing tho

I don't see a way to prove surjectivity from that

[pearl holly]

You can't use the pigeon hole principle on this right? Since if A is a subset of B, then pi(A) doesn't need to be a subset of pi(B) I think

[reef shore]

How would you usually show surjectivity of some map using pigeonhole?

It talks about non injectivity

[pearl holly]

Oh that might be right

I thought that if B is a subset of A, then a map f: A -> B must be surjective, but this might be wrong

[marsh forge]

pi(A) and be much larger than pi(B) is A is a subset of B

[marsh forge]

this is not true

[reef shore]

what is f here?

[marsh forge]

Let B be the unit circle and A be the 2-unit circle so B is inside of A

disk***

not circle

then you can shrink A and send it back into B

non-surjectively

[pearl holly]

f is just a random map from A to B

Okay I see now

[reef shore]

You can just take constant map then

and B a non singleton

Also A is a subset of itself

so that would suggest that all maps from a space to itself are surjective

[pearl holly]

[marsh forge]

just to encourage a little elegance, toki would you mind if I wrote up the result using the same approach but a little cleaner?

[pearl holly]

Yeah sure, go ahead! I wrote it in a way using symbols because I just don't want to skip any steps and I want to make sure that I understand everything

[sullen ether]

Guys, how do you define product topology on set of all functions from A to B?

[marsh forge]

Let $\omega$ be the generator of $\pi_1 S^1=\bZ$ and let $\gamma$ be the generator of $\pi_1 X=\bZ$. Then the inclusion $S^1\hookrightarrow X$ sends $\omega$ to $2\gamma$. Now, we know that if $r:X\to S^1$ is a retraction of the original inclusion, then $r\circ i\simeq id$ and therefore $(r\circ i)*=id$. But that must mean that $\omega = (r\circ i)(\omega) = r_(2\gamma)$ but no such homomorphism can exist as this would amount to a homomorphism $\bZ\to\bZ$ sending $1$ to $1/2$.

[gentle osprey]

Max's Talk is on Sunday

[marsh forge]

do you mean the topology on Maps(A,B) where A and B are topological spaces?

Its the "compact open" topology

[sullen ether]

A and B are topological groups and it asks me to prove that B^A is topological group with respect to the product topology :/

[marsh forge]

uhhhhh

can you post the actual source

like a screenshot

[sullen ether]

yes give me a minute

[reef shore]

I am guessing B^|A|

[marsh forge]

@pearl holly after a little practice, you will probably stop writing things like gamma and omega and just write like

1 and 2

[pearl holly]

Okay I see now! Thank you so much!

[sullen ether]

Yea I guess it means the same, all functions from A to B

[marsh forge]

uhhhh

that looks like either a typo

or i have no clue

[sullen ether]

Any idea how to define topology on this set?

Its probably the most natural way but I have never saw this before

[reef shore]

By that I meant you identify the space of such functions with the product of B with itself |A| times

and then give the subspace topology since you dont have all set functions

[sullen ether]

This makes sense thanks 😄

[reef shore]

Only a guess though

[sullen ether]

yes, thats what Moldilocks pointed out. I figured it 😄 Thanks 😄

[bold canopy]

are isotopy classes of torus in S^3 the same as knots? You can go from knots -> torus by thickening, but there doesn't seem to be canonical 'unthickening', however I had some vague memory that they are actually supposed to be the same thing?

[pearl holly]

@frigid river

[polar oak]

Just gunna leave this dini surface here. The product of a tractrix thursday.

[pearl holly]

The heck is that??? It looks super cool

[gritty widget]

toe pology

[delicate hollow]

a dingy surface

[ivory dragon]

#topology-and-feet when

[gritty widget]

you're the moderator

go ahead

[woeful oasis]

The standard solid torus in R^3 deformation retracts onto the circle. Via the embedding of the solid torus into S^3, you get a deformation retraction of the embedded solid torus onto a knot.

[regal sonnet]

is it like Calabi-yau Manifold?

[gritty widget]

compute its first chern class and tell me

(it's not compact)

[sharp reef]

If I were to have a set S to be the set of all continuous functions f:X->R with the property that each of them is non-negative and there exists a function that is also nonzero with X compact....can I conclude that the sum of n functions of this set has to be positive?

I mean f(X) would be bounded too due to continuity

[viral halo]

What do you mean by "positive"? Strictly positive everywhere?

[polar oak]

Dini surface, aka a twisted pseudosphere aka a revolved and mirrored twisted tractrix. Tractrix solid of revolution on acid 😉

[marsh forge]

its rare for me to enter this channel and not recognize a single word in a sentence lmao

[hollow harbor]

it's very nice. is it still constant negative curvature or is the twisting not an isometry?

oh yes it's still constant negative curvature. how pleasant.

log(tan(x/2))

[regal sonnet]

Thanks, got it!

[polar oak]

The twisting gets really strange, I used to have a good mathematica file on my old computer that can show it even disappearing with the right coefficients. Constant negative curvature.

[hollow harbor]

[regal sonnet]

Have you forgotten it now?(the solution)

[polar oak]

Yeah, haha, did it as undergrad. Trying to find the file. I had it in spherical and cyllindrical coordinates aswell >.>

[sharp reef]

Yes

[viral halo]

nonzero + nonnegative doesn't mean positive in that sense, so no

[sharp reef]

If it is nonzero and nonnegative ,what else could it be then?

Not a<0 nor a=0 is the same as saying a>0...

[viral halo]

at a point, sure

for a function, no

x² is nonnegative and nonzero, yet 0²=0

[gritty widget]

Hey, hey, hey. Anyone has a pdf of Algebraic Algebra? If yes, could you send it to me?

[viral halo]

unless by "nonzero" you mean "different from 0 everywhere", which isn't the usual meaning afaict

[sharp reef]

This is what I meant actually , and what was taught too

[viral halo]

the terminology on this is ambiguous

[sharp reef]

A function is nonzero means it is never 0 at any point

[viral halo]

f > 0 means "f(x)>0 for all x" but f = 0 means f is the zero function, i.e null everywhere

[sharp reef]

@empty grove explain your position

[empty grove]

I agree with syst3ms here

[gritty widget]

[sharp reef]

[viral halo]

woops

[empty grove]

[sharp reef]

Well

[gritty widget]

toe pology

[sharp reef]

I hate you with passion

I cannot unsee it now

Lol

[empty grove]

I agree with terra here.

[sharp reef]

You agree with everyone beside me here.

Interesting.

[empty grove]

[sharp reef]

[gritty widget]

#chill

[swift fjord]

This made me think

What's the dual of support

For homomorphisms and things of that nature you have the kernel, but I guess the dual of the support would live in the range

[fading vale]

complement of the kernel

or

closure

of that

[gritty widget]

it's the closure of the set of points where the function is non zero

[delicate hollow]

Me want more perspective, what is best explanation given for a mapping cylinder
AK

[marsh forge]

i think viewing it as just a specific homotopy pushout is fairly natural

but you need to understand homotopy pushouts for that to be helpful

[delicate hollow]

yea ill learn soon
AK

[bright acorn]

Could someone give me an intuitive explanation of what a uniform space is?

Because yeah, I get what the definition formally is.

But idk how it is connected to the fact that somehow "uniform spaces are those topological spaces where talking about cauchy sequences and uniform convergence makes sense"

[fathom cave]

I cant think of a space which is locally euclidean but not hausdorff.

[reef shore]

Take 2 copies of R, quotient by the equivalence relation which identifies all corresponding points except for the 0s

It is called line with 2 origins

at either origin, you have euclidean space neighbourhoods but any neighbourhood of one origin intersects all of the other

[bright acorn]

This is an exercise at Tu

[plain raven]

another good example is provided by sheaf theory

if you look up the construction of the "espace etale" associated to a sheaf

then for most sheaves over a manifold (for example the sheaf of C^\infty functions over a smooth manifold)

the espace etale of that sheaf would be locally homeomorphic to R^n but not Hausdorff

[sleek thicket]

huh, when will the espace etale even be hausdorff? That feels weird

[plain raven]

it holds when you have a sheaf of extremely rigid functions which are totally controlled on a neighborhood by their germ at a point. so like, the sheaf of holomorphic or meromorphic functions on C or on a riemann surface

in this case the space will be more or less a manifold except that it won't be second countable. but i think each connected component will be second countable, and you'll be able to put a complex analytic structure on it

[cedar pebble]

damn I haven't thought about it like this before that's really nice

[plain raven]

[summer jolt]

Are questions about functional analysis appropriate here or shall I post in #advanced-analysis ?

[gritty widget]

better in analysis channel

[summer jolt]

ok thanks I'll post it there

[swift fjord]

@gritty widget just like how the kernel has a dual which is the cokernel

Wondering if the support has a dual objecy

[sleek thicket]

stupid answer incoming:
in AG, the support of a (finitely generated) module is the set of primes at which its localization nonzero. this is an closed subscheme (so affine) and thus it's dual to the ring of functions on it

anyways someone just told me that the category of LCH spaces with proper maps and the category of pointed compact hausdorff spaces are equivalent via the one point compactification and it's neat! not that strange but I'd never thought about it that way

do you not think this is cool tterra

[gritty widget]

it is 🤨

i share your opinion

makes sense but ive literally never thought of it

[sleek thicket]

i am reading about k theory

this is so weird

that's what this book claims!

I think the other direction should be removing a point

ah sorry compact hausdorff

[gritty widget]

[sleek thicket]

okay, so say you have a pointed compact hausdorff space (X, pt)

let Y = X \ pt

first question is whether this is even locally compact

right, any neighborhood of a point in Y must contain a compact subset (in X)

yeah

you think it's more plausible now?

the maps in the first category must be proper and the maps in the second must be based

[gritty widget]

based

[sleek thicket]

otherwise it's not even a functor

i decided to try looking at a thing on operator k theory

so it would be very surprising if there was an obvious objection based on C* algebras

https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.649.9955&rep=rep1&type=pdf
just skimming the section on topological k theory rn to see if I can look further

yup

that's why this came up

yup!

not sure if i will stick with it but what we were talking about last night seemed interesting

and I wanted something new to look ahead to as motivation

right

It's neat

Haven't gotten to the new-to-me stuff yet

well, besides that equivalence

but the quick topological k theory section is nice and I like the writing

will do!

i have come across my second surprise equivalence of categories

no no no

categories of left and right modules over a fixed ring being equivalent

[tight agate]

this isnt true

what

[sleek thicket]

uhoh

[tight agate]

it should be left modules over R is equiv to right modules over R^op

[sleek thicket]

i mean i do agree with this brofib

what inverse?

[tight agate]

how

inverse?

[sleek thicket]

right

so left and right G-sets are equivalent

[tight agate]

but this is for general rings

[sleek thicket]

but how do you do the same for modules over a ring?

R and R^op certainly don't need to be isomorphic

[tight agate]

you can do it for hopf algebras

[sleek thicket]

those are jsut self important groups

page 8 of blackader

does anyone have some nice noncommutative rings we can try

M_n(A) for A commutative doesn't work (I think, at least, since any ring is morita equivalent to a matrix ring over it)

hopf algebras don't work

[tight agate]

steenrod algebra

/s

[sleek thicket]

hmmmmmmm

what rings are there...

quaternions i guess

[tight agate]

what are you trying to prove/disprove

[sleek thicket]

i wanted to try and figure out if the categories of left and right modules are equivalent [for a specific example]

but that's probably too hard even with a specific example

ahh that condition would make sense here

and it seems a little more believable

yeah i don't see a functor either way

maybe you want to pass through idempotents?

i also just realized that looking at K^0 of a suspension is less weird than i had previously thought

like, a bundle E on SX should be determined by its restriction to the two cones and the gluing data on their intersection

which is like, two trivial bundles and an isomorphism between bundles on X

something like that

[tight agate]

it probably works for the quaternions

[sleek thicket]

i agree

quaternions are too nice

i was thinking maybe division rings are too nice

[tight agate]

or just quaternion algebras in general

[sleek thicket]

doesn't linear algebra work pretty well for them?

[tight agate]

they are 2-torsion in the brauer group

so Q and Q^op are brauer equivalent

[sleek thicket]

are modules over a division ring free?

i feel like ive worked this out

[tight agate]

yup

[sleek thicket]

ah okay sick

so the category of left modules is equivalent to N with matrices

and the category of right modules is equivalent to N with matrices^op

or something

(this is over a generic division ring)

[tight agate]

there is a unique simple left module

for any CSA

[sleek thicket]

like find a counterexample?

[tight agate]

for division rings the inverse thing works

[sleek thicket]

I was just thinking about whether it's true for division rings

oh nice

[tight agate]

but it is not true in general

wait

[sleek thicket]

Right

taking like a colimit over the n?

hm i have one concern

[tight agate]

so all division rings have order 2 in the brauer group?

[sleek thicket]

it's not true that every divison ring is iso to its opposite if that's what you're asking

sorry i had two replies two different people

[tight agate]

oh ok nvm

[sleek thicket]

i figured out my concern

with your thing ultra

so i think this might work

ah neat!

wait...

if A^2 = A then (A^T)^2 = (A^2)^T = A^T

wouldn't this be the orthogonal complement?

wtf there are like two complements

oh

[gritty widget]

C*

[sleek thicket]

hmm this is not required in Blackadar

well, not yet at least(?)

oh but we are taking similarity classes right

and A is similar to A^T

i am C*ing myself tterra

ah and we handle the nonunital case by adjoining a unit

like how we compactified in the LCH case

oh right if you want to look at the algebra of like, compactly supported functions or something then you'll only get a unit in the compact case

interesting....

oh dear, the book has just said there is also an "algebraic" k theory for C* algebras and it can be different from the "topological" one

i do not like that

ah okay sick

okay back to non-C* stuff for a sec bc i want to write down my thoughts

suppose i wanted to motivate or define K^-1(X) more directly in terms of X

well clutching(?) should(?) tell you that a bundle on S X is like two bundles on CX with an isomorphism between their restrictions to the equator X

but CX is contractible!

so it's like two trivial bundles and an isomorphism between their restrictions

this should(?) be the data of a number n, the rank of the trivial bundle, and an automorphism of the trivial bundle of rank n on X

or something

so it's something like, K^0(X) "is" bundles on X and K^-1(X) "is" automorphisms of trivial bundles on X

or something???

okay sorry ultra you can post now

[sleek thicket]

i saw "ultraornithologist was typing" before I said this

ah okay

it is very rough right now

I'm just thinking about this picture

yup

so K^-1(X) "is" bundles on this, but you can decompose a bundle on the suspension into bundles on the two cones (maybe you want to add a little open set fuzz, or take the complement of the poles)

but those are contractible spaces!

so bundles over them are trivial

so my next step, which i am the least comfortable with

is saying that this means a rank n bundle on the suspension "is just" the data of an isomorphism between two trivial bundles of rank n on X

but it feels like a nicer interpretation of K^-1(X)

what do you mean by asymptotically?

Sure, although i will likely have to ask you to stop and explain things

hmm

this feels more like an anologue of taking the cylinder

okay, so we're sort of looking at functions on the suspension which are forced to be zero at the poles, I guess?

feel free to continue though

I think the blackadar book did mention this

suspension for LCH spaces is multiplying with R

yeah, they corrected above

Right

Okay, that makes sense to me

M_n(C) is free on matrices with exactly one 1 and zeroes elsewhere in the entries with some relations between the generators specifying multiplication

tensoring preserves those relations and generators

but now you're looking at function-linear-combinations

ooh

weird

hahaha

it does give me another interpretation of K_1

right so a K_1 class should be a (similarity class of but let's ignore this) idemptotents in M_n(C_c(0,1))

oh and the functions are compactly supported!

and there's finitely many!

so you can find like, a start and end time between 0 and 1

view those as initial and terminal idempotents

oooh

er wait... that isn't right

hmm

it is still a continuously varying family of idempotents

which is neat

does the zero matrix correspond to a trivial bundle?

then yeah, it interpolates from the zero matrix to the zero matrix!

hahaha

no you didn't

It's another perspective

the way we got there is very different, at least

and as we've just seen

that's the whole reason K^1(X) is interesting

:P

kk

ohhh

OHHHH

[tight agate]

cant you motivate K^{-n} by just: this makes the LES work

[sleek thicket]

i think i get why multiplying by (0,1) works

if you think of it for LCH spaces

then you're going to compactify it

and the endpoints are the same!

[tight agate]

the puppe sequence pretty much forces this to be the defintion

[sleek thicket]

something like that

[tight agate]

if you want it to be representable

[sleek thicket]

yeah we were talking about this earlier

no no you're good

[tight agate]

sure, but if youre triangulated-brained this would be the first thing to do

[sleek thicket]

okay i have finished the topological k theory review

now to look at the next couple sections

and then learn topological k theory

all math subjects should be called calculus

[tight agate]

why brofibration.

[sleek thicket]

not many

ive been in a bit of a math-funk for the past couple weeks, talking about K stuff on here today and yesterday has been good

I think I'm going to try to go to more of them

[tight agate]

did you go to the diff geo ones

[sleek thicket]

I was really hyped for peter's lectures and then I absolutely hated them

cannot stand his lecturing style

[sleek thicket]

sadly nope

should have

did not

[tight agate]

they were fun

[sleek thicket]

i wish I had seen the gauss bonnet explanation

[tight agate]

he did not prove gauss bonnet

[sleek thicket]

oh lol

well then

i already took a class that did that

i meant to go to the Introduction to Quantitative Topology one but I woke up too late

yup

[tight agate]

yupo\

[sleek thicket]

same for Borel–Weil and Beyond

[tight agate]

yeah I missed borel-weil-bott

[sleek thicket]

do you know waht the topics next week are?

[tight agate]

nope

[sleek thicket]

the one lecture I'd been going to super consistently was the ultrafilters one

but it's ended

[tight agate]

but the topology talks are going to be on something more basic I think

[sleek thicket]

i think i made a good impression though

[tight agate]

Shame.

[sleek thicket]

okay goodbye everyone

i am going to get food and maybe watch some fireworks

[tight agate]

lol logic

[sleek thicket]

sure but fireworks are pretty

My family always used to watch fireworks in my parents bedroom

But then they moved down to California

Maybe I should see if my brother wants to watch fireworks with me

And break into our old home

Yes

Have been for 58 days

wtf

Oh

Right

The sex number

[tight agate]

lol

[sleek thicket]

In 25 days I will turn 21

[tight agate]

hermitian sex operator

In 25 days I will turn 21

[sleek thicket]

I don't believe you

[tight agate]

turn 25 days into what

[sleek thicket]

???

profit

[tight agate]

That makes sense.

[cedar pebble]

sham tell us about K0 and K-1

[sleek thicket]

You can find it in the recent chat history

[cedar pebble]

[sleek thicket]

I said I was going to take a break and get food

[cedar pebble]

oh sick I see it

I will read thanks

[sleek thicket]

I was thinking about bundles on a suspension being determined by the like, clutching construction

Lmk if you find anything that sounds sus

[sleek thicket]

K Theory? What's next, DN theory?

[cedar pebble]

Deligne Numford stack

[tight agate]

derived category of nixed notives

[sleek thicket]

Can you recover the n skeleton of a CW complex from the underlying topological space?

Put another way, if X is a topological space with two CW complexes structures Y, Z, is Y^n = Z^n?

[marsh forge]

no take the two most common spheres

but just like more generally

i can always add extra 0 cells

[gritty widget]

sussy

[sleek thicket]

That makes sense

[unborn lotus]

hello

does calculus on manifolds talk go in here

or is that for #multivariable-calculus

[nimble jolt]

it goes in #cats

[gritty widget]

[unborn lotus]

to classify all angle preserving transformations

its just scaling and rotating uniformly

right

[viral atlas]

Translations would work too, no?

[unborn lotus]

er

linear transformations

[viral atlas]

Oh

Okay

[unborn lotus]

😌

yes

so scale all vectors by some nonzero k

[viral atlas]

Sounds right then

[unborn lotus]

and then also consider for every angle that the vectors can rotate in, each angle has to be a fixed angle

[nimble jolt]

depends if you want to preserve orientation as well, otherwise you can reflect as well (about hyperplanes through the origin)

[unborn lotus]

ah reflections

i did not consider that one

[nimble jolt]

https://tenor.com/view/cat-mirror-play-confused-gif-7745917

regretting not just posting this instead of writing out that answer.

[sleek thicket]

cute!!!!

[sacred laurel]

Please, does someone knows the explicit formula, nor a reference to find it, which tells us explicitly what is the interior product between a 1-form and k-form in local coordinates ?

Before I kill myself doing computations

[plain raven]

https://en.wikipedia.org/wiki/Interior_product

are you talking about this?

I'm a little confused because you mention the "interior product of a 1-form and a k-form." Do you mean the interior product of a vector field with a k-form?

[sacred laurel]

If you prefer, yes

[plain raven]

$i_X ( a_{j_1,\dots, j_k}(x_1,\dots, x_n) dx_{j_1} \land dx_{j_2}\land\dots\land dx_{j_k}) =\
a_{j_1,\dots, j_k}(x_1,\dots, x_n) \sum_{i=1}^k (-1)^i X(x_{j_i}) dx_{j_1} \land\dots\hat{dx}{j_i} \dots\land dx{j_k}$

should be this, right? interior product is a graded derivation

ah wait i forgot the sign. it should be an alternating sum

[gentle osprey]

diligentClerk

[plain raven]

i usually look this stuff up in warner's foundations of differentiable manifolds

[sacred laurel]

thanks a lot

it is cursed as fuck

but thanks

sighs go back to PDE's, since i'm aware of it now

Wait I have to count tensor's coordinates that don't vanish under my vector field

Oh god

lemme die

Adieu

[flint cove]

Are there topological vector spaces which are T1, but not T2 (Hausdorff)?
In the case of prenormed spaces, that clearly holds, because 0 being closed implies that we have a norm.

[bleak helm]

T2 implies T1

[flint cove]

That is why I'm asking whether that inclusion is strict.

[bleak helm]

Ohhh

I misread

Cofinite topology on R.

Oh

I'm very bad at reading today

I will show myself out

[reef shore]

Luna star for Luna?

[flint cove]

Can you summarize the idea? I mean we have a natural transitive action by homeomorphisms, but how does that help us

And does it hold in generality, or only when we're locally compact?

(will read up on that later)

Thanks!

Oh, (x,y)↦xy⁻¹ is continuous and preimage of {e} is {(x,x)}=Δ

how cool

First time I've used the „diagonal is closed“ characterization of T2

[uncut surge]

hi friends, how obvious is it that the dimension of the space of roots associated to a complex semisimple lie algebra is equal to the rank of the lie algebra

(if it's even true)

(kinda wish there was a lie theory channel because these questions are always somewhere on the blurry line between #groups-rings-fields and #point-set-topology)

(my definition of rank of a lie algebra is equal to the dimension of a cartan subalgebra, for example)

oh my

it's just the dimension of the cartan subalgebra on the nose

wonderful, i am garbage, thank you

[flint cove]

You're not garbage, you're at most a garbage can. Maybe there is some garbage inside of you right now, but you clearly aim at throwing that out and replacing it with valuable knowledge.

[pseudo crane]

inspiring

[sleek thicket]

I disagree, lartomato is garbage

I would recommend a ban if possible

[sleek thicket]

Okay so the k theory stuff I was talking about yesterday is simultaneously more and less complicated

I'm having trouble thinking about basepoints

But if I ignore them

$\mathrm{Vect}^n(\Sigma X) \cong [\Sigma X, \mathrm{Gr}_n(\C)] \cong [X, \Omega \mathrm{Gr}_n(\C)] \cong [X, \Omega B \mathrm{GL}_n(\C)] \cong [X, \mathrm{GL}_n(\C)]$

[gentle osprey]

nonprincipal shamrock

[sleek thicket]

I'm not sure if the first isomorphism actually holds, since the maps on the right hand side have to preserve basepoint and the left ones don't...

(I'm not sure what the right basepoint is either, but that shouldn't matter?)

Anyways if this works then it gives a nice description of K^-1(X) I think

[flint cove]

would based and unbased homotopy classes of maps coincide if we had a path-connected cw complex?

[sleek thicket]

hm

So BG should be path connected

since it's a quotient of EG, which is contractible

I think if you take the right construction then BG is a CW complex(?)

Why am I working in this generality lol

The infinite grassmannian is a path connected cw complex

I guess I'm worried about losing generality since I was considering an arbitrary compact hausdorff space X

but anyways

You have (X, x0) and (Y, y0)

For each y in Y, choose a path p_y : y0 -> y

I guess

Oh hm I've become more confused

we're allowing rank to be only locally constant

instead of globally constant

Which means vector bundles aren't quite classified by maps into the grassmannians

Maybe you can take Y to be the disjoint union of all infinite grassmannians and say [X, Y] is iso classes of vector bundles? But then based stuff becomes even worse!

Ignore that

[hazy nexus]

Will do

[gritty widget]

[sleek thicket]

I am s truggling

This feels like it should be clean and nice but no

[hazy nexus]

I can't find a counterexample, but for some reason a part of me really wants to say no to this

[sleek thicket]

Same

My deleted thing was a fake counterexample

what if you look at maps from a point into a circle

There's only one based map

[hazy nexus]

I thought I had just found a counterexample too, fuck

Oh, I think you're onto something

[sleek thicket]

Can you just contracr all unbased maps?

Yeah I think so

You just like, connect them

Via a path

And that's your homotopy

[hazy nexus]

Yeah, what you said definitely works, since unbased homotopies of that map would correspond to loops in S^1

[sleek thicket]

wait what

I don't think it works

So there's one based map right?

[hazy nexus]

Have I forgotten all definitions

[sleek thicket]

An unbased map is just a point

A homotopy between two maps is just a path

[hazy nexus]

A based homotopy is a homotopy that always fixes the basepoint, no?

[sleek thicket]

Right but the based case is trivial anyways

[hazy nexus]

And we're looking at unbased homotopies v.s. based homotopies of based maps?

[sleek thicket]

I was looking at unbased homotopies of unbased maps vs based homotopies of based maps

But there's only one based function anyways

[hazy nexus]

Yeah, so in that case it's obvious

[sleek thicket]

Right

So you need to show there are at least 2 unbased maps that aren't homotopic

I think they're all homotopic

[hazy nexus]

Which is why I thought lux was asking about what I said

[sleek thicket]

oh

[sleek thicket]

Wait now I'm confused by this

I thought you meant it works as a counterexample

[hazy nexus]

I think I'm also confused, lmao

[sleek thicket]

hahaha

[hazy nexus]

I think I see where I got confused

Wait no

So lux was asking whether it is true that if f and f are two based maps into a path-connected CW complex, same domain blah blah, then there is an unbased homotopy from f to g iff there is a based homotopy from f to g

I feel like I formulated it in a stupid way, but oh well

[sleek thicket]

Oh

Well if lux was asking that, it's not that relevant to me

[hazy nexus]

So wait, what's Vect^n(X)?

Is it defined as the homotopy classes of based maps from X to GL_n(C), or is that a consequence of something?

[sleek thicket]

It's the set of isomorphism classes of rank n vector bundles on X

It's true but not obvious that this is in bijection with the set of homotopy classes of maps from X to the n-grassmannian in infinite dimensions

(X must be paracompact hausdorff, I'm even assuming compact, and no basepoints are present here)

me waiting patiently for someone who already understands k theory to explain what's going on to me:

[hazy nexus]

Ah, I see

Well not really, but I haven't been keeping up with math

[sleek thicket]

I haven't either! I'm trying to get focused because I'm a quarter of the way through this reu and haven't done anything

[hazy nexus]

In any case, I'm pretty sure that there's a loop space-suspension adjunction that doesn't involve basepoints, so maybe there's a way for you to use that instead?

[sleek thicket]

Oh that would be very helpful

Although I'm not sure how you would even define loop spaces there

[hazy nexus]

I mean, just all continuous maps S^1 -> X

Although I can't seem to find a source for what I'm saying

So maybe I'm completely wrong here, lmao

[sleek thicket]

oh hm I think compact(ly generated) hausdorff spaces should just have like, Hom(X×Y, Z) = Hom(X, Hom(Y, Z))

Or something

But suspension isn't X × S^1

hnnng

Yesterday we were talking about multiplying by (0,1) as an analogue of suspension

This then that Hom(X×I, Y) = Hom(X, path space of Y)

Or something

[marsh forge]

@sleek thicket if youre talking based maps that tensor hom works with smash in which case you get suspension with S1

[sleek thicket]

I'm confusing myself quite a bit

I am not max

[plain raven]

I think I have something that is relevant to the based/unbased maps question

[sleek thicket]

Consider the following chain of isomorphisms:

this doesn't make sense because it plays fast and loose with based vs unbased maps

I'm trying to make sense of it

Gr_n(C) here is the grassmannian of n dimensional subspaces of C^infty

Yes

And the right hand side of the first iso should be unbased maps

I have no idea if this is salvageable, to be clear. I wrote it down and then thought better and I'm trying to see if it's workable in any way

[plain raven]

Let $(X,x_0)$ be a space with basepoint, and $Y$ be a space, then if $y_0$, $y_1$ are two points in $X$, and you draw any path $\omega$ from $y_0$ to $y_1$, then there is an induced map $[(X,x_0), (Y,y_0)]\to [(X,x_0),(Y,y_1)]$, assuming that the inclusion $x_0\to X$ is a cofibration.

[gentle osprey]

diligentClerk

[plain raven]

I know that there are good examples of cases where this action depends on $\omega$. For example, if you take $(X,x_0) = (S^1,x_0)$ and $y_0 = y_1$

[gentle osprey]

diligentClerk

[plain raven]

Are you talking to me?

I often don't finish my arguments in a single message.

I'm just thinking out loud here. We know that the induced endomorphism of the fundamental group may not be trivial, in general this will be conjugation. If $f : X\to Y$ is a map, then we can always tweak this so that it is based by giving a path from $y_0$ to $f(x_0)$ and homotoping $f$ along this path. But in general this correspondence will depend on the choice of path, so there's not a canonical way to make it based.

[gentle osprey]

diligentClerk

[plain raven]

Yeah

[sleek thicket]

Ultra, I thought I could simplify my thoughts from yesterday like this

But I no longer think so

I also think it might be that the description I wanted yesterday is more like maps X × (0,1) -> GL_n(C)

Well yesterday I thought I could just homotopy out the factor of (0,1)

But I was having trouble doing that

right

But I was looking at this from the topological pov

And saying like

"a bundle on the suspension is determined by the bundle on the two cones, each of which is trivial, and the gluing data on the intersection

[plain raven]

So what this shows is that two distinct elements of the fundamental group may be freely homotopic - because any two elements of the fundamental group identified under the monodromy action are freely homotopic. This means that the forgetful map from based maps to unbased maps is not injective, in general. Which I think gives an answer to the question.

@hazy nexus does this make sense?

[sleek thicket]

But it's no longer clear to me that you can just look at the intersection instead of including some factor like (0,1)

Sick

I am not confident in this

[sleek thicket]

Also my project mentor wrote down a computation like this last Tuesday

So I am confused

Something to ask about at least

It was this:

$\mathrm{Vect}^n(\mathbb{S}^k) \cong [\Sigma \mathbb{S}^{k-1}, \mathrm{Gr}_n(\C)] \cong [\mathbb{S}^{k-1}, \Omega \mathrm{Gr}_n(\C)] \cong [\mathbb{S}^{k-1}, \Omega B \mathrm{GL}_n(\C)] \cong [\mathbb{S}^{k-1}, \mathrm{GL}_n(\C)]$

[gentle osprey]

nonprincipal shamrock

[sleek thicket]

As a way to explain the clutching construction

indeed

[plain raven]

What resources are you using to study this stuff

It seems interesting

[sleek thicket]

Maybe I just need to remember how to justify the clutching construction (specifically that you only need to look at the equator, not any open overlap) and then rejustify it for the suspension

[sleek thicket]

I took a class on bundles in the winter using an unreleased book

[plain raven]

ok

[sleek thicket]

Now I'm trying to learn k theory

[plain raven]

nice

[sleek thicket]

using Hatcher's book

And this other one for operator algebras stuff

Uhh

Blackadar

Wym ultra?

[plain raven]

Nice

[sleek thicket]

Oh it's true for spheres, sorry

That's why I didn't think hard about it at the time

I've already seen the clutching construction and justified it in a different way

[hazy nexus]

ye, I think so

[sleek thicket]

Oh dear, I went to sit in the sun and they're are bees everywhere

Fear

Going to sit in a patch of dead, dry grass

[hazy nexus]

Get on my level

[sleek thicket]

dog!!

[hazy nexus]

He says hi

[marsh forge]

i love him

[hazy nexus]

He's sleepy, yeah

Although that's his natural state

He never refuses a nap

[gritty widget]

sweet dog

[plain raven]

what is the name of this dog

[gritty widget]

dog

[plain raven]

beautiful.

"Dog."

[gritty widget]

that's what ultraproduct would name it

[hazy nexus]

His name is Gaston

[plain raven]

What a lovely name

[sleek thicket]

Does he eat eggs?

(a large small barge though)

[hazy nexus]

Yes

And no one is as lovely as him

[flint cove]

okay, so to pop the question, where did you get this from? your head or karoubi?

[sleek thicket]

what's karoubi?

[flint cove]

the only text I know that does vector bundles in full generality with all the suspension stuff

[sleek thicket]

oh

well it wasn't that

[flint cove]

(and I kinda did not read it and switched to may because of that, I'm scared of capital Σs)

[sleek thicket]

From my head, I was trying to shortcut some thoughts I'd had yesterday

But also in the process I remembered my reu mentor writing something similar

[gritty widget]

Hello, I am trying to understand this formula for intersection number when we flip joint orientation $K \circ L = (-1)^{k*l} L \circ K$

That k*l as I understand it is referring to flipping the basis vectors of the tangent space for each manifold K and L until we've transposed the joint orientation at an intersection. My question is why do the flipping in what seems like the most possible steps?

[gentle osprey]

Abdul Aziz

[gritty widget]

So why do this

Rather than this which takes fewer steps and achieves the same result?

A video I was watching on this used k * l steps to conduct the transposition so I was wondering if that was necessary, for ultimately you'd arrive at the same result with a better transposition algorithm.

[plain raven]

Well, you have to have a consistent sign change procedure. Like, I can understand your reasoning but whatever transposition algorithm you come up with, you would have to make sure that it yields the same sign each way

Maybe that's not helpful. Think of like, working in R^n for some large n, and you have a bunch of vectors v1.... vn that span some kind of n-dimensional paralellogram.

Say that we assign this thing volume V.

Now you are asked to come up with some systematic way of assigning volumes in a linear way to the parallelogram spanned by (+/- v1, +/- v2, ... +/- vn)

always +/- V, but the question is what sign convention you should give it

The determinant of the vectors is pretty much the only good answer to this question

[gritty widget]

That makes sense to have some consistency with the sign changing you need something like the determinant.

[shut moat]

is there a classification of the measure preserving diffeomorphisms of R^n/ a way to "write them down"?

[lean marten]

Maybe matrices with determinant 1?

At least that should encompass the linear case

And in fact the requirement is probably the diffeomorphism has derivative with determinant 1 at each point

So it preserves local measure or someone

[empty grove]

Determinant should be ±1 yeah

[lean marten]

Something *

Oh lol forgot -1

[empty grove]

Hmm there do seem to be non linear ones as well

I think shearing by some smooth function could work

Like (x,y) → (x, y + e^x)

Since shearing preserves area of shapes at least. You could probably prove that it does preserve measures using fubini

[sleek thicket]

$\mu(E) = \int_E |\det DF|$ for any measurable set $E$ right?

[gentle osprey]

nonprincipal shamrock

[sleek thicket]

I want to like

Take E shrinking down

And say |det DF| = 1 everywhere

Something something lebesgue differentiation

@empty grove this actually proved your example works I think. The jacobian of your thing should be like
1 hjekeksj
0 1

[empty grove]

Yeah seems reasonable

[sleek thicket]

and this always has det 1

[empty grove]

Neat

But is this sufficient?

[sleek thicket]

It's def sufficient

I'm just not 100% sure it's necessary (it almost certainly is I'm just having some brain fog)

[sleek thicket]

Take μ(F(E))

Write as an integral

Use multivariate change of variables

[empty grove]

Right yeah

Necessity should come from |det DF| > 1 at x → > 1 in a neighborhood of x

And same for < 1

[sleek thicket]

right

yeah I was being silly

about using lebesgue diff

DF is just continuous lmao

So yeah the condition is equivalent to having jacobian det ±1 everywhere

[empty grove]

Let me google lebesgue differentation

[sleek thicket]

But like

Still very broad

[empty grove]

Broad?

[sleek thicket]

The class of all measure preserving diffeomorphisms

[empty grove]

Oh yeah I expected special linear

[sleek thicket]

yee not quite

[shut moat]

yeah, i was wondering if there was a characterization of functions with |det dF|=1

more than just that condition on the jacobian

[sleek thicket]

hm, doesn't really feel like it

[shut moat]

rip, i agree it seems hard

i ran into a paper that showed that the measure preserving diffeos of the torus are unclassifiable

while digging around

https://arxiv.org/abs/1705.04414

[barren sleet]

we dont understand this definition of corresponding vector fields, because the vector fields on M and N are taken to be in the ambient space, but the derivative of f is from the tanfgent space of M to the tsngent space of N. how does this definition work when the vectors in the vector field arent tangent to the manifold?

[gritty widget]

i feel like you would have to assume they're actually tangent to the submanifolds

does milnor say explicitly?

[barren sleet]

well the intended use of this is when they are perpenicular to the manifold

hes doing gauss mapping

my idea was to extend the map to a smooth map between the ambient spaces

[gritty widget]

yeah, but then you have to check that it doesn't depend on the extension somehow

hmm

[barren sleet]

yeah thats whats annoying

and i doubt it is

[gritty widget]

i have a copy of this book, give me some time and ill see if i can figure it out. what page?

[barren sleet]

32

can you ping @drifting sundial or me when you figure it out? im gonna sleep pretty soon

[gritty widget]

yeah @barren sleet @drifting sundial im pretty sure milnor is only talking about tangent vector fields here

[barren sleet]

:(

but it makes a hell of a lot more sense

[gritty widget]

i don't see why the definition of vector field correspondence should make sense if the vector fields aren't necessarily tangent to the manifold

[barren sleet]

yeah I see

[gritty widget]

btw this condition is sometimes called "f-related" in modern books, and it's only ever defined for tangent vector fields

[barren sleet]

ok yeah then I definitely misinterpreted it

[drifting sundial]

just for context, I was originally trying to prove that the degree of the gauss mapping is invariant under diffeo

is that true in general?

[barren sleet]

well it's pointless now

the unsigned degree*

[drifting sundial]

oh I forgot to clarify that I'm only considering boundaries of open subsets of R^n

things get a bit iffy in general manifolds

[gritty widget]

let me stare at the book to make sure im not talking out my ass lol

ive had this book on my shelf and have been meaning to read it for a week now

[rare storm]

x will be in atleast one of the open sets

but we cannot take a neighbourhood x that is a subset if x actually isnt in the set?

[bleak helm]

True

But if x is in H, then x is in each set G_i

By definition of intersection

[rare storm]

oh yikes

i read it as union

[bleak helm]

Ah, okay. Yeah, there is a different proof for union. Then you just know it's in one of the G_i

And that will be enough

[rare storm]

ty

[barren sleet]

he meant the degree

[gritty widget]

yeah sorry about that

i didn't get a lot of sleep

[flint cove]

Is there a binary coproduct of real topological vector spaces?
Or, explicitly: is there a coarsest topology that makes both injections as well as addition/subtraction/scalar multiplication continuous?

I would presume that the usual product top would suffice but I'm not sure why that would be initial in this respect

Not always, only if blank is a collection of arrows into our space, not sure whether that works with self-referential stuff like X\times X\to X like we need for the multiplication map

[gritty widget]

[flint cove]

Or is my top game just too weak lol

In what way?
I am dragging most of this out of my nose btw, in my analysis literature TVSes are not talked about per se, and the topological groups texts I recall did not care about vector spaces at all.

I mean we can be sure that the underlying VS is just the product because the forgetful functor to Vec has both adjoints, so preserves colimits

A friend of mine just found https://mathoverflow.net/a/345202/66999 and that sounds a bit complicated, would need to work through that. Perhaps the binary case becomes way simpler.

But man, This TopGrp/TVS stuff is all so intricate! I just wanted to know whether there is an obvious „simplest“ ctbl-dim TVS (hence infinite colimit) and whether that is normable. But hey, even binary stuff is non-obvious, lol.
But I guess that is more #advanced-analysis .

[sleek thicket]

this seems to give a simpler construction in the case of a coproduct instead of more general colimits https://mathoverflow.net/a/345160

same question, different answer

[flint cove]

Oh, something something attention span and lack of sleep

Wow, I've never seen such a construction. So the Idea seems to be that for large enough quotients, we definitely know a VS-top exists, and then we represent E as this subspace of that big product for all admissible quotients.
Not quite sure whether making the product smaller coarsens or refines the resulting top we get on E

e.g. restricting N to be of cofinite dim would still give us an embedding from E into the product

[verbal wraith]

The co-product for vector spaces is just the direct sum which is the same as the co-product for abelian groups. right?

Wouldn't the co-product for topological vector spaces also then be the co-product for topological abelian groups?

I don't think you need to make such weird constructions.

Oh I guess this is just how he's constructing the co-product for topological abelian groups.

It does seem too complicated.

G * H x G * H -> G * H is continuous iff the four maps G x G -> G * H, H x H -> G * H, H x G -> G * H, G x H -> G * H are continuous.

and all of those maps are continuous

[pastel thistle]

guys any help with question

dont mind the spanish is just. We have those subsets of S2 and is asking about the posibility of a continious function from S2 to [0, 1] ^ N

where the function maps A1 to (0, 0, 0, ...) and A2 to (1, 1/2, 1/3, ...)