point-set-topology

[sleek thicket]

maybe i should just forget about the affine case

[tough imp]

Okay suppose we know it for affines

[sleek thicket]

oh or not

yup

[tough imp]

We have a V < Y such that f^-1(V) is a finite union of affine opens U_i which are finite type blah blah so they're finite

[sleek thicket]

wait what

We can immediately wlog Y affine

[tough imp]

Okay sure

so f^-1(Y) is a finite union of affine opens U_i which are finite

over Y

[sleek thicket]

Yup

No

Finite type over Y

[tough imp]

well if we know the affine case

oh I guess you're right

[sleek thicket]

Yeah

[tough imp]

well okay, just pick out some affine open, call it U

[sleek thicket]

You get subsets of Y for each i

[tough imp]

and suppose you have a V for that U

[sleek thicket]

In X?

[tough imp]

yeah

[sleek thicket]

Sure

[tough imp]

Does U contain all fibers of eta?

[sleek thicket]

So cover X by finitely many affines

Then for each apply the affine case

this gives Vi in Y with f^-1(Vi) intersect Ui -> Vi finite

We intersect because you're applying the affine case to the restricted map Ui -> X -> Y

[sleek thicket]

I don't see why it would have to

But I'm also not sure why you're asking

Oh wait

It does

Sorry

[tough imp]

oh it does?

[sleek thicket]

No

It doesn't

My bad

[tough imp]

can you force one to?

[sleek thicket]

For a second I thought it did but I don't see why it would

Why are you asking it to? How does that help?

[tough imp]

If you can, you can look at its complement

it doesn't map to eta because no fibers of it are in it

call this thing Z for now

[sleek thicket]

sure

[tough imp]

Z is also quasicompact

[sleek thicket]

Because it's closed in a qcompact space?

[tough imp]

well, its a finite union of affine opens right?

err

[sleek thicket]

Lets go with what I said

[tough imp]

sure

anyway

if you have a qc morphism (so automatic when the source is qc)

f: X -> Y such that it misses the generic point

you can find a nonempty open of Y, containing the generic point with empty inverse image

[sleek thicket]

wait so are we giving Z a closed subscheme structure?

[tough imp]

sure

give it the reduced structure

[sleek thicket]

sure

Also yes I agree with this

Just take the complement of the closure

I guess this is totally topological

[tough imp]

yeah, basically

anyway, this solves the problem

because this gave an open of eta

which has empty inverse image in the complelemtn of U

aka the inverse image is in U

so now by being local on the target

you have f^-1(V) < U

is finite to V

[sleek thicket]

Solvrss the problem as in, reduces to the affine case?

[tough imp]

yeah

[sleek thicket]

I think I follow this

[tough imp]

like, if we know the affine case right

[sleek thicket]

or do I

So back up

[tough imp]

We have a U < X, and a V such that U -> Y is finite

U is affine open

[sleek thicket]

right

[tough imp]

look at the complement of U inside X

then you can grab an open V < Y

[sleek thicket]

(assume U can be chosen to contain the fiber of the generic point)

[tough imp]

such that f|_U^c^-1(V) is empty

[sleek thicket]

Yes okay

[tough imp]

so this says f^-1(V) < U

[sleek thicket]

Right

And now we're just restricting a finite map

[tough imp]

yeah

[sleek thicket]

So really no part of this is about U Adrien

*affine

[tough imp]

well

it sorta is

[sleek thicket]

We can just replace X with any neighborhood of the fiber of η

Howso?

[tough imp]

sure

right

Well I mean it does need an affine containing all fibers of eta

but it doesn't use affineness all that much

[sleek thicket]

I don't see what you mean

Oh you're saying to apply the affine case

[tough imp]

yeah

[sleek thicket]

Sorry I just meant like

This reduction isn't really a reduction to the affine case

It gives us a way to shrink X

And maybe we can find an affine fitting the shrinking conditions

[tough imp]

sure

also I think the proof I gave was a tiny bit off

[sleek thicket]

Rip

[tough imp]

but I think it still works

no like, I just wasn't careful about shrinking

[sleek thicket]

Oh okay

[tough imp]

so like, you have U -> Y, both affine

[sleek thicket]

Oh wait yeah

How do you know the smaller map is affine still

I guess you just choose V to be affine

And then you're fine

[tough imp]

right

okay so

all we need is

to find an open U < X, such that f|U: U -> Y is finite

and that all fibers of eta are in U

[sleek thicket]

Right, I agree

Ah

This isn't possible

[tough imp]

And we can also shrink Y if needed

[sleek thicket]

Yeah you will need to shrink Y in some cases

[tough imp]

to any open

[sleek thicket]

A^1 \ {0} -> A^1

[tough imp]

So this almost directly follows from the affine case

in that case we have U -> Y finite type

[sleek thicket]

?

Don't tell me how to do the affine case

I haven't thought about it yet

[tough imp]

get a V < Y such that f^-1(V) -> V is finite

[sleek thicket]

Sure

[tough imp]

oh wait, but then f^-1(V) automatically has all fibers of eta?

as long as V contains eta

[sleek thicket]

No

it's not f^-1(V) -> V

It's f^-1(V) intersect U -> V

[tough imp]

oh sure

[sleek thicket]

You apply the affine case to the restricted map

[tough imp]

right

[sleek thicket]

I can't think of a case where the fiber has multiple points lol

[tough imp]

I think they exist

something like umm

[sleek thicket]

Oh I guess a dvr going into the localization maybe?

Or even Z -> Q?

No

That's backwards sorry

Z -> Z/pZ?

[tough imp]

k[x] and k[x,y] I think?

[sleek thicket]

nope still backwards

Ah hm

[tough imp]

oh that's just not genericdally finite

[sleek thicket]

Oh right

[tough imp]

maybe if it's F_p?

[sleek thicket]

From or to what?

[tough imp]

for k = F_p

maybe it's generically finite

but idk

probably not

[sleek thicket]

Oh for k[x] -> k[x, y]?

[tough imp]

yeah

[sleek thicket]

I think it's still not generically finite

Spec F_p[x, y] still has a lot of primes

[tough imp]

yeah

[sleek thicket]

Wait actually

Why isn't it finite for k = C?

Oh I guess like the generic point of a horizontal line

Yes

[tough imp]

yeah

[sleek thicket]

(y+c)

I was thinking of vertical lines and getting confused

[tough imp]

generic points of irreducible curves usually sit over 0

I think

When I workeed through it earlier

[sleek thicket]

Makes sense

hmm

I'm thinking of primes whcih split

So in Z -> Z[i], the prime (5) is the preimage of the primes (1+2i) and (1-2i)

We can localize both at the multiplicative set S = Z\(5)

So we have Z_(5) -> Z_(5)[i]

Oh but rip the generic point of the left is (0)

Not (5)

[tough imp]

Are the fibers of eta

generic points of irreducible components of X?

wait

X is also integral

[sleek thicket]

Yeah

It's not that

[tough imp]

but then only one of them

There's only 1 irreducible component

[sleek thicket]

Oh what about like

y^2 = x ^2(x+1) projecting onto the y axis nope

[tough imp]

wait wait

[tough imp]

is what I originally asked true?

[sleek thicket]

Which part?

[tough imp]

fibers of eta

[sleek thicket]

X has exactly one irreducible component

[tough imp]

yeah

[sleek thicket]

So uh

Probably not?

[tough imp]

so like, no

[sleek thicket]

What does this even mean...

So if we assume X, Y affine

[tough imp]

time for affine groups

algebraic groups

[sleek thicket]

Oh shit

I totally forgot

[summer jolt]

I'm trying to understand the following calculation using the tetrad formalism

However for the coefficient of gamma_2[12] I don't quite get why is there a factor of 2 in the denominator

Anyone have any idea?

[gentle osprey]

Moth In Shambles

[fading vale]

The first direction is trivial: if f extends to f' then pf' extends pf

The second is harder though

I'm not sure where being a cofibration comes into this

I guess we have $q \colon Q \to P$ with $H \colon P \times I \to P$ and $K \colon Q \times I \to Q$ where $H(x, 0) = qp(x), H(x, 1) = x, K(x, 0) = pq(x), K(x, 1) = x$

[gentle osprey]

Moth In Shambles

[silver umbra]

I'm self studying basic topology atm

considering R with it's usual topology

are closed sets the union of intervals?

not necessarily closed intervals--just intervals in general

and also we let singletons be intervals

[gritty widget]

As in, you would want to include open intervals?

[cerulean oriole]

Then anything is a union of intervals.

[gritty widget]

(I think the answer is yes, closed sets are a countable union of closed intervals, including singletons and rays, but I'm not sure)

It is true that open sets are a countable union of disjoint open intervals, maybe there is an easy argument to prove the thing about closed sets

[silver umbra]

okay hmm, perhaps not including singletons

[empty grove]

They arent

the cantor set is an uncountable totally disconnected closed set

[gritty widget]

Ah, damn

[silver umbra]

oof

[gritty widget]

Wait

[long gorge]

The corresponding statement for open sets would be that they're countable intersections of open intervals, which isn't true

[empty grove]

It is, R is second countable

with only intervals in the countable basis

we also get disjointness as 8da said

[long gorge]

I don't see how that applies to intersections

[empty grove]

Oh sorry I misread

[long gorge]

Yeah I wasn't saying anything much of substance, but I feel like these sorts of intuition traps are avoided better by thinking of open sets rather than closed sets

[empty grove]

Wait Cantor set is an uncountable totally disconnected subset of R, so it seems intuitive that it would divide R into uncountably many components, but the complement is open so must be a union of countably many path connected open sets

Wtf

[gritty widget]

Yeah, that is weird to me

[empty grove]

[gritty widget]

It kind of feels like I'm misunderstanding something

[empty grove]

Could be legit

Would definitely not be legit if Cantor set were discrete

But I don't know how to think about totally disconnected sets that are not discrete

Oh yeah countable cover of complement by open intervals is also kinda obvious

Following the construction of the Cantor set

By removing the middle thirds

[empty grove]

It feels unintuitive because it seems that you could take any countable cover of the complement by intervals and get a surjection onto the Cantor set by taking infimums of all the open intervals, but that doesn't work because there are decreasing convergent sequences in the Cantor set

[verbal wraith]

One thing you can do is you can define discrete partitions of R. A discrete region = a union of parts in a discrete partition. Closed sets are exactly the intersections of discrete regions. [By discrete partition I actually mean a cover of R by intervals which intersect at end-points]

[honest terrace]

the same thing happens here than with R\Q, no ?

[empty grove]

Q won't be closed

[honest terrace]

oh huh yeah, but still

like I'm trying to relate this counter-intuitive fact to something I understand a bit better

that Q is countable, but dense in R (so rational between irrationnals and conversely) which is uncountable

the unintuitive part in both claims seems related

(not sure if I'm being clear here)

[empty grove]

I mean Q complement can't be a union of open intervals because it's not open

[honest terrace]

yes, I get that

I'm not speaking about the open/closed stuff here

just that you can fit uncountably many elements between countably many elements in R, or the reverse

[empty grove]

Oh yeah

That's true

R is weird

[honest terrace]

A friend of mine always says that R isn't nice. As I learn more maths, I more and more lean toward agreeing with him

[sharp yoke]

what's nicer

[unborn lotus]

C

idk

[bleak helm]

F_2

[chrome dew]

true, in R you can't always do sqrt(x), but in F_2 you can always do sqrt(x)

[sleek thicket]

in fact you can do nth roots for all n

which, by definition, means thst F2 is algebraically closed

[strong heron]

I have written this proof of orientability of the cover of an orientable manifold. Can someone verify it and let me know? Thanks.

[cursive flume]

why is the set of all smooth vector fields on a manifold infinite?

i.e. why is the vector space of all smooth vector fields an infinite dimensional R vector space?

I can see it is a vector space,but can't see why inf dimensional

[chrome dew]

try to list out its basis vectors

[gritty widget]

Hm, a guess: the space of tangent (cotangent?) vectors at a point is finite dimensional (the dimension of the manifold). Along these lines, if you could extend a set of k tangent vectors at various points on the manifold to a full smooth vector field, then letting k go to infinity would show that it is infinite dimensional.

On the other hand, maybe another approach is to consider some path on the manifold, and consider various parametrizations of it. All the various parametrizations should give ride to an infinite dimensional vector space of vector fields along the path, which can probably be extended to the whole manifold

(I am just sort of bullshitting out loud, so no guarantees this is correct. And I do not even remember enough about manifolds to know tangent vs cotangent 🙂 )

[cursive flume]

so the tangent space has the same dimension as the manifold,yes

I see this

[wanton marsh]

that's like asking why the space of smooth functions is infinite dimensional

[cursive flume]

yes this is what I can't see

[chrome dew]

yeah, sounds like you're conflating it with the tangent space or something

[cursive flume]

a vector field is a smooth map $X:M \to TM$

[gentle osprey]

ProphetX

[cursive flume]

but why is the collection of all such objects infinite dimensional?

[wanton marsh]

if you pick M = R^n then it's a smooth map R^n -> R^n

and like

you can have polynomial maps

that's already infinite dimensional

[gritty widget]

Ah yeah, I guess it's easier to just reduce to the Rn case immediately

[uncut surge]

Yeah it's a fun exercise to try and show why already all the monomials 1, x, x^2, ... on an open interval of R are linearly independent

[chrome dew]

honestly a picture might help

https://www.sculpteo.com/blog/wp-content/uploads/2017/10/Webp.net-resizeimage-3.png

there's a lot of ways to put vectors on a manifold

@cursive flume don't delete your messages

[cursive flume]

I thought it was very stupid what I said that's why I deleted

I wanted to ask whether thinknig of assigning a vector to each point of the manifold would make this infintie dimensional

but that's not true,because by definition one vector field is such an assignment

[versed pivot]

You can just pick an infinite sequence of disjoint open balls on the manifold and put a smooth vector field on each of them that extends to zero outside its little ball

[versed pivot]

Kind of the same idea as this except it's smooth

[wanton marsh]

the space of all functions from R to R is.. like... cardinality-of-P(R)-dimensional, so very very big infinite

for smooth functions I don't know if it's cardinality-of-R-dimensional or cardinality-of-N-dimensional

most likely cardinality-of-R-dimensional

so very big infinite

[marble socket]

yep constant functions are smooth

oh wait you said dimensional

[gritty widget]

Handsome

[cursive flume]

silly question, but when we define the commutator of vector fields, when we write $[X,Y]:=XY-YX$, do we mean $[X,Y]:=XY+(YX)^{-1}$, where - denotes the additive inverse of YX?

[gentle osprey]

ProphetX

[cursive flume]

since in particular there is a vector space structure,we can make sense of an additive inverse

thanks I was just not sure about this and wanted to clear it down

😮 this is enlightening

I never thought about it like this

but now it makes sense

[gritty widget]

commutator in the uhhh ring of linear operators on smooth functions

to be pedantic

ring commutator and group commutator are slightly different, which is why i butt in

[cursive flume]

but nevertheless this holds in ring case too?

or not anymore

I think it is just a notation to make life simpler and not write (YX)^{-1} always

what I meant is the remark here

cause if I just write XY-YX i'm not sure what - means

[chrome dew]

rings have additive inverses

[cursive flume]

ahh right

so the ^-1 denotes the additive inverse always

but not in cae of rings

[cursive flume]

if I write out explicitly additive inverse then even though we ahve algebra structure the sentence is correct,right?

ahh so it's the convention

I see

[gritty widget]

lie bracket

[pearl holly]

Wait, if a set is both closed and open then the closure of that set is the set itself, right?

[gritty widget]

if a set is closed then its closure is itself, so yes

[pearl holly]

okay good, thank you!

[digital wraith]

this is probably an extremely stupid question but if A is a retract (though not necessarily a deformation retract) of X why does the inclusion map induce an injective homomorphism of the fundamental groups

couldn't there be 2 paths that are homotopic through points in $X\setminus A$ so they are separate elements of $\pi_1(A,x_0)$ but the same element in $\pi_1(X,x_0)$

[gentle osprey]

Lagrange the Multifarious

[empty grove]

The injectivity says that exactly that thing cant happen

[digital wraith]

injectivity of the inclusion map?

or of the homomorphism between fundamental groups

[empty grove]

injectivity of the homomorphism

[digital wraith]

well yeah that's what I'm asking. Why is the map injective

saying "it's injective because it's injective" doesn't really help

[empty grove]

ah so if r is the retraction, i is the inclusion, then ri is identity, so r*i* = id, so i* must be injective

[digital wraith]

oh right that makes sense I think

[cursive flume]

is U(1) simply connected?

I found that U(n) is not simply connected,but it did not specify if n>1 or n>=1

[gritty widget]

U(1) is the circle S^1 right

[cursive flume]

ah

[gritty widget]

that's not simply connected

its pi_1 is Z

[cursive flume]

then why are we allowed to get the group U(1) from its lie algebra u(1)?

[gentle osprey]

ProphetX

[cursive flume]

but then the exponential map won't be surjective, or?

also is the statement right that u(1)=purely imaginary numbers? I didn't find it anywhere in books/internet but I computed it myself and seems to be so

[gritty widget]

i believe so

[gritty widget]

could you elaborate? i'm not too comfortable with lie theory, but i know that if we have a lie algebra g then we ought to be able to find a simply-connected lie group whose lie algebra is g, but i don't think that theorem says anything about not-simply-connected lie groups

i am interested in knowing the answer

[cursive flume]

yes this is my question too

in physics we always simply write u(1)->exp(u(1))

I know that we can lift any rep of a lie algebra to a rep of a smiply connected lie group

what would be the simply connected lie group to u(1)?

[gritty widget]

u(1) itself?

lol

[cursive flume]

its generators

$U(1)=e^{i \alpha}$

[gentle osprey]

ProphetX

[cursive flume]

this is what we always write in physics

the lie algebra of $U(1)$, i.e. $\text{u}(1)=i\alpha$, then why can we 'just exponentiate it'?

[gentle osprey]

ProphetX

[cursive flume]

or we actually are not allowed but we are sloppily doing this?

[gritty widget]

[cursive flume]

lol wtf

[gentle osprey]

ProphetX

ProphetX

[cursive flume]

is this coincidence?

i'm mega confused lol

is this e that e?

[gritty widget]

if it helps, im pretty sure this is the example the exponential map in lie theory is named after

[cursive flume]

but U(1) is not simply connected

why is U(1)=exp(u(1))

[gritty widget]

[cursive flume]

this is so confusing

[thorny nova]

Hi, I have a question about metric spaces

If I have a compact space and a convergence sequence. Is the set of point of the sequence a close set?

[gritty widget]

include the limit and the answer is yes

[empty grove]

Only if you include its limit

[gritty widget]

hmmCat

[thorny nova]

Ty

[gritty widget]

for example, the sequence 1/n in [0, 1] (a compact metric space!!) is convergent, but the set {1, 1/2, 1/3, ...} is not closed (after all, it doesn't contain the limit point 0). append 0 and it is

[wanton marsh]

isn't the simply connected lie group with lie algebra u(1) just (R,+)

[gritty widget]

that's what i tried to say, but yours is clearer ig

[summer jolt]

Hi guys, I have a question about the calculation of the Riemann curvature tensor. To calculate the Riemann coefficient for a metric $g$, one can employ the second Cartan's structure equation:

$$\frac{1}{2} \Omega_{ab} (\theta^a \wedge \theta^b) = -\frac{1}{4} R_{ijkl} (dx^i \wedge dx^j)(dx^k \wedge dx^l)$$

and using the tetrad formalism to compute the coefficients of the curvature tensor.

Now I'm trying to properly understand this method, I was doing this exercise for which I obtained:

$$\frac{1}{2} \Omega_{ab} (\theta^a \wedge \theta^b) = -\frac{1}{4} A (dx \wedge du)^2 -\frac{1}{2}B(dx \wedge du) (dy \wedge du) - \frac{1}{4} (dy \wedge du)^2$$

However, from here I'm not quite how I would read the coefficient for the Riemann tensor. From here it seems for example that we have:

$$-\frac{1}{4} A (dx \wedge du)^2 = -\frac{1}{4} R_{xuxu}(dx \wedge du)^2$$
$$-\frac{1}{2}B(dx \wedge du) (dy \wedge du) = -\frac{1}{4}R_{xuyu}(dx \wedge du)(dy \wedge du)$$

The answer is supposed to be $R_{xuxu} = \frac{1}{2}A$ and $R_{xuyu}= \frac{1}{2} B$, however I don't quite understand how this would be obtained from the equation above. I assume it has something to do with the symmetries of the Riemann tensor but not quite sure.

[gentle osprey]

snypehype

[tight agate]

not being simply connected isnt a problem

not being connected is

the conditions you want are compact and connected I think

[cursive flume]

so the exponential map is surjective iff the group is connected and compact?

[tight agate]

not iff

for example the exponential map R --> R_+^* is surjective

but the space of positive real numbers is not compact

[cursive flume]

okay, so it is true that if the group is compact and connected then the exponential map is surjective

but that's not the only case when it is

[tight agate]

and there are tons of examples where the exponential map surjects onto a non simply connected thing

all compact connected complex lie groups

that's how you prove all compact connected complex lie groups are quotients of C^n by some lattice

the exponential map is surjective, with discrete kernel

[cursive flume]

do you by chance know a self contained reference where I can look this up?

so far all the books on compact lie groups I found are so advanced

[tight agate]

I think mumford's abelian varieties does it

but it depends on what you mean by self-contained

[cursive flume]

for someone who hasn't done proper math undergrad courses I try to catch up by learning what I need in between

[tight agate]

if you know what a complex manifold is you should be fine

and some complex analysis

[strong heron]

When we write S1× S1 as the torus, which component is the meridian? Is there a standard convention?

[cursive flume]

if two lie algebras have the same structure constants, they are isomorphic

is the converse true aswell?

is this an iff statement?

[gentle osprey]

ττερρα

[cursive flume]

how can I see this?

[gentle osprey]

ττερρα

[gritty widget]

first equality b/c phi is a lie alg homomorphism

[cursive flume]

ohh and second equal because this is a linear map

by def of iso

[gritty widget]

gotta be careful when you say "structure constants of a lie algebra"

structure constants are always relative to a basis

[cursive flume]

why is $\dot A(0)$ in o(n)?

[gentle osprey]

ProphetX

[cursive flume]

I thought tangent vectors to curves live in the tangent space, this lives in the tangent space at identity

it is true that the tangent space at identity is isomorphic to the lie algebra,but it's not strictly the same

the tangent vectors live in a different space,not in the lie algebra

[gritty widget]

what's your precise definition of "lie algebra of a lie group?"

some people define it to be the tangent at the identity

[cursive flume]

lie algebra of a group=set of left invariant vector fields

[gritty widget]

they are isomorphic

[cursive flume]

yes,isomorphic but not the same

[gritty widget]

right

[cursive flume]

so strictly speaking M above isn't an element of o(n), rather its image under the isomorphism is, right?

[gritty widget]

sure

it usually doesn't hurt to just pretend they're the same, but it's good to recognize there's two equivalent definitions

if $G$ is a lie group then the isomorphism between $T_eG$ and the space of left-invariant vector fields is given as follows: if $v \in T_eG$, then associate to $v$ the vector field $V$ defined by $V_g=d(L_g)_e(v)$, $g\in G$.

[gentle osprey]

ττερρα

[gritty widget]

you're just using the left-multiplication to push tangent vectors at the identity around. this gives left-invariant vector fields, and conversely, every left-invariant vector field arises from this

[cursive flume]

Z_6 is compact right? ( the finite group)

[gritty widget]

finite spaces are compact

(no matter the topology)

[cursive flume]

then the product of SU(3)xSU(2)xU(1)xZ_6 is compact,right?

i.e. product of compcat groups is compact even if some are lie groups,some are discrete

[gentle osprey]

Results provided by WolframAlpha

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Requested by slimvesus#9286

[cursive flume]

wow this theorem seems to be powerful

I bet it's hard to prove

oh this works for infinite products too

where is Tychonoff's theoem usually done? in point-set topology in undegrad?

is the proof for 'every finite group is compact' difficult aswell?

I am not sure how to relate the group elements to topology or how to put any topology on a finite group outside of the trivial ones

[gritty widget]

No, it is trivial. "Compact" means "every open cover has a finite subcover", basically finiteness of the group directly transfers to compactness

[cursive flume]

yes,but how can I 'put a cover' or 'put a topology' on a finite group?

if I have that,then sure,there always will be a finite subcover

[gritty widget]

it's probably with the discrete topology

Maybe it is not too hard to characterize all topologies on a finite set (maybe harder for a group requiring multiplicatjon/inverse to be continuous)

Actually maybe it is not so easy

But anyhow you can write down some topologies for finite sets. For finite groups, you can start with some sets that you say should be open, then close it under the usual topology operations and also under inverse image of group multiplication and group inverse

[cursive flume]

I can use the discrete topology and in that topology the statement will always work I thnk

is the product of a lie group with a finite group a lie group?

[cursive flume]

i.e. is this a lie group?

[gritty widget]

product of lie groups should be another lie group with the group operations done componentwise

[cursive flume]

yes,but Z_6 is not a Lie group

[gritty widget]

[cursive flume]

is it?

[gritty widget]

maybe

i don't think about zero-dimensional smooth manifolds often

[cursive flume]

wait lol so every finite group is a 0 dimensional lie group?

[gritty widget]

yeah i guess so

[cursive flume]

https://math.stackexchange.com/questions/2236267/group-g-endowed-with-discrete-topology-is-a-zero-dimensional-lie-group wow

ok I will have to go through this proof

[sleek thicket]

It is very hard

Iirc every cw complex has the weak homotopy type of a finite space

I guess characterize doesn't mean classify

But my point is finite topologies can be complicated

[obtuse meteor]

This fact haunts my soul

[cursive flume]

is a finite group connected?

elated to yesterday's discussion I found the theoem @tight agate mentioned and I saw that connectedness is also a property required

[wanton marsh]

no

unless it's the trivial group

[cursive flume]

so Z_6 is not connected?

[wanton marsh]

it is not

[cursive flume]

then i'm not sure why I could say that the exp map for SU(3)xSU(2)xU(1)xZ_6 is surjective

[wanton marsh]

any topological space with discrete topology is going to be totally disconnected

[wanton marsh]

the image of the exp map is always going to be connected iirc

since a lie algebra is connected

and the map is continuous

its image will be the connected component of the group that contains the identity

it's going to be a normal subgroup H of G, and G/H can be any finite group

or even infinite if your lie group is not compact ? idk if that's allowed

[cursive flume]

so the fact that the image of the exp map is surjective doesn't follow from the torus theorem?

in order for the tous theorem to hold we need the group itself to be connected

but if we have z_6,which is not connected,then thist heoem does not hold

[wanton marsh]

yeah

by Z_6 I assume you mean a cyclic group of order 6 ?

[flint cove]

Wdym topological spaces over finite sets are completely given by their specialization preorder

[sleek thicket]

Yeah, I realized I was getting at something different than what was being said

"topologies on finite sets can be complicated" in that "the homotopy theory of cw complexes can be pretty complicated"

[flint cove]

Ah, so „homotopical characterization “ 🙃

[cursive flume]

yes

but if this theorem doesn't hold,how can I conclude that the exp is surjective for SU(3)xSU(2)xU(1)xZ_6

or is it not?

[wanton marsh]

it is not

[flint cove]

The reason would be that exp can only map into the connected component of the identity, right?

[uncut surge]

The image of exp will be connected, but it's not in general equal to the connected component of the identity; SL2(R) is connected, but exp(sl2(R)) is not all of SL2(R)

But yeah, it is definitely always contained in that connected component, so if your group is not connected (like here) then exp will not be surjective

If that above question is cleared up: What is happening here

In the calculation at the bottom of the screenshot: Okay, in the first equation we're using that the volume form is left-invariant, so we can pull an L_g^* out of nowhere. This doesn't necessarily agree with the next equation, though, where we're just pulling the L_g into the summand with the R_g

This is still fine -- we can just split up the R_g and the 1 before pulling out the L_g in one of the summands. Okay, cool. But then the next equation is totally lost

[sleek thicket]

sorry, what's R_g - I?

[uncut surge]

R and L denote the right and left multiplication, so $L_g h = gh$ and $R_g h = hg$ for group elements g and h

[gentle osprey]

Lartomato

[sleek thicket]

Yup, got that

I'm just not sure what the subtraction is

Oh is G a matrix group?

[uncut surge]

Ah, I see, that might be another notational oddity: We want something like $R_g^* \text{vol} = \text{vol}$, so it should rather be $R_g^* \text{vol} - \text{vol}$

[gentle osprey]

Lartomato

[sleek thicket]

Ahh I see

[uncut surge]

But I guess they ""simplified"" that to a pullback of R_g - id which doesn't exist lmao

[sleek thicket]

So it's like, R_g*- I

That's so dumb lmao

[uncut surge]

I guess I'm really just here to rant about how badly this is written

[sleek thicket]

lol

My thinking is maybe they mean L_g^-1 instead of L_g

Then Ad showing up makes more sense

[uncut surge]

Yeah, that's probably true too

[sleek thicket]

But then what's with the subtraction...

[uncut surge]

And then the last thing: I think if you had a volume form $\omega$, then $f^* \omega = \det Df \omega$ or something like that, which is probably roughly what we want here

[gentle osprey]

Lartomato

[sleek thicket]

Yeah I agree

I was thinking of something hazy like that

But then what's with the subtraction??

Maybe they just dropped a - vol_G(e) at the end?

[uncut surge]

Oh my god yeah

You don't need the subtraction at all, all you want is $R_g^* \text{vol} = \det(\text{Ad}(g)) \text{vol}$

[sleek thicket]

l o l

[gentle osprey]

Lartomato

[sleek thicket]

Yeah!

I was like

What?

This is so dumb lol

[uncut surge]

How many mistakes can you get away with in a single calculation for a published paper

Incredible

Thank you! That helped

[flint cove]

That is the most cursed notation I've seen in a while

[pearl holly]

So I just learned what a continuous function is in topology and it doesn't quite reach me intuitively. Why is the definition like it is, i.e $f: X \rightarrow Y$ is continuous if for each open subset $V$ och $Y$, the set $f^{-1}(V)$ is an open subset of $X$. It almost feels more natural to define it not in terms of $f^{-1}$ but in term of $f$ instead. Could someone just briefly give me an intuitive description of this?

[gentle osprey]

older sister

[rugged swan]

try to translate f(xn) -> f(xà if xn -> x in metric topoly, you will see that this is the right definition

[long gorge]

Maybe the epsilon delta definition is a bit easier to see

[pearl holly]

Yeah I searched that up and that makes sense

[rugged swan]

Or take for instance the 1st projection f : R² -> R² (x,y) ->(x,0), this is continuous since a tiny perturbation of x,y stay a tiny perturbation of (x,0) but the image of an open set isn't open

[pearl holly]

Yeah that's true

[rugged swan]

having the condition f(U) open is too restrictive dimension-wise imo

[pearl holly]

Okay but let's look at an example real quick, hold on

Okay so this one is discontinuous. Is it because if I take a open set/open interval on the Y-axis and "project" it down to the X-axis by the function, I will just have a singleton since there's an asymptote there?

Meaning that if I take a "big" enough open interval that lies high enough in the Y-axis and then do f^-1 on that I will just get a singleton, which in closed in R?

Well "big" enough is the wrong word, but I mean an open interval (a, b) where a is super big and b is also super big

Okay so why is it discontinuous then?

[gritty widget]

||this function is continuous on its domain||

[pearl holly]

Hol up, so why does the figure say that it's not?

[fathom cave]

can u give another picture
that image looks vague

[pearl holly]

Okay hold on

[gritty widget]

it'll be discontinuous as a function from R to R if you prescribe it an arbitrary value at a, but otherwise, it's a perfectly fine continuous function R\{a} -> R

[frosty sundial]

What is the source of this picture?

(Except a instead of 0)

[gritty widget]

oops

same thing

[pearl holly]

Here's another one. So, using the definition, how can I see that this is discontinuous?

[frosty sundial]

Agreed but didnt want older sister to be confused by what you meanr

[pearl holly]

https://www.math24.net/discontinuous-functions

[fathom cave]

take inverse image of y(a)-1, y(a)+1

[long gorge]

It's the same reason why it's intuitively discontinuous - there are points arbitrarily close to a on the left such that f(those points) aren't close to f(a)

[frosty sundial]

Im not sure what your definition of continuity is

[fathom cave]

it is [a, a+something) which is not open

[frosty sundial]

Is it using limits?

[fathom cave]

open sets

[pearl holly]

I am refering to the topological definition. I mentioned it above

[frosty sundial]

Take an open set around y(a)

Its preimage will be [a,b) for some b

[pearl holly]

Ohhh

[frosty sundial]

Because there are no points less than a for which f(c) is near a

The asymptote example is continuous on its domain as @gritty widget wrote. It doesnt make sense to ask if it’s continuous on all of R becsuse it’s not defined on all of R

[pearl holly]

Yeah that's true

[frosty sundial]

However, it’s a common convention in calculus classes to say it’s discontinuous

Even though that’s being a little handwavy with domain issues

[pearl holly]

yeah I see the problem. But I am starting to get it now, thank you so much!

And also, a homeomoprhism is basically just an isomophism we see in algebra right? Instead of operations being preserved, we now have an bijection between the open sets in the domain and in the image, right?

[fathom cave]

homeomorphism is
bijective+ continuous maping with continuous inverse mapping

[pearl holly]

And since open sets kind of tells us about how all points are connected, two spaces are homeomorphic if they can be "formed" into the other like the coffee cup and donut thing, right?

[fathom cave]

an open set is mapped to an open set and since this mapping is also bijective
it means ur just changing that names of the points. its literally means both spaces are same in a topological sense

[pearl holly]

Okay great, so it is equivalent to an isomorphism! Thank you so much!

[fathom cave]

i dont know much about isomorphism 😔

looks like that way to me but im not sure

[pearl holly]

Well two (for example) groups are isomorphic if they have the same algebraic structure. So I can "rename" all the elements in one group for it to become the other group, just like you described homeomorhpisms!

[fathom cave]

right

isomorphism is also a bijection right?

[pearl holly]

Yes!

[honest terrace]

Yeah, homeomorphisms are the isomorphisms of topological spaces @pearl holly

[sweet wing]

homemorphisms/diffeomorphisms are isomorphisms for some reason :<

[honest terrace]

This is the right notion to speak about "same" top spaces because it preserves pretty much any topological property you'd want to preserve if two spaces are the same

imagine a world where Top's isomorphisms would just be bijections

[pearl holly]

So to show that two "geometric figures" are homeomorphic I want to work with manifolds and somehow come up with some homeomorphic function between those manifolds right?

[long gorge]

You don't need to work with manifolds specifically - manifolds carry a lot more structure than topological spaces. You want to find a continuous function with a continuous inverse.

[pearl holly]

Okay! Manifolds sounds interesting tho, do you know if Munkres talks about it? From the "contents" section it seems like he doesn't, but he might at least mention them in a later chapter

[honest terrace]

But your geometric structures as top spaces will end up being manifolds anyway

[fathom cave]

yes munkres does talk bout manifolds

[pearl holly]

Okay great!

[long gorge]

You could also take a look at Lee's book on topological manifolds

[pearl holly]

Yeah I am going to take a note of that book! Thank you so much!

[uncut surge]

Given a Lie group $G$ acting on its Lie algebra $\mathfrak{g}$ via the adjoint action $\text{Ad}_g$ for $g \in G$, the orbits are immersed submanifolds. In this nice situation, are they even embedded submanifolds?

[gentle osprey]

Lartomato

[uncut surge]

uuuuh if you don't have that this stuff is semisimple, you can probably do freaky things with irrational tori somehow, so feel free to assume semisimplicity

i guess i just need to show that it's a proper action

[obtuse meteor]

It very much depends on what structure you're considering.

An algebraic topologist might say things like "the punctured plane is the same as a circle" bc they work up to homotopy equivalence

but this is nonsense to a differential geometer, a punctured plane is a 2-dimensional manifold (I think it's diffeomorphic to a band/cylinder actually) whereas a circle is a 1-dimensional manifold

and it's even nonsense to a general topologist, since if you remove any two points from a punctured plane you get something connected whereas if you remove any two points from a circle you get something disconnected

[pearl holly]

Okay I get kind of get that, but I still haven't learned what a homotopy is. I'll get there soon I hope! Thank you so much!

[gritty widget]

gaytopy

[obtuse meteor]

homo in homotopy

^ that's me

[obtuse meteor]

homotopy equivalence allows you to squish things more, so for example, a band/cylinder is homotopy equivalent to a circle becuase you can squish the band infinitely down along its axis

[pearl holly]

"Yeah I've spend decades researching in gaytopy"

[obtuse meteor]

@gritty widget did I get it right that a twice-punctured plane is diffeomorphic to a band. I'll be so proud of myself

[pearl holly]

ohhh okay!

[obtuse meteor]

(I have the mental picture)

Basically the picture is map it onto a twice-punctured sphere, then put a cylinder around the sphere to project to from the origin

maybe I should have specified an infinitely tall cylinder/band

boundary is fake lol

[pearl holly]

Okay that sounds really interesting, hope I'll learn that in depth soon! Thank you!

[sweet wing]

by any chance anyone has like a visual way to see that $\left(S^1\times S^1\right)\wedge S^1$ is homotopic equiv $S^2\vee S^2\vee S^3$
all i can figure so far is that since $S^1\vee S^1\subset S^1\times S^1$ is a cofibration we have some coexact sequence blahblah yoneda hence homotopic

[gentle osprey]

ari 十年生死两茫茫，不思量，自难忘。

[verbal wraith]

Remember, topologies are just glorified semi-lattices.

If you have two semi-lattices X and Y, and a monotone function f from X to Y then an element a of X is a sufficient factor for b in Y if for any refinement of X W, refinement of Y Z and monotone function f': W -> Z that extends f, for any element w of W, w subs a => f'(w) subs b. Likewise an element a of X is a necessary factor for b in Y if for any refinement of X W, refinement of Y Z and monotone function f': W -> Z that extends f, for any element w of W, w subs a <= f'(w) subs b. An element a of X is a determining factor for b in Y if it is a necessary and sufficient factor. The map f is factorable if every element of Y has a determining factor in X. This means that there exists a function f*: Y -> X.

What it means in topology for a map F: X to Y to be continuous is that the induced map f = cl o image_F, from the closed sets of X to the closed sets of Y is a factorable map.

[pearl holly]

I'll read that once I've become better at this, but thanks!

[verbal wraith]

I mean if she has the patience to read through Munkres it's plainly appropriate

Learning topology from Munkres is like learning Quantum Mechanics from NewScientist, Munkres is basically popsci for nerds.

[pearl holly]

I don't know if that's good or bad lmao but Munkres is all I had to choose from. My teacher only has Munkres so he let me borrow it

[empty grove]

Munkres is pretty good lol

[gritty widget]

Remember, topologies are just glorified semi-lattices.

.

[empty grove]

Nikita and I had the exact same discussion last time older sister asked about how to study top

[pearl holly]

Yeah I remember that

[empty grove]

[gritty widget]

[verbal wraith]

It's honestly hard to find any good book on topology. I want to learn more topology too, but the most of it is over my head.

[sacred quail]

sutherland is good and succint

[empty grove]

Starts with metric spaces tho

[sacred quail]

the portions are quite disjoint iirc, though I'm not sure why you wouldn't want to see anything about metric spaces first given the motivation for a topological space

[empty grove]

You're right, that makes sense to me

[sleek thicket]

What the fuck

[sleek thicket]

.pin

[sleek thicket]

This is the worst post I've seen since I returned

Jesus christ

Learning topology from Munkres is like learning Quantum Mechanics from NewScientist, Munkres is basically popsci for nerds.

[gritty widget]

Remember, topologies are just glorified semi-lattices.

[sleek thicket]

I want to learn more topology too, but the most of it is over my head.

[tight agate]

You can't estimate my level of thinking if you don't know even elementary terminology, so this is just emotions.

[sleek thicket]

Gee I fucking wonder why

[verbal wraith]

Yoikes

[tight agate]

lol

[sleek thicket]

The internet has had disastrous consequences for mathematics

[empty grove]

Return to pen and paper

[cloud owl]

no euler?

[empty grove]

Rocks and hieroglyphs?

[cloud owl]

how did people even live before $e^{i\theta}=cis(\theta)$

[gentle osprey]

Kaisheng21

[sleek thicket]

down with e^(iθ)

[empty grove]

They've played us for absolute fools!

[sleek thicket]

Lowmath was a mistake

ajfhsjakajhsj

[gritty widget]

[pseudo crane]

lmfao

[shut moat]

Amazing

[honest terrace]

That is the best one I've seen here personally, I couldn't stop laughing for a whole minute (like seriously, I'm still giggling )

actually nevermind

That is funnier

[sleek thicket]

Wait @fading vale have you seen the lattice thing yet?

[fading vale]

No

LMAO

the pins

holy shit

[sleek thicket]

lmfao

[fading vale]

Perhaps if it is all over your head you should not be commenting on whether or not Munkres is rigorous or "serious math."

[honest terrace]

@fading vale ok popsci-nerd

[tough imp]

Lol nerd, must have sucked to be in that class

[sleek thicket]

it did!

[tough imp]

If it sucked so bad why did I enjoy it

[cosmic beacon]

From Munkres, we have the following theorem: 

\bigskip

\textbf{Theorem 82.1}.\quad Let $B$ be path connected, locally path connected, and semilocally simply connected. Let $b_0\in B$. Given a subgroup $H$ of $\pi_1(B,b_0)$, there exists a covering map $p\colon E\to B$ and a point $e_0\in p^{-1}(b_0)$ such that
\[p_*(\pi_1(E,e_0))=H.\]

\bigskip

Given the space $B=\bR P^2\times\bR P^2$, we have that $\pi_1(B,b_0)=\bZ/2\bZ\oplus\bZ/2\bZ$. Consider the subgroup $\{(0,0),(1,1)\}$ of $\pi_1(B,b_0)$. Find the covering space corresponding to this subgroup.

\bigskip

I have been trying to solve this problem for hours. Can anyone who's knowledgeable give me any pointers/hints? I'm completely stuck. I found covering spaces corresponding to every other subgroup of $\pi_1(B,b_0)$ besides this one.

[gentle osprey]

Isaiah

[empty grove]

Have you found the universal cover of the space?

[cosmic beacon]

yes

S^2 x S^2

[empty grove]

Right

[cosmic beacon]

Is the idea to use some orbit space of S^2 x S^2?

[empty grove]

yeah

follow the construction of the cover corresponding to any subgroup H

as a quotient of the universal cover

[cosmic beacon]

So my confusion there, like wouldn't any orbit space of S^2 x S^2 correspond to the trivial group since S^2 x S^2 is simply connected?

[empty grove]

No, you look at the action of pi_1(B,b_0) on S^2 x S^2

[cosmic beacon]

Like for any group G, the map

q : S^2 x S^2 ---> (S^2 x S^2) / G

would induce the trivial homomorphism, namely,

q_*(pi1(S^2 x S^2)) = 0

[empty grove]

yeah, but that only means that the range is the zero group

the codomain could be larger

[cosmic beacon]

Should I be looking to use this theorem?

[empty grove]

Yeah, B will be the space you want

and G will be that subgroup

[cosmic beacon]

B would be RP^2 x RP^2?

and X would be S^2 x S^2

[empty grove]

Nope

The covering space that you want to construct

[cosmic beacon]

oh wait I see

right

[empty grove]

Since you would get that its fundamental group is the subgroup you quotient by

[cosmic beacon]

[\begin{tikzcd}
{S^2\times S^2} \
& {(S^2\times S^2)/(\pm\mathrm{id})} \
{\bR P^2\times\bR P^2} && {}
\arrow[from=1-1, to=3-1]
\arrow[from=1-1, to=2-2]
\arrow[from=2-2, to=3-1]
\end{tikzcd}]

[gentle osprey]

Isaiah

[empty grove]

Yes

[cosmic beacon]

awesome thank you so much!

[empty grove]

[empty grove]

Getting stuck here even after having gone through homology and cohomology twice before in the proof of im \partial \subset ker i_*, how do we know that \partial b = 0?

For context, [c] is in H_n(c), b is a pre image of c under j, and a is a preimage of \partial b under i

oh nevermind this was very stupid

[\partial b] = 0 since it is a boundary

[gritty widget]

i want a girlfriend that i can do a homology with before bed

[empty grove]

Seems like too big a restriction of the domain

[true robin]

Maybe that’s why they call it homology

[empty grove]

[sleek thicket]

Why do they call it homology when Hom oligy

[summer jolt]

Suppose you have a metric like this

THen I want to find an orthonormal frame that is dual $dt + e^x dy$, $dx$ and $\frac{1}{2}e^{x} dy$

[gentle osprey]

snypehype

[summer jolt]

Obviously $\frac{\partial}{\partial x}$ is dual to $dx$

[gentle osprey]

snypehype

[summer jolt]

but I don't how get to compute the dual to $dt + e^x dy$

[gentle osprey]

snypehype

[gilded shell]

Given a topology τ and a set S. We have τ=(smallest topology containing S as a subset).
Is there a name for an S with this property?

Eg a basis, an analytical basis, a whatever.

That might be the only name but i thought it had some other one. Ok, ty

[cerulean oriole]

Sub base IIRC

[honest terrace]

yeah I think that's a sub-basis 🤔

[cerulean oriole]

🤔

[honest terrace]

wait it's not clear. Is it S € tau or S \subset tau ?

[bleak helm]

Google is telling me a sub basis must cover your space, and if it doesn't cover your space, it is called a synthetic sub basis. This is wack tho, never used these terms before.

[cerulean oriole]

Synthetic sub-basis I've never heard of

Although the covering requirement makes sense

Although you can just add X

[bleak helm]

These are just the definitions from the one site I looked, lol. I've never used them so idk.

[cerulean oriole]

@empty grove I finally worked this out based on one of those MathSE posts. You don't need U compact but you do need neighbourhood.
Let c > 3; for example c = 5. Let V be a union of open balls covering E with volume < e/c^n.
For each x in E, choose a radius r > 0 around x st
(i) B(x,r) subset E (volume purposes)
(ii) B(x,cr) subset U (cover-inside-U purposes)
In fact, we can avoid Choice here (although needed later) by picking r maximal st
B(x,cr) subset interior(U) intersect V
as you did earlier.
Now use Vitali's Covering Lemma to take a countable subcollection of the above balls which is
(i) disjoint; thus total volume less than e/c^n as it is subset V
(ii) each ball dilated by c is inside U
Then the countable subcollection, each dilated by c, is within U with volume < e.

The same should work if E has non-zero measure, but the tight cover will have volume > 3^n the measure of E.

Hope there are no more mistakes 🤞.

[gilded shell]

And the definitions are equivalent because intersections distribute over unions. Therefore the finite intersection of a unions of finite interasections is a union of finite interasections.
Understandable. Have a nice day

[cursive flume]

since finite groups are 0 dimensional manifolds,are they also lie groups?

[cloud owl]

wikipedia says lie groups are groups that are differentiable manifolds

[cursive flume]

so finite groups are 0-dimensional Lie groups

[cloud owl]

if 0-dimensional manifolds are differentiable, sure

[cursive flume]

actually,how is R^0 formally defined?

[cloud owl]

idk that much topology, so should not be talking

[empty grove]

In (i), I think you want B(x,r) subset V instead of E. Also c probably has to be ≥5, because that's what you need in the infinite version of Vitali's lemma.
That thing in the last line is pretty nice, at least there's a finite constant you multiply by which feels intuitive if all balls are scaled up by upto a constant

[empty grove]

Singleton

[cursive flume]

what does that mean?

[long gorge]

1 point

[cursive flume]

so it has one element

but which is that?

[empty grove]

1 point, and the topology is that all (2) subsets are open

What the element is doesn't matter

If we call it 1 or * or whatever the topology will still behave in exactly the same way

[sleek thicket]

It's a single point, 0

[empty grove]

But you can take it to be 0 if you want

[cursive flume]

so i can take any real number and that is R^0?

R^0={x} ,where x can be any real number?

[sleek thicket]

Eh, depends on how much structure you want on your R^0

If it's a vector space then it should be 0, or you need to redefine addition and multiplication or whatever

[empty grove]

Yeah if you're only looking at the topology, take it to be whatever (not necessarily even real)

[cursive flume]

if I want now to make this into a topological manifolds,I choose charts: {x}->R^0 for any x

[true robin]

R={R}

[sleek thicket]

Yes

[cursive flume]

there will be as many charts as many group elements

how can I see this is smooth?

[long gorge]

To answer your question you can give any set with the discrete topology the structure of a 0-dimensional manifold by taking the charts to map each point to R^0

[cursive flume]

how is smooth defined on R^0?

[empty grove]

Reject regularity

[sleek thicket]

There are no overlaps phophetx

So the only thing you need to worry about is smoothness of the identity map

How so

[long gorge]

my bad

[cursive flume]

how is smooth defined in this sense? I don't know how to define a partial derivative which should exist and be continuous

continuity is clear from the definition with open sets

[long gorge]

you don't need to in this case

[sleek thicket]

Proohetx, there's only ever one map

[empty grove]

Everything is smooth lol

[sleek thicket]

Yeah

[gritty widget]

you're smooth

[sleek thicket]

Hello legosi

[long gorge]

yeah so any finite group is trivially a compact lie group

[empty grove]

[cursive flume]

aren't there group elements many maps as charts?

and all map to {0}

[long gorge]

and it's instructive to think about this specialization, since the representation theory of finite groups generalizes to compact groups by defining the haar measure

[cursive flume]

i.e. {e}->{0},{g}->{0},{g^2}->{0}

[long gorge]

and you give any finite set a measure by the counting measure

[sleek thicket]

There's only one map R^0 -> R^0

So the transition functions will all be this one map

[cursive flume]

so there is just 1 chart?

[sleek thicket]

Those are what you need to check are smooth for something to be a smooth manifold

No

There are |G| charts, you're correct

But smoothness of an atlas says that the overlap maps are smooth

[cursive flume]

right

[sleek thicket]

And there's only one of those maps

Because it's a function R^0 -> R^0

[cursive flume]

and why is this map smooth?

it maps a point to a point

[sleek thicket]

We just declare it to be so

It's the identity map

Any reasonable definition of smooth would call it smooth

[cursive flume]

so smoothness is not to be understood in the sense of partial derivatives being continuous

[long gorge]

it is, but this is kind of a relative sense

[sleek thicket]

Well, you could actually define partial derivatives here

R^0 is a vector space

you can make sense of smoothness of maps V -> V intrinsically by taking limits

But whatever definition you work through it'll end up being smooth

[empty grove]

There are no partial derivatives to check so it's vaccuously smooth

[sleek thicket]

Haha I was about to say that

There's no directions

[cursive flume]

okay,it makes sense now, thanks

[cursive flume]

vector space over which field? can be any?

[sleek thicket]

R

I abstain from clarifying whether R^0 = C^0

So one thing is like

Being a vector space or a manifold isn't a property you can have

It's structure

We have to add extra data

Like a smooth atlas, or a scalar multiplication

[cursive flume]

right,we need to define an addition and s multiplication to make it into a VS

[sleek thicket]

I'm considering R^0 as an R vector space here

But it's just a one point set

[cursive flume]

but how can one add elements,if there's just 1 element?

[sleek thicket]

0 + 0 = 0

[long gorge]

everything is 0 so we're good

[cursive flume]

what do I add it with?

ah

lol

similarly anything*0=0

[empty grove]

Even an empty addition won't be a problem lol apart from losing existence of an additive identity

[cursive flume]

but now comes the strange point, if we say a chart is {x}->{0}

how is {x} still R^0?

we said R^0={0}

[sleek thicket]

There is an empty semigroup, it's true

[cursive flume]

or R^0={0} and anything else {x} at the same time?

[sleek thicket]

charts aren't equalities prophetx

[empty grove]

Right, and here that anything has to be real because you're only defining scalar mult by reals.

[sleek thicket]

a little circle on a sphere isn't the disk

[true robin]

Remember, definitions are upto isomorphism

[empty grove]

That's why it's a real vector space

[sleek thicket]

No

What?

Why do people keep mentioning 0 dim vector spaces being iso

That's not relevant here

They asked why these are charts on a manifold

For a discrete group

[cursive flume]

so my confusion is,we said we have |G| many charts {x}->{0}, therefore this is a map from R^{0} to R^{0} right?

[sleek thicket]

The manifold has no linear structure

no

It's a map from a subset of G to R^0

Namely the subset {x}

But the transition maps are maps R^0 -> R^0

And smoothness of the atlas means that these are smooth

[true robin]

Well he asked wether R^{0} was every singleton, so I just mentioned that you could take any singleton as they would all be linearly isomorphic

[cursive flume]

and there's just 1 function because there's just 1 way to map 1 element to 1 element

right?

[sleek thicket]

Yes, exactly

[cursive flume]

math shocks me everyday this is very nice

I didn't find this in Lee's smooth mfds book it was an exercise

[sleek thicket]

He talks about 0 dim manifolds

[gritty widget]

no author wants to write about these things

[sleek thicket]

Jack is careful

I've only caught him on the empty manifolds once (maybe twice?) across 3 books

[gritty widget]

if jack is so careful why did i spend so long on 3-10 in irm

[sleek thicket]

lol

[gritty widget]

fuck

it's 3-9

not 3-10

[sleek thicket]

You were not careful

[gritty widget]

fuck that exercise

[sleek thicket]

Isn't it interesting...how the tteppa fails to consider?

[gritty widget]

pain

if you add the necessary assumption

yes

it is

[sleek thicket]

I would simply assume as much as I needed to prove it

[cursive flume]

is it useful for anything to view finite groups as 0 dimensional manifolds?

[sleek thicket]

Eh

[gritty widget]

you can make your statements say all natural n instead of n >= 1

[sleek thicket]

Everything about 0 dim manifolds is ultimately not using the 0 dim manifold structure

But it can be useful

Like, you can say that the isometry group of a riemannian manifold has a lie group structure

how incredibly useful

[cursive flume]

Bourbaki style most general way possible

I heard memes about Bourbaki books being super abstract and most general

[sleek thicket]

I think it is nice to not have to think too hard about corner cases

I have this game where I try to gotcha my friend with them (and him me)

It's a cold war so far

and nobody's had a victory

[cursive flume]

just a sanity check,SO(N) is compact and connected for any N right?

therefore I can obtain SO(N) by exponentiating so(n)?

I can certainly do this for SU(N), but i'm not sure about SO(N)

[empty grove]

These are pretty nice lol

[sleek thicket]

Sounds correct to me prophetx

Hm I'm not sure how you can easily show SO(n) is connected

[cursive flume]

1

[wanton marsh]

1

[sleek thicket]

Exactly one

You can gram schmidt to show SO(n) is homotopy eqv with the matrices of positive determinant and then find a path in that space, I think

But that's so ugly lol

[cursive flume]

is it similarly hard to show for SU(n)?

( I took it for granted )

but for SO(n) i was not sure the statement is even true

[sleek thicket]

I think it's easy for both

I just don't see it lol

God this is just a day for me to do linear algebra I guess

I was helping someone else in #groups-rings-fields earlier

[gritty widget]

differential geometry is just linear algebra locally

do geometry instead

[sleek thicket]

On it

Oh I have a meme proof

There's a bundle SO(n) -> SO(n+1) -> S^n

[cursive flume]

define trace coordinate-free

[sleek thicket]

Since SO(n+1) acts transitively on S^n with stabilizers SO(n)

So to show SO(n+1) is connected it suffices that SO(n) and S^n are connected

SO(1) = {1} is obviously connected and spheres are connected

[gritty widget]

linear term of characteristic polynomial (or is it the x^{n-1} term? idr)

[sleek thicket]

Derivative of the determinant at the identity

[gritty widget]

[cursive flume]

[gritty widget]

the trace is the trace.

[sleek thicket]

ttracepa

[shut moat]

https://math.mit.edu/~cohn/Thoughts/trace.html

This was a neat algebraic defn, although I prefer the derivative defn myself

[sleek thicket]

Ah yeah I love this one

[true robin]

Maybe a really stupid question, is there a way of showing a Vector space is isomorphic to its dual without choosing a basis?

[sleek thicket]

No, but I'm not sure how you could make this formal

Notably it fails in infinite dimensions

[shut moat]

I think the question was about finite dimensional vector spaces for that reason

[sleek thicket]

Right, I'm saying this might suggest you need to choose a basis

[shut moat]

ooh

[true robin]

Yeah but i mean the question asked for a coordinate free description, that’s why I asked

[shut moat]

that makes sense, if you could do it without reference to a basis then dimension wouldn't come into the picture

[true robin]

Well, is that necessarily true? I don’t even know how to properly ask the question in some sense

[sleek thicket]

Wait this is so dumb

I don't know why that link is doing it like this

Being iso to your dual is a red herring

You need to pick a basis to show that the canonical map V (×) V^* -> Hom(V, V) is an isomorphism

then there's a map V (×) V^* -> k given by the bilinear evaluation map V × V^* -> k

idk why you would go through the dual of Hom(V, V)

[honest terrace]

I think you can't, and that's why we say that V and V* aren't canonically isomorphic while V and V** are ?

🤔

(oh I'm very active now )

[sleek thicket]

I don't really agree

Idk how you would set it up categorically

It's true that the variances are different, so maybe category theory tells us it'd be weird if they're iso

[shut moat]

proving the existance of an isomorphism without reference to a basis is different from constructing a natural isomorphism though

[honest terrace]

the fact that V and V** are canonically (or more precisely naturally) isosomorphic is definitely something you can express in the categories language

but I may be missing your point shamrock

[sleek thicket]

Sure, and the iso V (×) V^* -> Hom(V, V) is natural in this sense

You don't need to pick a basis to construct it like you do for the dual

[shut moat]

yeah sry I was responding to shika blyat's thing

[sleek thicket]

Yes, but what does that have to do with the informal statement "you need to pick a basis to show V and V^* are isomorphic"

[sleek thicket]

No it's a good point, I think it highlights how the duality thing is just a red herring

So I guess I'm saying you have a single map Hom(V, V) <- V (×) V^* -> k

But to show the first map is invertible you need to pick a basis

Just like with the map V -> V**

Even though it's defined defined independent of a basis

[bright acorn]

What kind of information can you get more about a manifold by computing its compactly supported De Rham cohomology rather than the usual De Rham cohomology?

I don't really get the motivation that much

[honest terrace]

idk tbh, I was just thinking about the fact that there's some functor that makes V and V** naturally isomorphic but not V and V* because the functor is contravariant, but yeah now that you ask, I'm not sure how this would relate with the question, even though I've always assumed that both facts are related

so nvm me

[sleek thicket]

No it's okay

I think that's where slim's head was at too

But I think it's making a sort of jump