point-set-topology

[sleek thicket]

I think you can get a noncompact version of poincare duality by using compactly supported cohomology

And/or you need to go through compactly supported stuff in the proof

But it's been a while and I'm not sure

[bright acorn]

Didn't know Poincaré's duality was more general

[sleek thicket]

I think this is an exercise at the end of a chapter in ISM?

I don't have it on me, sorry

[honest terrace]

(ISM ?)

[bright acorn]

Lmao same tbh

I am still reading Tu and Spivak

OHHHH

Lee's book

Introduction to Smooth Manifolds

Yup

Haven't read on that one yet

[sleek thicket]

Ah sorry

This MSE post seems to support my hazy memory https://math.stackexchange.com/q/3592421/408287

[honest terrace]

Not formal at all, but here's some thoughts about saketh's question.
If you to assign to each vector v € V a linear functional phi_v, the most canonical way you can do this is by defining some linear functional such that phi_v(v) = 1, ig ?
But to define such a functional on V entirely, you'd need to pick some supplementary subspace to Vect(v) (or equivalently to add vectors to the family (v) to make it a basis), and to extend the partially defined phi_v linearly by making it 0 on the supplementary space.
(You'd also need to be sure to pick the same supplementary subspace while defining phi_w for every w € Vect(v)).
This enforces choices of the supplementary spaces

That doesn't prove at all that you can't do it using another way, but I think that's a reasonable assumption to say that if such a choice-free map existed, then the map would be a "nice" and intuitive map

Now I definitely need some sleep, the problem I spent the whole day searching in #groups-rings-fields killed me

[true robin]

I agree with this intuitively i suppose, but the real problem is that i have no clue how to formally phrase “choosing a basis” I suppose if it had a proper definition we would be able to easily prove that it is impossible.

[honest terrace]

yeah right

I've always assumed that this could be formally stated are "not naturally isomorphic" as I said earlier, but now that I thought about it, idk why

[cerulean oriole]

Yes 😅.
Unless there was a mistake I didn't notice in the proof on Wikipedia, it should hold for any c > 3 although it is usually stated for c = 5. (But not for c=3 except in the case of finitely many initial balls apparently.)

If you insist of knowing, you could make it the empty tuple IG?
But it really, really doesn't matter.

I think it may break some common "well-behavedness" assumptions if K is too big (separable? countably many connected components? Are these common assumptions?), but true.

[long gorge]

I think the size of K doesn't end up being an issue, since each connected component is still a single point.

[empty grove]

manifolds are second countable by definition I think

[empty grove]

hmm I see

[long gorge]

Yeah the original context was making a set with the discrete topology into a 0-dimensional manifold

And second-countability isn't really an issue for that

[empty grove]

yeah, so youll need the discrete topology to be countable

uncountable discrete space wont be second countable

[long gorge]

Do you? If you care about partitions of unity it would make sense to only care about each connected component being second-countable

But I don't really remember the details of this

[empty grove]

oh each connected component needs to be second countable?

[long gorge]

Yeah because what you really want is paracompactness

[empty grove]

hmm yeah ig we only care about local properties

ye idk much mani stuff, just saw their definition in a course as an example of some shit and there they were defined to be second countable lol

[long gorge]

The embedded submanifold example is troubling

[empty grove]

ty

[lean marten]

my supervisor last summer actually did her thesis on "non-metrizable manifolds" which are when you remove the 2nd countability condition from the definition

from memory you can get some real wacky manifolds

like higher dimensional versions of the long line I believe?

[honest terrace]

Why is everyone discriminating the long line 😔

[lean marten]

it's too big for my tiny brain lol

[tough imp]

No one is discriminating against it

We all love it

🥰

[river granite]

long cat line is long

[winter prawn]

can I say a "torus" is a type of quadric surface?

[ivory dragon]

no

it doesnt meet the definition

assuming by "torus" you mean T^2

in fact, the quadric surfaces of dimension 2 are classified entirely

a simple way to disprove it: note that we can draw a line that intersects a torus at 4 points

this contradicts the fact that projective quadrics over any field only intersect a given line 0, 1, 2, or infinitely many times

[winter prawn]

so a torus is a torus, it is none of any kind of "surface", "curves" or "planes"?

[ivory dragon]

uh... a torus is definitely a surface

its just not a quadric surface

[winter prawn]

would it be closer to "parametric surface"?

[ivory dragon]

you can define a torus parametrically, yes

[winter prawn]

i see, very helpful, thanks!

[ivory dragon]

im not sure what itd mean for a given surface in euclidean space to NOT be a parametric surface though

surely you can parametrize any closed surface through a sufficiently contrived function?

not that itll be nice

(though even if you define "parametric surface" more restrictively, any reasonable definition will fit the torus, so i guess thats a moot point)

(its certainly "more parametric" than your typical surface)

[winter prawn]

appreciate your explanation, thanks

[shadow tendon]

,tex Is there another notation for the space of $E$-valued k-forms? $E$ is a bundle over $M$. My class uses $\Omega^k(E)$ for $E$-valued k-forms, and $\Omega^k(M)$ for k-forms over $M$, so I got confused.

[gentle osprey]

pronk

[shadow tendon]

,tex oop nvm\the class uses $\Omega^k(M,E)$ too

[gentle osprey]

pronk

[gritty widget]

just write the whole $\Gamma((\bigwedge^kT^*M)\otimes E)$

[gentle osprey]

ττερρα

[gritty widget]

every single time

[sleek thicket]

Don't be obtuse

they should just write "set of E-valued k-forms"

no need for opaque symbols

[uncut surge]

gamer trick: always write something like "Consider the space of E-valued k-forms" and then the opaque symbol of your choice, then for the next couple paragraphs you're free to use the opaque symbol as much as you like without having to refer back to the full "E-valued k forms"

(and \Omega^k(M,E) is pretty standard anyway)

[sleek thicket]

Sorry that was a meme

I thought it was clearly unreasonable to not have a symbol for that

[gentle osprey]

slimevesus

[gritty widget]

this will be my new symbol for differential forms

$\qed$

[gentle osprey]

ττερρα

slimevesus

[sweet oasis]

Is the graph of a continues function on a closed interval which is compact, always rope connected?

I always forget its called path connected in English, but the straight translation is close enough

Like intuitively I think the answer is probably yes

But then I think of x*sin(1/x) on (0,1] and 0 at x=0

[gritty widget]

this is more or less the definition of path connected, right?

[sweet oasis]

Probably continues functions that aren't of bounded variation gonna not work

Oh, nvm then, I just know that like sin(1/x) x [-1,1] isn't path connected

I know

But I thought it wouldn't matter

If its a closure of that or if it started at 0 and was continues

Yes

No but you do add the final point

I guess you probably answered it

Wait what isn't

[gentle osprey]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by slimvesus#9286

[sweet oasis]

Yes so I just added 0 at x=0

So it is continues

One of my favorite functions actually is that function from [0,1] after you added the point at 0, composed with the cantor set

What you get is a continues function from [0,1] to [0,1] which isn't of bounded variation but has a derivative 0 almost everywhere

Pretty cool

Like you wouldn't think such a thing exists

Being continues and having derivative 0 almost everywhere means its very tame, but it is also not of bounded variation, which makes it very wild

Also very important it's continues in a compact domain

Such a beauty, wish I could see its graph but I don't have mathematica

[cerulean oriole]

g(x) := (x, f(x)) should be a path through the whole space if I'm not wrong.

The domain is of the form [a, b] with a < b, right?

[honest terrace]

How would one show that the x axis + the y axis isn't homeomorphic to the x axis ? (I think I have a proof, but I'm curious about how y'all would do it )

[sleek thicket]

if you remove the origin from the union of axes you get 4 connected components

If you remove any point from a line you get 2 connected components

[honest terrace]

Oh lol, I have the same thing but saying it in a fancier way

the first has 4 ends, the second 2 ends

[gritty widget]

what is the end of a line

[sleek thicket]

I bet you can distinguish the one point compactifications

It should be like, two intersecting circles versus a single circle

First one is homotopy equivalent to the wedge sum of 3 circles, which isn't homotopy equivalent to a single circle

You can check this by looking at fundamental groups (or just like compute the fundamental groups of the 1 point compactifications explicitly)

[honest terrace]

https://en.m.wikipedia.org/wiki/End_(topology)

[fathom cave]

https://cdn.discordapp.com/attachments/772499943902412813/846030732123701268/tangent_space_basis.png

${\frac{\partial }{\partial x^1}, \frac{\partial }{\partial x^2}, \cdots, \frac{\partial}{\partial x^n}}$ forms a basis for the vector space $D_p(\mathbf{R^n})$. \ $T_p(\mathbf{R})^n$ is isomorprhic to $D_p(\mathbf{R^n})$. If $v\in T_p(\mathbf{R^n})$ then $v$ in the standard basis ${e_1,e_2, \cdots,e_n}$ is $v= \sum_i^n a^ie_i$. What confuses me is why a vector $v$ in $T_p(\mathbf{R^n})$ is written as $v= \sum v^i\frac{\partial }{\partial x^i}$ which is an element of $D_p(\mathbf{R^n})$ unless we are implicitly referring to the vector $D_v \in D_p(\mathbf{R^n}) $, a vector associated with the vector $v$ since two spaces are isomorphic

[gentle osprey]

Zero0

[gritty widget]

"from now on, we will make this identification and write a tangent vector v..."

yes, they really mean D_v

but since you know that T_p(R^n) and D_p(R^n) are isomorphic via v <-> D_v, it doesn't hurt to pretend that v is actually D_v, and vice versa

[fathom cave]

right
it just felt a little odd to me since two vectors belong to the different spaces

ty

[raw sedge]

Here's a thing I noticed. Consider the informal idea of infinitely many lines converging in a point. For the record, let I be an infinite set to index the lines. One way to define this is using a metric, as such: Define the underlying set as (R+ cross I) union {0} where 0 is added freely, and then the distances between two elements different from 0 is either the distance between them as real numbers if their entry in I is the same, or their sum if it's not, whereas the distances between 0 to any element is its entry in R+ (or 0 if we're talking about d(0, 0), obviously)

But here's another way to define it

Consider I cross R>=0 in the product topology, where I is taken to be discrete. Then glue together all points of the form (i, 0).

So far, both of these are pretty intuitive, and you'd expect them to be homeomorphic, which I think they are although I can't construct an explicit homeomorphism right now

but, and here's the interesting part

the obvious homeomorphism actually doesn't work

Let X be the metric space we defined, and let Y be the quotient space we defined.

You might think of the set map X -> Y where 0 is taken to the point that resulted from gluing all of the (i, 0), and where (x, i) in X is taken to (i, x) in Y, which is an unambiguous point due to the way the quotient is constructed.

But it actually isn't even continuous.

For example, let the indexing set be I = { 1/n | n a positive integer }, and consider the set {(i, x) in Y | x < i} (verify that this is unambiguous considering the quotient, and also open)

then the pullback under the map I defined earlier isn't open

I hope my exposition wasn't needlessly complicated, I just encountered it and thought it was pretty cool and wanted to share.

[pastel linden]

trying to prove this biconditional

but I'm having some trouble with the implication

I tried doing something with induction on k but it's not working out

[gritty widget]

alpha is non-zero, so you could try extending it to a basis of the dual to V and messing around with that

(speaking for the -> direction, since the other direction is almost trivial)

@pastel linden

[cursive flume]

is the product of arbitrarily many connected topological spaces connected?

or only finite?

[raw sedge]

arbitrarily many work, I think

[empty grove]

Yes

[raw sedge]

let me just try to put an argument together

[cursive flume]

with respect to the product topology,right?

[gritty widget]

[empty grove]

Ye

[gritty widget]

no, the discrete topology

[raw sedge]

yeah

[empty grove]

Is this teppas mom

[raw sedge]

they probably meant as opposed to like the box or uniform topology

[pastel linden]

inconsiderate tteppa

[raw sedge]

but I think the product topology is the coarsest of these, so if any are connected, it is

[cursive flume]

memer tteppa

[empty grove]

Box isn't connected

The set of bounded sequence in R^omega is clopen

Under box

[raw sedge]

interesting

[empty grove]

Right so you can find a collection of connected sets which all intersect at one point such that their union is dense

The union will be connected, so its closure would be too

(for product topology)

So if the space is product of X_alpha's, fix a tuple (x_alpha) at which you'll make everything inteesect

And each connected set will be a set of tuples in which are equal to x_alpha in all entries except for some fixed collection of finitely many, which can vary over the whole set

(the finite collection of entries which can vary is fixed for each set, and you have a set for each finite collection)

You can prove from here that the union of these sets will be dense and connected

[gritty widget]

ew it squished my math

blah blah do it fiberwise and you get a statement about differential forms on manifolds

and you can get a cute version of e.g. frobenius' theorem using this

[gentle osprey]

ττερρα

[gritty widget]

i hate this bot

[obtuse meteor]

the words "extend a basis" are so common in difftop that it's illegal

[pastel linden]

now that I am actually doing the exercises in tu and know some analysis and linear algebra this is fun stuff

long live smooth manifolds

[gritty widget]

[cerulean oriole]

Huh.

[cerulean oriole]

(Assume I is infinite, as your map is a homeomorphism otherwise.)
In both spaces, any point except the central point has a neighbourhood homeomorphic to (-1, 1). Also, removing the central point should leave a disjoint union of |I| open rays, whereas removing any other point should leave only a copy of the original space. So any homeomorphism should map the central points to each other.
Removing the central points, we get a homeomorphism of the disjoint union of |I| rays. By looking at the connected components, there must be a permutation s of I such that it is a homeomorphism of ray i in the metric space to ray s(i) in the glue-space. Now consider the closed ray i, its image is the closed ray s(i), so it must also be a homeomorphism of those two.
Now take a open set [0, epsilon_i) in the metric space for every i in I such that epsilon_i has no positive lower bound (so it's not a neighbourhood of the central point) and consider the image in the glue-space. Since each ray-i to ray-s(i) is a homeomorphism, [0, epsilon_i) is mapped to a neighbourhood of 0 in ray s(i), so you have a union of (neighbourhood of 0 in ray s(i)), which is open in the glue-space (its preimage in the product space includes a open set containing 0 in every copy of R_{>= 0}, whose union is an open set including all the 0's), whose preimage is as mentioned before not open.

In short, there is no homeomorphism from the metric space to the glue-space.
Unless I made a mistake above, which I probably did.

This does seem similar to uniform convergence v.s. pointwise convergence, but I can't tell how.

[raw sedge]

I wasn't looking at it very rigorously, but this proof seems sound

it's really surprising that these constructions give different spaces then

[cerulean oriole]

Yes, like I said, it vaguely seems like some sort of uniform/pointwise distinction
But I can't say exactly how

[empty grove]

Why is the pullback of that set not open in the glue-space?

[raw sedge]

because given 0, an element of the set, for every positive epsilon there is some x distance less than epsilon away from it not in the set

which is a consequence of epsilon_i not having a lower bound

[cerulean oriole]

It is---it's in the metric space that it's not open.

[empty grove]

ohh

I see

Interesting

[cerulean oriole]

IKR

[raw sedge]

by the way I came up with this construction because I came across the concept of local compactness, and tried to generate examples to understand it

so, for example, in the metric space here, let U be some neighborhood of 0 and S be some set containing U. Then let epsilon be such that the ball of radius epsilon around 0, B(0, epsilon) is contained in U. Then consider the collection of open sets {s_i union B(0, epsilon/2) | i in I} where s_i is the ith open ray. Then this collection covers the whole space, and this covers S in it, but any finite subcollection fails to contain some element of U and thus some element of S. Covering a subset with open sets in the whole space like this, when considered with respect to the subspace topology, is equivalent to the normal definition of compactness.

then in a tired ADHD haze I just looked at that space, saw it could be defined in two different ways, and tried to see why they should be homeomorphic, and realized the obvious map isn't a homeomorphism

[cursive flume]

can someone explain me what this means? the surface of my brain is actually the 1-point compactification of the product of the (0, infty) topologists sine curve with itself

what is a compactification and the topologists sine curve?

smh this should be a meme but i'm so bad at topology i don't get it

[empty grove]

topologists sine curve is the graph of sin(1/x) as a subspace of R^2

ie the set of points (x,sin(1/x)), and i think the (0,infty) is supposed to mean x is positive

[honest terrace]

I think you usually add 0 to it, no ?

else it lose some "badness" properties

[empty grove]

Do you? idk either way doesnt really matter lmao

[honest terrace]

like it becomes path connected I think

[empty grove]

yeah thats why i embed it in R^2

take closures or whatever there lol

[honest terrace]

right yeah, same thing

[empty grove]

1-point compactification of a space is when you add a point at infinity to make the space compact. Like the real line isnt compact, so you just add a point at infinity and going infinitely far in either direction gets you to that point 1-pt compactification of R is homeomorphic to the circle, if you want a more precise thing then look at wikipedia

idk what the meme is trying to say tho

[chrome dew]

more brain wrinkles is supposed to mean you're smarter

[empty grove]

meppa more brain wrinkles confirmf?

[chrome dew]

help me cut open my skull to peek inside

idk tbh

[empty grove]

💇‍♂️

[honest terrace]

the important part of the meme is that topology is bad

[sweet wing]

important meme is that this is weakly homotopic to some 4 point space

[honest terrace]

I surely am weakly homotopic to some singleton equiped with its unique Hausdorff, connected, discrete, non-discrete topology

[sweet wing]

I surely am weakly homotopic to some singleton equiped with its unique T0, T1, T2, T3, T4, T5, T6, connected, path connected, locally path connected, semilocally simply connected, discrete, non-discrete topology

[cursive flume]

T 3/2 memes

[sweet wing]

ohgod

[empty grove]

[sweet wing]

add more to the list

damn this topology must be interesting

[honest terrace]

I wonder what topological properties the topology on a singleton don't have 🤔

[empty grove]

having 2 elements

[sweet wing]

fake property

[empty grove]

define property

[cursive flume]

[honest terrace]

I mean, it is indeed a topological property in the sense that its preserved under homeomorphisms, but that's not what I meant

[chrome dew]

is that fredric schuller's chalkboard lol

[cursive flume]

yes how do you know lol

[chrome dew]

brame rinkels

[honest terrace]

it's also the only topology that makes it a dim 0 variety !!

[cursive flume]

he's my fav mathphys teacher

[sweet wing]

i cant actually think of some property

[honest terrace]

same lol

[sweet wing]

<@&681259184582688842> help with this qn

[honest terrace]

(except cardinal properties obvsly)

[sweet wing]

those are fake

[honest terrace]

it is also metrizable and completely metrizable

[sweet wing]

what property does the unique topology on {} not have as well

[honest terrace]

with a countable basis

it is a baire space, too

[sweet wing]

its locally euclidean

hausdorff

[cursive flume]

compact,paracompact

[sweet wing]

second countable

[cursive flume]

everything

[honest terrace]

we already said hausdorff, but yeah

[sweet wing]

thats a lot of properties

(manifold)

lemme dig up the paper

this tells us it has some 100+ properties

[honest terrace]

Every property here that isn't a cardinal-ish property is answered by the unique topology on a singleton
https://en.wikipedia.org/wiki/Topological_property

[sweet wing]

[honest terrace]

[cursive flume]

memes apart,why are singletons interesting in math?

[honest terrace]

I checked all of them

one by one

(didn't take too much time tbh )

[sweet wing]

tru wiki list is lame

[sweet wing]

because they arent

they may be interesting when embedded in something

[cursive flume]

TIL

sober space

[honest terrace]

there's actually a whole kind of such spaces and properties

I think it has to do with tonnelled space or smth like that

[sweet wing]

i think theres some funny characterization of spaces

that are spec R

and sober was one of the criteria

[cursive flume]

time for me to learn some proper topology

[honest terrace]

there are some specific functions between sober spaces whose technical name is something like alcoolic-test

[cursive flume]

wait really?

lol

[honest terrace]

or tonelled space

I can't remember exactly

but smth like that

[cursive flume]

[sweet wing]

https://topology.jdabbs.com/spaces/S000162/properties

lmfaooooo

[honest terrace]

"has distinct points: X"

[empty grove]

its not non-metrizable 🤡

[honest terrace]

I mean

the space is discrete

and indiscrete

at the same time

why are we even asking for more

We're gifted with such a topological space and we don't accept it as it is 😔

[empty grove]

not even an initial object whats the point

add the point in and it becomes a 0 object

[cursive flume]

can one actually define singletons more formally than just saying they are sets which contain 1 element?

[empty grove]

more formally as in more abstractly?

[cursive flume]

i mean is this the precise definition?

A is singleton iff |A|=1?

[empty grove]

containing 1 element is a precise definition

[wanton marsh]

they are terminal objects in the category of sets

[sweet wing]

singletons are whatever you want

as long as they are some form of terminal object

[honest terrace]

I mean for any set X, can't we find a category containing X such that X is a terminal object in it ?

[honest terrace]

(yeah)

[cursive flume]

and is R^0 by def. a singleton,or this is a consequence?

[sweet wing]

it is a counterexample to discrete -> not (not discrete) as well!

[honest terrace]

[sweet wing]

id assume the way to motivate it

is

R^m x R^n = R^{m+n}

so R^0 must singleton

time to form the hypothetical spaces R^-1 R^-2 ...

[empty grove]

Or the definition of dimension also just gives singletonnessitivity

[cursive flume]

wait if we do R^{0}xR, why do we get R? we get pairs of {(a,x)|x \in R}

[sweet wing]

which is R

[honest terrace]

another way of motivating it is that you can think of R^n as the set of functions from a set of n elements (like [[1,n]]) to R

[empty grove]

That's isomorphic to R 👀

[cursive flume]

ah it's iso to R but not strictly R

yes makes sense

[honest terrace]

so like, if n = 0, then [[1,n]] = \varnothing

[empty grove]

Depends on what you mean by R

[cursive flume]

in math do we define everything up to iso/structure preserving maps?

[honest terrace]

and there's only one function f: \varnothing -> R

[empty grove]

Yes usually

[honest terrace]

depends on the context, but usually yes

[cursive flume]

okay then it makes sense

[sweet wing]

as long as it is unique iso tbh

if iso isn't unique then ehh

[empty grove]

I'm gonna take real numbers to be equivalence classes of Cauchy seqeunces of equivalence classes of pairs of equivalence classes of pairs of von Neumann ordinals in my next assignment

[sweet wing]

[honest terrace]

[cursive flume]

I tried following a course on bordisms and topoligcal quantum field theory this term and after 30 minutes i just saw drawings

[sweet wing]

[cursive flume]

I did not understand what do they even mean

and the teacher kept saying natural maps and that a qft is just a functor

[honest terrace]

do you think the drawer understands ?

[cursive flume]

I was lost

[honest terrace]

how naive

[empty grove]

Let r = [([([(a1,a2)])],[([(b1,b2)])],[([(c1,c2)])],...)]

[honest terrace]

the point is just to draw

can't they have fun drawing ?

[empty grove]

Why did I decide that it was a good idea to type this on phone

[honest terrace]

[cursive flume]

physics for mathematicians

i don't see where topology comes in the play though in this definition

these are supposed to be topological qfts

[sweet wing]

nonononono

you see

tqft is about that functor

it isnt about topology+qft

it is clickbait

it is actually about bordisms

[cursive flume]

yea the lecture is 'bordisms and tqft'

[sweet wing]

[cursive flume]

the good old meme my prof said ' a yellow penguin need not be yellow nor a penguin. in mathematics you can define new objects as you wish as'

[sweet wing]

facts

[gentle osprey]

ProphetX

ProphetX

[stray bane]

Does anyone have a nice intuitive understanding of what a quotient space is? I am stuggling to gain a inuition of what exactly they are, and why they are useful

[sweet wing]

quotient = glue essentially

[stray bane]

Alright, as far as i have read in the topology book by munkres, it requires a quotient map. I am however unsure about the notation $X/A$ where $A={\frac{1}{n}}\subseteq\mathbb{R}$

[gentle osprey]

Fro

[stray bane]

Alright cool, this explains something

What i am trying to do is with the set $A={\frac{1}{n}\mid n\in\mathbb{N}}$ and $\mathcal{B}$ beign the basis of a topology on $X$ consisting of all of the open sets of $\mathbb{R}$ and also the sets on the form $(a,b)\setminus(A\cap(a,b)), a,b\in\mathbb{R},a<b$. To show that $X/A$ is $T_1$ but not hausdorff

[gentle osprey]

Fro

[cursive flume]

there is the notion of complexifying Lie algebras. we know that for compact connected lie groups we can obtain the lie group by exponentiating the LIe algebra

we also know that SO(n) is compact and connected. we can reproduce SO(n) as exponentiating so(n) the real lie algebra. what do we get if we exponentiate its complexified version?

because SO(n) is strictly defined over R,not C

for sl(n) I could say that I get either SL(n,R) or SL(n,C) but for SO(n) i'm not sure what to say, neither for SU(n) what happens if I exponentiate its real form

[obtuse meteor]

It’s good to think about where hausdorffness will fail

[stray bane]

My initial guess would be around 0

there could be issues

[obtuse meteor]

For sure :)

In particular, you should look at open neighborhoods of 0 in your space and what they look like

[obtuse meteor]

(Also for onlookers, to me this seems like a Hawaiian earring with a “double origin” at the juncture point)

[stray bane]

Okay so i have been reading a bit, and my best understanding now is that i am essentially trying to pick up all of the points of $A$ and gluing them on top of eachother right? So now i should basically just show that this is $T_1$ but not $T_2$

[gentle osprey]

Fro

[obtuse meteor]

ye

so first like

imagine doing this without 0

what does it actually look like?

[obtuse meteor]

sorry, this is only the local picture with X = [0, 1]. Extending it to the global picture is not too bad though

[stray bane]

Hm my intuition says the real line spun around in rings on a single point that will represent the equivalence class of A

I think what i am stuggling with is what the sets of the topology will look like

Maybe i don't quite understand how the open sets in this quotient space will look

[obtuse meteor]

i think you're pretty spot on. Can you draw it?

[stray bane]

quite a bad sketch, but this is what i am thinking

Where each of the circles represent an interval $(1/n_i, 1/(n_{i+1}))$

[gentle osprey]

Fro

[stray bane]

Oh i see that i actually made this incorrectly

the circles should spin around 0 and not 1

It says in munkres that "an open set of this space is a collection of equivalence classes whose union is an open set of X"

[obtuse meteor]

that's true

I prefer to think of it in terms of inverse images

let me draw the kind of picture I have in mind for this

so how to think about this

the black point that all these circles meet up at is 1

and the black line is [1, infty)

the blue line is (-infty, 0]

now importantly

[obtuse meteor]

or rather, the equivalence class A

it's really hard to distinguish between 0 and 1

and they kinda "look" like they're at the same place

but 0 is the one that's actually touching the circles

[stray bane]

Yeah because they are essentially the same in this case right? Because they are both in A

[obtuse meteor]

whereas 0 is kinda not directly on the circles

[stray bane]

or well yes

[obtuse meteor]

they're not both in A

[stray bane]

0 is not in there right

[obtuse meteor]

but 0 is arbitrarily close to stuff in A

yeah

it's kinda fuzzy there

hm let me draw an extra thing

[stray bane]

How would i for example go about showing that this quotient space is T_1, it seems fairly obvious that i can pick out neighbourhoods for two distinct points that do not contain eachother, but i am not sure what look at if i want to do it rigourously

[obtuse meteor]

If we were allowed to like, specifically "zoom infinitely in"

the blue point would be 0

and the black point would be 1

[obtuse meteor]

so like

the first thing is to realize that if either point is not 0 or A

you can just do what you would do in R essentially

to pick out neighborhoods

[stray bane]

Alright yep, so how would i go about picking out an open neighborhood around 0 for example

Can i cross over from the blue line to the circles? My guess would be yes since i can union those equivalence classes to an open set in X

or actually no? because i would hit the equivalence class on A, and then the union would not be open in X

[obtuse meteor]

I agree something like this

any open set containing 0 should contain A right?

(think on this)

so maybe you want to go with an open set around A instead

[stray bane]

perhaps i can union a bunch of neighbourhoods of A together to get a neighborhood around 0?

or a neighbourhood around any point of A for that fact

i guess this makes intuitive sense since we glued them together

[obtuse meteor]

hmmm

so what are you trying to do?

like what neighborhood are you trying to construct

with what properties

[stray bane]

Well my initial problem was how to make it open

in the first place

[obtuse meteor]

ah I'm like trying to clarify your question

so ok we want to show it's T1 right?

[stray bane]

Yep

[obtuse meteor]

you agree that if we have points x, y so that neither x nor y are 0 or A

this is simple

[stray bane]

Yes definetly

[obtuse meteor]

just do the thing you'd do in R essentially

ok

what if x is 0 and y is not A

is it still simple?

[stray bane]

Well if i could find an open neigborhood around 0 then i would say so yes

because either its on the blue or black line going out, in which case it is simple. Or it is in one of the "rings" in which case i can simply pick an interval on one of the rings that doesn't go all the way to A

[obtuse meteor]

I think the key is to choose a neighborhood that's not around 0

remember the definition of T1

ah nvm I can very definitely read lol

boop

I'm dumb

wait

this space feels not T1

[stray bane]

Yes that was what i was thinking

[obtuse meteor]

because like

[stray bane]

however it is one of our exercises to try and prove that it is T_1 but not T_2

[obtuse meteor]

set U to be some open neighborhood of 0 in R

U always contains some 1/n

then for any open neighborhood V in R/A of 0

the union of all the equivalence classes in V is an open set around 0 in R

so it contains a 1/n

so some equivalence class in V contains 1/n

so A is in V

and so 0, A cannot be separated in R/A

It is T0

but not T1

so maybe there's some indexing issues?

[stray bane]

this is the exact definition of A

[obtuse meteor]

yeah I think this question is busted

[stray bane]

then it says B is all of the open intervals on R aswell as the sets on the form

then we let T dentoe the topology genereated by B and let X be the top space obtained from equipping R with T

[obtuse meteor]

ohhhhhhh

this isn't the standard topology on R

I see why I was confused now

my fault I led you on a goose chase

[stray bane]

oh wait i realise too now

because some of these sets are open

since we cut out parts of A

[obtuse meteor]

yeah

I missed that sentence in your description oops

[stray bane]

oh my god, i have been trying to wrap my head around that for 4 hours now

somehow i kinda forgot all about that too

but this would mean that for example (-1,1) should be open EVEN in X/A

right?

or to be more precise it would be under p

[empty grove]

Is this the question? (and do you still need help?)

[stray bane]

Its still the question, i think i got it mostly worked out, all i am missing now is why it is not hausdorff, but i think it has to do with the neighbourhoods around 0 and any point of A

I think we spent most of the time looking at the wrong problem which caused some confusion

[empty grove]

Neighborhoods around 0 right?

[stray bane]

yep!"

edited

[empty grove]

Yeah that will give the counterexample

Any open set containing 0 in X would intersect every neighborhood of A, so when you take quotients, the neighborhoods of [0] (which are images of specific neighborhoods of 0 in X) will intersect any neighborhood of the point [A]

[stray bane]

Yes exactly, this was what i came up with aswell

thanks!

[empty grove]

[stray bane]

Also thanks for the help @obtuse meteor this is my second math course ever so i am not the best yet, but your visualizations really helped me understand

[obtuse meteor]

Thank you so much!!!

[gritty widget]

homology

cohomology

[tepid depot]

cocohomology

hocomology

counting

cohocohococomology

this is just making me want hot cocoa

[cerulean oriole]

coco

[tepid depot]

pie is finite

[honest terrace]

🥥homology

It's the doubledual of homology for crazy people

[thin bramble]

@gritty widget I passed transformtaion geometry with a C+. Thanks to everyone who wished they could help me in homework.

[gritty widget]

i am not sure why you pinged me, but congrats on the pass

[sleek thicket]

Lmfao

[polar whale]

Anyone help me in 2.

[gritty widget]

cohomotopy

[ivory dragon]

a hint: consider a constant function

[true robin]

?

[ivory dragon]

this is the easiest ||counterexample||

to 2

[empty grove]

oof that is an annoying thing lol

why are limits defined that way

[ivory dragon]

life sucks

[true robin]

Oh damn, i forgot limit points are defined like that, goddamn that is annoying

[ivory dragon]

yea it feels kinda hacky

but whatever

[empty grove]

how can i copyright strike discord messages

[polar whale]

But then f(A) is a singleton set. If i take x a limit point of A then f(x) = that constant...so f(x) belongs to the singleton set. How so is then f(x) not a limit point of f(A)

[empty grove]

does your definition of limit point make every point of a set A a limit point of A?

ie are you allowed to take constant sequences/nets in A to demonstrate that a point a limit?

[ivory dragon]

how are you defining limit point

[polar whale]

Wait did i forget the definition ? Let me check

[ivory dragon]

the definition im most familiar with is:

x is a limit point of S iff every neighbourhood of x contains a point in S \ {x}

but if S = {x} then S \ {x} is empty

so....

[empty grove]

There are 2 non equivalent definitions lol

[polar whale]

Ahh cluster point right?!

[polar whale]

Cluster and limit points are diff.

[empty grove]

Not in all books, some books define limit points as how namington defined it

[empty grove]

But yeah then you can prove that statement

[polar whale]

Okay if i consider this then okay

[empty grove]

Many books make a distinction lol

[ivory dragon]

if you use different definitions then the statement is true i believe

[bleak helm]

You should use the definitions from your book

[ivory dragon]

and just definition pushing

[empty grove]

Yeah it's true

[polar whale]

I am reading Munkres

[empty grove]

(trivial using net continuity)

[polar whale]

But i have another book where cluster & lim pts are diff....anyway thanks a lot @empty grove @ivory dragon

[empty grove]

But yeah just try and write what it means for f(x) to be a limit point of f(A)

[polar whale]

Yepp

[ivory dragon]

lmao the wikipedia article i posted just cites stackexchange 3 times

brilliant

(i clicked the first link to make sure it was actually an SE citation)

i guess its better than no citations at all

[empty grove]

I blame the French, they come up with such rubbish terminology that other people start correcting it/using the terminology they were using before and then we are forever stuck in a state between corrected and uncorrected terminology

[sharp frost]

does anyone know why the differential is 0 at the singularity

I've only seen the exterior derivative defined on smooth things

[viral atlas]

@honest terrace Fix this mess baguette

[honest terrace]

eeeeeh

ok tbh that's not false

[empty grove]

[honest terrace]

idk if here's that's the case too

but yes, french mathematicians and terminology is.. something

[empty grove]

dont care, I have already found someone to blame

[honest terrace]

based

[pearl holly]

Okay so this is the exercise that I am working with: Let $x_1, x_2, \cdots$ be a sequence of the points of the product space $\prod X_\alpha$. Show that this sequence converges to the point $x$ if and only if the sequence $\pi_\alpha(x_1), \pi_\alpha(x_2), \cdots$ converges to $\pi_\alpha(x)$ for each $\alpha$. Is this fact true if one uses the box topology instead of the product topology? (Here, $\pi_\alpha(x_1)$ basically means the $alpha$ coordinate of $x_1$). I have proceeded as follows: So for the first implication, we know that $x_1, x_2$ converges to $x$. This means that, for any n greater than a natural number, $x_n$ is in $U_x$ for every neighbourhood $U_x$ of x. But then we know that $\pi_\alpha(x_n) \in \pi_\alpha(U_x)$ and $\pi_\alpha(U_x)$ is a neighbourhood of $\pi_\alpha(x)$ and every such neighbourhood is of this form. For the other implication, we know that $\pi_\alpha(x_n) \in U_{\pi(x)}$ where $U_{\pi(x)}$ is a neighbourhood of $\pi(x)$. This can be written as $\pi_\alpha(U_x)$ and so $x_n$ is in $U_x$ and we are done. There must be something wrong here but I just can't figure it out...

[gentle osprey]

older sister

[honest terrace]

first "This means for any n greater than some integer, x_n € U_x for every nbrd U_x of x" this is wrong.
It means that for any neighborhood U_x of x, for any n after some integer, x_n € U_x, that's different

[empty grove]

^ and the other implication part is wrong

That's where you need product topology

[pearl holly]

Yeah I figured that. It feels really wrong

[empty grove]

Right, try to say it in full and you'll see the problem

Like why does x_n converge to x

[honest terrace]

(if I were you I'd rewrite the proof of the first implication with more details, as it is I think it's right but kinda.. confusing ? Idk that may just be me tho )

[empty grove]

Yeah it kinda flows in the wrong direction

[pearl holly]

Like honestly, I am just confused of the product topology. In the book it says "The product topology on $\prod X_\alpha$ has as basis all sets of the form $\prod U_\alpha$ where $U_\alpha$ is open in $X_\alpha$ for each $\alpha$ and $U_\alpha$ equals $X_\alpha$ except for finitely many values of $\alpha$. What does even the "except for finitely many values of $\alpha$" even mean?

[gentle osprey]

older sister

[empty grove]

You should pick a neighborhood of pi_alpha(x) and say why that contains a tail of pi_alpha(x_n)

[honest terrace]

I think the more intuitive way to think about this is to first speak about the box topology and why it is not good enough

[pearl holly]

Yeah Munkres always says that "ohh there's a difference and we will come back to it" or something similar

[empty grove]

It means that almost all the factors will be the whole set, just some finitely many can be "interesting" open sets

In case the question is just about the definition

[honest terrace]

So let $(X_\alpha, T_\alpha){\alpha \in I}$ be some family of topological spaces.\
One "obvious" way to define a topology on $\prod{\alpha \in I}{X_\alpha}$ is to take as open sets $\prod_{\alpha \in I}{U_\alpha}$ where each $U_\alpha \in T_\alpha$

[gentle osprey]

Shika-Blyat

[honest terrace]

Do you agree with that ?

[pearl holly]

Yes sir I do agree with that

[honest terrace]

Ok so now the thing is that this topology doesn't have some properties we'd like it to have

[pearl holly]

Okay and that's the part where the product topology comes in right?

[honest terrace]

Yes exactly. So an example of where the box topology fails (the "obvious" one I just described) is in infinite product of compact spaces

If you let, for each $\alpha \in I$, $X_\alpha = {0,1}$ with the discrete topology, then each $X_\alpha$ is compact but $\prod{X_\alpha}$ isn't, because discrete spaces are compact iff they are finite

[gentle osprey]

Shika-Blyat

[empty grove]

(or one really nice example is the problem you are trying to solve )

[honest terrace]

Not sure if you've encountered compactness yet though, that may be why munkres postponed a bit the explanation

[pearl holly]

Okay spare the writing, I don't even know what a compact space is

[honest terrace]

[pearl holly]

But Munkres goes over an example where the product topology is "better"

[empty grove]

The exact thing we need from the product topology is the property that a function into the product should be continuous iff each coordinate function is continuous

[pearl holly]

Some sort of homeomorphism between cartesian products

[pearl holly]

Oh yeah, that's exacly what Munkres covered

[empty grove]

Yes, that's the defining property of a product

[honest terrace]

wikipedia gives a nice example of where this fails with the box topology

[pearl holly]

Oh yeah, Munkres had a similar example I think

[empty grove]

It's called "universal property of products". The idea is, that you want to define things not by their elements, but by their properties. You see products of groups, vector spaces, sets, etc and they all have the same name "product"

[honest terrace]

cat theory time

[empty grove]

The reason is that they all have this exact same property, except you replace continuous by homomorphism or whatever is appropriate

That's the main idea, and the reason why product topology is so useful is not because of open sets look nicer or whatever

Its because of that property

[pearl holly]

Oh okay, now I see!

[empty grove]

(the exact property is slightly different, you can see wikipedia if interested)

[honest terrace]

the wiki (just read the equational part if you're not familiar with diagrams, although it's less nice)

[empty grove]

Also a tip, you can often solve a lot of problems involving product spaces using just that property, without using the construction itself, and this makes the proof neater

So you can keep that property in mind and try both approaches whenever you see problems with products involved

[pearl holly]

Okay that's great to know! I will keep this in mind!

[honest terrace]

and more importantly, you can write "by the universal property of the product" while proving it, which is the ultimate flexing level

[pearl holly]

Okay good, thank you so much! But you guys mentioned something about the first implication of my "proof" of the exercise. Is that even right or is something wrong there?

[honest terrace]

the first implication is right if the reader knows what he expects to see in the proof, ig, but it's hard to read and doesn't "flow in the right direction" as moldi said

[empty grove]

You took an open set U of x, then showed that pi_alpha(x_n) is eventually in pi_alpha(U)

[empty grove]

It's actually somewhat wrong/incomplete too actuallu

[empty grove]

Can you see what's wrong with this?

[honest terrace]

they didn't just said that

[empty grove]

Oh wait

[honest terrace]

they also said that every nbrd of pi_alpha(x) will be of this form

[empty grove]

I missed the last part where you said every neighborhood is of that form

[honest terrace]

which is what is missing, I think ?

yeah right

[empty grove]

Yeah sorry, it's correct

[honest terrace]

(but that illustrates my point that it's hard to read )

[empty grove]

But usually you'd start from an open set around pi_alpha(x)

Then work your way back to the product

Because the statement you want to show is that "given any neighborhood of pi(x), pi(x_n) is eventually in that neighborhood"

So it's usually better to start your proof with the first half of that statement

And end with the second half

[honest terrace]

(or atleast, if you want to use that the projections are open, put the arguments in the right order)

[pearl holly]

Yeah that sounds right. I will keep this in mind, thank you!

[empty grove]

[pearl holly]

[honest terrace]

(I'm saying that even though I always forget to do it when I write a proof )

[empty grove]

based

[honest terrace]

(I just throw every arguments that needs to be said and I let the reader guess how to deal with them )

[pearl holly]

Okay but thank you guys so much! I will try to do the other implication now!

[honest terrace]

[honest terrace]

😂 😂 😂👌 hey hey @empty grove there exists balls that are.. squares 👌😂 😂 😂

(and that's how we truly shitpost, kids )

[empty grove]

based

[cursive flume]

why is SU(n) a real lie group?

I thought it's complex

[long gorge]

It's because complex conjugation isn't holomorphic - you can't write it down in complex equations.

[uncut surge]

Yeah, its matrices contain complex entries, but you can't give it a holomorphic manifold structure, so the best you can do is consider it as a real manifold (kind of ignoring the presence of complex entries via the identification C = R^2)

[bright acorn]

Ok so

I was trying to read something on algebraic topology these days

And it dealed with operads

I tried to look it up on places I could potentially learn about operads

But I couldn't

Are there any recommendations?

Also

When do they start to come up in algebraic topology?

I am curious

Ah yes

When you define the loop space of a pointed topological space.

That's what I have heard

And somewhat of what I could get out of all the stuff I have read about it so far lmao

But thanks a lot

[sleek thicket]

Tai-Danae Bradley's blog posts give a nice intuitive introduction imo

[native raptor]

I'm a little confused about the process of building a 3-manifold from a Heegaard diagram, I asked my advisor but didn't totally follow the explanation and didn't have time to ask more questions so I was wondering if someone else here could help shed a little more light?

[bold canopy]

https://web.math.princeton.edu/~petero/Introduction.pdf
^this pdf seems to give a nice introduction to heegard diagrams, but here is a kind of gist: a handlebody of genus g is a 3-manifold with boundary given by a surface of genus g (this is unique up to diffeomorphism) (you can think of this as a thickening of a bouquet of g-circles in R^3). Given two handlebodies of genus g, you can glue them along their boundaries and end up with a closed 3-manifold without boundary; and it turns out that all 3-manifolds come from such a construction. The question now is, how do you specify the data of this gluing map? It turns out that you can specify the gluing map by these red and blue circles, the red circles bound disks in one of the handlebodies, the blue bounds disks in the other handlebody

you can think of the topological information of how many times the blue curve crosses the red one as somehow encapsulating the information of the 'winding number' of the gluing map (you can think of it as gluing it on with some number of twists, that these curves encapsulate)

then - if you want to go forth and use this technology to do 3-manifold invariants; the whole business becomes identifying "when are two heegard diagrams give rise to the same 3-manifold?" (exactly like analysing reidemaster moves if you were doing knot theory, or kirby calculus if you were doing the surgery presentations)

[native raptor]

Thank you!! I appreciate the explanation

[vast yoke]

I have a question regarding blow ups of varieties and the points on exceptional divisors (the line where you're blowing up)
so if I have a curve in C[x,y], with a singularity at 0
and, let's say the tangents at that singularity are y=0, y=ix, y=-ix

The consider a blow up at 0. We're looking at points on zariski closure(pi^-1(variety\{(0,0)})) intersection Exceptional divisor
Then, as far as I understand, this will have 3 points, namely [1,0], [1,i], [1,-i]; correct? Well, assuming you identify the divisor with P¹, otherwise all of them have an affine (0,0) in front.
Now assuming that's fine, my question is why. Intuitively it makes sense, but I can't quite get there

So what I've tried is attempted to calculate that blowup, first without closure.
You say that it's equal to {((x,y), [x,y] | f(x,y) = 0 and x != 0} U same with y!= 0 (meaning you split the projective cases in half: One open set where x non-zero, the other when y non-zero. Then you can divide)
then, upon simplifying I get that it's isomorphic to V(x-y-y^3) for x!= 0 and V(yx^4-x^2-1) for y != 0
is there any way to see those tangent singularities from here? The points on the exceptional divisor.

[polar whale]

The box topology is finer than the product topology. Then why do we prefer product topology ?

[ivory dragon]

munkres has an entire chapter dedicated to basically this question

tl;dr it has a lot of not-so-nice behaviour

and is a pain to prove things about

https://en.wikipedia.org/wiki/Box_topology#Properties

its such a messy space that its main source of fame is being a fantastic source of counterexamples.

[empty grove]

discrete topology is finer than any topology yet we dont prefer it

[honest terrace]

finer doesn't mean nicer or more interesting, as moldilocks example shows ^

[ivory dragon]

in fact im not sure why you think it does?

theres basically no relation

[polar whale]

Actually i just started out on topology so it all seems very complex...sorry to bother with simple questions and ty for nice answers everyone

[cosmic wedge]

what does the notation, Top(X,Y) mean for topological spaces, X and Y.
For some reason the source I'm using only states, Top is the category of topological spaces and continuous maps between them, so I'm assuming it's the category that contains object X and Y with a continuous map f between them, please correct me if I'm wrong.

[marble socket]

I don't know a lot of category theory, but I think for a category C, and objects A and B.
C(A, B) or Hom_C(A, B) means the set of morphisms from A to B in C.
(I only encountered locally small categories, so idk how to handle when Hom_C(A, B) is bigger than what's allowed for it to be a set)

so i think Top(X, Y) just means set of continuous functions for X to Y.

[empty grove]

If you use grothendieck universes to found category theory on ZFC, then hom_C(A, B) is always a set, just not necessarily a small set

[marble socket]

what's a small set 😶

i know small categories are the ones where objects form a set

[empty grove]

Take a model of ZFC, this may have submodels of ZFC. Some of those submodels may themselves be sets in the original model (ie if the model is M, you may have a submodel N such that there is some U in M, such that everything in N ∈ᴹ U)

∈ᴹ meaning the interpretation of ∈ in M

Such a set U is called a universe, and it's a set in M which itself is also a model of ZFC in the obvious sense

So you fix a universe U, and then a small set S is any set that is an element of U

(nothing to do with cardinality)

So the category of all small sets for eg, is the category whose objects are all small sets and morphisms all set functions between small sets

But the object and morphism set of that category itself is then too large to be a small set

(in particular the object set is U and U ∉ U can't be a small set)

So it's a large category

[empty grove]

So here you should say "...form a small set"

Because if you found cat thy on set thy, every category is a tuple of the sets of objects and morphisms

Only issue is, existence of universes is independent of ZFC. So there are larger theories which just add that as an axiom (some use existence of a single universe, or you can add Tarski's axiom which says that for every set S, there is some universe U containing S. This extension is called Tarski grothendieck set thy)

[marble socket]

but then isn't definition of small set dependent on the chosen universe?

[empty grove]

It is

Not really an issue, because the sets whose existence you can prove in ZFC (and hence the only sets with which you'll ever work with explicitly) will be there in any universe

[marble socket]

oh okie

[verbal wraith]

The better approach is to just call it a collection.

[cerulean oriole]

And every M-element of U is in N?

[empty grove]

Yes lol

[cerulean oriole]

Hmm

Doesn't this depend on M as well as U?

[empty grove]

Yes

[cerulean oriole]

🤔

[empty grove]

You fix your universe before you do anything in cat thy

[cerulean oriole]

Can we take M to be the "model that we are ourselves working in"?

[cerulean oriole]

Oh

This is not how I'm used to thinking about "small" and "large"

[empty grove]

Yeah the name is kinda misleading

[cerulean oriole]

I think of it as predicates
And the predicates which are given by some set are "small"

And the rest are "large" or "proper classes" or whatever

[empty grove]

Also you can of course do this purely syntactically without referring to models, because you can state whether a set is a model of ZFC within ZFC

[empty grove]

I didn't get this

[cerulean oriole]

So any predicate you can write is a "collection"

Where you think of "exists y : y in x and forall z: z in x ==> z = y" as
{ x | exists y : y in x and ... }

Even though the latter notation isn't well-defined

And if you call that A, so
A := { x | exists y : y in x and ... }
Then any statement of the form

forall x in A : something else(x)
Is translated into
forall x : (exists y : y in x and ...) => something else(x)
and
exists x in A : P(x)
becomes
exists x : (exists y : y in x and ...) and P(x)

[empty grove]

So you're taking the collection of all singletons?

[cerulean oriole]

Yes
As an example

[empty grove]

ok

Right yeah

That's how you deal with some proper classes within ZFC

But large and small aren't used in that situation

They're just called classes I think

[cerulean oriole]

ic

[thin jewel]

Does anyone know of any generalisations of the cross ratio to higher dimensions?

or it is basically the only projective invariant left?

[summer jolt]

Does anyone know what is the intuition behind Borel functions and Borel sets?

[cerulean oriole]

I don't know for sure but if you require all open sets (or all closed sets) to be measurable you end up with all Borel sets measurable
I'm guessing that's where they come from?

[cerulean oriole]

Preimages 😍

[summer jolt]

I see. I saw them being used in the context of classical mechanics so want to check if in the context they were just a technicality or had a more profound meaning

[wheat gulch]

I think I have found a way to represent a D8×C2 group (D8 is the symmetries of the square, the dihedral group order 8) as a combination of two C2×C2×C2×C2 groups (the elementary group order 16). One of my friends told me it might be a cohomology but I am unsure. Has anyone ever heard of something like this or could point me towards key words I could google?

These are the two elementary groups order 16 as Fano pyramids. https://youtu.be/3RJQBSmJoRs

[feral copper]

Hello! Dumb question, because I don't remember how one calls this operation on spaces...
What is $X\widetilde{\times}Y$ ?

[gentle osprey]

Matplotlib

[feral copper]

Like $S^2\widetilde{\times}S^2$ for instance

[gentle osprey]

Matplotlib

[sleek thicket]

I have never seen this notation

[tough imp]

What does it do

[feral copper]

That's what I'd like to know 😛 I keep seeing it in papers, and I know I knew what it was at some point, but this knowledge flew out of my brain xD

[gritty widget]

Send the paper then

[neat mango]

[feral copper]

I didn't say there was one in particular, but here a few that use the notation :
─ TRISECTIONS OF NON-ORIENTABLE 4-MANIFOLDS, MAGGIE MILLER AND PATRICK NAYLOR
─ Reflections on trisection genus, Michelle Chu and Stephan Tillmann
─ SMOOTH STRUCTURES ON NON-ORIENTABLE
FOUR-MANIFOLDS AND FREE INVOLUTIONS, RAFAEL TORRES

(these are only three, but as soon as you dive into non-orientable topology (and especially 4-dim), you get those)

[sleek thicket]

the second paper calls S_g \tilde{\times} S^2 a "twisted bundle"

where S_g is the orientable surface of genus g

[feral copper]

But it doesn't seem like it's that... It's so weird, this notations seems to popup in lots of papers and it's nowhere defined :\

[sleek thicket]

why do you think it's not that?

also, I don't actually know what a "twisted bundle" is

[feral copper]

Because I once knew what this tilde product was, and I don't know anything about the classifying spaces that pop up on a google search on twisted bundle

[sleek thicket]

this also seems to use (S^1 x S^1) \tilde{x} I to mean a twisted I bundle over the torus https://pi.math.cornell.edu/~hatcher/3M/3Mdoublepage.pdf

yeah I think twisted bundle might be being used in a different way than the nlab

[cosmic socket]

Does there exist a topology in which there exists an open set with no closed subsets

[feral copper]

I looked at nlab but also these MO posts too

[sleek thicket]

no, the empty set is always closed

[feral copper]

X and O are closed

[cosmic socket]

(Nonempty subset)

[sleek thicket]

sure, take the sierpinski space

this is {0, 1} where the open sets are {}, {0}, {0,1}

the open set {0} has no nonempty closed subspaces

[cosmic socket]

Ok good point

[sleek thicket]

it's impossible in a locally compact hausdorff space, which is most of the spaces you would see in analysis or topology or whatever

non-pathological

[cosmic socket]

By definition right

[sleek thicket]

yup, pretty much

[feral copper]

About my thingy, Idk if it may help or not, but I think I once read that $$S^2\widetilde{\times}S^2\cong\mathbb{CP}^2#\overline{\mathbb{CP}}^2$$

[gentle osprey]

Matplotlib

[sleek thicket]

definition+compact sets in a hausdorff space are closed

[cosmic socket]

Yup

[sleek thicket]

it's also impossible in a metric (metrizable) space, any nonempty open set contains a nonempty open ball, take the closed ball of half the radius

[feral copper]

Or at least it was for their intersection form

[sleek thicket]

oh wait

im being silly

it's impossible if points are closed

[cosmic socket]

Metric spaces are locally compact and hausdorff

Oh

[sleek thicket]

because you can just take any point

so eg any hausdorff space

[cosmic socket]

Oh yea I’m being dumb

Lol thanks

[bold canopy]

it seems like $\tilde{\times}$ is used just as the twisted bundle lke you said? The definition (from ths book 'wild world of 4-manifolds', also from the hatcher link sent) was $X \tilde{\times} Y$ is the twisted X-bundle over Y? (it seems like in all the cases the notation is used, this is unambiguous?)

[gentle osprey]

jmask

[bold canopy]

some words, from the 4-manifold book

the motivation behind the notation being that $\times$ gives you the trivial bundle, $\tilde{\times}$ the nontrvial one when it exists and is unique?

[gentle osprey]

jmask

[sleek thicket]

oh, I see

I thought twisted bundle meant something fancier than just "nontrivial bundle"

[cerulean oriole]

Even in a T1 space right?

[sleek thicket]

Yeah we realized later on

[cerulean oriole]

Oh oops

[sleek thicket]

No it's okay

In your defense I was being really stupid

[gritty widget]

[sleek thicket]

hello T

t

e

r

r

a

[fathom cave]

I am little confused as to which algebra it is talkng bout. the only algebra im familiar with is $C_p^{\infty}$

[gentle osprey]

Zero0

[fathom cave]

is it algebra of smooth functions on U over R?

[tough imp]

That is what I think that stands for yeah

If smooth means infinitely differentiable

[fathom cave]

ye

[supple locust]

are regular point theorem and constant rank theorem equivalent?

[empty grove]

Can you find the distance between f_n and a constant function g_c(x) = c?

It should be f in A, g in B

Also it shouldn't be sup |f-g|

It should be d(f,g), which is equal to sup |f(x)-g(x)|

Yeah I mean sup |f-g| doesn't make sense

It should be sup |f(x)-g(x)|

f is a function while f(x) is a real number

But yeah notation aside

With that expanded definition, can you compute the distance?

Start from the inside

Fix some f and g in A and B respectively

And compute the distance between them

To get you started, let g be the constant function g(x) = c, f be f_n(x) = x^n

And the domain of both is [0, 1]

[empty grove]

x varies

so like try to draw the graph of x^n in [0,1]

and the graph of g

and just visually see what the maximum vertical distances between these 2 graphs is

nice

so x^n has range [0,1]

do you agree?

[gritty widget]

yeah

[empty grove]

so suppose c>=1

what is the distance then

[gentle osprey]

Chmoldilocks

[empty grove]

c>=x doesnt make sense

yes, but the reasoning is wrong

so in the supremum

set y = x^n

[gentle osprey]

Chmoldilocks

[empty grove]

do you agree?

can you compute that supremum now?

for the 3 cases c>1, c in [0,1], c<0

[thin jewel]

Does anyone know much about projective invariants?

They've come up in something I'm doing and there don't seem to be a lot of modern references for them

[violet sun]

What if instead of topology,we made a thing called a pseudo topology with the following properties:
1)X and phi in psuedotop
2)union of pseudtop elements is in psuedotop
(i.e.,we ignore finite intersection);
Is this interesting to study?

[fathom cave]

so we take a set X and form a bunch of subsets of X, this pesudotop contains X , phi and bunch of subsets we jsut formed and their unions?

[violet sun]

Yes

Mostly doing this because I don't see how finite intersections make topology "nice"

[verbal wraith]

this is a good take, the finite operation does nothing

[fathom cave]

wdym by does nothing?

[verbal wraith]

like you can just remove it and it makes no difference

[gritty widget]

set of things it does is empty

[fathom cave]

@violet sun this is the closest i can find as a motivation as to why to consider finite intersection. We keep finite intersection because this is how open sets in metric space behave and topology is a generalisation of that

page 78/79

[gritty widget]

was this made in word

[fathom cave]

it doesnot look like tho

layout looks a bit weird i agree

[cerulean oriole]

If you take the complements, you get a family closed under intersection but not under any unions
Which is similar to many subthings (subgroups, ideals, subfields, subspaces, ...) and other "closed under ____" sets
IDK if that is helpful but 🤷

[honest terrace]

that's not really convincing though

why do we keep general union of open sets in the def then

why don't we enforce connectedness of the space, since that's how R^n is

why don't we enforce local compactness, since that's how R^n is

I don't think saying that "it works well with the nice spaces" is convincing, because general topology is literally what we consider when the spaces aren't nice enough anymore

(to be clear I don't have any answer to Buncho's question, I'm just answering to your answer)

[pearl holly]

Okay so let's say that $g(t) = (t, t, t \cdots)$ is a function from $\mathbb{R}$ to $\mathbb{R}^\omega$ (the Cartesian product of $\mathbb{R}$ with itself $|\mathbb{Z}_+|$ times). I want to determine whether $g$ is continuous over the uniform topology on $\mathbb{R}^\omega$.

Let $\rho$ be the uniform metric on $\mathbb{R}^\omega$ and let $B_\rho(X, \epsilon)$ be a basis element for the uniform topology. Let $a \in g^{-1}(B_\rho(X, \epsilon))$. Then $g(a) \in B_\rho(X, \epsilon)$, so $\rho(g(a), X) < \epsilon$ which means that $\overline{d}(a, x_\alpha) < \epsilon$ where $\overline{d}$ is the standard bounded metric corresponding to $d$. This gives us two cases, either $|a - x_\alpha| < \epsilon$ or $d(a, x_\alpha) > 1$. In the first case we see that $a \in (x_\alpha - \epsilon, x_\alpha + \epsilon)$ and that this open interval is a subset of $B_\rho(X, \epsilon)$. This shows that $g$ is continuous for this case. But what about the other case, where $d(a, x_\alpha) > 1$? Then what? This argument no longer works, right?

[gentle osprey]

older sister

[pearl holly]

Here $\overline{d}(a, b) = \text{min}{|a-b|, 1}$ and $\rho(X, Y) = \text{sup}{\overline{d}(x_\alpha, y_\alpha) ; | ; \alpha \in J}$ where $J$ is the indexing set

[gentle osprey]

older sister

[pearl holly]

(I meant to say that rho is the uniform metric, not the topology itself. Latex won't update now for some reason)

[fathom cave]

I meant to say metric spaces. And that's the best answer I have gotten so far. @honest terrace

Like how we have defined continuous function

[honest terrace]

because the latex is too old, after some time texit stop caring about old messages, because that would be too much to track all the latex messages that ever got sent

[fathom cave]

Also I don't know compactness yet in topology

[honest terrace]

I'm confused about your notations @pearl holly, what does "Let $\rho(X, \varepsilon)$ be the uniform topology on $\bR^\omega$" mean ?

[gentle osprey]

Shika

[honest terrace]

like what's $\rho(X, \varepsilon)$ here

[gentle osprey]

Shika

[pearl holly]

Yeah sorry I meant to say that rho is the uniform metric on $\mathbb{R}^\omega$, not the topology itself. I have defined rho in a post earlier

[gentle osprey]

older sister

[honest terrace]

oh ok sry I could've inferred that

what's d @pearl holly

[pearl holly]

Oh, d(a, b) = |a - b|, so it's the "standard" metric on R

[honest terrace]

[gritty widget]

shika

@pearl holly Take that back.

[pearl holly]

no

[gritty widget]

ok

[honest terrace]

I'm really really confused by what you wrote, so I won't comment directly what you wrote and just give you an hint:\
Let $x_n, y_n \in \bR^\bN$, we define the uniform metric on $\bR^\bN$: $d_u(x_n, y_n) = \sup_{n \in \bN} \min {1, |x_n - y_n|}$

[gentle osprey]

Shika

[honest terrace]

Let $f: \bR \to \bR^\bN, x \mapsto (x, x, \cdots)$

[gentle osprey]

Shika

[honest terrace]

first, both of those spaces are metric spaces

so, as you probably already showed, saying that $f$ is continuous with the usual def of continuous topology is equivalent to saying that:

[gentle osprey]

Shika

[honest terrace]

$\forall a \in \bR, \forall \varepsilon > 0, \exists \delta > 0, \forall x \in \bR, |x - a| \leq \delta \implies |f(x) - f(a)| \leq \varepsilon$

[gentle osprey]

Shika

[honest terrace]

So take $a \in \bR, \varepsilon \in \bR^{+*}$

[gentle osprey]

Shika

[honest terrace]

Can you find a $\delta$ such that if $x \in ]a-\delta, a+\delta[$, then $f(x) \in B_u(f(a), \varepsilon)$, or equivalently $\sup_{n \in \bN} \min {1, |f(x)_n - f(a)_n|}$

[gentle osprey]

Shika

[honest terrace]

($f(x)_n$ meaning the $n$-th term of the sequence to which $x$ is mapped through $f$)

[gentle osprey]

Shika

[honest terrace]

@pearl holly

[pearl holly]

Oh, so this method uses the epsilon delta definition?

[honest terrace]

yes

and in metric spaces, it's often just easier to use the epsilon delta definition or the sequential characterization (a map f: (E, d_E) -> (F, d_f) is continuous at a point x € E iff for any converging sequences x_n -> x, we have lim_n f(x_n) = f(x)) than the continuity definition usually used in general topology (preimage of open sets are open)

[pearl holly]

My approach was to use the definition of a basis. So first it is sufficient to show that the inverse image of a basis element is open. Let's choose a basis, $B\rho(X, \epsilon)$ and we need to show that $g^{-1}(B\rho(X, \epsilon))$ is open, i.e for every $x$ in this set, there exists an open interval (a, b) in R such that $a \in (a, b) \subset g^{-1}(B\rho(X, \epsilon))$

[pearl holly]

yeah I actually completely forgot about that definition lmao

[gentle osprey]

older sister

[honest terrace]

doing it like that should work, but it'll be a bit more ugly and it's probably not what's expected

[pearl holly]

yeah that's true. Let me think about your hind

[honest terrace]

I forgot something here @pearl holly

I meant:

Can you find a $\delta$ such that if $x \in ]a-\delta, a+\delta[$, then $f(x) \in B_u(f(a), \varepsilon)$, or equivalently $\sup_{n \in \bN} \min {1, |f(x)_n - f(a)_n|} \leq \varepsilon$

[gentle osprey]

Shika

[honest terrace]

(I forgot the <= eps at the end)

[pearl holly]

oh yeah lol. Thank you! Let me think about this

Wait what is ]a, b[?

is it (a, b)?

[fathom cave]

Yes

[honest terrace]

yes sorry, french notations

[pearl holly]

lmao it's okay! I think that any delta between 0 and sigma will do the job, right?

Since for every $x, y \in \mathbb{R}$ such that $|x-y| < \delta$ we will have that $\rho(g(x), g(y)) = \text{sup min} {|x-y|, 1} < \epsilon$

(in your notation the rho will be the d_u)

[gentle osprey]

older sister

[pearl holly]

but what if min{|x-y|, 1} = 1? Then what? I come back to the same question as I had before lmao

No wait. If $\text{min} {|x-y|, 1} = 1$ then $1 < |x-y|<\delta < \epsilon$ so we would still have $|x-y| < \epsilon$

[gentle osprey]

older sister

[pearl holly]

right?

[honest terrace]

what's sigma ?

ig you mean epsilon ?

[pearl holly]

oh yeah sorry lmao

[honest terrace]

so yes, delta = epsilon work

[pearl holly]

I mean epsilon of course

[honest terrace]

or any smaller delta

It's easier like that, isn't it ?

[pearl holly]

Okay thank you so much, honestly! You helped me a lot! I was stuck on this problem for a while now and I repeated the same argument (the one I had above) like 100 times and got nothing lmao. But this method was a lot better!

I will try to keep the epsilon delta definition somewhere in the back of my mind when I do these kinds of problems! Thank you so much!

[honest terrace]

Glad it helped

[sharp yoke]

how would you do the second part

[sharp yoke]

i guess you can just pick two distinct points x and y in S then define a map sending everything in S to x except x which gets sent to y

[wanton marsh]

that wouldn't be a homeomorphism though

[sharp yoke]

true

[sharp frost]

Hi guys, I'm not experienced with working with tensors in their array representation so I'm not sure what mistake I'm making here. I have a tensor in one set of coordinates and all I need to do is work out what its components are in another set of coordinates. Here is what the book claims

but when I do the calculation I get something wacky

[flint cove]

A red flag for me is that you sum over two lower i's and j's (einstein summation contracts upper and lower for a reason)
Have you tried the inverse jacobi matrices here (del t over del s)?

(@sharp frost )

[hollow harbor]

These aren't two lower indices. A lower index in the denominator / as the variable of differentiation is counted as an upper index (it's somehow "doubly lower"). So this is fine. One can check this with dimensional analysis.

[sharp frost]

I did try inverting the Jacobi matrices and it seemed to transform the tensor into something with all zeros in the top row, which was weird

Thanks for the suggestion though

The book might just be wrong

[flint cove]

Wait, but aren't coordinate functions usually denoted with upper indices? Sorry if this is stupid

[sharp frost]

Ah yeah if I were following normal index conventions those it would look like s^i and t^i, but everything is contravariant in the cases I’m looking at

[hollow harbor]

I agree with you in general, but they aren't here

Yeah

I just don't see how you're getting these expressions

Like I havent computed them fully

[sharp frost]

https://media.discordapp.net/attachments/806125011357532182/848486371220127774/unknown.png

Some very ugly Mathematica code

[hollow harbor]

But the eta'_1,1, where is a minus sign even coming from?

Oh, from eta's bottom right

Hmm

[sharp frost]

Assuming the author made an error, what would be the best way to go about finding a change of coordinates that DOES turn eta into that antidiagonal matrix?

[hollow harbor]

Yeah I'm worried about what lux was worried about now, they may have been right. I feel like if you do this with the inverse jacobian you get the right result

So either (for some reason) the indices are being written wrong and some are actually covariant, or the transformation is supposed to be the inverse of the one you're given