point-set-topology

[floral gust]

My prof said CW is the most intuitive one so no simplicial or anything just directly CW ):

[wanton marsh]

can you find the place where they define the boundary of a cell

[floral gust]

Sure

[midnight jewel]

look at the picture, you have two hemispheres and from the perspective of each hemisphere one goes around clockwise, the other counterclockwise

[wanton marsh]

and the orientations

[midnight jewel]

so basically for one, the boundary is oriented “along” the orientation of the hemisphere, for the other “against”

[midnight jewel]

which is basically the analogue to being in front/back of the line

[floral gust]

Oh you mean, to find d2 u1 I should look from above

and for d2 l1 I should look from below

[midnight jewel]

intuitively, yea

[floral gust]

Ahhhhhhhhh

[midnight jewel]

actually, you have to use those ugly degree computations there, which’ll give you the same result ultimately

because for one you have a disk with positively oriented boundary (degree 1) and for the other one negative (degree -1)

so much for “intuitive”

[floral gust]

Exactly lmao.......

I dunno what my prof was thinking

[midnight jewel]

I mean like, it is intuitive to me in these cases

but Δ-complexes are way easier to introduce imo cause you just have like, triangles

[floral gust]

So I perhaps you use the same look from above look from below for d1 as well?

[wanton marsh]

and the boundary maps are super easy to define on the Δ complexes

[floral gust]

yeah but is there any iNtuItIoN, the part that sucks is almost any algtop book introduces homology using simplicial

@midnight jewel One last question, why wasnt d1 u1 = u0 - l0 instead of l0-u0?

[midnight jewel]

can I just say because it is that way and not the other way

I’m pretty sure it would all work out fine if you (consistently) flipped all the signs

[wanton marsh]

u1 is depicted as an arrow from u0 to l0

I would have written l0-u0 too

[floral gust]

Oh so the orientations at d1 doesnt effect the orientations at d2?

In later at the homology computations part

[midnight jewel]

I’ve just lost all intuition for signs of 1-simplices sry&thx

[floral gust]

):

[wanton marsh]

I don't understand that theorem cuz I don't know what those S(n-1)(alpha) and S(n-1)(beta) are

[floral gust]

we have cells labelled as e^1_alpha and e^1_beta (In my case it will be e^1_alpha = u1...). The boundary of these is called S^0_alpha

[wanton marsh]

to compute the degree you need to choose some orientation for all those boundaries and all those complexes ?

[floral gust]

Yes

[wanton marsh]

concretely, what is an orientation ?

you have to peel at so many definitions to understand the thing completely lol

[floral gust]

Not sure, but I think you can define initially the degree of maps from S1 to S1 by defining it to be deg(f) = !f(t+1) -!f(t) where !f is the lift of the map. Denote the orientation to be positive if the degree above is positive. From there maybe build the definition of orientation inductively

[marsh forge]

Thinking out loud

Ok

So I have a vector bundle of dim n over some connected top space B

So we have R^n -> E -> B

Then we can define a new bundle Ehat whose fibers are collections of orthonormal bases

for the original fibers

this can be topologized by the correspondence with O(n)

Then BO(n) will classify this bad boy

so let's say I start with an O(n) principal bundle

Oh, no, thats not what Iw ant

what I want is to recover a vector bundle

up to iso

once I've taken its o-frame bundle

[little radish]

Say we have a Convex Polytope in k dimensional space is there an efficent algorithm to determine if a point is inside the polytope?

I am trying somethign out but literally the only things I see people talk about are planer polygons and at most R3.

[marsh forge]

It feels like the naive thing shouldn't be that bad?

[little radish]

My gut says just testing if the point is in each supported half space from the constraints of the polytope

but that is generally a O(kn) cost with k being dim and n being number of half spaces

The downside is that I am looking at k~n~1000

I was wondering if there was an algo that made at least one term logarithmic or at least sqrt

[lime dune]

Is there a standard notation for one-point-compactification?

[marsh forge]

Yes

[lime dune]

Makes sense

[marsh forge]

And then theres a standard topology on that

[past apex]

https://gyazo.com/4155711b2b2138362c16196f03699538

can anyone help answer this

[sonic hill]

Can I just say this is the first time I've ever seen an accented character as a variable

$\sin è$

[gentle osprey]

Icy001:

[marsh forge]

this is the wrong channel

[floral gust]

Can someone explain the note that part where a and b are 1-cells and alpha is the two cell and the bracket denotes the equivalence class? How does the "note that" implies "so that"?

[raw sedge]

should I learn analysis properly before topology?

[vocal wharf]

some analysis is nice i guess, but mostly to obtain the required mathematical maturity

to me, my topology class was taking the good parts of real analysis and then going from there

[floral gust]

@raw sedge No need. I would say my analysis got complemented more from topology than topology got complemented from analysis.

[raw sedge]

thanks

do y'all know a good textbook to begin from? preferable something motivated rather than dry

[floral gust]

Munkres for text. Topology of surfaces for pictures

[sweet wing]

i feel munkres is kinda analysis oriented with the later chapters before the alg top part

[vocal wharf]

as a different suggestion as munkres is the standard: i enjoyed topology: an introduction by waldmann

but maybe that is because i know the author

[sweet wing]

counterexamples in topology is nice too once you get the definitions

[vocal wharf]

yeah, good suggestion to have around

[marsh forge]

@floral gust

Can you send a pic of the cw complex

I think I know what its saying but I would like to be certain

[marsh forge]

Ok cool

Have you written down the computation and seen why the algebra works out?

like if you write down dalpha

[floral gust]

Yes I know why the boundary maps are they way they are. I just dont know why "note that" implies "so that"

[marsh forge]

take a negative 1 out of the LHS

you get -(a+b)

then the two chains in question differ by a boundary (dalpha)

If two things differ by a boundary at the level of the chain complex

they are equal in homology

[floral gust]

Yeah that much I inferred from the statement by assuming it is true. I just want to see a demonstration of why that actually is true.

[marsh forge]

Think about what homology is

you take a quotient

by the image of d2

so "differ by an element of the image of d2"

means "equal"

[floral gust]

ohhhhh true true...

[floral gust]

The canonical CW structure obtained by identifying opposing sides of 1-torus (isomorphic to S1) is 1 0-cell and 1 1-cell.
Similarly, for 2-torus we obtain, 1 0-cell 2 1-cell and 1 2-cell.
Similarly, for 3-torus we obtain, 1 0-cell 3 1-cell and 3 2-cell and 1 3-cell.
Does this claim generalise? That is the number of cells in the k-th skeleton of an n dimensional torus equals n choose k?

[marsh forge]

That would not surprise me

But you can also just try to prove it

Inductively

You know the cw on S1

So you can compute the obvious cw on the n fold product

Also its important to note that bc everything we are doing is only canonical that the question is telling us only something about the cw on S1

Like we dont make any choices after that so its really a question of arithmetic

[floral gust]

True true. I mean yeah the main purpose of homology is just to calculate the groups which I can just do by using the Kunneth

[lime dune]

what's the intuition behind the types of space where connected iff path connected?

what property of R^n and C^n is it that makes this so, intuitively?

[marsh forge]

Uh local path connected works

thats the intuition

[floral gust]

What makes irrational plane to be path-connected in R^2 but not the rational plane? If its merely uncountability than is it true that dense uncountable subset of path-connected spaces are path-connected?

[marsh forge]

We can construct paths in the first

But not in the second

We can only guarantee that complements of countable sets are pc

Because we can demonstrate by contradiction that it is impossible for such a removal to kill all paths

The latter can be seen bc Q is a totally disconnected subspace of R

[floral gust]

So does the claim generalise then I suppose? Because removal of countably many points cannot kill all paths and thus uncountable dense subsets of pc are pc

oh ffs true true

Thanks

But I mean the unit circle is uncountable

Oh right I should have phrased as complement of countable

[dense heart]

Can anyone pls help with the intuition behind this

[marsh forge]

Can you latex it please

thats unreadable

[dense heart]

damn i thought my handwriting was neat 😦

hold on ill try to find a link somewhere

[honest narwhal]

@orchid forge a universal cover is a simply connected covering space

[dire grail]

i dont get what the space is

ah you are gluing it to itself

the circle is identified on itself

so I wouldn't describe it like that

anyway what's the fundamental group you computed?

generated by a, right

so now the universal cover will be 3 copies of the space glued together

do you recall the construction of the universal cover?

read through the proof

that every nice space has a universal cover

the construction is fairly explicit

essentially it's the space of paths in your space

where going along an element of the fundamental group takes you somewhere else

so it's like (point, element in fundamental group)

yes

gluing means that, if you take the boundary out, then your thing is contractible so it's homeo to its 3 fibers in the universal cover

and those three have their boundaries glued

yeah it's a compact simply connected surface so it has to be the sphere

wait

is this thing a surface

it should be

oh

no

of course it's not

no surface has fundamental group Z/3

so yeah it's gonna be something weird

i can't tell, it's not obvious

i think it's like 3 bulbs

joined at a point

nah not that

im rusty on this stuff

but you should definitely go through the general construction first

and then speicalize

for this one uhh

i guess it's this one

so the question is how did I come up with it

and again the answer is the construction of the universal cover

the construction is more general

basically you have n copies of your point

and loops which are nontrivial on the fundamental group are now paths connecting different points, right? since that's how deck transformations work

so we're gonna have 6 points and a bunch of paths

I chose a presentation where b has order 3 and a has order 2

but this is arbitrary

so each point has 4 paths going out

because it has to be locally homeo to the figure 8

b forward, b backwards, a forward, a backwards (not drawn)

a backwards are just parallel to the drawn a

and you can see by my labeling that there's an action of S3 happening here that respects the quotient

so it's what you want

I don't remember where I saw the construction for a general group

rather than just for the universal cover

but you might be able to find it

i'd read the trivial group case first though, which is the construction of the universal cover

you might be able to figure out how the general one goes

also

I sketched out the first one

and I think you're right

it's a sphere with a slab in the middle

where the two hemispheres and the slab are the triangles

there's a twist happening though

so the projection isn't just taking a line through them

i think

should be a rotation involved

but yeah that seems right

[floral gust]

Got an A in topology. Thanks everyone for their help! Special thanks to @marsh forge.

[raw sedge]

congratulations

[livid narwhal]

How would you map a cube to a 3-torus?

I need the mapping to embed periodicity into a representation of a unit cell of a crystal

the specific 3-torus would be a unit horn torus

[jade forge]

Why is this channel the 'liquid echo chamber'?

[bitter yoke]

@dim meadow

[jade forge]

Huh?

[next eagle]

Can someone compute me tangent vector from this definition? I don't get it.
Let:
r(t) = (t, t^2 | t in R)
v = [1,1]
x = (2,4)

[red atlas]

I'm taking algebraic geometry next term. Do you guys have some suggestions for material I should review before taking it to help succeed?

[honest narwhal]

What book is it using?/what is it covering?

[red atlas]

@honest narwhal Hartshorne

[honest narwhal]

You should have commutative algebra down well, some basic category theory

And probably you want to get an idea of how sheaves work from elsewhere

[red atlas]

Okay, I'm moderately comfortable with commutative algebra

[honest narwhal]

Apparently there's a notion of an "Etale space" that has to do with sheaves and which Hartshorne uses a lot without being explicit about it

And it may or may not help to get an idea of singular cohomology

Disclaimer: this is mostly hearsay, I'm also gonna start schemes next semester

[red atlas]

Okay, I'll look into this stuff. Thank you

[gritty widget]

you probably want some basic homological algebra down as well (for ch 3 hartshorne and onwards). derived functor formalisms especially.

[red atlas]

@gritty widget any resource you'd recommend for homological?

I'm guessing just the appendix in eisenbud

[honest narwhal]

I'll actually be taking a class on homological algebra next quarter using Weibel!

Though that's far more than what you'd need for Hartshorne

Also semester not quarter

[gritty widget]

@red atlas : weibel's intro to homological algebra is a student favorite, but gelfand & manin's methods in homological is good too. If you're feeling particularly ambitious there's Categories & Sheaves by Kashiwara and Schapira

There's no need to know "everything" in these books. Hartshorne also goes over the basic facts about derived functors. For a lot of the concrete exercises you'll do in Hartshorne, you'll most likely work with noetherian schemes and hence cech cohomology is good enough

(cats and sheaves by kashiwara and schapira also subsumes derived functors in the treatment of derived categories in one go)

[gritty widget]

Just bought one of Cliff Stoll's tiny Klein bottles as a Christmas present for myself

Gonna make it into a necklace with a Möbius loop haha

[gritty widget]

The Klein Stein is my fave haha imagine getting drunk out, off rather, of something with 0 volume

[gritty widget]

So, I'm looking to find the packing efficiency of 9-D hyperspheres. I know I know it hasn't been proven yet - but I don't need an exact number, I need a close approximation

I've been thinking about running tests on various sized radius spheres in a hyper-cube space and seeing what it might be through brute force

my question is - has anyone else tried this yet?

Im thinking someone has GOT to have tried this before

someone suggested that I try a physics simulation with 9-dimensions. Does anyone know of an open source physics simulation in X-dimensions?

[midnight jewel]

hey I have one of those

a klein bottle I mean

[vocal wharf]

my gf gifted me one last christmas

they're neat

[midnight jewel]

bought one to celebrate passing my first year at uni and my switching to mathematics ^^

[vocal wharf]

especially the pics stoll sent us of himself with my bottle

[midnight jewel]

yea same

[vocal wharf]

we answered with a pic of us in front of the christmas tree adorned with the bottle

[midnight jewel]

(yes I’ve solved that thing. it’s not very difficult, just tedious)

[fathom shadow]

oh nice

the most complicated one i have is a megaminx

which is again just tedious not difficult

[sweet wing]

I have a 2x4x6
to this day i havent figured a decent way to solve it

[rugged swan]

Hey, I have an issue understanding how to compute some pushout. Here I want to show that the fundamental group of base point O of the real plane without n points X is $<[\sigma_1],...,[\sigma_n]>$ and is isomorphic to $\mathbb{Z}^{*n}$ where those are the canonical loops around one point.
This is done by induction using Van Kampen theorem. After choosing correct open spaces U and V, we have the following pushout :
$\begin{tikzcd}
1 \ar[d] \ar[r]&<[\sigma_{p+1}]_V,...,[\sigma_n]_V> \ar[d]\
<[\sigma_1]_U,...,[\sigma_p]_U> \ar[r]&\pi_1(X,O)
\end{tikzcd}$
One can prove that it is isomorphic to $\mathbb{Z}^{*n}$ because it is the sum of $\mathbb{Z}^{p}$ and $\mathbb{Z}^{(n-p)}$.
But I can't figure out how I can show the equality $\pi_1(X,O) =<[\sigma_1],...,[\sigma_n]>$ by using only the Van Kampen theorem.

[gentle osprey]

Zak:

[rugged swan]

Oh I've just figured it out. We have an isomorphism, but by universal property this isomorphism is canonical in the sense that it commutes with injections. That's all.

[gritty widget]

@midnight jewel is that a genuine Cliff Stoll, Acme Klein bottle? He just shipped mine! I got a tiny one I'll make into a necklace :)

What a cool guy though am I right? I loved the pics he took!!

[midnight jewel]

@gritty widget it sure is!

and he sure is too

[sick jewel]

Hey anyone, I'm taking topology next quarter. Anyone know the standard text for undergrad (or at least what you used)? I wanna try to get a little feel for it. Thanks!

[vocal wharf]

standard text seems to be munkres for general topology

[floral gust]

Is a countable composition of continuous functions continuous?

[chrome dew]

I'll say no, it won't necessarily converge

imagining things like iterating the same function over and over for instance pretty much lands you in 3 spots, things push apart, you have an isometry, or you have fixed point

[sweet wing]

^ just consider the logistic map at like r=3 to 3.57ish

[chrome dew]

funny example just came to me, f(x)=-x just compose it with itself, definitely won't converge

[sweet wing]

oh lol yea

[floral gust]

And what if I provide it converges?

[chrome dew]

I wanna say it's possible to break it into a discontinuous function but I'd have to think about it

I think if you have a family of functions that you compose you can make it do something like how a fourier series can create a square wave

is it cheating if I actually just define my family of functions to be the partial sums?

each composition would then just be adding on the extra term

that would do it I think

with some clever composition I think

[vocal wharf]

x^2 on [0, 1] works with self composition

the limit is discontinuous

[chrome dew]

oh thanks, that's much better haha

[floral gust]

ahhhhh true true

[small obsidian]

@sick jewel
I liked topology without tears, free online

[sick jewel]

thanks @small obsidian and @vocal wharf

[shrewd lagoon]

what's the reason for the definition of a topological space where only a finite intersection of open sets is also an open set? why would an infinite intersection break this notion?

i guess i was looking for the right example for intuition, is a good example just a set of open intervals on the real line that tighten around a point, effectively making it closed?

if by singleton you mean one point, yeah

i've only heard singleton used in programming context

[gentle osprey]

Token:

[real notch]

you can do $\bigcup_{n \in \mathbb N,,n>a} [a, n)$

[gentle osprey]

Darkrifts:

[real notch]

and infinite unions work in topologies

[west spindle]

here's a question that just popped into my head

call a topological space X pseudocompact if every continuous function from X to R is bounded

it's obvious that all compact spaces are also pseudocompact

but does there exist a pseudocompact space that isn't compact?

[midnight jewel]

I can tell you that pseudocompact metric spaces are compact

Proof: Let $X$ be a non-compact metric space. Then there exists an infinite sequence $(x_1, x_2, \dots) \subseteq X$ without accumulation point. For each $x_n$, let $d_n > 0$ be such that no two $B_n := B(x_n, d_n)$ (balls of radius $d_n$ around $x_n$) intersect one-another. Now define $f\colon X \to \R$ by
[
f(x) = \begin{cases}
0 & x \notin \bigcup_{n \in \N} B_n \
n\left( 1 - \frac{\mathrm{d}(x,x_n)}{d_n}\right) & x \in B_n
\end{cases}
]
Then $f$ is unbounded and continuous.

[gentle osprey]

Sascha Baer:

[midnight jewel]

(intuitively, the function has a cone of height n around each x_n)

[west spindle]

what if X isn't metrizable

[midnight jewel]

then I don’t know, my point was merely to show that you’d have to look precisely at non-metrizable spaces to find a counterexample

if there is one

[grave egret]

I have a question about germs

According to this definition, two sections s1 s2 on the same open U have the same germ iff (s1,U)~(s2,U), i.e. s1|U = s2|U, i.e. s1=s2.
This would mean that the germ at only one point p determine entirely the section, but in the examples after it is not the case. Where is the flaw is my reasoning?

[fleet trench]

Consider $\mathbb{R}$ with the particular point topology, where $U\subset \mathbb{R}$ is open iff $U = \emptyset$ or $0\in U$. I'll call the space $X$ to avoid confusion with the standard topology on $\mathbb{R}$.
\begin{itemize}
\item $X$ is pseudocompact: consider a continuous $f\colon X\to \mathbb{R}$ then $f^{-1}(\mathbb{R}\setminus {f(0)})$ is an open set not containing 0, so it must be empty. Thus, $f$ has to be constant, which means it's bounded.
\item $X$ is not compact, however: consider the open cover of all the sets ${p, 0}$ for all $p\in\mathbb{R}$. It admits no finite subcover.
\end{itemize}

@west spindle

[west spindle]

you mean f^-1(R \ { f(0) }) right

[fleet trench]

yeah

[fleet trench]

just changed that lol

[west spindle]

huh

ok

thank you

hmm ok so this brings up another question

and i'm about to commit some terminological sins here, but w/e.

call a space X strongly pseudocompact if every continuous function from X to R is not only bounded but constant, and weakly pseudocompact if it's pseudocompact but not strongly

does there exist a space that's weakly pseudocompact but not compact

wait no

of course

just give R a... two-particular-points topology ig

[fleet trench]

haha

[west spindle]

also the sin is that compactness doesn't imply strong pseudocompactness

okay so conclusion

so far we have

compact $\subset$ pseudocompact

[gentle osprey]

Ann:

[west spindle]

obviously, compact Hausdorff $\subseteq$ pseudocompact Hausdorff

[gentle osprey]

Ann:

[west spindle]

so then, will the other inclusion hold

or can we find a hausdorff space that's pseudocompact but not compact

[tepid totem]

Counterexample space that is both Hausdorff and pseudocompact but not compact: ||positive integers with the (relatively) prime integer topology||

It's in Counterexamples in Topology, page 82

[west spindle]

what's the coprime integer topology

[west spindle]

huh

[tough hamlet]

[rugged swan]

Hey, I'm reading a book of algebraic topology. It's written "if Rn is the covering space kn : S1 -> S1, we see that the total space of km^*(Rn)..." where kn : z -> z^n. I don't get what's km^*.

[floral gust]

km^* might be the induced homomorphism on the fundamental groups. Would need to see the actual wording to be precisely sure.

@rugged swan

[rugged swan]

In my book, this morphism is written f_*. And what would be the sense of km^*(Rn) if it was this morphism ?

[floral gust]

@rugged swan Do you mind sending a screenshot?

[rugged swan]

It's in french but ok @floral gust

in the Remarque

it's written "If Rn is the covering space kn, we see that the total space of km*(Rn) is compunded of d disjoint circles. And kn*(Rnà is trivial, bc d = n. If m and n are coprime, km*(Rn) is a unique circle".

d = gcd(m,n) here

[floral gust]

What is km? The lift of kn?

I am guessing it has something to do with the fact that for every point, there exists a neighourhood that is evenly covered by the covering map. This will give you the disjoint circles in Rn (there will be disjoint sets in Rn which are locally homeomorphic to their image)

[rugged swan]

km : z -> z^m

[floral gust]

What is R_n? I suppose it is not R^n as in n dimensional R?

[rugged swan]

The exercise consists in showing that :
$\begin{tikzcd}
S¹\times \mathbb{U}_d \ar[d,"\phi"] \ar[r,"\psi"] & S¹ \ar[d,"k_n"]\
S¹ \ar[r,"k_m"]&S¹
\end{tikzcd}
$ is a pullback.

[gentle osprey]

Zak:

[rugged swan]

where $m = d\mu, n = d\nu, a\mu+b\nu = 1$, $\psi : (z,\zeta) \mapsto z^\mu\zeta^{-b}$ $\phi : (z,\zeta) \mapsto z^\nu\zeta^a$

[gentle osprey]

Zak:

[rugged swan]

$R_n = (S¹,S¹,k_n,F)$ the covering space $k_n$, F is the fiber $k_n^{-1}({1})$

[gentle osprey]

Zak:

[winged viper]

Why aren't closedness and boundedness topological properties?

[marsh forge]

they are

well, closedness is

and boundedness is arguably one

[winged viper]

;-;

[marsh forge]

oh

thats just a dumb definition

but based on that definition its obvious that they aren't

[winged viper]

counterexample?

[marsh forge]

there's a homeomorphism of R and (a,b)

only one of the two is bounded

[next eagle]

Hi, I wonder - is there some way to visualize, imagine, comprehend metric tensor and Christoffel symbols? Is there some geometric interpretation? Is there any method to "draw" metric tensor in 3d?

[marsh forge]

<@&646917276561309707>

@honest narwhal

[winged viper]

wouldn't that go from unbounded -> bounded? that doesn't show that its possible to go from bounded -> unbounded?

[marsh forge]

uh

homeomorphisms

go both ways

[gritty widget]

@marsh forge you summoned me

i am a christoffel symbol entusiast

[marsh forge]

why tf do you have that role

[gritty widget]

i dunno

[honest narwhal]

Yeah you just asked for it for no reason lol

[gritty widget]

[marsh forge]

btw dami

I did @ you for a reason

[winged viper]

@marsh forge sorry for the ping but do you have a counterexample for closedness being a topological property?

[marsh forge]

There are lots of examples on stack overflow

but here's a cheeky one

Let X be a space and let U be an open and not closed subset of X

U is a topological space in its own right with the subspace topology

under this topology U is obviously closed

So the inclusion of U into X is a continuous map sending a closed set to an unclosed one

[winged viper]

ah ok ty

[marsh forge]

it should be preserved by htpies

in which case barely anything is topological

we still got connected components babyyyy

[marsh forge]

Wait dami

Why does your natural numbers example work

Oh duh

Wait no duh

[honest narwhal]

N is its own closure in R

[marsh forge]

Yes

[honest narwhal]

And it's not a subgroup

[marsh forge]

Oh fuck me

[honest narwhal]

Lmfoa

[marsh forge]

I forgot N wasnt a group lmao

Long day

Anyway ok

Im w you now

[honest narwhal]

So A is closed under multiplication

[marsh forge]

Your proof for that

Is handwavy

But fine

So taking inverses is continuous

[honest narwhal]

No it can be made rigorous

A is closed

[marsh forge]

No I know

[honest narwhal]

So let's look at A closure

GxG->G

Preimage of A contains A x A

So it contains the closure

[marsh forge]

Ok so let a be in A closure

Inverses is tricky

I would need more than my phone

[honest narwhal]

We need to use compactness somehow

[nimble jolt]

Let h be the inverse of g

[honest narwhal]

Compact Hausdorff (all top groups here are Hausdorff) is normal

[nimble jolt]

Consider the sequence g^-n h

Extract a convergent subsequence

Ah wait, I might be setting this up slightly wrong.

No I think this works actually. You can use this sequence to cook up a sequence of g^n's that converges to g^-1.

[honest narwhal]

Oh you mean like

[nimble jolt]

I have written it in an ugly way though.

[honest narwhal]

g^n -> x

Or subsequence

Wait hmm

[nimble jolt]

$g^{n_k}\rightarrow L$ so $g^{n_k-n_{k-1}-1}\rightarrow g^{-1}$

[gentle osprey]

gomez:

[nimble jolt]

something like that

[honest narwhal]

Okay I'll use a fact from earlier in the chapter

[marsh forge]

Which one

[honest narwhal]

Oh okay I see your thing in the second countable case

If G is a topological group and U is a neighborhood of g, then there's a symmetric neighborhood V of the identity such that VgV^{-1} is in U

This feels applicable maybe?

[nimble jolt]

Perhaps I am getting confused, but where is second countability being used?

[honest narwhal]

To take a subsequence

If you're second countable + compact + Hausdorff you're metric

But for a general compact Hausdorff space you might need some net fuckery

[nimble jolt]

Oh I see what you mean, yes.

[honest narwhal]

I'll just say we proved this for Lie groups and move on

Might come back later

Repost so no scroll

So symmetric neighborhoods of the identity form a basis

So let's say U is symmetric. Should be something about intersecting conjugates

Ahhhh

I think it's like

Finitely many gU cover G

[honest narwhal]

Okay so

For that problem 3

The idea in general is this

[gentle osprey]

Daminark:

Daminark:

[honest narwhal]

Now the idea should be to just repeat this. g^{n-2} \in g^{-2}V and in \overline{A}

So we're done

Okay so now

[gentle osprey]

Daminark:

gustavn64:

gustavn64:

[honest narwhal]

Guess I'll continue

6 idk, might need Hausdorff, 5 was fine just used that the quotient map is open so if you have a disconnect you actually partition the cosets since gH is connected so lives in one of the two sets in the split

7 is like, part of a more general thing

SO(n) acts transitively on S^{n-1}

Why? You can take a point on the sphere, extend that to an orthonormal basis of R^n

And negate a vector if necessary so the matrix has det = 1

Well, stabilizer of the north pole? Well that's just the copy of SO(n-1)

So by topological orbit-stabilizer, SO(n)/SO(n-1) is homeomorphic to S^{n-1}

Hermitian Gram Schmidt is a thing too

So SU(n)/SU(n-1) ≈ S^{2n-1}

Prob same for quaternions but that's for nerds

That also handles problem 8 thanks to problem 5

Also it's clopen in O(n)

SU(n) is obv connected

Now we wanna say U(n)/SU(n)

Oh lol

U(n) has |det|=1

So act on S^1 by multiply by determinant

Boom

Center of a topological group is closed, I'm gonna be a fuck

G acts on itself by conjugation

Center is the kernel of this action

So this follows from problem 6

🙃

@marsh forge

Anyway so O(n+1) acts on S^n, can we make it an RP^n action?

Okay act by homogeneous coordinates I guess

?

Yeah

Lol

So write RP^n = R^{n+1}-0/~

A[x] = [Ax]

What's Kochen?

Anyway yeah this makes it clear, transitivity is lol and stabilizer of [0:...:0:1] is O(n)xO(1)

Same for CP^n

GL(n,H) is open in M_n(H)

Ehhhhh quaternions are for dorks

[honest narwhal]

Also probably von Neumann series

Anyway problem 4 is false for upper diagonal SL_2

Topological group is regular. Well gimme U,V symmetric neighborhoods of 1 such that V^2 \subset U

Is \overline{V} \subset U?

Something like that yeah

Eh von Neumann is the Neumann that matters tbh

I'll allow some Neumann washing

[next eagle]

Hello. Im a bit confused. I tried to found, but there is no strict geometric interpretation of metric tensor. I know how to calculate this one, both from dot product and linear element. So I got one question - is metric tensor just some kind of "shortcut"? Could it be theoretically written in other way instead of matrix? Is it just a convention?

[fleet trench]

there are a couple different ways to think about the metric tensor. im gonna assume you mean riemannian; some of the requirements can be loosened though to get other types of manifolds

• one is as a choice of inner product at each point (so a function that takes in two vectors and gives you a real number satisfying symmetry, bilinearity, and positive definiteness) that does so in a smooth way. its properties mean that it is a tensor, of rank (0, 2), and thus it is a geometric entity
• if you pick a coordinate chart, which gives you a basis for the tangent space at each point, you can then represent the metric by a matrix, since it's just a quadratic form. this is maybe a more physics-y way to think about it, since physicists do all their geometry in coordinates. you lose a bit of the actual idea of the metric as an inner product this way though imo

[pine heath]

English?

[next eagle]

@fleet trench quadratic form is something new in my exploration. I'm gonna study this thing.

[honest narwhal]

So actually I used to think that physicists did geometry in coordinates until I met one who really gets upset at this

He complained once that Milnor's book and lectures on differential topology has caused too many mathematicians to get addicted to embedding manifolds in R^n and/or working in charts instead of tensors

[marsh forge]

Imagine working in charts instead of sheaves

I refuse to think about what ive done

Forgive me for I have sinned

[wicked thorn]

a 1-dimensional measurement is called distance
a 2-dimensional measurement is called area
a 3-dimensional measurement is called volume
a 4-dimensional measurement is called what?
and is there a general name for n dimensions?

[sweet wing]

volume

[wicked thorn]

but how do you specify which dimension it applies to?

prepend a number? 4-volume? 6-volume?

do I have to spell it out as "4-dimensional volume"?

[marsh forge]

If youre working in a n-dimensional context

Then the term volume refers to n-volume

Like generally its obvious from context

[wicked thorn]

ohh okie

follow up question: what about applying lower dimension measurements to higher dimensions?
for example, measuring area on a three dimensional object becomes "surface area". Is there a generalization for that?

would saying something like "the 5-volume of our 8-sphere" make sense for that concept?

[gritty widget]

I've always just seen n-volume.

[wicked thorn]

okay

and last question: what's a decent introductory textbook for this?

[sweet wing]

ive always seen volume and surface area for nD space

[marsh forge]

This is only half true

Volume generalizes also to the theory of differential forms

Where n-volume refers to integrating over ‘volume forms’

[frigid patrol]

@bitter kestrel hi 🅱️

[marsh forge]

There might be some kind of representation theorem of this sort idk

[honest narwhal]

Okay so

You have k-forms on n-manifolds

When k isn't n I don't have much of a connection (heh) between the two concepts in my mind

You can't even integrate in that case

But

Let's say you have an orientable n-manifold, then there's a volume form \omega

I can now make sense of the integral of a continuous function f on the manifold: I just take \int_M f\omega in the sense of differential forms

Now I can do Riesz rep

[gritty widget]

@bitter kestrel differential forms do correspond to measures (in the background you use the lebesgue measure on R^n and then take a partition of unity)

[honest narwhal]

Okay so

We have that V and I become inverse bijections between radical ideals in k[x_1,...,x_n] and algebraic subsets of A^n

Well, irreducible varieties end up corresponding to prime ideals

And points to maximal ideals

So if X is an algebraic set, well we can consider the coordinate ring k[X] = k[x_1,...,x_n]/I(X)

And then the Nullstellensatz applies also to the coordinate ring. So subvarieties are radical ideals, irreducible subvarieties are prime ideals, and points are prime ideals

But this means X is naturally in bijection with mSpec(k[X])

And in fact when you put the Zariski topology on X, well closed sets are subvarieties, this gives us a topology on mSpec(k[X])

Namely you take the set of maximal ideals containing a given ideal I

(The set of points contained in V(I)\subset X)

Now it turns out that for various reasons it ends up being cleaner to work with prime ideals than maximal ideals, I don't know of a satisfactory/a priori explanation for why you'd wanna do this aside from "you guess randomly that prime ideals work nicer and they do"

But okay now given an arbitrary ring we have the set of prime ideals Spec(R)

And we put the analogous topology, namely you let V(I) = {prime ideals containing I}, and this should restrict to the earlier topology on mSpec

[gritty widget]

Prime ideals work better for technical reasons

Consider spec(DVR)

Mspec would look like spec k

Now the valuative criterion for properness relies on spec(DVR)

Also it cleans up definitions

Consider an elliptic surface over P^1

We want to define a mordell-Weil group of this

Either you have a definition that relies on sections of the elliptic fibration, or you can just define it as MW(fiber above generic point)

[honest narwhal]

So this is perhaps somewhat above my head, my AG at the moment is still kindergarten level

I guess I was aware that prime ideals work nicer insofar as like, e.g. you don't even get an induced map on mSpec out of a hom while you do get one on Spec

That comment I guess was more, I don't know if I have some sweeping "Ah this just ends up being the completion with respect to this obvious construction" answer that makes you feel warm inside

[lime dune]

In [-infinity,+infinity] ,do you have compact iff sequentially compact?

I know very little about two-point compactifications

[honest narwhal]

But yeah so I guess continuing, the business about being a ringed space is basically this. So V(I) = {prime ideals containing I}. Well, those are the closed sets. Complement of that is, {prime ideals not containing I}. You can write that set as \bigcup_{f\in I} {prime ideals not containing f}

So if you define D(f) = {prime ideals not containing f}, that's a basis for the topology

If you revert back to the varieties picture, it's basically saying where is a given polynomial f non-zero?

The nice thing about that is that this is precisely the open set where it makes sense to talk about the function 1/f, so these basis open sets are ones where we can speak about more than just polynomial functions, but rational functions

[gritty widget]

Well at that open set you can localize at f

[honest narwhal]

Yeah that's what I'm getting to

But yeah so once you define what a regular function is on a general open set, it turns out these basis open sets are isomorphic to affine varieties themselves, and the coordinate ring is k[X]_f

[marsh forge]

Question

So when we want to talk about the local ring at a point

We can either take the local ring and invert at all the nonvanishing polynomials

Or we can do the generic stalks thing

Do these match up

It seems nontrivial that they do

[honest narwhal]

So the definition of regular gives a sheaf of rings on the variety, and also an analogous sheaf of Spec, namely Spec(D(f)) = A_f. So this leads to the whole scheme thing. You want a space with a sheaf of rings, and locally you want this sheaf of rings to just be Spec(R) for some R

I'm not familiar with that business as much, sorry. Maybe abstract nonsense can answer

[gritty widget]

there are two ways of defining $\Gamma(D(f))$: either by hartshorne's "espace etale" method or by simply declaring that it be $A_f$ and check the sheaf glueing axioms (this is how vakil does it)

[gentle osprey]

hochs:

[gritty widget]

if you're doing it the hartshorne way, then the statement "generic stalk thing equals local ring at inverting all nonvanishing polynomials" is consequence of the fact that an element $a \in A$ is $0$ iff it's $0$ in all $A_p$ for $p$ in Spec $A$.

[gentle osprey]

hochs:

[honest narwhal]

Err yeah I meant Gamma rather than Spec of D(f)

[gritty widget]

If you're doing it the vakil way, it's really the statement that $\lim_{f \notin \mathfrak{p}} A[1/f] \cong A_{\mathfrak{p}}$.

[gentle osprey]

hochs:

[real notch]

Gamma being your ring sheaf?

[gritty widget]

this is direct limit

yes that's the section (of the structure sheaf)

[honest narwhal]

Ah hah

That makes sense

[gritty widget]

There are other ways (there's such a thing called "plus construction" that some algebraic geometers like to use to introduce sheaves)

(you'd need to do the "plus" construction twice to arrive at a sheaf from a presheaf assigning on the basis set though)

the way that hartshorne constructs sheaf is pretty concrete. he defines a section $s$ over an open set $U \subset \textnormal{Spec }A$ to be a function $s: U \to \coprod_{p \in \textnormal{Spec }A} A_p$ satisfying a certain "locally constant" type condition. so the stalk at $p \in \textnormal{Spec }A$ say represented by $\langle s , U \rangle$ with $p \in U$ can be mapped to $s(p) \in A_p$ and you can check that this is well-defined and is an isomorphism (the little commutative algebra fact that I mentioned above is the only nontrivial part of checking this isomorphism)

[gentle osprey]

hochs:

[gritty widget]

(which itself isn't hard. One looks at the ideal $\textnormal{ann}_A , a$, the annihilator of $a$ and if that doesn't contain $1$ then you take any prime ideal containing it and localize there)

[gentle osprey]

hochs:

[honest narwhal]

Interesting

[robust topaz]

hey guys, can someone explain this example? i can't see how the quotient spaces are different

[wanton marsh]

I am not seeing it either

[dim meadow]

I'm not sure what the equivalance classes are

[real notch]

It doesn't seem like they actually state it

[dim meadow]

Do they mean submanifolds with boundary?

[real notch]

i think the equivalence classes in question might be something like the two diagrams below?

[dim meadow]

I guess you're looking at a foliation of the open strip

Yeah it's not so clear what's happening here

[robust topaz]

i think the equivalence classes are represented by the lines, so for the first example it would be something like
$\newline x=t , t\in(-\infty, -\frac{\pi}{2})\cup(\frac{\pi}{2}, \infty);
\newline y(x, \alpha)=\tan(x)+\alpha , \alpha\in(-\infty, \infty), x\in(-\frac{\pi}{2}, \frac{\pi}{2})
$

[gentle osprey]

p-zombie:

[gritty widget]

$\dfrac{1}{2\pi} \int_0^L \dfrac{x(t)y'(t) - y(t)x'(t)}{x(t)^2+y(t)^2} dt$ if your curve is parameterized by [0,L]

[gentle osprey]

trex:

[gritty widget]

@buoyant comet you could draw the curve and then use one of the known algorithms to compute the winding number in each region

[gritty widget]

I was trying to show that the Affine Group is a Lie Group. I wanted to so showing that i can embeed Aff(R,n) into GL(R, n+1) and show that it would be a closed subspace, the point is that i can't show that is closed, any tips?

for reference this is the theorem that i'm trying to use https://en.wikipedia.org/wiki/Closed-subgroup_theorem

[gritty widget]

isn't this a bit overkill ?

I mean multiplication and inversion on matrices is rational component-wise, hence smooth; you don't need the closed subgroup theorem (which is pretty elaborate)

either way, the reason why the affine group is closed is that you can write it as the set of matrices of GL(n+1, R) with the first n coefficients of the last line equal to zero and n+1-st coefficient equal to 1, all of which are closed conditions

[nimble jolt]

Hadn't actually seen the affine group written in that way before haha.

[gritty widget]

It's how you visualize quadric as hyperplane in a space of higher dimension, kind of a cool trick

Their as smooth but I should find the open diffeomorphic to any matrix

*to an open nbhood of any Matrix

[ivory dragon]

i agree

[gritty widget]

I think the theorem is kinda of the easiest thing you can use

I considered showing that it is the preimage of some regular value (as you do with SL(n) or O(n) ) but could not find any cool function that does so

[gritty widget]

Well you can also say that since Aff(n,R) is the intersection of GL(n+1, R) with the affine subspace {matrices with last line (0, ..., 0, 1)}, it is open in that affine subspace, which is itself a submanifold of Mat(n+1, R), hence Aff(n,R) is a submanifold of Mat(n+1,R)

and then use the argument that matrix product and inverse are rational in the components of the matrices involved to claim that they are smooth maps

[tranquil vector]

topologically, is there a difference between a normal torus and a torus generated from a ring instead of a circle?

imagine a donut but instead of being filled with bread, it's empty on the inside, but on the outside it still looks like a donut

I assume there is, but just to be sure

[marsh forge]

Wait

Sorry

Are you saying the torus is ‘thick’

Like you take an annulus

And rotate it?

If so they are homotopic but not homeomorphic

So there is a topological difference

@tranquil vector

[tranquil vector]

yes

sorry was googling what "annulus" was lol

thanks!

[marsh forge]

(One of them is a manifold while the other isnt)

[tranquil vector]

oooh

[warm hedge]

anyone can help me how to show that if C={(-∞,a):a∈ℝ}U{(a,+∞):a∈ℝ} then every open cover of [0,1] from elements of C has an finite subcover?

[gloomy plover]

Are you familiar with the (or any) standard cover compactness proof of the unit interval?

[warm hedge]

is it a similar proof ?

[gloomy plover]

I looked one up just now and imo it's very easily modified

https://math.stackexchange.com/questions/2089317/compactness-of-the-open-and-closed-unit-intervals

[warm hedge]

oh ok thanks

[gloomy plover]

np

[warm hedge]

for the question above ... if i said that for an open cover {(-∞,ai)}i∈I U {(ai,+∞)}i∈I
if sup(ai) > 1 or inf(ai) < 0 then [0,1]c(-∞,supai) a finite subcover or [0,1]c(infai,∞) a finite subcover
else [0,1]c(-∞,supai)U(infai,∞) a finite subcover
is right ?

[gritty widget]

@warm hedge any two distinct elements of C cover all of R

[warm hedge]

not any two

[gloomy plover]

any two distinct elements

Also if $\sup(a_i)>1$ or $\inf(a_i)<0$ isn't the case, then you cannot generally say that $[0,1]\subset(-\infty,\sup(a_i))\cup(\inf(a_i),\infty)$

[gentle osprey]

Rijinaru:

[gloomy plover]

This is because it might still happen that $\sup(a_i)>0$ and $\inf(a_i)<1$, in which case $[0,1]$ isn't completely covered

[gentle osprey]

Rijinaru:

[gloomy plover]

If the supremum and the infimum of your index set $I$ are even finite

[gentle osprey]

Rijinaru:

[gloomy plover]

Which I highly doubt, because you are supposed to take any open cover of elements in $C$

[gentle osprey]

Rijinaru:

[gritty widget]

From $\mathbb R ^2$ plane with standard metric countably many circles have been removed. Does this set have to be uncountable?

[gentle osprey]

Godel:

[midnight jewel]

hint: ||consider a line in ℝ².||
second hint: ||if you can show that line still has uncountably many points after removing the circles from the plane, then you’re done||

[gritty widget]

yeah that's what I was thinking after a while

i'll try to prove it although seems obvious

ok cant cover the line because one circle can cover at most 2 points from a line

is that a fine argument?

although Im not sure how that implies can't cover R^2

cuz like you could do it with this kinda thing

[marsh forge]

Thats a fine arg godel

Any placement of circles removes a finite number from a line

Hence countable many remove countable many

[nimble jolt]

Topology exercise: Show that $T^n$ does not embed in $S^1 \times S^{n-1}$ for $n>2$.

[gentle osprey]

gomez:

[tacit stratus]

I have a question related to integrability of distributions.
When can you “integrate” distributions with submanifolds of higher dimension?
i.e given a distribution D, can one find a submanifold N of greater dimension where D ⊂ TN.
Or, For the sake of explicitness, given a smooth vector field on R3, can one always find a dimension 2 submanifold that is everywhere tangent with that vector field?
Intuitively, the extra degree of freedom from the normal integrability problem makes me suspect you can, but I’m not sure.

[gritty widget]

Fun, first off if $T^n$ embeds into $S^1 \times S^{n-1}$, then the embedding is a diffeomorphism (indeed, the image is closed because $T^n$ is compact, and open because of the inverse function theorem). Now, it only remains to prove that they cannot be homeomorphic. But, if $n > 2$, then by the Künneth formula $H_1(T^n, \mathbb R) = \mathbb R^n$ and $H_1(S^1 \times S^{n-1}, \mathbb R) = \mathbb R$

[gentle osprey]

trex:

[gritty widget]

Now I'm not sure how to adapt this if the embedding is just a topological one (can we still conclude that the embedding is a homeomorphism ?)

[dire grail]

topological maps are homotopic to smooth maps

on manifolds

so just reduce using that

[gritty widget]

Ah, very nice I didn't know that!

[nimble jolt]

Nice. Yeah I just meant topologically. A topological embedding is a homeomorphism onto its image by definition so you can argue just as you did, replacing the role of the inverse function theorem with invariance of domain :).

[gritty widget]

oooh right, right

my topology is rusty lol

[west spindle]

is the following enough to characterize the real line:
"a connected second-countable hausdorff space which upon the removal of any single point disconnects into two components"

[dire warren]

Yes @west spindle

I asked that on math overflow

[west spindle]

oh really

can this be weakened

[dire warren]

In fact you don’t need the second countable Hausdorff

Just connected. I beleive

Let me find tge thread

O okay you need the Hausdorff second countable

In fact you need more

https://mathoverflow.net/questions/319870/homeomorphic-characterization-of-the-real-line

Connected separable metric spac3 is the assumption

[warm hedge]

what it means when we are saying second countable ?

[nimble jolt]

It means the topology has a countable basis. Eg balls with rational coordinates and rational radius form a countable basis for the standard topology on R^n.

First countable by comparison means that each point has a countable local base, which is weaker.

[warm hedge]

oh i see

thank you

[spice egret]

Anyone know how to solve this?

[real notch]

[marsh forge]

Youll@have better luck in #geometry-and-trigonometry

@honest narwhal weekly reminder to change the name of this channel somehow

And to give me honorable

[honest narwhal]

There's no good answer, the problem is once people see "geometry", which kinda has to be in the name, they jump to it. And this isn't exactly frequent enough anymore to warrant such a change (since people see "topology" first, in contrast to when this was "advanced geometry")

[rugged swan]

hey, I'm currently studying homology and I can't figure out why, if I take the two canonical generators of the fundamental group of the torus c1 and c2, the chain c1+c2 isn't a border

because I have an homeomorphism between the torus without this chain and the square [0,1]^2

and the square is homeomorphic to the 2-simplex

then the torus without c1+c2 is a singular 2-simplex

and the border of this singular 2-simplex seems to be c1+c2 visually

[marsh forge]

Trace ur pen

Around ur drawing

While trying to follow the arrows

@rugged swan

It’s also obvious if you write it down explicitly from the fundamental square of the torus

What the boundary of that square is

You can’t forget abt orientation

@ me if you still want me to be more in depth but I think its important to figure it out urself

[rugged swan]

if I write the suqare like this :

is the border 2(b - c + a) ?

[marsh forge]

No

Literally what’s the border of the square

[rugged swan]

yeah I've just understand what did you mean, the border of the square is an oriented circle, but here it's not a circle since I can't follow one direction with a pen

thanks

[marsh forge]

Well

I was hoping you’d just write it down

And see that the border is like

a-b-a+b

=0

=/= a+ b

[rugged swan]

that's weird

I mean, that seems the right reasoning, but I can't see rigorously what's the structure of 2-chain of the fundamental square

[marsh forge]

Boundaries are oriented

The boundary of the square you are talking about would never be a+b anyway

If we didnt care about sign (like over Z/2) we’d have that the boundary is a + a + b + b

Ie 2a+2b

But we are working Z/2 so thats 0 anyway

[rugged swan]

Hm in fact it is logical since it's the border of the torus x). I was wrong on my previous drawing, to have a 2-chain structure I must orient c in the other sense. And the I would get border = a+b+c-a-b-c

[midnight jewel]

I’m being a bit dumb here. this should be elementary pointset topology but I’m stuck (I need this to finish a proof of the meier-vietoris sequence):

Assume $A, B \subseteq X$ such that $X = \mathrm{int}(A) \cup \mathrm{int}(B)$. Show that $\mathrm{cl}(B \setminus A) \subseteq \mathrm{int}(B)$

[gentle osprey]

Sascha Baer:

[midnight jewel]

my attempted proof was $$\mathrm{int}(B) \supseteq X \setminus \mathrm{int}(A) \supseteq \mathrm{cl}(X) \setminus \mathrm{cl}(A) = \mathrm{cl}(X \setminus A) \supseteq \mathrm{cl}(B \setminus A),$$ but it appears that that equality is not true in general

[gentle osprey]

Sascha Baer:

[midnight jewel]

so idk how to fix this

[marsh forge]

Should be like

Any limit point of B/A that isnt in int B must be in Int A, but removing A means you remove some neighborhood of the point

So its not a limit point of B/A after all

@midnight jewel

[midnight jewel]

ah that sounds sensible yea

thanks

[marsh forge]

Wait

Sorry

You also need that B/A itself is a subset

But if x is in B/A but not int B then its also in int A

Then its the same arg

It was never in B/A to begin with

Yeah that works

[midnight jewel]

yea I see thanks

[livid narwhal]

my apologies if this doesnt come under topology and geometry

I have a system defined by periodic boundary conditions, the unit cell of a crystal.

I have a metric tensor defining the basis of the "cell" and thus the shape of the periodic boundary conditions

I've also converted all of the points inside the cell to complex numbers from (a/A,b/B,c/C) or fractional co-ordinates (with respect to the basis) into (e^2ipia/A,e^2ipib/B,e^2ipic/C)

So at this point I have a metric tensor describing the periodic boundary conditions and the vectors that define the 'bounding box' and points within it that have a complex phase rather than a real number co-ordinate

My hope is that I can combine these in some way to create some form of canonical co-ordinate system where information about the periodic boundary conditions as well as the position within the cell is present in each point, does this sound like a reasonable way of going about it?

[little hemlock]

In the standard topology on R, would a disjoint union of open intervals like (1,2)U(3,4) be open? I can't think of a reason why not, but it seems unnatural for intervals with gaping holes to be open.

[honest narwhal]

You're thinking of connectedness

Which is separate from openness

You can be open and disconnected, like (1,2) \cup (3,4)

[little hemlock]

ah, okay.

[marsh forge]

(If it helps, you can remember that open iff completment of closed

And its not hard to believe [-infty,2] cup [2,3] cup [4,infty] contains all of its limit points

And not hard to see how this point of view allows open sets to be kinda all over the place

[floral gust]

@little hemlock Try using the definition. Can for every point you can find an open neighbourhood contained inside the set?

[snow gull]

hey can someone explain me the difference between afine conics and projective conics?

one of the is the afine and the other is the projective, but I dont understand which is which and why

[thick ferry]

How do I prove that for two connected sets A, B at least one of AnB and AUB is connected?

[sweet wing]

to get started could try to show if the intersection is nonempty, then the union is connected

[thick ferry]

ah thanks

that could be the start

[gritty widget]

hey folks

im really having a hard time trying to understand external algebras

i have already learned about dual bases

and now im trying to illustrate what the pairwise wedging of the dual base would look like

[plain panther]

So I have this ultra truncated isocohedron.

This was created by using code from https://github.com/mikolalysenko/conway-hart.

And then constructed by creating each face from a hexagon mesh for each set of 6 points.

However, if you look closely.

There's a lot of distortion around the pentagons.

So my question is does the dual and truncate operations keep regularity?

Do they deform it?

Or is this a numerical issue.

[marsh forge]

I've done it boys

Ok so by a theorem I don't fully understand at the moment

if we have a cohomology class $c$ in $H^{2k+1}(M)$ for $M$ a $4k+2$-manifold, then there is a map $f:M\to \Omega S^{2k+2}$ such that if $e_1,e_2$ are the generators of middle and top homology of the loops space, respectively

[gentle osprey]

MaxJ:

[marsh forge]

$f^*(e_1)=c$

[gentle osprey]

MaxJ:

[marsh forge]

Then we define a quadratic form by putting $q(c):=f^*(e_2)$ which lives in $H^{4k+2}(M)\cong \mathbb{Z}/2$

[gentle osprey]

MaxJ:

[marsh forge]

Then again, for reasons I do not totally understand yet it turns out that $q(c_1+c_2)=q(c_1)+q(c_2)+c_1c_2$ where that last product is the cup product

[gentle osprey]

MaxJ:

[marsh forge]

This means we have a quadratic form!

We can take a symplectic basis to compute it's Arf Invariant

And this is the kervaire invariant!

Theres also a way to do this with steenrod quares

but thats a lot of machinery

We actually know even more

bc we know that the associated bilinear to this quadratic

is in fact the Poincare Pairing

Once I understand those two things I mentioned

Im gonna be ready for my blog post

Ok ok

so the first thing

This actually clears up a lot of what I was missing ebcause the KI is defined by highly connected manifolds

and its like

whys that?

Ok heres why

(*) We want a cohomology class c to be a map M->LoopsS

The obstructions to this

are maps from pi_k(M) to pi_k(LoopsS)

Ok so let M be 4k+2 manifold again

then the first nontrivial htpy group of M has to be $2k+1$ in order to define the kervaire invariant

[gentle osprey]

MaxJ:

[marsh forge]

The reason for this is that obstructions to (*)

are maps from htpy groups of M to htpy groups of LoopsS

(Where S is S^{2k+1})

Well, the first obstruction possible is at pi_2k+1(M)->pi_2k+1(LoopsS)=Z

But this can be shown to not be a problem

Wait okay no this is sketchy

okay well I get the idea

and I won't go into htpy stuff in this post

And for the second thing

it turns out to just be a computation

So @honest narwhal I've figured out the kervaire invariant

give me a medal of honor

[honest narwhal]

@marsh forge okay

[rugged swan]

hi, why does the space of couple (L,x) where L is a linear line of R^2 and x is in L can be identified with the quotient space R^2/~ where (a,b)~(a',b') iff it exists k in Z s.t. (a',b') = (a+k,(-1)^kb) ?

[eternal anvil]

anyone know if Rn diff geo classes usually require an R3 class as a prerequisite

like, would taking a class on curves and surfaces benefit me much later on in an Rn class

[honest narwhal]

Probably helps for the sake of having examples which you can very directly visualize, subsets of R^3

But it's not necessarily a strict prerequisite

[warm hedge]

if (X,T) compact and T2 topological space and S an other topology in X and TcS and T≠S
i know that the (X,S) isnt always compact , but is there a chance that it could be compact ?
there could be an S bigger than T that would make the space compact ?

[warm hedge]

oh nice

thanks

[tidal cedar]

Quick question from a noob: Is this topologically equivalent to a Klein bottle?

Ignoring the fact they accidentally punctured all the way through on the left

Or wait no I'm dumb, that needs to be punctured through.

[midnight jewel]

it’s not, this is orientable (it has two sides)

@tidal cedar

I reckon it’s equivalent to a torus

here’s a prettier picture

[tidal cedar]

Ah, so that specifically is a torus model?

Huh

[midnight jewel]

yea just looked it up it’s homeomorphic to a torus

[tidal cedar]

What if the bottom section was closed up and it was extended into another spatial dimension

[midnight jewel]

wdym? how you embed it isn’t relevant

[midnight jewel]

this is only a torus if you pretend it doesn’t intersect itself

the same way you pretend a klein bottle doesn’t

[tidal cedar]

Ah

[midnight jewel]

https://www.youtube.com/watch?v=ixduANVe0gg the beginning of this video shows a deformation of a torus that goes through this double klein bottle

unrelatedly, had my algebraic topology exam today. unsure how it went

we had four exercises and had to solve three

more than enough time to work on all four though

I got one down quickly

and then on the other three got about 2/3s of the way through each before getting stuck

managed another one after a lot of thinking

at which point I was left with two exercises. one I couldn’t solve and one I probably did all wrong but I at least tried

(the one I couldn’t do was show that the line with two zeroes is not homotopy equivalent to S¹; the one I tried was to show that for an open cover U, the singular homology restricted to chains with each simplex contained in one of the covering sets is isomorphic to the regular one - and I was allowed to use anything we proved about barycentric subdivision in class

[tidal cedar]

Ah

[gritty widget]

Any book/notes recommendation to learn point set? Especially complete metric spaces and intro homotopies

[midnight jewel]

the usual recommendation for point set is either munkres’ topology or hatcher’s lecture notes, but the latter doesn’t cover either of those two things in detail. I’m not sure how much in detail munkres goes with metric spaces, the study of complete metric spaces is basically analysis

[gritty widget]

Yeah I have munkres but I want something else to read as well

[tepid totem]

my class used Runde's A Taste of Topology, i thought it was pretty nice

[honest narwhal]

Both of those are topics you'll learn from elsewhere. Like the main features of complete metric spaces are gonna be what

Baire Category is probably the most important

There's the business about compact = complete + totally bounded for metric spaces

[gritty widget]

Yep, mostly need to learn homotopies tho

Cause Ive been having trouble to understand those

[marsh forge]

homotopies are

pages 1-10

of any intro AT text

but you won't see them earlier

(in general)

Until you want to start doing serious topology

you don't need homotopy

And to be honest you should not learn homotopy and homeomorphisms at the same time probably

your intuition for both will suffer

[gritty widget]

Lol

If I dont Ill have to retake the course mate

And not to offend but I dont know if I should listen to you about the order of learning something

[marsh forge]

You should on this, but if its for a course you don't have a choice

Anyway, pages likes 1-10 of hatcher or something

@midnight jewel Did you figure out the S^1 thing

Now Im nerd sniped by it

it has fundamental group Z, so H1 is also Z

H2 should vanish

[midnight jewel]

yea, the homologies of the line with two zeroes are ℤ for 0,1 and 0 otherwise

I didn’t try anymore but I’m told the thing you have to do is first show that the two zeroes have to map to the same point in S¹

and then you can get a contradiction by that

[marsh forge]

Oh interesting

So you do explicit stuff w the homotopies

thats a cute problem I like it Ill think more

[midnight jewel]

the whole problem was:
Let X be the line with two zeroes

1. compute the singular homologies of X
2. is X homotopy equivalent to ℝⁿ for any n≥1?
3. is X homotopy equivalent to Sⁿ for any n≥0?

you can do 2 and all of 3 except S¹ with the solution to 1 right away, but S¹ has the same homology groups as X so you have to try sth else

and I couldn’t think of it

for 1 I just used Meyer-Vietoris with the sets being “all but one of the zeroes” (so both sets are homeo to ℝ)

from that you get it pretty quickly

(not immediate but not hard either)

[marsh forge]

ok so

I think you can compute the universal cover (after reading some MO)

and I think that the universal cover is contractible

in particular has vanishing higher htpy groups

[honest narwhal]

@marsh forge well not page 1-10 of Concise

[marsh forge]

fuck off lmao

anyway s^1 has horrible htpy groups

and we are done

[honest narwhal]

Page 10 of Concise you're on cofibrations

[marsh forge]

page 10 of concise you gotta buy More Concise

[honest narwhal]

S^n you mean?

S^1 is p easy lol

[marsh forge]

Fuck I was being dumb

ok my proof strat doesnt work

I forgot S^1 wasn't covered by S^1 lmao

[honest narwhal]

Well it is but that's not universal 😛

[marsh forge]

yes

[honest narwhal]

So if X has pi_1 = Z then like, X is weak homotopy equivalent to S^1

So you basically need to show that X doesn't have the homotopy type of a CW complex at all

[warm hedge]

T3+Lindelöf=T4 or it needs to be also T2 ?

or T3 ->T2

oh ok i see

thanks

yeah thats true

yeah

oh wait

i think not always

yeah

[warm hedge]

i have a (X,T) T3 second countable topological space and i want to poove that every open subset of X is countable union of close subsets of X .... any help ?

[midnight jewel]

@marsh forge we didn’t cover higher homotopy groups and covering theory should not be examinable for this exam

the only covering stuff we did was in the previous topology course (which is not exam-relevant) and there we only did as much as we needed to prove that the fundamental group of S¹ is ℤ

I don’t think the word “universal cover” was ever mentioned outside the non-examinable last lecture of last year where he showed some additional neat stuff

so it should be solvable without it

[marsh forge]

Well

My proof strat didnt work anwyay lmao

@honest narwhal that is correct but it is not clear to me how to prove that other than the vague idea sasha gave out

That you literally cant construct a reasonable homotopy

My guess is like, you take some path in the double origin line from one of its 0s to the other

And show that this has to become a loop in S^1

But Idk why

[warm hedge]

of T4 or T3 ?

i think i will use that for x ∈ V open there is a W open that x∈ W c closeureW c V

and maybe in place of W i will put a B open from X's base and V will be = with union of closure of Bx for every x ∈ V

maybe something like that ? @gritty widget

[warm hedge]

👍

[gritty widget]

Yo I don't understand one problem cause seems too ez: For all $n \in \mathbb Z $ let $C_n$ be a closed boundary set in the interval $\left[n,n+1\right]$. Let $D= \bigcup _{n \in \mathbb Z} C_n$. Show there exists $t \in \mathbb R$ such that $t + D \subset \mathbb R \setminus \mathbb Q$

[gentle osprey]

Godel:

[gritty widget]

So I was thinking - can't we take an irrational number from every [n,n+1] as C_n? Then t=0 will work

cause a single point is a boundary closed set right?

[marsh forge]

Well, for one, I assume this is for any choice of C_n

for two I am not familiar w the term 'boundary closed'

[heavy gull]

Right, I assume boundary closed refers to the closed boundaries, [], so that n and n+1 are included, right?

Can't really wrap my mind around the intent of the last statement.
I also have a simple question myself. Does anyone know anything about arcpolygons?

[gritty widget]

boundary closed I meant a boundary set (empty interior) and closed

but yeah, I misunderstood the question, ifc its for any choice C_n hmm

[heavy gull]

Oh... I think I might get it, now that you said the interior is empty it makes much more sense. In that case taking any irrational number really should do it. Say pi, it's irrational, so a subset of the reals, not a subset of the rationals, and so still a subset of the reals without the rationals. Since the same counts for pi +1, or +2 and so on, yeah.

[gritty widget]

yeah but its for any choice of C_n

so if I took 1/2 from C_0 and sqrt(2) form C_1 I still can pick such t (for example sqrt(3))

I guess I would have to somehow characterize what those C_n look like...

(I guess those need to be sets of points that have 'discontinuities' in some sense, can't have intervals, cause then it wouldn't have empty interior)

[heavy gull]

Yeah, any solid interval would break it. t=0 should work, but other integers too from what I can think of

[gritty widget]

no, cause, what if C_0 is a rational number

then t would have to be irrational

[heavy gull]

Well yeah, but I see nothing that would force a mix of rational and irration C_ns

[gritty widget]

I need to show that no matter how I choose what C_n is there still exists such t that works

so showing one example doesnt prove it obviously

[heavy gull]

Wouldn't it be enough then to show there is a t that keeps an irrational C_n irrational, and a t that makes any rational C_n irrational?

[gritty widget]

yep probably, I mean, it would be necesarry to show that for any number of irrationals (denote them x_i, i in I) there exists another irrational y such that x_i + y is irrational for all i in I

and that sounds pretty hard tbh

not sure if induction works because set I of irrationals might be uncountable

probably would?

[heavy gull]

Well... I mean if you pick transcendentals than adding any integer or even other rational will never make them rational. Likewise adding any transcendental to any rational would still be irrational, right?

What I'm unsure of is why there's got to be a specific irrational y that satisfies that requirement. It also doesn't sound satisfiable.

[gritty widget]

well, why doesnt it sound satisfable?

[real notch]

If you take the full set of irrationals

Uh

[gritty widget]

y needs to be irrational cause if it was rational then for some C_n that has only rationals it wouldnt satisfy our conditions

and Im not sure about transcendentals just because I don't know their properties 😛

and not sure if I believe it