point-set-topology

[digital wraith]

That is, given an action of pi_1(X,x_0), you want to construct a covering space such that the natural action on the fibers is that action.

I think I know the construction, I just can't visualize it or understand why it works.

[pastel linden]

dumb question about covering spaces

this is munkres' definition

In the case of p : R to S^1 by e^2 pi i x, wouldn't p^-1(S^1) just be R itself?

[orchid forge]

[pastel linden]

what im confused about is that then the preimage is not a disjoint union of open sets

I guess R itself is a single open set

[orchid forge]

every point has a neighborhood that is evenly covered

[pastel linden]

I am aware, I am trying to do a proof of the following

the approach I was thinking was showing that either S = set of points with preimage having k elements is clopen

[plain raven]

the prototypical $k$-fold covering is $f(z)=z^k$ as a map from $S^1\to S^1$

[gentle osprey]

diligentClerk

[plain raven]

the definition of a covering space does not require that the preimage of every open is of this form.

It requires only that the base admit a sufficiently fine open cover {U_i}, where the preimage of each U_i (or any open subset thereof) is of this form

[pastel linden]

Oh, its that every point has an evenly covered neighborhood

[plain raven]

yeah.

[orchid forge]

[pastel linden]

sorry I am wrangling with the actual technical details for the first time

Hatcher just skips over them initially

[orchid forge]

lol np

[plain raven]

kogasa sorry for saying the same thing you said lol

sometimes saying it twice helps to drive the point home

[pastel linden]

three times the charm

[gritty widget]

every point has a neighbourhood that is evenly covered

make me

[orchid forge]

too far

[shy moss]

in this $\Delta-$complex structure of klein bottle, is it right that $\partial(U)=a+b-c$ and $\partial(L)=c+a-b$?

[gentle osprey]

Or x1

[orchid forge]

assuming U and L are oriented in that way, yes

[loud scarab]

does a second countable space mean that the number of basis elements is countable or the number of elements in each basis element is countable?

[honest terrace]

what do you mean each basis @loud scarab ?

A top space is second countable if it admits a countable basis of open sets

[loud scarab]

I mean, what object exactly am I saying is countable

the number of basis elements in the base of the topology

or the number of elements in each basis element

I got very confused

[honest terrace]

Oh

The first

Let me rephrase

A top space (X, T) is secound countable if it admits a basis (O_i)_{i \in I} of open sets where I is countable

That is, if there are only countably many O_i

The number of elements in O_i don't matter

For example

R is second countable

Because a basis is given by (]p, q[)_{(p, q) € Q²}

Which is a countable set

But ]p, q[ is uncountable whenever p != q

Is that clearer ? @loud scarab

[loud scarab]

Yes that is very clear, but what does this mean?

[honest terrace]

what does what mean ?

]p, q[ ?

The set of reals x such that p < x < q

(p, q) if you prefer

[loud scarab]

what does it mean uncountable whenever p!=q

[honest terrace]

Well, the set ]p, q[ is uncountable whenever p and q are distinct rationals

I'm not getting what exactly is confusing you, could you be more specific ?

[loud scarab]

I understand this very well

[loud scarab]

and obviously this in itself is uncountable

but I dont care about that, right?

if I'm going to consider the second countable property

[honest terrace]

Yes you don't

[loud scarab]

okay very well!

[honest terrace]

I was emphasizing that ]p, q[ is uncountable

but that we don't care about that

[loud scarab]

yeah, thank you

[digital wraith]

A top space is second countable if it's countable but in a weird way so you don't figure out the proof until your second try

[true garden]

Hello. Why is the existence of a apace-filling curve counter-intuitive?

[rancid umbra]

cus how did somebody get a line with no width to fill up space?

[true garden]

The set of parallel segments y=x for 0<x<1 fill the space. Now, connect the segments to each other.

[rancid umbra]

how

and make it a function

[marble socket]

can only connect countably many pieces nicely

[true garden]

connect its end points.

[empty grove]

end points

[rancid umbra]

lol

[empty grove]

Line segment at y_0 connects to which line segment is also a problem

[true garden]

Ok, it might have some technical problems. But, I cannot understand why such existence is counter-intuitive.

[marble socket]

curves are very 1-dimensional thingies

spaces are higher dimensional

how did we make the jump?

[rancid umbra]

like, how are you going to get a continuous function from R to R^2 so that you connect all of your uncountably many line segments together

[empty grove]

If it's not counter intuitive then that's good because it is true

[true garden]

But I cannot understand why it is counter-intuitive for all people other than me.

[empty grove]

Lower life forms

[rancid umbra]

fr

[empty grove]

git gud det and c²

[rancid umbra]

nah but fr, we're trying to tell you that this idea wont work. fixing some technical probs wont fix this

[empty grove]

cope

[true garden]

What is "fr"?

[rancid umbra]

for real

[sturdy torrent]

🇫🇷 🇫🇷 🇫🇷

[rancid umbra]

thought it was gonna be friend💀

[fading vale]

Fomenko and fuchs moment

[uncut surge]

fuchs can't write

fuck fuchs

[fading vale]

Woke and based

[uncut surge]

100% cringe

no being too lazy to explain your math is cringe

15 minutes of thinking
obvious

idk fam

if you waste 15 minutes for every single person in that classroom, you've wasted like 3 hours of collective human lifetime

okay maybe fair

i still feel like there's always something better to write than "obvious"

[wise panther]

I still don't understand how this exact sequence shows theres an isomorphism between $H_2(S^1\times S^1)$ and $H_2(S^1\times S^1, S^1\vee S^1)$

[gentle osprey]

Pan

[wise panther]

I get that $H_2(S^1\vee S^1)$ is 0

[gentle osprey]

Pan

[wise panther]

But then you have
$0\to H_2(S^1\times S^1) \to H_2(S^1\times S^1, S^1\vee S^1) \to \mathbb{Z}\times \mathbb{Z} \to \mathbb{Z}\times\mathbb{Z}$

[gentle osprey]

Pan

[wise panther]

So how do you conclude that $H_2(S^1\times S^1, S^1\vee S^1) \to \mathbb{Z}\times \mathbb{Z}$ is the trivial mapping?

[gentle osprey]

Pan

[wise panther]

nvm

I figured it out

thanks the help guys

[plain raven]

ok lol

i was gonna say maybe shift your focus to proving that $H_1(S^1\lor S^1)\to H_1(S^1\times S^1)$ is injective, which is equivalent

[gentle osprey]

diligentClerk

[plain raven]

did this work

or did you do something else

[wise panther]

That's basically it

[plain raven]

ok nice

[wise panther]

the next term in the sequence is 0

so you know it's surjective

[plain raven]

oh $H_1(S^1\lor S^1)$ is zero?

[gentle osprey]

diligentClerk

[wise panther]

no

that's ZxZ

[plain raven]

ok you said the next term in the sequence was zero so i was confused

i guess idk what "next term" you mean

the next map?

[wise panther]

yeah the term $H_1(S^1\times S^1, S^1\vee S^1)$

[gentle osprey]

Pan

[wise panther]

it's not shown

The sequence is then $0\to H_2(S^1\times S^1) \to H_2(S^1\times S^1, S^1\vee S^1) \to \mathbb{Z}\times \mathbb{Z} \to \mathbb{Z}\times\mathbb{Z}\to 0$

[gentle osprey]

Pan

[wise panther]

and if I'm correct, because the map ZxZ->ZxZ, it maps generators onto generators

which means it's injective right

[plain raven]

Ok. Yeah. I am out of practice with geometrically intuiting what those maps should be. but that sounds about right.

[wise panther]

Although is that correct reasoning? That a surjective map from ZxZ to ZxZ is also injective

It's like a linear transformation

[plain raven]

uhh this should hold in this case but it does require nontrivial amounts of commutative algebra to show this

https://math.stackexchange.com/questions/153516/a-surjective-homomorphism-between-finite-free-modules-of-the-same-rank

[rocky olive]

https://math.stackexchange.com/questions/4285268/what-is-this-branch-of-mathematics

Does topology have this topic?

Packing problems but the containers are malleable

[orchid forge]

I feel like that would simplify things quite a good bit

[rocky olive]

The person's comment, you are saying?

[orchid forge]

no, I mean just allowing the objects in a packing problem to be malleable

as long as they're malleable enough i guess

[tough imp]

This is a bad proof

You can WLOG assume that it’s an endomorphism and then this follows very easily by Nakayama

Well… not easily but it’s a common proof

Any endomorpjism of a finite module M which is surjective is a bijection

[plain raven]

Thanks for checking. I did not read it.

I just linked it to prove to this person it was true

[tough imp]

Proof, let phi be the surjection, then define an A[x] module structure by letting x act as phi. Then (x)M = M by assumption that phi is surjective

[plain raven]

Ah, yes.

[tough imp]

Then by Nakayama there is something 1 mod (x) which kills M, write this as 1 + xf(x)

Assume m in ker phi

Then 0= (1 + xf(x))m = m

So phi is injective

It doesn’t really require much commutative algebra, the given solution says the word projective

Which is way more than you need

[plain raven]

Ok. It seems you are responding to what I said, "nontrivial commutative algebra." My assumption was that many people who are working through hatcher have genuinely never picked up Altman-Kleiman, Atiyah-Macdonald, Eisenbud, etc, so I literally meant nontrivial as in "you should have picked up a commutative algebra book at least once in your life."

[tough imp]

I feel like Nakayama is seen even outside a commutative algebra context

But idk

[plain raven]

Yeah, it's surely an important theorem.

[tough imp]

For me I guess non-trivial means a lot

Which is not what non-trivial means

[plain raven]

It usually does mean a lot, you're right.

This is the common usage.

Really all I meant was "if you're going to look up the proof of Nakayama's lemma and you haven't studied any comm alg at all, it may take you some time to understand it."

[little hemlock]

There's memorable proof of this I learned in A-M which barely uses comm alg. If E -> F is a surjection of free R-modules of same finite rank, then take a maximal ideal m and tensor with R/m, this gives you a surjection E/mE -> F/mF of finite dim vector spaces of the same rank

[plain raven]

That's nice.

Ok, yeah

[fading vale]

Thats the AM proof yeah

[rocky olive]

I don't follow...

:(

[orchid forge]

well the tricky thing about packing problems is the fact that the objects don't stack nicely right?

[rocky olive]

Right

[orchid forge]

if they're perfectly malleable then you can just make them into the most space efficient shape

[rocky olive]

Ah I see, but these bags aren't

Hmm

[tough imp]

How does this prove it?

Without Nakayama I mean

you can conclude that K/mK = 0 if K is the kernel

Although the only way I can see that is using Tor

And then you’d still need to conclude K is 0 right?

[little hemlock]

oh do you need nakayama to show if rank(E) = rank(F) = n, then dim(E/mE) = dim(F/mE) = n?

[tough imp]

No

But I mean wasn’t the goal to show a surjection is also injective?

If you know the map is injective after tensoring with A/m why is it injective to begin with?

That’s where I think you’d need Nakayama, plus Tor (actually you don’t NEED Tor you can prove what you need using snake lemma) and you even need a Noetherian hypothesis to conclude that the kernel of E -> F is finitely generated

If that’s how you want to approach it that is

Okay well I guess there’s a proof using matrices that basically is this idea kekw

Define the original map in a matrix

Reduce mod m, that’s invertible so the determinant isn’t in m

Oh hey original matrix has invertible determinant

GG has an inverse

[little hemlock]

oh yea hmm, youre right what I said doesn't work. i didn't think think too hard about this. I think the statement in AM is like: if rank(E) <= rank(F) and there is a surjection E -> F, then it is an isomorphism, which isn't quite the same

[tough imp]

I mean this would be a special case

Cuz we’re assuming the rank is equal

I just don’t see how tensorinf with A/m let’s you conclude it without more arguments

[little hemlock]

agh no ive really messed up, the actual statement is that if E -> F is a surjection then rank(E) >= rank(F).

[tough imp]

Lmao yes this is easy after tensorinf

[plain raven]

God damn it.

I was tired and out of mental energy so I didn't want to carefully read what you said and think through it, but I also didn't want to seem inappreciative by just bouncing. So I just assumed it was correct and said "Ah yes I see, everything is clear to me now." But then immediately afterwards i was like "This is going to be turn out to be wrong and I'm going to look like an idiot. But whatever, YOLO."

If only I were a strict Kantian, all of this could have been avoided. I could have just said "I don't read SHIT unless i'm getting paid for it."

Saving this screenshot of a reminder to myself of what it looks like when I'm pretending to follow along

[empty grove]

Haha yeah that must be clerk alone because I would never do that

[loud scarab]

I just started studying the quotient topology so I have some questions:

1. What does the slash notation here mean?
2. Does this statement mean exactly as it says or does it mean that the quotient space is homemorphic to S1xS1?
3. Do I have to use group theory in these kind of problems or not?

[orchid forge]

it's the quotient group R^2 / Z^2

not sure what you mean by 2, it's telling you the topology on both sets, which you need to know in order to ask if a map is a homeomorphism

you shouldn't need any group theory beyond understanding what the quotient group is and how to get a quotient topology out of this one

[loud scarab]

the quotient topology is the topology induced by the quotient map, so it technically means I'm considering the partition set of R2/Z2, is that correct?

[orchid forge]

i'm not sure what you mean

[loud scarab]

maybe I'm just having gaps because I dont know what a quotient group is

what is a good reference for quotient groups? Is it in Munkres?

[orchid forge]

dummit and foote i guess? but in this case, you are just taking R^2 and identifying two points if their difference is in Z x Z, so for example (0, 1/2) = (1, 3/2)

each equivalence class is represented by an element of the unit square [0,1] x [0,1], which is unique except on the boundary

[honest terrace]

Yeah any algebra text would do it.

[honest terrace]

Well just [0,1) x [0,1), then, no ?

[orchid forge]

sure, but it doesn't change much

it's important to recognize that you are identifying the edges of the unit square

[honest terrace]

Yeah, fair

[empty grove]

You don't need any group theory for this

[honest terrace]

It probably makes more sense with group theory tho, but yeah I guess you don't strictly need it if you view everything topologically and forget the group structures

[empty grove]

But the group structure are irrelevant though

They're only talking about topology and homeomorphism

[orchid forge]

yeah you just need to know what the elements of R^2 / Z^2 are

and what the quotient map is I guess

[loud scarab]

anything that happens outside the unit square is irrelevant?

[orchid forge]

every point outside the unit square is equivalent to one inside the unit square

so you can think of the quotient group R^2 / Z^2 as just being the unit square, where the left and right edges are identified and the bottom/top edges are identified

then to figure out what the quotient topology is, you should look at what happens to open sets under the quotient map

[honest terrace]

Well they technically are but the fact that ||R/Z ~ S^1|| is really made obvious by the group structure imo, but again it's not strictly needed at all I know

[empty grove]

The group isomorphism isn't necessarily a homeomorphism though

[honest terrace]

well in this case it is

[sonic hill]

Damn, the notation R^2 / Z^2 could also totally mean R^2 with Z^2 collapsed to a point. Especially in topology. The problem didn't clarify!

[honest terrace]

I'm not saying group theory strictly comes into play

I'm not saying group theory strictly comes into play

I'm just saying it helps with intuition, here

[orchid forge]

such a thing is probably not defined yet

also, there is little risk of confusion since that would not be isomorphic to S1 x S1

[lunar yoke]

Im a bit late but this is wrong, see https://math.stackexchange.com/questions/3272545/are-countable-topological-spaces-second-countable/3272560#3272560

Cost me a good few hours trying to prove this

[vocal anchor]

Suppose $p: E \to B$ is an open surjection and $q: X \to Y$ a quotient map of spaces over $B$. Is it necessarily the case that $id_E \times_B q$ is a quotient map? continuity and surjectivity are ofc clear, but I don't know how to see that $(id \times_B q)^{-1}(U)$ open implies U open. :/

[gentle osprey]

expectTheUnexpected

[bleak path]

Hello, have a question about 1-forms

I'm given a 1-form \alpha, and have shown that d\alpha = 0. Now I'm expected to find a function f such that \alpha = df. I believe that I am supposed to integrate \alpha somehow, and by integrating \alpha then I also am integrating df, and so I get f. Does what I say make sense, or have I gone off somewhere?

[wise sigil]

integrating alpha just yields a number doesnt it?

[empty grove]

Try fixing a point x and the value of f at y is defined by integrating alpha along the path from x to y

[bleak path]

I don't know this for sure, and I don't think this is true

[empty grove]

1 forms can only be integrated over paths

Over a single path they give a number

[bleak path]

Right, I understand this

I think I will need to look at my question more though, since I don't think I have a path right now, maybe the path is hidden in the question somewhere

[gritty widget]

you need the domain of your form to be simply connected to do this

[empty grove]

I'm assuming they have a particular form in the question not a general thing

[bleak path]

To be more explicit about my question, i'm given a 1-form \alpha, and a curve c: [a, b]. I'm asked to find a smooth function f from R^3 to R such that \alpha = df, and hence evaluate \alpha along c without explicit integration

[bleak path]

They do, if it helps this is the 1-form

And yes, the domain is simply connected

[empty grove]

Yeah then that thing works

You have to do a few checks along the way and you'll see why you need simply connected

[bleak path]

Thanks! I'll try making more headway and ask again if I need more help

[obtuse meteor]

sanity checking myself for a homework

is it true that pi_2(S^1 vee S^2) is Z^Z (that is like Z direct summed with itself Z times).

that is the free abelian group on Z.

Argument roughly goes

take universal cover

contract the R portion bc its contractible

wedge of spheres

so H_2 is free abelian on Z

hurewicz theorem implies pi_2 is the same

[bleak path]

Managed to solve it, thanks guys 🙂

[plain raven]

Sorry, Faye. Hurewicz theorem? I don't follow.

IIRC the Hurewicz theorem says that at the first n for which one of pi_n, H_n is nonzero, the Hurewicz map is an isomorphism.

And maybe there are some slight strengthenings where we can say under weaker hypotheses that it's surjective or something

But wouldn't you need pi_1 to be zero? And I would think pi_1(S^1 vee S^2) is just Z.

Not to say you couldn't use the Hurewicz map here, because the map is natural and commutes with the boundary maps of the LES, somehow that could be used to relate pi_2 to pi_1.

[sonic hill]

Why are you considering homology? Covering maps already induce isomorphisms on higher homotopy groups

[obtuse meteor]

You lift to universal cover then apply hurewicz to compute pi2 of the universal cover

@plain raven @sonic hill

This makes pi1 0 so it’s not bullshit

[plain raven]

oh, i see.

[sonic hill]

Oh I suppose pi_2 of a wedge of spheres isn't obviously Z^Z

I didn't think too closely about how I'd compute pi_2 of that

[plain raven]

Yeah, Faye, I agree.

It should be correct.

How bizarre

I find it hard to get intuition for the higher homotopy groups but ... ugh. just seems so strange to me that every map S^2 -> S^1 is nullhomotopic and pi_2(S^2)=Z, and yet when you take their wedge sum you get the free abelian group on Z

[sonic hill]

Any explicit descriptions of some non-homotopic maps from S^2 to the wedge sum?

ok I went backwards and I think I got it

take a sphere with a line segment sticking out of it and map it so that the line segment wraps around the circle some number of times

It's hard to take that and get an actual map from the sphere to the wedge sum though 😒

in terms of visualizing it

[plain raven]

Ok, nice.

Yeah.

[sonic hill]

There's just one thing that's bothering me

The inclusion of S^2 to the upside-down lollipop is a homotopy equivalence yeah? But if you pre-compose the above maps by that inclusion, you get the same maps?

Something's going wrong and I can't find it

I'm taking the basepoints to be the north pole of the sphere and the intersection of the sphere and circle

It goes away if I take the tip of the lollipop as my base point hmm....

but why did I have to do that?

Ohh yes I figured it out

For anyone else who got stuck on the same issue

That's nice
Ok, yeah

[obtuse meteor]

It’s very strange and cool

[plain raven]

Hahaha yep

but in this case i do understand and i think it's correct

so we are all good

[obtuse meteor]

It’s actually true for anything which has pi1 trivial that pi2(S^1 wedge X) is direct sun of Z copies of pi2(X)

Same proof

Fucked up

[hallow swan]

is there a mapping between topological spaces that is bijective and continuous, but is not a homeomorphisms?

i.e. it's inverse isn't continuous

[obtuse meteor]

Yes

Take the identity mapping from R with discrete topology to R with standard topology

[sonic hill]

[0,1) to the unit circle R/Z

Advantage of this one is no weird topologies

[obtuse meteor]

Cringe

[sonic hill]

⁉️

Fine let's do it your way

Two point set with discrete topology to two point set with indiscrete topology

😤

[hallow swan]

holy

dumb question

or maybe i'm dumb

[gritty widget]

Can people give me some examples of

cohomology theories that obey all the eilenberg steenrod axioms

including the dimension axiom

but are not isomorphic to singular cohomolog

[tight agate]

theyre all iso to singular

so no

[gritty widget]

It seems

this sully react is not always used to mean

"i am sad about this"

but that is how i mean it

[sonic hill]

More like "bro... do you know your math?"

[hallow swan]

do you even math bro?

[obtuse meteor]

Only on things homotopy equiv to a CW complex

Unless you take preserving weak equivalence as an axiom

Iirc

[plain raven]

Cech cohomology.

Cech cohomology behaves better than singular theory with respect to issues of "dimension". I think it also has better colimiting properties, like when you want to express a space as the colimit of subspaces, but i haven't studied that stuff in much detail.

Cech homology and cech cohomology are also a more appropriate language to describe some things in Poincare duality/Alexander duality

I think in the compact case you only need the singular version, or rather there's no real difference between them.

but outside of a compact manifold there are some differences.

[gritty widget]

ah thats a good one thank you

[gentle osprey]

grist bundle

[quartz edge]

Grant me chain rule so that the exterior derivative distributes over composition of maps.

[gentle osprey]

grist bundle

grist bundle

[quartz edge]

Any ideas on where to go from here?

Guillemin & Pollack (page 10) says it is necessary to note that dphi_0 and dpsi_0 are isomorphisms (is this solely due to diffeomorphism of phi, psi? why?) so I guess I could play with the domain and image of df_x, but what property of isomorphism is useful in showing that df_x doesn't change when you vary phi, psi?

[gentle osprey]

grist bundle

[sonic hill]

I have a random question about manifolds that I thought of. If you take a manifold $M$ and define a vector bundle $E$ of the same rank as the dimension of $M$, with transition maps being the same as the transition maps of $M$, do you get a well defined and meaningful vector bundle?

[gentle osprey]

Icy001

[gritty widget]

that the image of dphi_0^-1 is the entirety of the domain of dh_0

sorry i meant to reply to the texit post by gristle

[sonic hill]

No worries, I'm still thinking about my question honestly, the construction might not even make sense

OK, I got why it doesn't make sense. Transition maps on a vector bundle must be fiberwise linear

Hohoho

[quartz edge]

@gritty widget yes, i use that to obtain this

[sonic hill]

So taking the derivative of the transition maps makes sense, and this is exactly the tangent bundle

[quartz edge]

but i do not see where to go from here. the possibility of dim(Y) > dim(X) is disturbing to me; i suppose df_x depends only on the definition of dh_0

The part that is worrying to me is that it says that df_x itself does not depend on phi, psi. Is that really true?

I would think it would just be the case that the image of f doesn't depend on phi, psi

not the actual element-wise mapping

and even still, it seems wrong

[quartz edge]

after this it just seems like (especially if k < l) even the image of df_x isn't undisturbed by choice of phi

oh shit

sorry

psi

,,\Im(df_x) = d\psi_0(\Im(dh_0))

[gentle osprey]

grist bundle

[quartz edge]

because we don't have that dh_0 is surjective; we only have that it's smooth

[echo elk]

Oh sorry, nevermind then.

[quartz edge]

it's not

well it was worth posting because the question is extremely similar but the answer is not satisfactory despite being accepted

for reference: https://math.stackexchange.com/questions/961341/a-differentiable-map-doesnt-depend-on-the-parametrization

also

the question is regarding differentiable maps between surfaces, i.e. 2-manifolds

this one is about smooth maps from a k-manifold to an l-manifold

i truly don't see how the property the authors suggest can be true

actually instead of emailing anyone i will just post on mathexchange

[orchid forge]

Given a different set of charts, we want to show that the induced maps $TR^m \to TR^n$ are the same after pulling them up to a map $TX \to TY$ i.e., they differ by conjugation by a pair of charts. If you don't even know that $df$ is well defined yet, then writing $ f$ as a composition and computing $df$ in 2 different ways will be circular. Instead, write out the map $h_0$, pick different charts and get $h_1$ and show that $dh_0$ and $dh_1$ differ by conjugation by a pair of transition maps. Then you can define $df$ by $d\psi \circ dh_0 \circ d\phi^{-1}$

[gentle osprey]

Kogasa

[orchid forge]

knowing that the choice of charts is arbitrary

[quartz edge]

what's h_0?

h(0)?

oh

sorry

dh_0 is referring to the 0 taken from U

[orchid forge]

Oh, I was using them as indices

[quartz edge]

ye

[orchid forge]

Implicitly these are all charts for some fixed point, and you can take them to be centered at the origin if you want

[quartz edge]

ok so

you're doing a pullback of f to h and showing that that pullback has to be unique

[orchid forge]

Up to conjugation by a pair of charts

[quartz edge]

i don't understand

[orchid forge]

Which is exactly what you need to define a map of manifolds

f : X --> Y induces a map U --> V

Given different charts, it induces a different map U' --> V'

These two maps are related

[quartz edge]

yes, by the definition of h

[orchid forge]

In the sense that $\psi_1 f \phi_1^{-1} = (\psi_1 \psi_0^{-1}) \psi_0 f \phi_0^{-1} (\phi_0 \phi_1^{-1})$

[quartz edge]

wait

you mean

[gentle osprey]

Kogasa

[quartz edge]

ohh

ok

[orchid forge]

A smooth map of manifolds is equivalent (by definition) to a class of smooth maps from (a subset of) R^m to (a subset of) R^n that differ by conjugation by transition maps in this way

[quartz edge]

pyep

[orchid forge]

So to show df is well defined, we pass to R^n by picking charts, define it like we want to, then show if we pick different charts, the two functions differ by conjugation like this

[quartz edge]

well

the book says they don't differ

[orchid forge]

What do you mean? The resulting function df is well defined but of course if you take different charts, the induced maps will be different

[quartz edge]

rofl

i don't understand why the book claims this

maybe i'm misinterpreting

[orchid forge]

You said earlier the book says df_x is well defined

Which is what I'm saying

[quartz edge]

hmm

"As $d\phi_0$ is an isomorphism, there is only one acceptable definition for $df_xb$, namely,

$$df_x = d\psi_0 \circ dh_0 \circ d\phi_0^{-1}.$$

Of course, in order to use this definition of $df_x$, we must verify that it does not depend on the particular parametrizations $\phi$ and $\psi$ used."

[gentle osprey]

grist bundle

[orchid forge]

Yes, this is exactly what I mean

[quartz edge]

I think I have misinterpreted

I think I was taking this to mean "df_x as a map does not depend on the parametrizations."

[orchid forge]

It does not

[quartz edge]

thank you

so just to get this straight

they are saying

[orchid forge]

A map of manifolds is equivalent to a collection of maps on R^n --> R^m which differ by transition maps

[quartz edge]

df_x is well-defined no matter the parametrization

or rather

well

[orchid forge]

They are saying if you want to define df in terms of a map R^n --> R^m, you must verify that the definition doesn't depend on the charts

[quartz edge]

hmm

b-

but it does?

i mean it's quite literally a function of both charts

[orchid forge]

It does not

[quartz edge]

in what sense are you saying "depend"

[orchid forge]

If you use different charts, you get a different map df : TM --> TN

That would be bad

We want to show this doesn't happen

[quartz edge]

but we just said that we do get a different map

TM still maps to TN

but the way in which it does depends on the parametrizations

[orchid forge]

No, you get the same df, represented by two different induced maps U --> V and U' --> V' which are related by conjugation

The fact that the two induced maps differ by conjugation by transition maps is equivalent to the fact that df is well defined

[quartz edge]

i agree that it's well defined for a given h, phi, psi

or wait

dh, dphi, dpsi

[orchid forge]

I'm saying if you pick a different phi and psi, and get the corresponding different h, then that h and the original h differ by conjugation by transition maps

[quartz edge]

ah yeah

that's definitely true

[orchid forge]

That is exactly what you need for this formula $\psi h \phi^{-1}$ to give a map of manifolds

[gentle osprey]

Kogasa

[quartz edge]

ok, agreed

so we elevate up to tangent spaces

[orchid forge]

The tangent space is the manifold we're talking about here, so you need to bear in mind what the transition maps are

[quartz edge]

i have not seen the term transition map

is that phi, psi?

or rather

f

[orchid forge]

Technically $\phi_1 \phi_2^{-1}$ where $\phi_i$ are parametrizations

[gentle osprey]

Kogasa

[orchid forge]

You can equivalently just talk about charts or parametrizations

[quartz edge]

ok so we can safely say

f = psi h phi^-1

[orchid forge]

You just need to know what they are for TX and TY

[quartz edge]

that is a thing we can say now

i can't see why that could possibly be an issue since we defined h by psi^-1 f phi

and we know phi, psi are isomorphisms so their inverses are good

now by chain rule you get

df = dpsi dh dphi^-1

the really confusing thing is that guillemin says "this verification is just like the one given to establish that the definition of T_x X does not depend on parametrization"

which is just

ridiculous

the two proofs cannot look anything alike

[orchid forge]

we can't use the chain rule if we don't even know what df is yet

we are trying to define df

[quartz edge]

he took parametrizations (U, phi) and (V, psi) and showed that h : U -> V given by psi^-1 phi admits dh = dpsi^-1 dphi

by chain rule

(actually he just said "we assume chain rule" and thus we have that by default)

then he said

dphi = dpsi dh

then

Image(dphi) = Image(dpsi dh) < Image(dpsi)

then he chose g : V -> U = phi^-1 psi

dg = dphi^-1 dpsi

dpsi = dphi dg

Image(dpsi) = Image(dphi dg) < Image(dphi)

hence Image(dphi) = Image(dpsi)

and he defines T_x X as that image

this strategy does not seem to work for df_x

i don't understand how a similar approach can be possible

[orchid forge]

If you are given a definition of $d\phi$ and $d \psi$ such that the chain rule $d(\phi \psi^{-1}) = d \phi \circ d\psi^{-1}$ holds, then this proof makes sense

[gentle osprey]

Kogasa

[quartz edge]

ya, he makes that assumption indeed

[orchid forge]

It seems a bit backwards to me since $d\phi : R^n \to TX$, so we should really define the tangent bundle first

[gentle osprey]

Kogasa

[quartz edge]

well this is just on page 10 of the text

we haven't yet defined TX

(but yeah)

[orchid forge]

but the idea is the same, you are defining a map (or a subspace) in terms of a manifold by defining it in terms of a chart, and showing that your definition does not depend on the chart you pick

[quartz edge]

how would it?

if i define it in terms of a chart then like

are you saying i need to show it doesn't for some reason fail given certain charts?

[orchid forge]

no, I'm saying that if you apply your definition to two different pairs of charts, you get the same map

[quartz edge]

my proof then would look like "suppose we have a chart that breaks this definition. let's examine what the ways in which that might happen are"

ah-ha

ok

right, right

i tried examining the images of the maps

like

Image(df_x) = Image(dpsi_0 dh_0) < Image(dpsi_0)

[orchid forge]

that won't work, because you're not trying to show two spaces / sets are the same (i.e. contain the same elements), you're trying to show that two maps are the same, so if you put the same point $(p, v) \in T_pX$ you get the same point $(f(p), w) \in T_{f(p)}Y$

[gentle osprey]

Kogasa

[quartz edge]

indeed

i just thought, though, that it's a necessary condition

even if not sufficient

i think this whole thing is starting to make sense to me now

thank you for your help

i'm going to take another crack at writing out the proof properly in the morning

[orchid forge]

no problem

[gentle osprey]

Hausdorff

[quartz edge]

the equation is correct hausdorff

what about the curvature definition is confusing to you?

think only about the arclength parametrization of the curve (unit speed). we can consider this WLOG because any regular curve can be arclength-reparametrized

[vast estuary]

why is there a ||gamma'(t)|| in the denominator

[quartz edge]

so just look at

ah yeah

ok

[orchid forge]

if it helps, it is the same as the equation $\kappa = \frac{dT}{ds} = \frac{dT}{dt} \left(\frac{ds}{dt}\right)^{-1}$

[gentle osprey]

Kogasa

[quartz edge]

so are you fine with the unit speed version?

like that one makes sense?

[vast estuary]

yep!

that's ok

[quartz edge]

ok

well

[orchid forge]

since $\frac{ds}{dt} = |\gamma'|$, you get $\kappa = \frac{|T'|}{|\gamma'|}$ as shown

[gentle osprey]

Kogasa

[quartz edge]

,,\kappa = \left|\left|\frac{dT(t)}{ds}\right|\right|

[gentle osprey]

grist bundle

Hausdorff

[quartz edge]

you don't need that

[orchid forge]

yes, so here $s$ is the arclength, which you can use as a parameter

[vast estuary]

oh ok so this is the definition?

[orchid forge]

then the curvature is more simply expressed as the derivative with respect to that parameter of the unit tangent vector

[quartz edge]

yeah, it is

chain rule grants you the other equation you have

[orchid forge]

the chain rule gives your definition without passing to the unit speed parametrization

ye

[quartz edge]

^

[vast estuary]

then all is good lol

which book can i see for this?

[quartz edge]

erwin kreyzsig - differential geometry

[vast estuary]

rn i'm only relying on my prof's lecture notes and he seems to have defined curvature differently in different places, which sucks

[quartz edge]

do carmo - differential geometry of curves and surfaces

theodore shifrin - a first course in differential geometry

^ the one i learned from

kreyszig is just as good but is the cheapest

[orchid forge]

this would also be in a calc 3 book

[vast estuary]

ah cool! thanks

btw now that we're talking let me confirm one more thing

how is torsion defined for y'all?

[gritty widget]

nabla x y - nabla y x - [x, y]

[orchid forge]

the non-free part of the homology group

i may have done one computation with the torsion you are talking about in my life

[gentle osprey]

Hausdorff

[vast estuary]

curve may or may not be unit speed, is this right? we take B' = (tau)N in the frenet serret formulae for unit speed curves

[orchid forge]

oh wait that is what this is

i didn't connect it to calc 3 at all

[quartz edge]

the frenet serret equations are

,,\begin{pmatrix} T' \ N' \ B' \end{pmatrix} = \begin{pmatrix} 0 & \kappa & 0 \ -\kappa & 0 & \tau \ 0 & -\tau & 0 \end{pmatrix} \begin{pmatrix} T \ N \ B \end{pmatrix}

[gentle osprey]

grist bundle

[quartz edge]

in a unit speed p.arametrization

who cares about non unit speed curves

silly

[orchid forge]

I mean you can always replace $\frac{dV}{ds}$ with $\frac{dV}{dt} \cdot \frac{1}{|\gamma'(t)|}$ as above

[gentle osprey]

Kogasa

[orchid forge]

so it should be easy to get from the unit speed definition to the general one

[vast estuary]

yup cool

thanks a lot!!

[vast estuary]

Regardless of whether $\gamma$ is unit speed or not, we can define $N(t) = \frac{\gamma''(t)}{|\gamma''(t)|}$ right? $N(t)$ is the normal to $\gamma$ at $t$.

[gentle osprey]

Hausdorff

[gentle osprey]

Hausdorff

[vast estuary]

Ah wait. I think if $\gamma(t) = (x(t), y(t))$, then $\gamma' = (x', y')$ and we make the choice $N(t) = (-y'(t), x'(t))$. With that, if $\gamma$ has unit speed, we get $\gamma'' = (x'', y'') = \kappa_s(-y',x')$. I still can't figure out $N(t)$ in terms of $\gamma''$ alone.

[gentle osprey]

Hausdorff

[vast estuary]

@quartz edge or @orchid forge Could you help?

[void ingot]

I think you cant do it Generally with a + or - sign here. If my curve gamma moves to left the $\gamma''$ is in the same direction as $N(t)$, but if gamma moves to the right, $\gamma''$ is of opposite direction.

[gentle osprey]

KIl

[vast estuary]

Yeah, you're right.

So we just define it as (-y', x') then?

(Algebraic Topology) Why is p^{-1}(b) in one-one correspondence with pi_1(X,b)? Could someone tell me the map?

[void ingot]

Jup

[lunar yoke]

you fix a point x in the fiber over b. Then you choose a representative of the homotopy class of a loop, lift it to x, and look at where the lifted path stops. This will be an element of p^{-1}(b). The kernel of this will be the characteristic subgroup (so trivial in the case of the universal cover) and the image will be all points in the fiber over b that lie in the path component of x (so again everything here since the universal cover is pathconnected)

[vast estuary]

Thanks, that works!

[quasi forum]

I am not so sure I understand how the hint helps.

[marble socket]

have you seen some properties that are preserved by homeomorphisms?

[quasi forum]

Well, connectedness/path conmectedness is certainly preserved

[empty grove]

Restriction of a homeomorphism is also a homeomorphism onto its image

That's what they mean by removing a point

[quasi forum]

Hmmm. Well, I am not sure what property det is referring to. 🤔

[empty grove]

What you said

[quasi forum]

Hmm, well no space is path connected if we do that.

[empty grove]

Not true

[quasi forum]

Wait, what?

[empty grove]

ends 😌

[quasi forum]

I'm not sure I understand what you're getting at :/

I know the end points are the key interest here, but how does that make some spaces path connected when you take out a point?

[marble socket]

say you have a homeomorphism (0,1] -> (0, 1)
and say 1 maps to some real number r

what does removing that point do?

[quasi forum]

Removing that point means 1 does not map to anything

[marble socket]

oh, we're also going to remove its image... else the map won't even be surjective

[quasi forum]

Ah, good call.

So this restriction isn't homeomorphic because it isnt even a function on [0,1]

[marble socket]

yea... so we need to fix it

[quasi forum]

How so?

[marble socket]

so what's the image of (0,1) under the map?

[quasi forum]

Well, that'd just be (0,r)U(r,1)

[marble socket]

wait... this is a little ....

[marble socket]

right

so do we get a homeomorphism from (0, 1) --> (0, r) u (r, 1)?

[quasi forum]

No, because (0,r)U(r,1) is not path connected, but (0,1) is

[marble socket]

exactly... but we got this map after restricting a homeomorphism

[quasi forum]

So it should be homeomorphic

Okay, let me see if I understood what happened.
So we assumed a homeomorphism existed, say f, between [0,1] and (0,1). With that, we extracted a point in the image for f([0,1]), which gives the image of (0,r)U(r,1).
So now what you are saying is that (0,1) and (0,r)U(r,1) should be homeomorphic to one another?

[marble socket]

oh i started with (0, 1] and removed only one point

[quasi forum]

True, [0,1] would be the same except you remove two points

[quasi forum]

I just wanna make sure I understand this last part right.

[marble socket]

um.. so if we start with [0, 1] --> (0, 1)
after removing one point will get something like (0, 1] --> (0, r) u (r, 1)

not from (0, 1)

so more generally you have a map f : X --> Y and it's inverse g : Y --> X such that both are continuous.
now say A is a subset of X. And say B = f(A). this is same as A = g(B)
so you can restrict the map f to get the map f' : A --> B and restrict g to get g' : B --> A. These have be inverses of each other!

Restriction of a homeomorphism is also a homeomorphism onto its image
this is what moldi meant

[quasi forum]

Ohhh, and since we removed the point in the image, we also removed the point in the domain

So (0,1] cuts down to (0,1) when we remove f(1)

[marble socket]

[quasi forum]

Okay, I see what you're getting at now.

There is another part to this problem that wants me to show R and R^n are not homeomorphic if n>1.
And this approach does the job, since removing a point in R^n keeps it path connected, but it's image in R is not.

[marble socket]

[quasi forum]

Okay, quick question. What is an imbedding?

[marble socket]

a continuous injective map f : X --> Y is an imbedding/embedding of X into Y, if f : X --> f(X) is a homeomorphism

[quasi forum]

Ah okay, that makes sense.

Ah okay. So the question is to show by example that just because f:X->Y and g:Y->X are imbeddings does not mean they are homeomorphic.
I was thinking about just using (0,1) and (0,1] but I am not sure of an injective function for both f and g

[marble socket]

that's a great start!

so can you find a copy of (0, 1) inside (0, 1]?

[quasi forum]

An easy imbedding from (0,1) to (0,1] is just the identity map

[marble socket]

... okeii... let's call it inclusion map tho.. 😛 identity is reserved for the actually identity map X --> X

[quasi forum]

Okay okay sure. You know what I am getting at though

[marble socket]

anyway... can you find a copy of (0, 1] inside (0, 1) now?

[quasi forum]

Oh, what about (0,1/2], so just divide everything by 2

[marble socket]

[quasi forum]

And there it is

So I'm ngl, I tend to really struggle finding examples/counter examples of things. There are just too many options to siften through.
When asked to find such an example, what things should I be thinking about to start limiting my options?

This one was obvious because it was right in front of me :p

[marble socket]

first give away is you need to look for two non-homeomorphic spaces

but yea in general stuff can be hard...

[quasi forum]

That's definitely a thing I need to work on.
For instance, you helped me find a counter example for an abstract algebra question (A,B being subgroups does not mean AB is a subgroup), and I know I definitely would not have thought to look in S_3.

I was trying to look through all the things I know, and the permutation groups completely slipped past me

[marble socket]

yea so you see in that problem, we wanted to find subgroups A and B such that AB is not a subgroup.. so the first thing that would come to my mind is some sort of criterion that would easily rule out when something is not a subgroup. and you know what it was... after that it's usually a smooth ride.

[quasi forum]

I was also thinking D_3, and I think the same thing would work by looking at the powers of F and R

[marble socket]

D3 = S3

[quasi forum]

Oh yeah. When you're right you're right :p

[marble socket]

you sure about R? that has order 3. so you'll get the whole group.

[quasi forum]

Oh true. Yeah...disregard

Are inclusion maps just restricted identities?

[marble socket]

yea

maybe backwards lol

[quasi forum]

Okay. So if I say, the inclusion map f:(0,1)->(0,1], then there should be no confusion as to what I am talking about?

[marble socket]

to restrict a map f : X --> Y on a subset A of X, you precompose with the inclusion map A --> X

but yea not a serious difference

[marble socket]

yep

[quasi forum]

Okay cool!

[marble socket]

if A is a subset of B, then you get the inclusion map!

this turns out to be continuous for simple reasons

[quasi forum]

Okay, so quick question (Pretty sure I know how to do it, just wanna make sure).
So the question states, let f:S^1->R be a continuous map, show that there exists an x so that f(x)=f(-x).

Okay, pretty sure this just uses IVT. WLOG, we know there is some y so that f(y)-f(-y)>=0, but then, f(-y)-f(y)<=0.
f(x)-f(-x) is continuous and the image of the comnected space, S^1, is connected, and since we found a negative and positive value for this function, we know 0 must be a value in the function by IVT.

Does that all sound right?

[orchid forge]

seems good to me

[gritty widget]

is the cardinality of any open set in R^n = the cardinality of R^n?

[orchid forge]

any nonempty open set yes, and that is the same as the cardinality of R

[pearl holly]

is it correct to say that $\Lambda[\alpha_1, \cdots \alpha_n]$ is to the direct sum of all $\oplus_{i = 1}^n \otimes_{i = 1}^n \Lambda[\alpha_i]$?

[gentle osprey]

Tokidoki ✓

[pearl holly]

where \Lambda is the exterior algebra

Hatcher writes "the exterior algebra \Lambda_R(a_1, ... a_n) is the graded tensor product over R of the one-variable exterior algebras \Lambda[\alpha_i]". Does he mean the that I wrote above?

[orchid forge]

If you change the order in your tensor product, how do you get a sign out of it

[pearl holly]

idk I'm just guessing now lmao

[pearl holly]

this makes me feel like $\Lambda_R(\alpha_1, \ldots, \alpha_n) = \otimes_i \Lambda[\alpha_i]$ but at the same time you can write $\Lambda_R(\alpha_1, \ldots, \alpha_n) = \oplus_i \Lambda[\alpha_i]$

[gentle osprey]

Tokidoki ✓

[pearl holly]

edited this btw

[orchid forge]

if you're taking the tensor product of graded algebras, the tensor product is also a graded algebra right

the latter expression should not be true, since you should not be able to freely commute stuff in the exterior algebra

[pearl holly]

ah yeah right

[pearl holly]

so Hatcher means the first expression?

[orchid forge]

yes I think so

I think the grading on the tensor product is exactly the thing you need to say e.g. $v \otimes w = -(w \otimes v)$ in $\Lambda_R(\alpha_1) \otimes_R \Lambda_R(\alpha_2)$

[gentle osprey]

Kogasa

[pearl holly]

yeah okay I see. So now $\Lambda_\mathbb{Z}[\alpha_1] \otimes \Lambda_\mathbb{Z}[\alpha_2] = \Lambda_\mathbb{Z}[\alpha_1, \alpha_2]$ right?

[gentle osprey]

Tokidoki ✓

[pearl holly]

more generally I guess it doesn't have to be over Z

[orchid forge]

yes, the multiplication on the graded algebra $\Lambda[\alpha_1] \otimes \Lambda[\alpha_2]$ should give this

[gentle osprey]

Kogasa

[pearl holly]

ye okay I see. Thank you so much!

[quartz edge]

Thank you kogasa for what you did yesterday

I finished up the proof and it's very nice

[gritty widget]

I was trying to solve an exercise from a book but i got stuck in my calculation.
I'm trying to show that foundamental groups of a class of orientable 3-manifold is not left orderable where by this i mean that they admits a strict ordere invariant for left multiplication.
The groups are Seifert manifold built over a surface where the surface is $\mathbb{R}P^{2}.$

They admit the following presentation: $$\pi{1}(M)=\langle\gamma{1},...,\gamma{n},y,h;|$$
$$yhy^{-1}=h^{-1},\gamma{j}^{\alpha{j}}=h^{-\beta{j}},\gamma{j}h\gamma{j}^{-1}=h,y^{2}\gamma{1}\cdots\gamma{n}=1\rangle.$$

With some calculation I got to show that if $k>0,h>1$ the following holds: $h^{-k}<y^2<h^k$ and $h^{-k}<y^{-2}<h^{k}.$ In the book I was given the hint to show that $yhy^{-1}>1()$ and this in facts concludes as it is a contradiction with the assumption that $h>1$ given the first relation but I could not find a way to prove the key fact $()$. Any suggestion in what kind of relations I should look for?
Thanks in advance.

[gentle osprey]

Stephen

[pearl holly]

sorry for the ping but I was just thinking about the exterior algebra stuff and remembered that we had a chat about this. I just learned what tensor products are and now I'm curious on how you would define this but with tensor products

[marble socket]

Oh i see

[pearl holly]

I'm guessing that you maybe define a^b to be a tensor b or something like that?

[gentle osprey]

det

[marble socket]

This satisfies the nice universal property which you can guess

[pearl holly]

oh okay, so K has the "relations" that for example a tensor a = 0?

[marble socket]

not quite... it's not an algebra yet... just an R-module

[gentle osprey]

det

[marble socket]

This is just the wedge/exterior power like i defined last time

you can define the exterior algebra as taking the nth exterior power as the component corresponding to degree n

which was the equation i wrote last time

[pearl holly]

ah yeah okay I see

so I assume that the coset [a \tensor b] is a ^ b now?

(a wedge b)

[marble socket]

yea to define the algebra multiplication here, just put the things side by side

you can also directly construct the exterior algebra

basically this

okie should have sent this first

[pearl holly]

okay I see. But if the ideal is generated by $(m \otimes n - n \otimes m)$, wouldn't the wedge sum be commutative?

[gentle osprey]

Tokidoki ✓

[pearl holly]

not the wedge sum lmao

I mean the wedge product or whatever that was called

[marble socket]

yea this is the symmetric flavor of tensor... so called symmetric algebra

we're interested in I_Wedge here

[pearl holly]

but shouldn't a^b = -b^a in the exterior algebra

ye but isn't the a^b = [a tensor b]?

[pearl holly]

I mean this

[marble socket]

only if these are like smol things

so you know that (a + b) ^ (a+b) = 0

expand and use a ^ a = b ^ b = 0

but if a had degree i and b has degree j, then it becomes a ^ b = (-1)^ij b ^ a

[marble socket]

by smol i meant degree 1 lol

these are like weird constructions so only nice way of working with them is via their universal properties

[pearl holly]

yeah okay I see. Just to make sure I get stuff right, $\Lambda^{n+m}(M) = \Lambda^{n}(M) \otimes \Lambda^m(M)$

[pearl holly]

using this definition

[gentle osprey]

Tokidoki ✓

[pearl holly]

is that right?

[marble socket]

not quite

[pearl holly]

frick

[marble socket]

think about it like this... the left and right part don't interact with each other

if we had a repeat, the corresponding pure tensor should be 0

but left and right don't know about each other

[pearl holly]

okay I see, then I have no clue what Hatcher says when he says that the exterior algebra is the graded tensor product over R of the one variable exterior algebras

[marble socket]

the wedge powers satisfies this simple universal property as you might guess... giving a multilinear alternating map from M^l --> P is "same" as giving a liner map from the exterior power to P

[pearl holly]

ye okay I kind of see that

[marble socket]

proof are pretty simple just from the very definitions

we'll get a linear map from nth tensor power of M to P

but this dies on the submodule written above

this one

so the map factors through exterior powers

[pearl holly]

yeah okay I see. Thank you so much for the help!

[marble socket]

lol sorry i just finished an assignment a few minutes before you asked something, so was a little excited and tired at the same time

prolly i didn't say a lot of things nicely

this is the universal property of the exterior algebra

[pearl holly]

no it's fine lmao

[quasi forum]

Looking at #3. Correct me if I am wrong, but shouldn't all intervals have a fixed point?

If not, I am not sure I understand why

And x/2+1/4 should be a counter example for both (0,1) and (0,1].
So what exactly makes this fail

[little hemlock]

what book is that from?

[marble socket]

Aluffi algebra chapter 0

the only algebra book i read lol

[tough imp]

Swag

[marble socket]

more like that book was so long that didn't have time to read other things 😂

[plain raven]

@orchid forge i still haven't read anything in iversen this week, RIP. however i am posting this here as a public reminder to myself that i still have to do this. I think we agreed to discuss this like, last wednesday, so whoops. anyway i will read it for a few hours this weekend and then just say some stuff here on monday

[orchid forge]

lol it's ok, i've been busy as well

i did take a look at the first ~50 pages so i might be able to say something though

[quasi forum]

Wait, x/2+1/4 does not do it :(

[plain raven]

huh this is weird. makes sense but i've never seen this version of it before

Usually i see people stress that it's the free graded graded-commutative algebra on A

er, on M

[marble socket]

skew-commutative... right?

[plain raven]

Not sure if you got an answer to this, but this is what he means. If $M$ and $N$ are $\mathbb{N}$-graded modules, then their tensor product is also itself naturally $\mathbb{N}$- graded. The tensor product distributes over direct sum, so like, $A\otimes \left(\bigoplus_i B_i\right) \cong \bigoplus_i A\otimes B_i$, and $\left( \bigoplus_i A_i\right) \otimes \left(\bigoplus_j B_j\right)\cong \bigoplus_{(i,j)\in I\times J}A_i\otimes B_j$

[gentle osprey]

diligentClerk

[plain raven]

So if we have $A$ and $B$ $\mathbb{N}$-graded, so $A = \bigoplus_{n\in \mathbb{N}}A_n$ and $B = \bigoplus_{m\in\mathbb{N}} B_n$, then $A\otimes B \cong \bigoplus_{(m,n)\in \mathbb{N}\times\mathbb{N}} A_n\otimes B_m$

[gentle osprey]

diligentClerk

[plain raven]

so this tensor product module is sort of "bigraded" by $\mathbb{N}\times\mathbb{N}$

[gentle osprey]

diligentClerk

[plain raven]

and we traditionally simplify this down to an $\mathbb{N}$-graded module by rewriting it as $\bigoplus_{n\in\mathbb{N}}\left(\bigoplus_{p+q=n}A_p\otimes B_q\right)$

[gentle osprey]

diligentClerk

[quasi forum]

When you get a chance, I could really use help on #9

[plain raven]

i.e. the grading is $(A\otimes B)n = \bigoplus{p+q=n}A_p\otimes B_q$

[gentle osprey]

diligentClerk

[plain raven]

We sometimes refer to this as the "total grading"

it's the conventional choice of grading

[marble socket]

oh this looks like the tensor product of complexes i was doing for the AG assignment...

[plain raven]

so, $\Lambda_R[\alpha]$ is the graded algebra sum $R\cdot 1 \oplus R\cdot \alpha$, where the degree $0$ component is $R\cdot 1$ and is generated as an $R$-module by $1$ and the degree $1$ component is $R\cdot \alpha$ which is generated as an $R$-module by $\alpha$. And Hatcher is saying that if you take the tensor product in the graded sense $\Lambda_R[\alpha_1]\otimes \Lambda_R[\alpha_2]\otimes\dots \Lambda_R[\alpha_n]$ you should get the exterior algebra $\Lambda_R[\alpha_1,\dots,\alpha_n]$ with the grading such that monomials of degree $k$ generate the degree $k$ component.

[gentle osprey]

diligentClerk

[pearl holly]

oohh okay right I see

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

[plain raven]

neat.

[tidal cedar]

kinda weird / dumb question

the one point compactification of a countably infinite number of open intervals is apparently the Hawaiian Earring space

not the infinite wedge sum of circles

and ofc the Hawaiian Earring space is different from the wedge sum because every neighborhood of the base point contains all but finitely many of the circles

infinite wedge sum has contractible basis of the point

and I'm specially wondering how the one point compactification has this property, IG

why compactifying countable disjoint union of intervals makes all nbhds of the point at infinity include all but finitely many of the intervals

wait

[plain raven]

hmmm

[tidal cedar]

ok so the definition has that you're adding all sets of the form $U = (X - C) \cup {\infty}$

[gentle osprey]

1 wug 2 wugs

[quasi forum]

If anyone gets the chance, may I please have help with this?

[tidal cedar]

where C is closed and compact

and so the closed and compact sets here are finite numbers of intervals

ok so IG there's my answer

uhh additional question, would someone have an AT way of showing this

like

fundamental wedge sum of countably many circles is the free group on countable generators

can we use a generalisation of Seifert van Kampen to show the compactification here has a pi_1 that isn't that, for example

[plain raven]

hi dackid

let $p,q$ be two points in the plane. Let $L$ be the set of all points in $\mathbb{R}^2$ which are equidistant from $p$ and $q$.

[gentle osprey]

diligentClerk

[tidal cedar]

so d(x,p) = d(x,q)?

[plain raven]

Yes. That's what i mean

[tidal cedar]

Ahh

[plain raven]

for any point $t$ in $L$, we denote by $\sigma_t$ the path from $p$ to $q$ in $\mathbb{R}^2$ which is the composition of the following two paths: first we proceed from $p$ to $t$ in a straight line, and then we proceed from $t$ to $q$ in a straight line.

[gentle osprey]

diligentClerk

[obtuse meteor]

Can anyone help me with some Quillen + construction shit

my brain is running on sick fumes

[plain raven]

Now prove that it's impossible that every $\sigma_t$ intersects $A$ nontrivially, i.e. there is some $t$ so that $\sigma_t$ is in $\mathbb{R}^2-A$.

[gentle osprey]

diligentClerk

[obtuse meteor]

why are people sad reacting quillen + construction ;-;

[quasi forum]

Hmmm, I was pretty much with you up to the last part. This is about where I am at too

[obtuse meteor]

it's a series of "Uncountable thing - Countable Thing = Uncountable Thing"

[quasi forum]

But I am unsure how to prove it is impossible. Or even how to go about thinking about the proof

[tidal cedar]

IDK if this necessarily show existence of a path, roughly speaking, the failure could be not at some t \in L?

[obtuse meteor]

doesn't matter fiona you can still win

[quasi forum]

Not at some t in L? What do you mean?

Or are you saying that the point does not matter

[obtuse meteor]

Hint: The # of t in L is uncountable. And the # of lines from p crossing through some point in A is countable

likewise the # of lines from q crossing through some point in A is countable

[quasi forum]

Okay, why is the # of lines crossing through a point countable? (That's what I need to get to).

[tidal cedar]

number of lines or number of paths?

[plain raven]

Let $L'\subset L$ be the set of $t$ such that the path $\sigma_t$ meets $A$ nontrivially.

[gentle osprey]

diligentClerk

[tidal cedar]

you can show number of paths through X is uncountable by this construction

[plain raven]

By the axiom of choice, for each $t\in L'$, choose some $a\in A$ which lies in the image of $\sigma_t$.

[gentle osprey]

diligentClerk

[plain raven]

This defines a function $L'\to A$.

[gentle osprey]

diligentClerk

[plain raven]

The function is injective, because it $t\neq t'$, then the lines $\sigma_t$ and $\sigma_{t'}$ have no common points in their image, except for the endpoints at $p$ and $q$, which are not in $A$ by assumption.

[gentle osprey]

diligentClerk

[plain raven]

Thus, $L'$ embeds into a countable set, and is thus countable

[gentle osprey]

diligentClerk

[tidal cedar]

uncountable - countable = uncountable

so there are uncountably many paths

does this suffice

[obtuse meteor]

so

quillen plus

👉 👈

[quasi forum]

So here you specified they were equidistant. Why?

[tidal cedar]

the equidistant points are a line, hence uncountable

i.e. points equidistant from p and q forms a line bisecting the direct line segment from p to q

[quasi forum]

Ah, I see.

Okay and since there are only countably many paths that pass through A, we know that there is some path in L that does not pass through A

[tidal cedar]

yeah

well

only countably many paths that do the L thing that pass through A

since all the paths in L here have no point in common

[quasi forum]

And that's because we are dealing with lines here, right?

[tidal cedar]

that's a property of how those are constructed yeah

p -> some point in L -> q only has common points at p and q

[gritty widget]

Let X be a space with an action of a group G

what is a good name for

X - F

where F are the fixed points

so what is a good name for the points that are not fixed

coinvariants seems like it probably means something else?

[obtuse meteor]

variants

[orchid forge]

Nonfixed points

[plain raven]

lol

i like 'variants'

[obtuse meteor]

clerk

are you familiar with dumb ways of stating quillen +

I've read the good ways of stating it

but my prof is making us do a dumb way on this homework and I'm fucking dying

[plain raven]

No, I wish. Sorry, no idea how to help you.

[obtuse meteor]

;-;

rip

[plain raven]

sometimes 'coinvariants' means something like the quotient of X by the action of G, or some related quotienting

[obtuse meteor]

I'm sick bitch so my brain is running on windows vista rn

[gritty widget]

this is good

is anyone here into

equivariant homotopy theory

if so

how do these so called

geometric fixed points

of spectra

correspond to fixed points of topological spaces

[quasi forum]

This seems like a silly question, but is the empty set connected?

[orchid forge]

yes, remember "connected" means "there exists no separation"

[quasi forum]

Okay cool. There is a Question that asks if bd(A) and int(A) is connected, then is it true that A is connected. The answer would be know, and Q is the counter example

[orchid forge]

on the other hand, the empty set and itself are a separation

[quasi forum]

Well the sets need to be distinct

[orchid forge]

wait no, the sets in a separation are nonempty

so yeah i was right the first time

[quasi forum]

Okay cool!

Okay, so here is a question. If A and B are connected, I know AuB is connected if they share a common point. But what if they only share common limit points?

[coral pivot]

do you mean union of A and B

[quasi forum]

Yes, sorry about that

[coral pivot]

consider open disk and a translate of it

[orchid forge]

"only share limit points" meaning they are disjoint?

[quasi forum]

Yes

They are disjoint

[orchid forge]

then they would be a pair of disjoint nonempty open subsets that cover A union B

[quasi forum]

I also need to check the converse.

Ah, so we found a separation

[cursive flume]

didn't manage to latex, but here it is in case you are interested. thanks a lot for the help!

[quasi forum]

Oh good lord, this is obviously false -_-

Given a,b in R, (a,b) is connected. But the boundary {a,b} is clearly not

[quasi forum]

Okay, so I am working on 3. Iirc, the ordered square is [0,1]x[0,1] with the dictionary order topology.

I am having a hard time seeing where local path connectedness fails

[gentle osprey]

Hausdorff

[empty grove]

Just check that the diagram commutes

Each triangle commutes

So the diagram does too

[empty grove]

Hint: there is never a path from a to b if their x coordinates differ

[vast estuary]

Yeah, I had noticed that!

For tilde p' circ tilde p we'll need a different diagram right

[empty grove]

The top arrow composition is exactly that

They're saying that the top row is a lift of p' along p'

The diagram would be exactly this but with X tilde removed

[vast estuary]

The top arrow composition is tilde p circ tilde p'

I'm saying that for tilde p' circ tilde p we need a different diagram

[empty grove]

oh

Yes you're right, then we glue these triangles in the other order

[vast estuary]

Yup thanks!

[vocal anchor]

Dumb question, but to check that a function is continuous it's enough to consider opens from a base, right?

[gritty widget]

not a dumb question. yes

[vocal anchor]

it's dumb because I forgot all my topology I ever knew haha

But thanks 🙂 (Also you're right, it's not a productive attitude)

[mellow prairie]

i need to show why Z is incomplete with the 2 adic metric.

the sequence 9,99,999.. converges to -1 so i said 3,33,333... converges to -1/3 which is not in Z so Z is incomplete

(as the sequence is cauchy)

but it seems wrong

to make the argument more precise can instead consider the completion of Z with the 2 adic metric (whatever the completion is) and then use the uniqueness of limits

[viral halo]

i believe that's correct, that's what completeness is

[mellow prairie]

yeah but the argument itself is a little lose

[viral halo]

it's complete iff every cauchy sequence with values in Z_2 has a limit in Z_2

which is clearly not the case

[mellow prairie]

i need to show that there is no integer in Z that the sequnce above converges to

showing that it converges to -1/3 doesnt do that so maybe i need to invoke the uniquess of limits by considering the completion

alternatively i can show that 3s+1 tends to 0

[viral halo]

what?

[mellow prairie]

so if there is a limit c Z then 3c+1 tends to 0

[viral halo]

unless -1/3 = 2 with the 2-adic metric i don't know what it the problem

[empty grove]

Not tends, equals. The limit is just a number

[viral halo]

but then again i never studied p-adic metrics so take it with a grain of salt

[mellow prairie]

oh yeah thats what i meant xd

[empty grove]

So you know that the metric plays nicely with the ring operations (ie addition and multiplication and negation are all continuous functions) so for a sequence x_n
lim 3x_n = 3 lim x_n. And you are arguing that since lim 3x_n is -1, lim x_n doesn't exist, because there is no integer which times 3 is -1

This is how I'd argue

Yeah that's exactly why I'm avoiding saying 1/3 in my argument

Strictly speaking it's not wrong because division by 3 is injective in Q, so if a solution exists then it is unique, so if a solution exists in Q\Z then none exist in Z

But you see that this is just an annoying extra line

[mellow prairie]

yeah makes sense xD

[empty grove]

You could also say that 1/3 is shorthand for a number such that 3 times it is 1

Instead of the 1/3 in Q

[mellow prairie]

thats why i wanted to do this

[empty grove]

But yeah like this is just a pedantic remark

Yeah, uniqueness of limit as well as of solutions to 3x = -1

Did you lose marks for this though?

[mellow prairie]

thats sneaky xDDD

[mellow prairie]

no it was just a problem sheet, so didnt have any marks to lose or gain

[empty grove]

Ah cool, pretty based

[mellow prairie]

thanks!