point-set-topology

[sleek thicket]

we get $a_i X^j dx^i\left(\frac{\partial}{\partial x^j}\right) = a_i X^j \delta^i_j$

right?

[gentle osprey]

shamroc$\overline{k}$

[summer jolt]

Why do the Xj come out?

[sleek thicket]

dx^i is a linear map

[summer jolt]

Linear in what?

[sleek thicket]

I'm not sure what you're asking

Oh sorry I see

It's a field of linear maps

[gritty widget]

C^\infty-linear

[sleek thicket]

You can think of it as a function on vector fields which is linear over the ring of smooth functions

lol tterra posted my real answer as his meme answer

[summer jolt]

Ah I see so covector fields are linear over the module of smooth functions on a manifold

[sleek thicket]

basically if $\omega$ is a covector field and $X$ a vector field, and $f$ is a smooth function, then $\omega(f X) = f\omega(X)$

[gentle osprey]

shamroc$\overline{k}$

[sleek thicket]

yup, that's exactly it

$\Gamma(T^* M) = \Gamma(TM)^*$

[gentle osprey]

shamroc$\overline{k}$

[sleek thicket]

where Γ is the C^infty module of global sections

The first * takes the dual bundle and the second one takes the dual as a C^infty module

this is true for tensor bundles in general btw

[gritty widget]

flexing your bundles course i see

[sleek thicket]

nah lol we did this in my manifolds course last year

[gritty widget]

i figured lmao just memeing

[sleek thicket]

oh but I did want to flex my bundles course

on you

Very cool problem

how many isomorphism classes of real line bundles are there on the circle?

guess

[gritty widget]

1

[sleek thicket]

no

[gritty widget]

sully

dan

[sleek thicket]

there's the tirival one yeah

but then also

Mobius bundle

right?

[gritty widget]

i guess so

you're the bundles expert

[sleek thicket]

become bundlebrained ttera

the mobius bundle is just like

Take a mobius strip

Extend the line segment infinitely outward

tada

now it's a vector bundle

anyways the proof is very cool

You should learn bundles is the main point

[gritty widget]

from what book

[sleek thicket]

not this book tbh

I'm not a huge fan

so far

It could use more category theory

unironically

It does stuff with functors

but then it doesn't do it well

it doesn't prove enough things about the stuff it does with functors

[cedar pebble]

[sleek thicket]

So the exercises are really tedious

[sinful pecan]

if you haven't looked at it, Natural operations in diff geo is quite nice

[gritty widget]

this one?

[sinful pecan]

yes

[gritty widget]

had a pdf sitting in my geometry folder

[shut moat]

oh shit am I supposed to know what a module is before I do lee

[gritty widget]

it helps to know the definition for when you get to vector fields / covector fields / tensor fields

but like

i don't think you need to have an in depth knowledge of modules to do lee

[shut moat]

phew

[gritty widget]

basically, the set of all tensor fields on a manifold is a module over the ring of smooth functions

what else...

[tight agate]

pi_0 (R*) = 2?

[gritty widget]

[tight agate]

yeah that should do it

as bundles on spheres are in one to one correspondence with homotopy classes of functions from the equator to the structure group

and the equator of the circle is just 2 points

and the structure group is R*

[sleek thicket]

¯\_(ツ)_/¯

I don't know anything about classifying spaces

you can argue that real line bundles on the interval are all trivial by like, covering with finitely many trivializations which go from the start to the end, each only overlapping with the next

via the lebesgue number lemma

then you can define a consistent global orientation

It's trivial on each little bit and when you hit the next patch you can flip the orientation to compensate

now you can cover the circle by two line segments, their overlap has two connected components

by choosing a metric and an orthonormal frame we get that the bundle is determined by a transition function landing in O(1), and this is just an assignment of ±1 to each component

Flipping the sign of the transition function gives the same bundle, so there's only two options

[tight agate]

yeah the argument is pretty much the same

it isnt using anything about classifying spaces

the reason why the bundle it determined by that function is because the upper and lower hemispheres are both contractible, so the bundle is trivial

and now all you need to know is how to identify fibers

which is the same data as a map from the equator to R*

[sleek thicket]

ah okay, I saw homotopy classes of maps and assumed it was classifying space stuff

[tight agate]

yeah it's called a "clutching function"

so by the same argument fiber bundles over a sphere S^n with a given structure group G are determined by homotopy classes of maps from S^n-1 to G

so elements of pi_(n-1)(G)

for low dimensions and groups like SO(n), O(n), ..., you can actually compute the homotopy groups

[sleek thicket]

Yeah I think this argument is using some facts I didn't know about bundles, eg all bundles on a contractible space are trivial or gluing data/transition functions only matter up to homotopy

[tight agate]

stuff on that is in May's concise course in alg top

[sleek thicket]

well the next chapter of my textbook is on "bundles and homotopy", so presumably I'll learn it soon

[tight agate]

yeah it's all pretty straightforward

the main point is the functor that assigns rank n vector bundles (up to iso) to each factors through the homotopy category

and now you can start using homotopical methods to study them

like representability follows from abstract nonsense

Milnor and Stasheff is also nice for bundle stuff

[placid thorn]

so $f(x,y) = (x/(\sqrt{x^2+y^2}), y/(\sqrt{x^2+y^2}))$ maps points on a square to points on a circle

[gentle osprey]

Corgwn

[placid thorn]

so, will solving x = x/(\sqrt{x^2+y^2}) for y, do I retrieve the y component of the formula for mapping a circle to a square?

[placid thorn]

@gritty widget sorry to bother you but could you take a look at this?

[gritty widget]

you may not get something going to a square

this map takes the entire plane minus the origin to the circle

er

why are you solving x = x / sqrt(x^2 + y^2)? how did you get that?

[gentle osprey]

slimvesus

[placid thorn]

Yeah I’m trying to come up with a map from the unit circle to the unit square

[placid thorn]

For a point on a unit square, it gets the x coord of it mapped to a unit circle

Uhhh no I’m just curious about what it would look like

But I know it should exist, given that circles and squares are topologically congruent

Yeah

[gritty widget]

you need to restrict the domain

and then it will be a homeomorphism

computing an inverse probably isn't so bad

[placid thorn]

But in that case solving the x= whatever equation above will work yea?

How do you even go about finding the inverse for a multi variate eq?

[fathom cave]

what exactly is going on here?

so we have defined $T^{ij}_k$ is some odd way wrt some coordinate system $\bar{Z}^i$ and showed that this definition is in fact a tensor?

[gentle osprey]

Zero0

[shut moat]

@placid thorn if you're still working on it I recommend looking at each side of the square individually

then coming up with a nice function that patches it all together

[placid thorn]

oo okay

f(x,y) = (x/max(|x|,|y|), y/max(|x|,|y|))

[shut moat]

ding ding ding

[placid thorn]

I think that works

yea

[obtuse meteor]

Yeah that works well. Another good way to do it is to parameterize each of them and then do convex combination

[tough imp]

AAAA

what a word lol, I used that on a complex analysis hw last year haha

I just asserted stuff about convexity using that I learned from computational geometry crap I've done and the TA just was like "sure"

[obtuse meteor]

Nice

[placid thorn]

hmmm

how could I prove that in isometric transformations, vertexes map to vertexes?

intuitively I feel like that should be try

*true

yes that

what I want to prove is that a square and a triangle aren't isometric

my current argument has to do with vertexes

basically, suppose that for 3 of the four vertexes, the distances between them in the isometry is fixed

they'd have to be right?

d(x,y) = d(f(x), f(y))

right, and in a square

for one vertex the euclidian distance to 2 other vertexes is a, and then to the last one is sqrt{2}*a

yeah so let's assume that the we map the two closest vertex to a triangle

getting an isoceles

and since the last vertex which should be length a * sqrt{2} will have to lie somewhere on this isoceles triangle

it's going to have to be closer to the vertex than the other two which have their distances fixed

so it can't possibly hold here that d(x,y) = d(f(x), f(y)) for this vertex

I feel like it's not a good approach

yeah

and therefore none for vice versa

well where?

is there a more elegant approach?

I feel like there’s a more elegant way to prove this

hm...

could I use the triangle inequality to prove that for isometries between polygons, vertexes map to vertexes?

[tough imp]

Can you show that an isometry preserves angles

Like show that they’re conformal maps or whatever

Then it’s automatic I think

[placid thorn]

"Suppose that in an isometry between two polygons, there existed at least one mapping of a vertex, $a$, to a point that isn't a vertex, $e$. Consider that on a polygon, points are either on vertexes, or lay on line segments that connect them, so $c$ must be on a point between two vertexes in this new isometry, $g$ and $h$. Suppose that $f(b) = h$ and $f(c) = g$, for points on the polygon where $a$ lies $b$ and $c$, and thus $d(b,c) = d(h,g)$. Thus, because they lie in a straight line with eachother, the property holds that $d(h,e) + d(e,g) = d(h,g)$, so $d(b,c) = d(h,e) + d(e,g)$. Thus, $d(b,c) = d(f(b), f(a)) + d(f(a),f(c))$, or just $d(b,c) = d(b,a) + d(a,c)$."

I feel like I'm on to something here

[gentle osprey]

Corgwn

[placid thorn]

"Suppose that in an isometry between two polygons, there existed at least one mapping of a vertex, $a$, to a point that isn't a vertex, $e$. Consider that on a polygon, points are either on vertexes, or lay on line segments that connect them, so $c$ must be on a point between two vertexes in this new isometry, $g$ and $h$. Suppose that $f(b) = h$ and $f(c) = g$, for points on the polygon where $a$ lies $b$ and $c$, and thus $d(b,c) = d(h,g)$. Thus, because they lie in a straight line with eachother, the property holds that $d(h,e) + d(e,g) = d(h,g)$, so $d(b,c) = d(h,e) + d(e,g)$. Thus, $d(b,c) = d(f(b), f(a)) + d(f(a),f(c))$, or just $d(b,c) = d(b,a) + d(a,c)$. Consider that this equality can hold if and only if $a$ was a point along the shortest line between $b$ and $c$. This implies that $a$ is a point that lies along the line connecting $b$ and $c$, and is therefore not a vertex, but this is a contradiction of our assumptions that it was. Thus, in an isometric function, all vertexes must map to other vertexes. "

Here we go!

[gentle osprey]

Corgwn

[cloud owl]

would this work? f(x) = x if x is rational and -x otherwise

[gritty widget]

yes, because |f(x)| = |x| for any x, so if x goes to 0, f(x) goes to 0 = f(0)

(just a quick way to see continuity at 0)

[cloud owl]

awwww yeah

[marsh forge]

if youre supposed to prove it dont forget to talk about discontinuity elsewhere

[gentle osprey]

Moth | not male

[fading vale]

so its easy here to show that Z^p is continuous and that f cont implies f Z^p cont

the hard(ish?) part is f circ Z^p cont implies f cont

where f is an arbitrary map into F(Y, Z)

i dont really understand how the preimage of a compact set = compact helps or what map it is supposed to apply to here cuz if f: A -> F(Y, Z) idk anything about the topology on A or how compactness relates to openness

unless its supposed to be like

continuity at ap oint or something

but if u pass thru Z^p from an open set of f(a) to F(X, Z) and then go back into A via (f circ Z^p)^-1 i dont think this would necessarily pass under f into a subset of our original open set

and it doesnt seem to take preimage of compact set is compact into account anyway

[wanton marsh]

what are the topologies on F(Y,Z) and F(X,Z) ?

[marsh forge]

compact open i assume

[broken berry]

Hey, could anyone pls verify if this 4d coordinate system projected to 3d is correct. I did the calculations of angles and modeling of it, so I don't know if it works. If someone could help me please send me a message.

[shy moss]

In the plane, is the topology induced by the discret metric equal to the usual topology?

[gritty widget]

in the discrete metric, points are open sets, and they certainly aren't in the usual topology

phrased more suggestively

in the discrete metric all sets are open

[gentle osprey]

TTerra

[gritty widget]

(and "all sets are open" follows from the fact that arbitrary unions of open sets are open, and that any set can be written as a union of singletons)

(and really, by "points are open," i meant "singletons are open," just to be 100% clear)

[sleek thicket]

am I crazy or does this definition not make sense

j < i
i >= j

The function isn't defined if j > i????

I'm pretty sure it should be
$s^i(j) = \begin{cases} j & \text{if } j \leq i \ j-1 &\text{if } j > i\end{cases}$

[gentle osprey]

shamroc$\overline{k}$

[sleek thicket]

But would like confirmation from someone else

@gilded shell i can try to help you determine the lie algebra

[gilded shell]

Ok, am I fine?

So

I'm not sure where I'm effing up

I have two ways to calculate it and it seems contradictory

G=GL(n), with real matrices.

[sleek thicket]

mhm

[gilded shell]

Let L_g multiplication on the left by g.

Now the vector V at T_I G, corresponds to what left invariant vector field?

edit: i->I

Yes

If we can identify the Matrices and the tangent space T_I G, as in, if: V=V_ij d_ij , we can say V = [V_ij]

[sleek thicket]

Sorry, what do you mean by d_ij?

Is this the matrix with a 1 in the ijth spot and 0s elsewhere?

[gilded shell]

I mean the partial derivative with respect to x^ij

[sleek thicket]

ah, gotcha

then I agree

(sorry, I wanted to clarify because d_ij looks to me like the covector dx^ij instead of the vector partial/partial x^IJ)

[gilded shell]

So two ways of calculating the desired object:

$L_g _*,e (V)$

Not sure why I'm not latexing

[sleek thicket]

Is * here the total derivative, ie pushforward?

And it gets confused when you have spaces at the start or end

[gilded shell]

Yeah, makes sense. But I dont have the symbol for partial here.

[gentle osprey]

Professor M'Oats
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[sleek thicket]

$(L_g)_*(V)$

This?

[gentle osprey]

shamroc$\overline{k}$

[gilded shell]

Yeah, I'll write it like that then

Thanks

[gilded shell]

It is

[sleek thicket]

Sure, so this is one way

[gilded shell]

So according to books we'll have:

$(L_{g})_{*}(V) = gV$

[sleek thicket]

Yup

[gentle osprey]

Professor M'Oats

[sleek thicket]

Do you understand why this is?

[gilded shell]

I dont know, because I'm getting a different answer someway else.

But maybe it's cuz I need sleep but I really need to finish this right now so no sleep allowed

[sleek thicket]

$L_g$ is the restriction of a linear map $M(n) \to M(n)$, namely left multiplication by $g$. The differential of a linear map is that map itself

[gilded shell]

But I think I do, to a degree.

[gentle osprey]

shamroc$\overline{k}$

[sleek thicket]

Also oof :(

We're basically working in M(n) instead of GL(n)

Since M(n) is a vector space

That's how we identify the tangent space with M(n) too, right?

[gilded shell]

Yes

[sleek thicket]

Alright so what's the other way?

[gilded shell]

Ok

So let's have g be the respective function G->G, g(h)=gh.

Another way to calculate our vector field should be this

$(L_{g}){*}(V) = V{ij} d_{ij} (L_{g})$

[sleek thicket]

hmm

[gentle osprey]

Professor M'Oats

[gilded shell]

By definition of pushforward

no?

[sleek thicket]

I think I agree with this, give me one sec

[gilded shell]

I have done a bit of identification

[sleek thicket]

so we get L_g(partial/partial x^ij)

Hmm I don't think I understand this

[gilded shell]

Isn't this literally the definition?

[sleek thicket]

Wait, is d_ij the tangent vector?

[gilded shell]

More rigorously
$(L_{g}){*}(V) {ab} = V{ij} d{ij} (L_{g}) {ab} d{ab}}$

[sleek thicket]

I got confused by the notational clash with covectors

[gentle osprey]

Professor M'Oats
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[gilded shell]

NO FORMS EVER

Always vectors

[sleek thicket]

lmfao

So if we think of partial/partial x^ij as a derivation on the algebra of smooth functions

[gilded shell]

Sorry for the confusion, btw, hope I solve this. ;-; Thanks

[sleek thicket]

The pushfoward is prrcomposition

no you're fine! It's just been a while since I actually went down to definitions of this stuff

and there's a bunch of different ways to define everything

[gilded shell]

So, all agreed?

[sleek thicket]

Alright so when you say $d_{ij}(L_g)$ do you mean $\frac{\partial}{\partial x^ij} \circ L_g^*$?

I'm still catching up lol

[gentle osprey]

shamroc$\overline{k}$

[sleek thicket]

The notation d_ij(L_g) does not parse for me

[gritty widget]

since left translation by $g$ is a diffeo you have a well defined pushforward $(L_g)*:\Gamma(TG)\to\Gamma(TG)$ where $X\mapsto (L_g)(X)$ where $(L_g)_(X):G\to TG$ so $h\mapsto (L_g)*(X)(h)=(\ell_g)*(Xg^{-1}h)$

[gentle osprey]

spinsicle

[sleek thicket]

oh wait

Yes it does

[gilded shell]

maybe this guy helps ∂

[sleek thicket]

I'm being stupid, sorry

The partial acts on the function L_g

🤦

Yeah alright professor I agree with your statement now, sorry for the confusion

[gilded shell]

Nah, you're doing fine

Reading spins'

[sleek thicket]

anyways prof, continue

You can also compute it this way, I agree

[gilded shell]

The g(-1) is confusing me a bit in spinsicle but I guess it's to obtain an automorphism or sth

Anyway, I'l ldothe computation

[sleek thicket]

This should also work

Spin, I don't understand your last expression. What is \ell_g?

[gritty widget]

if you consider a $X\in\Gamma(TM)$ as a linear map $X:C^{\infty}(G)\to C^{\infty}(G)$ then you can just write $(L_g)_*(X)(f)=X(f\circ \ell_g)$ for any smooth $f$

[gentle osprey]

spinsicle

[gritty widget]

\ell_g is just the left translation map

[sleek thicket]

yup, eventually I realized this

[gilded shell]

$(L_{g}){*}(V) {ab} = V{ij} d{ij} (L_{g}) {ab} d{ab}} = $
$V_{ij} d_{ij} (L_{g}) {ab}$
$=V{ij} d_{ij} (g_{ak} x_{kb}$
$=V_{ij} (g_{ak} delta_{ki} delta_{jb}$
$=V_{ij} (g_{ai} delta_{jb}$

[gritty widget]

why

[sleek thicket]

?

[gilded shell]

Does the bot support multiple $ $$$ ?

[sleek thicket]

I was being thrown off by the use of lowercase d to mean a vector instead of a covector

[gritty widget]

components

ughhghg

[sleek thicket]

It does, but you'll want an align*

[gilded shell]

How?

[gritty widget]

why use components

why torture yourself this way

[gentle osprey]

Professor M'Oats
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[gilded shell]

Because I'm getting it wrong

Because the formula I got doesnt correspond to L_g_*(X)=gX.

And I have to be fucking up royally somewhere

[sleek thicket]

Actually prof I'm back to being confused lol. You write $L_* \frac{\partial}{\partial x^{ij}} = \frac{\partial}{\partial x^{ij}} L$ but this seems wrong to me. The left hand side is a vector field while the right is a function

[gentle osprey]

shamroc$\overline{k}$

[sleek thicket]

Unless I've misunderstood your computation using linearity of the pushfoward

[gritty widget]

i think there's a bit of confusion about what L_* actually is here

[sleek thicket]

I'm confused at least

[gilded shell]

So L_g is a smooth map G->G. Which allows us to pushforward a tangent vector at I, to a tangent vector at gI=g.

[sleek thicket]

Sure

[gilded shell]

And V is a tangent vector at I.

[sleek thicket]

But the pushforward of d_ij isn't d_ij(L_g), right?

d_ij(L_g) is some function

[gritty widget]

do you remember what the definition of the pushforward is prof

[gilded shell]

It should be $V_{ij} (g_{ai} \delta_{jb} d_{ab} |_{g}$

[sleek thicket]

trailing space

[gilded shell]

The pushforward by F of v applied to f, = v(foF)

[sleek thicket]

shouldn't it be the vector field $\left(L_* \frac{\partial}{\partial x^{ij}}\right)f = \frac{\partial}{\partial x^{ij}} (f \circ L_g)$?

oops

[gentle osprey]

shamroc$\overline{k}$

[sleek thicket]

I think this is what you said prof

But latexed

[gentle osprey]

Professor M'Oats

[gritty widget]

what

also why is there a metric here what

[sleek thicket]

I am also confused by this notation

[gilded shell]

Isn't that the same?

[sleek thicket]

This is right

[sleek thicket]

This does not make sense to me

[gilded shell]

It's not a metric, it's a kronecker delta.

[sleek thicket]

I think the metric is the g lol

[gilded shell]

Yes, it does not, but I'm not sure why atm.

[gritty widget]

^

[sleek thicket]

Which is an invertible matrix, spin

It's an element of the group G = GL(n, R)

[gilded shell]

Oh the g is a matrix so it has components

[gritty widget]

yes i know but this is just like

[sleek thicket]

Okay so prof, you have $(L_g)* V = V^{ij} (L_g)*\frac{\partial}{\partial x^{ij}}\Big|_{I}$

[gentle osprey]

shamroc$\overline{k}$

[sleek thicket]

This is what you started with

Correct?

[gilded shell]

Yes

[sleek thicket]

Then you tried to compute the pushfoward

And lost me

I think your computation of the pushfoward was wrong

[gilded shell]

The pushforward was just calculating the derivatives of gX, which is the multiplication of two matrices

[gritty widget]

why are you even trying to compute pushforward

[sleek thicket]

I don't think it makes sense to talk about the derivative of a product of matrices with respect to X

[gilded shell]

Because I'm computing it wrong so I'm messing up somewhere I don't udnerstand

[sleek thicket]

Also spin they tried to compute something two ways and got different answers

Even if one way is easier you should still try to resolve the contradiction

so yeah prof have you tried computing the pushfoward by a curve?

[gilded shell]

G is a a group of matrices. Multiplying the matrix g on the left gives a smooth map, which we can take the derivative of.

I meant this in reply to how it doesnt make sense to calculate the derivatives of so and so

[sleek thicket]

partial/partial x^ij is the tangent vector of a very easy to write down curve

I agree, but that derivative is the pushfoward map

I was saying you can't do like, d/dX gX

anyways, I would try to compute the pushfoward by using the curve γ(t) = I + tx^ij

The velocity of this at 0 is partial/partial x^ij

so the pushfoward can be computed as the velocity of the curve g * γ(t) at 0

[gilded shell]

The curve being the (I+e^ij t).

[sleek thicket]

Yes

[gilded shell]

Any reason why?

[gritty widget]

why don’t you just play around with this on the tangent bundle before you do all this 😦

[sleek thicket]

Say F : M -> N is a smooth map. What is the derivative of F?

Also spin they know how to compute it by identifying the tangent space with M(n)

[gilded shell]

Ah, I have to leave soon

[sleek thicket]

That's how this question started

Prof, if you want to compute it like this I would suggest using that curve

[gilded shell]

I will, can I ask for a specific case on how you'd do it?

[sleek thicket]

sure

[gilded shell]

I'll compute it, I just wanna say it because of time sorry

You have the block matrix with rows (R_s v)(0,1). Where R is the 2x2 matrix (cos(s) -sin(s))(sin(s) cos(s)) and v is 2x1 (x)(y).

It's really simple if you draw it out.

[sleek thicket]

uhhh

[gilded shell]

How do you prove X= partial/partial s is an invariant vector field?

[gritty widget]

by applying the definition of a left invariant vector field

look at this abstractly

[sleek thicket]

I do not know what you mean by this. What's v? The matrix
x x
y y?

[gritty widget]

and understand it then do the computation

[gilded shell]

I did it in two ways, none was direct direct computation.

[gilded shell]

No
The matrix is
( cos s -sin s x)
(sin s cos s y)
(0 0 1)

[sleek thicket]

I thought we were working in GL(n)? How are we getting a nonsquare matrix?

[gilded shell]

How so?
$(L_{g})_{*} ( \partial _{s})(f) = ( \partial _{s})(f o g)$
This is the one by computation.

[gilded shell]

It's 3x3

[sleek thicket]

Oh, I didn't see the third row somehow

So you have this matrix R_s

[sleek thicket]

This one

What's the vector field you want to show is left invariant?

[gilded shell]

partial _s

[gentle osprey]

Professor M'Oats

[gilded shell]

is wrote theta sorry

[sleek thicket]

I don't know what that means

is s a coordinate?

On GL(3)?

[gilded shell]

It's not a coordinate on GL(3). IT parameterizes the subgroup given by the subgroup i was trying to define

I think I need to take a break, still need to sleep and wake up

And I'm being a unclear

Thank you though

Sorry that I have to leave midway

[sleek thicket]

You should do that, it seems like you have stuff getting mixed up/not being communicated clearly

[gilded shell]

You too spin

[sleek thicket]

I hope you can get your homework done in time

[gilded shell]

thanks

[sleek thicket]

Would appreciate if someone looked at this question I asked about the simplex category earlier

[tight agate]

yes, you defined it the right way

[sleek thicket]

cool

this is not the first typo from this pdf lol

[tight agate]

Kerodon is a great reference for this stuff

[sleek thicket]

Oh yeah I remember hearing about that, thanks!

[tight agate]

and Goerss and Jardine

[sleek thicket]

I'm trying to focus on just what I need to understand dold kan, so I didn't want to use a more general homotopy theory reference

even weibel's presentation is a little more general than necessary for me atm

[tight agate]

what are y'all doing in your hom alg class right now?

[sleek thicket]

Just starting derived functors

A bunch of people in the course are concurrently taking it with the first year grad algebra sequence

So there's been more review than I would like

[tight agate]

review can be a good thing

[sleek thicket]

i do not want to relearn the definition of a natural transformation for the 20th time

[tight agate]

lol

there was a lot of overlap between the undergrad and first-year grad algebra sequence, but I think I got to learn the material a lot better due to it

but yeah, twice is fine

20 times is not

have fun simping

I should read through the simplicial stuff again

[sleek thicket]

I would like to learn more about this stuff

Haven't really had an opportunity

But presenting on dold kan seemed like a nice way to get into it

[tight agate]

it seems like it might be super important 🤷

with all the "brave new algebra" stuff that's coming up

[sleek thicket]

¯\_(ツ)_/¯

I've always been a bit of a category memer

so I just want to understand what all the higher category business is about

[tight agate]

I want to be a category memer

but it is so dry

[sleek thicket]

but until now I've suppressed my instinct to dive into simplicial stuff

this is true lol

It's weird because like

All the category theorists I talk to are the opposite of dry

They think about things at a less formal level than I'm comfortable with

But then the actual math is incredibly formal

I don't know how to square it lol

All the category theorists I've ever talked to are "ideas people" if that makes sense

[tight agate]

yeah I get what you're saying

we've got those over here too

[sleek thicket]

im not talking about like ugtc memers

[tight agate]

ik

[sleek thicket]

kk, sorry

[tight agate]

lol

a lot of the profs are category memers too

[sleek thicket]

For some reason "we've got those over here" makes me think of ugtc memes

[tight agate]

perhaps I should have worded it better

[sleek thicket]

I remember my algebra prof telling a story about like

back in the day he was a category theorist memer

he had to introduce the derived category at the start of his talks

and then as time went on he became much more concrete

Not because he'd changed, but because algebraic geometry had

[tight agate]

huh

[tough imp]

wait what

really?

[sleek thicket]

Sandor told me about this

Yeah

[tough imp]

I mean

I asked him if I need to know cat theory for algebra and he was like

[sleek thicket]

I remember specifically he said he used to need to introduce the derived category at the start of his talks because the audience might not be familiar

[tough imp]

"well I used the derived cat in my thesis"

lol

that's crazy

how times have changed

[tight agate]

well yeah no more derived cats

[sleek thicket]

lmfaoo

[tough imp]

hahaha

[sleek thicket]

Now it's infinity cats

[tight agate]

only E-infinity ring spectra /s

[tough imp]

Tfw $\mathbb{R}^i$

[gentle osprey]

Chmonkey 2.0

[tough imp]

that's such a meme name

"hyper derived functor"

I need that meme like "hyper derived functor? what is that. No, derived functor"

[tight agate]

anything with hyper sounds memey

[tough imp]

and it calls it like stuck up or whatever

[tight agate]

except hyperplane

or hypersurface

[tough imp]

hyperline

hyperline is a dim 1 subspace

aka a line

[tight agate]

hyperbole

[sleek thicket]

okay good night everybody. chmonkey we're meeting at 1pm

[tough imp]

yes

We are

[sleek thicket]

read the thingy

[tough imp]

yes

[sleek thicket]

before 1pm

[tough imp]

Read the thingy

[tight agate]

yes

[tough imp]

before the 1pm

do the manifolds

read the thingy

die

[sleek thicket]

:(

[tough imp]

👍

[obtuse meteor]

based

[shy moss]

How to show tha the fundamental group of an indiscrete space is trivial?

[gritty widget]

try showing that any loop in an indiscrete space is contractible

[cedar totem]

What is a tangent space? Is it like a tangent line but only a surface that is touching a single point in R^3

[gritty widget]

it's the best "linear" approximation to the curve/surface/manifold near that point

which should immediately make you think of derivatives

[shy moss]

A loop is contractible if it is homotopic to a point?

[gritty widget]

yes

in particular this implies that the fundamental group of the space (based at a point on your loop) is trivial

since then every loop based at that point is homotopic to the constant loop, i.e. its homotopy class in pi_1 is equal to the identity (homotopy class of constant loop)

(if i'm remembering my basic AT correctly)

so once you can construct that homotopy, you're good

[shy moss]

For a loop f in X, the homotopy is $f \cdot c=c$ where c is constant loop, right?

[gentle osprey]

Go

[marsh forge]

?

you want to show that if X has the triv topology that any map S^1->X is homotopic to a constant map to the basepoint

[shy moss]

I want to show that the fundamental group of a indiscrete space is trival

[marsh forge]

do you not see that those are the same thing?

Oh I guess sometimes this is first taught as special maps I->X

[shy moss]

No

[marsh forge]

rather than S^1

an element of pi_1(X,x) is a map I->X with both endpoints sent to the basepoint x

you want to show that any such map is homotopic to the constant map

[hushed plover]

I have a problem I'm working on

[gritty widget]

don't we all?

[marsh forge]

be patient someone else is asking a Q rn sorry

[hushed plover]

sorry

[shy moss]

This is because $f \cdot c=c$ right?

[gentle osprey]

Go

[marsh forge]

what do you mean by this

[shy moss]

The homotopy between f and c is $f \cdot c$?

[gentle osprey]

Go

[marsh forge]

thats not a homotopy

thats a composition

i still don't know what you mean

in general $f\cdot c \simeq f$ is true, but $f\cdot c\simeq c$ is equivalent to $f\simeq c$ which is what you want to show in the first place.

[gentle osprey]

doja max

[shy moss]

Ooooo, I get it

[marsh forge]

oh ok lol

[hushed plover]

is it alright if I ask a question now?

[marsh forge]

i guess? i definitley didnt answer Go's original

but go for it ig

[hushed plover]

I have two points on a sphere (x1, y1, z1) and (x2, y2, z2) and I want to have a quadratic bezier curve to go between them. I want the arc to go over the sphere instead of inside the spehere

I made the sphere transparent so you can see that it goes inside the sphere instead of on top of it

currently I am doing this to find the mid point for the bezier curve

midX = ((startPoint.x - endPoint.x) / 2);
midY = ((startPoint.y - endPoint.y) / 2);
midZ = ((startPoint.z - endPoint.z) / 2);

but how do I find the midpoint of the two points so that it is above the sphere

[chrome dew]

you can work out the midpoint on the sphere with some trigonometry, draw a circular cross section that passes through those two points to get started

well, actually no you don't need to do that if you already have the midpoint

just normalize the vector from the center to the midpoint

then multiply it by the radius of the sphere to put it on the sphere

[cedar totem]

What is TpM?

Tangent plane.

[gritty widget]

yes, T_pM is (one) notation for the tangent plane to M at p

are you asking what it represents?

are you asking how to define it?

[tough imp]

What’s T_pTTerra?

[gritty widget]

a line

i'm a curve

[sleek thicket]

imagine only being 1 dimensional

this post made by surface gang

hey ttera what are your sectional curvatures? Oh that's right you don't even have them

[gritty widget]

[shut moat]

is ur username some double tangent bundle meme

[gritty widget]

no

[shut moat]

[gritty widget]

there

now it's a tangent bundle meme

actually cotangent bundles are cooler

😌

i am now the cotangent bundle of Terra equipped with its canonical symplectic form

[cloud owl]

idk what that is but it sounds terrable

[gritty widget]

lmao they really gave that definition instead of just saying "union of duals to tangent spaces"

[shut moat]

NICE

symplectic geometry seems rly cool

fancy ass classical mechanics stuff

although I've heard that apparently some huge theorem might actually be wrong

[gritty widget]

do you know which?

[shut moat]

let me find it again one sec

yeah lol

[gritty widget]

gromov

i will read in a moment

finishing up a lie gp problem

[elder yew]

Entire theses are written off of sentences of gromov

Cuz he doesn't fucking fill in any details

[cloud owl]

i don't think it is

uh yes

but ctrl-f gromov gives one match, right at the start

[broken berry]

could someone please explain me what is double ortographic projction?e

I know what a ortographic projection is, but the double doesn't make sense, anybody could help me?

[sleek thicket]

@gritty widget so CP^1

Tangent spaces have the structure of complex multiplication

Right?

Like they're 1d complex vector spaces

[gritty widget]

sure

[sleek thicket]

Okay so

There's this very fancy line bundle on CP^1

Line bundle means 1d vector bundle

In this case complex

Define $L = { (\ell, v) \in \C\mathbb{P}^n \times \C^{n+1} : v \in \ell}$

[gentle osprey]

shamroc$\overline{k}$

[sleek thicket]

So the fiber over ell is the line ell itself

It's all pairs (ell, v) where v is a vector on the line ell

And it's a subbundle of the trivial bundle CP^n × C^(n+1)

Does this make sense?

[tight agate]

petthecat

[gritty widget]

[sleek thicket]

ttterrrrrra answer me

[gritty widget]

sure i guess so

[sleek thicket]

Okay so

Weird bundle

I'm supposed to show $\C\mathbb{P}^1 \cong L^\otimes L^$

[gentle osprey]

shamroc$\overline{k}$

[sleek thicket]

Wtf why is this true

Well it comes down to computing transition functions

[gritty widget]

check the errata

[sleek thicket]

Like you can cover by two affines patches

Where you normalize each coordinate

Yeah?

Turns out these are both trivial over those

So you just need to know they transform on the overlaps the same way

I ended up computing a transition function for the tangent bundle

It's just the jacobian of the change of coordinates map

Ez

Map is f(z) = 1/z

So f(x, y) = (x,-y)/(x^2 + y^2)

,w Jacobian matrix of (x, - y)/(x^2 + y^2)

wait this is wrong lol

[gritty widget]

is WA okay

[sleek thicket]

Fixed

[gentle osprey]

Results provided by WolframAlpha

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

Requested by shamrock#0598

[sleek thicket]

Right so

This is 1/|z|^4 times the matrix
y^2 - x^2 | - 2xy
2xy | y^2 - x^2

I go

"aha this is a complex number!"

Cause y^2 - x^2 is the real part of z^2 and 2xy the imaginary part

So this matrix is multiplication by z^2/|z|^2 = 1/(z conj)^2

But oh wait it's supposed to be 1/z^2

Fuck

I've been stuck on this for like a week mind you

today I realize (y^2 - x^2, 2xy) is not z^2

Re z^2 = x^2 - y^2

So it's actually -(z conj)^2

[gritty widget]

[sleek thicket]

And suddenly everything works out

Fml

I wasted a week on squaring a complex number wrong

[gritty widget]

[sleek thicket]

Such is math

[tough imp]

oof

[cedar totem]

is ttera a bot?

[tight agate]

yes

[shut moat]

yes

[gritty widget]

yes

[gritty widget]

[sleek thicket]

/TTera

oops, typo

/TTerra

[sleek thicket]

[sleek thicket]

im still working on this fucking stereographic projection problem

it is like

a hydra

every time I fix one sign error

or issue with flipped orientation

[honest narwhal]

Let us c

[sleek thicket]

or backwards transition function

another one pops up

i feel like I am being stereographically projected

conceptually it is easy

just check it on affines/stereographic projection coordinates

but holy shit it is unbelievably finnicky

[honest narwhal]

I'm guessing 15?

[sleek thicket]

sorry yeah it's 15 but the statement of 15 depends on 14

so you sort of need to prove 14 too

(15 was assigned, 14 wasn't lol)

anyways, i now have a minus sign

finally got rid of that conjugation

im gonna scream

i screamed

im also just not including any computations

because they are so fucking long

i have a sage notebook

checking things work as I go

wow isn't this insightful

P.<x,y,z> = ZZ['x, y, z']
R.<xx,yy,zz> = QuotientRing(P, P.ideal(x^2 + y^2 + z^2 - 1))

X1_1 = 1 + zz - xx^2
X1_2 = -xx*yy
X1_3 = -xx*(1 + zz)

X2_1 = -xx*yy
X2_2 = 1 + zz - yy^2
X2_3 = -yy*(1 + zz)

Y1_1 = 1 - zz - xx^2
Y1_2 = -xx*yy
Y1_3 = xx*(1-zz)

Y2_1 = xx*yy
Y2_2 = yy^2 - 1 + zz
Y2_3 = -yy*(1-zz)

print(bool(X1_1 * (1-zz)^2 == (yy^2 - xx^2) * Y1_1 + 2*x*y* Y2_1))
print(bool(X1_2 * (1-zz)^2 == (yy^2 - xx^2) * Y1_2 + 2*x*y* Y2_2))
print(bool(X1_3 * (1-zz)^2 == (yy^2 - xx^2) * Y1_3 + 2*x*y* Y2_3))

print(bool(X2_1 * (1-zz)^2 == (yy^2 - xx^2) * Y2_1 - 2*x*y* Y1_1))
print(bool(X2_2 * (1-zz)^2 == (yy^2 - xx^2) * Y2_2 - 2*x*y* Y1_2))
print(bool(X2_3 * (1-zz)^2 == (yy^2 - xx^2) * Y2_3 - 2*x*y* Y1_3))

[honest narwhal]

So L*\otimes L* pulls back to the holomorphic tangent bundle basically

[sleek thicket]

right

except i don't know anything about complex manifolds

so

i just have vague intuit that this should be thought of as the holomorphic tangent bundle

[honest narwhal]

I wonder if there's some weird sneak

Like oh knowing that they're isomorphic as real bundles and some not too hard extra condition

[sleek thicket]

maybe with more machinery ¯_(ツ)_/¯

i mean

[honest narwhal]

Gives that they're isomorphic as complex bundles

[sleek thicket]

it is not easier to show they're real as real bundles

tbh

there's just a lot of places to choose the wrong one of two options

or get the wrong result out of a computation

orientation, conjugation, signs

i know there's a lot of them because ive done it wrong at every possible step

ahhhhh

anyways, i have a minus sign and I must scream

[honest narwhal]

Yeah that's fair I guess. At least there are only two charts 🙃

[sleek thicket]

yee

except

I chose the wrong charts to start

my original charts had the flipped orientation

so I had to redo all of these horrendous computations in spherical coordinates

[honest narwhal]

Okay so I guess my conjecture that I am guessing is wrong because that's way too much plot armor is that if your iso of real bundles preserves orientation then it's an iso of complex bundles

[sleek thicket]

lol

okay so

assume this is true

this is what I checked with sage

so

probably true

do you see an error in this computation?

[honest narwhal]

Honestly my eyes are glossing over a bit

[sleek thicket]

yeah uh

mine too

🙃

tfw due tomorrow

[honest narwhal]

So what's our def of pullback again?

[sleek thicket]

i haven't gotten there yet lol

that's actually the easy half

but it's the pullback as topological spaces

[honest narwhal]

Or okay what's this computation doing conceptually exactly?

I just see two lines of notation

[sleek thicket]

figuring out the transition function

for the stereographic coordinate charts on S^2

[honest narwhal]

Oh it's just 1/z I think

[sleek thicket]

it is not

err

[honest narwhal]

z complex

Not S^2\subset R^3

[sleek thicket]

do you mean the transition function as a manifold?

I'm computing the transition function of the tangent bundle

[honest narwhal]

Right right

[sleek thicket]

so it is that

which means it should be -1/z^2

err wait

hang on

that's what I got

wait wtf

the sign should be here

but uhhhh

won't the other one not have this sign????

okay maybe I should proceed, talking about this has reassured me

I think my computations are right at least

[honest narwhal]

Does this need to be done? You know the tangent space to a point in S^2 is just the orthogonal complement of its span

[sleek thicket]

...yes?

i don't see the relevance

the tautological line bundle L is trivial over affine opens

which are the same as stereographic projection charts

so my instinct is to compute the transition functions of (TS^2)' on the stereographic projection charts

and then the transition functions of L^* (x) L^* on affines

[honest narwhal]

I'm just thinking you could work with the explicit description of things in S^2 \times R^3 such that blah blah blah

[sleek thicket]

i did that earlier

[honest narwhal]

And possibly just write down a map straight up

[sleek thicket]

oh, I had that thought

like

triyng to define a global map

but I don't see it

now I even have an isomorphism locally defined

and I still don't understand it

if you can explain to me why this isomorphism exists I will be very happy

[honest narwhal]

I'll think a bit and see if I can cheese it

[tough imp]

well clearly

the thing to do is to cry

:(

[sleek thicket]

ahh i think i see where the minus sign will come in

probably maybe

[honest narwhal]

Okay so what's the tautological bundle again? Is that just like

CP^1 is a quotient of C^2

[sleek thicket]

the fiber over ell is ell

[honest narwhal]

Okay yes

And now we're taking the dual of it

[sleek thicket]

yes

if you like AG

then apparently $L = O(-1)$

[gentle osprey]

shamroc$\overline{k}$

[sleek thicket]

but i think things are too floppy/differenteo-geometric to make sense of O(-1)

I thought about it

[honest narwhal]

So apparently tangent bundle is O(2) or smth

[sleek thicket]

right

[honest narwhal]

I mean complex setting you can actually make it work I'm pretty sure

GAGA

But that's not the intent

[sleek thicket]

yeah, probably

[honest narwhal]

Do we have a nice explicit description of L*?

[sleek thicket]

i couldn't think of one

[honest narwhal]

Oh it's literally the dual space at each point so I guess like

[sleek thicket]

yeah

[honest narwhal]

(\ell, \ell*) where \ell* is a linear functional on \ell

[sleek thicket]

so there is this https://en.wikipedia.org/wiki/Euler_sequence

but like

[honest narwhal]

I doubt we need AG here

[sleek thicket]

no i mean

an AG head told me this is morally how you should understand the iso

but it did not help me

[gentle osprey]

Sloth King Daminark

[honest narwhal]

I guess

I'm tired so implicit disclaimer in everything I say

[sleek thicket]

sure

i mean like

you can also think of it as bilinear forms

on x

[honest narwhal]

Tru

[sleek thicket]

also also

[honest narwhal]

So okay time for the nightmare lmao

[sleek thicket]

it suffices to find a map $T \otimes L \otimes L \to \C$

[gentle osprey]

shamroc$\overline{k}$

[sleek thicket]

probably with conditions on it idk

a bundle iso

but basically this is just

how do you evaluate

[honest narwhal]

God I hate that this map is piecewise defined it makes it painful to give a description of the pullback bundle in symbols

[sleek thicket]

yes

[honest narwhal]

So wait what exactly is this map? In the nice case if we say S^2 is C \cup infinity

Then there's the map we like to CP^1

[sleek thicket]

right so your question is really

[honest narwhal]

Is this that written funny or is it genuinely different?

[sleek thicket]

what is the local representation of this map

in stereographic vs affine coordinates

it is doing the thing you're thinking of

but like

there's some choices you could make

like is infinity [0:1] or [1:0]

is the north pole infinity or the south

etc

[honest narwhal]

Yeah... Okay so conceptually what do we want to happen right now? If you give me a point on the sphere I stereo project to a complex number, let's say north pole or smth idk, whichever makes it work right

[sleek thicket]

right, that's what it does in one of the two cases

and in the other you stereographically project onto a different plane

but get the same result

idk

[honest narwhal]

I'm basically trying to think of whether in words I can say "Oh this map is just apply the bilinear form to somebody!"

But it's weird for sure lol because we're hopping between real and complex

[sleek thicket]

yes

i spent a lot of time on this

before giving up and doing it locally

somehow the local proof is even worse than I was expecting

[honest narwhal]

Yeah you should probably do things locally, I'm gonna go to sleep soon anyway I was just hopeful. I like giving nice one shot descriptions

[sleek thicket]

👍

[honest narwhal]

I mean okay let's revert the complex description of the tautological bundle over CP^1 back to the real case right?

[sleek thicket]

this problem sucks so fucking bad lmao

im gonna

not do that

and actually work on this problem

let me know if there 'sa nic eproof

[honest narwhal]

I wonder if duals and pullbacks play nicely

And whether you can say something like

[sleek thicket]

They do

Err

I was thinning duals and tensors

[honest narwhal]

Oh the dual of this pullback bundle is the holomorphic cotangent bundle of S^2

[sleek thicket]

But they should all commute

Because f is a diffeomorphism

Yeah I was originally just working with TCP^1

But I decided I don't want to justify going back and forth

[honest narwhal]

Fair

Yeah that's all that 2-3AM Daminark's got, sorry

[sleek thicket]

np lol

[honest narwhal]

It might be the case that there's no way around just doing garbage

:/

[sleek thicket]

Hopefully I finish soon

I shouldn't have put this off

(it's due tomorrow)

man today was incredibly shitty lmao

[honest narwhal]

Good luck. Is that your last problem?

[sleek thicket]

Yup

[honest narwhal]

That's something. When do you have to wake up?

[sleek thicket]

9:30 class lol

so like

9am

[honest narwhal]

By virtue of the pset or smth else?

[sleek thicket]

Analysis class

[honest narwhal]

Disgusting, I have one of those too next semester

[sleek thicket]

I had a little bit of a meltdown in #advanced-analysis earlier

my grader is insane

I got 3/10 on a correct solution

They completely failed to understand my proof

The feedback was incredibly condescending

[honest narwhal]

That's obnoxious

[sleek thicket]

Yes

I'm going to go to the prof

and maybe drop the class

This is not the first time it's happened

[honest narwhal]

Lol maybe your grader's just a moron. It happens

[sleek thicket]

And the program dirextor is looking into the class

[honest narwhal]

I mean maybe your prof can be like

Please get a brain

[sleek thicket]

because of how many complaints there are with the class

[honest narwhal]

And then you don't have to drop. But idk that's prob a pipe dream

[sleek thicket]

Last quarter the prof addressed it and backed up the grader

[honest narwhal]

If you do have to drop what are you gonna do?

[sleek thicket]

But hopefully I can change her mind

Learn functional analysis in grad school and hope really hard this doesn't look bad on my grad school app

[honest narwhal]

Is it past the drop deadline?

[sleek thicket]

I got a 3.8 in the first quarter of this class and have like a full year of 4.0s in grad courses

So dropping this during covid times shouldn't look crazy

And yeah I think I'd get a W

[honest narwhal]

Chicago gives 3 weeks before giving Ws lol, that's unfortunate

[sleek thicket]

We just finished week 3

So maybe there's time if I do it tomorrow

[honest narwhal]

Did you start January 2nd or smth?

[sleek thicket]

Ah it looks like it's only the first two weeks

That you can drop without a W

We started on the 4th

[honest narwhal]

Jeez

Washington seems fucking annoying wrt technical matters

[sleek thicket]

yes

Very much so

[honest narwhal]

Well all I can say is good luck at this rate

And good night

[sleek thicket]

ty

and night

[red garden]

if X is a connected top space

then for any proper usbset of X , boundary is nonempty

[sleek thicket]

@honest narwhal I think I figured out why this problem is so fucking confusing/hard

The map f is orientation reversing

WHAT THE FUCK

Why would you do that!!!

Ended up getting the negative conjugate of what I should have gotten

It doesn't matter

I give up

[red garden]

help with number 2 please?

[bleak helm]

Can you do the forward direction?

[red garden]

no

@bleak helm

nothing

suppose X is connected? what then lmao xd

let x and y be the two points

[bleak helm]

$X \subseteq X$

[gentle osprey]

Lunasong

[red garden]

oh

okay

now how about the other way

can we

wait

can we like

'induct'?

like just keep running over all the two points

in the space

and then take unions

(union of connected is connected)?

[bleak helm]

1 sec, sorry

That can work. Do you have an exact result about the union of connected sets being connected?

Because they must have a point in common

[red garden]

idk umm

disjoint union?

idk

[bleak helm]

Do you or do you not have a result about unions of connected subspaces? Because if you don't, you can't use it without proving it first

[red garden]

i do

idk the proof tho

[bleak helm]

What does it say?

[red garden]

ohh

i need nonempty intersection

[bleak helm]

If you have already proved it, you don't need to prove it again, you can just use it

Yes

[bleak helm]

.

[red garden]

how can i find this point lol

[bleak helm]

Just fix any point in X

Say x

Then choose subspaces that all contain x so that you can take the union to give you X

[red garden]

yea

got it

tysm

the problem is done right?

[bleak helm]

Is it? Idk. What did you do?

What subspaces did you choose to take the union of?

[red garden]

[Y_n] = {Y_i is a subspace | Y_i contains x} ?

[bleak helm]

How do you know the union equals X?

[red garden]

i dont lmao thats the problem

umm

[bleak helm]

Let y be a point in X. Can you show that y belongs to the union?

[red garden]

ohh