This isn't a joke, I think a lot of math is justified in itself. Geometry is interesting because I like shapes. Number theory is interesting because the integers behave in bizarre but still vaguely predictable ways
#point-set-topology
sleek thicket
You can argue that this is bad and math needs external justification, but I think people tend to apply this principle selectively
gritty widget
how dare you do something because it's interesting to you!?
i don't like it therefore you can't do it
(/s)
sleek thicket
tterra I'm on your analysis TAs side now, you deserve it
sleek thicket
based
gritty widget
what i submitted on the final problem set is what im submitting
sleek thicket
> submitting a screenshot from the student guide
gritty widget
multiple people have already emailed the department and theyre going to look into it
shut moat
I'm still stuck on the "on which topology" thing
like there is no way the TA was serious
gritty widget
:)
shut moat
like that has to be spite
sleek thicket
:)
obtuse meteor
which topology of what?
gritty widget
this person wasnt specifically talking about that course but, it was in the course's group chat, so probably
sleek thicket
Which topology the determinant is continuous on
obtuse meteor
b-bruh
sleek thicket
Apparently this was unclear to the TA, faye
anyways tterra yall should start a Google doc or something
gritty widget
someone wrote "det is continuous" and lost marks for not specifying what topology
sleek thicket
Put everything on there
gritty widget
look i already started a list of infractions early in the semester but they picked up so quickly i had to stop
sleek thicket
email the dept and ask for a refund or whatever 
gritty widget
my last complex analysis problem set has three short problems
and then three extremely long problems
ugh i dont want to check these
gritty widget
sleek thicket
shut moat
do some nice diff geo to revive you 
sleek thicket
Nice!
gritty widget
one letter away from ugct
sleek thicket
I'm typing up a long 
post on the discussion board for my cs course
I just went to OH
Showed up at 6pm, the start
They didn't get to me by 7pm
And this was the 2nd OH I tried to go to today
So now I have to ask convince them that the homework is wrong via the discussion board
Which is going to be a huge pain
fading vale
someone wrote "det is continuous" and lost marks for not specifying what topolog...
????????????????????????
????????????????????????????????????????
????????????????????????????????????????????????????????????
gritty widget
are u gonna upgrade from "firebomb?"
gritty widget
:)
i need to add the disclaimer
that this person was not specifying what course this was in
but
it's highly likely it was analysis :)
fading vale
thats so fucking bad holy shit
gritty widget
im losing good things to say about this school's math undergrad lmao
fading vale
god.
gritty widget
hold on
fading vale
thats literally so fucked up holy shit
gritty widget
completely normal grade distribution on a problem 
ah yes
93% of the class made the same mistake
||"mistake"||
obtuse meteor
so
I have no idea how to compute
tensor product of modules lol
namely I need it over abelian groups
oh wait hm
I claim
and this feels right
that tensor product of G with Z/2Z is just G + G
no
G + G/2G?
nah but it's like a graded thing
idk how to write this with the structure theorem is the problem rip
this really belongs in #groups-rings-fields but I'm doing it for topology
tight agate
it should be just G/2G
obtuse meteor
sounds like effffort
tight agate
them R/I (x) M iso to M/IM
obtuse meteor
mhm
tight agate
where the tensor product is taken over R
tight agate
the proof is pretty much "choose the correct exact sequence and tensor with the right thing"
obtuse meteor
hmm
idk how tensor preserves exactness
or when it does
is tensor like always left exact?
tight agate
right exact
marsh forge
I believe it should be Z/2 in all degrees leq n
obtuse meteor
oh yeah I cannot compute lol one sec
sleek thicket
A more low tech proof would be to use yoneda
fading vale
for a moment i thought you meant the homology sham and i was so confused
sleek thicket
Oh lmao sorry
I meant show that M (×) R/I has the same universal property as M/IM
I guess this is just proving it preserves quotients, which is short of showing right exactness
fading vale
hm cant you take ||0 -> I -> R -> R/I -> 0 and tensor by M? you get in general that I otimes M -> A otimes M -> R/I otimes M -> 0 is exact so if you prove that the first map is injective you're done?||
fading vale
oh wait yes i am silly
tight agate
IM is the image of the map
tight agate
but yeah that's the proof
fading vale
I otimes M is not always isomorphic to IM anyway 
so if it was exact you would have some problems lol
obtuse meteor
@marsh forge I think you might be wrong when n = 1
then I think H_1(RP^1, Z/2Z) = Z/2Z + Z
frosty sundial
(co)homology with coefficients in a field F will always be a vector space over F
so you shouldn't have the Z term there
frosty sundial
i think that's the reason :(
shut moat
why tf isn't ultraproduct honorable btw
frosty sundial
we are trying
ive suggested him several times
the last reason was that they weren't going to promote anyone new while they were reworking the roles
now that the roles are reworked, maybe some action will actually happen soon
gentle ospreyBOT
obtuse meteor
it should be just Z/2
yeah I figured out the error in my calculation
dusk heron
The thing under the red line must surely be a typo, no?
The negative sign should be on the y^k instead of the x^k.
Well, I guess multiplication by $-i$ is also a complex structure on $\mathbb{C}^n$, but the standard complex structure is here referring to multiplication by $i$.
gentle ospreyBOT
gustavn64
dusk heron
@gritty widget Well, if the vector $(x^1,\dots,x^n,y^1,\dots,y^n)\in\mathbb{R}^{2n}$ corresponds to the vector $(x^1+iy^1,\dots,x^n+iy^n)\in\mathbb{C}^n$, then the complex structure should take it to $i(x^1+iy^1,\dots,x^n+iy^n)=(-y^1+ix^1,\dots,-y^n+ix^n)\in\mathbb{C}^n$, which corresponds to $(-y^1,\dots,-y^n,x^1,\dots,x^n)\in\mathbb{R}^{2n}$.
gentle ospreyBOT
gustavn64
dusk heron
Yeah, but on the previous page it seems like they're talking about i.
The book is "Foundations of Differential Geometry, Volume II" by Kobayashi and Nomizu
plush needle
If I have a space that is both arc and path connected, and given that hyperconnected spaces need not be path connected and ultraconnected spaces need not be arc connected, is there a stronger named condition that this space fulfils, like superhyperconnected?
gritty widget
strong heron
This is the standard delta-complex structure on RP^2. My question is what goes wrong if we reverse the direction of c in this diagram?
honest narwhal
Does something go wrong? Do a handstand and look at it
strong heron
I mean, will it still be a delta-complex structure? I'm not very comfortable understanding this.
gritty widget
it will be
and it'll still be RP² is what sloth said (just swap U and L after reversing c)
strong heron
Ah, I see. Thanks both!
fading vale
pull a francaise and just call it miconnected
long hornet
I want to show that, given any two points p, q on a connected topological manifold M, there is a chart domain containing both. I tried to use the fact that M is path connected and used a path p joining them, then I have a series of charts (V_t, g_t) for each point p(t). Now I let V be their union, and tried to show that it is a chart domain. But I am not sure about that.
There is no obvious way to patch the g_t's together
sleek thicket
A connected chart domain?
long hornet
Oh, well, by chart domain I really mean a "coordinate ball"
sleek thicket
Ah okay
long hornet
So yes
sleek thicket
In the terminology I'm familiar with the codomain of a chart can be any open subset of R^n
This seems very very hard
And possibly not true
long hornet
In the terminology I'm familiar with the codomain of a chart can be any open sub...
True, that's the same terminology I know.
sleek thicket
I want to show that, given any two points p, q on a connected topological manifo...
Is there a reason you think this is the case?
long hornet
Well, I know it's true for smooth manifolds (saw an answer online)
gritty widget
sleek thicket
long hornet
There is no obvious way to patch the g_t's together
Well, I thought of a Zorn's lemma approach where I take "maximal charts" (I am not sure these exist) V_p and V_q, and assuming the claim is false, use these to give a separation of M.
shut moat
oh wait this is smooth
sleek thicket
I don't see at all why a chain of charts would have union a chart
Np approx I was asking for thst
Yeah, I think this same argument works
long hornet
oh wait this is smooth
The argument works though
sleek thicket
So to prove there's such a homeomorphism
Take a path connecting them
Cover by finitely many charts
long hornet
Cover by finitely many charts
Brilliant
sleek thicket
Then do it in each chart
tight agate
woah
long hornet
was just looking at a stack exchange answer
This is hard to follow, but it assumes that we know how to construct connected neighborhoods.
shut moat
yeah I haven't gotten my head fully around it, but I'll think about it more after I finish this last min pset 
sleek thicket
I outlined how to do so, right?
long hornet
The proof I saw uses the "tubular neighborhood theorem". I think the idea is to take a path joining them and then a small neighborhood around it.
long hornet
I outlined how to do so, right?
Yes.
sleek thicket
Ah yeah I don't think that argument would generalize
tight agate
so we do not need to assume that mflds are hausdorff?
sleek thicket
tight agate
or does the proof that there is a function taking one point to another use the fact that mflds are hausdorff
gritty widget
I feel like the stack exchange answer just offloads the hard work to this theorem about f(a)=b (is it supposed to be obvious?)
sleek thicket
I think it does brofib
The proof I'm using would rely on compact sets being closed
long hornet
I feel like the stack exchange answer just offloads the hard work to this theore...
It isn't obvious to me, but it is not very hard to prove (I know that because I proved it!)
sleek thicket
So brofib
Connect the two points with a path
Take a finite union of overlapping balls
The union is your connected nbhd
It suffices to show that in any coordinate ball, you can find an automorphism of the manifold sending one point to the other which is the identity on the complement of the ball
Yeah?
I would want to send the point to the other in R^n via a homeomorphism which is the identity outside a compact set
These can be constructed in R^n by like, ODEs or whatever
Then the compact set is closed in the manifold, so the complement is open and we can define the map globally to be the identity on the complement of that compacr set
This is continuous by the gluing lemma
Yeah?
This uses that compact sets are closed in all of M
Yeah?
tight agate
ye
sleek thicket
Which relies on hausdorffness
tight agate
neat!
long hornet
Then the compact set is closed in the manifold, so the complement is open and we...
I know how to do that when the compact set is a coordinate ball.
But isn't the union of these overlapping balls homeomorphic to a ball?
sleek thicket
yeah
You take a point in the overlaps
Keep swapping
x -> x1 -> x2 ->...-> y
Each map is the identity outside the union of our balls
long hornet
Yeah
long hornet
yeah
So the tubular neighborhood theorem (if I understand it) also holds, I mean we can take a neighborhood containing the path
Namely, the interior of the union of these overlapping balls that cover the path
Is that right?
long hornet
So the tubular neighborhood theorem (if I understand it) also holds, I mean we c...
We can take a coordinate ball*
sleek thicket
What do you think the tubular neighborhood theorem says
long hornet
I think it says what I said in the second message
sleek thicket
That any path is contained in a connected neighborhood?
long hornet
Yes, and that this neighborhood can be taken to be a coordinate ball.
sleek thicket
this is not what the tubular neighborhood theorem says
sleek thicket
Yes, and that this neighborhood can be taken to be a coordinate ball.
And I don't think this is given by the argument in the MSE post
long hornet
Well, I am not surprised. But I don't think it matters (I am interested in the statement I gave)
sleek thicket
You don't know all of W is a a coordinate ball
long hornet
W = union of the balls covering the path?
sleek thicket
Wait I thought you were going off of the argument in the MSE post
sleek thicket
But isn't the union of these overlapping balls homeomorphic to a ball?
But this is wrong in general
Consider a path connecting the north and sole poles
Of a sphere
The union is a whole sphere
long hornet
The union of what? I mean, which coordinate balls did you use?
sleek thicket
The complements of the north and south poles
long hornet
We can use a single coordinate ball, of course, but I think you mean some choices don't work
long hornet
What if we take the minimal number?
long hornet
I agree. Too good to be true.
Is there a nice bound on the least number of charts needed to cover a compact manifolds?
sleek thicket
Do your charts need to be connected?
I think an n manifold can be covered by n charts
Let me find the reference
No, that's not right
long hornet
I think an n manifold can be covered by n charts
Does RP^n achieve this bound?
long hornet
Oh
sleek thicket
n+1 charts
long hornet
I see, thanks!
shut moat
homework submitted 
it took me a concerning while to figure out how to show that a retract of a hausdorff space is closed
shut moat
I think an n manifold can be covered by n charts
that's surprising
ok so this is unrelated to the main proof, but I've wondered why manifolds ppl more commonly define charts as maps from M to U
like i get that it's equivalent
but from an operational standpoint the way you use charts by, say, composing the inverse with functions suggests that embeddings from U to M might be more natural
like just good ole parametrizations
like even when defining vectors. Usually you take a chart phi, right. ∂i at p is the inverse dphi^-1(∂i_phi(p)), whereas with a parametrization it's more naturally just the pushforward of ∂_i
shut moat
same difference \o/
sleek thicket
also maybe it makes it easier to talk about the overlaps of charts?
I don't really know why we prefer charts to parameterization
shut moat
so many conventions to get to the bottom of
i rly want to be able to sit down at some point and motivate geometry/difftop with no ambiguity, but that's going to take a lot of experience
sleek thicket
it's pretty cool
shut moat
indeed
kind sluice
Hi :D
Ok so
Im studying it from a topological point of view rather than algebraic
Let $F$ be an infinite field
gentle ospreyBOT
Maikel
kind sluice
and let $\tau$ be the coarsest topology in $F^{2}$ such that every polynomial in F[x,y] is continuous
gentle ospreyBOT
Maikel
kind sluice
We are considering the cofinite topology in F btw
I want to prove that the sets of the form {(x,y): p(x,y) != 0} are a basis for tau
I used that fact to prove that $\tau$ is not the cofinite topology neither Hausdorff
gentle ospreyBOT
Maikel
kind sluice
I just havent been able to prove that tiny fact
Sure, im sorry, its an arbitrary polynomial in F[x,y]
Heck yes, its the preimage of F{0} by p
i meant F - {0}
which is an open set...
and thus the proof is done
How couldnt i see it
hahah
the set of all preimages by arbitrary p's in F[x,y] is a basis for the topology tau, right?
Now i want to prove that the support of polynomials in F[x,y] is also a basis
gritty widget
It is for sure a sub basis, I'm not sure how to show it is a basis
gritty widget
Wait
gritty widget
But p!=0 and q!=0 iff pq!=0
kind sluice
dw
sleek thicket
So you want to show the zariski topology on F^2 is neither cofinite nor hausdorff?
kind sluice
i already proved that, using this basis of the topology were talking about
sleek thicket
Ah okay
kind sluice
i just can prove its a basis
Sorry, i proved its actually a basis, i just cant prove its a basis for the zariski topology
The coarsest one that makes every polynomial continuous
idk if thats the correct definition, its probably a particular case
Im studying it as an interesting example in an undergrad topology course
remote beacon
Zariski is non hausdorff right?
kind sluice
Not if the field is infinite
I cant use that fact though, since i just proved that using this basis
sleek thicket
In which case?
sleek thicket
The coarsest one that makes every polynomial continuous
This is a good definition but kind of unwieldy
Slim's is easier to work with
kind sluice
yeah, im working with an infinite field
sleek thicket
Slim I think "not if the field is infinite" was agreement
Not a double negative
It's emphasizing that it is hausdorff in the finite case
kind sluice
Oh
Yes, i was saying if the field is infinite, it is not hausdorff
Basically, for any preimage U of any open set in F (cofinite top) by any polynomial in F[x,y], and any x in U, i want to find a set of the form V={(x,y):p(x,y)!=0} such that x in V c U
sleek thicket
kind sluice
i.e i want to show the sets like V form a basis for that topology which i know as zariski's
sleek thicket
Ah this is a good problem. These are called "distinguished opens"
So what does an arbitrary open set look like?
kind sluice
hmm
sleek thicket
Also I agree, this is a bad topological space. It doesn't even have a generic point for each irreducible subset!
kind sluice
i dont really know
sleek thicket
Sure, so how did you do the previous part of the problem?
Slim they said the definition wasn't in terms of closed sets
kind sluice
the non.hausdorff part?
sleek thicket
Yeah
kind sluice
i proved every pair of elements from that basis couldnt be disjoint
which in particular implies not-hausdorff
sleek thicket
Right, so what's the basis? The preimages of cofinite sets in F?
kind sluice
No, thats the thing. The basis i used is exactly the one i want to prove its actually a basis for the topology
kind sluice
Basically i just need to prove this to end the proof
sleek thicket
I want to make sure they know what the open and closed sets are
And can justify it
sleek thicket
the set of all preimages by arbitrary p's in F[x,y] is a basis for the topology ...
so this is what I was trying to aim you towards
kind sluice
Basically, for any preimage U of any open set in F (cofinite top) by any polynom...
and thus this
kind sluice
No worries!
sleek thicket
I think it would be best if we tried to think about what the general open sets of F^2 look like first
sleek thicket
Hahaha
They certainly are
Okay so
We know that they're all unions of the basis sets
Yeah?
kind sluice
Indeed they are... but whats the basis?
sleek thicket
The cofinite-preimage basis I mean
kind sluice
The cofinite-preimage basis I mean
btw thats a nice compact way of describing it lol
sleek thicket
I think it is a basis in this case but maybe that's not obvious
shut moat
the set of (x,y) st p(x,y) !=0 can be rewritten as p^-1(F-0), yea?
sleek thicket
So we would want to know that the intersection of any two basis set is a union of basis sets
Right?
shut moat
so doesn't that automatically grant that p^-1(F-0) is also open in the other topology bcz F-0 is cofinite
sleek thicket
so doesn't that automatically grant that p^-1(F-0) is also open in the other top...
The question is whether sets of this form are a basis for the topology
shut moat
right, and the topology is formed by the subbasis of preimages of cofinite sets
sleek thicket
wouldnt we like to know if its closed under finite intersections?
Nope, this is stronger than we require for a basis of a topological space
shut moat
so double inclusion would do the job, I thought (as in for all x in one basis element, there's an element of the other basis contained in it)
sleek thicket
Approx we're still working on the other inclusion
sleek thicket
Basically, for any preimage U of any open set in F (cofinite top) by any polynom...
This
I know you can do this directly, I just wanted to give a more complete description of the zariski top
kind sluice
what i said there is actually enough right?
sleek thicket
It would be sufficient, yes
kind sluice
So we would want to know that the intersection of any two basis set is a union o...
Ok. right i get it
its the definition of a basis
sleek thicket
what i said there is actually enough right?
Actually, I'm not sure
kind sluice
At least it implies it being a basis
digital peak
if it's closed under finite intersection it's a basis
kind sluice
Oh you were talking about an earlier message
digital peak
(I have no context here)
sleek thicket
My concern was with a different claim
Yes maikel, sorry for the confusion
What you said about bases is correct
Sorry, we sort of fractured into two topics
kind sluice
nice
digital peak
zariski top 
sleek thicket
Okay, let's back up
kind sluice
Thanks
sleek thicket
Basically, for any preimage U of any open set in F (cofinite top) by any polynom...
Suppose you know this is true
And know that the collection of sets of the form {x : p(x) ≠ 0} is a basis
sleek thicket
That you can cover intersections of any two elements by a union of others?
If so, nice
It is straightforward, p(x) q(x) ≠ 0 iff p(x) ≠ 0 and q(x) ≠ 0
So with these two facts we can say that the collection of sets of the form {x : p(x) ≠ 0} is a basis for the zariski topology
kind sluice
exactly
sleek thicket
Yeah?
kind sluice
The open set of pq does the job
sleek thicket
Basically, for any preimage U of any open set in F (cofinite top) by any polynom...
So we just need to prove this
I'm just making sure we're all on the same page
Is why I'm reiterating stuff
Because there were a couple discussions happening
sleek thicket
Right
kind sluice
we need to prove my earlier message
sleek thicket
So what does U look like?
kind sluice
Ok, it looks like {(x,y): p(x,y) in A for p in F[x,y] and A cofinite in F}
By definition... idk if youre seeking for a different answer
sleek thicket
That's all I'm asking
kind sluice
Ok nice
sleek thicket
Let's enumerate the complement of A
kind sluice
Ok
kind sluice
Perfect
sleek thicket
So what does p(x, y) in A mean in terms of these bi?
kind sluice
p(x,y) != b_i for i in {1,...,n}
sleek thicket
Right
sleek thicket
Right, so this is zero iff p = bi for some i
kind sluice
which is not
sleek thicket
So this is nonzero iff p ≠ bi for all i!
sleek thicket
Hahaha
kind sluice
So thats the V i was seeking for right?
kind sluice
Wow
sleek thicket
But maybe you need to union some of these together for varying p
To get every open set
kind sluice
so its the same V for any x in U...
sleek thicket
Yeah, but only because the U we started with is super special
sleek thicket
Yup
Here's a tricky problem
So let D(f) = { (x, y) : f(x, y) ≠ 0 }
We've shown this is a basis
It's closed under finite intersections and any open set is a union of things like this
yeah?
sleek thicket
Show that for any open set U, there are finitely many f1,...fn where U = D(f1) union ... union D(fn)
Well that's because it is AG!
This is saying that any open set is compact
kind sluice
Oh so its a special basis in which all open sets are generated by finite unions? :O
sleek thicket
Nah, it's about compactness
Not the basis itself
If you have a basis B for a space and a subspace Z, Z is compact iff every cover of Z by elements of B has a finite subcover
kind sluice
I mean i see the compactness def in there actually
Yeah, exactly that hehe
But is what i said wrong?
sleek thicket
It's not wrong but it's the same as saying "every open set is compact"
kind sluice
That means open sets are finite unions of the basis
sleek thicket
And it works for any basis
It's a property of the topological space which doesn't depend on any particular basis
sleek thicket
Oh so its a special basis in which all open sets are generated by finite unions?...
If this is true for one basis it is automatically true for all of them
Wait, no
I am lying
Yeah sorry you're right
If we required this for all possible unions of basis sets it would be compactness
But you're just saying there's some union
kind sluice
Yeah yeah
sleek thicket
Anyways, the property that all open subsets are compact is called being noetherian
It's pretty important in AG
sleek thicket
Yup! It's equivalent to the ascending chain condition on open sets
For rings it's the ascending chain condition on ideals
kind sluice
the one in PIDs?
kind sluice
PIDs are the ones in which that condition holds for ideals
If i remember correctly
Ooh i think i see
sleek thicket
No, this is not true
also mniip sorry for clogging chat
Feel free to ask your questions about spherical stuff
kind sluice
Hey thanks for all the help
it was quite easy actually, i was just too tired to think properly
gritty widget
kind sluice
lol hahahahaha
gritty widget
digital peak
kind sluice
now i gotta figure out some ergodic theory stuff
gritty widget
role collector
gritty widget
,ban mniip
gentle ospreyBOT
kind sluice
wanna help me too? hahah
digital peak
We have $SO(3)$ acting on $\mathbb{S}^2$ in the usual way, and is thus acting on the vector space of functions $\mathbb{S}^2 \to \bR$. I'm being asked to find invariant subspaces under this action. Now I'm vaguely aware of the spherical harmonics being related to this but I'm struggling to find a concrete connection. How does the laplacian come into play? Why do the polynomials have to be homogeneous? $\dv[2]{x}+\dv[2]{y}+\dv[2]{z}$ seems to suggest a connection to $x^2+y^2+z^2=1$?
sleek thicket
Rip
kind sluice
F
gentle ospreyBOT
mniip
digital peak
sure but polynomials are dense or something
sleek thicket
Only in continuous functions
sleek thicket
Seems hard
digital peak
thanks 
sleek thicket
digital peak
the thing is that this is a problem that's trying to like motivate irreducible representations of so(3)
and all I could find on the internet about this is "follows trivially from irreps of so(3)"
sleek thicket
lol
gritty widget
Huh, so3 cannot permute (x, y, z) to whatever right? Does this mean the invariant polynomials don't need to be symmetric?
digital peak
it's not about the specific invariant polynomials, rather spaces of them
gritty widget
Ah right
digital peak
for example, restrictions of linear functions to the shere is clearly an invariant subspace
sleek thicket
Any function gives an invariant subspace via it's orbit
and this will be pretty small
Seems really hard to classify all of them
I am thinking
digital peak
I suspect these are orbits of like x^l
sleek thicket
Sorry, I'm not sure what you mean
Like I don't think for every f there's such an l
digital peak
does it sound believable that any polynomial is a linear combination of rotations of x^l
sleek thicket
To show what?
digital peak
this is schauder basis vibes not hamel basis vibes
sleek thicket
I feel like moving will fuck up the subspace
like if you're in some invariant subspace
Which is like a proper subset
You can always move ε in some direction and leave
digital peak
yes
ok let's restrict our attention to just the polynomials
the other part is outside of the scope of the problem
so we have polynomials in x,y,z restricted to the sphere, with SO(3) acting on them
sleek thicket
Right
so any polynomial is a sum of things like a x^l
but you can't fold like the a into a rotation
digital peak
?
let A be the 45 deg rotation in the xy plane
then A(xy) = (x^2-y^2)/2
and A^2(x^2) = y^2
so we have A(xy) = A^-1(x^2)/2 - A(x^2)/2
so we have
$$px^2 + qxy + ry^2 = (p + \frac{A^{-1}q}{2} - \frac{Aq}{2}+rA^2)x^2$$
gentle ospreyBOT
mniip
sleek thicket
Hm you're right
digital peak
now how does this generalize to higher orders I have no idea
there's probably a similar connection in 2D, between fourier and polynomials and the laplacian in the circle
damn wish I did more in my first PDE class
tf is this notation
|.|^2 being r^2?
what is this
kind sluice
@sleek thicket are you still here?
sleek thicket
what is this
I think this is defining an inner product?
digital peak
yeah I think I get what it does now
sleek thicket
like evaluate the first poly at the laplacian to get a differential operator
👍
weird that it's like, conjugate-symmetric
digital peak
we're defining <x^k, x^m> = k! delta_km
sleek thicket
ah okay, less weird
digital peak
and <x^k, y^m> = 0
sleek thicket
anyways yes maikel i'm here
kind sluice
Right, so this is zero iff p = bi for some i
I was thinking, this shows that p^{-1}(U) is a subset of {(p-b1)...(p-bn)!=0}, but we actually needed the inverse inclusion
kind sluice
Right you said they were equal
sleek thicket
yup
kind sluice
i dont get why tho
kind sluice
Totally
sleek thicket
so (p(x)-b1)...(p(x)-bn) ≠ 0 iff p(x) ≠ bi for all i
iff p(x) in U iff x in p^{-1}(U)
p(x) ≠ bi for all i iff p(x) in U because U is the complement of b1,..,bn
kind sluice
Yup, i think its pretty clear put in that way
Let me write it to make sure i do get it
I do, thanks! :D
gritty widget
Does anyone have a good answer to this? Diffeological spaces seems like a very nice generalisation of manifolds. But they do not seem to be very popular?
Is this just because, the kind of stuff looked in differential geometry is inhernitaly non categorical, e.g. you often are looking at a specific manifold with a specific metric and really working in that manifold
or is it just that the theory isn't well developed
cold vine
Does anyone have a hint for the following: Given a cofibration i:A->X and Y contractible, any map f:A->Y has extension g:X->Y
This is the definition I'm working with
Thus it's quite difficult since it assumes a map X->Y which is what we want here. At least since Y is contractible and if we have that a map X->Y it is is nullhomotopic which gives us a homotopy to extend the restriction to A, but this doesn't immediately help
cold vine
Similarly I'm trying to show that fibration p:E->B where B path connected gives us from Y contractible f:Y->B has a lifting f:Y->E but here I have a similar problem
frosty sundial
Thus it's quite difficult since it assumes a map X->Y which is what we want here...
@cold vine I think the idea is to take the map X --> Y to just be the constant map to some point y in Y. The data of the map A --> Y isn't encoded there, it gets encoded in G.
Basically, G should be the homotopy which takes g: A --> {y} to f: A --> Y
the first map g is the restriction of the map X --> {y} \subset Y
cold vine
<@!140112579384049664> I think the idea is to take the map X --> Y to just be th...
Ahhhh thank you that makes complete sense! That should do it
frosty sundial
np and gl, sorry i cant look at your second question cuz i have to go run now
but maybe a similar idea works
cold vine
but maybe a similar idea works
Yeah no problem. It seems very similar (maybe dual even) so I think this helps enough.
scarlet mist
Not sure if this is the right place to ask this, but in Example 2.1.6 (b), why is there no square root on the LHS? Are they just not using the Euclidean metric?
gritty widget
lol
cold vine
Dont we have that maps from contractible space to path connected space are homotopic? I think this is true since isnt [X,Y] a one point space when X contr. and Y path conn.?
obtuse meteor
yes this is true
cold vine
great!
uncut surge
I guess the proof is as easy as you'd imagine; first you use contractibility to show that any map is homotopic to a constant one, and then path-connectedness implies that all constant maps are homotopic
shut moat
Advisor approved, I'm taking grad manifolds next sem 
gritty widget
sleek thicket
Nice!!!
plush needle
I am working on a paper, and I've realized that I can create "boxes" in order to investigate the particular space the box is embedded in. I know x_1 and x_2 are colinear, and y_1 and y_2 are colinear, and I can get expressions for the lengths of the sides of a box. Now I'd like to be able to, given this and possibly other (but few) conditions, figure out the metric. The image shows an example of a box in a familar space.
Note that the box is not drawn to scale. In this particular case, letting x_1 = y_1 = 0 will collapse the box to a triangle, for which the side lengths correspond to what you would get using Pythagoras. I would, however, like to find the metric without tricks like that, since I will be getting more exotic boxes than this example.
shut moat
I could be wrong but I don't think this is nearly enough information to determine a metric
because the metric can change completely arbitrarily away from the box you've drawn
well not "completely arbitrarily" but you know what I mean
viral atlas
I think the question here probably is: given fixed real values x_1,...,y_2, can we determine a metric which ensures the above dimensions correspond to a convex quadrilateral in R^2?
plush needle
@shut moat Indeed. But what if I could arbitrarily move such a box around the space and see how the lengths of the sides change?
It's as if I have two halves of a metric. I know how to measure distance between two points on the same straight line, and I know how to measure the distance between two points on two different straight lines. But, I don't know how to measure the distance between two arbitrary points.
uncut surge
Oh, hm, that actually sounds very interesting. So the mathematical version is something like "if I know all values d((x,0),(x',0)) and d((0,y),(0,y')), can i deduce what the metric is on general points d((x,y),(x',y'))"?
plush needle
Exactly.
uncut surge
It feels like the answer might be no, since metrics can get really freaky, but I'll think about this for a bit
Since in this kind of formulation, I will never be able to say anything specific about terms of the shape, d((x,0),(0,y)), except some very very rough estimates, since it's smaller than d((x,0),(0,0)) + d((0,0),(0,y))
The French metro metric comes to mind, not sure if that's an interesting example here
plush needle
But I know the side lengths too, d(x,y) and d(x',y'), but not d((x,y),(x'y')).
uncut surge
Alright yeah I think I getcha; so my original formulation wasn't quite right
https://mathworld.wolfram.com/FrenchMetroMetric.html but I think this is still a good example
cold vine
I found a material claiming that from the claim here follows that restriction of a fibration is a fibration. What is the pullback of the inclusion which makes this clear?
uncut surge
So this metric looks simply like d(x,y) = |x| + |y| on all sides of squares (outside of sides of squares which lie on coordinate axes), but if x is collinear to y, it turns into a different metric
cold vine
I found a material claiming that from the claim here follows that restriction of...
I mean I guess it would be that P is the subspace of E with upper horizontal arrow being inclusion and A being B with identity, but is this obviously a pullback
plush needle
@uncut surge Ah, indeed. So there would be a way to circumvent most attempts with some exotic metric then.
uncut surge
Maybe! I've not thought about all the details yet, but this metric breaks a lot of "intuitive" things that one would like to do with metrics
So it's definitely a good example to check
plush needle
What if it was sufficient to find a metric, and not the metric?
uncut surge
@cold vine I know this more in the context of bundles, but there, the restriction of a fibration E -> B corresponds to taking a subspace A of B, not a subspace of E
Then the fibration P -> A is also supposed to have the same fibres as E; so yeah, restrictions of bundles/fibrations are restrictions to smaller parts in the base space; https://en.wikipedia.org/wiki/Pullback_bundle maybe this article gives some better intuition
@plush needle If you just wanted some metric, you might be able to do something with infimum-constructions perhaps?
I'm thinking something like "d(x,y) = infimum of all the sums of sides-of-square-lengths which go from x to y"
cold vine
Ah alright! I guess this still helps me since I'm trying to show that p|E0: E0 -> p(E0) is a fibration, where E0 is a path component. I have already shown that image of a path component is a path component
uncut surge
So if you have a bundle $\pi: E \to B$ and a map $f: A \to B$, then the pullback bundle $f^* E \to A$ is defined as $f^* E := { (a,e) \in A \times E : f(a) = \pi(e)}$
gentle ospreyBOT
Lartomato
uncut surge
I suppose you can do exactly the same thing with pullback fibrations; now the only thing to do is to show that E_0 \to p(E_0) is the same as this definition of the pullback fibration under the inclusion map p(E_0) \to B
would be weird if it wasn't
cold vine
Ahh alright! thanks
obtuse meteor
we're spending the last week of my manifolds in R^d class talking about Riemannian metric
@sleek thicket would be happy
sleek thicket
fading vale
only the personality of his that likes riemannian geometry
his other personality is filled with rage
gritty widget
MINUS!
sleek thicket
COMPACT
gentle ospreyBOT
sleek thicket
me and the boys proving pi using the Levi civita connection on a compact manifold
gritty widget
a smooth function on M is just a section of the trivial bundle M x R 
everything is bundles
sleek thicket
correct
fair idol
Is convex geometry still an active area ?
obtuse meteor
me and the boys proving pi using the Levi civita connection on a compact manifol...
I need the context
sleek thicket
Tterra and I were very stupid about compactness in vc the other week
> cotangent bundle is the first example he gives me of a symplectic manifold
> on a compact symplectic manifold, all even cohomology groups are nontrivial, and in particular the symplectic form is exact
> the symplectic form on the cotangent bundle is defined as the exterior derivative of something
it took us like half an hour to realize that thr cotangent bundle isn't exact
It literally contains vector spaces in it as closed subsets. It's a fucking bundle!!!
I forget where proving pi is from
gritty widget
I forget where proving pi is from
jaboi
sleek thicket
oh lol yeah
cold vine
Hmm, does anyone have a tip on showing that if $Y^I$ is the free path space and $p:Y^I\to Y, \omega \mapsto \omega (1)$ then $p$ is a homotopy equivalence
gentle ospreyBOT
PAHUS
bright acorn
While the Eilenberg-Steenrod axioms give you a nice framework to study simplicial, singular homology and other homology and cohomology theories of topological spaces, I was thinking if there is a similar set of axioms used to study homotopy groups.
Is there such a thing?
I still haven't studied more advanced homotopy theory
So I got a bit curious
tawdry widget
How is the topology on Y^I defined?
obtuse meteor
Presumably C(I, Y), [I, Y], or similar
and usually we use the compact-open topology
so if you have a compact subset K of I and an open set U of Y, then the set of all functions f : X -> Y so that f(K) is contained in U is denoted V(K, U), and then V(K, U) is a subbasis for the topology on [I, Y]
tawdry widget
Hmm, does anyone have a tip on showing that if $Y^I$ is the free path space and ...
I proved your statement on C(I,Y).I think Y^I doesn’t work
Wait... I made a mistake at the end,there still remains one step I haven’t completed yet...:
done：
cold vine
Thanks so much! Yeah C(I,Y) should be good here since it's the free path space. Yeah I got some of that but it was super mechanical 😛
lyric wadi
Hey, so I'm trying to prove the universal coefficient theorem for cohomology with integral coefficients directly (so not invoking the theorem for general coefficients), and I've managed to reduce the problem to here but I'm having a hard time working out why the torsion of H^n(X;Z) should be contained in Ker(h)
I guess that since we have that every element of the kernel of h is a coboundary, we have that something messes up when we try to extend the element to work on any chain
rotund thicket
What is the intuition of having a continuous function f : X -> Y and Y being Hausdorff?
rotund thicket
For example, in a problem if f, g : X -> Y and f(x) = g(x) for all x in D where D is dense in X and Y is Hausdorff, show that f(x) = g(x) for all x in X.
I'm asking how to think about the importance of Y being Hausdorff
I guess if $x \in U$ and $y \in V$ both open sets, then $f^{-1}(U \cap V) = f^{-1}(U)\cap f^{-1}(V) = \emptyset$
that's the importance
gentle ospreyBOT
Trichloromethane
rotund thicket
i think i answered my own question lol
so $f^{-1}(x) \not= f^{-1}(y)$ if $x\not= y$?
gentle ospreyBOT
Trichloromethane
rotund thicket
or even stronger that $f^{-1}(x) \not\subset f^{-1}(V)$
gentle ospreyBOT
Trichloromethane
rotund thicket
I forgot im dealing with pre-images
oh yeah you're right lol
well im asking how to think about the assumption that the codomain is Hausdorff in the context of a continuous function
well for this one U and V are open sets containing x and y respectively
I understand that
yeah i guess that's sort of what im thinking
well my understanding is that for any two points in the codomain, there exists disjoint neighborhoods containing each
therefore the pre-image of these sets is disjoint (by prop. of pre-image)
i suppose the useful thing about this is that these are disjoint open (by continuity) pre-images
well if im trying to solve the problem then yeah
well, im asking somewhat of an open-ended question
because ive seen this assumption that the codomain being hausdorff quite often
oh yeah
uncut surge
Hausdorffness is just a very common property in general, and generally tells you that you can freely switch between points and open sets, in a sense
There's nothing super special about codomains being Hausdorff, since generally, Hausdorffness is not super special (in most areas of mathematics); but it is very useful, since it means that you can distinguish points, topologically
And the way you should use it in proofs is generally always the same: Try to somehow force two points to be unequal, and then Hausdorffness gives you disjoint open sets for free
And open sets are cool to combine with continuous maps; specifically, if you form their preimage, you get something open again. Hence, I guess, that's why your Hausdorffness condition may often show up in the codomain: Then the open sets you get in the codomain can be preimaged immediately
tru
rotund thicket
yeah it feels like most of it is just going back to the definitions
summer jolt
What is the meaning of the notation: $([-1,1] \times {-1,1})$?
gentle ospreyBOT
snypehype
summer jolt
No without the round brackets
obtuse meteor
obtuse meteor
drifting sundial
the way I like to think about Hausdorffness is that there cannot be "infinitely close" distinct points
but yeah to prove stuff you have to fall back on the precise defn
strong heron
Let X be the 2-sphere with two points {p,q} identified. I am supposed to find the homology groups of the pair (X,X-[p]), where [p] denotes the equivalence class {p,q}.
However, I'm not supposed to use the homology of X for my computation. I feel I must use excision somehow, but can't figure out how.
Won't a small neighbourhood of [p] be like a cone with vertex at [p], and then using excision we can conclude that the required homology is same as that of (D^2, D^2-{point})?
uncut surge
A neighbourhood of [p] won't satisfy the condition that you need for excision though, right? You need an open set in the sphere whose closure is also in the interior of X - [p]
I'm thinking if you take p and q to be the north- and southpole, excision of an open band around the equator might be bettah?
Also a neighbourhood of [p] is less like a cone and more like two cones glued together at their vertex
@strong heron wheee
strong heron
Also a neighbourhood of [p] is less like a cone and more like two cones glued to...
Hmm, that's right. But it's homotopic to a single cone, no? That's what I thought I could use. But maybe not.
strong heron
I'm thinking if you take p and q to be the north- and southpole, excision of an ...
Yeah, but how does that help? So, after gluing, we get a pair (U,U-[p]), where U is two cones joined together at [p], right?
uncut surge
Hmm, that's right. But it's homotopic to a single cone, no? That's what I though...
Yeah it is, I guess it's even contractible. If we instead cut out the complement of such a small neighrbourhood around [p], that works better, and that's sorta what we do when we do the equator-band thing
uncut surge
Yeah, but how does that help? So, after gluing, we get a pair (U,U-[p]), where U...
Yeah! So U is contractible, which is pretty good, and U - [p] is two cones with their vertices removed (so basically two circles)
I imagine the long exact sequence in homology might bring happiness and good luck then
strong heron
U-[p] is two disjoint circles, right?
uncut surge
Ye that's what I'd think
strong heron
So we get $H_n(U,U-[p]) \cong \tilde{H}_{n-1}(U-[p])$
gentle ospreyBOT
whysee
uncut surge
For n >= 2 ye
In the part of the long exact sequence where degree zero pops up you need to be careful but it doesn't get too complicated there either
strong heron
Yeah, that's right. But this means I seem to be getting H_2=Z+Z
But when I actually used the homology of X initially and calculated the homology of (X,X-[p]) using the long exact sequence, I got H_2 to be Z.
I'm confused 😦
strong heron
What's that? Did you figure it out?
strong heron
But when I actually used the homology of X initially and calculated the homology...
I'm pretty sure my this computation is correct. I actually got that H_n(X,X-[p]) is Z for n=2 and 0 otherwise.
uncut surge
Mhm, no, I'm not sure yet; One can think about X as a wedge sum of a sphere and a circle, but even knowing that, the Long exact sequence with the pair X, X - [p] doesn't become immediately conclusive
How did you get that?
At least H_2 = Z^2 and H_1 = Z would be consistent with the long exact sequence I think
but H_2 = Z and H_1 = 0 as well lol
strong heron
X-[p] is a cylinder, right? So, it is a circle upto homotopy.
uncut surge
Yea
strong heron
What I thought is that the map H_1(X-[p]) to H_1(X) in the l.e.s. will be an isomorphism, no?
uncut surge
Hm, don't think so actually; if you embed the cylinder back into the space X, then the nontrivial loop around the cylinder becomes contractible on the sphere
So I think that map has to be zero, and then the H_2 = Z^2 and H_1 = Z result must be true
strong heron
Yeah, I think I can see my error now.
uncut surge
But yeah, the analysis of that morphism was exactly the necessary step that I was missing; now I get it too
strong heron
So our H_2 is settled with the other method also now
uncut surge
Yeah, I believe so!
I'm always so happy when these mystical LES-calculations work out, lol
uncut surge
No problem, glad to help!
pastel thistle
Hi
Any way of proving that R^n\Q^n is connected ??
I prove it for n=2 and I would like to use induction but Im stuck
marble socket
show that its path connected!
pastel thistle
I know is kinda obvious but I wanted a formal proof
pastel thistle
show that its path connected!
Any idea?
marble socket
yea so take 2 points and consider a 2d-plane passing through the 2 points
there are uncountably infinite lines passing through the first point lying on the plane.
but there are only countably many points with rational coordinates
so you can find lines passing through the 2 points which don't contain any points with rational coordinates
but since there are many such lines, you can find a pair of intersecting lines, giving you a path between the two points
(the claim is false for n = 1)
like the same proof shows that R^n \ any countable set is (path) connected.
for n>=2
marble socket
uncut surge
holy shit i've never heard of that proof before this is beautiful
summer jolt
Is there a nice way formula for the deformation retraction of a filled square with its centre point removed onto its boundary?
I’ve been trying for quite a bit and couldn’t find it. What I can see is that the points should mapped along the line passing through that point and the centre onto the boundary.
strong heron
You can perhaps consider the unit square centered at the origin and then for a point (a,b) just consider the line bx-ay=0 and check where it meets the edges of the square.
You will get 4 regions corresponding to the 4 edges of the square from the lines x-y=0 and x+y=0.
This map is from the square to its boundary and it will descend, via quotient maps, to give a map from the torus to its boundary.
I think writing it out must not be very difficult.
summer jolt
@strong heron I see yes dividing the problem into 4 different parts seems reasonable as one the end coordinates of the line will be known and I would just have to determine the other one.
gritty widget
@obtuse meteor I love differential geometry and Topology. Do you love?
barren sleet
"Do you love?" what a deep question
gentle ospreyBOT
Sigma, o Liestinche
gritty widget
curvature and topology on complete riemannian manifolds <:flonshed:6298350714650...
Yes
Very pretty
sharp yoke
I think i proved the forward direction correctly
having a bit of trouble with the reverse direction
i tried some stuff with projections/component functions but nothing was shaking out
sleek thicket
Huh, that's my first instinct too
I thought that you could check smoothness locally and in each chart take f to be the coordinate functions
Then f ° F should be the components of the local representation of F, so if they're all smooth then F is smooth on the chart
What went wrong for you?
sharp yoke
i think im doing something wrong proving the coordinate functions are in $C^k(N)$
gentle ospreyBOT
『Urahairywizard』
sharp yoke
Around a point in N you have a chart containing that point. And then the chart composed with your coordinate function will be C^k on your coordinate neighborhood, not necessarily on your entire manifold N
i could be talking nonsense tho
im thinking that to apply the assumption, we need our small f to be defined on all of N right
delicate hollow
Quick question: I am told a subset A in R^2 is bounded, prove that A complement has only one unbounded component
But if I take an example like a closed disc in R2
that is bounded right?
but the complement of its components are both unbounded.
since any number not in a unit disc counts and that can any point where x^2 + y^2 > 1
or I dont understand what componenets are
I think I dont :/
sharp yoke
could be talking about connected components but idk
barren sleet
complements are everything not in the set
delicate hollow
yeah its connected componenets
barren sleet
the complement of the closed disc is one connected component
delicate hollow
that would only make sense
I thought components were (x,y) like the individual parts
barren sleet
yes
delicate hollow
ok
barren sleet
plenty
delicate hollow
components are kinda weird for me to visualize ngl
barren sleet
even simpler, [0,1)x[0,1)
delicate hollow
does open unit disc with point at (0,1) count?
barren sleet
wdym
delicate hollow
is it an example
barren sleet
yes
barren sleet
do you know a topological definition of connected?
barren sleet
using the definition of bounded and the definition of connected makes the exercise easier to prove
barren sleet
no clopen sets?
no proper clopen subsets
delicate hollow
yea
delicate hollow
i like thinking in terms of no seperation though
thats the definition im super comfortable using
barren sleet
so getting intuition for why the definition is that way?
there is some nice intuitiom for it
drifting sundial
@barren sleet we talked about this before
drifting sundial
something about sets touching each other
barren sleet
it's easy you first consider metric spaces
delicate hollow
but no clopen is weirder
drifting sundial
basically if it's disconnected, then there are disjoint sets that don't touch each other
barren sleet
I can talk later
drifting sundial
I.e. they have no boundary
barren sleet
I will explain
delicate hollow
wait
drifting sundial
and you can prove that clopen sets are precisely the ones with no boundary
delicate hollow
bruh
bounded means no boundary 🤦♂️
wait
bounded means there is a boundary
or that the boundary is included
drifting sundial
uh bounded and boundary are completely unrelated
delicate hollow
hmm
ok
so
my def of bounded
A is bounded iff for all x in A m<=x<=M
but
in R2
this means that for all (x,y) in A, m_x<=x<=M_x and m_y<=y<=M_y
thats what i remember atleast
and it makes the most sense
is it right?
bruh
i think I found an answer to my exercise by just browsing SE
i wasnt searching for answer though, this is unrelated
sharp yoke
literally found this in 2 seconds
@sleek thicket does my problem makes since btw?
sense*
sleek thicket
Ah yeah I see
This is a really good catch
But you can extend via a bump function
So say you want to check that F is smooth on a neighborhood of a point p
Choose a chart with coordinate x^1,...,x^n containing p
Choose a smaller neighborhood U of p (one with closure contained in the chart) and a C^infty bump function φ supported in the chart which is identically 1 on U
Then consider the global functions φx^1,...,φx^n
does that make sense?
So these functions aren't defined to be φ(q) x^1(q) if q is in our chart and 0 if q is outside
delicate hollow
https://math.stackexchange.com/questions/133290/complement-of-a-bounded-set-b-in...
wasnt trying to cheat
sleek thicket
@sharp yoke
These are smooth inside the chart because they're a product of smooth functions. And they're identically zero on the complement of the closure of U, so definitely smooth there. Thus they're smooth on an open cover of the manifold, so globally smooth
barren sleet
@delicate hollow are you still working on how to make intuitive sense of the clopen definition of connected?
barren sleet
do you want my explanation
delicate hollow
yes