WhyamIsohot?:
#point-set-topology
gentle ospreyBOT
spark acorn
I mean they are not inclusions
Like this was another way to define CW-complexes as taking colimit over all skeltons , But I couldn't exactly visualise the toplogy on the colimit .....
Yeah I mean so if I define CW complex as colimit will the topology you gave still hold?
I was just trying to prove things the other way . I mean I first learned CW complex in lee's book where he defined it as exactly you told before now I saw jeffrey stroms books he defines it as colimit and gave the definition you told as excercise ie "Sets in CW complexes are open/closed iff their intersection with each skeleton is open/closed" , But I'm having hard time to visulaize things in this direction ....
honest narwhal
Think of it using inclusions
You have an inclusion R^1 \cup R^2 \cup ... = R^{\infty}
So R^{\infty} is the set of sequences s.t. almost all the terms are 0
You have an inclusion R^1 -> R^{\infty}
spark acorn
You have an inclusion R^1 -> R^{\infty}
@honest narwhal So my topology on R^{\infty} is such that a set is open in the colimit iff it's inverse image is open in R_i for all i right ? (under the inclusion)
honest narwhal
Sorry I trailed off but yeah the idea is that the topology is that you take the disjoint union of the X_i
Modulo an equivalence relation
In the case of R^{\infty} you're identifying (1,0) with its image in R^3, (1,0,0), and so forth
And that's why this is the same thing as just taking infinite sequences
And you're putting the topology of disjoint union, then quotient topology
Then you have the map R^n -> R^{\infty} just by including in the disjoint union and passing through the equivalence relation, and it's what you think it's supposed to be
Now our topology was defined such that these maps are continuous!
Let's say i_n is the inclusion of R^n in R^{\infty}
Well, if B\subset R^{\infty} is open (closed), then by continuity i_n^{-1}(B), which is basically just B\cap R^n, is open (closed). As you said
And it turns out you're just defining the topology such that this is pretty much exactly enough to be open/closed
spark acorn
Aaaah thanks now it all makes sense yay!!!
honest narwhal
Sure thing fam
gentle ospreyBOT
slimvesus:
spark acorn
Wait why is M\subset R^k ?
I mean only compact manifolds have embedding into euclidean spaces ....
honest narwhal
You have objects called abstract manifolds in general
WhyamIsocold is saying wait, why is Milnor doing this? It's not perfectly general
But I think Whitney holds for non-compact manifolds even
dim meadow
Yes
honest narwhal
Maybe it's easier in the compact case
I know bringing down the dimension is easy for compact because you can just be like
Sard tho
honest narwhal
Everything here is implicit smooth
dim meadow
But there is an embedding result about topological manifolds in general
honest narwhal
Partially because this is Milnor's difftop book but also all manifolds are smooth
Arnold thinks a lot of things
honest narwhal
Those things making sense is a tougher condition
Arnold is like, Bourbaki goes too far so I'll go too far (like "I don't even think there's a valid case for going this far") in the other direction to counterbalance
You don't wanna have to go and verify that everything you touch has an embedding
The point of Whitney isn't "Oh I've invalidated abstract manifolds" so much as "I can work with abstract manifolds and without thinking stick them in R^n when I need to"
spark acorn
But I think Whitney holds for non-compact manifolds even
@honest narwhal Yeah in general every n-manifold admits an embedding into R^{2n+1}
dim meadow
There's a sharper result in terms of the binary expansion of n
If I remember correctly
honest narwhal
"I mean only compact manifolds have embedding into euclidean spaces ...." then where'd this come from?
Huh
I know 2n is a thing by the "Whitney trick"
But binary expansion jeez
spark acorn
"I mean only compact manifolds have embedding into euclidean spaces ...." then where'd this come from?
@honest narwhal I mean I only know the proof of this lol
honest narwhal
Oh lmfoa
dim meadow
It's in terms of the number of 1 digits when n is written in binary
Apparently there is a sharp bound written this way
honest narwhal
Slimvesus: I think Milnor didn't mean that M surjects onto y\times R^{m-n} in general, rather that hyperplane within that neighborhood V of (y,L(x))
spark acorn
Wait sure?
dim meadow
Lol I'm trying to find a reference
honest narwhal
It's poorly written but that's the thing which, at a glance, would make sense to me
Yes it is
I think that sentence honestly could be omitted in favor of the "In fact..."
spark acorn
It's in terms of the number of 1 digits when n is written in binary
@dim meadow Is there a bound on n or something ?
Like we can't embedd rp^2 in lesser dimensions ....
Wait
It's no of 1's in binary expansion of n ryt?
Oops I didn't see that lol
gritty widget
does div F describe a tangent bundle to the space of every coordinate transformation to F?
frosty sundial
@marsh forge I am kind of curious about this question I asked last night about this terrible topological space. well, ignoring the hawaiian earring space now, do you know any examples of compact spaces which have a piece of (singular, probably?) H_1 that is divisible?
like, you can always just like, write down generators and relations for Q and then build a CW complex with Q as its first homology, but that won't be compact
classes in H_1 are represented by loops (hurewicz) and so I want to know what such a loop could look like
marsh forge
I'm guessing you want coefficients in Z?
frosty sundial
also, just some terminology: an abelian group A is divisible if every element of A can be divided by n for all integers n
yeah I thinks o
marsh forge
bc obviously its not too bad if you just want H_1(X;Q)=Q hahaha
frosty sundial
yeah, I want like, standard singular homology
"the thing which is the abelianization of pi_1"
marsh forge
I have a meeting rn
but you won't be able to use a compact manifold
and i feel like the space will have to be ugly
frosty sundial
I mean that's not true
the hawaiian earring space is compact
and has a divisible piece in its homology
that was the example which led me to ask this
oh wait sorry
marsh forge
Thats not a manifold
frosty sundial
finite CW complexes have finitely generated homology
marsh forge
so yeah the space will be bad
frosty sundial
all I want is an example of a reasonable space and a reasonable loop in that space which is divisible in homology
marsh forge
oh
frosty sundial
where by "reasonable" I mean "as reasonable as possible"
for example, we could take the space to be the hawaiian earring
marsh forge
i will try to think abt this after my meeting w peter
my gut tells me such a space will suck
frosty sundial
maybe I can rephrase my request into something more productive. It is true that the hawaiian earring space has a divisible piece of its first singular homology
I would like to try to find a loop which lives in that piece
I also know that such a loop must go around each circle a "net" 0 times
my best guess for what's happening is something like:
consider the loop which goes around circles 1-10 and then backward around circle 1, then forward around circles 11-20, then backward around circle 2, then forward around circles 21-30, then backward around circle 3
or something like that
because of the compactness, that does define an element of pi_1, and I have a feeling that it's probably not 0 in homology
but I don't know how to show that or show that it is divisible (and it might not be divisible, there are also nondivisible pieces of the homology)
marsh forge
Maybe if you take apart the computation of its H1
it will be obvious what geometric things correspond to what you want?
frosty sundial
the computation of the H1 is "this is the pi_1 now we're going to do group theory to determine its abelianization"
and the group theory isn't explicit at all
marsh forge
is there not a way to keep track of a loop that is sent to what you want
after abelianization
maybe if you sent a reference i could try to think about it? I tend to skip over anything pathological lol
frosty sundial
like, what I mean is that hte paper proves that "the abelianization is abstractly isomorphic to this group"
with no explicit maps
like I said, it's not topological at all, it's just like abstract group theory
it's theorem 3.1
the paper really isn't that helpful for the topology
you don't have to think about it, I just want to put this out there for people who might have thought abotu this before
marsh forge
sometimes problems give me the vibe that i will understand something i care about better
if i can answer them
this problem gives me that vibe
sleek thicket
This problem is interesting
frosty sundial
yeah it's weird to me that like, a loop which goes around each circle in the hawaiian earring a net 0 times is still nontrivial in homology
(and in fact, there are such loops which are also divisible in homology)
marsh forge
Well you can do like aba^{-1}b^{-1} in S1vS1
is that what you mean by net 0?
oh duh
that dies in homology
i was thinking fg
yeah okay that is weird then
frosty sundial
yeah that was my whole point. a divisible element of homology necessarily goes around each loop a net 0 times, because for each loop there's a map to H1(that loop) which is Z and divisible elements must mat to 0
so like you said, the thing you wrote for S1vS1 is 0 in homology
but the hawaiian earring has nonzero homology classes which still are net 0 around each loop
sleek thicket
Here's something interesting which says you can't get Q or Q/Z
if you want a nice compact space with this fundamental group
frosty sundial
you can't get just Q
sleek thicket
Oh right, you just want a divisible element
frosty sundial
I just want to know what one ahs to look like
and I think that the hawaiian earring space is a good example because it's relatively concrete
I gave an example of an element of pi_1(H) which I believe is nonzero in H_1(H)
which is a candidate to being divisible
but I don't know how to show that (or if it's even possible to show)
sleek thicket
Oh that's a weird path
It's like the paradox where you put balls into an urn, I forget the name
Sorry I missed that above
frosty sundial
yeah that's how I got hte idea
like, I know that such a path must go around each loop a net 0 times, and I know that there are such paths which are nonzero in homology
and that was my first guess for what such a path could look like, but also, maybe that path is still "nice enough" that it actually is 0 in homology
honest narwhal
inb4 Liquid says something in here
dim meadow
Okay so suppose we have a hyperbolic surface (which is the quotient of H by a discrete subgroup of PSL(2, R))
dim meadow
So let's say we have an element of pi_1(S)
gritty widget
Get your disgusting fundamental group out of my face
This might be that I've upgraded to stimulants
dim meadow
We can convert the element of $pi_1(S)$ into a map $\alpha$ from R to S and lift that to a map $\tilde \alpha: \mathbb{R}\to H$
honest narwhal
To S^1 or to H?
gentle ospreyBOT
Liquid:
dim meadow
Oof
So first off let's talk about the end behavior of the lift
So we have a deck transformation which fixes the lift and is an element of the discrete subgroup of PSL(2, R)
So I guess that since we have a characterization of the elements of PSL(2,R) based on the elements they stabilize, that characterization should kind of talk about the end behavior of $\tilde \alpha$
gentle ospreyBOT
Liquid:
dim meadow
Like we have elliptic elements which fix one element in H, parabolic which fix one element on the boundary, and hyperbolic which fix 2 elements on the boundary
So if $\tilde \alpha$ has 2 limit points on the boundary then our deck transformation which stabilizes $\tilde \alpha$ should fix those points I think
gentle ospreyBOT
Liquid:
honest narwhal
Dead
dim meadow
Yeah I realized right after I tagged mods
honest narwhal
So possibly under some hypotheses the deck group should act transitively on the preimage of a point in S right? Obv this is a single deck map but I wonder if this is relevant
dim meadow
Yeah it does, that's fine
We're dealing with hyperbolic surfaces so everything is nice
I guess my reasoning is that if we view alpha tilde as a map from (0, 1) to H then if we take the extension from [0, 1] to \bar H then and our deck transformation fixes the image of alpha tilde then it should fix the extension as well
Our deck transformation is a hyperbolic isometry which also acts on \bar H
Honestly my reasoning is a bit loose rn
slow urchin
hi folks!
(okay...no preview)
Just curious what the prereqs are for such a book, if any of ya'll are familiar?
the prefaces doesn't make an appeal to analysis....so i'm hoping this book can be done independtly of analysis?
tough imp
It doesn’t look like you really need analysis
But I feel as though having familiarity with analysis would only be beneficial, as it would perhaps give a bit more intuition as you’d probably be familiar with basic point set topology, but it’s undoubtedly probably not required
I think for the homotopy part of the book you’ll need abstract algebra, it mentions the fundamental group so knowing group theory is a definite prereq (unless it does a crash course on group theory in the book)
Beyond that, I’m not sure how much algebra it presumes the reader has
slow urchin
point set topology is distinct from regular topology? (im not sure what is what here cuz i'm a noob, sozz)
tough imp
It just means the basic intro topology stuff
slow urchin
okay. cool.
ivory dragon
Topology is a broad subject area like algebra or analysis
tough imp
Like, before you do algebraic topology or manifolds or something
ivory dragon
Point set topology is the kind of material covered in a first topology course
tough imp
It’ll talk about open sets, closed sets, conmectedness, etc which the book will cover right at the start
ivory dragon
Like how real analysis is the material covered in a first analysis course
slow urchin
point set topo book would be like Lee's Intro to Topology, yeah?
ivory dragon
You'll also hear "general topology" which means the same thing as point-set
tough imp
I’m not sure if he has a basic point set book. I’m familiar with introduction to topological manifolds
Which does cover point set, but also introduces the idea of a topological manifold
Think maybe like Munkre’s
That book will surely cover point-set at the start, it’s just a differentiation from say algebraic topology or homotopy theory which both fall broadly under the umbrella of topology
slow urchin
tyty for the advice
wind portal
I'm gonna give munkres a shot while simultaneously refining my analysis knowledge.
How much analysis would I need to really get into it?
Or can I possibly do analysis concurrently with it?
spark acorn
You can but to properly motivate open sets and stuff knowing some analysis is better
tough imp
You don’t need any analysis to do topology, but t motivates it
Especially when you cover metric spaces
spark acorn
Btw Let's pretend that we don't know any analysis can we still motivate open sets and stuffs ?
i mean why should i need a cube and a sphere to be "same" ?
tough imp
I mean... idk it’s kinda hard
If you just
Accept topology as interesting in its own right then yeah
icy pine
what are the laws of trigonometric identities?
tough imp
This is the wrong channel for that, try #geometry-and-trigonometry
charred cosmos
How can a discrete topology have every point as an open set when they are actually singleton sets? Aren't singleton sets not open?
chrome dew
what's an open set?
tough imp
What does the discrete topology mean to you
charred cosmos
A subset O of R is open if for each point x belongs to O there exists an interval (a,b) that contains x and is contained in O. This is the def. Provided by the resource I'm currently reading.
tough imp
that's not the discrete topology
that's the standard metric topology on R
the discrete topology is that every single subset of the set is an open set
charred cosmos
the discrete topology is that every single subset of the set is an open set
@tough imp
Then doesn't that mean every set is an open set in discrete topology due to the fact that union of open sets is an open set?
tough imp
I mean
it's more direct
discrete is literally every set is open
That's its definition
Set of open sets = power set
It also means every set is closed
Since the complement of every set is again, a set, and thus open
charred cosmos
So the def. of open sets is different in different topologys.
tough imp
Yup
that's what makes them different
A topology on a set is just what sets you consider to be open
(Or closed, since the definition goes both ways)
charred cosmos
Can we choose whatever def. want for opennes?
tough imp
No
Because you might fail to have unions of opens be open
or finite intersections to be open
charred cosmos
Ah yes.
tough imp
You also require empty set, and whole set are open
but besides that you have free reign
It turns out for any set
the discrete and non-discrete topology is always a topology
which are every set is open, and only empty and whole set is open resp.
charred cosmos
So any collection of sets which obeys these
tough imp
Yes
Will define a topology on a set
You can also choose to define it in terms of closed sets if you wish
in which case you need finite unions, and infinite intersections
but they end up being exactly the same thing
charred cosmos
Like a compliment to all the defs. But they are not mutually exculsive as in the case with discrete all sets are both open and close right?
tough imp
Yup
in any topology
empty set and whole set are both open and closed
or "clopen"
as some people call it
conversely
some sets are neither open nor closed
which to my knowledge does not have some cute name to it
Take for example (a,b]
in that topology you described on R
Or, in the indiscrete topology (only open is empty and whole set), just any non-empty, proper subset
charred cosmos
Thanks it cleared up a lot of fog.
tough imp
NP
just to make a point
the discrete and indiscrete topologies are not very interesting
because literally every set has them
and for the discrete, any function out is continuous, and indiscrete any function in is continuous
So for almost every purpose, in the case of R (and other stuff too) it is beneficial to view singletons are closed
The point I'm trying to make is, for analysis
don't start overthinking it and thinking singletons are open haha
charred cosmos
Yeah I get it thanks, BTW whats a recommended resource for basic point set topology?
tough imp
I used Munkre's
I found it fine
some people don't like it
if your focus is on learning it for analysis, a lot of analysis books have a section on point set
Like the graduate Folland analysis textbook does (it's a bit into it oddly enough)
or you could try Lee's intro to topological manifolds
that has some extra stuff in it (like topological manifolds lol), but it's mainly point-set
and some stuff about like the fundamental group
the benefit is topological manifolds aren't really hard, so being acquainted with them will probably make your life easier when you go into smooth manifolds
to do your "calculus on manifolds" type classes
Also, check out #books-old
there's some compiled resources that the server as a whole (or at least whoever wrote it lmao) thought were nice
charred cosmos
I was already reading those found on web, seeing that's recommended that might be good, I will also try Lee's thanks.
tough imp
Np :)
sturdy basalt
If X has a countable base for its Topology, then X has a countable dense subset
how do you prove this?
please give some intuition for it as well
tough imp
Isn't the base dense?
I need to think about it, but take the closure
rather, sorry take any x in X
x is in the closure if every neighborhood of x intersects the base right?
Wait nvm
this is so dumb
The base can't be dense
but does it suffice to just pick an arbitrary point in each base element?
Oh yes, duh
@sturdy basalt If you take an arbitray element of each base element that's dense
Use choice to get this if you want to be specific
So for each base element B_alpha take some x_alpha in B_alpha
then throw all x_alpha in a set, call this S this has the same cardinality as the base so it's countable
given x in X, then consider an arbitrary nbd U of x
contained within U there exists some basis element B_alpha which contains x
Then B_alpha intersects S at the point x_alpha (maybe even more)
so this means U intersects S at x_alpha
so every nbd of x intersects S
so x is in the closure of S
since x was arbitrary, the closure of S is X
Intuition is, idk you have countable base and you need a countable dense set. You can intuitively get a countable set by taking a point from each element of the base, and to me it stands to reckon that that set has a good chance of being dense
sturdy basalt
Thanks!!
marsh forge
Inclusion
gritty widget
this is from problems of section locally compact sets
oh
ok thanks
how do I write that in latex :P
marsh forge
\hookrightarrow
tough imp
In general that is symbol used for injective functions too
And an arrow with two heads so it looks like a harpoon is a surjection
marsh forge
Injective functions are inclusions tho
tough imp
I feel I should point it out
since that's sort of a fuzzy term
Like some people say you can "include" a ring A into any localization S^{-1}A
but you don't know the map A -> S^{-1}A is actually injective
marsh forge
I think that is cursed
tough imp
But you can "see" A in there as the set {a/1}
Me too
but I think it's worth pointing out
west spindle
@gritty widget i will insist that we move here because i am sick of your channel stubbornness
given two topological spaces X and Y, we say that a function f: X -> Y is a homeomorphism if it is continuous, has an inverse, and f^-1 is also continuous.
gritty widget
wouldn't that describe a isomorphism under multiplication too?
west spindle
what's an "isomorphism under multiplication"
gritty widget
well
west spindle
and even ignoring that i don't know what you mean by that
west spindle
stop throwing terms around when you don't know their meanings. "isomorphism under multiplication" doesn't mean anything.
and if you wanted to take note of the rather irrelevant fact that composing my map with itself four times gives the identity
gritty widget
the nth roots of unity under multiplication are isomorphic to whatever ngon they represent
west spindle
what you said is wrong in the sense that you don't understand what the word "isomorphic" means and are trying to get me to give you the okay on a statement which misuses said word
and to continue my point...
and if you wanted to take note of the rather irrelevant fact that composing my map with itself four times gives the identity,
then i'll replace "quarter-turn" by "1 radian"
gritty widget
i am still confused how it's wrong
west spindle
also we're talking specifically about topological spaces and homeomorphisms right now, are we not?
are you confused about the nonsensicality of this? or what
the nth roots of unity under multiplication are isomorphic to whatever ngon they represent
gritty widget
yeah? sounds good to me?
west spindle
yea because again you don't know what isomorphic means
west spindle
the closest i could think of to making your statement sound correct is
the group of n'th roots of unity (under multiplication) is isomorphic to the group of rotational symmetries of a regular n-gon
and this is now a statement in group theory
a subfield of abstract algebra
and "isomorphic" now has the meaning of "isomorphic as groups", i.e. the existence of a group isomorphism between the two
west spindle
no
just because your statement could be amended into one that is correct doesn't mean the original was correct
the original was teetering on the edge of nonsense
and don't you fucking dare pull the approximation joke again
gritty widget
lmao
marsh forge
Topology uses homeomorphisms which are isomorphisms in the category of topological spaces
Isometry is not (in my opinion) a topological notion
modern trail
isometry is homeomorphism preserving distances
plucky veldt
isometry is a metric space notion
metric spaces are a subcategory of topological spaces
or rather metric spaces are a subcategory of topological spaces with additional structure
what's redundant?
I didn't mean it in the category theory sense
proven raven
say you have two seperate rubber bands. is it possible to loop one through the other without making any cuts?
proven raven
no
proven raven
I think so
oh it's not a problem i'm working on it's just a general question i had
oh, sorry
well like in the picture for a hopf link. say you had two seperate rings that can't pass through each other or be cut, would you be able to end up with one looping through the other?
does this make any sense
i haven't done topology yet so im not sure of any terminology that much
yeah
dusk heron
Why is the deformation retract in the last paragraph well-defined? More precisely, how do we know that $(x,(1-t)s+ta)\in\Phi_1(W)$ whenever $(x,s)\in\Phi_1(W)$ and $t\in[0,1]$?
gentle ospreyBOT
gustavn64:
dusk heron
By the way, is anyone here experienced with removing coordinate singularities in Riemannian manifolds? On wikipedia it is claimed that this Riemannian manifold has topology $\mathbb{R}^2\times S^2$, but it seems like this coordinate expression only gives a Riemannian metric on $(r_0,\infty)\times S^1\times (S^2\setminus{N,S})\cong (\mathbb{R}^2\setminus{0})\times (S^2\setminus{N,S})$, where $N$ and $S$ are the north and south poles. How would I go about showing that this metric extends smoothly to all of $\mathbb{R}^2\times S^2$? I would need to find some new coordinates in which the apparent singularities disappear, but how would I know which coordinate system to choose?
gentle ospreyBOT
gustavn64:
gentle ospreyBOT
slimvesus:
uncut geyser
is the fundamental theorem of curves that given two curves with the same curvature and absolute torsion, there exists an isometry between them, the reciprocal of that or both?(Ik both are true), but Id like to know what the theorem itself states
noble dock
I’m sorry if this is the wrong channel for this but
is it known the highest number of categories that a Venn Diagram can have on the euclidean plane?
Where all sections must be equally shaped and sized, but are allowed to reflect and rotate differently
For example, this is allowed
but this is not
Not just circles
frosty sundial
this says that you can only construct symmetric ones for prime number of sets
as in being prime is necessary to the existence
but it doesn't say it's sufficient
here, "symmetric" means rotationally symmetric about the center point
ah okay
is that saying you just need to add "connected" to the hypotheses?
noble dock
also just want to quickly verify this is the correct channel for this question?
k just checking
frosty sundial
also, yeah it seems that connectedness is the only issue
noble dock
Topology’s always interested me, but I don’t think I have the necessary background for it yet
what’s connectedness mean here?
frosty sundial
I think that the regions in the diagram are connected
like
you don't want to have the "A but not B" region be split into two parts
noble dock
how is that rigorously defined out of curiosity?
just that between any two points in a region there exists another point? no, that won’t work
why can’t you write R as the disjoint regions n<0 and n>=0? Or is that not what disjoint means here?
I know the word in terms of sets
ah
right
and I assume you mean two continuous disjoint regions?
actually first- do you mean exactly two or at least two?
or are they the same in practice?
intuitively I feel like there’s a counterexample, but I can’t think of one
gentle ospreyBOT
slimvesus:
noble dock
like, a way you could write a connected region that way
Can I ask what I should know before learning topology?
I assume a set is open if for anything in the set, there is something closer to the edge than it
or at least that’s how I’d formalize it, without much foreknowledge
and obviously “edge” isn’t like a line, but you know what I mean
right
any union of disjoint open sets would necessarily leave a point between them, so it wouldn’t be connected, right.
But how can you write, for example, [0,1] and [2,3] as disjoint open sets? Wouldn’t the closeness of the originals force sets that make them up to be closed?
yeah sorry I should’ve made that clear
And I assume you mean finitely many open sets
?
anything else?
I’m working through a book for real analysis right now
Tao’s
no- well, not as I understand topology at least
marsh forge
Your informal defn does not quite work
noble dock
mine?
noble dock
Is Rudin a textbook author?
marsh forge
A set is open if it you can draw some small ball around every point such that this ball is still in the set
The ball has to he completely in the set
Be*
noble dock
Oh so you mean Rudin’s analysis book, not a topology book by him?
plucky veldt
I mean, formally, a set is open if it's open, you can define a topology where any subset of a space is open
noble dock
I might see if I can find a way to get access to that book then
plucky veldt
but I think in most reasonable spaces aka manifolds you have to have a ball around every point
marsh forge
Replace ball with basis element and my claim is formal anyway
marsh forge
Point set topology isnt what most mathematicians mean when they say topology
But its a preteq
Prereq
plucky veldt
point-set is about the nitty gritty formal details of topology
noble dock
okay sure
marsh forge
Most of them useless
noble dock
Honestly, I’m fine with definitions
marsh forge
I only recently got a hold on like
What T0,T1,etc are
Bc no one uses them
In actual topology
noble dock
at least when something gets defined, I know we’re pretty much just declaring it to be that way, which instantly answers “why?”
plucky veldt
I wonder if any topology is done on non-metrizable spaces
noble dock
and since I tend to mostly get hung up on the “why?”s, definitions are great
marsh forge
Someone should write a good point set book that actually focuses on what like
Topologists care about
Rather than set theory memes
noble dock
I’m trying to dabble a bit in every main branch of math before heading to college so I can hopefully pick my classes to guide me down my preferred branch
marsh forge
Topology is best done after analysis as much as I dislike analysis
plucky veldt
I think point-set topology is pretty exciting
marsh forge
Thats partially true
But like
If you like topology you might not need to bother with pde
Anyway take whatever you want
noble dock
pde?
marsh forge
Partial diff equations
plucky veldt
point set topology is interesting because it formalizes intuitive ideas about spaces
marsh forge
Thats not exciting tho
You can formalize all types of stuff
Its usually a boring process
The whole point is to get a solid foundation on which to build exciting math
noble dock
eh, I’m enjoying it so far in my analysis book
marsh forge
And be confident your informal ideas can be made formal
Maybe in Tao2?
I have no idea
Also used rudin
noble dock
It is disappointing that Khan Academy-style resources always stop at Linear Algebra
marsh forge
A few people have tried
Idk if khan-style meshes well with proof based math
Yeah
Woke
Tbh idk why someone would want mayer vietoris
If they weren’t planning to actually learn stuff
Oh sure yeah
That wouldnt be a bad video series actually
maybe i should do that
Busy
noble dock
Even just a series that covers core concepts and what a course is about
noble dock
I’ll watch that when I get home for sure!
hopefully I can read whatever’s on the.... greenboard? what do you call those
dense sundial
I call them "blackboards that happen to be green"
plucky veldt
for point-set topology you don't need abstract algebra
willow spear
Wait why are metric spaces introduced first before introducing topologicla spaces
marsh forge
because topological spaces generalize them
uncut geyser
jeez
egregium gauss is a pain to demonstrate
chrome dew
@uncut geyser what do you mean?
uncut geyser
@chrome dew like getting the gaussian curvature through the connecting(is that how it is in english) form and dual forms
chrome dew
oh I see, I thought you meant demonstrate by examples
uncut geyser
well
here we use demonstrate as a synonym to prove
I guess the usage is slightly different
quiet pilot
I'm having some trouble following the proof of 3 implies 1
The map p is the map from R to S^1 which is just (cos 2pi *x,sin 2pi * x)
I thought I understood the proof, but then I realized I have no idea why it is important that p_0 generates the fundamental group
quiet pilot
This is Munkres topology btw
dim meadow
@quiet pilot if h_* is trivial, then you can lift h to a map S^1 to R. Then you can make a homotopy between the lift and a constant map (because R is contractible) and then compose that homotopy with the covering map to get a homotopy between h and a constant map
signal venture
F^+(U) here is defined as a set of functions from U to the union of the stalks, but how will that be a group? you need the codomain of the functions to be a group to use pointwise addition right? i'm looking at another source that defines the codomain as the direct sum of the stalks rather than the union
that picture is from hartshorne so seems doubtful it's incorrect
wanton marsh
this doesn't mention group anywhere
signal venture
everything he's done with sheaves has been in terms of abelian groups so far, and saying the definitions naturally generalize to other structures
wanton marsh
well then each stalk is a group, right ?
signal venture
yes. but the union of groups is not a group in general
wanton marsh
and when adding pointwise, you add germs from the same stalk
s(P) is in F_P, t(P) is in F_P too, and so you define (s+t)(P) = s(P) + t(P) where the addition is the one in F_P
wanton marsh
yep
signal venture
it's hard to keep this organized in my head
on a related note, is there some reason for the names of section, stalk, germ? i keep getting the terminology mixed up
wanton marsh
a germ is something small that can grow into a larger thing, or in this case possibly determines a larger thing, like a seed that grows
I don't think I have a nice image for stalks
a small vector space that begs to get bigger ?
maybe a bunch of lines pinched together
signal venture
hmm. so germs are classes of pairs <V,s> where V is an open set in the topological space and s is an element in F(V). but i don't see the analogy in how it determines a larger object
wanton marsh
also because you are looking at possible graphs of functions near a point
if you are familiar with the Zariski topology or complex analysis, germs of functions completely determine the function on a much much bigger region than just the small open defining it
well uh
in the Zariski case, the open themselves are very big
so maybe that wasn't the best choice
:s
signal venture
yeah i think i've seen that
also i think it's that s in that definition above is determined on a whole open set V by one element t
wanton marsh
so they carry a lot of information
in more boring cases like continuous functions R -> R
they don't carry much information at all
signal venture
gonna try to finish this problem i'm working on. thanks for the clarification
quiet pilot
@dim meadow yes I believe I understood those steps, but I'm not sure where we use that p_0 generates the fundamental group which Munkres points out
Is this a redundant part of the proof?
signal venture
ah, the ole double-proof. classic move
tough imp
Auvera, sheafification and all that is super duper hard to keep track of at first
Think about the functions as just tuples
So for each p in U, you have an element of the stalk F_p and the condition that f(p) in F_p is just saying it’s a tuple indexed by U
For uncountably many points this gets messy, but it’s the idea to have
FWIW it took me like 4 hours to prove the statements about the sheafification the first time, and about 1 hour the second time
But you’ll get used to it and start to think about it way more intuitively in time
Just my 2 cents
signal venture
yeah the first condition is clear enough to me i think. for each p in U, f(p) is some pair <V_p,s_p> in the stalk F_p
it's the second condition in that picture that hasn't clicked for me yet
sheafification is the process of turning a presheaf into a sheaf?
tough imp
Yeah so
Think local representability if that makes sense
Let me give an example, are you familiar with a bit of complex analysis?
@signal venture
If so I can maybe give an example that might make it make a little bit more sense. The sheafification is made basically of bits of local data, the germs in the stalk. The thing is if you want to glue germs together to get an actual element of some set you need to know how you can glue them
That’s where that condition comes into place
signal venture
only the basics, i took an intro course in undergrad but it wasn't super rigorous
tough imp
Okay that might be enough
So remember the punctured plane?
The function 1/x is analytic on the punctured plane
The thing is that there’s no function such that 1/x looks like e^f(x) on the entire punctured plane
I think that’s the right example
Sorry it should be the function z
What we’re looking for is the logarithm of z essentially to say that z = e^f(z)
But you can’t do this on the entire punctured plane because of branch cuts, you pick up this extra 2pi when you go around the punctured plane
So there will always be some place where you suddenly jump and log wouldn’t be continuous
So there’s a sheaf of analytic functions of the punctured plane, and z is in there, and there’s also a presheaf which is the functions of the form e^f(z) where f(z) is analytic one the punctured plane
They aren’t equal by this example
signal venture
so you're trying to invert 1/z, but strictly speaking an inverse doesn't exist
branch cuts always threw me off
signal venture
i have to go into a meeting now but i'll be back soon
tough imp
Because normally e^log(z) = z
No worries, I’ll finish typing up and you can ping me whenever you get a chance to look at it if you have questions
So one thing is that on any sufficiently small disk, any analytic function has a “logarithm”, the way you construct it is by means of a path integral
So for any z_0 in the punctured plane, there’s some nbd U along with a function f_U(z) such that z looks like e^f_U(z) on U
The thing is you just can’t define a global logarithm for z
marsh forge
side question: does the actual construction of sheafification come up
its determined by a universal property and the actual construction seems gross
tough imp
Sometimes when you want to work with it more directly yes
tough imp
Umm, yeah proving the universal property lol.
It also sort of matters for subjectivity of a morphism of sheaves
marsh forge
okay but thats proving that your construction works not that sheafification works
tough imp
Since you require the sheafification of the image sheaf to be isomorphic to the target by a specific map
tough imp
In AG though you usually work only with sheaves though
And often times if you get something that needs to be shesfified
You just go by means of universal property
The fact that it’s adjoint to the forgetful functor usually means if you want to show like the sheafification of something is some limit in sheaves
That it suffices to check the presheaf is that in presheaf land
Or colimit or whatever I forget which one haha
marsh forge
depends on which side adjoint and i forget which sheafification is
tough imp
It’s left adjoint
marsh forge
then colimits
tough imp
Yeah
marsh forge
RAPL
tough imp
Yup
I just didn’t want to think about which side it is lmfao
Anyway back t my example with complex analysis lol
So the fact that z looks like e^f_U(z) on these nbds U which cover the punctured plane actually means that z exists in the sheafification of that presheaf of functions of the the form e^f(z)
Note that for all those U’s, that e^f_U(z) lives inside of that presheaf on the set U
So z is sort of “locally representable” by elements of the presheaf, but you need to know that all those germs can glue
So for all these p in the punctured plane, you have that you have some U a nbd of p and z looks like e^f_U(z)
What’s important is that z looks e^f_U(z) on all of U, not just like at that point
Elements of the sheafification of F looks like this collection of germs, so you have like for all p in U you just have some little tiny local data for p but you need to know that basically all that local data actually looks similar
There’s an exercise in Hartshorne, II.1.16 e) where you consider the “sheaf of discontinuous sections” which is basically the same as the sheafification except you don’t force the second condition
I think something that helps to see why that condition matters is to ponder the following question:
Given a sheaf F, and an open set U, along with an element s_p in F_p for all p in U, how can we construct an s in F(U) such that the germ of s at p is equal to s_p for all p? You’ll find actually that if you pick some random s_p’s it’s impossible to do this, you need some open sets along which you can glue the s_p’s (rather a representation of s_p as an element <t,U>)
The only way to do this is to enforce a condition that is basically going to be equivalent to the second condition of the sheafification.
It’s a bit hard to put into words all my thoughts about this 😓, but hopefully this helps even a little bit. Try to see if you can’t solve that problem I proposed, and try to construct the map from the sheafification which Hartshorne mentions. I think you’ll get a better feel for why that condition is needed, as you’ll definitely need to use it
signal venture
okay this is a lot to think about
i'm trying to specifically identify the sheaf you're looking at here
your space X is the punctured plane
the sheaf sends the open sets in the punctured plane to some set of functions defined on that open set?
@tough imp
signal venture
oh ok
tough imp
{f:U->C| f is analytic}
And the presheaf is {e^f| f:U-> C is analytic}
On every open set U
signal venture
e^f because we're trying to define a logarithm here
tough imp
Yeah
So if e^f = g
Then we say f is a logarithm of g
So we want to find a logarithm for z
Which normally is just log(z)
But the fact that log picks up the change in argument when you wind around
Means you can’t define it on all of the punctured plane
(And have it be continuous)
signal venture
the sheaf F^+ associated to that presheaf F would strictly speaking be defined by F^+(U) = { f : U -> union of stalks | f(z) is in stalk F_z, and f has that second property }
i'm trying to see how that union of stalks turns into C
and i'm guessing that second property corresponds to f being analytic
tough imp
If by C you mean like the set of analytic functions on the punctured plane
signal venture
{f:U->C| f is analytic}
@tough imp i was comparing to this
tough imp
Oh the stalks isnt C
So you think of the function itself
As a tuple
So for any point p in U
It picks out some element in the stalk F_p right?
signal venture
yeah
tough imp
Think of that as saying that “close to p, f looks like that element of the stalk”
So like, basically the element of a stalk is an equivalence class
<g,U> where U is a nbd of p
What this says is just that f|_U = g
So suitably close to U, f looks like g
But that says that the germ f_p is the same as that germ in the stalk (here I’m saying let f be the analytic function you can make out of the f:U -> union of stalks)
f really is a tuple indexed by U
And in each index you have little local data that jjst says “close to p I look like this analytic function”
But if you didn’t enforce any sort of coherence (this is what that second condition is)
What’s to say that like at a point p I look like say sin(z)
But for points arbitrarily close to p I look like say, z^2 + 2z
There’d be no way for me to like, glue that together to get 1 function which is defined everywhere
I think something that helps to see why that condition matters is to ponder the following question:
Given a sheaf F, and an open set U, along with an element s_p in F_p for all p in U, how can we construct an s in F(U) such that the germ of s at p is equal to s_p for all p? You’ll find actually that if you pick some random s_p’s it’s impossible to do this, you need some open sets along which you can glue the s_p’s (rather a representation of s_p as an element <t,U>)
@tough imp this is what this is about
signal venture
damn it this is hard lmao
tough imp
It really is the first time
signal venture
we dont want to just any functions from f : U -> union F_p, p in U. i guess we want to restrict to ones that will make F into a sheaf
tough imp
I don’t remember if you need that to make F a sheaf actually
This is really important for the maps out of the sheafification though
signal venture
the universal property?
tough imp
Yup
So do you know the property of localization?
And how that proof went
Basically the requirement that the diagram commutes means that for some subset of S^-1A you’re forced to defined the map a certain way
signal venture
i probably saw it at one point
tough imp
This is all the elements a/1
But S^-1A has elements of the form a/s
So what about those?
Well actually s/1 is forced to map to something
By commutativity of the diagram
So actually 1/s has to map to that thing’s inverse
signal venture
yeah this sounds familiar actually
tough imp
Then multiplicativity locks in what a/saps to
So I’ll give you a little hack
So the map F-> F^+
Maps an element s in F(U)
To the function f:U -> union of stalks
By f(p) = s_p
So all it does is at every point p says “you look like s”
Now commutativity of that diagram means that function needs to map to something
Given a map F-> G
It’s locked in
signal venture
yeah actually when you said that before, i realized i did something similar in this problem i've been working on. hartshorne problem to show that the constant presheaf associates to the constant sheaf he defined. and the morphism i defined did something super similar
tough imp
Yeah!
signal venture
i haven't finished it yet so don't have all the details
tough imp
So we have a formula for what we map some subset of F^+ to right?
Somehow it needs to map to whatever the original s did
So the question is basically, how can we extend this to all of F^+ right?
Because not everything in F^+ looks like f(p) = s_p for some fixed s in F(U) right?
But locally everything in F^+ actually looks like that
So you actually leverage that to get both uniqueness of that map, and to even show it exists
signal venture
your terminology is confusing me. F^+ is the sheaf, so a subset is of sheaf is?
tough imp
Uhh
I use it colloquially
I just mean for some elements of F^+(U) for various U
Namely the elements that arise in the image of the maps F(U) -> F^+(U)
These are the ones that just say f(p) = s_p for all p in U, given an s in F(U)
I’m trying to draw a parallel to how it worked for localization
There we knew where had to map every element of the form a/1
But it turns out the conditions means that gives you the formula for everything
signal venture
damn i feel so dumb, i didn't even realize there was an inclusion there haha
tough imp
It’s hard
It took me like
30 minutes to figure out that’s the map haha
The first time I did it
Hartshorne calls this map theta
But do you see what I’m trying to get at maybe?
signal venture
this actually helps me a lot. so F^+ really extends F by enlarging each F(U) to some bigger set F^+(U)
tough imp
Yeah basically
Remember how I said like
Because z locally looks like e^f(z) for various f(z) it’s in the sheafification of that presheaf?
signal venture
i think i see what you're saying
signal venture
if you have a morphism from F to some G, then since F^+ extends F, you nearly already have a morphism from F^+ to G
tough imp
That sort of makes that rigorous
Yup
You have it on some like “subset”
But the fact that you require things of F^+ to locally look like stuff which falls into that subset
The conditions of a sheaf means that the map is unique first of all
Then the fact that you can glue in sheaves
signal venture
i'll have to iron out that inclusion F(U) -> F^+(U), that seems really important
tough imp
Means you actually can even define that map
Yeah, I showed you the formula for the map on each open
But you have to work out it’s actually a map of (pre)sheaves
But that follows kind of trivially
tough imp
Yeah it really is weird at first
It gets surprisingly intuitive
Basically like umm
Sheaves are different from presheaves because they’re more local right?
You can glue stuff if they agree
signal venture
right
tough imp
Ideally we want to be able to glue germs to get an element of like the sheaf
A section on some open
But we should only be able to glue stuff if they agree in some way. This is what it means for them to agree on intersections
This is the like
For a U, and all p in U, we have s_p in F_p
How do we get an s in F(U) whose germs are s_p?
We’re gluing the s_p
But we need to know they “agree”
For a U, and all p in U, we have s_p in F_p
@tough imp also this is exactly the same as a function f:U -> union of stalks such that f(p) in F_p
signal venture
oh yeah, i can see a connection there
tough imp
Yeah so trying to figure out how that condition actually lets you glue
That’s like one of the biggest hurdles
The other one is
consider F^+ right?
The elements like f(p), our local data
signal venture
seems like i really need to do the proof of showing this theta exists. i think that will help
tough imp
Are germs of a pre sheaf
How can I glue those?
You actually can’t
Buuuuuit
If you have a map F -> G and G is a sheaf
You can glue stuff together in G
Once you prove sheafification is a sheaf, and has the universal property
I think everything makes a lot more sense
It’ll be hard
And it’ll take time and seem like you have no idea what to do
But at the end you’ll see you used every single condition you needed
And that that’s exactly the least amount of stuff you need to assume to make it work
It’s actually really remarkable how it works out
tough imp
GL 👍
signal venture
this helped a lot i think
tough imp
Np, this stuff made me@super confused at first too haha
signal venture
i should have taken that exclamation point more seriously
west brook
http://www.polytope.net/hedrondude/primary6.htm
I found an error on this webpage. It says hop is the only self-dual but gaje is, too.
tough imp
Yeah @signal venture verify everything that guy says
And expect all of it to be completely non-obvious
signal venture
i'll remember that 🤣
honest narwhal
emails the author
"Tbh I forgot"
west brook
There’s actually one called “shit”.
spark acorn
And expect all of it to be completely non-obvious
@tough imp Exactly! he did the same thing when defining sheaf of rings on SpecA . I mean he said that it's easy to see that this a sheaf and done tf ? lmao
west brook
It’s the pentacosihexacontahexadecahecatondodecaexon.
tough imp
That has to be fake words right??
Like that page is just nonsense
Oh right
I skipped my 8th grade smskxkdoometry clas
I looked it up but apparently a peton is a real thing
A 5-d face in a polytope but I couldn’t find anything about a verf
True
honest narwhal
Some people didn't take smskxkdoometry in 8th grade and it shows
tough imp
Honestly
The AG people who do all the derived category shit sound the same as the stuff on there
Just they have an air of “this is high level math” where as that just reeks of crankery
But if you hear some people talk it might as well be talking about verfs and tacs
honest narwhal
That's pretty true lmao
west brook
A verf is a vertex figure (https://en.m.wikipedia.org/wiki/Vertex_figure) and tac is the 5-orthoplex (https://en.m.wikipedia.org/wiki/5-orthoplex).
slim pilot
can anyone help me out here?
honest narwhal
@slim pilot wrong channel, go to #geometry-and-trigonometry
slim pilot
ty
west brook
Am I allowed to post links to other Discord servers here?
‘Cause I’m on the Polytope Discord and can link that.
burnt spruce
@west brook there's a polytope discord?
west brook
@burnt spruce yes.
chrome dew
was it started by a guy named peter
west brook
It was started by spiritbackup. Is that Peter?
tired lotus
If a pyramid is to a Pentatope, and a cube is to a tesseract, what is a regular icosahedron to?
Can someone help me with this problem?
west brook
An icosahedron (ike) is to a hexacosichoron (ex).
cobalt crescent
hold up spiritbackup?
west brook
Yes.
thin bramble
Will topology be helpful for Applied math?
uncut geyser
@thin bramble yes
but it really depends on what youre gonna do
for areas such as optimization(esp nonlinear), it is really useful
anything over PDEs also requires some knowledge of topology
solving and modelling dynamical systems also does
numerical analysis requires some basic topology as well
thin bramble
@uncut geyser Thank you! Looking to do PDEs and applied analysis in grad school
uncut geyser
@thin bramble doing PDEs rn 
gentle ospreyBOT
slimvesus:
honest narwhal
What's vPhi^1 here?
Right tangent vectors are derivations sure
I'm still too used to thinking of embedded submanifolds of R^n
So v Phi^i = 0 by proposition 5.37
Wait I think that thing you said might just be true
gentle ospreyBOT
slimvesus:
honest narwhal
I need to look up Lee's definitions one second
But the definition I'd have to guess for df_p(v) where you think of v as a derivation
Would be basically that
gentle ospreyBOT
slimvesus:
honest narwhal
Well, if you map to R
Then we have specific choices for f
Identity maybe?
dF_p(v)(id) = v(F)
Okay that's prob not the idea but I think you just show that those two sides are equal for all f
Well okay dPhi^i_p(v) is a derivation not a real number right?
Well, R
Leibniz rule maybe
Phi^i
So wait no I think I'm getting the idea here
Assume v(Phi^i) = 0
For all i
Then dPhi^i_p(v)(f) = v(f \circ Phi^i)
Wait meh derivation should be with respect to product, not composition
What I don't like is that by these definitions that doesn't type check
I guess we're identifying points and tangent vectors on R is the thing
Lol manifold tech is obnoxious sometimes yeah
Okay so
Let f = Phi^i so I don't have to type 4 extra keys
So I'll call the derivation df_p(v) = T
v(f) = T(id) here
f(p) = 0 right? Because S is the level set
So T is a derivation at 0
Now T(g) = T(g id) = id(0) T(g) + g(0) T(id) = 0
@gritty widget does this check out?
I guess the key here is that if you give me a derivation at 0, if it's 0 on the identity it's just 0
gentle ospreyBOT
slimvesus:
honest narwhal
How do we know the second equality?
Is it v(Phi^i * f)
Wait but v(f o Phi^i) is a number
So then what's (vPhi^i)f
Sounds like a function
Oh wait
Times identity
I'm dumb
g isn't g times identity
v(Phi^i) = dPhi^i_p(v)(id), that much we can be sure of, right?
Well derivations on R are just scalar multiples of taking the derivative
Right?
I feel like something like that if you unpack it works
Chances are I'm being dumb and Liquid's just gonna come here and be like lol, anyway I should prob get back to my work
tough furnace
"The equation of a sphere can be described by the equation: x2 + y2 + z2 + Ax + By + Cz + D = 0" ==> What are A B C D?
Taken from this : http://ambrsoft.com/TrigoCalc/Sphere/TwoSpheres/Intersection.htm
lucid turret
Hey plz why when I have Riemannian connection on a surface in R^3 (the orthogon projection onto tangent space) the curvature tensor is zero?
fleet arch
I am attempting to show that the union of the $xy$- and $xz$-planes in $\mathbb{R}^3$ is not a surface with boundary. I know that im supposed to find a point such that every open set containing it cant be homeomorphically mapped to an open set of $\mathbb{H}^2$, but I dont know how to do this
gentle ospreyBOT
𝔸kash:
fleet arch
Probably the points where the two planes intersect would cause problems
Im not quite sure what the problem is though
marsh forge
Thats a good idea
Whats so different about these points?
Do they look euclidean?
Or do small neighborhoods of them?
fleet arch
umm, im not sure what you mean by look euclidean
marsh forge
Does it look like a plane?
fleet arch
no
marsh forge
Thats basically the entire idea
fleet arch
yep
marsh forge
Well here is how Id prove it
Let f be a homeo from a neighborhood of the origin to a plane. We can remove the x axis from the former, and f will still be a homeo onto its image. This consists of removing some line from R^2, which of course splits it into two connected components. However, removing the x axis splits the small neighborhood into 4 connected components, a contradiction
I think methods like this are how you normally prove these types of statements until you just stop proving them lol
marsh forge
In this context you can think of a connected component as the sets of points connected by paths in your space
So like if i remove the x axis from the union of two planes
I cant cross from one into the other anymore
marsh forge
Yes
marsh forge
Same proof works except you have two cases
Either you remove some random line in H^2 and have 2 components
Or you remove the boundary
And have 1
But both 1 and 2 are less than 4
fleet arch
I see
marsh forge
Tbh proving a homeo doesnt exist normally requires some trickery like this
So im surprised a prof would give it to you
Without any machinery
Unless the prof just wants you to look at it and be like ‘yeah not a manifold’
fleet arch
potentially silly question: how do you know what the image of the x-axis looks like under f?
how do you know it's a line?
marsh forge
Well
Its homeo to a line
By f
So i was being a bit loose
But theres no way to cut a plane in 4 with something homeo to a line
(You probably need some heavy machinery to prove that rigorously)
fleet arch
I see
gentle ospreyBOT
slimvesus:
honest narwhal
Okay good I had the ingredients!
dusk heron
Let $V_1,V_2\subseteq\mathbb{R}^3$ be open and rotationally symmetric around the $x_1$-axis, and consider the maps $\pi_i:V_i\to\mathbb{R}\times[0,\infty)$ defined by $\pi_i(x_1,x_2,x_3)=(x_1,x_2^2+x_3^2)$. Now let $\widehat{V}_i:=\pi(V_i)\subseteq \mathbb{R}\times[0,\infty)$, and let $f:V_1\to V_2$ be a diffeomorphism for which $\pi_2\circ f$ makes the same identifications as $\pi_1$. This induces a homeomorphism $\widehat{f}:\widehat{V}_1\to\widehat{V}_2$. Is $\widehat{f}$ smooth?
gentle ospreyBOT
gustavn64:
dusk heron
Note that $(\pi_2\circ f)(x_1,x_2,x_3)$ depends only on $x_1$ and $x_2^2+x_3^2$, and the crucial question is really whether the dependence on $x_2^2+x_3^2$ is smooth or not.
gentle ospreyBOT
gustavn64:
signal venture
@tough imp i finally finished the whole proof. that was quite the adventure
tough imp
Woohoo!
signal venture
i feel like i have a decent mechanical understanding of what's going on now. at least i feel comfortable with the definitions and how to use them to prove some stuff