(maybe after you pick a starting point)
#point-set-topology
marsh forge
pearl holly
So I'm trying to figure out what the path components of the lower limit topology on R are and then to find all continuous functions from the R to the lower limit topology on R. There is an earlier exercise that wanted me to conjecture what the continuous functions from the R to the lower limit topology on R are, and I answered that these need to be constant. So now I am trying to prove that the path components are singletons. To do so I can prove that the connected components are singletons because every path connected space is also connected. So let A be a connected subspace in R_l (the lower limit topology on R). Now A cap [x, x+e) is open in A. I want to show that it is also closed because then the set equals either the empty set or A. If A cap [x, x+e) = A then A needs to a singleton or the empty set. But the empty set can't be a component since it is contained in the singletons. But how do I show that A cap [x, x+e) is also closed? A - (A cap [x, x+e)) = A - [x, x + e). But then what? How do I know that this set is open?
Hmm but what if A cap [x, x+e) is the empty set? Then A doesn't need to be a singleton
empty grove
A intersection that set is closed in A because that set is closed in R_l
One way to prove that connected components are singletons (such a space is called totally disconnected btw) is to show that given any 2 distinct points, you can find a separation of the space which places them in different components
pearl holly
wait wait... A cap [x, x+e) = A doesn't imply that A is a singleton, right?
pearl holly
okay but if x is in A then? Then it implies that A = {x}, right?
empty grove
Why?
pearl holly
well I guess it doesn't have to be just {x} because [x, x+e) can still be a subset
empty grove
Yeah
lofty stratus
Thank you very much!
pearl holly
One way to prove that connected components are singletons (such a space is calle...
hmm how does this imply that the space is totally disconnected?
empty grove
hmm how does this imply that the space is totally disconnected?
Consider a connected component A. Suppose it had 2 distinct points. Then whole space can be separated while placing them into different components, and take the intersection of that separation with A to get a separation of A
pearl holly
"Then whole space can be separated while placing them into different components" what do you mean by this?
like what are the separations?
empty grove
oof let me be more precise with the claim itself
X is totally disconnected iff given any x and y, distinct, in X, you can find disjoint open sets U and V that cover X, such that x is in U and y is in V
So like, Hausdorff, but U and V must be a partition of the space, not just some small neighborhoods
pearl holly
wait wait, (a, b) is open in the lower limit topology, right?
empty grove
Yez
pearl holly
okay then it's ez using the fact that you gave me
empty grove
The intuition behind the fact I gave is that you can place any 2 points in different components
iff the components are all singletons
empty grove
strong heron
You can see Moldi's result quite directly also in this case. Let $A$ be path connected in lower-limit topology with at least two points, say $x$ and $y$. Then $[x,y)\subset A$ and $(-\infty,z)\cap A$ and $[z,\infty)\cap A$ constitute a separation of $A$, for some $x<z<y$.
gentle ospreyBOT
whysee
strong heron
Therefore, A must be a singleton.
pearl holly
You can see Moldi's result quite directly also in this case. Let $A$ be path con...
oh sorry that I didn't respond. I had to go somewhere. But now I see it, thank you!
It's so sad that :catlove: and :catheart: are gone. There's no way for me to emote a helpful post now
viral atlas
gritty widget
viral atlas
I suggest replacements.
empty grove
frigid river
Could any of you please explain to me what a mapping cylinder is? 0.0 I am not really getting any smarter from the wikipedia definition
empty grove
Consider the map from S^1 to itself which wraps the circle around itself twice (view S^1 as a subset of ℂ and take the squaring map). The mapping cylinder is S^1 x [0,1], but on S^1 x {1} you identify all the things that have the same image
pseudo crane
:hatcher:
empty grove
This gives one example of it, and here you can visually see the cylinder
strong heron
Consider the map from S^1 to itself which wraps the circle around itself twice (...
On an unrelated note, how did you write complex numbers like so in your message?
empty grove
On an unrelated note, how did you write complex numbers like so in your message?
GBoard allows you to import dictionaries, and you can find a GBoard latex dictionary on github
When I type \mathbbC it suggests ℂ in the completions
strong heron
which page is this?
You can browse via the index. Mapping cylinder is there in the index.
pseudo crane
is the first chapter that long
strong heron
When I type \mathbbC it suggests ℂ in the completions
Wow, cool!
strong heron
Page 2, Jon
pseudo crane
yah
empty grove
Gotta import dictionaries 
frigid river
i believe it's in chapter 0
Found it, thanks.
strong heron
Basically you are gluing the cylinder over X to Y via the map f. This is a specific example of what are called identification spaces. You take the union of two spaces and identify some subspace in both of them via a map.
In case of mapping cylinder, you take the union of X×I and Y and identify X×{0} \cong X with its image in Y under f, i.e. f(X) via f.
pastel thistle
some help in order to clarify myself, every point on the X axis is related to each other
all the rest of points in R2 are related to itself
I need to show that the quotient space is not N1
dim meadow
What does N1 mean?
pastel thistle
first countable axiom
there must be a point who does not have an open basis of neighbourhoods
Im being clear ?
Specifically, a space X is said to be first-countable if each point has a countable neighbourhood basis (local base).
dim meadow
I'm a bit confused about what the partition is here
pastel thistle
all the points in the X axis relate to each other
any other point relates to itself
pastel thistle
yhea I tried going by an absurd but I cant get it
I supposed there exist {V_n} open neighbourhoods around {A} (or the class of lets say (0,0) ) and I cant find the problem honestly
dim meadow
Okay I think I see what's going on here
Any neighborhood around that point is the projection of a neighborhood around the whole x axis
dim meadow
Okay so suppose we have a countable set of open sets around the x axis
pastel thistle
okey
dim meadow
We want to make an open set around the x axis which does not contain any of those sets
pastel thistle
(I already proved the projection is closed but not open and that there are (V_n) around {A} such that the intersection of all of them is {A} if that helps somehow)
pastel thistle
We want to make an open set around the x axis which does not contain any of thos...
yes
dim meadow
Okay so how can we avoid a single one of those sets?
Sorry I switched is contained by and contains above
pastel thistle
I see
im kinda lost wait im thinking
I understand your question but cant answer it
dim meadow
Let's say we are given a neighborhood of the x axis to avoid. We can pick a single point on the x axis and look at the largest epsilon so that the neighborhood contains a epsilon ball around that point
We know that such an epsilon will exist and be > 0
dim meadow
Yes it won't, if it did that would be very boring because it would be the whole plain
pastel thistle
you could unite open balls touching just on their "edges" (sorry my english)
dim meadow
Yes that's a way of building a neighborhood, but you have to build your neighborhood based on the list you are given
pastel thistle
Im struggling I follow but cant see any further
dim meadow
Okay so let's reiterate
We have a countable list of open sets containing the x axis
You are given an open set N_i on the list. You pick a point on the x axis. You build an open ball around that point which does not contain the largest open ball around that point contained in N_i
Therefore if you have an open set around the x axis whose largest open ball around your point is this ball, this open set will not contain N_i
pastel thistle
wait but an open set wich contains the whole x axis does not contain all open balls around any point in the axis if we pick small enough epsilons ?
dim meadow
Well the point is the definition of a neighborhood basis is any open set around the x axis contains an element of the neighborhood basis
An open set around the x axis will contain some open ball around every point on the x axis, but not all open balls around every point
pastel thistle
yes
dim meadow
So if we are given an open set A around the x axis and you don't want your open set B which you are building to contain A, then a nice way of doing this is for B to have a smaller maximal open ball around some point on the x axis than A has
If this is the case, B will certainly not contain A
pastel thistle
yes I see
dim meadow
And the nice thing is you can choose a countable number of points on the x axis which are far apart to do this for every element of a countable list
pastel thistle
yes
dim meadow
Okay so do you understand how you can do this now?
pastel thistle
mmmm yes I kind of understand
thank you 🙂 : ) :))))
Im reading again and thinking a little more I think I got the idea
so thank u again, for real
tight agate
gritty widget
tight agate
a bee has been chilling in the same spot in my room for over 4 hours 
gritty widget
free it
tight agate
kodaira embedding
that's a cool theorem
lefschetz hyperplane theorem
uniformization
gritty widget
hollow harbor
best differential geometry theorem
Hopf Rinow or Atiyah-Singer index (but that's a cop out choice)
pearl holly
So this is the exercise that I'm working with: Let X be locally path connected. Show that every connected open set in X is path connected. So let A be an open connected set in X. Then it is contained in a connected component. In a locally path connected space, the connected components are the same as the path connected components. So A is contained in a path connected component. But how do I proceed from here? If we assume that A is nonempty, then it is a neighbourhood of some x in X. Then there exist a path connected neighbourhood contained in A. So now A is "sandwiched" between a path connected set and a path connected component, but I don't see what this gives me. Maybe I'm completely on the wrong path here, I don't know. Any hints?
hazy nexus
Have you seen that a space X is locally path-connected iff for every open set U of X, the path-components of U are open?
pearl holly
yes I have, but I didn't see how I would use this
hazy nexus
Me neither, but it feels like it would be useful
pearl holly
but wait, A is a union of path connected neighbourhoods. These neighbourhoods can't be disjoint because that would contradict the connectives of A. So they all share a point. So the union of these path connected neighbourhoods must be path connected?
hazy nexus
Oh wait, no I do
Well, they can't share a point
Path-components are disjoint
Which means A can only have one path-component
Because having more than one would contradiction connectedness of A, by what you said
pearl holly
but wait, A is a union of path connected neighbourhoods. These neighbourhoods ca...
So this is wrong?
hazy nexus
It's not wrong per say, you just arrived at the wrong conclusion
pearl holly
But I was supposed to show that every connected open set in X is path connected
hazy nexus
And you did
Suppose A is not path-connected
Then it has a family ${ C_i }{i \in I}$ of path-components, and this family contains strictly more than one set. You know that the $C_i$ are pairwise disjoint, and $\bigcup{i \in I} C_i = A$, right?
gentle ospreyBOT
Tormeson
hazy nexus
However, $X$ is locally path-connected, and $A$ is open, which implies that each $C_i$ is open in $A$. But then, that means you just wrote $A$ as a disjoint union of open sets, which contradicts connectedness of $A$
gentle ospreyBOT
Tormeson
pearl holly
oh yeah I see. But how much does this actually differ from the thing that I wrote?
pearl holly
but wait, A is a union of path connected neighbourhoods. These neighbourhoods ca...
I'm referring to this
and the neighbourhoods come from the fact that X is locally path connected
hazy nexus
Oh I'm dumb
I misread your thing completely, lmao
Although you can't really conclude that they all share a point
Think of taking R² as the union of unit balls, one at each point
You're taking the union of path-connected open neighbourhoods, and the union is a connected set, but you can find disjoint neighbourhoods
pearl holly
so let's say that only 2 of these neighbourhoods share a point. Then the intersection of all of these neighbourhoods will still be empty, a contradiction. Right?
So therefore, all of these neighbourhoods must share a point. Otherwise it would contradict the connectives of A
hazy nexus
I don't see what the contradiction would be there. Again, consider $\R^2 = \bigcup_{x \in \R^2} B_1(x)$. I think it's pretty clear that $\bigcap_{x \in \R^2} B_1(x) = \varnothing$
gentle ospreyBOT
Tormeson
hazy nexus
There's probably an argument to be made with what you want to do here, but it will take a bit more work
pearl holly
ohhhh okay I get it now lmao. I thought that the unions and the intersections "behave nicely" in a way. But I forgot that the intersections are distributive over unions. So they all don't need to share a point
Anyway, thank you so much for the help!
I thought that $U_1 \cap U_2 \cdots = (U_1 \cap U_2) \cup (U_3 \cap U_4) \cdots$ lmao
gentle ospreyBOT
Tokidoki
reef shore
So this is the exercise that I'm working with: Let X be locally path connected. ...
A standard trick to showing that a property holds for all points in a connected space is to prove that the set of all points that satisfy that property is non empty and clopen. Here you can fix a point x, and consider the property "- is connected to x via a path". Local path connectedness tell you that the subset of points where this holds is open in U, and so is U - that subset. The former is non empty because x is trivially in it.
gritty widget
Taken from Lectures on Minimal Surface in $\mathbb{R}^3$ by Yi Fang. There are some questions that I could not understand. How does the Laplace operator appeared on last equation and why was $\pdv{\log (1+|g|^2)}{r}$ bounded?
gentle ospreyBOT
weilam06
gritty widget
i.e. my concern were the equation $\int^{2\pi}0 \pdv{\log |f_i|)}{r} R\ dt=\int{D^i_R}\Delta (\log |f_i|) \ dx \ dy$ and the boundedness of $\pdv{\log (1+|g|^2)}{r}$ that need to verify
gentle ospreyBOT
weilam06
sweet wing
best differential geometry theorem
is Mostow's rigidity theorem considered diff geo
cerulean oriole
So this is the exercise that I'm working with: Let X be locally path connected. ...
Is A not also locally path-connected?
pearl holly
Hmm I honestly don't know lol
cerulean oriole
Oh wait
I was thinking take open nbhds A, P of a where P is path-connected
Nvm
hazy nexus
Maybe open subsets of locally path-connected spaces are locally path-connected 👀
pearl holly
maybe open subsets of open subsets of locally path-connected spaces are locally path-connected?
heck, maybe every single space is path connected!
empty grove
empty grove
Maybe open subsets of locally path-connected spaces are locally path-connected �...
This is true yes
empty grove
maybe open subsets of open subsets of locally path-connected spaces are locally ...
This is also true
empty grove
It's just the previous thing applied twice
hazy nexus
This is true yes
yes I know, lmao
empty grove
Couldn't tell if you were being serious or not 
marsh forge
maybe open subsets of open subsets of locally path-connected spaces are locally ...
every open set is an open subset (except the whole space if you require it to be strict)
cerulean oriole
It's just the previous thing applied twice <:KEK:586240877358350341>
...
hazy nexus
I mean, it's true
honest terrace
...
hazy nexus
I mean, it's true
hazy nexus
It's not that much more roundabout, lmao
empty grove
,ban Raghuram
gentle ospreyBOT
gritty widget
,ban tterra
gentle ospreyBOT
pearl holly
,ban TeXit
gentle ospreyBOT
marsh forge
model categories are SO GOOD
ugh
I'd love to really sit down with Hovey
the book or the person
pearl holly
Okay so this is the exercise that I am working with: let p: X -> Y be a quotient map. Show that if X is locally connected, then Y is locally connected.
I have done this: my book defines a quotient map as a surjective map. So p(X) = Y, so we can show that p(X) is locally connected. Pick x in p(X). Then p^-1(x) is in X. Let U be a neighbourhood of p^-1(x). All neighbourhoods of p^-1(x) can be written as p^-1(U_x) where U_x is a neighbourhood of x. So U = p^-1(U_x). We also know that there is a connected neighbourhood V of p^-1(x) that is contained in U = p^-1(U_x). But then p(V) is contained in U_x and p(V) is connected since V is connected. Now we are done since U_x is an arbitrary neighbourhood of some arbitrary point in p(X).
This just feels really wrong. There's a hint under the problem that says "If C is a component of the open set U of Y, show that p^-1(C) is a union of components of p^-1(U)". I haven't showed nor used this hint so I feel like something is wrong
gritty widget
gritty widget
Okay so this is the exercise that I am working with: let p: X -> Y be a quotient...
i haven't read the whole thing, but p^{-1}(x) isn't guaranteed to be a point
p doesn't need to be bijective
pearl holly
It's okay, I won't snitch on you!
hmm but p is surjective. Doesn't this imply that p^-1({x}) is a point?
gritty widget
no
if p were also injective, it would
but multiple points could map to the same x
marsh forge
surjective implies it is nonempty
pearl holly
oh yeah, it could be a set
marsh forge
which is good enough
marsh forge
do you have a citation for
“every open set in X is the preimage of an open set in Y”
this is def not true in general but might be true for quotient maps
no it shouldnt be either
yeah its not true at all
(for example take two points and send them both to one point. this is a quotient map but the open set containing either of just one of the points is not the preimage of any open set in the one point space)
in general @pearl holly it seems like your proofs are way too short and you are making too many claims that make sense to you but are either unjustified or actually false
pearl holly
yeah lmao 
marsh forge
learning to write concise proofs is a good eventual skill but while you are still beginning you need to spell everything out
to stop yourself from lying to yourself
pearl holly
yeah that's actually good advice. I was stuck on this exercise and tried to prove the hint and then I just said "screw it!" and scribbled something that resembled a proof
I will think about this problem later. I'm busy eating popcorn
marsh forge
hahaha okay
just imagine me skeptically asking “why” every time you write a sentence
pearl holly
lmao will do that!
gritty widget
hard mode: imagine ultraproduct going "Why ." every time you write something
marsh forge
its hard to differentiate a genuine "why" and a skeptical "why" from ultra
marsh forge
Maybe what I mean is like
a "why" when you are (almost) certain the result is false
verses a "why" when you think it might well be true but you don't yet understand it
to explain the first question, I think that you just tend to be a bit reserved in general so the excited "why" and the sully "why" are less distinct
frigid river
marsh forge
lol no not at all
its not a big deal and the result for the explainer should basically be the seem either way
(although its more fun to be right when someone is sure you are wrong)
pearl holly
Okay I'm done eating my popcorn and I have thought about this and I think that I'm very close. I just need this to be checked: if C is a component of p^-1(U), then is C saturated?
It must be, right?
marsh forge
Can you define saturated real quick
pearl holly
A set is saturated if it equals the complete inverse image of a subset of Y
and since C is a subset of a complete inverse image, it must be equal to a complete inverse image I guess
marsh forge
So $A\subset X$ is saturated if $\exists U\subset Y$ such that $A=f^{-1}(U)$?
gentle ospreyBOT
MaxJ (Local Snitch)
pearl holly
yes
marsh forge
this is not true for similar reasons to what I said earlier
in fact the same counterexample works
pearl holly
yeah I figured that
marsh forge
Can you help me understand why you think you need this?
pearl holly
Because I have gotten to the point where I have this: $C = \bigcup_{i \in I} p(C_i)$ where $C_i$ are components of $p^{-1}(U)$ where $U$ is an open set and $C$ is a component of $U$. If I manage to show that $p(C_i)$ are open then I'm done
gentle ospreyBOT
Tokidoki
pearl holly
(and U is open in Y)
marsh forge
Im not sure I understand what C is
pearl holly
C is a component of U. So if I manage to show that C is open then by a theorem in my book, I have shown that Y is locally connected
marsh forge
Isn't C_i open by construction
(ill be fully honest we are entering the part of point-set topology where all the counterexamples are horrific spaces no one cares about so i might well make mistakes)
pearl holly
Isn't C_i open by construction
Well all the C_i's are open because they are components of p^-1(U), which is open. So X is locally connected and p^-1(U) is open so every component must be open in X
marsh forge
so p is a quotient map and C_i is open and you want to show p(C_i) is open?
pearl holly
Yes
marsh forge
Can you send the lemma you intend to use?
I'm going to think about a more elementary proof I have in mind
pearl holly
Wait what lemma?
marsh forge
C is a component of U. So if I manage to show that C is open then by a theorem i...
"theorem in my book"
btw i am pretty sure my elementary proof works if you would like a hint toward that, or you can send this lemma and I can think about it
pearl holly
Theorem 25.3
pearl holly
btw i am pretty sure my elementary proof works if you would like a hint toward t...
Oh that's interesting! I just used the hint in the problem that goes like this: If C is a component of the open set U of Y, show that p^-1(C) is a union of components of p^-1(U). I have proved this (I think) and I haven't made much progress since then
pearl holly
Oh that's interesting! I just used the hint in the problem that goes like this: ...
So from this hint we get this
pearl holly
**Tokidoki**
.
marsh forge
Oh I think I might know how to do it
yeah okay this is a bit finnicky
Let me review real quick.
$p:X\to Y$ is a quotient map, meaning that if $p^{-1}(A)$ is open then $A$ is open. The space $X$ is locally connected. This means that given an open subset $U$ of $X$ the connected components of $U$ are open in $X$.
gentle ospreyBOT
MaxJ (Local Snitch)
marsh forge
Your work so far:
Let $V\subset Y$ be open and let $C\subset U$ be a connected component. Then $p^{-1}(C)$ is a union of connected components $C_i$ in $X$.
gentle ospreyBOT
MaxJ (Local Snitch)
pearl holly
marsh forge
Oh now I am worried because it feels like I'm not using the hypotheses
Oh okay no
I am I guess
Okay so, we know that each connected component $C_i$ is open in $X$, do you agree?
gentle ospreyBOT
MaxJ (Local Snitch)
pearl holly
Yes sir
marsh forge
Okay so their union must be right
pearl holly
Yes
marsh forge
Okay, what's the only hypothesis I haven't used?
pearl holly
Well that X is locally connected maybe?
marsh forge
Apparently locally connectedness is why we know that C_i is open
pearl holly
Oh no, you used that one
marsh forge
hint: what I have used about "p"
marsh forge
okay in particular, what strong assumption does the exercise make about p
pearl holly
Hmm I honestly don't know. All I know about p is that it maps saturated open sets to open sets and that U is open iff p^-1(U) is open
marsh forge
hahahaha
marsh forge
Yep!
pearl holly
marsh forge
dont beat yourself up about it ive been doing this for years and this problem still managed to stump me for a bit
the issue with this problem is that like
every obstruction to proving it
is something that is true anyway for any decent topological space
so like
its just a pita
lol
pearl holly
lmao but seriously, thank you so much! I think that I managed to prove the hint as well and I'm almost certain that it is right, I'm just too lazy to write it out now. It's quite late here. But thank you again, I really appreciate it!
pearl holly
So manifolds are locally conencted?
gritty widget
don't sully me slim
gritty widget
So manifolds are locally conencted?
yes
they are very nice spaces
why are you sullying me
river granite
gritty widget
they're probably going to read the manifolds section of munkres at some point
might as well give them something to look forward to
pearl holly
You mean me?
pearl holly
yeah I don't know anything about manifolds lmao I've only read the wikipedia definition but it seems really interesting
gritty widget
he proves a theorem about embeddings of manifolds
they're topological spaces that are locally homeomorphic to (open subsets of) euclidean space
with that definition it shouldn't be a surprise that topological manifolds share with euclidean spaces most of the "local" topological properties you can think of
they're nice
pearl holly
Yeah that makes sense
marsh forge
"all spaces should be locally connected"
"manifold"
i think you skipped some steps
pearl holly
So a coffee cup is a manifold?
gritty widget
marsh forge
there are a lot of compromise spaces in between
I think a manifold should have literally every local property euclidean space has
(a coffee cup is a manifold potentially with boundary depending on how you shape it)
i.e. is the coffee cup hollow?
marsh forge
donuts can also be hollow or filled in
i.e. a torus or something htpy eq to a circle
pearl holly
Wait, is a torus hollow or filled in?
marsh forge
hollow
pearl holly
Oh okay
gritty widget
donut with no filling
pearl holly
Well anyway, I'm going to sleep now 
But thank you so much again!
frigid patrol
Can someone explain the fiber product of covering spaces?
shrewd moth
can someone help me a geometry problem
gritty widget
just ask
modern osprey
cream w no donut
pliant parcel
rock
swift fjord
disk
elder crescent
Hi, can anyone help me with the correct order of the 4th and 5th vertex.
What should we do after we reach the last case that I drew?
gritty widget
It says "topological" sort, so it must mean it belongs in #point-set-topology, right?
I'm not sure I understand the diagrams, but the valid orderings should be c b a e d and c b e a d
empty grove
In the third diagram you can pick either a or e
Both are valid orderings
And yeah this is more #numerical-analysis or #discrete-math
vocal anchor
Does anyone know either a reference or how to see the surgery link (in $S^3$) of $\Sigma_g \times S^1$, where $\Sigma_g$ is a genus $g$ surface (closed oriented)?
gentle ospreyBOT
expectTheUnexpected
fathom cave
I dont understand this problem.
obviously not every metric equivalent to a bounded metric. for example, euclidean metric is not equivalent to a discrete metric on R
hazy nexus
And? The question doesn't ask you to show that any given metric d is equivalent to all bounded metrics, only that there always some bounded metric such that d is equivalent to it
fathom cave
oo i should show give a metric, there is always a bounded metric equivalent to it?
my poor English comprehension is getting in my ways rip
empty grove
oo i should show give a metric, there is always a bounded metric equivalent to i...
Yes
fathom cave
ty
frigid river
If I have $\pi_1(\mathbb{R},x)$, it would make sense that this is the trivial group. But how would I write this? Just as 0? Or do I pick any function, since all are homotopic to each other?
empty grove
Write what? The trivial group?
gentle ospreyBOT
veryignorantperson
frigid river
yes
empty grove
You can write it as 0 or 1 ig
Or {1}
Or ({1}, {(1,1,1)}) of you wanna be really explicit
frigid river
If it had one, would I just write it as $\pi(X,x)={[function]}$, where the [] denotes the homotopy class?
empty grove
But usually you just say isomorphic to the trivial group 1
marsh forge
"it's trivial"
gentle ospreyBOT
veryignorantperson
empty grove
Yeah saying trivial is the easiest way out
frigid river
I still miss :catlove:, the heart ❤️ seems a little too much every time I press it
empty grove
cerulean oriole
Or ({1}, {(1,1,1)}) of you wanna be really explicit
👀
cerulean oriole
Or ({1}, {(1,1,1)}) of you wanna be really explicit
IG this could be useful if it's not easy to see that the group is trivial and you want to point out "here's an element, show that all the others are equal to it".
But otherwise ... no point really.
marsh forge
If it had one, would I just write it as $\pi(X,x)=\{[function]\}$, where the [] ...
oh also just noticed this
be careful to always write
$\pi_1$
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
not just $\pi$
gentle ospreyBOT
MaxJ (Sockoist)
frigid river
Is there pi_2?
marsh forge
There are $\pi_n$ for all $n\geq 0$
gentle ospreyBOT
MaxJ (Sockoist)
frigid river
what is pi_2 then? 0.0
marsh forge
Basically instead of maps $I\to X$
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
you think about maps $I\times I \to X$
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
where $I=[0,1]$
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
Then instead of forcing $0,1$ to be the same point
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
we force all of the border points on the square to get sent to the same point
frigid river
So instead of paths, you have planes?
marsh forge
well
loops are really circles
so these new things are spheres
and then you just use higher dimensional spheres
you have "planes where going to the edge of the plane always takes you to the same point"
just like loops are "paths where going to the edge of the interval takes you to the same point"
frigid river
Okay, thanks. And might I ask, what's a "path component"?
marsh forge
The path component of a point $x\in X$ is all the points $y\in X$ such that there is a path from x to y
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
if $y$ is in the path component of $x$ then the path component of $y$ is the same as the path component of $x$ (exercise)
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
so it makes sense to just talk about the path components of a space
without mentioning a point
fathom cave
i like that exercise part
marsh forge
i think its always good to include exercises in an explanation
forces people to actually think about what you say
frigid river
forces people to actually think about what you say
Intuitively, this makes total sense. But I haven't really done any topological proof yet, so I won't really be able to do this :/
gritty widget
frigid river
you should be doing exercises
I haven't reached the exercises yet 0.0 I am still on, like, the third page of the first chapter.
4th page now
marsh forge
ah well
the exercise i gave you shouldnt be any harder than the exercises in the book
so give it a shot whenever
frigid river
I still have 10 pages to go until the exercises though, therefore I can still procrasinate 0.0
gritty widget
do what im doing with lee and do the exercises before reading the chapter
/s
(this doesn't count as most of it is a review)
frigid river
do what im doing with lee and do the exercises before reading the chapter
I tried that with Munkres, didn't end well
strong heron
Exercises needn't only be the ones that are assigned as exercises in a particular chapter. You should try to make your own little exercises and solve them while you're reading the chapter itself. MaxJ's exercise is an example of one such exercise.
frigid river
that moment when you don't even understand the theorem
first, what exactly is S1? xD
Sphere 1, so any sphere? Or the unit sphere? Doesn't matter, probably
wanton marsh
S1 is the unit circle
frigid river
hmm, close enough
wanton marsh
S2 is the sphere
frigid river
and how does that function look like? Is that a spiral?
wanton marsh
well psi(0) is the constant loop, psi(1) is a loop that goes around the circle once, psi(2) is a loop that goes around the circle twice, etc
homotopy class of*
frigid river
Okay, and "based at (1,0)" just means that the loop starts, there, right?
wanton marsh
yeah it means the endpoints are at (1,0)
frigid river
Okay, and why does this matter?
Shouldn't this be true for all points within the sphere?
wanton marsh
because most likely pi1 is defined with a base point
frigid river
Thank you, and is there a way to visualise this?
If I just imagine a circle, and a loop going around within the circle, shouldn't it just have a trivial fundamental group?
wanton marsh
well the theorem is saying that the fundamental group of the circle is nontrivial
and is giving you an explicit representative
for each homotopy class
frigid river
I get what it's saying by now, but I still can't visualise it.
I think I get it now, thank you.
wanton marsh
and for any loop in the circle there is is an integer n such that that loop is homotopic to the loop that loops around the circle n times
it's called the winding number
marsh forge
S1 is the unit circle
S^1 is just defined up to homeomorphism usually
oh I see in this case it does matter tho
hollow harbor
when you forget the group structure
gritty widget
marsh forge
Before my meeting: "I can't wait for peter to clarify this subtle technical issue so I can move forward with research"
After my meeting: "What do you mean he doesn't know the answer to every roadblock i hit why do I have to think"
rancid umbra
rip research 😔
random slate
This might be kinda random but
. . . why "varieties"?
Has anyone ever heard the reason why that particular word was chosen?
queen cave
@verbal wraith what is a lattice, also can i become your apprentice latice understander
👉👈
verbal wraith
A semi-lattice: it's a poset with an intersection operation P(X) -> X.
$\forall, S \in P(X),, x \in X,, x \leq \cap, S ,\Longleftrightarrow, \forall, y \in S,, x \leq y$
gentle ospreyBOT
NikitaShlyapustin
verbal wraith
You can classify some collection of things by a semi-lattice of properties.
so like 'x in A', 'y in B'.
and the reason why intersections are useful is you have a function to the semi-lattice that tells you all the properties that some element x has
queen cave
and the reason why intersections are useful is you have a function to the semi-l...
I think i kinda sorta have the intuition for the concept, because i can see how properties would be "ordered" (e. g. x is zero, $x \in {0}$. x is zero or more, $x \in {y | \forall y \in R . y \geq 0}$. So in a sense the property ${0}$ is "less than" ${y | \forall y \in R . y \geq 0}.$) I'm a bit lost as to the meaning of the double implication you posted though
gentle ospreyBOT
jcob
verbal wraith
It's the definition of an intersection operation
queen cave
Oh ok!! So it's just saying that if a property is in the meet/intersection of a bigger property, it is that property or it's less than that property?
is the "is included in" what makes the partial order? This is really neat 😎
verbal wraith
Idk what you mean by intersection of a property. For properties you call the intersection operator and. Like A and B and C is the intersection of {A, B, C}.
queen cave
Idk what you mean by intersection of a property. For properties you call the int...
i guess i was saying $\cap {A, B, C}$ would itself be the property x is A and B and C
gentle ospreyBOT
jcob
verbal wraith
Something like that
frigid patrol
. . . why "varieties"?
The Italian translation of the German word "mannigfaltigkeit" (manifold) is "varietà" (variety)
Varieties are Italian manifolds
frigid patrol
Heres some more stuff
https://en.m.wikipedia.org/wiki/History_of_manifolds_and_varieties
empty grove
The French at it again
fading vale
this is kind of more based terminology
pearl holly
(I will write f instead of f with the weird ~ above. Likewise with w_n) So I am reading this proof and I am kind of confused with the very last sentence that I've marked with my cursor. Obviously, pf is not equal to pw_n, they are only homotopic. So the definition of psi does not match the original. So lets set a(n) = pf. Now it feels intuitive that if a(n) induces a isomorphism of pi_1(S^2) with Z, then so does psi because pf and pw_n are homotopic. But how do I prove this formally?
Like what does the author mean when he writes "and the new definition of psi(n) agrees with the old one"?
empty grove
Definition of psi does match the original, because in both cases psi maps n to the homotopy class of w_n
Psi maps n to [pf], and [pf] = [w_n]
pearl holly
oh lmao I completely missed the homotopy class part lol
marsh forge
i.e. two functions named psi are defined here which agree once you think about homotopy class but don’t necessarily agree before then
pearl holly
oh okay I see now. Thank you!
pearl holly
You here?
frigid river
yep
pearl holly
okay good
Okay so now I am reading the part that start with "to verify that psi is a homomorphism"
Do you understand everything above it?
frigid river
Let me read that first 0.0
pearl holly
(I will post this section here so that we are on the same page)
(ignore the blue thing)
frigid river
I think I understood most of it, I will notice if I didn't later anyway.
Still not sure what exactly a "lift" is though.
pearl holly
yeah I have no idea lmao. I just get used to the weird definition names lol
pearl holly
oh lol
oh, because the lift literally lifts up the other morphisms I think
Look at the diagram
frigid river
the one in hatcher?
pearl holly
no, I mean the one in the wikipedia page, sorry
so if we have f and g, then we can "lift them up" by letting h=g 0 h where 0 is composition or something idk
frigid river
so if we have f and g, then we can "lift them up" by letting h=g 0 h where 0 is ...
not sure what that means, but should we just try reading the next sentence? 0.0
pearl holly
so if we have f and g, then we can "lift them up" by letting h=g 0 h where 0 is ...
and you could draw a similar diagram for the scenario we have here
yeah I think that we shouldn't get caught up on definition names lol
okay anyway, so now we are in the sentence that starts with "to verify that psi is a homomorphism", right?=
frigid river
yep
frigid river
still trying to figure out what a translation is 0.0
pearl holly
You basically translate the path. So if the path is from 0 to 1, then after the translation, it will be a path from m to m+1
something like this
marsh forge
this dotted line when it exists is called a lift of the bottom map
the image is that like
you are literally trying to lift it up
and you cant always
this is called a “lifting problem”
pearl holly
marsh forge
you can take the bottom map or the vertical map to be an inclusion
if the vertical map is an inclusion
thats wrong whoops
pearl holly
okay yeah that explains the name lol
marsh forge
oh okay yes if the vertical map is an inclusion and the lift exists
that tells you that the original map didnt land in the rest of Y
and this is an iff
pearl holly
oh okay, good to know! Thank you!
marsh forge
sorry for derailing lol
frigid river
Sorry toki, I don't have any idea what's going on in that second part xD
Was $\tilde{\omega}$ the thing "going vertically"?
gentle ospreyBOT
veryignorantperson
pearl holly
Well I think that we can set aside all the category things
Just forget about all the lines and let's look at the definitions instead
(by lines I mean all the lines in the diagram that we talked about)
frigid river
yeah, I wasn't even trying to understand those 0.0
pearl holly
Okay so what are you stuck on now?
frigid river
I am trying to understand what the second part wants me to understand 0.0
pearl holly
hmm okay and what second part are you referring to?
frigid river
Then some greek symbols is a path in R from 0 to m+n.
at first, what's m and what's n?
pearl holly
they are just integers
frigid river
n is the winding number, right?
pearl holly
yes
frigid river
and what's m?
pearl holly
just another integer lol
frigid river
yeah, but what does it represent?
pearl holly
the winding number
frigid river
of what?
marsh forge
no
frigid river
0.0
marsh forge
its not a winding number haha
maybe it will help if i draw a picture? do@you mind toki?
i dont wanna barge in but i think this t_m business confuses most ppl seeing this for the first time
pearl holly
Yeah sure. I'm reading with veryignorantperson and we are trying to understand this proof together lol
empty grove
w_n does go around the circle n times tho doesnt it? 
pearl holly
yeah that's what I thought
empty grove
Like it's not the winding number in the case of t_n
pearl holly
oh yeah, there are two m's lol
empty grove
Yeah 
frigid river
You basically translate the path. So if the path is from 0 to 1, then after the ...
So you just put the path somewhere else?
marsh forge
okay the point here is that we want to combine w_n with w_m but they both start at the same place so we cant do one and then the other. but once we project to the circle, as long as we start at some integer the circle doesnt care which integer. hence we use tau_n to move w_m to the right so we can start with w_n and then do w_m
so alpha and alpha’ are the same path as far as the circle cares
and we can use alpha’ after we do w_n
where alpha is w_m and alpha’ is tau_nw_m
pearl holly
ah yeah, I see
frigid river
and the w_n, why is there a n anyway? Is this like the time when talking about homomorphisms?
marsh forge
w_n represents going around the circle n times
its just a path from 0 to n on the real line
but if we wrap each length one interval once around the circle
moving to the right n intervals is the same as going clockwise around the circle n times
I think maybe there is a little confusion about what you are trying to prove
hatcher has already shown that going around the circle n times is not homotopic to going around m times (m not equal to n)
so you know you have a copy of the integers living in pi_1S^1
because each n is its own homotopy class
hatcher is now proving that these are the only homotopy classes (and that this map is indeed not just a bijection but a homomorphism too)
I think its easy to see the homomorphism intuitively
combining a loop going n times around the circle with one going m times is clearly a loop going m+n times
empty grove
I think it might be good to be more explicit with the difference between paths on the circle and their liftings
marsh forge
well wait
give them a second to digest and ask bc i wanna make sure this at least makes sense first
but youre right
this is kind of a tricky point if you dont have a lecturer
frigid river
combining a loop going n times around the circle with one going m times is clear...
this makes perfect sense to me
marsh forge
this is so much easier to explain on a board
frigid river
the rest, maybe not so much xD
frigid river
let me just reread xD
marsh forge
no shame im not doing the worlds best explanation
i just need to know which parts are not making sense
pearl holly
hatcher is now proving that these are the *only* homotopy classes (and that this...
hmm but how does he prove it? By proving that psi is a homomorphism?
marsh forge
lets focus on the big picture first
|| the idea is that its hard to prove by only looking at the circle but not hard to prove once by lifting things to the real line first ||
pearl holly
ight are we ready?
pearl holly
okay so everyone understands the things that max talked about?
frigid river
pearl holly
should we move on?
frigid river
I think it's fine.
What a translation is is still unclear to me.
although he even drew a picture
marsh forge
translation is just a shift
like
you know how in high school they taught you that like
frigid river
and what are we shifting into what?
frigid river
then f(x+5) shifts the graph to the side?
sure
marsh forge
its that same idea
thats a translation
w_n by definition starts at 0
but we can shift it to the side
by just adding some interger
empty grove
w_n tilde
marsh forge
yes sorry
the lift is what im talking about
but im too lazy to write tilde
ill specify if im talking about the projection
frigid river
but wn tilde is just n*s
marsh forge
yes why is that a but
frigid river
I thought we were like shifting a path
empty grove
To get n*s + m
marsh forge
does anything I said not make sense then?
frigid river
so how could I imagine what's going on here? 0.0
marsh forge
okay
frigid river
does anything I said not make sense then?
I think I get it.
marsh forge
lets think through each term
w_m goes from 0 to m
w_n goes from 0 to n
tau_m takes w_n and turns it into the path from m to m+n
so conbining them
we get a path from 0 to m and then a path from m to m+n
you should think of w_n as being “n times around the circle” after projection and w_m as “m times around” after projection. then tau simply is a slight adjustment that allows us to build “n+m” times around
frigid river
yeah, that makes sense (if I am not lying to myself)
marsh forge
basically tau fixes the fact that once we lift the loops on the circle to paths on the real line, they no longer start and stop at the same place
this is all like
a long winded way of saying that fact you said made complete sense
marsh forge
this makes perfect sense to me
this
the real meat of the proof is what comes next esp the (a) and (b)
frigid river
the real meat of the proof is what comes next esp the (a) and (b)
calm down, we haven't reached that yet xD
I am still thinking about the next line
marsh forge
oh which one
pearl holly
so how could I imagine what's going on here? 0.0
Isn't this just w^tilde_{m+n}
frigid river
oh which one
cerulean oriole
Is this a group reading?
marsh forge
thats the point:)
frigid river
Is this a group reading?
yep xD
marsh forge
you want “combining w_m and w_n”
cerulean oriole
Cool!
marsh forge
to be the same as
w_m+n
i should run a hatcher reading group some say
day*
maybe ill test drive it with my reu kiddos
frigid river
what's p though
marsh forge
its defined at the beginning
frigid river
ah
marsh forge
its the projection from R to S^1
frigid river
found it
pearl holly
and remember that pw_n tilde = w_n
frigid river
and this? I thought we couldn't do that without a translation, since the paths don't have to be next to each other?
marsh forge
this is where
my ignoring the tildes
is gonna be an issue
lol
you only need tau to combine w tildes
like i said
tau fixes the fact that the lifts to the real line dont start and stop at the same place
this isnt a problem for the non-lifted maps on the circle
huh?
frigid river
this isnt a problem for the non-lifted maps on the circle
and what's a non lifted map?
frigid river
we have a circle
finally something I understand 0.0
marsh forge
we agree (do we) that going around the circle n times and m times are different homotopy classes of maps from the circle to itself
assuming n and m are different
okay
so write w_n for the map from the circle to itself n times
and similarly for w_m
frigid river
we agree (do we) that going around the circle n times and m times are different ...
I agree, but isn't this what we are trying to prove though? 0.0
that a circle doesn't have the trivial fundamental group?
marsh forge
Oh sorry youre completely right hatcher does this later in the proof I assumed he had already done it
okay so we dont know that they are different
thats not important for now
okay
so we have w_n and w_m
maybe they are homotopic
maybe not
not important right now
what is important is that we already know what w_n * w_m is
well we know it exists
but like
these are loops
and we know it makes sense to compose them
right?
frigid river
yes
marsh forge
okay
now we want to show that the map from Z to pi_1 S^1 is a homomorphism
that means we need to prove thay
w_{n+m} is homotopic to w_n*w_m
this is honestly
kind of obvious
which makes the proof confusing
bc maybe its not clear what even needs to be proven
but lets carry on
frigid river
w_{n+m} is homotopic to w_n*w_m
and w was just the spiral, right?
frigid river
gulp
pearl holly
but wait, don't we have to prove that [w_n+m] = [w_n]*[w_m]? Since psi maps stuff to the homotopy class?
marsh forge
right yeah sorry so once you get intuition for this these things all mean the same thing
but yes
but it suffices to prove what i said because the multiplication of the fundamental group is well defined up to homotopy
i.e. what is the rhs of thay equality you wrote?
well
you take any representative for [w_n] and [w_m]
say, w_n and w_m themselves
and then the product [w_n]*[w_m] is just [w_n * w_m] by definition
so if we show that w_n+m is homotopic to w_n*w_m
weve done what you wanted
im switching to my pc
so i can type properly
frigid river
have you been doing that all on your phone? 
marsh forge
yes
frigid river
marsh forge
i can basically type latex without looking at this point on my phone lol
anyway
im on my laptop now
@pearl holly did I address your concern there
pearl holly
yes you did!
marsh forge
okay great
pearl holly
I get that now!
marsh forge
Okay
so
Theres this gross trig definition of w_n that hatcher starts with
its bad
we want to give a better definition of w_n
hatcher does this in the following way
let $\tilde \omega_n$ be the straight line path from $0$ to $n$ in $\bR$
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
then there is a map $p:\bR \to S^1$ given by wraping each interval $[k,k+1]$ around the circle
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
then if you like picture this
then $p\circ \tilde \omega_n$ obviously gives you $\omega_n$ but now we have a better way of thinking about it
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
so we redefine $\omega_n:=p\tilde\omega_n$
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
but we havent actually changed anything
we just made our lives easier
does this all make sense?
pearl holly
Yes
frigid river
Yes.
marsh forge
Okay great
now we have a new definition of $\omega_n$ and $\omega_m$. We want to prove that $\omega_n\cdot \omega_m\simeq \omega_{n+m}$
gentle ospreyBOT
MaxJ (Sockoist)
frigid river
wait, what's the dot here?
marsh forge
rephrasing we need to prove that $p\tilde\omega_n\cdot p\tilde\omega_m \simeq p\tilde\omega_{n+m}$
gentle ospreyBOT
MaxJ (Sockoist)
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
i was only using that because I was being lazy
frigid river
is that path composition?
marsh forge
yes
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
so
frigid river
marsh forge
actually give me a second sorry let me think for a second
Sorry Hatcher is being kind of headass about this imo but its important to understand the machinery for the less obvious part of the proof
basically what hatcher shows is that $p\circ (\tilde\omega_m\cdot (\tau_m\tilde\omega_n))$ is the same thing as $p\tilde\omega_{n+m}=\omega_{n+m}$ which is clear because $\tilde\omega_{n+m}$ is exactly the same thing as $(\tilde\omega_m\cdot (\tau_m\tilde\omega_n))$
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
Basically all of this garbage notation says the following
If I start in R, and go from 0 to n+m, and then go to the circle
its the same thing as taking the 0 to n and 0 to m paths in R
going to the circle
and combining them there
frigid river
okay
frigid river
and now we know it's a homomorphism?
marsh forge
yes
pearl holly
frigid river
marsh forge
if you understand what I said and you understand the concept of the lifts and tau
then youre fine
the lifts are much more important
frigid river
so "lift" is just a mapping from the line to the circle, right?
frigid river
great
marsh forge
one second
frigid river
xD
marsh forge
When I say "lift"
what I mean is
we have a map that makes sense on the circle
we have a map from R to the circle
then the lift is finding a path in R
that projects down to the map we want on the circle
think of R as being "upstairs"
and S^1 as being "downstairs"
and the map p takes us from upstairs to downstairs
the lift is an upstairs version of the thing we want downstairs
namely $\tilde\omega_n$ is the lift ot $\omega_n$ because $p\tilde\omega_n=\omega_n$
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
of course we kinda made this a definition
so its a little confusing
but hatcher wrote this poorly
so blame him
i honestly hate this proof more every time i read it lol
does that language thing make sense
I can stop using the word lift if its confusing
it matters not at all to this proof
frigid river
still not sure, sorry
so a lift is a path being projected onto the circle?
marsh forge
In this specific case yes. The concept of lifting is more general but thats not helpful right now.
if I say a lift of something
frigid river
okay, great. Thank you
marsh forge
that means it has to project to that something
Here's the outline of hatcher's proof, vaguely
- Show that the map in question is a homomorphism. I think this is like, very intuitive and that hatcher somehow makes it more confusing by proving it in a roundabout way
- Show that it's injective (i.e. what you said early, show that $\omega_n\simeq \omega_m$ iff $m=n$)
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
- show that its surjective
and then we are done
he might not do it in that order I have to scroll up
but thats the idea
the idea of using the real line and then projecting down is honestly not necessary for (1) but whatever
it is necessary for the other two in hatcher's proof I believe
so thats why he introduces it
frigid river
❤️
marsh forge
Hatcher also does a common thing in the next part of the proof
he says
"We are going to prove (a) and (b) and then prove the main theorem" but he starts assuming (a) and (b) are true, finished the theorem, and then proves (a) and (b)
frigid river
Okay, that doesn't bother me 0.0
marsh forge
this is of course a valid strategy but I figured its worth warning since earlier math books avoid tricks like that
okay
great
pearl holly
ahh I see. So the proof of the theorem isn't actually that long. The only thing takes him awhile is to prove these three facts
I think
Idk, doesn't matter
marsh forge
the proof of the theorem is 2 lines if you know some complex analysis
okay
lets move on
can someone post the screenshot again lol
frigid river
marsh forge
Okay I am going to assume (a) and (b) like hatcher and sketch out the rest of the proof in a perhaps more intuitive or at least less wordy way.
So, (a) and (b) amount to showing you that, for any LOOP $f:I\to S^1$ there is a PATH $\tilde f:I\to \bR$ such that $p\circ\tilde f=f$
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
Moreover
it tells you that $f\simeq g$ if and only if $\tilde f \simeq \tilde g$ where by $\simeq$ i mean a homotopy that fixes the start and end points
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
now, because the lifts $\tilde f$ started life as loops, we know that they must start at $0$ (by my definition) and they must end at some integer $m$.
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
Suppose that $\tilde f$ ends at some integer $m$. Then because we are in $\bR$ we can easily imagine putting like pins at $0$ and $m$ and then dragging $\tilde f$ until it looks like $\tilde\omega_m$
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
What this says is that every $\tilde f$ is path homotopic to some $\tilde\omega_m$
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
so every $f$ is path (loop) homotopic to some $\omega_m$
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
This tells us that the map is surjective
Now consider $m\neq n$ and $\omega_n$ and $\omega_m$. Then we can lift to $\tilde\omega_m$ and $\tilde \omega_n$. The first map has a different end point than the second map, so there is no way they could possibly be path homotopic because we aren't allowed to move the endpoints!
gentle ospreyBOT
MaxJ (Sockoist)
marsh forge
So the map is injective
and we are done
I'm gonna stop here and let you ask questions
pearl holly
Yes, I'm just trying to read this carefully lol
marsh forge
we can break this up into parts its not going anywhere
frigid river
It's fine, it's just a lot
marsh forge
Yeah, just go line by line and lmk if you want extra explanation
when reading proofs on your own its good to try really hard to understand it
but honestly when working w someone in a dialogue
its okay to ask clarification questions without thinking that hard first
Im gonna go make lunch feel free to ping me
pearl holly
Okay I think that I understand it now
marsh forge
I am glad
pliant parcel
I'll give this another shot in here since I didn't get any answers on stack exchange 😬
Summary: The volume of a geodesic ball on an analytic riemannian manifold is given by:
$$ S(r) = \int_{S^{n-1}(1)} r^{n-1} \omega_{1...n} (\text{exp} (ru)) , du$$
Where S(r) is the volume, $\omega$ is the volume element and $exp(ru)$ is a change of variables factor coming from $S^{n-1}(1) \rightarrow exp(S^{n-1}(r))$.
From this equation I'm wondering how one might find the variational derivative in terms of the metric $\frac{\delta S}{\delta g^{ab}}$.
The one idea I've had so far is to let r be the magnitude of a generic curve $\gamma$ s.t.
$$|\vec{r}| = \int_0^{\lambda_f} \sqrt{g(\dot{\gamma}(\lambda),\dot{\gamma}(\lambda))} , d\lambda.$$
While $|\vec{r}|$ is constant on its own, we would have to vary $\lambda_f$ over $S^{n-1}(1)$ (or $u$) to get the proper expression. The geodesics themselves would just be straight lines that vary over $u$ with the restriction $\gamma(\lambda_f) \in S^{n-1}(r)$ (also maybe a restriction on $\dot{\gamma}$ being constant?). Put together, it might look something along the lines of:
$$S(r) = \int_{S^{n-1}(1)} \left(\int_0^{\lambda_f(u)} \sqrt{g(\dot{\gamma},\dot{\gamma})} , d\lambda\right)^{n-1}\sqrt{\text{det}(g)} , \text{exp}\left(u\cdot\int_0^{\lambda_f(u)} \sqrt{g(\dot{\gamma},\dot{\gamma})} , d\lambda \right) , du$$
And then we'd vary by $g$ ...
gentle ospreyBOT
Anteater
pliant parcel
Looks like some of it got cut off here hopefully it's clear. Fairly lost tbh. lots of ideas but the expressions are so daunting and whatnot
pliant parcel
I got some help on stack exchange 😅
gritty widget
Any motivic homotopy theory heads here?
gritty widget
what are you using?
tight agate
there's an article on it in the handbook of homotopy theory
the morel-voevodsky paper is super hard to read
there's also a thing that weibel wrote up
it's called Voevodsky's seattle lectures
or something like that
marsh forge
theres a book if multiple authors
by voevodsky et al
but the stuff i need is all like
“cellular” motivic homotopy
tight agate
the one on motivic cohomology?
marsh forge
so ive been learning that specifically
marsh forge
the one on motivic cohomology?
no “motivic homotopy theory” should be the title
voev didnt actually write any of it
its just based on his lectures
marsh forge
yes
gritty widget
i too am also trying to read this
quartz edge
how is there a unique geodesic on 3-manifolds for any point and initial velocity
take a unit 3-sphere
hmm
i guess the geodesics then terminate and don't wrap around nicely like in 2-manifolds
so there couldn't be any sectional curvature in a 3-manifold embedded in R3
maybe i'm wrong but i don't see any other way to do this
it should seem that i'm wrong
the geodesics have to be just straight lines
i guess my intuition is right.
the tangent space is 3 dimensional so the only way to parallel transport a velocity vector is to not move it at all
gritty widget
how is there a unique geodesic on 3-manifolds for any point and initial velocity
unless im misunderstanding, this has nothing to do with the dimension or embedding and is true for all riemannian manifolds. for any point and initial velocity at that point, there's a unique (maximal) geodesic with that initial position and velocity.
quartz edge
ye i was falsely expecting geodesics on the boundary of a voluminous unit sphere in R3 to wrap around in great circles
they wouldn't
gentle ospreyBOT
R2T2 ✓
gritty widget
(we need the extension to be normal to N to do the second equality)
gentle ospreyBOT
R2T2 ✓
