Hausdorff
#point-set-topology
gentle ospreyBOT
vast estuary
b instead of B, of course
quasi forum
That's what I was thinking it should be, but I am having a hard time seeing why.
empty grove
Try proving that any such path must hit each value between x and y (in the dictionary order). Since the set of values between x and y has uncountably many disjoint open subsets, such a path cannot exist 
By disjoint I mean pairwise disjoint
quasi forum
Man, I feel like I depend on y'all too much for this class. Like I understand all the explanations, and sometimes I just need a headstart, but I never would've figured out some of this stuff on my own. :/
empty grove
Well you are looking at connectedness etc for the first time right
You just have to get used to thinking about such properties, which are preserved under continuous maps
oh and dictionary order is kinda wacky so it's not too bad
quasi forum
Nah, I learned about connectedness in my first topology class (mainly metric), and we did a homework assignment with connectedness already (a couple weeks ago).
This one is all about path connectedness, IVT, and components.
empty grove
Like it will be locally path connected if you did (0,1)²
quasi forum
Like it will be locally path connected if you did (0,1)²
Wait...what do you mean?
quasi forum
Ah, and the ordered square is [0,1]^2
empty grove
Ye
And there's also some stuff like
[0,1]² in the subspace top of R² with dict order top is not the same as [0,1]² in dict order top
The former is also locally path connected
Whereas subspace of product = product of subspaces holds
So not the most intuitive
But I'd suggest, try to think about what connected sets look like in an order topology
Any order topology
You might be able to link it to the intermediate value theorem
quasi forum
Okay, back to the topic at hand. What does uncountability have to do with it?
empty grove
Take inverse image of that uncountable collection
quasi forum
Well, it should be uncountable. Open intervals are uncountable, so why would that be an issue?
empty grove
Each open interval contains a rational
empty grove
Well, it should be uncountable. Open intervals are uncountable, so why would tha...
These aren't open intervals, it's an uncountable collection of open sets
Pairwise disjoint
quasi forum
I remember this discussion was in our notes. But I didnt really understand why this was a contradiction (and I forgot to ask my teacher).
Oh, since each contains a distinct rational, we can count the Ux's, and that's the contradiction!
I get it now. That's funky
empty grove
Yep
R has a smallness property of never having uncountably many pairwise disjoint subsets
quasi forum
That makes a lot of sense!
empty grove
And this is very useful (it's called separability) and once you start seeing these words studied on their own then things should start to tie together
Well I'm not sure if that's separability, it's certainly implied by separability
ie has countable dense subset
quasi forum
This is a really interesting property. Thank you for helping Moldilocks, it makes a lot more sense now.
vast estuary
**Hausdorff**
Is this trivial or something
empty grove
Bijective group homomorphisms are isomorphisms lol
vast estuary
Bijective group homomorphisms are isomorphisms lol
ooh i was checking that it's a homomorphism, but it clearly is, lol. ok bye thanks oof
quasi forum
So moldilocks, you were saying (0,1)^2 is locally path connected. Why is that?
Oh, is it because basis sets are only in those strips?
empty grove
Yeah so for each element you can take the vertical strip as a neighborhood, and that's homeomorphic to R so path connected
quasi forum
Ayyy, I'm getting it. Thx Moldilocks
quasi forum
Okay, looking at 5 a). Path connectedness is super easy to show, but I am having a hard time with local connectedness at p/showing other points are not locally connected.
empty grove
For not locally connected elsewhere, here's how you can think (but you can prove this directly): removing the point p from this gives you an open subset. If you can prove that this isn't locally connected anywhere you'll be done. But after removing P this is just a bunch of disjoint lines at each integer (ie homeomorphic to Q x [0,1]) and the local connectedness of that can be treated the same way you treat the local connectedness of Q
Hopefully this give visual intuition for what you should do (though you can definitely also try this approach if you can't get a direct argument)
quasi forum
Okay, what about p itself? I know it is an accumulation point, so I understand intuitively why this is
empty grove
An open set around p contains a ball around p. A ball around p is a central point with spokes going out of it
swift fjord
How can I see that any two geometric realisations X_1,X_2 of an abstract simplicial complex Y are homeomorphic? I know that I have a bijection between the vertices, of Y and the vertices of the complexes, so I wanna take a bijection between the vertices of X_1,X_2 and I wanna extend it linearly somehow to an affine isomorphism that sends one to the other, but I need to choose a set of n+1 affinely independent vectors. I should have a set of n+1 affinely independent vectors where n is the dimension of Y right? Thing is, I don't know if when I choose any n+1 affinely independent point, this means that X_1 will map to X_2. I see why it must be an isomorphism but not why it must map X_1 to X_2
quasi forum
For not locally connected elsewhere, here's how you can think (but you can prove...
I gotta use homeomorphisms more often
swift fjord
Sorry I didn't mean to send that yet I wanted to wait till you're done
swift fjord
How can I see that any two geometric realisations X_1,X_2 of an abstract simpli...
also if my geometric realisations are in different dimensions (Even though they have the same dimension themselves they can be embedded) say one in R^d and one in R^m, how do I complete my set of vertices to an affine basis in both spaces such that it doesn't interfere with my homeomorphism
empty grove
Yeah at least in your head, use homeomorphisms for everything. It's hard to parameterize a coffis cup after all
quasi forum
But it's not hard to parameterize a Taurus 
quasi forum
Okay, so the homeomorphism here is between the line segments and Q?
I don't think the points themselves are homeomorphic to Q
empty grove
Not to Q
To Q x [0,1)
Which is very similar to Q in some ways 
Q is a deformation retract of this which you might see later
But the same argument for no local connectedness on Q will work here
empty grove
To Q x [0,1)
Just slanting the line segments
hollow harbor
Only if the space is like, second countable or something right?
I'm not 100% sure but you could imagine a paracompact space which is massive, but does the right thing locally. Maybe this is false though
empty grove
Take large disjoint unions of R
hollow harbor
Is this paracompact though
empty grove
Idk I just looked up paracompact
But it feels like you can intersect each element of cover with each copy of R
hollow harbor
If you order the disjoint R's, and then take your cover to be "sets including all R's below some point"
empty grove
Isn't each copy open
hollow harbor
I thought paracompactness meant "every cover has a locally finite subcover"
empty grove
Wait subcover 
hollow harbor
Isn't each copy open
Sure but they might not each be in the cover...
empty grove
Wikipedia says refinement
hollow harbor
Oh is it refinement
empty grove
quasi forum
So wait, why is Q not locally connected (seems like a silly question, but still)
empty grove
So wait, why is Q not locally connected (seems like a silly question, but still)
No non empty open subset is connected There are holes
quasi forum
Also, in R, arent the only connected spaces R and its intervals/rays?
empty grove
Yeah connected subsets are exactly intervals
quasi forum
No non empty open subset is connected <:catThink:526983414226747393> There are h...
But if a separation were to exist for some interval (a,b), say (a,p) and (r,b), we know there are actually rationals between them
So it was not a spearation
empty grove
p = r
quasi forum
Welp, I feel dumb :p
Okay, that makes sense. Can we go back to why p is locally connected?
Oh duh. If the separation contains p, it intersects with all the line segments.
Beautiful! Very last thing. I need to find a path connected space in R^2 that is not locally connected anywhere
gritty widget
i would recommend the book
counter examples in topology
you really do learn alot of topology
by trying to see what goes wrong
quasi forum
Well that reaction is concerning 🤣
gritty widget
Beautiful! Very last thing. I need to find a path connected space in R^2 that is...
you could google this and get an answer pretty quick
but its such a counter intuitive statment
quasi forum
I'm gonna speak with my prof today and ask him during office hours
gritty widget
if you are just given the counter example
quasi forum
Well, this is just a thought, what about R^2-Q^2?
gritty widget
you won't know really get the full benefit
quasi forum
Ooo, this is path connected (I've shown this already)
gritty widget
Q^2 is very disconncted
empty grove
Beautiful! Very last thing. I need to find a path connected space in R^2 that is...
A very dumb example and very annoying to write rigorously, but maybe someone can simplify:
Take the example you had, of the point 0 x 1 connected to each rational point in [0,1] x 0 by a straight line. The problem right now is local connectedness at p, so we would like to add "hairs around it", much like the other points don't have local connectedness because of hairiness. For this, continue all these line segments beyond the interval, and make them all curve back up and back to p, while keeping everything parallel kind of. Enumerate the lines with natural numbers. End the ith like segment at distance 1/i before p
quasi forum
Sure, but we are removing Q^2
gritty widget
R^2-Q^2 is like shooting R^2 with a very fine shot pellet shotgun
empty grove
Well, this is just a thought, what about R^2-Q^2?
Is locally connected
quasi forum
Damn! :(
empty grove
A very dumb example and very annoying to write rigorously, but maybe someone can...
I'm sure this works
gritty widget
the pellets are to fine to actually cut apart anythng
empty grove
But idek if I'm communicating the picture right
quasi forum
I think I'm just gonna talk to my prof first. I'll come back to y'all after if I have any questions.
gritty widget
A very dumb example and very annoying to write rigorously, but maybe someone can...
but read this before talking to your proffessor
and try think about what hairs do
sonic tusk
counter examples in topology
Sounds like something I'd like to read, unrelated to topology but are their any other books like that out there🤔
pearl holly
Sorry for the ping but I'm back now (it was getting late for me yesterday so I decided to go to sleep) and now I have a few questions. So $\Lambda_R [\alpha_1] \otimes_R \Lambda_R [\alpha_2] = (R \oplus_R R \alpha_1) \otimes (R \oplus_R R \alpha_2)$ which after expanding etc. I get $R \oplus R \alpha_2 \oplus R\alpha_1 \oplus ( R \alpha_1 \otimes R \alpha_2)$ which should be equal to $\Lambda[\alpha_1, \alpha_2]$? Here I just expanded stuff but you also said that $(A \otimes B)n = \bigoplus{p + q = n} A_p \otimes B_q$ where $A, B$ are graded. If I try to use this "formula" then I guess that I get
$$(\Lambda_R[\alpha_1] \otimes_R \Lambda_R[\alpha_2])_2 = R \alpha_1 \otimes_R R \alpha_2$$
since there's no "degree" 2 component in both $\Lambda_R[\alpha_1]$ and in $\Lambda_R[\alpha_2]$. So which one of these is the right one?
gentle ospreyBOT
Tokidoki ✓
pearl holly
I also just replaced \otimes with \otimes_R in everything I did. Is that right or should I stick to \otimes like you did?
quasi forum
A very dumb example and very annoying to write rigorously, but maybe someone can...
We got one, and I think it is in line with what you were saying (not exactly, but the idea of the 'hairs' is there).
plain raven
@pearl holly
It is true that $\Lambda_R[\alpha_1]\otimes_R \Lambda_R[\alpha_2] = (R \oplus_R R \alpha_1) \otimes (R \oplus_R R \alpha_2)$
gentle ospreyBOT
diligentClerk
plain raven
which is equal to
$R \oplus R \alpha_2 \oplus R\alpha_1 \oplus ( R \alpha_1 \otimes R \alpha_2)$ which should be equal to $\Lambda[\alpha_1, \alpha_2]$
gentle ospreyBOT
diligentClerk
plain raven
all that is correct.
quasi forum
@empty grove here is what we came up with. One broom fails, but two brooms don't!
pearl holly
car go broom broom
empty grove
ahh epic
plain raven
Now, in addition to everything you just said, $\Lambda_R[\alpha_1]$ and $\Lambda_R[\alpha_2]$ are equipped with a grading, and when two modules are graded, their tensor product is also naturally graded. In this case $\Lambda_R[\alpha_1,\alpha_2]$ thus inherits a grading where $\Lambda_R[\alpha_1,\alpha_2]_0 = R, \Lambda_R[\alpha_1,\alpha_2]_1 = R\alpha_1\oplus R\alpha_2, \Lambda_R[\alpha_1,\alpha_2]_2 = R\alpha_1\otimes R\alpha 2$
gentle ospreyBOT
diligentClerk
plain raven
it decomposes as the direct sum of these three modules
pearl holly
so the n here stands for the "n-th" component?
pearl holly
right right okay I see
plain raven
normally i'd use big parentheses around everything to avoid confusion but it's hard to type things quickly on discord
by the way this graded tensor product can be generalized to chain complexes. there is a graded tensor product of chain complexes whose underlying graded group is the graded tensor product of graded groups
if that makes sense. maybe that's unnecessarily verbose
pearl holly
ye okay I will read about those later but now I have another question real quick
plain raven
if you want a cool exercise in algebraic topology involving the graded tensor product (it's not hard) i have a recommendation
pearl holly
so now I know what $\Lambda_1[\alpha_1, \alpha_2]$ "looks like" in terms of the tensor products etc. But how do I know that $\alpha_i \alpha_j = -\alpha_j \alpha_i$? I guess that $\alpha_i^2$ comes from the fact that this holds in each $\Lambda[\alpha_i]$ by definition but how about the "anti commutativity" part?
gentle ospreyBOT
Tokidoki ✓
plain raven
oh, you mean how does it inherit the multiplication
let me think about that for a minute.
pearl holly
if you want a cool exercise in algebraic topology involving the graded tensor pr...
oof yeah sure!
pearl holly
oh, you mean how does it inherit the multiplication
wait I think Max said this, hold on
And the product of $(a\otimes b)(a'\otimes b'):=aa'\otimes bb'$
gentle ospreyBOT
Tokidoki ✓
pearl holly
maybe I should just ask Max at this point idk 
plain raven
ok. hmm
i have an educated guess but i haven't checked this out yet
suppose that $R,R'$ are two rings.
gentle ospreyBOT
diligentClerk
plain raven
Then $R\otimes R'$ can be equipped with the structure of a ring. The multiplication is given as follows:
gentle ospreyBOT
diligentClerk
plain raven
$(R\otimes R')\otimes (R\otimes R')\cong R\otimes R\otimes R'\otimes R'\xrightarrow{\mu\otimes \mu'} R\otimes R'$
gentle ospreyBOT
diligentClerk
pearl holly
yo
plain raven
here i'm using $\mu,\mu'$ for the multiplications of each ring individually.
gentle ospreyBOT
diligentClerk
plain raven
ok. but i want to point out something interesting here
in order for this to work we had to rely on something special about the tensor product of Abelian groups
namely that it's symmetric, i.e. we have an isomorphism $R\otimes R'\cong R'\otimes R$
gentle ospreyBOT
diligentClerk
plain raven
right?
pearl holly
when did we use this assumption?
plain raven
when we swapped things around in the first step
look at the map i drew
the middle two terms get swapped, that's the first step
pearl holly
okay yeah sure
plain raven
so this points something out to us: We might be able to equip the tensor product of graded rings with a multiplication making it into a graded ring. But before we can do that, we need to know that graded groups have a symmetry isomorphism, and we need to know what that is
this is the first major way in which graded groups differ from ordinary groups. there is a symmetry isomorphism $A\otimes B\cong B\otimes A$ but it's not the usual one
gentle ospreyBOT
diligentClerk
plain raven
instead it's defined as follows: if $a$ is homogeneous of degree $p$ in $A$ and $b$ is homogeneous of degree $q$ in $B$, then we send $a\otimes b$ to $(-1)^{p\cdot q}b\otimes a$
gentle ospreyBOT
diligentClerk
pearl holly
wait what does homogeneous mean?
plain raven
this might seem a little bizarre but this definition is the only one that works and makes sense for chain complexes (because in the case of chain complexes, if you just sent $a\otimes b$ to $b\otimes a$, it turns out this doesn't commute with the differentials) so we're kind of viewing graded groups here as chain complexes with all differentials equal to zero
gentle ospreyBOT
diligentClerk
plain raven
An element of a graded group $A=\bigoplus_iA_i$ is homogeneous if it lives in one of the $A_i$.
gentle ospreyBOT
diligentClerk
pearl holly
oh okay 
plain raven
speaking about grading by natural numbers, it's homogeneous of degree $p$ if it lives in $A_p$
gentle ospreyBOT
diligentClerk
pearl holly
Roger that
pearl holly
**diligentClerk**
okay so this is the definition of the multiplication?
plain raven
yeah so we should be able to use this to figure out the answer to your question.
so like
what is $\alpha_2 \cdot \alpha_1$, where we view each of these as being homogeneous of degree $1$?
gentle ospreyBOT
diligentClerk
plain raven
well, to be really annoying and pedantic it's convenient to write $\alpha_2$ as $1\otimes \alpha_2$, because it lives in the tensor product of $R$ (from the degree $0$ component in the first copy of the group) with $R\alpha_2$ from the degree 1 component of the second group
gentle ospreyBOT
diligentClerk
plain raven
similarly i prefer to write $\alpha_1 = \alpha_1\otimes 1$
gentle ospreyBOT
diligentClerk
plain raven
so the question is really, what is $(1\otimes \alpha_2)\cdot (\alpha_1\otimes 1)$
gentle ospreyBOT
diligentClerk
pearl holly
oh okay right
plain raven
now how do we do this? we have to first apply the symmetry isomorphism, so we send $1\otimes \alpha_2\otimes \alpha_1\otimes 1$ to $ (-1)^{1\cdot 1} 1\otimes \alpha_1\otimes\alpha_2\otimes 1$, where the term $(-1)^{1\cdot 1}$ comes from the symmetry isomorphism as we've defined it and the fact that both $\alpha_1,\alpha_2$ are of degree $1$
gentle ospreyBOT
diligentClerk
plain raven
then apply the multiplications to the first pair and last pair respectively, so $1\cdot \alpha_1 = \alpha_1,\alpha_2\cdot 1=\alpha_2$ so the final answer is $- \alpha_2\otimes\alpha_1$ in $R\alpha_1\otimes R\alpha_2$
gentle ospreyBOT
diligentClerk
plain raven
does that make sense?
pearl holly
yeah it does!
plain raven
cool.
pearl holly
okay let me just quickly re-read everything to make sure that I get everything and not just run away
plain raven
oh by the way you are usually working over some fixed ground ring $R$ which for the kind of stuff hatcher's doing is usually pretty nice, like a PID, and you can suppress constant reference to $R$ in the notation as it's assumed you're working over it. So i write $\otimes$ for $\otimes_R$ because i'm taking for granted we're tensoring over some ring which is given in a certain context
gentle ospreyBOT
diligentClerk
pearl holly
oh okay right I see
and working over PID's are nice because modules over it are free?
or did I get that wrong?
so like subgroups of free abelian groups are also free which is nice and this is kind of analogous to that?
plain raven
yes, submodules of a free module are free, every projective module is free
you should think of $R$ as being $\mathbb{Z}$ lol
gentle ospreyBOT
diligentClerk
plain raven
like PID's are extremely nice and linear algebra in the category of $R$ modules for $R$ a PID is basically the same as linear algebra over abelian groups
gentle ospreyBOT
diligentClerk
plain raven
subgroups of free modules being free is a dope property, i mean like
say $C, D$ are chain complexes, and $C$ is a complex of free groups.
gentle ospreyBOT
diligentClerk
plain raven
And say you have a map $C\to D$ which is surjective
gentle ospreyBOT
diligentClerk
plain raven
then its kernel is a submodule of $C$ which is free
gentle ospreyBOT
diligentClerk
plain raven
i can't remember where i was going with this example, but basically a lot of your short exact sequences turn out to be split degreewise
and this is very very nice
pearl holly
yeah I see. I remember Hatcher using the fact that subgroups of free abelian groups are also free a lot of times lol
and then you can just pass it to submodules and I guess the same stuff holds there too because of this property
plain raven
yeah so everything hatcher's doing will go through for a PID. i mean i'm not that much of a comm alg person so the only big PID i know is $k[x]$ for $k$ a field but
gentle ospreyBOT
diligentClerk
pearl holly
oof no I don't
plain raven
HA
this is irrelevant to you it's just me trying to score a point over chmonkey in a conversation we had like three days ago over whether nakayama's lemma is common knowledge
plain raven
anyway you can use nakayama's lemma to prove that any surjective map $\mathbb{Z}^n\to\mathbb{Z}^n$ is also injective and thus an isomorphism which you can see being important from an AT perspective because whenever you have a complex with finitely many cells in a given dimension then the associated chain complex will be free of degree n in each dimension
gentle ospreyBOT
diligentClerk
plain raven
ok let me give you that exercise i was going to give you and then you can peace.
so for the sake of this question i'm going to be annoying and ask you to adopt a slightly different convention for the usual functor from delta complexes to chain complexes.
usually you define: $C_0$ = free abelian group on points, $C_1$ free abelian group on edges, etc..
gentle ospreyBOT
diligentClerk
plain raven
so for some purposes in algebra, at least in the stuff i've been doing recently, it's convenient to define a different functor $C'$
gentle ospreyBOT
diligentClerk
plain raven
and $C'$ assigns gradings as follows. The empty topological space is the convex hull of the empty set of vertices, so $C'_0(X)$ is free Abelian on maps from the empty space into $X$. So $C'_0(X)$ is always just $\mathbb{Z}$
gentle ospreyBOT
diligentClerk
plain raven
$C'_1(X)$ is maps from the convex hull of $1$ point, which is a single point, so $C'_1(X)$ is the free Abelian group on vertices
gentle ospreyBOT
diligentClerk
plain raven
$C'_2(X)$ is the free ABelian group on maps from the convex hull of two points, i.e. edges
gentle ospreyBOT
diligentClerk
plain raven
does this make sense
pearl holly
ye it does
plain raven
so $C'_{n+1}(X) \cong C_n(X)$ for all $n\geq 0$ and $C'_0(X) =\mathbb{Z}$
gentle ospreyBOT
diligentClerk
plain raven
very simple
and actually much more convenient for certain tasks.
ok
so
take $X$ to be a simplicial complex which is just a single point.
gentle ospreyBOT
diligentClerk
plain raven
Then $C'(X)$ is concentrated in degrees $0$ and $1$. It's $\mathbb{Z}$ in both degrees. The boundary map $d : \mathbb{Z}\to\mathbb{Z}$ is the identity
gentle ospreyBOT
diligentClerk
plain raven
this boundary map does make sense because a singular $0$ simplex does have one face, which is the empty face; you get it by precomposing with the unique map $\emptyset\to 1$
gentle ospreyBOT
diligentClerk
pearl holly
hm yeah it does
plain raven
ok
so the exercise is this: Let $\Delta^n$ be the standard $n$-dimensional simplex. So it has one cell in dimension $n$, which means $C'(\Delta^n)_{n+1} = \mathbb{Z}$.
gentle ospreyBOT
diligentClerk
plain raven
sorry for throwing you off with the dimension stuff here i know it's weird
Prove that $C'(\Delta^n)$ is the $n+1$-fold tensor product of $C'(\Delta^0)$ with itself, where the tensor product is in the category of chain complexes. Now from what you just said you have not learned the tensor product of chain complexes yet, so just compute it on graded groups for now and prove that you have the same number of generators in each degree
gentle ospreyBOT
diligentClerk
plain raven
this illustrates that the tensor product of chain complexes does actually correspond to a certain geometric operation on simplicial complexes, which is neat
and i can go into more detail about that another time
i think this special chain complex $C'(\Delta^0)$ is very interesting, all other chain complexes can be built out of it in a canonical way
gentle ospreyBOT
diligentClerk
pearl holly
hmm okay I see
yeah okay I will give this a shot. Thank you so much the help I received today and yesterday, I really appreciate it! 
white peak
I have a question about regular cell complexes constructed from cycles in a graph. In a paper I am reading they claim that if I attach a closed 2d disc to all simple cycles in a graph I get a regular cell complex. For example:
However, let us consider this example of two triangles that share an edge:
If I now compute the intersection of the left small triangle and the right small triangle then it is non empty (at least in my opinion). Which contradicts the following requirement from regular cell complexes:
That is because none of the small triangles are a subset of the other.
I hope someone can tell me what I am misunderstanding, I don't have much experience with this
Well the intersection of the small triangles is non empty, hence by (2) one triangle should be a subset of the other?
Which it is not
I am obviously misunderstanding something, but I am not sure where my mistake lies
plain raven
catal maybe they mean open cells, like the interior of the triangle
or the open interval (0,1) for a 1-cell
this is sometimes used because a complex is the disjoint union of all its open cells
white peak
So I glue the disk to the triangles and afterwards their "boundary" becomes part of the edges and the triangle is the deformed open disc
The fact that cells need to be open follows from this:
right?
plain raven
Yeah. that's a big clue that they're talking about open cells. Every closed cell is compact, so clearly not homeomorphic to R^n.
white peak
Thanks!
If I stay with my graph problem: this means that my edges are homemorphic to $\mathbb{R}$ and thus the intersection between an edge and a vertex is empty?
gentle ospreyBOT
catal
plain raven
an open edge yeah. do they post the precise definitions everywhere?
white peak
That's pretty much all I have from them (paper is "Weisfeiler and Lehman Go Cellular: CW Networks" from https://www.researchgate.net/publication/353038849_Weisfeiler_and_Lehman_Go_Cellular_CW_Networks)
digital wraith
how should I visualize the product space $A\times S$ where $A$ is a topological space and $S$ is a finite set with the discrete topology.
gentle ospreyBOT
R is countable
plain raven
ok sweet
digital wraith
oh wait is this channel occupied
white peak
oh wait is this channel occupied
(sorry)
white peak
Thanks ❤️
Something the authors also claim later is that they can create cells based on "(induced) subgraphs". I am trying to understand this.
For something trivial: can I glue the boundary of a disc to a path while leaving an opening?
Or can I just take any subgraph and glue it on the boundary of a sphere (which might mean that I skip using disks at all)?
plain raven
i don't know exactly how to interpret your comment
but
the whole boundary of the disk has to be glued onto the existing structure
i would avoid trying to glue things onto the boundary of a sphere, only glue using the boundaries of disks, although you might be able to strategize here and figure out how to reconfigure that in terms of disks
CW complexes are treated in various sources. one good introduction to them is "Modern classical algebraic topology" by Strom
Hatcher also has an appendix on them
General CW complexes don't need to be "regular" but you can just add this as an extra axiom
white peak
Thank you for those sources, I'll be sure to check them out. I am definitely trying to run before I can walk right now
tawny jewel
Hey, if I have set of unit circles with centers at (1/n, 0) (for all natural n), that's a compact set right?
white peak
i don't know exactly how to interpret your comment
Yea my comment is very disjointed.
It is pretty easy to create cells from simple circles in the graph. But there is no trivial way to create cells from subgraphs, right?
If I understand it correctly, there is not even a way to create a cell from a single edge as I would need to overlap the vertices which would violate (2) (unless it is a selfloop).
gritty widget
In cohomology we have the long exact sequence of the pair (X,A)
$\begin{tikzcd}
& H^(X) \arrow[rd] & \
{H^(X,A)} \arrow[ru] & & H^*(A) \arrow[ll]
\end{tikzcd}$
gentle ospreyBOT
lime_soup
gritty widget
is it also true that we have this exact triangle
as rings
modulo maybe having to take reduced cohomology instead
oh nevermind i see
tight agate
@gritty widget Ring is triangulated?
before you pass to cohomology, a pair (X,A) gives you an exact triangle of chain complexes in the derived category
mossy ermine
No
cursive flume
how is the family of 'open metric balls' defined?
for each point,i have more metric balls
which one do I choose?
mossy ermine
All of them
cursive flume
so for each point, I have infinitely many metric balls?
😮
because I can make the radius as large as I like
mossy ermine
Yes
digital wraith
Why is it that any deck transformation neccesary corresponds to a path between basepoints in the covering space?
I feel like there's a really basic idea that I'm missing
I get how any path between basepoints in the cover corresponds to a deck transformation iff conjugation by that path stabilizes the image of the fundamental group of the cover
digital wraith
Bruh
pastel linden
sorry, didn't realize you just posted that
digital wraith
Also to answer your question I think it's literally just "the infinite cyclic group generated by A"
pastel linden
is it like, "A" is the generator
digital wraith
Yeah
pastel linden
that makes sense lol
digital wraith
You have an isomorphism into Z^+ that sends A to 1
digital wraith
Om since you're also reading Hatcher apparently, answer my question
pastel linden
anyway carry on, sorry
digital wraith
I'm looking at the first paragraph of the proof of proposition 1.39
pastel linden
I haven't finished everything in the pi_1 section yet, I'm just reading this for something else 
sorry
digital wraith
digital wraith
b r u h
finite heath
So
Y as a subspace of R with the standard topology
For (0, 1) to be open in the subspace topology of Y, it has to equal the intersection of an open set in the topological space and Y
So like
digital wraith
finite heath
<:bleak:877093280905592842>
?
digital wraith
ask question
instantly have 2 other people ask questions and burry mine
finite heath
Oh, you didn't say anything when I asked
digital wraith
I seem to have found an answer to my question online though
finite heath
I seem to have found an answer to my question online though
😮
digital wraith
there's a very good set of alg top lectures on youtube
so you can finish asking your question
finite heath
there's a very good set of alg top lectures on youtube
can u help me mister :]
digital wraith
you want the lectures?
also I'm not "mister"
the lectures are by Pierre Albin
Ok I have now successfully answered my own question
pastel linden
For (0, 1) to be open in the subspace topology of Y, it has to equal the interse...
this is the right idea
so Y is a subspace of R, and open sets in R look like balls
(or rather, open intervals)
so take some intersections and see what you get
vast estuary
hey i have a question
i have proved that the sphere is a regular surface in R^3
can i somehow use this to show that an ellipsoid is a regular surface in R^3
we can get an ellipsoid from a circle via a diffeomorphism
vast estuary
cool!
gritty widget
pedantic terminology note
"ellipsoid" usually refers to the two-dimensional surface (x/a)^2 + (y/b)^2 + (z/c)^2 = 1
you want "ellipse"
vast estuary
It was a typo, I meant R^3
anyway surface in R^2 makes no sense from what ik
like it's a two-dimensional manifold so
gritty widget
true
vast estuary
for fun let me ask
vast estuary
for fun let me ask
do surfaces in R^2 make sense by any other definition?
gritty widget
if "surface" means "two dimensional manifold" then sure
and i can't think of another way you'd define "surface" lol
i've seen "surface" just mean "level sets of smooth functions before" without any regard for regularity or dimension. if that works for you then definitely
vast estuary
ohh yes! thanks
vast estuary
i've seen "surface" just mean "level sets of smooth functions before" without an...
umm so in R^3 are these two definitions equivalent?
gentle ospreyBOT
Hausdorff
vast estuary
**Hausdorff**
if and only if S is the level set of a smooth function f: R^3 to R?
gritty widget
locally
and assuming grad f is never 0
$S \subset \bR^3$ is a regular surface if and only if, for every $p \in S$, there exists an open neighbourhood $U\subset\bR^3$ of $p$, and a smooth function $f\colon U \to \bR$ such that $f^{-1}({0}) = U \cap S$ and $\nabla f(x) \neq 0$ for all $x\in U\cap S$.
gentle ospreyBOT
TTerra
gritty widget
something like that
if you don't assume that the gradient of f is non-zero then you can get some silly things that can't be regular surfaces
gritty widget
probably like 1 person at most is here
if you have a question you should ask anyways please
swift fjord
This is probably very simple but i'm struggling to see why the geometric realisation of a simplicial map is continuous
Like I get it intuitively that you can take sufficiently small barycentric coordinates
but I'm not sure how to formalise it
I guess I can show that it's continuous on each simplex and then use gluing lemma (The simplex is finite), and I can use the fact that the map that takes a point to its barycentric coordinates is continuous, then I just need to show that the map that maps barycentric coordinates to the other simplex is continuous, but this is easy since you can always just take them to be sufficiently small
gentle ospreyBOT
ShiN
swift fjord
does this seem fine
and each simplex is of course closed and by definition of simplicial complexes this map is well-defined on intersections so by gluing lemma I have a continuous map on the entire complex
fair idol
Let S be the limit of a decreasing sequences of sets $S_1 \supset S_2 \supset ... = S $ . Apparently if each term of the sequence is connected and compact then S is connected. Any idea why?
gritty widget
<@456226577798135808> Ring is triangulated?
I’m not sure what this means
vast estuary
**Hausdorff**
here is the definition of a surface, and i have proved that the sphere in R^3 is a surface
void ingot
Let S be the limit of a decreasing sequences of sets $S_1 \supset S_2 \supset .....
In any Topological space or just R^n?
gentle ospreyBOT
Hausdorff
vast estuary
**Hausdorff**
can someone help me justify further why it is enough to take p = (0,0,1)? it may have something to do with the fact that rotations are diffeomorphisms (right?) - i have intuition for it, but not sure how to formalize it
fair idol
In any Topological space or just R^n?
Actually in my case R^n
Specifically I have a decreasing sequence of intervals. It says that the sequences converges to a single point
void ingot
Actually in my case R^n
Specifically I have a decreasing sequence of intervals....
But of intervalls, then you could try to show that the result is $[sup ai, inf bi]$ with $S_i= [ai, bi]$
gentle ospreyBOT
KIl
void ingot
Idk if it works, but that would be my guess
fair idol
Ahhh that's a good point actually
void ingot
Yas
In general R^n would it be more complex🥺
But maybe there would a exist a similar idea
fair idol
Well now that I think about it I even misspoke. I just need to say it about R. So that sup/inf argument works I think.
Perhaps I thought I needed more hardwear than I actually did
void ingot
🤷♂️
ornate sorrel
supai 
pearl holly
fbi 
pearl holly
I just read it as senpai 
ornate sorrel
i don't know i just didn't parse it right at the start
empty grove
bad arc, will skip on rewatch
vocal anchor
If fg is a quotient map and g is a quotient map, must f also be one?
gentle ospreyBOT
expectTheUnexpected
bleak path
Hi, have a few T/F questions that I would appreciate some help in figuring out
This is the first question, which I feel is true but can't justify
bleak path
What is that notation
lol it's the open set, don't know why it was written this way
digital wraith
So S is just all open sets (-infty,a) or (a,infty)
That's what I thought I just wanted to make sure
So do you see how S generates the standard topology on R?
Because that's basically all you need
bleak path
Is it because it can generate all closed subsets through the complement?
digital wraith
Are you familiar with the concept of a basis for a topology?
bleak path
I have just learnt about it and so while I do know it, it's not intutive yet
Uhh how should I put it
digital wraith
It's just a set that "generates" the topology under unions
bleak path
I know the definitions for a basis for a topology, I don't see how a basis generates a topology
digital wraith
And also is closed under intersections
Basis generates a topology because the open sets are unions of basis elements
vast estuary
can someone help me justify further why it is enough to take p = (0,0,1)? it may...
hey.. still need help here lol
bleak path
This is going a bit backwards, but there's a bit in my notes which I don't fully understand about bases for topologies (which I assume is the same idea as the basis you mean here?)
digital wraith
hey.. still need help here lol
I feel your pain
bleak path
Bases is just plural for basis
Ah it was base for me
digital wraith
I posted a question here last night and immediately got buried by 2 other questions
bleak path
This is the definition I have for a basis B, I'm not seeing why O is a system of open sets
digital wraith
So the V_is form a base?
pearl holly
maybe you can create threats and ask there?
digital wraith
Because that looks like subbasis to me
bleak path
Ah sorry you're right, that's the definition of subbasis
digital wraith
A basis is just you have some collection of open sets such that every other open set is a union of elements in the basis
And the basis is closed under intersection
Think of it this way
Suppose I have some collection of subsets of X
pearl holly
is a basis necessarily closed under intersection tho?
vast estuary
I posted a question here last night and immediately got buried by 2 other questi...
ah yeah i hope someone will check..
digital wraith
is a basis necessarily closed under intersection tho?
I think that's part of the definition
Since otherwise it wouldn't generate a topology
*Closed under finite intersection
If that's what you meant. Because I should have been more clear there
pearl holly
Isn't it that if U and V are basis elements then for every x in the intersection of U and V there should be a third basis element containing x such that that basis element is a subset of the intersection U and V?
digital wraith
Oh wait is that it?
pearl holly
that's the definition in Munkres if I remember correctly
digital wraith
Either that's the correct definition then or there are multiple definitions
A basis is such that the set of unions of basis elements forms a topology
That's what's important
bleak path
This is the definition I have for a base, if that helps
digital wraith
You might right though that closure under intersection isn't neccesary
bleak path
Base here should be basis in our conversation
digital wraith
Right so there are a lot of equivalent definitions you can give
bleak path
Okay, so would my intuition be correct that a base for X is just a collection of (open?) sets such that any (also open?) set in X can be formed by the union of these sets?
digital wraith
Basically yeah
bleak path
Is the open part necessary for it to be a basis?
bleak helm
That is exactly what a basis for a specific topology is, and yes the open part is necessary.
digital wraith
Well your basis essentially defines what's open
bleak path
Is there a reason why the definition is not suitable if we don't have the open part?
digital wraith
The key property of a basis is that the collection of unions of basis elements forms a topology
bleak helm
What can cause confusion is there are also definitions out there that describes a collection of sets as a basis if it is the basis for some topology on X, and not a specific given topology.
digital wraith
That is, it is closed under finite intersection and contains X
Then you say it's a basis for a specific topology if the topology the basis gives is that topology
bleak helm
Is there a reason why the definition is not suitable if we don't have the open p...
We want unions of basis sets to be the same as open sets, so the union of a single basis set (which is just the basis set) has to be open for that to be true.
digital wraith
For example take the collection of half open intervals [a,b)
This is a basis for a topology on R, but it's not a basis for the standard topology
bleak path
We want unions of basis sets to be the same as open sets, so the union of a sing...
And is that because topologies on a set can be described? defined? by the union of open sets?
That's a definition my lecturer mentioned could be used to define a topology, but he chose to define a topology in an alternate manner
digital wraith
Again, you're defining a topology with your basis.
bleak helm
I don't know what you mean now
bleak path
Again, you're defining a topology with your basis.
Okay I understand this now, and it is much more intuitive than before
bleak path
I don't know what you mean now
Let me try and find the specific wording
digital wraith
The collection of singleton sets {a: a in R} forms a basis but it certainly isn't a basis for the standard topology
Since it makes every set open
bleak helm
A basis is a subset of your topology that generates the whole topology if you take unions of the elements of the basis.
bleak path
This was the given definition for topology here
bleak helm
So the topology = {all unions of elements the basis}
bleak path
And this was the note mentioned about how a topology could be specified with open sets
bleak helm
Every element of the basis is itself a union of elements of the basis
So it wouldn't make sense for them to not be open
digital wraith
The collection of intervals (a,b) also forms a basis, but this one generates the standard topology
In fact, the standard topology on R is (as far as I know) defined to be the one generated by this basis
Since it's a metric topology
Have you covered metric spaces wOne?
digital wraith
Ah. Some topology classes start with them
bleak path
Mine didn't
bleak path
Since it's a metric topology
But would I be wrong in understanding that this means that its a topology with some notion of distance?
digital wraith
Sincd topologies are in a sense generalizations of metric spaces
Do you know what it means for a set to be open in a metric space?
bleak path
No lol
I think this is getting into a part of math that I need to understand more formally
The definition of open I was given was if its complement was closed
Is that sufficient?
bleak path
I didn't think about how it could be too weak a definition until now, but we were working with "A set S is closed if the closure of S is S"
bleak path
lol
digital wraith
Do you know the definition of a topology?
bleak path
This was the given definition for topology here
I know this
pearl holly
Question about naturality of the cross product in Kunneth formula
digital wraith
Never seen this definition
digital wraith
Your instructor is clearly trying to do something strange with this course
jaunty ferry
I don't matter, but neither have I seen a topology defined as a map
bleak path
This is the first question, which I feel is true but can't justify
I'm sorry for taking up your time btw, and I appreciate your help, but I think it would probably be best if we just went back to the main question here
jaunty ferry
There must be something in those 4 properties that make it equivalent to the standard union intersection definitions though
bleak path
Your instructor is clearly trying to do something strange with this course
I first had that inkling when there was a side note about how topologies were normally defined
digital wraith
No see this is going to be weird with with you working with that definition
It's actually really cool
bleak path
There must be something in those 4 properties that make it equivalent to the sta...
I think so too but it hasn't completely clicked yet
digital wraith
Because I kinda see how it would be equivalent to the normal definition and also possibly be more intuitive, but oh my what
bleak path
I will take that as "you can work with it and understand stuff but you probably won't be able to communicate easily with others"
digital wraith
No this is really interesting
digital wraith
Yes
It's a completely different way to think of a topological space
That might also be more intuitive
I think the tradeoff might be that it's less easy to formally work with
But I had a confusion the other day in topological that I think might have been lessened if I had leaned it this way
Anyways
The way you normally define a topology is take a collection of sets which you call open
bleak path
without giving away too much about my identity my lecturer has won the berwick prize in the london mathematics society so i'm inclined to think he probably knows what he's talking about
(and if you do figure out do keep it a secret please)
bleak path
The way you normally define a topology is take a collection of sets which you ca...
Yes, I've heard this
digital wraith
Such that it contains the empty set, X, and is closed under unions and finite intersections
Then you say a set is closed if it's compliment is open
bleak path
Which lines up with my understanding/definition that a set is open if its complement is closed
digital wraith
And you define the closure of a set A to be the set of all points p such that every open set containing p contains a point in A
Then prove that a set is closed iff it contains its closure
bleak path
yes, I can understand that
digital wraith
So the standard definition makes the definition of a basis much more intuitive
Which likely explains your confusion
bleak path
Probably, I can see how it would be easier to relate
Since this was the first structure/object in the course that was defined as a union of open sets
bleak path
And you define the closure of a set A to be the set of all points p such that ev...
here the
open set containing p
refers to a neighborhood of p right?
digital wraith
Not quite
I think
I believe a neighborhood of a point is a set that contains an open set containing the point
But you don't really care about the distinction in practice
bleak path
Alright, good to hear then
So when you said S can generate a basis for R, was that because S can form any open set in R?
Because in the case where b > a then we would just have R itself, and in the case where b < a then we would have any closed set through the complement
Am I right in my understanding or have I dropped the ball somewhere lol
digital wraith
Because every element of S is open in the standard topology on R and every basis element in R can be built out of S
Actually I believe S is a subbase for the topology on R
bleak path
For example, how would you generate the open interval (0, 1)?
digital wraith
Intersection of (-infinity, 1) and (0,infinity)
You need intersections not just unions though so it's a subbase
bleak path
Oh! So that's a distinction I missed
So subbases can use intersections and unions, but bases can only use unions?
I see
bleak path
Thanks for clarifying
digital wraith
And the unions of basis elements have to be closed under finite intersections
bleak path
right!
digital wraith
So not every collection of sets is a basis
But every collection of sets is a subbasis on the union of all of them
Though a collection of subsets of R for example will only be a subbase for a topology on R if it contains every point in R
But it will always be a subbase for something
bleak path
Thank you for clarifying 🙂
digital wraith
So do you see why S is a subbase for the standard topology on R?
bleak path
Yes, because you can use intersections to create intervals as well, and so every open interval (open set) on R can be created if not through unions then through intersections
The reason why it's a subbase and not a base is because we need intersections and cannot create every open set through unions
digital wraith
Right and then the rest normally follows from the fact that the standard topology on R is defined to be the one generated by the open intervals. But since your class is defining things in weird way, you might need to prove things are equivalent.
bleak path
It's a simple true/false question so in this case I don't need to
Thanks again for helping
Oh wait lol
I think its false
Since the question defines S as a union and not intersection
digital wraith
Yes but it says "generated by S"
Which i assume implies unions and finite intersections
bleak path
Hmmm, that part I will need to clarify
digital wraith
I would he very surprised if that wasn't what it meant
bleak path
Me too, but never hurts to check
digital wraith
Yeah agreed
what textbook are you using?
because this is an interesting approach to topology and I want to mention it to my professor
bleak path
what textbook are you using?
According to the lecturer he's just written it and so hasn't been formally published yet
I don't know how much topology a normal textbook should cover but this one goes until Trennungsiaxiome and Urysohn & Tietze, whatever those mean
digital wraith
According to the lecturer he's just written it and so hasn't been formally publi...
oh I see
fathom cave
Such a radical name lol
next spade
Does a connection on T*M map exact 1- forms to exact 1-forms ? What about closed 1- forms.
fair idol
Is there a nice visualization of a cotangent space?
gritty widget
maybe something with hyperplanes in the tangent space
algebraically, a non-zero 1-form determines a hyperplane, and the converse is true up to a non-zero scalar multiple. maybe that can get you somewhere geometrically
gritty widget
i don't see why it should be true. try a simple case, like the euclidean connection. then the dual connection takes on a particularly simple form, and you can use coordinates to see if this is true
next spade
Ohh okay ill try and see if i get it
gritty widget
the euclidean connection on R^2 is probably the simplest case where you can find a counterexample
next spade
the euclidean connection on R^2 is probably the simplest case where you can find...
For both exact and closed 1 form?
next spade
Kk ill try it later
tough imp
Compact Riemann surfaces of genus g are isomorphic if and only if their Jacobian varieties are isomorphic as p.p.a.v.
No need to thank me
frigid patrol
First idk what ppav means
frigid patrol
principally polarized abeliabn variety
tough imp
So hurbed
rancid umbra
hurbed?
tough imp
Hurbed
quartz edge
Is there a nice visualization of a cotangent space?
Covector stacks in the tangent space a la what tterra said with hyperplanes
rigid reef
I'm trying to tesselate some fundamental surfaces in 284kbps voice chat could I get some guidance please?
rigid reef
I'm trying to show what a AABB^-1 infinite tesselation looks like and I can't get it right
is there a diagram of how to do it?
fair idol
What are some big concepts I should get out of a chapter on cotangent bundles?
coral pivot
So I am a bit confused about this, what does it mean to be "tangent to the submanifold G"
fair idol
Hmm I know a vector field is said to be tangent to a submanifold if it's value is in the tangent space of the submanifold
vocal anchor
If
\begin{tikzcd}
X
\arrow[r, shift left=1, "f"] \arrow[r, swap, shift right=1, "g"]
& Y
\arrow[r, "h"] & Q
\end{tikzcd}
is a coequalizer and $f, g$ are open surjections, why must $h$ also be an open surjection?
It's somehow supposed to be obvious, but it's really not 🙈
gentle ospreyBOT
expectTheUnexpected
vocal anchor
I mean, the "open" part is not obvious
pearl holly
what does "stable" mean in context of algebraic topology?
I've heard that blabla is stable if blabla commutes with the suspension, but what does this mean?
gritty widget
Do you know what cohomology operations are
gritty widget
yeah so the blabla is stable if blahba commutes
if doing your cohomology operation commutes with taking the suspension
then there is also
stable homotopy groups
pearl holly
ye okay I see but I don't really understand how the cohomology operation should commute with the suspension. Like what exactly does this mean?
are we talking about reduced suspension or just "regular" suspension?
gritty widget
ye okay I see but I don't really understand how the cohomology operation should ...
so the cup product
does not commute with suspension
gritty widget
oh sorry
you make a square with
your cohomology operations as the verticle arrows
and taking suspecion as your horizontal arrow
and you want this square to commute
pearl holly
oooh okay I see
ye okay, thank you so much!
ye but I haven't gotten to that part yet lmao
gritty widget
or at least
pearl holly
I tried ctrl + f and search for that but I couldn't find anything, there were a lot of results
gritty widget
can someone maybe say a few words on this
In topological K theory we define K^0 as the group completion/grothendieck construction of equivalence classes of vector bundles
you can then define the negative K-theory groups K^-n by defining them as the K_0 of the nth suspension
Now, a posteriori, we don't need to define then from positive n because of Bott perodicity
Can someone say a bit about, how you would define them, and maybe the history and order of how these things went?
shy moss
is it right that for reduced cohomology $H^{i}(S^n;G)=G$ when $i=n$ and $H^{i}(S^n;G)=0$ when $i \neq n$?
gentle ospreyBOT
Or x1
pearl holly
I think that's true because of the universal coefficient theorem but I'm still noob so don't listen to me lmao
shy moss
i did the calculation with mayer vietoris and long exact secuences of a pair
for unreduced cohomology is it right that $H^0(S^n;G)=G*G$ when $n=0$ and $H^0(S^n;G)=G$ when $n \neq 0$?
gentle ospreyBOT
Or x1
pearl holly
sounds right to me but idk
shy moss
and , is cohomology the ring $H^*(S^n;G)=G \oplus G$?
gentle ospreyBOT
Or x1
pearl holly
so you're asking if that's the cohomology ring?
shy moss
so you're asking if that's the cohomology ring?
yes
pearl holly
okay then that isn't true I think
that's the group structure tho
but then you have to remember that you have multiplication as well
gentle ospreyBOT
Or x1
pearl holly
right but you also have to look at the cup product, right?
pearl holly
so let's look at the case when $R$ is just $\mathbb{Z}$. So let 1 be the generator of the degree 0 part. Then we know that 1 must be the unit in the the cohomology ring since it lives in degree zero $\smile H^0 \times H^x = H^x$. Now let $x$ be a non-trivial element in the other copy of the group. If you square it (take the cup product of itself) then you land in $H^{2n} = 0$ because we are looking at $S^n$. So the cohomology ring will be $\mathbb{Z} \cdot 1 \oplus \mathbb{Z} \cdot x \cong \mathbb{Z}[x]/(x^2)$
gentle ospreyBOT
Tokidoki ✓
shy moss
thank you very much
pearl holly
and you want this square to commute
okay so now I know what a stable cohomology operation is but the definition seems very random for me. Why would we want (or wouldn't want) cohomology operations to commute with suspensions?
Heck, why are even cohomology operations defined as they are? Is it always the case that they make up some sort of multiplication in a ring or something?
obtuse meteor
suspensions and cohomology are closely related
a generalized cohomology theory is determined by a spectrum, which is like a list of spaces each of which is a suspension of the one just below it (up to homotopy)
orchid forge
and
obtuse meteor
and to get something stable under suspension it should be that it commutes with suspension
orchid forge
every (additive reduced) generalized cohomology theory is represented in this way
stone cipher
help me pls. I have 0 idea.
pearl holly
oh crap
so spectra are important I guess?
Okay I see. But about the cohomology operations, is it the case that every cohomology operation induces some sort of multiplication in a ring or module or whatever?
right so wikipedia says this: given a cohomology theory $\epsilon^*: CW^op \to Ab$ there exist spaces $E^k$ such that $\epsilon^k(X) \cong [X, E^k]$ and I guess that the spaces $E^k$ are the spaces in my spectra?
gentle ospreyBOT
Tokidoki ✓
pearl holly
and this I guess generalizes cohomology since $H^n(X; G) \cong [X, K(G, n)]$ if I recall correctly?
gentle ospreyBOT
Tokidoki ✓
pearl holly
I guess that cohomology operations are just another way of adding additional structure to cohomology that induced maps must satisfy and that could be used to prove that a map between spaces can't be surjective/injective like the case with the disk and its boundary
livid drift
can anyone share some materials on
parallel transport on the unit sphere for levi-civita connection along the equator(the curve (cost,sint,0))
shy moss
what is collapsability of CW complexes?
slender plank
Hello does someone have a proof of Ritz theorem ?
stone cipher
Hello does someone have a proof of Ritz theorem ?
which one?
slender plank
The one that says if B(0,1) is compact then E is finite
hollow harbor
What is E 
fair idol
Can someone explain to me what a flow of a vector field is? I understand this to be the set of integrals curves of a vector field.
I read somewhere that if I have a global flow defined by a (compact) vector field then I there is a diffeomorphism. But I'm confused. How is a flow a diffeomorphism
stone cipher
I can't see which are the points that don't separate in disjoint open neighbourhoods (on e))
gentle ospreyBOT
Todd Male (#talks Nov 6)
stone cipher
I can't see which are the points that don't separate in disjoint open neighbourh...
ok. I solved it 😐
fair idol
Holy moly that is a mind trip todd-meister. Thank you for that thorough explaination
hollow harbor
plain raven
i like the explanation in the book by uhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
Kolar, Michor and Slovak
bronze lake
Kolar, Michor and Slovak
my mind wants to read that as spivak lol
tight agate
Holy moly that is a mind trip todd-meister. Thank you for that thorough explaination
cedar pebble
Holy moly that is a mind trip todd-meister. Thank you for that thorough explaination
cursive flume
if I define a set to be finite if it is bijective with any subset of the power set of the natural numbers, is that an okay definition?
hollow harbor
Subsets of the power set of the natural numbers are definitely not usually finite
tight agate
infinte
tough imp
Wut
tight agate
probably a typo
cursive flume
how does one define finite then?
cursive flume
why is this not equiv to what I wrote?
ahhhhhhhh
cause the power set contains N itself
hollow harbor
Yeah, but also
cursive flume
which is infinite
hollow harbor
You're thinking of elements of the power set
tight agate
3 people answering the question simultaneously
hollow harbor
Subsets of the power set are sets of subsets of N
Why are you guys being assholes 
Chmonkey needs help if thats his definition of R
cursive flume
Finite is just bijective with {1, ..., N} for some N
so this is the proper def
hollow harbor
Anyway, the point is that each element of the power set of N is a subset of N
tough imp
With subset of powerset of N
cursive flume
With subset of powerset of N
yes,right
hollow harbor
The subsets of the power set of N are sets which CONTAIN subsets of N
Not sets which contain elements of N
(those are just subsets of N, not subsets of the power set of N)
But as you saw even subsets of N dont need to be finite.
Besides N itself, you could take the prime numbers, or the square numbers, or whatever
Inb4 there are finitely many primes
Nope, it was worse
hollow harbor
You pick an n
cursive flume
I can always say n larger
hollow harbor
Yeah
hollow harbor
Yep
cursive flume
so one of them would not be finite?
if n neq m
if i apply the definition, {1,...,m} is finite if it can be put into a bijection wit {1,...,n} for some n
i need to choose n=m for this t be true
if n neq m,how are they finite?
cursive flume
**Todd Male (#talks Nov 6)**
right,so how are both finite
intuitively they are,by definition not
i'd need a bijection
but if n neq m,it doesn't exist
but then I have to choose n=m
hollow harbor
Not any two finite sets are in bijection
gentle ospreyBOT
Todd Male (#talks Nov 6)
hollow harbor
You can have finite sets which dont have a bijection between them, that's fine
cursive flume
ahh I see
life's hard,we have a mathphys class,which this year is taught by a pure mathemathician and it's pure math and we're 40 students of phys
he doesn't define things and expects us to know naturally
tutor said this is gonna be a hard semester for both him and us
he gave us two topology problems and...when we showed him our proof
he was like:this is not a proof
😢
fair idol
Anyone know of a school that publically posts their old qualifying exams for geometry and topology? (Smooth manifolds by John lee stuff)
cursive flume
he gave us two topology problems and...when we showed him our proof
he desperately asked: is there any mathematics student in the room?
how could I start proving this?
tough imp
The program at UW pretty much just does only Lee, since he’s a professor here at UW
plain raven
why is this person asking for the definition of "finite" in #point-set-topology lmao
unless you are asking what is meant by a finite cell complex
fair idol
Ah yeah. Now that I think about it you actually know the inverse exists but you just need that it's continuous.
Todd, why is closed sets to closed sets sufficient I'm curious and this ain't even my problem
echo elk
Ah yeah. Now that I think about it you actually know the inverse exists but you ...
I believe it's from the contrapositive of the original definition. Correction: Let (M, O₁), (N, O₂) be topological spaces. φ:M→N, is continuous iff V₁∈O₂ ⇒ preimᵩ(V₁)∈O₁. One may let V₁=N\V₂ for a V₂ in N. This means N\V₂ ∈O₂⇒ preimᵩ(N\V₂)∈O₁, but preimᵩ(N\V₂)=preimᵩ(N)\preimᵩ(V₂). This is finally the same as saying V₂ being closed implies the preimage of V₂ under phi is closed.
fair idol
Yeah it does seem like that actually
echo elk
Nevermind my statement as I see it being incomplete, and likely wrong.
fair idol
No worries castle I thought the same 🤣
gentle ospreyBOT
Todd Male (#talks Nov 6)
coral pivot
We haven’t talked about stuff like tangent bundles yet
He defines smooth manifolds in the next chapter for example
Right so this is for matrix groups only
Which can be identified with sub spaces of Euclidean spaces
So he is saying submanifold if each neighborhood has a smooth map to Rn with smooth inverse
Hmm
In that defination ig I don’t get why the path being contained in G means the derivative would be
orchid forge
In that defination ig I don’t get why the path being contained in G means the de...
If the path $\gamma : [0,1] \to \mathrm{Mat}_n(\mathbb C))$ has image contained in $G \subset \mathrm{Mat}_n(\mathbb{C})$, then it restricts to a path $\gamma' : [0,1] \to G$, so it defines a tangent vector in $T_0G \subset T_0(GL(n)) \subset \mathrm{Mat}_n(\mathbb{C})$. So, if $\gamma$ is contained in $G$ for all $t$, then it is tangent to G at the identity.
gentle ospreyBOT
Kogasa
coral pivot
hmm im probably gonna go read the next chapter and learn the definition of tangent spaces and such to appreciate this
ty for help slim and kogasa
frigid patrol
@cedar pebble what's the difference between period matrix and Riemann matrix of a Riemann surface?
cedar pebble
Riemann matrix?
frigid patrol
It's referenced in the Sage documentation
cedar pebble
it sounds like this is a period matrix with respect to a certain basis for plane curves
frigid patrol
Its also mentioned on Wikipedia but not defined
Sage documentation kinda whack
Therefore the complex Schottky problem becomes the question of characterising the period matrices of compact Riemann surfaces of genus g, formed by integrating a basis for the abelian integrals round a basis for the first homology group, amongst all Riemann matrices
cedar pebble
yes
frigid patrol
This sentence uses both terms again
cedar pebble
so Riemannian matrix means a specific thing
symmetric with positive definite imaginary part
frigid patrol
Okay
I'm a bit confused
A 2g dim lattice in C^g is specified by a g by 2g matrix of complex numbers
What does it mean for this to be symmetric and have positive definite imaginary part
cedar pebble
you can "complete" this to a 2gx2g matrix
cedar pebble
yes
sorry I think you want gxg matrices
but same idea holds
if you have half the matrix you can "complete" it to get a matrix in Sp_2g(R) mod U(g)
frigid patrol
Conjugate under the action fo GL(g,C) means that you can choose a suitable vector space basis of C^g such that your original lattice has its first n vectors being the n vectors of your basis
And that's clearly true
cedar pebble
sure
frigid patrol
Alright I know what a Riemann matrix is now
Riemann matrices are just those bases of lattices in C^g such that the quotient is a projective variety.
Period matrices are those Riemann matrices that arise from the period lattice of an Riemann surface
cedar pebble
right
geometrically this is asking about the Jacobian locus in A_g
since A_g is H_g/Sp_2g(Z) where H_g is the space of Riemann matrices
and the Jacobian locus in here consists of those matrices, up to change of integral homology basis, which occur as period matrices of curves of genus g
these are Riemann matrices by Hodge-Riemann bilinear relations
this is the Schottky problem: identifying the Jacobian locus in A_g
the first case where this is really nontrivial is g=4 and the Jacobian locus in A_4 is smaller than what one would expect, and this is witnessed by what is called the Schottky form which is a Siegel modular form for Sp_8(Z)
frigid patrol
Wow
The existence of certain modular forms accounts for the fact that the Jocobian locus is not all of A_g?
cedar pebble
rather it accounts for the fact that not all ppav's of dimension 4 are products of Jacobians of curves yea
frigid patrol
I see
Thats pretty cool
I do t have the topology/AG background to make sense of this
limpid leaf
hi everyone. im able to show that cl U cap V is empty and that there is some open neighbourhood U'' of x and of clV so that U'' cap cl V is empty
but i cannot figure out how to combine these facts
cedar pebble
limpid leaf
what ng. this is the topology chat
anyways i figured id need to do some shit like show that for the open neighbourhood U' containing cl U and the open neighbourhood U'' from above, since htey are both neighbourhoods of x, one there must be containment in either direction, and that will imply result. but i cant figure out how to get U'' subset U' to give me that cl U subset U''
which means im surely doing something wrong
after that complex midterm u r definitely right
i just odnt know if im supposed to be proving this containment part
or if it just falls out of what ive already shown
hollow harbor
anyways i figured id need to do some shit like show that for the open neighbourh...
maybe you can intersect U' and U''?
after all, the intersection of two open sets is open.
i'm not sure
limpid leaf
hollow harbor
oh, the problem then is that cl(U) is not in U''
limpid leaf
yes
hollow harbor
wait but then cl(U intersect U'') is
is it?
idk
something like that
it's not quite i guess
limpid leaf
well ideally we would have that clU subset (U' cap U'') i guess
but idk how to do hat
dont think there is enough information
limpid leaf
no we can guarantee that because regular
U' is the open set containing x that has empty intersection with (an open neighbourhood of) cl V
oh sorry
yes you're right
hollow harbor
i thought that was U''