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00:04
wlog $G = \mathbb{Z}[J^{-1}]$ for a set of primes $J$ or $G = \bigoplus_{p \in J} \mathbb{Z} / p$, but I'm not sure how much that helps
(this is a claim from a paper by Bousfield btw.)
that should simplify things a lot
but there is some interaction between divisibility and unique divisibility that I'm not immediately grasping
wait
if we restrict to the special cases you just mentioned, doesn't same acyclicity imply equal
the first group is all torsion-free and $J$ is characterized as the set of primes at which it is uniquely divisible, the second group is all torsion and $J$ is characterized as the complement of the set of primes at which it is uniquely divisible
If $G' = \mathbb{Z}(p^\infty)$, what's the corresponding $G$? It has to be torsion and all primes $\neq p$ have to act by isomorphism, so it should be... $\mathbb{Z} / p$, but that cannot be since $\mathbb{Z} / p \otimes \mathbb{Z} / p \cong \mathbb{Z} / p$ but $\mathbb{Z}(p^\infty) \otimes \mathbb{Z} / p = 0$ (??)
@Thorgott what special case? That's a reduction!
You may assume one group is of that form, by definition
oh I misread
00:12
that's why I don't think it's necessarily much help :)
@BenSteffan and why is that not a counter-example?
I would like to know that as well
that just came into my head
I suppose $0$ and $\mathbb{Q}$ also do the job
though you may have different preferences as to whether $0$ counts as torsion-free
00:14
I would have to ask Bousfield
but evidently something is fishy
are the two claims perhaps made dependently?
i.e. (A o G = 0 and Tor(A,G) = 0) iff (A o G' = 0 and Tor(A,G') = 0)
...yes, indeed
so for A=Z/p, this is equivalent to unique divisibility at p
which checks out
ah, right
it's the same for Z/p^n, so you get all f.g. A by taking direct sums
i really wanna take a colimit, but it doesn't seem like it quite does the job
00:20
yeah, that's the first thing that came to mind as well
but how are you going to get control of the maps
hm, if you take a presentation $\bigoplus_I \mathbb{Z} \to \bigoplus_J \mathbb{Z} \to A \to 0$ and tensor with $G$, you get an exact sequence $0 \to \mathrm{Tor}(A, G) \to \bigoplus_I G \to \bigoplus_J G \to A \otimes G \to 0$, and similar for $G'$.
can we say that the middle map is an isomorphism iff it is an isomorphism in the sequence for $G'$...
@BenSteffan can I suggest a binder?
bind 'er? I barely even know 'er!
Three hole punch the maps.
Then you can flip between them.
That'll give you a lot of control.
01:25
I decided to post this on the main site math.stackexchange.com/q/5037801/681666
thanks for the help @Thorgott :>
01:54
hoping this gets answered
it's probably a stupid trick, but I also got stuck on it
apparently it was so obvious to Bousfield he doesn't even mention this direction of the argument at all
02:06
something something (homological) algebra is beneath an algebraic topologist something
unless the algebra itself is somehow interesting
 
3 hours later…
05:23
6
Q: Is a positive degree self map on a Riemann surface homotopic to a holomorphic self map?

Ali TaghaviLet $S$ be a compact Riemann surface and $f:S\to S$ be a continuous self map of positive degree. Is $f$ homotopic to a holomorphic map on $S$? Motivation: I had intention to consider this question for every map $f:S\to S^2$ where $S$ is an arbitrary complex manifold. The homotopy class of a ...

The motivation in the post is interesting
 
2 hours later…
07:04
i was under the impression, that a tuple is a finite ordering (a,b,c,d)
but we usually refer to tensors as tuples (t_ijkl)
even though no ordering seems to be given?
 
5 hours later…
11:52
Fun question: Suppose we have a chess team of 6 players. Before the start of round one, we have to submit an ordered list of the players. At each round, we can send any 4 of them to play. But they have to maintain the order. So if A>B in the list then A has to be at a higher board than B( the boards are ordered as 1>2>3>4).
Now suppose A and B are our best players and also assume that before each round, we can accurately predict which 4 players of the opponent team would play. Our goal is to place A and B in the original list so that we have the most flexibility. So where should we place them?
 
1 hour later…
12:59
@Madder a tuple is just an indexed collection of stuff
your indexing set can be ordered, it can also not be ordered
but why would no ordering be given? tensor products are ordered, even if they are commutative
the ordering is extra data in any case, it's not relevant to the tuple
I would implicitly assume a partial ordering, yeah, if you're gonna tell me you will totally order it lexographically, I will give you a weird look
 
2 hours later…
15:24
Bousfield writes $SG$ for the Moore of type $G$ and $\Sigma X$ for the suspension of a spectrum $X$. Question for the audience: What would be the most logical choice of notation for desuspension/loops of a spectrum $X$ in this setting?
Answer 1: $\Omega X$
Answer 2: $\Sigma^{-1} X$
Answer 3: $S^{-1} X$
Guess which one Bousfield chose
a winner is you
bonus points if you can explain to me why on earth he would do that
No idea lol
15:51
@BenSteffan I have no idea what any of this means, but $S^{-1}$ is the obvious choice.
Greek letters are dumb.
awful
@XanderHenderson well he could have opted to write suspension as $SX$ and choose something else for the Moore spectrum
but he woke up that day and chose chaos
As we all should.
Huzzah for chaos!
$\mathfrak{S}X$
be careful what you wish for
@BenSteffan If I think Greek letters are dumb, what do you think I am going to say about that?
(It is sonderful!)
(and I love it!)
16:08
If $(X, d)$ is a metric space, then apparently it becomes a topological space if we call $U \subset X$ open when each element has an epsilon neighborhood.
But why can we be sure that each element of $X$ has such an epsilon neighborhood? Only if we can surely say this can we make the above claim.
That's a very strange way to phrase it
...but ok: $B_x(\epsilon)$ is an $\epsilon$-neighborhood of $x$, for any $x$...
Usually you would say that $X$ becomes a topological space by declaring "metric open" sets to be open
The second and third conditions for a topological space are clear from this, I'm just asking about the first which says the empty set and $X$ itself have to be elements
or by endowing it with the topological which has the collection of all $\epsilon$-balls, some $\epsilon$ as a subbase
@BenSteffan Yes but how do we know $X$ itself is open by our definition?
$X$ is just some random metric space, right?
That's immediate
$X$ contains arbitrary $\epsilon$-neighborhoods of all its elements
Obviously every ball is contained in the whole space
I struggle to see what the issue is supposed to be :)
16:15
@BenSteffan Is the $U \subset X$ strict or not?
Why is that relevant?
You wouldn't want it to be strict, generally speaking
Otherwise things like the one-point space will give you trouble
@BenSteffan Yeah exactly that's what I was thinking about!
Ok so yeah if it's not strict we don't have to think about that and everything is fine
Thank you!
...if you have a concrete issue in mind please make that clear (the next time)!
but you're welcome :)
17:05
all inclusions are non-strict unless specified otherwise
I know a few people I'd like to send a box of "$\subseteq$'s" to
17:20
Just now I saw that Laila Podlesny, Oksana Gimmel, OlegK and Cleo have all been suspended for about 100 years!! I believe they have made great contributions to the site by presenting intriguing and challenging integrals; although these were effectively sockpuppets, the intention of the person behind them wasn't bad according to me. Also, these accounts haven't been active since years. I don't see the point in suspending them for a ridiculously long period.
i don't have strong feelings about it either way, but it's now a famous example of sockpuppeting, and one point in suspending the accounts might be to avoid someone else getting caught sockpuppeting, and pointing at those and saying "when that guy did it, it was hilarious and nothing happened"
so (purely speculating here) the thinking might be, it's bad from the point of view of general site policy if looks like there's a "if you become a famous meme, it's different" exception to site rules
does it make any actual difference to those specific accounts, as you say, probably not
Note that mods can't give suspensions that long. That can only be done by SE staff.
it wouldn't surprise me if SE staff would like to discourage someone from "doing a cleo" on other network sites, even if we wanted to have a local culture where it was permissible
(which is very much an open question, haha)
Exactly
A century isn't that long ... ;) physics.stackexchange.com/users/4864/ron-maimon
given that these accounts haven't been active in years, I also don't think it matters that they did get banned
17:33
Right. It's just to discourage similar behaviour in the future.
17:44
What is the weak sequential closure of unit sphere in $\ell^\infty$?
18:28
@RakshithPL Yeah, Vladimir is the creator of Cleo. Bro lost 14 k points today. This reddit thread seems reliable
18:38
Hey Sine
18:49
@SoumikMukherjee hi
19:08
Just as a general note, speculating about the suspensions of other users is discouraged.
19:20
$S^{-1} X$ for the desuspension
 
3 hours later…
22:29
I love using $B(\overline{\underline{I}}(x))_{s\in|\Delta^l|_e}$ to denote a single space
dear god
underline and overline??
and that's the less confusing part
$\underline{I}$ is an (ordered) collection of intervals and $\overline{\underline{I}}$ is the smallest interval containing them (plus we then remove the outmost endpoints)
fantastic
wait no, I'm wrong
that's what the $B$ is for
overline underline is just the name for the collection
but it's doubly indexed, so I suppose that's why there's two lines
idk man, its horrible
if the notation is impenetrable then no one can check your work for errors... :)
22:42
it's not my notation unfortunately, but of a paper that I wish to understand
sometimes I long back to the middle of the last century when this sort of notational assault was impossible due to the limitations of the humble typewriter
im not sure if that would've made the math prettier
but probably times were better before people wrote down infty-categories

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