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12:03 AM
This looks like a partially ordered set
 
well, it'll be an equivalence relation if "can be exchanged for" satisfies the defining properties of an equivalence relation. when you say it's a "one way process" this suggests maybe it's not an equivalence relation. e.g. if some token t can be exchanged for ice cream, but ice cream cannot be exchanged for t.
 
@leslietownes It is more like the latter. You wouldn't be able to sell the ice cream for coupons.
 
so it might be something weaker than that. you should try to identify what abstract properties your relation has. en.wikipedia.org/wiki/Order_theory might be an OK start (and the suggestion to look at partial orders might be a good one) depending. sometimes a relation will acquire nicer properties after you introduce some weak notion of equivalence.
 
@EthakkaappamwithChai I think it's a partially ordering relation, but I'm rather weak and rusty in this field. Does it make sense to take a quotient by a non-equivalence class relation?
@leslietownes I know it is not reflexive or transitive in the cases that I care about.
However, identity is preserved, so $\forall x, x \geq x$
 
wikipedia for order theory is kinda disjointed. if you go around the various pages, different places tend to variously state/restate stuff, notation is not consistent. there's a whole thing of strict vs. non-strict partial orders... if it seems confusing, probably ignore it.
what do you want to do with this relation?
 
12:11 AM
I want to meaningfully describe the set of things $P_\alpha$ such that if you drew randomly from $P$, you would effectively have anything in $P_\alpha \subseteq P$
 
huh. is this a software thing or a theoretical thing?
 
Combinatorics
 
so you're not feeding a set of adjacency relationships into an algorithm and wanting it to spit out a set of elements or ordered pairs? you just want to... study this?
 
My math vocabulary and notation skills are very bad since I've been mostly working in coding for the past two years, I haven't done much math.
Yeah, I want to do analysis over a problem modeled by this.
I already have a program that simulates a process modeled by this, but the time complexity makes it take ages. I'm hoping to better model the problem mathematically so that I can find some more optimal algorithm later
But I'm more fighting to just find a way to model the problem in the first place
 
i dunno. i guess i would start by trying to pin down formal properties of the relation that hold all the time. i guess transitivity is not on the list. :(
 
12:17 AM
I'm looking to expand the intuition behind Gian-Carlo Rota's Twelvefold Way to other cases of relations over functions $f : P \to X$. In this case, $P$ has a relevant relation over it $\geq$. I'm trying to figure out how that affects the count of functions $f$, which in turn will decide how many configurations there are.
 
i am out of my depth now.
 
Normal combinatoric patterns seem to be rather difficult to model for very exotic relations, but I'm running into those very exotic relations in my work. @_@
 
This problem comes up a lot in economics
you often want to order things where a uniform comparison doesn't make sense
mass collel's microeconomic theory
see chapter-1
 
Is the book free anywhere? Or will I have to order it?
 
It's a pretty standard book so you should be able to find it by searching the pdf online, let me know if it helps
 
12:28 AM
I'll find a copy and read though it. Thanks @EthakkaappamwithChai
 
For the sake of your question, I think the concept of Hasse diagram and poset will suffice
perhaps with some modification
Did the excerpt sound like what you were looking for?
 
It's very close, but it's a lot to digest at the moment. I'm coming at it from a very different context.
The hard part isn't in modeling the relation, because that is definitely obvious, but in the problem of cardinality of configurations, which may be likened to the cardinality of choices as described in this book.
 
here is some more
the book goes over pretty much all the way you can run into trouble when trying to use poset ideas to model real world behaviour
 
I did misspeak earlier. The relations I'm working with are not symmetric but are reflexive
It seems it may take a bit of work to work within this framework, given that it seems the fundamental difference between my case and the preffered choice case is that in mine "choosing nothing" is a valid option.
 
Ah
So I think what you want is a rational preference
 
12:35 AM
I do not have completeness or transitivity in my case.
 
oof
that's tough
there is another book you can try check out
might be harder to find a pdf of online
but also glosses over stuff like the above
 
I'll give it a look. I think my case might be even more general than these, which unfortunately puts me in a new frontier, as far as I can tell
 
the book is sort of old stuff
not like the cutting edge stuff of economics
it links some papers, maybe you can check out arxiv or something and check if something like what you're doing has been done b4
 
12:55 AM
What's the proper way to denote a choice function? (One that picks an element from each set $S_\alpha$ of a set of sets $S$)?
 
no standard notation that i know of.
 
@copper.hat: I left a comment on a not yet flagged answer.
 
1:28 AM
I just deleted it.
 
Is robjohn hounding poor copper for PSQ?
 
Would it be weird to consider a function a relation? $f = \{(x, y) \mid f(x) = y \}$ seems like a common interpretation.
Assuming I'm allowing myself to quotient by non-equivalence class relations, I should be able to quotient by functions as well.
 
2:11 AM
@TedShifrin they did a little work. i deleted my answer and added a leading comment instead. i am a pushover.
 
@TedShifrin woof! woof!
@copper.hat I didn't want someone to flag your post.
 
Some people whose answers used to be so good have nonetheless been banned from network I see
 
i am striving to achieve the banned goodness
 
@DevanshBhardwaj It is not necessarily the quality of the answer. The answer may be excellent, but if the question is poorly stated (according to site guidelines), it is considered bad to answer.
 
oh yes some of them were doing so
 
2:24 AM
did the $H^1_0$ question get resolved?
 
i'm not sure it ever got asked. i couldn't make out what the question was.
 
@leslietownes the same with many questions of a similar source.
 
yeah. shrug.
 
Sometimes the questions are quite basic, and sometimes they are much more elevated.
it's hard to tell where they are coming from
@copper.hat I see that if they answer your hint, they will have answered my hint
 
2:41 AM
@robjohn i find $\cap,\cup$ notation a little opaque at times, perhaps my digital logic background :-). i find this much easier to evaluate $(A+B)(\bar{B}+\bar{C}) = A\bar{B} + B \bar{C} + A \bar{C}= A\bar{B} + B \bar{C} + A B\bar{C}+ A \bar{B} \bar{C} = A\bar{B} + B \bar{C}$
 
That shows that $A\bar C=\color{#090}{AB\bar C}+\color{#C00}{A\bar B\bar C}\subset \color{#C00}{A\bar B}+\color{#090}{B\bar C}$ which is what my hint suggested
@copper.hat I did find the work they did marginally sufficient, since it is on the same path I took, but with the downvotes and flags, I was afraid your post would be flagged.
rather than flagging or downvoting, I try to suggest ways to improve the post.
 
why are there so many different definitions for the essential spectrum? I'm guessing they're not always equivalent...
 
3:00 AM
i think they coincide for operators on hilbert space and probably yeah not in general.
 
@robjohn I was surprised that Asaf edited the post.
 
@leslietownes a likely story
 
@copper.hat I believe he gets annoyed when is misused, and perhaps he thought the work presented was sufficient, as well (though, don't quote me on that).
 
calvin: i'm thinking because one of the definitions in the hilbert case, spectrum of the image in B(H)/compacts, seems like it could get weird in the banach space case, because compact operators can be so weird there.
 
@leslietownes wait im just now walking into all this functional analysis...whats so weird about compact operators
arent they supposed to be the pretty ones you can bring home to meet your parents
 
3:08 AM
well, nothing's weird about them on hilbert space. they're norm limits of finite rank operators. that's what you'd expect them to be.
on a banach space there can be compact operators that aren't norm limits of finite rank operators. approximating operators with other operators is a pain in the ass.
 
oh right. i remember this faintly\
enflo?
dang ok
 
another weird fact, on hilbert space an operator is invertible mod compact operators iff it's invertible mod finite rank operators. on at least some banach spaces, those are probably different notions.
 
yikes all round
 
Compact Operators is a decent overview.
 
haha, wikipedia gives a list of five candidates for essential spectrum for an operator on a banach space (none of which are obviously equivalent to what i was thinking about), and says they're not generally equivalent. [[citation needed]]
 
3:19 AM
> The essential spectrum ${\displaystyle \sigma _{\mathrm {ess} ,k}(T)}$ is invariant under compact perturbations for k = 1,2,3,4, but not for k = 5.
 
 
1 hour later…
4:33 AM
gonna have to shelf this for now, spent an hour going down the citations looking for proofs that the definitions could be different
 
5:10 AM
banach space theory is too weird even for me
 
@leslietownes I dunno, it seems pretty normal to me.
 
booooo
 
then there're banachek spaces
 
5:29 AM
@BalarkaSen I think I finally get why you get a torus
I'll tell you in the morning though
(the Borromean surgery thing)
 
 
2 hours later…
7:04 AM
@BalarkaSen sure but I mean this is not in $|z|<1$ where it converges uniformly.
 
 
1 hour later…
8:10 AM
Ted is asleep i guess
Sad!
 
 
1 hour later…
9:23 AM
@BalarkaSen couldn't you prove convergence on $S^1$ like this: $(1-x)^{\frac{1}{2}} = 1 - \sum_{n=0}^{\infty} c_n x^n$, as a maclaurin series where $c_n > 0$, then $\lim_{x > 0 ; x \uparrow 1 } \sum_{n=0}^{\infty} c_n x^n = \sum_{n=0}^{\infty} c_n$ by the MCT, so $1 - \sum_{n=0}^{\infty} c_n = 1 - \lim_{x > 0 ; x \uparrow 1 } c_n x^n = \lim_{x > 0 ; x \uparrow 1} (1-x)^{\frac{1}{2})}= 0$, i.e. the maclaurin series converges at $1$
here MCT used with the usual counting measure on $\{1,2,3,.... \}$
where we use the usual principal branch of the log as well
 
 
1 hour later…
10:28 AM
And it's up again.
 
 
1 hour later…
11:47 AM
Hello
I need some help in algebra please
If d divise n×m
d/n×m
If ans only if
There exist a in N* and b in N* such that a/n and b/m and ab=d
 
12:37 PM
Can someone give me a hint how to compute $[M_{g} ,S^{2}]$ where $M_{g}$ is the surface of genus g and $[M_{g} ,S^{2}]$ is the set of continuous maps from $M_{g}$ to the sphere modulo (free) homotopy ? My hint is to use the Cellular Approximation theorem. I know both are CW complexes.
 
12:57 PM
Is N* the set of natural numbers?
@Vrouvrou
 
 
1 hour later…
2:17 PM
Is this reasoning correct to show that assuming the limit definition for $0<\epsilon<a$ is WLOG for proving a limit statement for any $\epsilon>0$?
"By hypothesis, for any $0<\epsilon<a$ there exists $\delta(\epsilon)>0$ such that for any $x \in \text{dom}(f) \cap (x_0-\delta(\epsilon),x_0+\delta(\epsilon)) \setminus \{x_0\}$, $|x-x_0|<\delta(\epsilon) \implies |f(x)-l|<\epsilon$.

Suppose $\epsilon' \ge a$. Using the limit hypothesis with $0<\frac{a-\epsilon}{2}<a$, there exists $\delta'=\delta((a-\epsilon)/2)>0$ such that for any $x \in \text{dom}(f) \cap (x_0-\delta',x_0+\delta') \setminus \{x_0\}$, $|x-x_0|<\delta' \implies |f(x)-l|<(a-\epsilon)/2<a\le\epsilon'$. Since $\epsilon' \ge a$ is arbitrary, this holds for any $\epsilon' \
That is, for any $\epsilon' \ge a$ there exists $\delta'>0$ such that for any $x \in \text{dom}(f) \cap (x_0-\delta',x_0+\delta') \setminus \{x_0\}$, $|x-x_0|<\delta' \implies |f(x)-l|<\epsilon'$."
 
2:55 PM
@TymaGaidash yes natural number not equal to 0
 
3:34 PM
@leslietownes I'm reading about Banach space theory, actually. I'm interested in their geometry
exploring the topic of weak topologies for now, things like polars and Alaoglu-Bourbaki theorem
 
If i am calculating $Kernel(\varphi - a*id )$ Can i just calculate $Kernel(\varphi^2-a*\varphi)$ and say it is equal?
It seems okay to do so in my brain
 
why would it be equal?
 
I thought something like $ \varphi = a id \Leftrightarrow \varphi^2= a \varphi$
Same functions, should have same kernel, am i not right?
 
why?
 
They are the same object.
 
3:43 PM
they are the same picture
 
You mean image, and i have no idea what you mean
 
Theoretically, $e^{\pi i}=(1+(\sqrt e-1))^{2\pi i}=\sum_{n\ge0}\binom{2\pi i}n(\sqrt e-1)^n$, where $\binom rn=r(r-1)\cdots(r-n+1)/n!$ for any complex number $r$. Does this actually converge, and to the right thing?
 
@MadSpaces I was just joking. I find your claims funny
 
So what should i understand now? If you are not serious, please refrain from answering!
You are only confusing me even more.
 
@AkivaWeinberger can you help me in algebra please
 
3:48 PM
@MadSpaces well, you're not explaining your way of thought
I don't know what you're expecting
 
4:01 PM
Does anybody have a suggestion how to work out the second part?
2
Q: Difficulty with verifying Stokes' theorem for a given domain and a specific differential $(n-1)$-form

Invisible Let $f:\Bbb R^{n-1}\to\Bbb R$ be an everywhere positive $C^\infty$ function and let $U$ be a bounded open subset of $\Bbb R^{n-1}.$ Verify directly that Stokes' theorem is true if $D$ is the domain defined by $$0<x_n<f(x_1,\ldots,x_{n-1}),\quad (x_1,\ldots,x_{n-1})\in U$$ and $\mu$ an $(n-1)$-fo...

 
@Myprofileissaved follow the hint you've been given. you need homotopy classes of cellular maps $M_g\rightarrow S^2$. what can you say about such a cellular map? think about the $1$-skeleton specifically.
 
4:39 PM
@AkivaWeinberger it actually does converge to the right thing
i.e. -1
the question basically comes down to asking whether the series $\sum_{n \geq 0 } \binom{2 \pi i}{n} (x)^n$, which converges absolutely for $|x| < 1$ , is really equal to $(1+x)^{2 \pi i}$, which it is because it satisfies the relevant diff eq. characterizing $(1+x)^{\alpha}$, $\alpha \in \mathbb{C}$, with initial condition $1^{\alpha} = 1$
 
does $\overline{f(x)} = f(\bar x) $ imply holomorphicity
 
the wikipedia page on binomial series gives details of exactly when you get absolute convergence of that power series: en.wikipedia.org/wiki/Binomial_series, it may be of interest to you
@MadSpaces no. take $f(x) = \overline{x}$, $\overline{f(x)} = \overline{\overline{x}} = x = f(\overline{x}) = \overline{\overline{x}}$
$f$ is far from holomorphic
 
Okay thanks apparently if f is holomorhpic and the restrictions on the reals is real then the statement is true but only in one direction.
 
if $f$ is holomorphic in $\Omega$, then $\overline{f(\overline{x})}$ defined in $\overline{\Omega}$ is holomorphic
you can see this by applying the chain rule with the wirtinger derivatives
what you just said is a corollary of this
its called the schwarz reflection principle
 
Thank you i had no idea such thing existed. I learned something new.
 
4:53 PM
the chain rule for the wirtinger derivatives is a handy thing, might be a good exercise to also try and derive it
and no problem
by the way, the full statement is essentially that if $f$ is holomorphic on a domain $\Omega$, and continuously extends to $\Omega \cap \mathbb{R}$ where it is real, then its analytic continuation to $\Omega \cup \overline{\Omega}$ is determined by that formula
 
what if we said that if what i wrote applied, then there is some function with restriction which fullfills that statement
Without me thinking abot it long just quick inquiry, if you think it would work ?
 
what is 'the statement', if what you wrote applied, we cannot conclude holomorphicity at all
 
Alright.
 
the function $f(z) = \overline{z}$ satisfies 'the statement' and is not holomorphic
 
Alright thanks i was thinking something else. like some kind of weaker condition , but never mind thats out of the scope and speculative
 
4:58 PM
see , your statement is just saying that $f$ commutes with reflecting across the real axis
that is what it is saying visually
conjugation is reflection across the real axis
and your statement is saying $r \circ f = f \circ r$, where $r$ denotes the reflection
you can start with any random function $f$ defined in the upper half plane , continuous but certainly not holomorphic, and real on the real axis
and then force its definition in the lower half plane so it satisfies the above
 
@Wave Ted was talking about analyticity at points on the boundary of the unit disk.
@porridgemathematics First of all, the sign of the coefficients of the Taylor series at $0$ of $\sqrt{1 - x}$ alternates, so I don't see why $c_n > 0$ is true. You can try to run your argument with $\sqrt{1 + x}$ instead. Even then, I don't see why the MCT implies your limit claim. We can do the same thing with $1/(1 - x) = \sum x^n$. Does MCT imply $\sum (-1)^n = \lim_{x \to {-1}} \sum x^n = 1/2$? No.
 
what is the size of the smallest square that can fit two non-overlapping unit radius circles?
is it 2+2√2 ?
 
@Thorgott there's an awful way to do this and im curious if you see it lol
or if you want to
 
@BalarkaSen huh, the maclaurin series literally has that form? wolframalpha.com/… , is that not $1 - (\sum_{n \geq } c_n x^n )$, $c_n \geq 0$?
 
Ah OK maybe you're right.
 
5:13 PM
im only using the MCT on the RHS summand
 
But it still doesn't prove what you want it to prove.
 
Why doesn't the same argument prove $\sum (-1)^n = 1/2$?
@AkivaWeinberger OK tell me the torus thing
 
because you cant apply the MCT?
 
Why not?
 
5:14 PM
when you go to $-1$
your series isnt monotonic
 
$-1$ is in $S^1$.
 
i.e. go to -1 on the real line
when you go to $+1$
on the real line
 
I don't care about the real line. I want to show that the complex power series converges on points of $S^1$.
 
all you need to show is that the maclaurin series converges at some point of $S^1$ though
I showed it converges at $1$
 
No, I want to show the Taylor series at $0$ of $\sqrt{1 - z}$ converges at all points of $S^1$.
That's the claim.
 
5:17 PM
oh sorry, you are right
for some dumb reason I thought if a power series converges at some point, it converges uniformly in the closed disk of that points radius
but its just the open disk
my bad :(
(uniformly on compact subsets of *)
maybe what I said can be adapted somehow though
 
Maybe what you're suggesting is true. If $\sum |c_n|$ converges, certainly $\sum c_n z^n$ converges on $|z| \leq 1$, right? Absolute convergence.
In fact, convergence of $\sum |c_n|$ is what I showed by Raabe.
 
ah yes
indeed!
but its not quite what I thought in any case
in this case it turns out the absolute value sum coincides
 
Maybe this works, then.
Interesting. Nice proof.
 
ill try and formalize it a bit further, may have spoke too soon
ah wait, i think its basically just what you wrote, isnt this just the M-test?
 
Looks fine to me now that I understand the strategy. You're saying: It is enough to show $\sum |c_n|$ converges, so screw the complex power series and consider $\sum |c_n| x^n$ as a function on the interval $[0, 1] \subset \Bbb R$, which is magically $1 - \sqrt{1 - x}$ or something on $[0, 1)$, so by MCT you're done.
The key point is that the coefficients do not alternate sign or anything. I never checked that.
 
5:25 PM
yeah, they're all the same sign besides the constant term
 
Cool proof.
 
i think the magic is that the alternation in negative sign for ln(1+x) occurs in even powers
 
Ah.
 
so negating and then swapping x with its negative gives the same sign
 
Pretty neat.
@TedShifrin: Here's an elementary proof ^
 
5:30 PM
thanks, i had never heard of raabes before as well - looks useful
 
i had to get Raabes shots once
 
Ratio test compares with the geometric series $\sum \lambda^n$ whereas Raabe compares with $\sum 1/n^\lambda$. I only remember this because I had this in my first year analysis exam
 
@porridgemathematics not sure if this is relevant: math.stackexchange.com/a/288796/27978
 
This is a feature of 1D complex analysis; every open set is a domain of holomorphy
The basic idea is clear; take the model $\sum z^{2^n}$ which has the unit disk as a domain of holomorphy. You try the same thing on a general open sets; on the boundary put lots of poles. By lots of, I mean a dense set of
In higher dimensions, it's a very very hard problem. Domains of holomorphies are called Stein domains
(Tangential to copper's MSE post)
 
oh I see, that is a good way of conceptualizing the test
@copper.hat thanks, will take a look
 
5:42 PM
@BalarkaSen are you insinuating it's more complicated than what i suggested?
oh wait, you mean that $S^2$ is the $2$-skeleton of a $K(\mathbb{Z},2)$, whence $[M_g,S^2]=[M_g,K(\mathbb{Z},2)]=H^2(M_g,\mathbb{Z})$?
 
I want to show that if I have a möbius transformation f such that $f(\Bbb{R}\cup \{\infty\})=\Bbb{R}\cup \{\infty\}$ then $f=\frac{az+b}{cz+d}$ where $a,b,c,d\in \Bbb{R}$. In the solutions they considered only the following cases:
1. c=0
2. c,d\neq 0
3. d=0

Why don't we need further cases? So I mean a=0, b=0, a,b\neq 0?
 
Hello please i need help in algebra
 
@Thorgott yeah lol
your hint is the most reasonable
 
@BalarkaSen abstract nonsense gang
 
yo, neda
 
5:55 PM
guess it's not too bad cause i can write down a $K(\mathbb{Z},2)$ at least
but going through $M_g/(M_g)^1$ is cuter
 
wave: they seem to be considered? i.e. whatever a and b are, the map still falls into the various categories 1, 2, 3?
 
@leslietownes sorry I don't see why for example a=0 falls into one of the categories 1,2,3
 
wave: you aren't just randomly being given the number a; you have the whole tuple (a,b,c,d)
it falls into one of the categories 1, 2, 3, depending on what c and d are
maybe chase through the argument to see for example that the map f(z) = z is covered by the case c = 0
 
but does this mean that a FLT only depends on the choices of c,d?
 
6:00 PM
or that f(z) = 1/z is covered by the case d = 0
you're stuck on something basic here. why they split it up into those cases, i don't know. but those cases do cover everything
the way they split up the argument only depends on the values of c and d
you don't need to know what the arguments are to be able to convince yourself of this
 
@leslietownes don't we also need here that b=0
 
imagine you had an argument about ordered pairs of integers. and some argument dealt with it in the cases 1. "okay, what if the first integer is odd" and 2. "okay, and what if the first integer is even"
 
@leslietownes Yes I don't see how to get to those cases.
 
that covers every conceivable case
and then you're basically asking "well, what about the second integer." what about it? the case analysis depends on something other than that
if you want to know what happens to a pair of integers in that argument, look at the first one and run through that case
if you want to know what happens if a or b is whatever it is, look at what c and d are and run through that case
try the examples z and 1/z
for f(z) = z, "b" is 0. but guess what? so is "c". so it's covered by case 1. for f(z) = 1/z, "a" is 0. and "d" is too. so that example is case 3
etc
whatever (a,b,c,d) is, the argument seems to be organized just by what c and d are
 
But I mean for f(z)=z can't we also say i'ts covered by case 3
 
6:08 PM
can you identify a tuple (a,b,c,d) corresponding to f(z)=z
i would doublecheck the denominator of that
 
So only (1,0,0,1)
 
yeah (or any (t,0,0,t) for nonzero t)
 
ah okey here I see why it's not covered by 3
 
6:28 PM
note that the argument says that you can choose real $a,b,c,d$. you cannot conclude that $a,b,c,d$ are real without fixing at least one of the $a,b,c,d$.
 
6:39 PM
right
 
7:08 PM
@BalarkaSen OK so
When I do zero-surgery on a circle, it means a surface in the space can go through that transformation
(the stuff on the left is what's happening inside the torus that's glued inside)
Notice that all three surfaces in that image are genus 1
Now (ignore the surface on the bottom) ^here's the complement of one of the rings in the Borromean rings
 
@BalarkaSen Really? That doesn't sound right. Domain of holomorphy = pseudoconvex, but Stein manifolds are properly embedded.
 
To do 0-surgery on the Borromean rings, do 0-surgery on the other two rings as well as gluing a torus onto exactly the same space occupied by the drawing
Now I'm gonna foliate that space with a circle of (2-)tori
I drew one of them on the bottom, resembling the middle strip of the comic
 
@AkivaWeinberger how do you do your drawings?
 
@copper.hat An iPhone app called Sketch Pad
 
Thanks!
 
7:14 PM
@BalarkaSen Do the transformation from the comic as you push the surface around the space
Because a genus-1 surface is the equidistant surface of a Hopf link, it can kinda just switch which ring it's on
(At this point my drawing skills fail)
Something like that I dunno
I guess I also have to check that this isn't a nontrivial mapping torus of the torus, but I can do that by keeping track of a longitude and meridian of the foliated surface as it goes around
(Actually, if you do that, I think the longitude and meridian swap as the foliating surface travels from the bottom to the top, but they swap a second time as it continues going around)
 
@Balarka: Hmm, apparently I never knew this fact. I'll have to think more.
 
7:37 PM
28 mins ago, by Akiva Weinberger
(the stuff on the left is what's happening inside the torus that's glued inside)
*on the right @BalarkaSen
 
7:51 PM
Please, $gcd(d,n)\times gcd(d,m)= d $?
 
@TedShifrin Stein manifolds are properly embedded where?
You can holomorphically re-embed an open subset in some other Euclidean space so that it is properly embedded.
For example, $\Bbb C \setminus \{0\} \subseteq \Bbb C$ can be re-embedded as $\{zw = 1\}$ in $\Bbb C^2$.
 
Right.
 
If $m$ and $n$ are coprime and $d|mn$ then I think yes? @Vrouvrou
That means $m$ and $n$ are disjoint products of primes, so for any prime $p$, one of them has all the $p$s $d$ needs
 
8:09 PM
@Vrouvrou what would $\gcd(d,d)$ be?
 
d
 
so clearly you need more conditions for the above to be true
 
I have d /nm and gcd(n.m)=1
 
@AkivaWeinberger Finally understood your pictures. That's an interesting way of thinking about it, let me see if it's correct.
Looks good to me!
@AkivaWeinberger This is right. Let $L$ denote the link inside the solid torus you drew. When the surface $S \subset S^1 \times D^2 \setminus L$ crosses one of the "hooks" of the link, it's like turning a punctured torus inside out.
That switches the meridian and the longitude.
This happens twice as the boundary $\partial S$ circle of the surfaces revolves around the solid torus.
 
8:36 PM
@copper.hat please if d/nm how to choose a and b such that a/n and b/m and d=ab?
I choose a=gcd(d,n) but I don't how to choose b?
I need a and b natural
 
8:54 PM
In more detail, just think of the complement of the link $L$ inside $S^1 \times D^2$. What you drew here then is a literal surface $S$ bounding $\{\theta\} \times \partial D^2$ in this link complement. Once you do a $0$-surgery on the two link components, you can "push" $S$ inside one of the added torii post-surgery -- which one will it go inside, as there are two? Well, it's the one which has both the circles $(\{\theta\} \times D^2) \cap \partial\nu(L)$ as a meridian (assuming $\theta$ is one for which $(\{\theta\} \times D^2) \cap L$ is $2$ points), where $\nu(L)$ is a thickening of $L$
This is a very pretty proof, what I had in mind was the following simpleton reasoning: Doing $0$-surgery on an unknot in $S^3 = \partial D^4$ is the same as attaching a $2$-handle (thick disk) along the unknot to $D^4$, or scooping out a thickened $2$-disk bounding the unknot (aka, attaching a $1$-handle, or a thick bridge) in $D^4$, and then taking a boundary. Do two scoops, and one attachment.
The third component in a Borromean link is $aba^{-1}b^{-1}$, where $a, b$ are the meridians of the two others. So at the level of the cores of the handles, what we have done is attached two intervals (core of the $1$-handles) $a, b$ to a point (core of $D^4$), and glued in a disk (core of the $2$-handle) along $aba^{-1}b^{-1}$.
This is exactly a torus $T^2$. Since everything is thickened to be a $4$-manifold, we have $T^2 \times D^2$.
Boundary is $T^2 \times S^1 = T^3$.
I think what you have drawn above, as one of the surfaces of the foliation, is this first factor $T^2$.
 
Someone have an idea?
 
9:44 PM
@Vrouvrou Suppose none of $d,n,m$ are zero. Let $a = \gcd(d,n)$ then $d = a {d \over a} \mid a {n \over a} m$ and so ${d \over a} \mid {n \over a} m$, with $\gcd({d \over a}, \mid {n \over a}) = 1$. Hence ${d \over a} \mid m$.
too late to edit: remove the $\mid$ in the last $gcd$.
 
10:04 PM
So b=d÷a
 
10:15 PM
?
 
10:35 PM
@BalarkaSen I guess I didn't really get that digging out a $2$-handle is the same as attaching a dual $1$-handle
I'm still not 100% convinced tbh
 
10:52 PM
OK I think I get it now
Suppose our circle is the equator of some sphere
If we cut out the "3-disk" bounded by that sphere, that fully cuts $D^4$ into two pieces
Now add back the two hemispheres of that sphere (sans equator). That's like adding in two bumps, as well as half of $D^4$ joining them together
Those bumps plus the half of $D^4$ is homotopy equivalent to a path, so it's a 1-handle
@BalarkaSen If $M^n$ is an $n$-manifold with boundary and it's homeomorphic to $N^d$, $d<n$, a $d$-manifold without boundary, is it necessarily the case that $M^n\simeq N^d\times D^{n-d}$?
I guess not actually (let $M^n$=Möbius band, $N^d$=circle)
So how do we justify that last step, where we see that it's homotopic to a torus, and therefore $T^2\times D^2$?
This may be a nitpick but I wanna tie up loose ends
I do like that the key element was that both a torus and the Borromean rings have to do with the group element $aba^{-1}b^{-1}$
We also see from both proofs that it still works even if we put a bunch of twists in one of the links
 
11:26 PM
@AkivaWeinberger $D^4$ with two disks bounding unlinked circles on $\partial D^4 = S^3$ carved in is basically $S^1 \times D^3 \# S^1 \times D^3$. This is actually homeomorphic to $X \times D^2$ where $X \subset T^2$ is a $\varepsilon$-neighborhood of $S^1 \times \{\theta_0\} \cup \{\theta_0\} \times S^1$.
If the meridians of the unlinked circles on $S^3$ are $a, b$, then it is easily verified that the curve $C \subset X$ where $C$ is the boundary circles of $X$, is isotopic to $aba^{-1}b^{-1}$.
 
@BalarkaSen Meridians of the… circles? Not of tori containing the circles?
I suppose simply generators of $\pi_1$ of the complements
 
Yes, of course. I call meridian of a link component to be the circle which has linking number 1 with that component.
We need to glue in a 2-handle that we're supposed to glue to $D^4$ along the third component $L_3$ of the Borromean link by the identity map $\partial D^2 \times D^2 \to \nu(L_3)$. Under the homeomorphism {0-surgery on $L_1$, $L_2$} -> $S^1 \times D^3 \# S^1 \times D^3$ -> $X \times D^2$, $L_3$ maps to $C$.
This restricts to a homeomorphism $\nu(L_3) \to C \times D^2$. So the final manifold is $D^2 \times D^2$ glued to $X \times D^2$ by identifying $\partial D^2 \times D^2$ with $C \times D^2$.
The last thing to be checked is that the attaching map $\partial D^2 \times D^2 \to C \times D^2$ really is $S^1 \times D^2 \to S^1 \times D^2$, $(z, w) \mapsto (z, w)$, and not $(z, w) \mapsto (z, zw)$, say. If we know this, this manifold is ($D^2$ glued to $X$ by identifying $\partial D^2$ with $C$) $\times D^2$
i.e., $T^2 \times D^2$.
How do we check that the attaching map is really the standard one? It's actually true that any self-homeomorphism of $S^1 \times D^2$ is isotopic to $(z, w) \mapsto (z, z^n w)$, a "$n$-fold Dehn twist". To see if a diffeomorphism indeed is this or not, just compute linking number of the image of $S^1 \times \{0\}$ and $S^1 \times \{x\}$ for some $x \in \partial D^2$.
If it's $n$, that's the $n$ for your map.
Chasing the homeomorphism, it's obvious that image of a nearby pushoff of $L_3$ in $C \times D^2$ is has linking number $0$ with $C \times \{0\}$.
@AkivaWeinberger This is true. There is an algorithmic way to see this, want to hear?
 
11:43 PM
Sure
I will say, it's really hard to see the pictures from the notation
I would appreciate some drawings
 
You're right.
I'm feeling lazy so I will just say a few words and give more details out later. There is a way to "slide" one handle over another handle.
This is only to slide a $2$-handle over another $2$-handle. Usually that's the most straightforward thing.
 
@BalarkaSen glomp
 
@AkivaWeinberger If I attach a 2-handle along all three, then slide one over the other as indicated by the cyan arrow, I untwist the twist in the middle
Slide multiple times to untwist everything
Handle-slides do not change topology. So these 0-surgeries are all the same manifolds, $T^3$.
 
Ah, that makes sense
You could also cut at the disk bounded by the rightmost circle in your drawing and twist the link complement
 
Before I know it, Bskarka and DigAteMy will be drawing Kirby calculus all over, board after board.
 
11:57 PM
This is basically him introducing me to Kirby calculus, I'm pretty sure
I think it's because he's reading a book on it?
 
@AkivaWeinberger Yeah, this is known as a Rolfsen twist. This is also a good move.
Right, this is basically Kirby calculus. Soon Akiva will know everything I know.
 
I saw hundreds of these in my 5 years of grad school.
 
Then it'll be hard to keep up
 
Colored chalk helps :)
 

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