12:08 AM
@TedShifrin Let $R$ be a finite ring with identity. Since $R$ is closed under addition,mult. we have for any $a \in R$, $ab = 1_R$ or $ab = 0_R$ where $b$ is either a zero divisor or unit. If it weren't either, then we'd have no such element $b \in R$ such that $ab = 1$ or $ab = 0$ which implies $R$ isn't closed under multiplication (idk how to show this part)
I know $0_R$ and $1_R$ need to be the products of some other elements multiplication for some reason, like maybe existing in some equivalence relation
like $(\forall a,b\in R), (\exists ab \in R)$ and $[1_R,0_R,a \in R \implies (\exists b\in R, s.t. ab = 1_R \lor ab = 0_R)]$ I gotta prove that part

Sloppy quantifiers for sure. If $ab=0$ implies $b=0$, why must $ac=1$ for some $c$?
Yes, precisely.
Oops.
If $a$ is not a zero divisor, then you want to show …

I know it means it is or isn't cancellable i just forget why
time to look at my notes
I have $a\neq 0$ and $a \in R$ is not a zero divisor $\iff$ a is cancellable, i.e, $\forall x,y \in R: ax = ay \implies x=y$
@TedShifrin that for all $c \in R$, $ac \neq 0$ except for $c = 0$
and.. so..etc.. \qed
the center of 2x2 matrices over R is the identity $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and the zero matrix right
center meaning the elements of $M_2(\mathbb{R})$ that commute multiplication with all other elements in the ring

12:31 AM
the center is among other things an additive subgroup so e.g. I+I will be in there
maybe consider the case R = R (can't resist, meaning real numbers)

center of reals is all the reals tho, wait what do u mean consider R = R

i mean consider the 2x2 real matrices
there are enough edits going on that i can't tell what you originally wrote, but if you were computing the center of M_n(R) for arbitrary n and (commutative?) ring R, i suggest first the case of the real field that we all know and love

thats what I meant by M_2(R) so I guess $\begin{pmatrix}n & 0 \\ 0 & n\end{pmatrix}$ for all $n \in \mathbb{R}$ is in its center

not just "i guess," but yes :)

noice
i cant tell whats harder, math or the new york times wordle

12:37 AM
this is not the entire center, to be clear

what am I missing @Thorgott
what other elements of M_2(R) commute with every other element?

think about what it means to be a linear transformation
commuting with a specific kind of transformation is baked into the definition

Oh so $\begin{pmatrix}n & m \\ m & n\end{pmatrix}$ for all $n,m \in R$ ?

i don't know exactly what thorgott has in mind, but obliv, there are various computations of the center of M_n(F), F a field, on MSE
maybe more generally than F a field

ya no what I just wrote doesn't work
maybe $\begin{pmatrix} 0 & n \\ n & 0\end{pmatrix}$ idk im just guessing

12:43 AM
don't guess, think

von neumann algebras with trivial (in this sense) center are called "factors" and you sometimes see weirdos asking questions about them on main

Yeah I'm at a loss, $\begin{pmatrix}a & b\\c & d\end{pmatrix}\begin{pmatrix}e & f\\g & h\end{pmatrix}$ requires $ae + bg = ea + fc$
which is only possible for $g,f = 0$
since $a,b,c,d$ can be anything
what other elements are there? can I get a hint @Thorgott
am I being trolled

youre not missing any

1:01 AM
obliv, a modified version of thorgotts observation is that all linear transformations commute with all 'scalar matrices' by definition, so the fact that the center contains all 'scalar matrices' (and not e.g. just 0 and I) is more or less immediate
there's a tiny amount of effort in proving it doesn't contain anything else, the MSE question up above collects some fairly good arguments
if you study less than full subsets of M_n(F), including e.g. subalgebras of M_n(F), that other inclusion [center subset {scalars}] may not hold at all and you can find other stuff in the center

noted, I gotta show that $U(R_1\times R_2) = U(R_1)\times U(R_2)$ for any two rings with identity $R_1,R_2$ and $U$ just means units of
I gotta show both ways $\subseteq$ i guess
nvm this is trivial

1:17 AM
@leslietownes yeah, this is what i had in mind
i just misread $n$ to be an integer

@XanderHenderson I love veggie soup
I don't even care about meat in my soup
I could survive on veggie soup alone
honestly meat is way overrated
it doesn't even taste that good

@Jakobian I wanna eventually transition to being vegan but Idk how people can say this if they've had a really good steak/burger
I guess to each their own

I don't want to be vegan at all. Vegetarian at best

I suppose by itself no, not really but cooked properly meat can be glorious
does that include fish and eggs?

Yes it does

1:25 AM
sushi would be hard to give up :[
@Jakobian I wouldn't mind eating meat if the cattle wasn't put into horrendous conditions. Like, if it were raised by someone I knew treated them well etc

@Obliv I'll never actually stop eating meat. Maybe eat less of it, but not stop eating it

but also veganism can just be very healthy
what if you had unlimited good veggie soup tho

@Obliv that doesn't matter to me, you're still killing the cow/chicken/pig

yeah true, but it'd be less terrible than their current conditions maybe.. like partially ordered morality

conditions are whatever at the end of the day you're still killing the animal so who cares really

1:30 AM
I mean on the one hand if you stop eating them, then they'd just go extinct or live in the wild where they're no longer safe from predators
if you don't kill them, they die anyway.

humans will never stop eating animals

Let $R$ be a commutative ring with identity and $r \in R$. Define $L_r:R\to R$ by $L_r(x) = rx$ for all $x \in R$. Prove that $L_r$ is injective iff $r$ is not a zero divisor. Prove that $L_r$ is surjective iff $r$ is a unit.
I think I have to use $r$ is cancellable

I don't know, maybe its shocking to hear this for someone that doesn't view death like I do
I see death of an organism as way worse than anything that could happen to it while it was still alive

I was gonna say it's not shocking but then you followed up your somewhat normal statement with something absurd
@Jakobian so you think eternal suffering > death?

I don't think its absurd at all
@Obliv there's no such thing as eternal suffering, because eternal things don't exist

1:36 AM
Such confidence :P
so then what is death?

stop of any function

stop w.r.t what?
if you claim death is worse than anything an animal can experience while alive but you don't believe in eternity then what even is death

@Obliv what are you trying to ask me here exactly

if something were to "stop", but not forever, then it has to resume at some point?
so w.r.t. what is it stopping/starting
some parameter $t$? lol

it won't resume, it simply won't be

1:42 AM
so you believe in a finite universe? then what was before the big bang @Jakobian
I don't know enough physics but basically what is before nothing, if it's just nothing then you believe in infinity

why do you presume big bang ever happened

unless we don't agree on time being an infinite parameter

its just a popularized theory with no real evidence

I mean there is evidence but you can choose to believe what you want. I don't think you should take anything for granted either but I find it strange that you don't believe in eternity IRL but use $\infty$ in your mathematics

also you're mentioning me like I'm not here
@Obliv no there isn't

1:47 AM
@Jakobian et tu

untrue that there is no real evidence for the big bang

there is merely a suggestion based on an object that no one even knows if it exists

Maybe his standards for evidence are different/his philosophy presupposes something that isn't met

called dark matter

afaik big bang has nothing to do with dark matter
it's just extrapolating the expansion of space backwards

1:50 AM
@Obliv you are wrong then
astronomers justify their errors in calculations by existence of dark matter
that never was proven to exist
and everyone seems to know that much but suddenly when it comes to big bang, everyone seems to think it must be true

if anything, it'd support your finite universe ideas more since there'd be a start and then you can imagine an end

@Obliv straw man

Then explain how you can have a finite universe? by definition it has to have a beginning & end to be finite no?

straw man

you cant prove that things exist

1:54 AM
@Jakobian BRO
@Thorgott I'm sidestepping the philosophy for now just to understand where he's coming from

1. We weren't talking about universe but about human beings (so I'm allowed some slack about what I'm saying here)
2. You are equating infinite to eternal for some reason

so eternal is $(-\infty,\infty)$ but infinite can be $(0,\infty)$?
can have an end or beginning but eternal requires having neither
also I didn't know you meant human beings, you said "eternal things don't exist"
so I'm not giving you slack bud

we're not buds
@Thorgott irrelevant point. Somehow we proved that Higgs boson exists.

not in a truly objective sense, but i dont care to discuss epistemology

you're just not good willed enough to interpret it appropriately what it means to prove something, maybe

2:05 AM
my point is simply that dismissing observational evidence altogether is not accurate as to how physics works

to this context
@Thorgott astronomy deals with massive errors so observational evidence when it comes to it doesn't matter that much. The errors are massive so the explanation "dark matter exists" may just as well be false
it wasn't proved. We need more concrete evidence

dangit wish there was an undelete. what I wrote was right I had to show that $L(a) = L(r)$ for $x = r,a$ respectively so yeah not injective.

it's a good thing that they created such theory and that it exists, based on some unproven things. But people assume it must be true

I only showed contrapositive in one direction so I need the other as well i think

I'm not really dismissing observational evidence. Conclusion that big bang is true is like few steps ahead of the evidence

2:16 AM
Can a ring consist of only zero divisors (not including $0_R$) seems obviously false

I realize evidence is subjective. Like existence of horses implying existence of unicorns
sure there's "evidence"
probably no one would call it evidence. But some people might
conspiracy theorists come up with evidence all the time. Are they right?

19

I believe there is a straightforward formula for the abelianization of a semi-direct product: if $G$ acts on $H$, and we form the semi-direct product of $G$ and $H$ in the usual way, and the abelianization of this semi-direct product is the product $G^{ab}\times (H^{ab})_{G}$. (Here the subscrip...

iiinteresting

you're shifting a lot of goalposts. the big bang theory and dark matter theory are not logically equivalent and nobody said anything like that either "must be true". physics is not that definite and variants of the big bang are still in contention, but in its general features, the big bang is accepted by physicists on the basis of multiple layers of evidence. no amount of your mental gymnastics can change that.

I don't think so. If such evidence existed you'd provide me with it, surely
I never said they are equivalent in any sense. But the conclusion of big bang theory is based on dark matter, as I know
that's what they justify their gap in reasoning as far as I know
well maybe there really is some evidence that I'm not aware of. No one even hinted at what it could be, though

2:33 AM
let $R$ and $S$ be rings. showing $R\times S$ is isomorphic to $S\times R$ does that mean I have to show $(a,b)+(c,d)=f(a,b)+f(c,d) = f(b,a)+f(d,c)=(b,a)+(d,c)$ from ring axioms
and it's injective,surjective
also the multiplication property

@Obliv $1$ is not a zero divisor
but it can happen if you don't assume $1$ to exist
Take any commutative group and set $xy = 0$ for all $x, y$
@Obliv that doesn't look right
you are supposed to construct an isomorphism $f:R\times S\to S\times R$

Thanks, but I already figured out the proof I needed. Right now trying to construct an isomorphism from RxS to SxR. I'm thinking if I define $f:(R\times S) \to (S\times R)$ by $f(a,b) = (c,d)$ where $a \mapsto d$ and $b \mapsto c$ that's good?
yeah what I wrote before wasn't right

@Obliv and what are $c$ and $d$?

$a,d \in R$ and $b,c \in S$?
identity map?

Oh okay so you're thinking of general maps $a\mapsto d$ and $b\mapsto c$

2:41 AM
oh right then $f(a,b) = (b,a)$
then since $a \mapsto a$ and $b \mapsto b$ we have bijection and I can do all the homomorphism properties etc

Yes. But your approach is also good. If you had $f(a, b) = (k(b), j(a))$ where $j:R\to R$ and $k:S\to S$ are isomorphism then $f$ is also an isomorphism

ah okay
let $R$ be an integral domain, then $R\times R$ is not isomorphic to $R$ for some reason

not that one what I wrote wasn't one (in removed)
@Obliv $R\times R$ has zero divisors

oh okay so for some nonzero $(a,b),(c,d) \in (R\times R)$ we have $(ac,bd)=(0,0)$

A better observation is that $R$ is not isomorphic to $R_1\times R_2$ for non-trivial $R_1, R_2$.
Pete Clarks calls such $R$ "connected"
(not sure if standard)
This should be equivalent to induced topological space $\text{Spec}(R)$ being connected, hence name

2:48 AM
wait idk why I wrote that, if it's an integral domain then there is no zero divisors..
@Jakobian wait how come $R\times R$ has zero divisors but $R$ doesnt

@Obliv $(1, 0)(0, 1) = (0, 0)$

oh

$R$ doesn't have them from definition

so how come there are no isomorphisms connecting them

Huh?

2:50 AM
@TedShifrin Hi

nods

If $R_1$ is an integral domain and $f:R_1\to R_2$ is an isomorphism, then so is $R_2$
its a property preserved by isomorphisms

I think it wants me to prove that

left to the Obliv

I'm guessing it has something to do with injectivity

2:52 AM
well, do so. No, you need homomorphism properties.

for me, and probably everyone here, that this property is preserved by isomorphisms is obvious
and hopefully you will see it once you write it down on paper
Note that an isomorphism $f:R_1\to R_2$ is a homomorphism such that $f^{-1}:R_2\to R_1$ exists and also is a homomorphism

That’s the point of isomorphisms. They preserve all properties in the category.

(this should be introduced as definition, but you might have heard that isomorphism is a bijective homomorphism instead)
(that also true but maybe less appropriate)

I think it's because (a,b)(c,d) mapping to 0 inevitably contradicts R being an integral domain

focus on arbitrary $R_1, R_2$
not on $R$ and $R\times R$
there is no reason for pairs $(a, b)$ and so on
0

What's a good reference for the classification of finite-dimensional algebraic representations of $\mathrm{GL}(n,k)$ when $k$ is an arbitrary field of characteristic zero? For $k = \mathbb{C}$, Fulton does it using Young diagrams in Chapter 8 of his book Young Tableaux. Other treatments for $... John Baez 2:59 AM I'm gonna stick to the direct product because I'm dumb and can't think about anything in generality. for$f:(R\times R)\to R$to define an isomorphism we need$(a,b)\cdot(c,d)=f(a,b)\cdot f(c,d)f(a,b) = g$and$f(c,d) = h$for some$g,h \in R$then$(ac,bd) = gh$? But$R$is arbitrary so you clearly can since$R$is an integral domain we have$gh \neq 0$for nonzero$g,h$and similarly$ac,bd$but suppose for$c=0$we have$(0_R,bd) = gh$for any$a \in R$which means$f$isn't injective? @Jakobian sorry, i need to work on my self deprecation you're only obfuscating notation right now probably confusing yourself in the process$(a,b),(c,d) \in (R\times R)$and$g,h \in R$@Jakobian Cool dude. I wonder what he is working on. The reference he is looking for seems somewhat far from what he's been doing for the last decade. 3:14 AM @AlessandroCodenotti Mycielski's theorem from graph theory? I searched on google and it's about graph theory Weird that a theorem from graph theory is about graph theory. Balarka's approach is interesting though I think$\Lambda$should be$S^1$. I don't have a deep understanding about modular group and I think number theoriest (who uses hyperbolic geometry) knows much better I'd never heard of that theorem and I thought it's a theorem in analysis or number theory but weirdly graph theory came out and I asked if I searched correctly well it says algebraic independence and cantor set @XanderHenderson bad Xander 5 hours later… 7:52 AM @Shaun 8:03 AM @onepotatotwopotato No, Mycielski's theorem about equivalence relations on Polish spaces (even though you need a slightly more general version here, the one presented in Kechris's book for example) 8:33 AM I don't even know what Polish space is about. So yeah... that's an interesting result! @Jakobian I wish she does something interesting in the 2nd half of the semester. Would be nice if we studied actual probability theory 9:36 AM @onepotatotwopotato Thanks, I shall take a look @onepotatotwopotato$\Lambda$above is the orbit of$0$under$\mathrm{PSL}_2(\Bbb Z)$; I agree the limit set of$\mathrm{PSL}_2(\Bbb Z)$is the entire circle. 9:57 AM @DanielDonnelly Hi :) 10:22 AM @onepotatotwopotato a Polish space is a completely metrizable and separable space. Basically a very good space 11:14 AM @AlessandroCodenotti This one? @SoumikMukherjee yes 12:22 PM In any semigroup, D-class is defined to be there is c such that a L c and c L b If we permutate all the possible a,b in S, we get abstractly, a structure consists of all possible pairs of (aLc, cLb) which is almost like a automorphism between pairs of elements. Question ia, is this always have the be the same as the automorphism semigroup of a given semigroup? 1:05 PM @Secret a L c and c R b @Jakobian typo yeah @Secret for$a, b\in S$with$a D b$how do you choose$c$? Do you just choose random one$c_{a, b}$for each$a, b$with$a D b$? yeah,$c_{a,b}$for each$a,b$with$aDb$, so one c each time Now what do you mean by taking pair$(a L c, c R b)$, when we write$a L c$it means$(a, c)\in L$, which can be true or false I needed some help with 3 basic questions from several variable calculus, is this the right place to ask? 1:18 PM @Jakobian$(a,c) \in L$and$(c,b) \in R$, so we basically have pairs$((a,c),(c,b))$defining a map from the left argument to the right one @Tapi Sure @Secret is that a map sending$(a, c)$to$(c, b)$? @Jakobian yep 1:43 PM @Secret if$S$is the zero semigroup i.e.$xy = 0$for all$x, y$then$x D y$iff$x = y$. Then your map sends pairs$(a, a)$to themselves. But if$f$is any bijection from$S$to itself with$f(0) = 0$then$f$is an automorphism of$S$1:57 PM @Jakobian Got it, thanks 2:17 PM @Balarka sanity check: does every pair$(X,A)$have the weak homotopy type of a CW-pair? its standard to build a weak homotopy equivalence$A'\rightarrow A$with$A'$a CW-complex cell-by-cell and then extend the composite$A'\rightarrow A\subseteq X$to a weak homotopy equivalence$X'\rightarrow X$cell-by-cell s.t.$(X',A')$is a relative CW-complex. now, the cell attachments are out-of-order, but we can use cellular approximation to homotope the attaching map of, say, a$(k+1)$-cell of$X$into the$k$-skeleton of$A$and doing this inductively over the skeleta of the relative CW-complex @Jakobian hi 2:30 PM @Sahaj hi you asked a riddle the other day i can't find its answer can you link it please 2 days ago, by Jakobian When I speak, most don't listen. When they listen, they don't understand. Who am I? A mathematician. :P oh that can really be anyone who deals with very technical subjects it's like cause and effect 2:45 PM @Sahaj not really. Any science subject you can explain to someone simply. But not with math I dont really find science that easy to understand At least you can give a rough idea in science subjects That's true, but for advanced topics like quantum mechanics and whatnot I don't think it's possible to grasp it as easily as elementary scientific ideas There's plenty of exposition for quantum mechanics phenomena online Sometime particle sometime wave, mood swings at the atomic level. 2:52 PM @SoumikMukherjee As a layman thats really all I know about the subject @Sahaj What do you know about advanced math as a layman? nothing I think that's case in point Science topics are mostly related to the physical universe, so people(may) have a vague understanding about them. 3:08 PM agreed @Thorgott works yeah thanks 3:24 PM @Thorgott What's the easiest proof that an infinite cyclic cover of any compact connected oriented$3$-manifold$M$is obtained from cutting$M$along a compact surface$S \subset M$, and pasting infinitely many copies of these end-to-end?$3$is not essential above, you can replace surface by a codimension$1$submanifold. The proof I have in mind is that such covers are classified by$\mathrm{Hom}(\pi_1(M), \Bbb Z) = [M, S^1]$, one can take a regular value of$M\to S^1$and its preimage is$S$. Surely there's a by-hand proof where you look at the infinite cyclic cover$\widetilde{M}$, consider a fundamental domain, and modify the action slightly so that the fundamental domain is a nice manifold with corners? 3:38 PM @Jakobian A crackpot. 4:31 PM @XanderHenderson he's what I want to be roughly. A mathematician and a physicist :) His blogs pretty cool as well @nickbros123 He was on my committee. @XanderHenderson Your PhD defense? That must've been cool He was on my PhD committee. So, yes at the defense. can someone give me a hint for this? @Tapi maybe they want u to use the directional derivative and arguments from analysis on$R^1$4:39 PM yeah I know a similar result for R^1 can be proved using Mean Value Theorem,.. I'm just unable to use those ideas here Halmos on the standard convention regarding "sets with structure" My question is a bit irrelevant, but when he says that the second coordinate (i.e. just the order relation) encodes all of the information in the ordered pair I am not sure I agree Surely$dom \, \leq \subset X$and it can be a proper subset in general (the order is partial) 5:01 PM Did someone hear a mention of a classification of subgroups of Q² ? I can't seem to figure out whether it has a neat classification like subgroups of Q do, or if the set of subgroups has a much bigger cardinality than the dim 1 case. @AdrienZabat There is continuum many subgroups of Q so it can't have larger cardindality For LaTeX in chat see channel description @Tapi For starters,$f$is differentiable everywhere in$S$and$\nabla f(a,b)\cdot (h,k)>0$for all$(h,k)\in S$. But I don't think the question quite makes sense. Why is$(a+th,b+tk)\in S$for all$(h,k)\in S$and all$-\delta<t<0$? fair, there's still the question of whether a classification is tractable whats the classification for$\mathbb{Q}$? The classification is that such subgroups have the form$\{r, v_p(r) \geq n_p\}$, where$n_p$is a sequence in$\mathbb{Z} \cup \{-\infty\}$indexed by primes. 5:07 PM @EE18$\text{dom}(\leq) = X$because$x\leq x$for all$x\in X$oy i forgot reflexivity thank you I was picturing arbitrary relations but I guess for this particular structure it's OK @EE18 I think reflexivity forgot you. Happens too often it seems But yeah we usually think of set$X$itself. Still, the important part is$(X, \leq)$. No one think in terms of$\leq$alone unless you're an alien that you can recover$X$from$\leq$is not a useful observation very fair and agreed it is only useful to someone like me who is stil lgrappling with the relevant definitions :) 5:15 PM No its not really useful to anybody I've been looking at triplets$(a,b,c) \in \mathbb N^3$such that$\gcd(a,b,c)=1$and no two elements of the triplet are pairwise co-prime. For all the triplets I've noticed I can always find a common prime divisor$p$of any two elements (WLOG say$a$and$b$so that$p | a,b$) such that we have$\pmod p: \gcd(c,a) \neq \gcd(c,b)$. I'm not sure if this always holds though... can someone help me find counterexamples or hint for a proof?? @Sahaj doesn't this just follow by assumption$(a,b)\neq 1, (b,c)\neq 1, (a,c) \neq 1$so there exists$p_1 \mid a,b$and$p_2\mid b,c$and$p_3 \mid a,c$for the "no two elements of the triplet are pairwise co-prime" part$p_1,p_2,p_3$are gcds greater than 1 idk why I labeled as if they were prime and it follows if$p_1 \mid a,b$then$p_1 \nmid c$and$p_2 \mid b,c \implies p_2 \nmid a$etc so why isn't$R\times R$isomorphic to$R$if$R$is an integral domain? last night I tried to argue injectivity fails but @TedShifrin said it's the homomorphism properties let$f:(R\times R)\to R$and$(x,y) \in (R\times R)$then for$f$to define a hom. we need$(x_1,y_1)+(x_2,y_2) = f(x_1,y_1)+f(x_2,y_2)$and$(x_1,y_1)\cdot (x_2,y_2)=f(x_1,y_1)\cdot f(x_2,y_2)$@Obliv so with that reasoning we have that$p \nmid \gcd(c,b)$and$p\nmid \gcd(c,a)$. But its not clear to me how that means$p \nmid \gcd(c,b) - \gcd(c,a)$.. For example$3\nmid 5$and$3 \nmid 2$but$3\mid (5-2)$... since$R$is an ID we have nonzero$a,b \in R$s.t.$ab = 0$I totally misread your problem, disregard @sahaj 5:32 PM @BalarkaSen i dont think this is clear at all modulating a map to$S^1$seems like a reasonable approach yeah i couldnt make it work at least the geometry here is too subtle for me to understand take a knot complement for example any Seifert surface gives rise to the same cover not clear to me at all, geometrically That's because any two Seifert surfaces are isotopic by handle moves inside the complement though *and attaching handles @Sahaj if$(c,a) \equiv (c,b) \pmod p$then does that mean their gcds share a common factor (I'm writing (a,b) to mean gcd(a,b)) Attaching handles ambiently don't affect the exterior, because it's like a cancelling pair 5:36 PM If they share a common factor, then$(a,b,c) \neq 1$you should try to prove this part i said the geometry is hard for me, not impossible :P the more general question i should raise is the following given only a two-sided codimension$1$surface$S$in a manifold$M$and letting$\tilde{M}$denote the cover obtained by cutting$M$along$S$and gluing infinitely many copies, how do we determine the subgroup of$\pi_1(M)$corresponding to$\tilde{M}$? i dont have a good answer in particular, the corollary question is how to tell if two surfaces determine the same cover it's only easy for knot complements cause they only have one subgroup that is co-$\mathbb{Z}$I assume$S$is not closed, otherwise$M \setminus S$is disconnected. Let$\mu$be a circle transverse to$S$, cutting it at exactly one point. Then$\pi_1(\tilde{M})$is amalgamated product of infinitely many copies of$\pi_1(M \setminus S)$where you identify a loop$\lambda_{+}$given by a positive pushoff of a loop$\lambda$in$S$with$\mu \lambda_{-} \mu^{-1}$where$\lambda_{-}$is the negative pushoff @Thorgott This is not difficult, the surface is related to the cover by the map$M \to S^1$. Any two regular values determine the same cover, so the answer is that the surfaces must be related by an ambient (cyclic) cobordism.$S$should definitely be closed @Obliv Huh? @Thorgott That's not the case for Seifert surfaces... 5:49 PM also, your description sounds reasonable (though idk why exactly$\mu$exists), but the issue is that$\pi_1(M\setminus S)$is not at all easy to understand. it's not straightforwardly clear how to realize any co-\mathbb{Z}$ subgroup of $\pi_1(M)$ by this process.
@BalarkaSen oh i meant closed as a subset

Oh thats fine

whether $S$ has boundary or not depends on whether $M$ does

@Obliv it isn’t necessarily the case. For example take $(6,10,15)$. Note that $\bmod 5: gcd(6,10)\neq \gcd(6,15)$. Whereas $\gcd(6,10)=2$ and $\gcd(6,15)=3$

@Thorgott $S^3$ has no boundary, Seifert surfaces do...

the knot complement has a boundary

5:50 PM
Ah, OK, sure
Works

you don't create infinite cyclic covers of $S^3$

Yep, fair, fair.
@Thorgott $\mu$ is the Poincare dual of $S$

Actually. I found a counterexample. I think this conjecture is resolved mpw

$[(S, \partial S)]$ gives a second homology class in $H_2(M, \partial M; \Bbb Z) = H^1(M; \Bbb Z) = H_1(M; \Bbb Z)$, pick $\mu$ to be a loop representing the element in the last group

where does that last equality come from
they should be dual

5:53 PM
universal coefficients theorem?

what if $H_1$ has torsion

@TedShifrin Huh?

Map it to the non-torsion part. The PD cannot be a torsion element anyway

still non-canonical, but i guess it works anyway?

5:54 PM
Oh, wait, this works only if $S$ is incompressible in $M$
If it's homologically trivial, then this doesn't work

@Sahaj what's the counterexample?

also, given $S$, can you explicitly construct a corresponding $M\rightarrow S^1$ s.t. $S$ is a regular fiber?

@TedShifrin DOH

ig we can take the map on $M\setminus S$ by algebraic considerations and modify it in a collar around $S$ somehow?

5:57 PM
The point is that $R$ has no zero divisors and $R\times R$ has many. An isomorphism must preserve products (and sends $0$ to $0$), and so an isomorphism — or its inverse, depending on the direction you use — would produce zero divisors in $R$.

Does anyone know where I can find a proof of the above theorem? The forward direction is pretty trivial, however, the backward not so.
AC stands for absolute continuous.

@Thorgott Seems reasonable

still, what if two regular values of $M\rightarrow S^1$ don't lie in the same component of the regular values in $S^1$? how do you get a cobordism then?

@psie $L_{\mathrm{loc}}^1$ are the locally integrable functions.

@psie That's basically the generalization of the fundamental theorem of calculus you find in most graduate real analysis books, I think. I no longer have any of the books, but check Folland or papa Rudin.