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3:00 AM
Well it all boils down to my quest to try to understand everything, which caused me to reach out to areas outside of chemistry. In order:
2013: 4D visualisation, 2014: Time travel, 2016: Fine arts, 2016: Philosophy and Magic, 2017: Politics, 2018: Law, Culture, Language etc.
 
Sanity check: is $e^{s/\ln(x)}=e^{s/\ln(1-x)}$ equivalent to $x^s=(1-x)^s$.
 
each of these give me new ways of thinking, it is these new ways to comprehend things interested and drive me
 
(In that case, I must simply be performing some poor arithmetic)
 
A big change, from 4D to politics
 
yeah, it just explodes and explodes
 
3:03 AM
Your interests have been becoming progressively non-geeky/nerdy. You must be doing something wrong.
 
@Ultradark Should be $\frac{1}{x^8}=\frac{1}{(1-x)^8}$, no?
 
@Secret you will never understand everything in a finite amount of time. But if there is life after death then you can continue your pursuit and learn everything in a countably infinite amount of time maybe
@Secret you should try to solve the problem of consciousness
 
My interest didn't change as much; computer science->physics->computer science->analysis->topology->set theory. Math was always in the back of my mind but school never taught how beautiful math really is so I start getting into only later than I would expect
 
@Symposium actually everyone lives in their own world viewed through their eyes. but i don't know what i wanted to say, but great i typed it :d
 
Well there is a possible explanation to that: I am trying to understand nonanalytic thinking as I complained myself being too reductive and thus unable to comprehend something like emotions and gut feelings
 
3:05 AM
@Rithaniel why should it be that?
 
sinxe 2013
 
@secret have you ever heard of relational biology?
 
(Wait, yeah, not that.)
 
probably not the term itself, but it sounds familiar, let me check
 
I tried to explore more things like literature, but although I love to read, I hate analysing it
 
3:07 AM
I'm only interested in politics for comedic values. I had a great time for the past 2 years or so.
 
politics was never my thing either..
too complex for basically "living on the same planet under some rules"
 
Yeah, I don't think $e^\frac{8}{lnx}$ can be simplified, actually.
 
Basically the only things I do outside of studying is for comedy sake, like someone somewhen said "tragedy+time=comedy"
 
Has anyone seen Lee-Yang theorem in a textbook?
Google books only returns books on physics (statistical mechanics).
 
Ok this is very weird. Never expect seeing category theory here
 
3:13 AM
@Rithaniel no it cannot be simplified
@Secret lol
 
but then it's bioinformatics, so it is not surprising that level of abstraction will be involved
 
I should start studying category theory...
 
does anyone know if there's a database of known solvable transcendental equations
 
@Secret What result in set theory surprised you the most?
 
But the moment I think to study something that is not set theory my motivation goes to 0...
 
3:16 AM
Same question goes for @Holo as well.
 
@Rithaniel do you know what kind of object $e^\frac{1}{lnx}$ is
sorry for the multiple pings :)
 
You're okay, but I might not respond quickly.
What do you mean "object?" Are you wanting to know the general shape of the graph?
 
I'm not sure if it's an algebraic function
or a transcendental function
 
Ah, hmmm
That's something I've let to learn about.
 
oh okay
it has to be transcendental
 
3:21 AM
Hmm @Symposium , it is hard question, maybe the consistency of the existence of amorphous sets?
 
guys I'm going crzy because nobody checked my sanity check
 
@Symposium That $\omega_1$ is profoundly nonconstructive, to the point we can say almost nothing about its internal structure
 
@Ultradark what? Exp(1/ln(x)) is the best you can get I think
 
@Holo the equation $e^{s/\ln(x)}=e^{s/\ln(1-x)}$ and $x^s=(1-x)^s$
By taking the natural log of both sides and doing some algebra one can get to the equation on the right
I'm just confused
 
also this:
Feb 3 '17 at 13:03, by Secret
Actually, I do find it surprising that the box counting method that we learn in high school when we were taught permutations and combinations work quite well up into the infinite cardinals
 
3:28 AM
Interesting @Secret and @Holo. These answers sent me through a rabbit-hole.
 
I'm confused because the first equation is comprised of setting two transcendental functions equal to each other, and after simplifying one gets an algebraic object in the form of a polynomial
 
Taking ln is a transcendental operation, and composition of transcendental functions need not be transcendental
 
@Ultradark well, this is algebra, the equations have the same solutions. What it means is that f(x)=0 implies g(x)=0, but f(x) not necessarily equal g(x)
 
it is the same as how binary operations between transcendental numbers can be algebraic
 
oh okay
 
3:31 AM
@Secret box counting method?
 
recall how counting combinations, you have n boxes and m objects. The first box has m choices, the second has m-1 choices and so on. Thus there are a total of $m!$ permutations
The same principle can be used to determine the largest cardinality of the image of some function given the cardinality of the domain
 
Never heard of this method
Actually I did in programming
But not at school
(btw, the school I went to was bad)
 
@Secret that relates to $n!$ is the number of ways one can order a set of $n$ objects
?
 
@Ultradark yeah, except you can have less boxes than objects, thus you need to divide by some factor
 
3:36 AM
Finite sets are weird...
Let's stick with infinite ones
 
weirdness does not confine to infinite sets
large finite sets can get pretty weird as googology and computational science showed, but they are what makes the world go round
 
People always takes finite sets as matter of course, but hell they are more confusing than d-infinite
@Secret do you know about large countable ordinals?
 
@Holo yup and simpleart taught me all about ordinal notations 2 years ago
 
nice, user21820 shows me how to get to Γ0 few weeks ago
 
simpleart show me how to get to bachmann
and beyond
 
3:43 AM
I can solve an infinite number of transcendental equations muhaha @Secret
 
So I guess you have a really good idea on how to prove trascendence of $e^e$?
(The fact this question is still open bothers me like there is no tomorrow for some reason, to the point every time I saw exp^{complicated} popping up I always have the instinct to plug e^e into it)
 
hmm
that is very interesting!
 
Hmm, @Secret I don't know this ordinal, I am guessing bachmann > Feferman–Schütte
 
In mathematics, the Bachmann–Howard ordinal (or Howard ordinal) is a large countable ordinal. It is the proof-theoretic ordinal of several mathematical theories, such as Kripke–Platek set theory (with the axiom of infinity) and the system CZF of constructive set theory. It was introduced by Heinz Bachmann (1950) and William Alvin Howard (1972). == Definition == The Bachmann–Howard ordinal is defined using an ordinal collapsing function: εα enumerates the epsilon numbers, the ordinals ε such that ωε = ε. Ω = ω1 is the first uncountable ordinal. εΩ+1 is the first epsilon number after Ω = εΩ. ψ...
 
OwO
 
3:49 AM
It's impredicative this 21820 does not talk much about it except for one conversation with simpleart as they try to figure out how to make ordinal collapsing function predicative
 
I see
 
vzn
in MathOverflow, 4 mins ago, by vzn
hi all any experts? an apparently very sharp anonymous commenter has posted what looks like some very substantial code revealing a remarkable property... am wondering if it mentioned/ known in the (rather vast) literature somewhere... https://vzn1.wordpress.com/2018/10/03/collatz-fusion/#comment-8023
 
Ordinal collapsing function is a way to write large ordinals by projecting down from some inaccessible ordinal
 
But doesn't it requir the existence of the inaccessible ordinal?
Wait ordinal, not cardinal
No, sorry I'm stupid, continue
 
Usually that ordinal is called $\Omega$ and it is assumed to exist (this is why it is impredicative, you cannot get that from the natural numbers)
If we only care about countable ordinals, then $\Omega$ can be $\omega_1$ or $\omega_1^{CK}$
 
3:58 AM
I see
 
@Secret $e^k$ for $k=1,2,3,...$ is transcendental right?
 
Lindemann weistrass said if the exponent x is algebraic, then e^x is transcendental
 
cool that's the lemma I need
 
@Ultradark Yes, suppose exp(k) is algebraic then exists P such that P(exp(k))=0, then define Q(x)=P(x^k) and you'll get Q(e)=0 which means e is algebraic
I'm too slow
 
hmm...
exp(e), P(exp(e))=0, Q(x)=P(x^e), Q(e)=P(e^e)=0 ??????????
 
4:07 AM
Your Q is not polynomial...
 
ooops
 
Lol
 
4:24 AM
Hi chat.
 
4:34 AM
yo
 
4:53 AM
howdy
 
5:30 AM
The empty set a subspace topology of every possible topological space, correct? Or is there some distinction to keep in mind which I have missed?
 
5:47 AM
Hi
 
Did you make progress @Ultradark?
 
6:14 AM
@Rithaniel it is. It is even an open subspace
 
Excellent, then my proof is valid.
 
@user602338 hi
 
6:50 AM
Is any finite $T_3$ topological space automatically also $T_0$?
 
7:10 AM
Not just finite. Any T3 space is a subclass of T0
since any T3 must be T0 and regular
 
Yeah, I was reading up on it, and apparently there's some mix up on the definitions we've been given. Apparently the common consensus of regularity is what I was taught $T_3$ was, and that regularity is $T_0$ and $T_3$.
 
7:36 AM
Checking: $X/F$ is the quotient topology and $X \setminus F$ is "$X$ but not $F$," correct?
 
Yes
 
(Had to struggle with that $\setminus$)
And so, if $F$ is a closed subset of $X$, then $X/F$ is $X$, but with all points in $F$ collapsed to a single point, correct?
 
Also if $F$ isn't closed
 
Mornin' all
 
Hi @ÍgjøgnumMeg
 
7:45 AM
Yoooo
How's it going?
 
I survived the first week of lectures so pretty well :P
A couple of the courses are a bit over my level so I have some catching up to do, but that's what the weekend is for...
What about you?
 
That's fair, i'd assume that's to be expected
I'm just at work, reading Fesenko's notes, complaining about the rain
It seems like now term has started some of the conversation in here has plummeted
lol
 
I don't have internet in the dorm at the moment so I'm surely less active here now
 
That sucks :(
How are the dorms?
 
The one I'm in is pretty bad, it's a private one, not a uni one
But I'll find something else for the next semester
 
7:58 AM
I can't figure out how this proof of $\limsup_{n\to\infty} \frac1{x_n} = \frac1{\liminf_{n\to\infty} x_n}$ works. by pluffing $y_n=1/x_n$, I get $\frac{1}{y+\varepsilon}<x_n$ for all sufficiently large $n$ and $x_n<\frac{1}{y-\varepsilon}$ for infinitely many $n$. What should i do after that?
 
@Alessandro fair enough, at least you have a roof over your head!
 
Another question about RSA algorithm. A computer program generates two big primes P and Q, and defines two variables such that A = (p*q)xor(p+q) and B = (p*q)xor(p-q). Now I have three numbers, A, B and E (where E=pow(M, 65537)mod(p*q), and xor and pow are bitwise exclusive OR and power). Is it possible to revert the unencrypted message M?
 
@LeakyNun will you please look at my question above?
 
If anyone can help, the numbers A, B and E are here in the HTML comment section
 
Zee
8:15 AM
My god
Just took the meanest one , felt so good
 
8:47 AM
Why does wat get 5 stars?
 
@WillHunting because it described the feeling in the chatroom at the time
lol
4
my name jeff
4
 
Not starring that one
 
sad
was testing to see how far I could go
 
9:17 AM
Could anyone tell me what I'm trying to prove in the context of this question for one-to-one?
 
@PrashinJeevaganth as=at implies s=t
 
This is the suggested proof for this question but I dont understand the solution
@LeakyNun how did u come to that conclusion?
 
by thinking about it a bit
 
Does equating both of the strings have something to do with proving one-to-one?
@LeakyNun
 
9:32 AM
If I'm trying to show that on the rational numbers, trying to get a measure based on µ([a, b)) = b - a will not be countably additive, is it sufficient to say that a sequence of disjoint subsets can have their measure sum to an irrational, while their union could have rational measure, which is a contradiction?
 
@bjb568 why must their union have a rational measure?
 
well, if the union is also like [a, b), b - a is rational.
I guess I should just write out an example where this is the case then, thanks.
 
@bjb568 that's not how it works
you'll just get [0,sqrt(2))
sqrt(2)-0 is not rational
 
Hi @MatheinBoulomenos
 
hi @BalarkaSen
 
9:47 AM
Hi @Leaky
 
Hi @BalarkaSen @LeakyNun
 
Hi @Mathein @Leaky @Balarka
 
Hi @ÍgjøgnumMeg
 
Given measure spaces $(X, \Sigma_X, \mu_X)$ and $(X, \Sigma_Y, \mu_Y)$, how does one define the product? I suppose we take the sigma algebra on $X \times Y$ to be the one generated by $\{E_X \times E_Y : E_X \in \Sigma_X, E_Y \in \Sigma_Y\}$?
 
you need the Caratheodory extension theorem
 
9:56 AM
$\mu_{X \times Y}$ is naturally defined on those by $\mu_{X \times Y}(E_X \times E_Y) = \mu_X(E_X) \mu_Y(Y)$
 
right
 
Then those sets I think gives a semialgebra with a premeasure $\mu_{X \times Y}$, upon which we extend
 
speak of the devil @AlessandroCodenotti
 
@BalarkaSen you often want the completion of this $\sigma$-algebra
 
@LeakyNun Ya that's obvious, but I am asking if that is the correct semialgebra I should look at
 
9:58 AM
(also iirc you need $\sigma$-finite spaces to have a unique product measure)
 
@BalarkaSen *$\mu_X(E_X) \mu_Y(E_Y)$
@Alessandro The measure needs to be $\sigma$-finite, yeah.
I think this falls apart if you have an infinite collection $\{(X_\alpha, \Sigma_\alpha, \mu_\alpha)\}$ of $\sigma$-finite measure spaces. There isn't a unique way to get a product measure, right?
Because the premeasure on the semialgebra I wrote down isn't going to be $\sigma$-finite.
 
Infinite products of measures are hard
You should look at Kolmogorov's extension theorem at the end of Tao's book if you're interested
 
I think it's in the appendix of Durrett. I'll check it out, thanks!
If $(X, \mathcal{B}_X, \mu_X), (Y, \mathcal{B}_Y, \mu_Y)$ are topological spaces with Borel measures, then $\mu_{X \times Y}$ as I defined is a Borel measure on $(X \times Y, \mathcal{B}_{X \times Y})$, right?
Because in the product topology open sets of the form $U \times V$, where $U$ is open in $X$ and $V$ is open in $Y$ forms a basis for the topology on $X \times Y$
So $\mathcal{B}_{X \times Y} = \langle \mathcal{B}_X \times \mathcal{B}_Y \rangle$, I think.
Writing down details for all of these are annoying, of course. This is "clearly true" :P
 
but don't you get in trouble with uncountable unions? you want your spaces second-countable I think
 
$\sigma$-algebras are closed under countable unions.
How does uncountable union play a role?
 
10:13 AM
yes, but in the definition of the product topology, you allow uncountable unions of the basis elements
 
Oh.
Good point
 
hello everyone. Apart from Rudin analysis book, are there some other books that have thought provoking problems on real analysis. Thanks in advance. P.S. This is my message in the chat. I hope it is allowed here. Cheers and have a good day
 
I do want a countable basis.
@MatheinBoulomenos So in general $\mathcal{B}_{X \times Y}$ would be "finer" (i.e. have more elements) than $\langle \mathcal{B}_X \times \mathcal{B}_Y \rangle$, is the upshot of your observation.
 
Cool point.
 
10:19 AM
there's a notion of product measures for two outer Radon measures on locally compact spaces that makes Fubini work (even when we don't have $\sigma$-finiteness),
basically if $X$ and $Y$ are locally compact spaces with outer Radon measures $\mu$ and $\nu$ respectively, then there's a unique outer Radon measure on $X \times Y$ that makes some form of Fubini work
 
Huh
 
I learned this from the appendix of Principles of Harmonic Analysis by Deitmar/Echterhoff
 
I borrowed Federer from the library, it just started Radon measures. :3
Measure theory is too technical man
 
yeah, it's too technical for me and that's coming from someone who enjoys group cohomology
 
Hah
@Alessandro Oh, Kolmogorov's extension theorem is really nice. It's precisely what I have been looking for.
 
10:36 AM
It is! There are generalisations to projective families of measure spaces or something like that, not only products, but I don't know anything about that
 
10:46 AM
If x1,x2,...,x6 are the roots of x^6+2x^5+4x^4+8x^3+16x^2+32x+64 then |x_i|=2 according to my book for all values of i... putting x=2 does not satisfy this. I don't get what the book tries to say
 
@tatan Why would you put $x=2$?
 
@Alessandro I wanted to understand what "Simple random walk on Z starting at 0 returns to 0 infinitely often with probability 1" meant. You can say something as follows: Model a simple random walk of length $n$ starting at $0$ by an i.i.d. sequence of random variables $(X_1, X_2, \cdots, X_n)$ where $X_i = \pm 1$ uniformly with probability 1/2. Then $\{S_k = \sum_{i = 1}^k X_i\}_{1 \leq k \leq n}$ is a Markov chain that represents the walk. Call $p_n = \Bbb P(S_n = 0)$. Then you can observe that $\sum_n p_n = \infty$ - this somehow models the fact that the random walk returns to $0$ infinit
 
The book is saying that any root will have norm $2$, not that $2$ will be a root
 
@TobiasKildetoft What is norm 2?
 
@tatan Norm as a complex number
 
10:54 AM
Can you explain?
 
But I think Kolmogorov extension theorem gives a way to define the appropriate probability space $(\Omega, \Bbb P)$.
It's $\Bbb Z^\Bbb N$ with some appropriate product measure.
 
@TobiasKildetoft Okay... do you mean that the if a+ib is a root, then sqrt{a^2+b^2}=2 as given here?
 
@tatan Yes
 
How do I find a+ib then? I mean how do i prove that |a+ib|=2?@TobiasKildetoft
 
@tatan No idea, I didn't think about that part.
 
11:00 AM
I think it's this: Give $\Bbb Z$ the uniform measure $\mu$ supported on $\{-1, +1\}$. Then consider $(\Bbb Z^\Bbb N, \mu^\Bbb N)$ where $\mu^\Bbb N$ is the unique measure you get that is compatible with $\mu^n$ for any finite $n$ using Kolmogorov(??). Let $\Phi : \Bbb Z^\Bbb N \to \Bbb Z^\Bbb N$ be the map $\Phi(a_1, a_2, a_3, \cdots) = (a_1, a_1 + a_2, a_1 + a_2 + a_3, \cdots)$. Define the pushforward measure $\Phi_*(\mu^\Bbb N)$ on the range.
Then the hypothetical $"(\Omega, \Bbb P)"$ should be $(\Bbb Z^\Bbb N, \Phi_*(\mu^\Bbb N))$
I'll look into all this in greater detail and think about it
 
Just left a slightly lengthy comment giving the dimensions of irreps of $GL_n(\mathbb{F}_2)$ over $\overline{\mathbb{F}}_2$ for all $n\leq 4$. Now I am wondering how feasible it is to actually calculate the $n=5$ case.
@MatheinBoulomenos Maybe that is a good exercise for you.
 
i don't know if this is a mathematical question or programming one. Can there be a case where a shorter password provides more strength(or equal) than a longer. Other things being equal, like used characters(digits,chars,special symbols).
 
11:31 AM
@Alucard An individual password does not really have a strength until you specify a method of attack. But once you do, sure there are plenty of cases where some shorter passwords will be much stronger (if for example the attack involves dictionary and the longer password is a single word)
 
@TobiasKildetoft ah didn't thought about 1 long word vs 2 short ones :)
 
@Alucard Or just something that is not a word
 
even tho vs bruteforce the longer word has it's merits
 
Right, the strength of a password depends heavily on the method of attack
 
12:29 PM
Does someone know if there is a general formula for all $(g,n,g',n')$ such that the compact orientable surface $\Sigma_{g,n}$ of genus $g$ with $n$ punctures is a finite covering space of $\Sigma_{g',n'}$?
 
Good question. I don't. Euler characteristic considerations say that $2-2g' -n$ divides $2-2g-n$.
 
For example, the twice-punctured torus $\Sigma_{1,2}$ is a $2$-sheeted covering space of the once-punctured torus $\Sigma_{1,1}$.
 
But I don't believe that's sufficient.
 
Yeah, I have the same feeling.
 
For tori Sigma_{1,dk} k-fold covers Sigma_{1,d}.
 
12:41 PM
Do you think one can make the genus arbitrarily large in some finite covering of $\Sigma_{g,n}$, at least in the hyperbolic case of $2-2g-n < 0$?
Because then the fundamental group is residually finite and so there seem to be many finite covering spaces.
 
Yes, the only difficulty is getting to genus at least 2. Once you're there, you know a genus 2 surface is covered by a genus d surface for any d > 2. So cap off your discs, do the covering, delete the preimages of those discs.
This is the easy covering Sigma_{dg-d + 1, dk} -> Sigma_{g,k}
 
Mhmm, so you mean if $Y \to X$ is a finite covering, then it gives me a covering $Y \setminus \{\ldots\} \to X \setminus \{p_1,\ldots,p_n\}$ for any $p_1,\ldots,p_n \in X$?
 
Ah, you're literally just puncturing. Yeah. I was thinking of the corresponding surfaces with boundary, where you delete tiny open discs.
 
Yeah, sorry, I wasn't clear enough. I mean $\Sigma_{g,n} := \Sigma_g \setminus \{ n \text{ points }\}$.
 
No harm done. They're basically the same notion.
Note that the 'easy formula' above includes all the torus examples.
 
12:48 PM
Does it still work if I have punctures instead of removed disks?
 
However, it's definitely not everything. In particular it includes no interesting covers of punctures spheres.
Yes, what you just said is fine.
 
Ah right, I guess there are actually not much requirements on the covering $f \colon Y \to X$ for it to work out. For any $p_1,\ldots,p_n \in X$, I get a covering $\widetilde{f} \colon Y \setminus f^{-1}(\{p_1,\ldots,p_n\}) \to X \setminus \{p_1,\ldots,p_n\}, y \mapsto f(y)$.
 
@user600999 yeah. So basically, perturbing the algebraic structure in question on a fixed topological surface, leads to the notion of moduli. The moduli spaces of curves are algebraic objects that parametrize all such algebraic structures. So the natural place to start for me, is to perturb the curves on structure of genus $1$, like the torus
 
I think there is a degree 2 covering Sigma_{g,2g+2} -> Sigma_{0,2g+2}. Note that the Euler characteristic formula works out.
3
Q: Degree 2 branched map from the torus to the sphere

Randy BrownAlgebraic geometry predicts a degree 2 branched cover from an elliptic curve to the projective line. What does this map look like topologically?

 
@Ultradark That makes very little sense to me
@Ultradark How have you decoupled the algebraic and topological structure for an algebraic curve?
 
12:53 PM
Oh, I've just realized something. First, are you willing to trust me that the answer to "which punctured surfaces cover each other" is the same, both in your delete a point model, and my model with boundary circles?
 
I believe you, through I probably would have to think about why this is true.
 
To go from the boundary to the punctured version, just delete the boundary circles. (The interior of my version is homeomorphic to your version.)
The other way is a little trickier. Anyway.
 
Ah, okay. I get the idea.
 
@user600999 just think of an algebraic curve flowing along a fixed topological surface. then take multiple of these curves, maybe a family of functions, all flowing along the surface. And what do you mean by the second point?
 
Given any covering map of surfaces with boundary, each boundary circle 'downstairs' is covered by some number of circles 'upstairs'.
Restricted to a single one of those upstairs boundary circles, it is a covering map of some degree, say k.
 
12:57 PM
@Ultradark You are saying you are only dealing with subvarieties of a given variety?
 
You can then patch in a disc at both the upstairs and downstairs boundary circles and extend the covering map, by defining it on the disc as $z \mapsto z^k$.
This is a branched covering map, with degree $k$ near that point.
 
@Ultradark Saying fixed topological surface makes little sense to me in this context. For my second comment, I mean that the algebraic and topological structure are highly intrinsically related for an algebraic curve
 
Doing this for all boundary discs, you get a branched covering map between closed surfaces.
 
So I get a branched covering $\Sigma_{g',0} \to \Sigma_{g,0}$ from any covering $\Sigma_{g',n'} \to \Sigma_{g,n}$.
 
So the question is the same as: do we know what pairs (g,n') -> (g,n) admit a branched covering with n' branch points upstairs and n downstairs?
This is related to the Riemann-Hurwitz formula: to get that theorem you go in the opposite direction and delete branch points.
I anticipate this question should be well known...
 
1:03 PM
Very interesting.
How did you come up with the formula for the covering $\Sigma_{dg-d + 1, dk} \to \Sigma_{g,k}$?
Is this taken from a covering $\Sigma_{dg-d + 1,0} \to \Sigma_{g,0}$?
 
Yes
 
If the left term is g' and this is a degree d cover, I need 2 - 2g' = d(2-2g) to be true. So I just solved.
And yup.
 
Just $d$ surfaces of genus $g$ arranged circularly like the picture in Hatcher
 
Ah thanks @BalarkaSen
Thats the one with the "rotational symmetry"
 
Yup
Then you just introduce a puncture and unwrap the puncture upstairs to get many punctures
I want to see an example where the $\chi$ restriction is satisfied, and $n > n'$, but there is no such covering map. $\pi_1$ considerations are not enough because everything is a free group once $n, n' > 0$.
Some local bollocks will happen if you choose $n$ and $n'$ cleverly, I guess. Near some punctures the map is a connected cover of the circle but near other punctures it's a disconnected cover of the circle...
 
1:11 PM
The link was to a paper on this for a sphere
There is a restriction we missed: if the cover is degree $d$, then $n' = dn$ mod 2
This is because you can recover the punctures mod 2 from the Euler characteristic
 
@user600999 my approach is basically, to define the algebraic curve as a 2-dimensional object in a subset of euclidean space, then extend this definition to a family of algebraic curves. Honestly I don't know how to decouple the algebraic and topological structure for the algebraic curve
 
@MikeMiller Ah fair
 
I am having trouble coming up with a case this actually rules out.
I mean, if the Euler characteristic formula holds, this does too ...
 
yeah meh just $\chi mod 2$
 
Oh I see. It's nontrivial when you're not orientable.
 
1:19 PM
So is an $n$-times punctured torus $\Sigma_{1,n}$ for $n \geq 1$ covered by a surface of higher genus?
 
But I will just stick with the orientable case.
All of them are, since $\Sigma_{1,1}$ is triple covered by $\Sigma_{2,1}$, I believe.
Does that sound right @Balarka?
 
I can see the cover on the level of 1-skeleton but not the full thing
 
Theorem 2.1 of that paper seems like it says that the Euler characteristic restriction is the only restriction, as long as the base is positive genus.
 
In fact you can even specify the branch degree at each point.
 
1:25 PM
Weird
 
@user600999 still a lot of work to do :)
 
@Ultradark You want a curve to be $2$-dimensional?
I thought you were happy with the family of schemes I gave you last night?
 
Ok, so let's resolve the easier question for a sphere.
 
Anyone know any good Number Theory texts?
 
A degree d cover of Sigma_{0,n} must have Euler characteristic 2d-dn. So it must be Sigma_{g, dn + 2(g-d) + 2}.
Nor very helpful.
 
1:30 PM
Maybe one can use that the fundamental group is residually finite, so we always get some proper finite index subgroup, this means either the genus must increase or/and the number of punctures.
Doing this over and over again, maybe its impossible for only the number of punctures to increase?
 
The covers of Sigma_1,n by Sigma_1,n' are always necessarily cyclic group actions. You just need a noncyclic representation of the fundamental group on a finite set.
It's certainly better to just find some covers.
 
@TobiasKildetoft maybe I'll try if I find time
 
I think my formula for number of punctures above is wrong.
 
@MatheinBoulomenos As I said, I am not sure how feasible it is. There are $16$ of them, and if I recall correctly, $10$ of them should be straightforward.
Probably one of the remaining $6$ is dealt with in Jantzen's Habilitation thesis from the 70's.
 
yeah I'm not sure how one would compute that
this is some algebraic groups stuff, right?
 
1:40 PM
Well, the first step should be Jantzen's sum formula, which limits the options, but might well leave the question of some multiplicities
yeah
 
Brb
 
I'm afraid I'm not of any help then
One of my friends wrote his bachelor thesis on representations of $\mathrm{GL}_2(\Bbb F_q)$ in characteristic $p$ (where $q=p^n$)
 
@MatheinBoulomenos Neat. That case is pretty much completely understood.
 
Yes, it should be Sigma_{g,dn-2(g+d)+2}.
 
(at least when the field is algebraically closed. I am not actually sure how much is known about reps over finite fields, though probably quite a bit is still known).
 
1:51 PM
This accommodates the hyperelliptic involution example, we get a degree 2 cover of Sigma_{0,2g+2} by Sigma_{g,2g+2}, and 2(2g+2) - 2(g+2)+2=2g+2.
 
Anybody familiar with Witt group
 
2:04 PM
@MikeMiller this is kind of embarrassing: https://math.stackexchange.com/questions/896589/if-b-is-a-regular-value-of-f-f-1-infty-b-is-a-regular-domain

I don't see how you automatically get a chart to $\mathbb H^n$ from the projection.
I think I am missing something obvious.
Also, hi @BalarkaSen!!!!
 
2:19 PM
Hi @anakhro!!
 
Maybe you have insight into the question I asked mike, too. :P
@BalarkaSen my oft-repeated joke about your whereabouts has been that the dean of mathematics has you locked in his basement, constantly forcing you to calculate various Laplace transforms.
 
@anakhro Amazing
I lol'd out loud
 
@BalarkaSen Isn't "out loud" idempotent?
 
The dean of mathematics here is a functional analyst. Not too far off.
 
Have you made a name for yourself at your university already, Balarka?
 
2:25 PM
@Tobias lol is a word, not an acronym
 
@BalarkaSen Then I think it should have been lolled, not lol'd
And it is a very strange thing to do out loud (how do you even make that make a sound?)
 
@TobiasKildetoft do you know any good books on complex geometry + representation theory + Kirilov method?
Like coadjoint orbit stuff
 
@anakhro No idea about anything in that intersection
 
Like the symplectic structure that arises from a Lie group $G$ acting on $\mathfrak g^*$.
On the orbits.
And these orbits are in correspondence with unitary irreducible representations of $G$, I think was the point.
Maybe the complex geometry request is not needed.
Just the latter two.
 
never studied that sort of thing
 
2:29 PM
Do you do Lie theory at all? Or is your representation theory somewhere else?
Or maybe I am mistaking you for someone else and you do neither.
thinks very hard.
 
I do representation theory but not really Lie theory
I do Lie algebras, but from the algebraic side of things, so category $\mathcal{O}$ stuff
plus algebraic groups in positive characteristic
 
Ah, I see!
 
@anakhro Eh, it's whatever
@TobiasKildetoft With gusto!
 
What math have you been up to, @BalarkaSen?
 
Too many things at once.
 
2:42 PM
The usual reference for these things in general i know is ‘representation theory and complex geometry’ but i suspect you know this already
 
Ah!!!
@loch this was the book I was looking for !!!
Thanks!
Or at least I think it was.
A professor suggested a book on the subject to me a long time ago and I couldn't remember it.
 
@anakhro $f^{-1}(b) \subset M$ is a submanifold by inverse function theorem. Then there is a chart $U$ around $x$ for any $x \in f^{-1}(b)$ such that there is a diffeomorphism $(U, U \cap f^{-1}(b)) \to (\Bbb R^m, \Bbb R^{m-1})$.
That's all
 
@BalarkaSen I don't see how you get the chart to halfspace from that.
I know the regular value theorem gets you that f^{-1}(b) is a (regular) submanifold. But I am having difficulties constructing the chart around $x\in f^{-1}(b)$ to $\mathbb H^n$
 
Restrict the diffeomorphism I wrote to $U \cap f^{-1}((-\infty, b])$.
That is a diffeomorphism onto $\Bbb H^n = \{\mathbf{x} \in \Bbb R^m : x_n\geq 0\}$
The "top half" of $\Bbb R^m - \Bbb R^{m-1}$
I'm leaving in a bit. If you still have questions we can talk about it later
 
Oh glad it helped!
 
2:54 PM
How is this diffeomorphism defined?
 
What's up chatters
 

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