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6:12 PM
@TedShifrin .... your input and suggestions have always been extremely helpful! Could you help me with my current problem about isometric embeddings (I have been stuck for days)? I would be super grateful for your support: chat.stackexchange.com/transcript/message/52623421#52623421
 
@eigenvalue: Ugh. You should post stuff actually typed up. All right ... I have the handwritten thing. So what are your specific questions? Where does eqn (1) come from? Is that your first question?
There are presumably initial conditions being imposed which explain the next stuff.
What's the original question? You're given a spacelike surface with the metric $dx^2+t\,dt^2$ and want to isometrically locally imbed in Minkowski?
 
@TedShifrin yes, the two lines with the initial conditions. Is there an explanation why exactly those two conditions were picked?
 
LOL, you didn't tell me the entire original question. I can't tell who picked what why.
 
Thats all I have. I have the two metrics and the 2D metric should be isometrically embedded into Minkowski space
 
For what values of $t$? All values, including negative?
Then I was wrong about spacelike surface.
 
6:19 PM
yes, all agues of t, negative ones and zero included
 
The metric is degenerate at $t=0$, so this isn't a good metric.
Hmm ...
 
yes, it's a signature-changing type metric
 
Interesting — this is not something I've ever thought about.
OK, so you do understand equation (1), right?
 
I kind of can explain the first requirement, but have no clue why the second line makes sense (with the two derivatives >=0)
yes, (1) is clear
 
how many days have you been stuck :^)
 
6:23 PM
@skullpetrol at least 7
 
We're just trying to guess a solution. So let's do something simple.
The question is: can we find something that works for all $t$?
To get an embedding, they're leaving $x$ alone, and we have to give $(\nu,\xi)$ as a function of $t$ with nowhere-zero derivative.
I guess something funny might happen when $t=0$ since the metric degenerates there; I dunno.
 
I agree, maybe a simple straightforward approach will clarify what's going on here
 
So the initial values of $\xi$ and $\nu$ are no big deal. Pick anything you want. That's why they did $0$ for both.
The more interesting thing to think about is giving a reasonable curve that satisfies $\xi'^2 - \nu'^2 = t$.
 
When I check the literature, they only show how to set up the embedding functions and the equations. But nowhere they show how to set up initial conditions or how to actually calculate the solutions.
 
6:29 PM
When I see Minkowski, my immediate impulse is to ask what physics problem is this :P
 
Initial conditions are irrelevant here. Don't worry about that. You can translate anywhere .
 
(e.g. a particle undergoing constant proper acceleration)
 
So note that when $t<0$, $|\nu'|$ has to be larger than $|\xi'|$, and the opposite when $t>0$, and they cancel at $t=0$.
The fact that $\sqrt t$ is very non-smooth at the origin makes this more challenging.
So are they trying to do just a local isometric embedding, or do we really want to try to do it globally (for all $t$)?
@Semiclassic: Have you seen the changing signature with $t$ before?
 
I believe we need a global embedding? Otherwise we only have an embedding for a neighborhood
 
hrm. nope.
a changing metric seems more like something out of GR, and I'm no good for that
 
6:34 PM
@TedShifrin Dray uses signature changing metrics in many of his papers
 
Ah, he's an old friend of mine.
 
when you say it's changing signature, you mean specifically at $t=0$?
that seems gross.
 
Yeah. It's a definite singularity.
 
@Semiclassical yes, in this particular case (this metric) the signature changes at t=0
 
So their solution is fine, but I'm more curious to understand if we can find a global solution.
 
6:36 PM
I'd be a lot happier with $\xi'^2-\nu'^2=t^2$
 
Well, then you just set $\xi'=t, \nu'=0$. Duh.
 
@TedShifrin and the solution I have in my example is global, isn't it?
 
No, only for $t\ge -1$.
 
@TedShifrin no sorry, I am wrong
 
6:38 PM
But I think that using bump functions we might be able to do this globally (but not with an explicit algebraic function).
 
@TedShifrin I know there is a global solution, but I first wanted to start with this one... and understand it before I move forward
 
OK, so you just try to make intelligent choices to make things as simple as possible.
To make it global, I think we just want a smoothing of the sgn function.
 
closest analogy I'm seeing in physics is with the FRW metric in GR
but then you've got something like $a(t)^2$ rather than $t$
which, again, doesn't change signature
 
In the literature they only set up the embedding maps and show how to get the equations. But I have never seen an exact calculation so that I can understand it ... is there a good source about isometric embeddings with some examples?
 
I've done research work with singularities in various geometric settings, but not this one.
@eigenvalue: There's nothing relevant about isometric embeddings here. In general, that's a very hard overdetermined PDE question. Here they're reducing to ODE by having the $x$ variable disappear. So it's just about solving that ODE.
This one is underdetermined so you just have to try to think of functions so that you can get $\xi'^2 = t + \nu'^2$. This means that we need $\nu'^2 \ge -t$.
This is obviously no issue when $t>0$, but we need to think of it for $t<0$.
What if we take $\nu'(t)=\cosh t$, for example.
 
6:46 PM
@TedShifrin OK... Let me think about this... I am going to further deal with this problem today.... maybe we can talk later again?
 
OK ... There's no geometry left ... It's just calculus/ODE.
 
7:27 PM
An odd component in a graph is just a component with an odd number of vertices right?
This is the definition I have: An odd component is a connected component of odd size.

Unfortunately no examples though
 
If $X$ is a locally compact metric space must every bounded subset be contained in a compact one? This is obvious for proper metric spaces, and a locally compact one is something like "locally proper"
 
If $\mathcal{F}_\Gamma$ is a fundamental domain for the action of $\Gamma$ on $\frak h$ where $\Gamma \subset \operatorname{SL}_2(\Bbb Z)$ is a congruence subgroup then

$$\frac{1}{[\overline{\operatorname{SL}_2(\Bbb Z)}:\overline{\Gamma}]}\int_{\mathcal{F}_\Gamma}f(M\langle z\rangle)\overline{g(M\langle z\rangle})(\Im(M\langle z\rangle))^kd\omega(z) = \frac{1}{[\overline{\operatorname{SL}_2(\Bbb Z)}:M\overline{\Gamma}M^{-1}]}\int_{\mathcal{F}_{M\Gamma M^{-1}}}f(z)\overline{g(z)}\Im(z)^k d\omega(z)$$
hoping someone tells me I'm wrong
 
7:48 PM
You're wrong @ÍgjøgnumMeg
 
@ÍgjøgnumMeg, what is $M$?
A fixed element of $SL_2$?
 
@Ted $M \in \operatorname{GL}_2(\Bbb Q)^+$, forgot to mention that
and $d\omega(z)$ is
 
So the change of variables is off by a factor of $1/\det M$.
So that's $\pm 1$.
 
Hmm but I think that comes from $\Im(M\langle z\rangle)^k$
 
7:54 PM
But you take absolute value, so no foul.
No, it would be clearer to make the substitution $u=Mz$.
I think you act on the fundamental domain, though, just on the left, not by conjugation.
Hmm ...
Dunno why you're using $\langle z\rangle$ when you're restricting to a fundamental domain.
But I don't know what you're doing, so never mind.
 
What do you understand by $\langle z \rangle$? I just mean $Mz$
my notes just write it that way
 
I thought you had a function on the quotient space and were choosing a representative of the equivalence class in the fundamental domain.
 
and yeah the main thing I'm not understanding is why $\mathcal{F}_\Gamma$ becomes $\mathcal{F}_{M\Gamma M^{-1}}$
I see, no it's just M acting on z
 
The change of variables is given by left multiplication, not conjugation.
So it should be $M\cdot\mathcal F_\Gamma$.
 
Fair
Thanks :)
 
8:01 PM
Now, is that $\mathcal F_{M\Gamma M^{-1}}$?
What worries me is that action by $GL_2$ might not preserve the half-plane, right? But conjugation does.
 
well.. if $A_1, \dots, A_m$ are representatives for $\Gamma \setminus \operatorname{SL}_2(\Bbb Z)$ then $\mathcal{F}_\Gamma = \bigcup_{\mu = 1}^m A_\mu \mathcal{F}$
loool union
 
Are you sure that you don't have $M\in SL_2$?
 
no
$M \in \operatorname{GL}_2(\Bbb Q)^+$
 
Oh, $+$.
 
right
 
8:03 PM
Same thing.
What does $+$ mean?
 
positive determinant
 
Oh, but with $\Bbb Q$.
 
Right, but $M$ is also chosen such that $M\Gamma M^{-1} \subseteq \operatorname{SL}_2(\Bbb Z)$
so that it's also a congruence subgroup
and the index of $M\Gamma M^{-1}$ in $\operatorname{SL}_2(\Bbb Z)$ is the same as that of $\Gamma$
 
Yeah, this reminds me of the fact that the stabilizer of $g\cdot s$ is the $g$-conjugate of the stabilizer of $s$.
 
which is nice because this coefficient in front of the integral then changes back to the index of $\Gamma$ in $\operatorname{SL}_2(\Bbb Z)$
 
8:05 PM
I'm just saying that change of variables would give you $M\cdot\mathcal F$, and you have to sort out why that should be a fundamental domain of the conjugate.
 
Yeah that sounds familiar
 
Fill me in later :P
 
@Ted actually, I don't think I need to do that at all; the integral is independent of the choice of fundamental domain I think, so I just need to show that $M \cdot \mathcal{F}_\Gamma$ is a fundamental domain for the action of $\Gamma$ on the upper half plane and then it's fine
I htink
think*
(this is a theorem)
anyway the point is that I'm showing that the Petersson inner product on the space of weight $2k$ cusp forms is somehow invariant under the action of the Petersson slash operator
 
8:20 PM
There are some people in the room learning to differentiate polynomials and it's making me jealous
 
9:13 PM
@ÍgjøgnumMeg maybe they're jealous of you too
the grass is always greener
 
but I wanna just sit and differentiate polynomials blindly
@Ted I think I have an argument for $M\cdot \mathcal{F}$ being a fundamental domain for the action of $M \Gamma M^{-1}$
which is nice
actually quick question: do analytic functions preserve closedness/connectedness
actually wait, a möbius transformation is a homeomorphism
so that's fine
 
you also have the open mapping theorem
and continuous function preserves connectedness
 
Ah nice :)
 
Suppose I look at the polynomial $x^n - \sqrt{2}$. I can argue this is irreducible over $K = \Bbb Q(\sqrt{2})$ by noting that $K$ is the fraction field of the UFD $A = \Bbb Z[\sqrt{2}]$ so that by Gauss lemma it suffices to show it's irreducible over $A$ but $\sqrt{2}$ is an irreducible element by a norm argument. Then I just apply Eisenstein's criterion.
But what if I have something nasty like $x^n - \sqrt{5}$?
 
Nasty because $5 \equiv 1 \bmod 4$?
 
9:26 PM
I don't actually know when $\mathcal{O}_{\Bbb Q(\sqrt{d})}$ is a UFD. This stuff is real class field theory, right?
 
No not at all
$\mathcal{O}_K$ is a UFD iff PID, so you compute the ideal class group
which can be done "reasonably" easily for small $\sqrt{d}$
 
@ÍgjøgnumMeg Yeah, firstly I can't pass to $\Bbb Z[\sqrt{5}]$, and that's not a UFD. Is $\Bbb Z[(1 + \sqrt{5})/2]$ a UFD?
No clue how to see something like that. I know the result for imaginary quadratic fields.
 
Hi @Balarka
 
@Balarka you look at the Minkowski bound $M_K = \frac{n^n}{n!}\left( \frac{4}{\pi}\right)^{r_2}\sqrt{\lvert d_K\rvert}$
 
@ÍgjøgnumMeg Yeah, these are all 1-dimensional Noetherian domains, which are PID iff UFD
 
9:28 PM
where $n = [K : \Bbb Q]$, $r_2$ is the number of complex embeddings of $K \hookrightarrow \Bbb C$ and $d_K$ is the discriminant of your number field
 
I can prove this by ideal factorization, I think
Hi @Alessandro
 
then to compute the ideal class group you only need to look at prime ideals of norm less than $M_K$ (by Minkowski theory and stuff)
(class group is generated by classes of prime ideals by ideal factorisation)
 
I have vaguely heard these things from some seniors taking algebraic number theory. Dunno much about it. What's the smallest example of a real quadratic field with ring of integers not a PID?
 
Uhhhh I can't remember, but you need a number field that gives a large enough Minkowski bound at least
 
Mmk
 
9:33 PM
@Balarka $\Bbb Z[(1+\sqrt{5})/2]$ has class number one though
so it's a UFD
 
Huh, cool
And $\sqrt{5}$ is indeed irreducible there, by the Pell norm stuff again, I imagine?
 
shouldn't be hard to prove using the class group, since the minkowski bound is ~4.5
something like that
the norm is slightly different in that ring though
 
Yeah
 
@BalarkaSen it's completely classified, just look it up
 
9:36 PM
oops it isn't completely classified
 
it is for imaginary quadratic fields lo
l
 
Isn't it like wide open or something lol
Yeah I know Stark-Heegner
 
K is called real quadratic if d > 0. K has class number 1 for the following values of d (sequence A003172 in the OEIS):
2*, 3, 5*, 6, 7, 11, 13*, 14, 17*, 19, 21, 22, 23, 29*, 31, 33, 37*, 38, 41*, 43, 46, 47, 53*, 57, 59, 61*, 62, 67, 69, 71, 73*, 77, 83, 86, 89*, 93, 94, 97*, ...

K has class number 1 exactly for the following negative values of d:
−1, −2, −3, −7, −11, −19, −43, −67, −163
 
in fact it's not even known if there are even infinitely many number fields at all with class number 1
 
9:37 PM
damn
 
but ok $x^n - \sqrt5$
I think we know that $x^{2n} - 5$ is irreducible over $\Bbb Q$
 
Just Eisenstein, sure
 
and the galois group acts transitively on the factors
over $\Bbb Q(\sqrt5)$ the polynomial becomes $(x^n - \sqrt5)(x^n + \sqrt5)$
it has at most $2$ factors
so $x^n - \sqrt5$ must be irreducible over $\Bbb Q(\sqrt5)$
 
Yeah so I guess I'm basically asking if degree of $\Bbb Q(5^{1/2n})$ over $\Bbb Q(\sqrt{5})$ is $n$
 
now if you ask it this way it becomes easier lol
by tower law
 
9:41 PM
But degree of the former over $\Bbb Q$ is $2n$ by what you said
Yup, cool argument
Thanks!
 
np
 
It's fun how complicated this looks if someone doesn't know field theory
 
@ÍgjøgnumMeg have you learnt L-functions of modular forms
 
@Leaky nope lol, I'm learning modular forms right now.
 
it's fun how complicated math looks if someone doesn't know maths
@ÍgjøgnumMeg have you heard of them
 
9:46 PM
yeah I've heard of them from the L-Functions course
 
what's the L-functions course
 
Construction of L-Functions
 
of what
 
Several different things lol
 
like what
 
9:48 PM
We had Dirichlet L-Functions, Dedekind L-Functions, Hecke L-Functions, Artin L-Functions
L-Functions for elliptic curves
L-Functions for modular forms
etc.
I stopped attending because I didn't really have the background for a lot of the stuff
and needed to concentrate on modular forms :P
 
nice
 
actually I think Hecke L-Functions are the ones for modular forms lol
Artin L-Functions are somehow attached to Galois representations
 
I forget the example of a polynomial which is reducible modulo every prime but irreducible over $\Bbb Z$
$x^4 + 1$?
 
What was the argument, though? If $p$ is an odd prime then $p^2 - 1$ is divisible by $8$, so $x^8 - 1$ has a solution over $\Bbb F_{p^2}$
 
9:54 PM
Is this $\Bbb H^2$ or $\Bbb H^3$????
 
Taimina's crocheted hyperbolic plane! :D
H2
 
@Balarka I don't remember, but I looked it up and there's a cool proof on the site
 
I think you can find images online where they'd stitched a pentagonal tiling on that (where each vertex has 90 degrees)
 
@BalarkaSen there's also $x^4 - 10x^2 + 1$
 
Yeah I think I might be trying to remember that argument
 
9:56 PM
and the proof uses quadratic residues
 
Ah, OK, just take an element of order $8$ in $\Bbb F_{p^2}$ by Lagrange. Then it has to satisfy $x^8 - 1$ and cannot satisfy $x^4 - 1$, so is a root of $x^4 + 1$
 
There you go (much less curvature that in your pic - presumably for yours the pentagons would have to be smaller)
^Here's two parallels to a given line through a point
 
But degree of that element over $\Bbb F_p$ is $2$, so some quadratic polynomial appears as a factor of $x^4 + 1$ in $\Bbb F_p$
 
(Those are indeed geodesics because you can fold the surface along them)
 
9:58 PM
@AkivaWeinberger is there a ball model of $\Bbb R^3?$
 
@LeakyNun Oh yeah I saw this in Artin but didn't see how to prove this.
 
Yes. There is also a disc model of $\Bbb R^2$
 
I know 0 number theory
 
$0$ is a number, for example
 
where is this ball model of $\Bbb R^3$?
there seem to be a plethora of models of hyperbolic space
 
10:00 PM
If I remember right you basically map $(x,y,z)\mapsto(x,y,z)/\sqrt{1+r^2}$
 
so everything gets mapped inside a 2 sphere?
 
so $\left(\dfrac x{\sqrt{1+x^2+y^2+z^2}},\dfrac y{\sqrt{1+x^2+y^2+z^2}},\dfrac z{\sqrt{1+x^2+y^2+z^2}}\right)$
Yeah
@Ultradark To do this geometrically for $\Bbb R^2$
 
I don't even think anyone has made a physical model of that
 
First do the gnomonic projection from the plane to a hemisphere
(The gnomonic projection is projection from the center of the sphere)
 
plane to hemisphere or the other way around?
 
10:03 PM
Plane to hemisphere
and then we do hemisphere to disc
by ignoring the height axis and projecting straight down
If I remember right, geodesics in this model correspond to ellipses tangent to the boundary
with centers at the center of the disc
Googling says this is called the Gans projection, but unfortunately I can't find images
Wait @Ultradark
I found an image in a JSTOR article, give me a moment
 
okay
 
Are you still here? @Balarka
 
@AkivaWeinberger okay so that is for $\Bbb R^2$
 
The lines in this model are half-ellipses whose major axes are diameters of the disc, as well as the diameters themselves
 
10:11 PM
any luck on a physical model of $\Bbb R^3$
 
Physical? I dunno
Maybe pester Henry Segerman on Twitter until he 3D prints one
 
maybe an artist made one?
 
Yeah @Alessandro
 
I have a topology doubt, I also have a feeling it should be straightforward
Do you want to think about it?
 
howdy, a @Balarka, demonic @Alessandro
 
10:12 PM
Hi @Ted
 
@Alessandro Is it point-set topology? :P
 
Let $X$ be a locally compact metric space. Is every bounded subset of $X$ contained in a compact one?
 
No.
 
This is obvious for proper metric spaces and locally compact metric spaces look like a proper one locally
But this is not a local statement
 
I have an immediate counterexample.
(But I'm just chopped liver.)
 
10:14 PM
What's your example?
 
Well, the first thing I thought of was a punctured disk. But an open interval will do.
 
The punctured disk is more interesting because you have counterexamples which are proper subspaces
 
You need some sort of completeness for it to work.
Same in the interval, silly.
Take a quarter interval inside.
 
I was about to say take $\Bbb R$ and make the metric bounded, but I thought I was misunderstanding the question.
 
Oh of course, just go to the boundary
 
10:17 PM
What if he says totally bounded?
 
I've been thinking about asymptotic topology a lot lately, I didn't remember there are bounded spaces, ok?
 
Then it's true, I think.
But I'm not stopping to think.
 
Wait doesn't the interval example still work?
 
Do you want it to be complete?
 
10:19 PM
Well, is $(0,1/2)$ a totally bounded subspace of $(0,1)$?
 
I think so with the metric induced from $\Bbb R$
 
Oh, yeah, it is, isn't it. Hmm.
Hmm, but compact = closed + totally bounded.
Something's fishy.
 
compact = complete + totally bounded
 
Oh, oops. bad Ted
 
And $(0,1/2]$ is closed in $(0,1)$ but not complete so everything is fine
 
10:21 PM
I think I should consider retiring for real. I was typing up an answer earlier today without being careful and totally messed it up until I realized.
 
@TedShifrin I think I found a good way to do that problem I noted earlier, one which entirely reduces it to Vandermonde
 
Yes, @Alessandro, that same counterexample came to me, which is why I deleted.
 
and which works for any sequence of monic polynomials of increasing degree $f_0,f_1,\ldots, f_n$
 
Twitter oneboxing is broken in chat
 
10:23 PM
Interesting, @Semiclassic.
 
It's kinda annoying when you have to deal with metric rather than metrizable spaces
 
It's actually pretty trivial once you see it:
 
I should start a blog
I have at least two things I could write a decent *cardinality of words about
 
Number of words, not amount.
 
including the thing currently on my mind, Gödel's incompleteness theorem
 
10:24 PM
(One of my pet peeves in English syntax)
 
$(f_0(x),f_1(x),\cdots, f_n(x))=(1,x,x^2,\cdots, x^n)F$ where $F$ is some upper triangular matrix with ones on the diagonal.
So therefore if I make a matrix with entries $f_j(x_k)$, then I automatically can factorize it as the vandermonde matrix times $F$
and since $\det F=1$, the determinant of the matrix of $f$'s is just the Vandermonde determinant
Easy peasy.
(Now watch me have done something silly.)
It comes down to linearity, so no big shock.
 
Are they really 1's on the diagonal, or just some nonzero constants?
 
Monic polynomials.
 
OK, I thought you were working with an orthonormal basis with respect to $L^2$ or something.
Irrelephant, anyhow.
 
It's a simple thing but I'm proud of it now
 
10:32 PM
Does that give it immediately for his matrix with cosines?
 
@AkivaWeinberger thanks, I don't have twitter lol
 
@TedShifrin Not quite immediately, since Chebyshev polynomials aren't monic
but they're close enough to being monic that it's easy to account for it
 
As I suggested, monic is irrelephant.
As long as they are of increasing degrees. Are these?
 
I find it to be irgiraffe
 
yep.
the only reason monic matters is to get the prefactor.
 
10:36 PM
OK, then it's just the hippopotamus and we're fine.
 
doesn't change whether the matrix is singular
 
Nah, @semiclassic, all you need is nonzero leading coefficients.
Oh, I see what you mean.
 
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