@TedShifrin Thank you, sir. Now, since my book didn't make their definitions explicit, I'm actually trying to understand what $\Bbb Q[\alpha] $ and $\Bbb Q(\alpha)$ look like, with $\alpha\in \Bbb C $. I've come to know that $\Bbb Q(\alpha)$ is the smallest field that contains both $\Bbb Q $ and $\alpha $. Is it true that $\Bbb Q[\alpha]= \{ p(\alpha)| p\in\Bbb Q[x]\}? $ Can they be equal?

any hints on how to calculate the discriminant of $x^7+x+1$?

Hey @PVAL!

@TedShifrin bu really anyone who may want to give me a hand: For example it seems to me that $\Bbb Q[\sqrt(2)]=\Bbb Q(\sqrt(2))=\{a+\sqrt {2}b|a,b\in\Bbb Q\}, but if the definition of $\Bbb Q[\alpha] $ I gave is correct, why is $\Bbb Q[\sqrt2] $ like this? And when are $\Bbb Q[\alpha] $ and $\Bbb Q(\alpha)$ equal in general?

@Richard $\Bbb Q[\alpha]=\Bbb Q(\alpha)$ iff $\alpha$ is algebraic.

$\Bbb Q[\pi]$ does not contain $\frac1\pi$ (or something like $\frac{\pi+1}{3\pi^2+\pi}$); $~\Bbb Q(\pi)$ does.

it's easy to show the condition Akiva mentions to, just consider the ring homomorphism $\mathbb{Q}[X] \to \mathbb{Q}(\alpha)$ given by $p(X) \mapsto p(\alpha)$ and think about factoring

$\sqrt[{\Large 3}]2$ is algebraic, so $\Bbb Q(\sqrt[{\Large 3}]2)$ should equal $\Bbb Q[\sqrt[{\Large 3}]2]$, according to what I wrote above. That means that we should be able to write:$$\dfrac{\sqrt[{\Large 3}]2+1}{3\cdot\sqrt[{\Large 3}]2{}^2+\sqrt[{\Large 3}]2}$$in the form $a+b\sqrt[{\Large 3}]2+c\sqrt[{\Large 3}]2{}^2$. And if you give me a minute to play with Wolfram Alpha, I can tell you exactly what that form is.

(Try to think about we could derive it)

OK, if $\alpha=\sqrt[3]2$, then $\dfrac{\alpha+1}{3\alpha^2+\alpha}=-\dfrac2{55}+\dfrac6{55} \alpha+\dfrac{19}{110}\alpha^2$.

Alright, thank you @AkivaWeinberger @EricSilva

I just had a strange thought pertaining to fractional calculus

@SimplyBeautifulArt which is?

Well, we have $f(x)\in C^k$ if $\frac{d^k}{dx^k}f(x)$ exists and is continuous.

uuuh

syre

you are aware that I never retained fractional calc right?

Let us define $f(x)\in C^\omega$ if $\forall k\in\mathbb N(f(x)\in C^k)$

too many advanced functions for my ability to wrap my head around

Yeah yeah, I'm just talking out loud

@SimplyBeautifulArt exists and is continuous*

@SimplyBeautifulArt any thoughts of a version of the implied integral but for periodic functions?

i feel like it is incredibly important

yet not trivial to write

also usually $C^{\omega}$ is used to mean the analytic functions while $C^{\infty}$ is used for functions who derivatives of all orders exist

Let $D^s_xf(x)=\frac{d^s}{dx^s}f(x)=f^{(s)}(x)$ be defined for any $s\in\mathbb R$.

@EricSilva Yeah, but I need the $\omega$ becomes I'm going to $\omega+1$ and stuff

ok

Let us have $f(x)\in C^{\omega+1}$ if $\frac d{ds}f^{(s)}(x)$ exists and is continuous.

More generally, let us have $f(x)\in C^{\omega+k}$ if $\frac{d^k}{ds^k}f^{(s)}(x)$ exists and is continuous

And let us have $f(x)\in C^{\omega2}$ if $\forall k\in\mathbb N(f(x)\in C^{\omega+k})$

Let us define $\frac{d^t}{ds^t}f^{(s)}(x)=f^{(t,s)}(x)$

$f(x)\in C^{\omega2+1}\iff\frac d{dt}f^{(t,s)}(x)$ exists and is continuous

etc.

I'm gonna name it the fractional hierarchy of super smoothness.

Of course, we must define which fractional derivative we are interested in as well

I am wondering if such functions are unique i.e. does there exists $f(x)$ such that $f(x)\in C^\omega\land f(x)\notin C^{\omega+1}$

I'm gonna post this question tomorrow, for now good night

@AkivaWeinberger The above might interest you... :D

Huh, interesting idea

See if you can find interesting examples of functions in those classes

i feel that there aren't many

i would think that fractional derivatives of power functions are already sometimes poorly behaved because they have gamma functions popping up in them

I thought there was more than one way to define fractional derivatives

One has $D^sx^n$ being an expression with factorials (Gamma functions), and another has $D^se^{ax}=a^se^{ax}$

hi chat

there are multiple

@Semiclassical o/

i have only seen a couple ways, but idk what kind you'd want to answer such a question for if any

in the contexts ive seen people do fractional calc typically you really wanted robust results that didn't require a lot of regularity so this kind of question is weird to me

anyway im out for the day, bye chat

So I saw an interesting object on MSE called a "flow"

apparently used in dynamical systems or something

(I know nothing about dynamical systems)

But imaging we had all the points of a space $X$ "flowing" around or something

and $f:X\to X$ is an arbitrary function, where we're interpreting $f(x)$ to mean "where the point $x$ will have flowed to in $0.1$ seconds"

so $f^{\circ10}(x)$ is "where the point $x$ will have flowed to in $1$ second"

So a flow is like that but continuous

A flow is a function $\pi:X\times\Bbb R\to X$, interpreting $\pi(x,t)$ as "where $x$ will have flowed to in $t$ seconds"

and to makes sure that makes sense, we make it satisfy the following conditions

$\pi(x,0)=x$, and $\pi(\pi(x,t),s)=\pi(x,t+s)$

I'm only mentioning this here because I thought the idea looked cool

This is from the textbook excerpts from this question

(I guess it's a special type of homotopy.)

go go Hamiltonian flow @AkivaWeinberger

I kinda want to interpret $\pi(x,t)$ as $\pi^{\circ t}(x)$ where $\pi(x)=\pi(x,1)$, but that's not rigorous because that's not defined for fractional $t$

@Semiclassical What is this used for?

Physics, basically.

In Newtonian mechanics, motion is determined by forces through Newton's second law.

In Lagrangian mechanics, it comes via the Euler-Lagrange equations.

In Hamiltonian mechanics, it comes via Hamilton's equations.

That last one can be formulated in terms of trajectories in phase space, aka flows on a symplectic manifold.

Symplectic?

Yeah.

What does that mean

Symplectic

Well, Hamilton's equations are $\frac{dq}{dt}=\frac{\partial H}{\partial p}$, $\frac{dp}{dt}=-\frac{\partial H}{\partial q}$

in the one-variable case, where $H=H(q,p)$ is the so-called Hamiltonian.

That minus sign is more-or-less what makes it symplectic.

For the origin of the word: en.wikipedia.org/wiki/Symplectic_geometry#Name

I should also point out, I guess, that for any function $f=f(p,q)$ one has $$\frac{df}{dt}=\frac{dp}{dt}\frac{\partial f}{\partial p}+\frac{dq}{dt}\frac{\partial f}{\partial p}=\frac{\partial H}{\partial q}\frac{\partial f}{\partial p}-\frac{\partial f}{\partial q}\frac{\partial H}{\partial p}\equiv \{ H,f\}$$

where in the last expression I've introduced the so-called Poisson bracket.

It's manifestly bilinear and antisymmetric, so in particular $\frac{dH}{dt}=\{H,H\}=0$

So the Hamiltonian is preserved under the Hamiltonian flow, yay.

Alternatively you can write this using vector fields as $\frac{df}{dt}=X_H f$ where $X_H=\frac{\partial H}{\partial q}\frac{\partial}{\partial p}-\frac{\partial H}{\partial p}\frac{\partial}{\partial q}$.

Typically in symplectic geometry, a Hamiltonian is extra data.

Is there any physical reason it should be fixed as something in particular?

Eh, not really. Though the usual one that comes up in physics is $H=p^2/2m+U(q)$ where $m$ is the particle's mass and $U(q)$ is the potential energy function

In that case $\frac{dq}{dt}=\frac{p}{m}\implies p=m\dfrac{dq}{dt}$ (which matches how linear momentum is defined in intro physics) and $\frac{dp}{dt}=m\frac{d^2 q}{dt^2}=-\frac{dU}{dq}$ (i.e. Newton's second law, with a conservative force $F=-dU/dq$.)

@Waiting Sorry, I was out taking a friend and her dog to the vet. We were out all day. I am fine, but busy; how are you?

It depends on the problem, is really the point.

So somehow $q$ is really a generalization of Newtonian momentum

Nah. That's $p$.

and $p$ is really always position in the standard sense?

sorry

other way around then

$q$ is the generalized coordinate, yeah.

That's the terminology that gets used, in fact.

generalized coordinates and generalized momenta

You also have cases where $H$ is explicitly time-dependent, but that's inherently harder.

The way I learned Hamilton's equations is $i_(X_H) \omega= dH$, where $\omega= dp \wedge dq$. (or sum of $dp_i \wedge dq_i$ in higher dimensions).

then $X_H$ is as I've written it above, I think.

A lot of the questions in symplectic topology are on the fixed points of Hamiltonian flows. All of which are completely trivial in the autonomous (time-independent) case.

Right.

I think one talks about Hamilton-Jacobi stuff in the time-dependent case a lot.

I don't know of many interesting questions involving an autonomous Hamiltonian.

heh, yeah. time-independent hamiltonians are a bit too nice to have a lot of open questions left.

In some sense, they aren't nice.

There's a very cute proof that the set of time-independent hamiltonians is never a group.

hmm.

Namely, they are closed under conjugation inside the full hamiltonian group, but the hamiltonian group is simple.

If you write the natural inverse operation of an autonomous Hamiltonian, you get an idea why there's no reason as to why there shouldn

The main reason Hamiltonian mechanics are interesting in physics, I think, is because they provide the straightest jump to Schrodinger's equation

t be an autonomous inverse.

as compared with Lagrangian or Newtonian mechanics, I mean.

I guess one has to understand quantization to understand that jump.

with the canonical commutation relations $[\hat{q},\hat{p}]=i\hbar$ serving as a quantum version of the Poisson bracket $\{q,p\}=1$.

In fact, there's a pretty quick transition. If you suppose $\hat{q}=q$ and $\hat{p}=-i\hbar \partial_q$, then $[\hat{q},\hat{p}]=i\hbar$ is fulfilled automatically.

And then the Hamiltonian becomes $\hat{H}=\hat{p}^2/2m+V(q)$. The time-independent Schrodinger's equation is then just $\hat{H}\Psi=E\Psi$.

It's hard for me to really 'motivate' this, though. I'm so used to just using them.

I sort of suspect people have "classical" physical reasons for studying Hamiltonian mechanics as well.

Eh, yes and no.

I don't know if I know enough to have any convincing evidence.

The trouble is that from an applications PoV, one ofoten might as well just do Lagrangian mechanics.

You get one over on Newtonian mechanics, in that you get to work in terms of energies rather than forces, but you don't have to figure out momenta like you do in Hamiltonian mechanics.

There is at least one (math) person here who does statistical physics from a Hamiltonian perspective though I don't know much about their work.

Neat.

I actually need to do some rereading in statistical physics to remember some stuff I saw in undergrad.

(Landau theory of phase transitions.)

Here's his preprint list ma.utexas.edu/cgi-bin/mps?src=a&key=koch

ah, renormalization group

Not surprised to see KAM theorem stuff either.

The stuff I need to reread in stat physics is pretty simple in its basics, amusingly

namely: If $f(x,t)=x^4+tx^2$, then for $t>0$ this has a single global minimum at $x=0$

but if $t<0$ then that minimum becomes unstable and there's two global minima

this is intended as a very simplified picture of a symmetry-breaking phase transition e.g. magnetism as a function of temperature.

at high temperatures, there's no net magnetization to the system because things are too random to order.

but if you cool it down past a certain temperature then it'll have a net magnetization.

blah blah blah

@AkivaWeinberger "Pero en inglés se enseñamos evitar la voz pasiva"... "se" is the third-person reflexive object pronoun, which you definitely can't use in a first-person plural context (enseñamos)

Pero en inglés se enseña a evitar la voz pasiva, I think

but then I might be wrong

when does the solutions of a second order differential equation have common zeros

@robjohn I put him in the math mods' office and he won't come out.

@LeakyNun You could use it in a different sense, to show the passive voice

It's one of the ways to do the passive voice

@AkivaWeinberger no you can't use se ..mos

@LasVegasRaiders eh?

Apparently Daniel Fischer has taken this chat room off his favourites list since me no longer comes here regularly @robjohn

I asked him about it, in the math mods' office.

@AkivaWeinberger "se" is used for impersonal passives such as "it is taught"

\o @BalarkaSen

h

@AkivaW Usually you want $\pi(-, t)$ to be a homeomorphism of $X$

You could flow along vector fields, for example. If $M$ is a compact manifold and $V$ is a vector field on it, then define $f(x, t) : M \times \Bbb R \to M$ by the solution to $f'(x_0, t) = V(f(x_0, t))$ for any specific $x_0$ (with initial condition $f(x_0, 0) = x_0$).

That's like every point on $M$ following the vector field at any specific instance of time.

If $V$ has a zero, $V(x) = 0$, then $f(x_0, t)$ just stays at $x_0$ with respect to the flow of time. So a zero of the vector field is a fixed point of the flow.

This can be used to prove Poincare-Hopf theorem: Suppose you have a bunch of isolated, index 1 zeroes of $V$ on $M$. You want to prove the number of those zeroes is $\chi(M)$. Note that if I flow up along $V$ a bit I get a diffeomorphism $f$ with fixed points precisely the zeroes of $V$, so it suffices to count the fixed points.

But Lefschetz fixed point theorem (well, the general version of it) just says that's $\chi(M)$.

[To be done] Construct a mindmap for linear algebra, in order to visualise the subject in its entirely

But here's the catch. The mindmap has to look as similar to a categorical diagram as possible, including the major relations between each theorm and proposition in the subject

After all, for the big picture of a subject, often the topology of the proof structure is crucial to know when to use what theorem

and most importantly, why

Another way to put this, is a network digraph of linear algebra

Relevant transcript

[Matrix product integrals]

Let $A(x) \in M_n(\Bbb{C})$. Consider:

$$\prod_a^bA(x)^{dx}$$

This describes a continuum matrix product of the matrix function $A(x)$ in the interval $(a,b)$

Other possible generalisation include:

$$\prod_{k\in I} A^{dX(k)}$$

Transcript items that inspired the above as the concept "linear algebra" fused with the concept "integrals"

Theoretically, we can generalise this to a path integral like thing which involve integrating a continuum of tensor contractions

However details are to be worked out

In abstract algebra we have similar concepts called the direct integral

en.m.wikipedia.org/wiki/Direct_integral

This integral has an integrals that reminds of a sphere. Imagine doing some kinda of geometric evaluation where you have cos (sphere), that could be interesting

Wow! You're really digging up some old gems pal.

:-D

And some really random free associations when reading these generates a host of strange looking maths as well, any one of these can be a full fledged PhD project

The above integral, combined with the scrolling list of iPhones, inspired the following idea:

Consider a function with a strange argument:

$$\left( x\frac{(x^2-\pi^2)^{(x^3-\pi^3)^{\cdots}}}{(x^2-e^2)_{(x^3-e^3)_{\cdots}}})\righ‌​t$$

$$\left( x\frac{(x^2-\pi^2)^{(x^3-\pi^3)^{\cdots}}}{(x^2-e^2)_{(x^3-e^3)_{ \cdots}}} \right)$$

One way to interpret this is assume that the stuff in the argument is some manifold such as a sphere with a field of functions at each point. We can then describe this sphere with a metric. The result is that the smaller the symbols are, the further away they are and thus their value is weighted correspondingly by the metric

Hey everyone!

@Balarka @Fargle get on discord!

Hi Daminark

Ok

Realising that distributive law is the same as saying multiplication by some element is a homomorphiam on + is really handy in tracking how your structure is affected when you vary your axioms. In general. Homomorphisms are useful to said something about an algebraic structure. If you have isomorphism, you can easily translate one structure into a different form which might be easier to see what hapoened

The way I partially bypass this difficulty is to pool my research with others, so that as worldview should mix, new questions and answers pop up that push the project forward. While it might be true that some small fraction of stuff only I know, I can try to phrase it in the context of the existing literature so people can pick up on it quickly and generate their thought

Put it simply, I have a question asking algorithm that make people to engage in my projects and contribute to the research together, even if they have no background

It does not matter if they have no background, because it will be built over time

This is one reason I seemed to be learning all sorts of subjects, because I am actually obtaining the knowledge to translate one field of study into another

I am not satisfied with that, just as I am not satisfied with drug discovery research found a drug that cures (insert disease). For me, why is a more important question. If I don't understand why, then all I have there is just a bunch of uncorrelated facts

For the drug discovery case, I am more interested in its mechanism of action. The why is the thing that interested me more because it connects to the big picture

For closed forms, we have the integral project, which not only help us map the relatively uncharted territory of non elementary integrals, but to tell us the overall structure of it as much as possibke

Of course, without how and what, we cannot ask why. The what give us ideas on how to ask why, and the answer to why, if it exists, explains the what

@robjohn Not that bad. Working here pretty much. I discovered some new results lately. They are so fun!

hmm, I need to check whether there is an analogue of Jordan normal forms in tensors

So the interior is in some sense "far away" from its boundary...?

@Secret in the sense that there is always another point "between" the boundary and a point in the interior

or in the sense that it is possible to construct a (countably) infinite list of points from the interior to the boundary and never reaching it

Ah that makes a lot more sense

Btw I heard the Fong bong today, you doing ok?

If two subgroups $U$ and $V$ of a group $G$ are commensurable (i.e. their intersection has finite index in both $U$ and $V$), and $G$ is isomorphic to a group $G'$, does it follow that the images $U'$ and $V'$ are commensurable?

@abenthy Yes, isomorphisms preserve pretty much everything

But I read that $\mathbb{Z}$ is an arithmetic subgroup of $\mathbb{R}$ when considered as the group of matrices $\begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}$. But clearly, $\mathbb{Z}$ does not usually have finite index in $\mathbb{R}$.

@abenthy What does arithmetic mean here?

For a linear algebraic $\mathbb{Q}$-group $G \subseteq GL(n,\mathbb{C})$, a subgroup $\Gamma$ of $G \cap GL(n,\mathbb{Q})$ is an arithmetic subgroup if $\Gamma \cap GL(n,\mathbb{Z})$ has finite index in both $\Gamma$ and $G \cap GL(n,\mathbb{Z})$.

And how did the reals enter into this then?

$(\mathbb{R},+)$ can be identified with a subgroup of $GL(n,\mathbb{R})$ via $a \mapsto \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}$

sure, but being arithmetic required it to be a subgroup of the rational points

well, $\mathbb{Q}$-rational I should say

yeah, the arithmetic subgroup would be $\mathbb{Z}$ under the above identification

which is a subgroup of the group of rational points

@abenthy But none of the groups it should have finite index in can be the reals

(from math.lsa.umich.edu/~lji/a-tale-of-two-groups.pdf )

@abenthy That does not fit into the definition you gave

Oh, so he uses a more general one?

That could be

In your definition everything was inside the general linear group over the rationals

I'm trying to find some examples of arithmetic subgroups in the sense I defined them above.

So $(\mathbb{R},+)$ does not admit any in the sense I defined it?

@abenthy It makes no sense to ask for them since by definition they were subgroups of a smaller group than that

They have to be subgroups of $(\mathbb{Q},+)$, is that what you mean?

no, subgroups of $GL_n(\mathbb{Q})$

ah and you mean since $GL_2(\mathbb{Q})$ has infinite index in $GL_2(\mathbb{R})$, there cannot be any arithmetic subgroups of $(\mathbb{R},+)$ as embedded in $GL_2(\mathbb{R})$?

@abenthy No, I mean simply by the definition you gave, an arithmetic subgroup was a subgroup of $GL_n(\mathbb{Q})$

Heh, just realized that all three papers written by this guy have titles starting with "Simple transitive $2$-representations of"

if f is a holomorphic function on the strip { -1 < Im z < 1} such that limit x -> \infty (along real axis) f(x) = 0 , is it true that limit x -> \infty (along real axis) f(x + iy) = 0 for all y ?

Hey @Eric, @Steamy, and @Tobias!

@Daminark Hi

How's everything going?

Good. Finally getting back into doing some research

Nice

Though I am still also catching up on what has been put on the arXiv and organizing those papers so I know what to read and where to quickly find them

(I have a whole system set up for this that is fairly quick to keep up to date, but takes a bit of work when it has not been kept for a while)

Hmm, how does it go?

I have some menus in my top bar. One for papers I have yet to read and then a set of menus split into ranges of letters where I put all papers I might need to look up stuff in sorted by author

I see

I have to learn chemistry

:(

More organic?

Not quite. Physical chemistry, which is good news

Indeed.

sup

inf

@Daminark did Marianna prove to you that monotone functions are diff a.e.

limsup

sliminf

@Eric Yeah

coliminf?

did she use the rising sun lemma and all that

sure

So that was subsumed because we were like, BV functions are differentiable almost everywhere

wait hold on what?

there's this lemma that gets used to prove it some times

see im lecturing tmr and i need to show that lipschitz functions are diff a.e. but i dont have a lot of time

so i wanted to see if i could slim down the result

that's a theorem i always wanted to know the proof of

Do you not have compactness at your disposal?

I was thinking about this sum $\sum_{n=1}^{\infty}\sin(nx)$

@TobiasKildetoft kek

compactness of what

Domain

anyway my idea was to use Riesz Representation + Lebesgue differentiation bc they're big sticks and maybe i could go fast with them

differentiation is local so yes i do have compactness of the domain

Oh wait a sec you're not necessarily working with functions mapping from $\mathbb{R}$

I was thinking to use that Lipschitz functions are absolutely continuous

the space is nice enough

i don't wanna talk about absolutely continuous functions

for the purpose of geometric things we stick to Lipschitz cause it's basically like $C^{1}$ + bounded derivative

@Daminark ohi

actually that depends on $x$ though!

I see. Well, aside from our having proven (or was it just asserted?) that this was true for BV functions and just saying it followed

Proving it for monotone functions probably invokes Vitali

anyway so my idea was: use Lipschtiz -> define bdd linear functional on step functions -> use density to extend to $L^{2}$ -> bdd linear functionals on $L^{2}$ are given by integrating against an $L^{2}$ function say $f$ -> integrating indicator of interval against it shows that $F$ is given (locally) by the integral of some $L^{2}$ function -> lebesgue differentiation tells us that $F$ is differentiable a.e. (its an integral)

all differentiation theorems basicaslly invoke vitali or besicovitch

Oh good lord

no, for monotone functions you need vitali (I mean I invoked it by invoking Lebesgue differentiation), but the usual proofs are really long

@Steamy how's it going?

not that long but don't have much time

@Daminark Well, academic year is completely over now, so research research research :D

My "oh good lord" was to the chain

i never responded to the oh good lord

but it's not that long

there are 5 steps

I mean it just feels like a really roundabout way of doing things, like using Lebesgue differentiation and Riesz-representation (presumably the one that gets you Radon measures out of linear functionals on compact stuff?)

no the baby Riesz rep

the one for Hilbert spaces that's easy to prove

Oh that

@SteamyRoot Noice

it's not roundabout, the usual proofs of monotonic functions being diff a.e. is way longer

bc you don't need hilbert space stuff

It's rare to see this much spooky scary analysis in chat :O

Is it clear that $SL(2, \mathbb{Z}[\sqrt2])$ has finite index in $SL(2,\mathbb{Z})$?

Yeah true we need to bring balance. So, uh, what do you guys think of primitive groups?

@abenthy Do you mean the other way around?

@Daminark i meant to ask if you could see an error in what i wrote cuz it seems too quick

Sure!

bc to me riesz for hilbert spaces is like not a big hammer

@Daminark They need to develop some culture

@TobiasKildetoft Oh right... what I actually mean is, is it clear that $SL(2, \mathbb{Z}[\sqrt2])$ and $SL(2,\mathbb{Z})$ are commensurable as subgroups of $SL(2,\mathbb{Q})$?

Oh I mean yeah, I guess I assumed the other Riesz because my apartment-mate is doing rep/harmonic stuff for the REU and his book uses bigger Riesz to define Haar measure, so I've been hearing quite a bit about it recently

@TobiasKildetoft So yeah, the other way around.

@TobiasKildetoft Oh yikes

we need moar topology here

@abenthy But the first one is not a subgroup of $SL(2,\mathbb{Q})$