Mathematics - 2019-12-20 (page 1 of 2)

Lucas Henrique

00:00

You're right, although that's a very specific case. How can I generally solve this kind of problem?

The next problem is to find $[\mathbb{Q}(\sqrt{2}, \sqrt[3]{2}) : \mathbb{Q}]$, in which this argument wouldn't work.

I could use that $\mathbb{Q}[$number$] \cong \mathbb{Q}[X]/$(min poly of number), but the book didn't get there

Thorgott

Writing down minimal polynomials is a generally good approach. Arguments like the above can often help simplify things. So can embedding everything in $\mathbb{C}$ (in case you're dealing with extensions of $\mathbb{Q}$).

Lucas Henrique

Oh, there's this theorem that assures that $\exists \gamma: \,F[\alpha_1, \ldots, \alpha_n] \cong F[\gamma]$ using embbedings. So once you find such gamma, you're done (just get the first powers of gamma)

Thorgott

If you are referring to the primitive element theorem, that requires your extension to be finite and separable (the latter being an integral condition)

Also, finding primitive elements can, in general, get quite ugly

Lucas Henrique

I'm using S. Lang's book, and this comes literally after this theorem (he doesn't give a name)

Maybe he gave these problems just to apply the theorem.

Thorgott

00:16

A fairly nice result is that if $K$ is a field of characteristic $0$ and $K(\alpha,\beta)$ is a finite extension such that $K(\alpha)/K$ and $K(\beta)/K$ are Galois extensions, s.t. $K(\alpha)\cap K(\beta)=K$, then $K(\alpha,\beta)=K(\alpha+\beta)$.

which theorem in Lang?

Lucas Henrique

@Thorgott I'll learn about Galois extensions literally in the next page :)

@Thorgott it's in his Undergraduate Algebra; theorem 6 in the Field theory chapter.

Thorgott

Ah, I see. He circumvents separability for a bit by only discussing characteristic $0$. This is addressed in a Remark further down.

Ultradark

02:05

$(x,y) \mapsto (y,x)$

This is an inverse map

it reflects $(x,y)$ across the line $y=x$

Semiclassical

do you mean that it's a map which is its own inverse?

if so, the more typical name for that is an involution

(no idea why tbh)

Ultradark

what about $(x,y) \mapsto (-y,x) $

Leaky Nun

that has order 4

Ultradark

so it's a $4-$ involution?

Semiclassical

eh, that's not how the terminology works

all involution deals with is "is it its own inverse", not "if I act enough times, is it the identity map"

in group theory, you refer to that exponent as the "order" of a group element (hence Leaky's comment)

Ultradark

02:17

oh yeah

$f(x,y) \mapsto f(-y,x) $

Semiclassical

that's not a map.

well

not unless you consider it as "invert f to get (x,y) from f(x,y), then apply the rotation, then apply f again"

Ultradark

I guess what I'm trying to say is a function that maps all points $(x,y) \mapsto (-y,x)$

Semiclassical

that'd just be $f(x,y)=(-y,x)$

Ultradark

ah yes

Semiclassical

a fun variation on this is to consider the linear transformation with matrix $M=\begin{pmatrix} \cos\theta & \sin \theta \\ -\sin \theta & \cos \theta\end{pmatrix}$

If $\theta=\pi/2$, then that gives $(x,y)\mapsto (-y,x)$ as before

and if you consider $\theta=2\pi/n$ with odd $n$, then $M^n=I$

so you'd have a mapping of order $n$.

more generally, if $\theta = 2\pi p/q$ with $p,q$ in least terms, then it's a map of order $q$

best part, though, is when you replace $p/q$ with an irrational number

in that case, you'll never find an $n$ such that $M^n=I$. so in that case it'll have infinite order

Ultradark

02:27

that is a fun variation

Semiclassical

even better is this: If you write that mapping as $T$ (as in, what it does to x/y rather than writing it as a matrix)

then you can ask how much different $T^n(x,y)$ is from $(x,y)$ after $n$ steps

if you have finite order, then eventually the two are the same for a certain $n$

if you have infinite order, you can't do that. but you can do this: for any (x,y), you can always find $n$ large enough that the distance between $T^n(x,y)$ and $(x,y)$ is arbitrarily small

Ultradark

I don't undertand that notation

$f(x,y)=(-y,x)$

that's a function of two variables

Semiclassical

$f^{2}(x,y)=f(f(x,y)) = f(-y,x) = (x,y)$

is how I mean it

(could disambiguate that as $f^{\circ n}(x,y)$ I guess, i.e., the $n$-fold composition)

Ultradark

and the map $(x,y) \mapsto (-y,x)$ can be expressed as $M$

do you reckon $(x,y) \mapsto (\frac{1}{y}, \frac{1}{x})$ can be written as a matrix

this is an order two transformation (or an involution) as I've learned

Semiclassical

as $$\begin{pmatrix} x \\ y \end{pmatrix} \mapsto \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$

@Ultradark not as a matrix acting on the vector (x,y)

it's not a linear function of x,y, so no matrix will work

that said, it is a linear function in the variables $\ln x,\ln y$

Ultradark

02:36

so can yone write it as a matrix doing a little trickery with that knowledge?

Semiclassical

so you could write it as $$\begin{pmatrix} \ln x \\ \ln y\end{pmatrix} \mapsto \begin{pmatrix} 0 & -1 \\ -1 & 0\end{pmatrix} \begin{pmatrix} \ln x \\ \ln y\end{pmatrix}$$

Ultradark

and this transformation is not linear. Is it a dilation?

Semiclassical

it's not linear in $x,y$

it is linear in $\ln x,\ln y$

it probably doesn't look like anything nice, though

Ultradark

that's weird lol

Semiclassical

at least, not in (x,y) space

note, though, that plotting curves in terms of logs is actually a pretty common trick for scientists

it's called a log plot if you plot x vs log(y) for (x,y) data

Ultradark

02:39

so I could plot this on a log-log plot

Semiclassical

right. suppose you had some point set for which you made a log-log plot

if you then applied that mapping to those points, and plotted them, I think what that would do is inversion through the origin (here understood as the point $(\ln x,\ln y)=(0,0)$

Ultradark

what do you mean by "through the origin"

Semiclassical

let me get a picture

hmm, looks like my interpretation was wrong:

better

Ultradark

what if you plot log(x) vs log(y)

Semiclassical

that's a log-log plot of y=0.02x^(1/2) (blue line), and the orange line is what happens if you apply your map to all of the blue points

evidently, it flips it along the 45-degree line in that graph. that corresponds to the line log(y) = - log(x)

Ultradark

02:51

that's cool

Semiclassical

yeah

Ultradark

can I test it out with more functions

Semiclassical

what do you have in mind? log-log plots are mostly useful for power law relations like y=ax^r

since those come out as straight lines in log-log plots

Ultradark

is $r$ negative

Semiclassical

it can be, yeah

the slope of the 'line' you get in a log-log plot is basically just r

so it can be either positive or negative

Ultradark

02:55

so when you do the change of variables $x=\ln(x)$ and $y=\ln(y)$ the now, linear, map is a reflection

Semiclassical

right

well, you wouldn't write it like that

because it looks like you're doing equations not substitutions :P

in more colloquial terms, I think you'd say that the mapping $(x,y)\to (\ln x,\ln y)$ "linearizes" the mapping $(x,y)\to (1/y,1/x)$

that's really useful when the 'linearizing' mapping is simple and invertible

typically, though, you can't make things exactly linear

best you can do is some kind of approximation

so it's great when you can pull it off, but one is not always so fortunate

Ultradark

I will remember this useful map if I need to make something linear sometime in the future

Semiclassical

it's very problem-dependent, unfortunately

(there is theory about how one finds such mappings, but I think that falls under Lie theory and that's outside my realm of expertise)

Ultradark

mappings that linearize nonlinear maps?

Semiclassical

right

which is not surprising, really

it's not unlike the situation with integrals: it's easy to start with a simple integral and, by integrating by parts and substituting, get a complicated-looking one which is still solvable

what's hard is to look at two similar integrals and detect that one is easily solvable

Ultradark

03:12

can $(x,y)\to (1/y,1/x)$ be a projective map?

Ted E

@ShineOnYouCrazyDiamond Good proof +1

anakhro

Love that ted E

Shine On You Crazy Diamond

@Ultradark math.stackexchange.com/questions/3482435/…

Ultradark

good job that looks like a nice question

Shine On You Crazy Diamond

03:28

If $am^2 = bn^2$ where all vars are integers.

Let $a = a'x^2$ where $a'$ is square-free

$b = b'y^2$.

Then $a' = b'$ and $x^2 m^2 = y^2 n^2$

Then $\text{lcm} (a,b) = a'\text{lcm}(x,y)^2$

@Ultradark can you figure it out?

Lucas Henrique

0

Q: General way to determine $\mathbb{Q}(\gamma) = \mathbb{Q}(\alpha,\beta)$ given $\alpha$ and $\beta$

I'm currently reading S. Lang's "Undergraduate Algebra". After the primitive root element theorem (Field theory chapter), there are a bunch of exercises to find one primitive element of extensions and, then, their degrees. However, I don't even know how I should start. They are as below: I...

Shine On You Crazy Diamond

03:43

$z =7\cdot 2^2\cdot (3 \cdot 5)^2 = 7 \cdot 3^2 (2\cdot 5)^2 \implies z\in \text{lcm}(a,b) \gcd(x,y)^2$.

Shine On You Crazy Diamond

04:54

0

Q: Is $\Bbb{Z}$ compact under the evenly spaced integer topology?

The topology is generated by the basis $U_{a,b} = a \Bbb{Z} + b$. $\{U_{a,x}\}_{x = 1\dots a}$ covers $\Bbb{Z}$ for any $a \in \Bbb{Z}$ finitely since $x$ ranges over $1..a$. Does this somehow imply that any open cover of $\Bbb{Z}$ contains a finite subcover?

general-topology compactness ideals integers

@Ultradark

topology in action

Ultradark

@shi

nice

can you help me with a simple quotient space example

Shine On You Crazy Diamond

@Ultradark sure

Lucas Henrique

05:18

How can I find the minimal polynomial of $\sqrt{2} + \sqrt[3]{2}$ ?

Akiva Weinberger

05:40

> The concept 'computable' is in a certain definite sense 'absolute', while practically all other familiar metamathematical concepts (e.g. provable, definable, etc.) depend quite essentially on the system to which they are defined. - Gödel

(From here)

Alessandro Codenotti

@BalarkaSen I don't follow the details of the covering argument, but it makes sense, thanks!

Akiva Weinberger

06:09

Post-modernism is named after Emil Post

Martin Sleziak

@LucasHenrique There are a few posts on main that can help you. You could check Finding the minimal polynomial of $\sqrt 2 + \sqrt[3] 2$ over $\mathbb Q$. and other posts linked there

I found it using Approach0, for more tips on searching, see: How to search on this site?

Shine On You Crazy Diamond

07:16

0

Q: If $U_a = \{a n^2 - 1 : n \in \Bbb{Z}\}$ is a topological basis on $\Bbb{Z}$ then $U_a$ is clopen?

Let $P = $ the prime numbers in $\Bbb{N}$. Define $P^i = \{ \pm q_1 \cdots q_i : q_j \in P\}$. These sets $\{ P^j\}_{j\geq 0}$ with $P^0 := \{\pm 1\}, P^{-1} := \{0\}$ form a basis for a topological space on $\Bbb{Z}$ turning it into a topological monoid. The basis sets $P^i$ are both open and...

general-topology elementary-number-theory prime-numbers integers

Astyx

13:10

Hi chat

ÍgjøgnumMeg

13:52

Afternoon

Akiva Weinberger

What do you call it when you bake 256 cookies

Snack overflow

ÍgjøgnumMeg

heh

Secret

14:53

plato.stanford.edu/entries/identity

cannot even make "=" noncircular => philosophers are screwed

mick

Hi all

0

Q: $ b_n = \frac{\exp(\ln(b_{n-1})^3)} {b_{n-2}} $ Conjectured $\sup \inf$

My mentor tommy1729 told me : Let $b_1 = 1.$ Let $ b_2 = x.$ Also $ \frac{1}{2} < x < 2 $ Define for $ n>2 $: $$ b_n = \frac{\exp(\ln(b_{n-1})^3)}{b_{n-2}} $$ Then $$ \sup_{n>2} b_n = x , \inf_{n>2} b_n = \frac{1}{x} $$ Is this true ? How to show that ?

logarithms recurrence-relations inverse closed-form supremum-and-infimum

Lucas Henrique

15:44

Hi chat!

@MartinSleziak thank you! I checked this search engine yesterday but I wasn’t sure how I should use it.

Secret

16:12

the philosophy of the number one, is the smallest unit of identity

Secret

16:38

in Logic, 7 mins ago, by Secret

3

Q: Circularity in identity definition

It is written in the SEP: Identity is often said to be a relation each thing bears to itself and to no other thing (e.g., Zalabardo 2000). This characterization is clearly circular (“no other thing”)'. The question is: what is circular here? Deutsch, Harry and Garbacz, Pawel, "Relativ...

identity

but what is identity, I have no idea

Akiva Weinberger

16:55

This is why category theorists want to burn the idea of "equality" to the ground and only allow isomorphisms

or so I'm told

In axiomatic set theory, we call two sets "equal" if there's nothing in one that's not in the other. For this reason, $\{1,2,2\}$ and $\{1,2\}$ are considered equal: for every object, if it's in one, it's in the other

Thus, we can kinda argue that "equal" in the theory doesn't mean identical but means close enough

Identity of indiscernibles

The identity of indiscernibles is an ontological principle that states that there cannot be separate objects or entities that have all their properties in common. That is, entities x and y are identical if every predicate possessed by x is also possessed by y and vice versa; to suppose two things indiscernible is to suppose the same thing under two names. It states that no two distinct things (such as snowflakes) can be exactly alike, but this is intended as a metaphysical principle rather than one of natural science. A related principle is the indiscernibility of identicals, discussed below. A...

@Secret You're familiar with this^ I'm guessing

I like their Clark Kent example

Similarly

If I write an article and don't sign my name, the article's author is anonymous

If I write a second article and sign my name, that article's author is not anonymous

But I am the author of both articles. Am I anonymous or not?

Are "the first article's author" and "the second article's author" not identical, because one is anonymous and the other isn't?

@Secret Here's a screenshot from Jeff Week's Curved Spaces program

We can either interpret this as a triply periodic universe shaped like Euclidean space, or a compact universe shaped like the 3-torus

Are the Earths identical?

It depends on the interpretation, but there's no experiment that can determine which interpretation is correct, which means that both interpretations represent the same theory

Secret

yeah, if there is no indication that you are the author (including choice of words, handwriting etc.), there seemed to be no way to tell identical. The Jeff Week's case is even worse, because one of the most crazy interpretation is it is the same earth despite it looks like it is a copy at different spacetime locations

Akiva Weinberger

17:11

The only conclusion I can get is, since the concept of "identical" isn't determined by the theory, it's not a real property of the universe, only our understanding of it

Incidentally

It's been years since I've read that but I remember it being good

@Secret There's also a neat linguistic thing with how we use adjectives

Naïvely, we'd model them as conjunction: "This is a red ball" = "This is red and this is a ball"

But consider the example "He's a good person but a bad pilot"

The model fails ("He's good and a person and bad and a pilot?")

Secret

well, the "but" will probably mean you cannot resolve that as "and"

at least not directly

Akiva Weinberger

I don't actually quite understand the semantics of "but"

> introducing a clause contrary to prior belief or in contrast with the preceding clause or sentence

Is that semantics or pragmatics

(TODO: learn what pragmatics is)

Secret

"but" is supposed to convey some notion of contrast, so like the first half is related to a notion of desirable, and the latter half relates to a notion of undesirable, even though all 4 clauses are true description of the man

Akiva Weinberger

> Pragmatics is a subfield of linguistics and semiotics that studies the ways in which context contributes to meaning.

In any case I can say "He's a good person and a bad pilot" and still make sense

Entirely unrelated, but there's a sense in which the sentence "The boy threw the ball" is ungrammatical: it makes no sense unless we've established what boy and ball we're talking about (they have to already be "common knowledge" between us)

On the other hand, "A boy threw a ball" is grammatical whether or not we have a boy and ball in our common knowledge

Similarly, the sentence "I did." I can't start a conversation with that, so it's essentially "ungrammatical" in that context

We need to have established a verb

Correction: I think the technical term is "common ground" not "common knowledge"

Hold on let me watch this video

How Do We Create a Shared World in Conversation? Common Ground

►

thelingspace.com/episode-58

I don't think that's the right one; I vaguely remember one of their episodes going more into this topic

You know, I've heard lots of explanations of the Coriolis effect

I've never had it explained to me why the centrifugal and Coriolis forces are the only fictitious forces you get in a rotating reference frame

Secret

17:54

For your last question, it is because the centrifugal force is the axial component and the Coriolis force is the tangential component and a rotating frame is basically a 2D coordinate system which you can resolve any force into two orthogonal components in that system

Akiva Weinberger

Doesn't Coriolis depend on your velocity and centrifugal depend on your position?

Kinda like if friction pushed you to the side rather than backwards

Semiclassical

I remain disappointed that this recent SMBV strip didn’t go full Nietzche

smbc-comics.com/comic/santa

“Santa is dead, and we have killed him.”

Akiva Weinberger

18:11

Wat

geometrygames.org/Maze4D/index.html

Jeff Weeks made a 4D maze

@Semiclassical Santa is in a box with a Geiger counter and poison

(The box, naturally, has festive wrapping paper around it and a bow)

Also, nice hat

Semiclassical

Thx

Akiva Weinberger

Fractal zoom

Cohomology fractal zoom

►

Explanation

Cohomology fractals

►

Akiva Weinberger

18:29

More explanation:

icerm.brown.edu/video_archive/?play=2037

@BalarkaSen

Balarka Sen

@AkivaWeinberger Thanks! Watching

Akiva Weinberger

In the half-space model, the points at infinity (the ideal points) form a plane, except for one. What's the name of that point?

I guess "ideal point at infinity" works

Balarka Sen

I just think of it as a sphere at infinity

Akiva Weinberger

It's basically saying it's at infinity at infinity

Balarka Sen

I don't think you assign any name to a specific point on the sphere

Akiva Weinberger

18:43

@BalarkaSen Yeah but in the half-space model

Balarka Sen

The ideal boundary is still a sphere

Akiva Weinberger

When you choose the model, one of the ideal points becomes special

Balarka Sen

They become special in the sense it's not embedded as the metric closure of the half-space in $\Bbb R^3$ anymore. So what

No geometric speciality is there

Akiva Weinberger

Yes but does it have a name

Balarka Sen

Shrug. Who cares

Feels like it shouldnt have

Akiva Weinberger

18:45

People who are thinking of the model as a function $f:\Bbb H^3\to\Bbb R^3_+$ that extends to a function $\bar f:\bar{\Bbb H^3}\to\bar{\Bbb R^3_+}$ and want to think about $\bar f^{-1}(\infty)$

Shootforthemoon

Hi, I quite do not get how we could rewrite $F(x,y,z)=0$ as $z=f(x,y)$ if $F= yz^2+7+4xz-3x^2y$. I mean, we could use the quadratic formula, but we would get two distinct roots, and if z is plus/minus something, this is no more a function.

Anyone can help?

Balarka Sen

Why is that any different from thinking about $\overline{f}^{-1}(p)$ for any point $p$ on the ideal boundary lol

Akiva Weinberger

$p$ isn't a point on the ideal boundary

It's a Euclidean point

Type mismatch error

Balarka Sen

Oh your function maps to R^3_+

Why would you do that seems like a random thing to do

Akiva Weinberger

That's literally what the half-space model is

Balarka Sen

18:47

Which makes sense if "people" means singleton set Akiva

Akiva Weinberger

@Shootforthemoon That's not a function in $x$ and $y$. If you graph it, you'll see it fails the vertical line test

Balarka Sen

The half space model is a metric on R^3_+ which is isometric to H^3. It doesn't make sense to compare the metric boundary and the ideal boundary to me

Akiva Weinberger

It's similar to if you wanted to rewrite $x^2+y^2=1$ (whose graph is a circle) as $y={}$. You can write $y=\pm\sqrt{1-x^2}$ but there's no single formula because the circle fails the vertical line test

Shootforthemoon

@AkivaWeinberger Ah, thanks what a relief

Akiva Weinberger

@BalarkaSen What if you want to draw it

Thorgott

18:50

You should be able to do it locally though

Akiva Weinberger

You'll want to think about how the hyperbolic elements interact with the Euclidean ones

Like, "in the Poincare model, hyperbolic circles are Euclidean circles. In the Beltrami–Klein model, they are ellipses"

That sort of statement

Balarka Sen

Fair enough

Akiva Weinberger

Not all properties are intrinsic properties

Balarka Sen

You might be interested in something I was thinking about here

and subsequent possible explanation here

About how geometry of the ideal boundary is inherently Euclidean in some sense

The natural "perception measures" on the ideal boundary are a family of Lebesgue measures

Akiva Weinberger

Hm. I feel like there shouldn't be an intrinsic measure

If you choose a basepoint and then centrally project it onto a sphere around the basepoint, you get a measure

but it depends on the basepoint

Balarka Sen

18:55

Yes, so you get a family of measures

And their consistency is given by the Radon-Nikodym derivatives; they are all absolutely continuous wrt each other

Akiva Weinberger

What's "absolutely continuous"

Balarka Sen

For measures $\mu$ and $\nu$ are a.c. wrt each other if $\mu = \int f d\nu$ for some $\nu$-measurable function $f$. It's like having a pdf

That's the origin of the term "absolutely continuous random variables", those which have a pdf

This is one of the many equivalent criterion for a.c.ness (assuming $\sigma$-finiteness and all)

Thorgott

That only implies $\mu$ is a.c. wrt to $\nu$, not the other way round, no?

Balarka Sen

Yeah I meant to say that

Ultradark

19:11

how do you do a one point compactification of a hyperbola

Balarka Sen

@Thorgott I saw your question about symmetric groups. The answer is not true, $Q_8$ is not a subgroup of $S_7$ even though $8$ divides $7!$

Akiva Weinberger

@Ultradark $\infty$

Ultradark

is it just the same topologically as a one point compactification of the real line

Akiva Weinberger

No

It's the shape of $\infty$

A lemniscate

Ultradark

oh lol

Akiva Weinberger

19:17

@BalarkaSen Finally beat that maze game

The hyperbolic orb thing

Those realms were massive

Balarka Sen

The boundary strategy is surprisingly useful to me

Yeah

Akiva Weinberger

Yeah

Thorgott

@BalarkaSen But $S_7$ does have a subgroup of order $8$, namely an isomorphic copy of $D_4$.

Akiva Weinberger

I had bread crumbs on - it was surprising how rarely I saw them

Thorgott

Turns out the statement isn't true either way though. $S_5$ has no subgroup of order $30$, it seems.

Balarka Sen

19:17

Oh, I misunderstood the question.

Akiva Weinberger

(like how rarely I ended up where I was before)

> If one prefers, the maze lives on a compact Riemann surface of genus 8812 (having 352440 symmetries)

Wow

Balarka Sen

@Akiva I was able to draw essential closed curves using the bread crumbs

I could prove it's essential by Gauss-Bonnet

Tesseract

@akiva What maze game?

Akiva Weinberger

madore.org/~david/math/hyperbolic-maze-2.html

Hyperbolic maze

Thorgott

I have not yet figured out why $S_5$ has no subgroup of order $30$, however

Balarka Sen

19:19

Here

@Thorgott $A_5$ has no index $2$ subgroup, I feel, is the way to go

Tesseract

@Akiva Thank you!

Balarka Sen

I mean if $H$ is a subgroup of order $30$ in $S_5$ then $H \cap A_5$ is either of order $15$ or $30$ in $A_5$.

30 is ruled out

Why is 15 not possible? Let's see

Akiva Weinberger

There's also this one on the same site which is mocking me with its info on the side

Balarka Sen

Oh, groups of order 15 are cyclic, that's why

Akiva Weinberger

@BalarkaSen Hard mode: play it on Beltrami–Klein

Balarka Sen

19:25

I am really unfamiliar with that model

It looks huge in B-K

Akiva Weinberger

You can barely see anything

Balarka Sen

What's the deal with Beltrami-Klein? I don't really know what the precise relation between the geometry of $\Bbb H^2$ and that is

It's like some stereographic projection, maybe?

Akiva Weinberger

@BalarkaSen You can conformally map H2 onto a hemisphere

Balarka Sen

Ah, it's conformal, right

Makes sense

Akiva Weinberger

To go from that to Poincaré, imagine it's the southern hemisphere and stereographically project from the north pole

To go from that to Beltrami–Klein, orthographically project downwards

So no it's not conformal

Balarka Sen

19:29

Huh

What do the geodesics look like once you conformally project to hemisphere

Akiva Weinberger

Slice it with vertical planes

Balarka Sen

Also of course it's not conformal because angles work like Euclidean geometry in B-K

Akiva Weinberger

so they're perpendicular to the boundary

Balarka Sen

in particular sum of angles of a triangle is still 180 degrees

Akiva Weinberger

Mhm

Balarka Sen

19:31

@AkivaWeinberger Aight

Akiva Weinberger

This is why geodesics in Poincaré are arcs perpendicular to the boundary, and in Beltrami–Klein are straight lines

Thorgott

Why can the order of $H\cap A_5$ not be smaller than $15$?

Balarka Sen

@Thorgott This is a half-lives-half-dies fact. For every subgroup of $S_n$, if you intersect that with $A_n$, the order either remains the same or becomes halved

It's a good exercise

Akiva Weinberger

A whatnow

Balarka Sen

Essentially 2nd isomorphism theorem and index of $A_n$ in $S_n$ is $2$

@Akiva cool name, right?

Akiva Weinberger

19:33

Perfectly balanced

Very snappy

Shine On You Crazy Diamond

19:50

@Ul

Shootforthemoon

I'm struggling to find an example showing the implicit function theorem is not a double implication; anyone can help?

Balarka Sen

@Akiva Very cool how the simplest example of a Cannon-Thurston map arises out of lifting the map $T^2 \setminus pt \to S^3 \setminus K$ embedding a Seifert surface in the trefoil EDIT figure eight complement

Akiva Weinberger

The trefoil isn't a hyperbolic knot

You'd probably want the figure-eight knot

Balarka Sen

Yeah that

Akiva Weinberger

(There's no uniform hyperbolic metric on its complement)

Balarka Sen

19:53

Trefoil is a torus knot so has an essential annulus

So can't admit a negatively curved metric

Akiva Weinberger

What's an essential annulus

Balarka Sen

If you look at the torus the trefoil is sitting on, then remove the trefoil from the torus, you can an annulus

Akiva Weinberger

Ah

I was gonna say "two annuluses" (-i) but yeah it's just one going around twice isn't it

Balarka Sen

yeah u cant make a torus disconnected by removing one curve

betti number problems

Akiva Weinberger

Oh. Right. Genus

or that

Balarka Sen

19:57

Same difference yeah

Akiva Weinberger

I'd ask why that means you can't give it a hyperbolic metric but I feel like it might be in do Carmo and I forgot

Is the problem that it has 0 total curvature?

Balarka Sen

No I don't think it's trivial, but I'd be very interested if you can give me an easy proof

Akiva Weinberger

Wait that's not enough

Yeah I dunno I'll review do Carmo

Wait no I left it in New Haven!

(I'm back home for winter break)

I'll think about it

I don't actually think it covered cusped manifolds now that I think about it

Transcript for

$Mathematics$ Mathematics