Mathematics - 2018-08-01 (page 1 of 2)

But also it's fundamental insofar as you have two things, area and linearization, which you don't think should have anything to do with each other, and yet they do

user2236

5:59 AM

Leaky Nun

u w0t m8

isn't this how the left hand side is defined

Rudi_Birnbaum

7:22 AM

Hi all there (good morning)!

Tobias Kildetoft

@Rudi_Birnbaum Hi

Rudi_Birnbaum

The number of primes less then $n$ is $\pi(n) = \sum_{j=2}^{n} \frac{sin^2 \pi \frac{((j-1)!)^2}{j}}{sin^2 \frac{\pi}{j} }$. I cannot see how exactly. But I see that $\frac{(j-1)!}{j}$ is integer iff $j$ is non-prime.

Alessandro Codenotti

7:43 AM

@Leaky Differentiation is continuous if you use the usual norms (the one mentioned by @Dami)

Leaky Nun

@AlessandroCodenotti perche?

Alessandro Codenotti

It is bounded (by $1$)

Leaky Nun

oh wait, I was using the wrong norm

I used the subspace topology

if I use the subspace topology for $C^1$ then the map $C^1 \to C^0$ isn't continuous right

Alessandro Codenotti

Nope

It's discontinuous even with the $||\cdot||_{\infty}$ norm on both $C^1$ and $C^0$

(Which is the same as the inherited one, derp)

Leaky Nun

lol

mercio

7:52 AM

@Rudi_Birnbaum if $j$ is prime you can pair every factor in a $(j-1)!$ with its inverse mod $j$ in the other $(j-1)!$ so that the whole $(j-1)!^2$ is $1$ mod $j$, and if $j = ab$ then picking $a$ in $(j-1)!$ and $b$ in the other you get that $(j-1)!^2 = 0$ mod $j$

Daminark

Oh right I'm dumb lmao

user1732

in Homotopy Theory, Jun 28 at 21:08, by Eric Peterson

@skd the whole mathematical community would feel better, as individuals & collectively, if everyone stopped calling him- or herself an idiot / ridiculously stupid / etc

Rudi_Birnbaum

@mercio Ah ok! $(j-1)!^2$ is either $1$ mod j for $j$ prime or $0$ for $j$ composite. Ah and then we have $\frac{sin^2 \pi \frac{1}{j}}{sin^2 \pi \frac{1}{j}}$ for prime $j$ ... Thanks!!

mercio

it's not a very useful formula

Rudi_Birnbaum

Since you have to calculate a lot more then only factorize??

Or why do you think so?

Leaky Nun

8:19 AM

@Daminark what?

Daminark

Regarding continuity wrt my norm

Leaky Nun

why do you add the derivatives together in your norm?

Daminark

It's the usual norm I've seen on that space since it keeps track of all the derivatives

Leaky Nun

what's the topology on $C^\infty$?

Daminark

Hmm, probably that all derivatives converge uniformly? That or I dunno

Leaky Nun

8:22 AM

I've heard that a set is closed in $C^\infty$ iff it is closed in $C^n$ for all $n$

Daminark

Oh yeah I guess you could play that game

MatheinBoulomenos

9:04 AM

@LeakyNun $C^\infty = \cap_k C^k = \varprojlim C^k$, take the limit topology

Leaky Nun

ok

MatheinBoulomenos

(this is the same as what you said)

Leaky Nun

why closed sets though

Iza_lazet

9:33 AM

one can define topology either by open sets or by closed sets

is there a problem there?

Rudi_Birnbaum

Just had a look at some climate change 'discussion' and wonder about to which degree 'trust' plays a role in learning science. Maths would be a nice 'baby model', since you can get about anyone to agree that maths is 'correct'. So my basic hypothesis is you cannot learn anything without trust in a common bases.

Leaky Nun

> you can get about anyone to agree that maths is 'correct'

you'll be surprised

Iza_lazet

well, how do you know mount Everest exists?

Rudi_Birnbaum

@leaky I know there are counterexamples or open questions

Iza_lazet

Have you ever been there?

Leaky Nun

9:37 AM

I haven't

and I don't know, with 100% absolute certainty, that mount Everest exists

Rudi_Birnbaum

@Iza_lazet exactly.

Tobias Kildetoft

@Rudi_Birnbaum Given the number of people who still propose to trisect angles, there are certainly those who do not agree with math

Iza_lazet

ah, but that is the darnedest thing. "The trouble with the world is that the stupid are cocksure and the intelligent are full of doubt."

Rudi_Birnbaum

@LeakyNun Is there anything in maths you consider you know with 100% certainty?

Leaky Nun

no

everything is under axioms

Daminark

9:39 AM

What does "know" mean in math?

Rudi_Birnbaum

@LeakyNun But then what about the conditional "under the axioms"? Is that conditionally 100% true?

Daminark

Do you know that X is true from the axioms? Well, you do. The axioms define something, and you know about it. Do the axioms define something empty? If they can hold arithmetic, you don't know

Iza_lazet

"know" means something can be derived from commonly accepted axioms and accepted logical deduction methods. The keyword is "accepted."

Rudi_Birnbaum

@Daminark But then how do you know that you conclusions were all correct?

Daminark

Well, everything's correct about the empty model, and if you've proven something you've proven it

Are you asking whether the "axioms are correct"?

Alessandro Codenotti

9:42 AM

@Daminark That works even if they can interpret PA, if the axioms prove X you know that X follows from the axioms

Rudi_Birnbaum

@Daminark No I don't.

MatheinBoulomenos

remember guys, Galois theory is just a theory

Alessandro Codenotti

What you don't know is whether everything follows from those axioms

Daminark

That is a much trickier and more philosophical question. If there's some idea (the Platonic form of X, perhaps) and you're wondering whether the axioms truly capture that idea... Honestly that's not something I've ever thought about, how certain you are about such matters

Rudi_Birnbaum

But there is that princple of the excluded third and still more fundamental stuff like that.

Alessandro Codenotti

9:43 AM

@MatheinBoulomenos Fun fact: you can spot the algebraists at a conference because of the tinfoil hats

MatheinBoulomenos

@AlessandroCodenotti what

Leaky Nun

so the question is whether we "know" that "given [logical axioms], [deduction rules], [assumptions], [conclusion] is true"?

Rudi_Birnbaum

Sounds funny, how comes @AlessandroCodenotti

Tobias Kildetoft

@AlessandroCodenotti You mean my tinfoil hat can also spot algebraists?

Daminark

I kinda feel you "know" that in the same way you "know" that that truth values behave as deductive reasoning suggests. Of course in principle there could be an evil demon making you think that X is true => not X is false

Alessandro Codenotti

9:44 AM

Don't you know about the conspiracy theory of the government trying to control minds and people wearing tinfoil hats to avoid the influence of the mind controlling ray?

MatheinBoulomenos

how is that related to algebra?

Daminark

@Alessandro more what I was going for earlier was whether or not Peano was consistent

Rudi_Birnbaum

@LeakyNun Yes one of the questions.

@AlessandroCodenotti "how is that related to algebra?" yes i also wonder.

Daminark

"X is just a theory" is talk that's often associated with groups of people which overlap very significantly with conspiracy theorists

And conspiracy theorists are jokingly identified by tinfoil hats

Alessandro Codenotti

^

MatheinBoulomenos

9:47 AM

oh I thought it was associated with hardcore Christians who talk about evolution

Leaky Nun

same here ^

Daminark

Well... they often intersect pretty reasonably (or so the stereotype says) with conspiracy theorists

But yeah so finishing up my thought earlier, course there's the practical question of being literally unable to answer those questions (does logic even check out? does $d(x,x_0) = c$ truly capture the idea of a circle?) and there being only one real "actionable" answer

MatheinBoulomenos

personally I'm an ultrainfinitist, I don't think sets of cardinality smaller than $\aleph_{17}$ exist

Daminark

Gotta let that go and say "You know deep in your soul that LEM is true and the constructivists are just trying to take away your contradictions"

Alessandro Codenotti

Are there two schools of ultrainfinitism split on whether $\Bbb R$ exists?

Rudi_Birnbaum

9:50 AM

@Daminark thats why i think its necessary to trust on a common basis.

MatheinBoulomenos

@AlessandroCodenotti well yeah it depends on CH of course

Rudi_Birnbaum

Without that you cannot learn.

Daminark

"$2^{2^{\mathbb{R}}}$ is the smallest set"

MatheinBoulomenos

I also think you need a basis

Alessandro Codenotti

Well anyway as I said before we should clearly have $2^{\aleph_0}=\aleph_{46}$ so that's fine

MatheinBoulomenos

9:52 AM

every vector space needs a basis, therefore we need AC (which implies LEM)

Leaky Nun

hi @loch

Rudi_Birnbaum

Exactly, and here I see some larger problem in modern societies.

When people start on the basis of mistrust to read any kind of stuff, nothing good can come out. All sorts of people comment on all sorts of things and have their "own opinion" completely ignoring large bodies of human knowledge.

Daminark

Yeah I mean, mistrust is a bit of a killer, even aside from all the abstract stuff we have, it means you're not able to black box anyone else's work, so as far as pursuit of knowledge goes you're starting from scratch

Or you find other people who are less reliable that you do trust and end up believing incorrect things

I'm not sure exactly where public mistrust of science comes from. I think part of it is public belief that science is under political influence from folk who don't have the truth/the needs of the people in mind. In principle that's fair, though it becomes a question as to whether the probability is higher that there's some elaborate set up or if it's genuine

(I say "in principle that's fair" not as a defense so much as, government probably has many reasons to lie, and if they were to see it as being worth the effort I could very much see various governments try to manipulate science like so, much as in this case it's probably wrong)

Rudi_Birnbaum

@Daminark: Yes sure. I think its important to make clear what the difference between a healthy "scientific scepticism" and this kind of "distrust" is.

Daminark

Anyway, returning to math :P

Rudi_Birnbaum

10:03 AM

Ok

Daminark

(Oh I just meant I'll be returning to math with this next comment, you can continue discussing philosophy and whatnot, I just had nothing more to say :P)

@Mathein one thing that happens to me a bunch with Neukirch is that he often suppresses a kinda simple argument that takes me so long to see, but once I do it's just like, okay I kinda see why you didn't say anything

Leaky Nun

@Daminark so why?

Daminark

So why... did he not say anything? I mean it ends up being the case that the argument is simple/not too different from previous types of arguments so there's no need to do it out. If anything it's good practice for the reader to be like, fam wut?... oic

user1732

10:37 AM

that takes "concise" to a whole new level

(proof left to the reader)

Leaky Nun

10:52 AM

when can a continuous function $A \to C$ be extended to a continuous function $B \to C$ if $A$ is a subspace of $B$?

Iza_lazet

Regarding Neukirch, mathematics requires a person to be an active learner. There are a million things that can not be fully explained, but only worked out. It is frustrating sometimes, but overcoming the challenges and connecting the dots feels worthwhile. It's like playing Dark Souls, if people here understand that reference.

@LeakyNun: Either use Tietze theorem or when the function is Lipschitz (in metric spaces)

Leaky Nun

@Iza_lazet can you elaborate on your last point?

Iza_lazet

or for a trivial case: in functional analysis , when A is a dense vector subspace of B and the function is linearly bounded function between Banach spaces

Leaky Nun

but thanks for your first point

Iza_lazet

Tietze extension theorem

In topology, the Tietze extension theorem (also known as the Tietze–Urysohn–Brouwer extension theorem) states that continuous functions on a closed subset of a normal topological space can be extended to the entire space, preserving boundedness if necessary. == Formal statement == If X is a normal topological space and f : A → R {\displaystyle f:A\to \mathbb {R} } is a continuous map from a closed subset A of X into the real numbers carrying the standard topology, then there exists a continuous map ...

see variations

Leaky Nun

11:00 AM

wow

Iza_lazet

we can define the extension explicitly in the Lipschitz case, though the formula is kinda magical

Leaky Nun

for the Lipschitz one the space doesn't even need to be clsoed?

Iza_lazet

yupe. Magical formula exists.

Leaky Nun

what is the formula?

Mats Granvik

11:25 AM

"Beware of geeks bearing formulas." Warren Buffet.

user2236

11:36 AM

"Beware of buying Facebook stock."

Rudi_Birnbaum

"Beware of geeks bewaring of geeks bearing formulas." R. B.

12:05 PM

Are the implicit function theorem and the inverse function theorem equivalent? I know that the implicit function theorem follows from the inverse function theorem but does it work the other way round?

Alessandro Codenotti

@user2236 Is it an issue on my end or is the audio problematic for everyone?

MatheinBoulomenos

@Alessandro it's problematic for me as well

user2236

It is a problem for me also.

Mike Miller

m.imgur.com/a/OxBYJEt

user2236

Star studded line up.

Leaky Nun

12:25 PM

es ist auch ein Problem per me

Iza_lazet

@JannikPitt: they are equivalent

I'd call it the fundamental theorem of differential geometry

TT Farreo

2/3

Leaky Nun

audio is horrible

Alessandro Codenotti

Yeah I didn't quite get the name for the Gauss medal

Iza_lazet

12:46 PM

the video is getting pretty cheesy

loch

birkar got fields medal?

MatheinBoulomenos

yes

Iza_lazet

who won?

loch

wow i see - i just woke up :p

Alessandro Codenotti

@Iza_lazet While the audio is getting worse

Mike Miller

12:48 PM

Alessio Figali, Akshay Venkatesh, Peter Scholze, and Caucher Birkar

3

loch

i can check "attending a class from a fields medalist" off my list :p

Alessandro Codenotti

@loch You have to attend it after the medal or it doesn't count

loch

i can check "attending a class from a guy who went on to become a fields medalist" off my list

Alessandro Codenotti

The closest I got was attending a class from a student of a student of a fields medalist :P

loch

it'll probably be easier when you get to Bonn

Alessandro Codenotti

12:52 PM

Very likely

abenthy

I have this question which I still haven't been able to solve. If $A,B \subset M$ are submanifolds with $A$ compact and $B$ closed (as a subset of $M$), does $A \cap B$ have only finitely many path connected components?

(If $A \cap B$ is locally path connected, then the path connected components are open and compactness of $A \cap B$ implies that it must have finitely of them)

Alessandro Codenotti

Figalli is the second Italian to ever win a fields medal I think

abenthy

1:10 PM

If someone has ideas or speculations about the question, they are highly appreciated :)

user2236

We are watching this

1 hour ago, by user2236

ICM2018

►

quallenjäger

who won fileds?

figali?

abenthy

Oh interesting, I didn't realize its still running. Thanks user2236.

user2236

26 mins ago, by Mike Miller

Alessio Figali, Akshay Venkatesh, Peter Scholze, and Caucher Birkar

quallenjäger

We have 4 winners this year?

user2236

1:15 PM

yep

Leaky Nun

@abenthy I think $A \cap B$ is compact as well, being a closed subspace of a compact Hausdorff space?

user2236

"In math we tend to be obsessive"

abenthy

@LeakyNun Yes correct, but that does not directly imply that it has only finitely many path connected components. For example $\{0,1,\frac{1}{2},\frac{1}{3},\frac{1}{4}\ldots\} \subset \mathbb{R}$ is compact, Hausdorff and has infinitely many path connected components.

Leaky Nun

I see

Arrow

Hello. Given a field $\Bbbk$, a group $G$, and actions of $G$ on finite sets $X,Y$, I am trying to find a basis for $\mathrm{Hom}_{\Bbbk[G]}(\Bbbk[X],\Bbbk[Y])$. I was hinted this set should be in bijection with the set of orbits of the product action of $G$ on $X\times Y$. I am a bit lost.

abenthy

1:24 PM

Sadly the only condition for a compact space $X$ to guarantee that $X$ has only finitely many path connected components that I know of is $X$ being locally path connected. But I actually doubt that the intersection $A \cap B$ has this property.

Leaky Nun

@abenthy $A = \{0,1,\frac12,\frac13,\cdots\}$, $B = M = \Bbb R$

abenthy

But is $A$ really a compact submanifold of $\mathbb{R}$?

Leaky Nun

what is a submanifold?

Mike Miller

Take the north and southern hemisphere of $S^2$, and wiggle the north one so that it is bounded by a slightly different Jordan curve. You can picture the Jordan curves as in the plane, one as the unit circle. (cont)

For the second one, take what looks like a sine function going around the circle, with increasing frequency but lower amplitude. If the amplitude dampens quickly enough this is a smooth curve.

Leaky Nun

@Arrow why should a basis exist?

Arrow

1:29 PM

@LeakyNun I mean a basis for the underlying $\Bbbk$-linear space, which is finite dimensional as a subspace of a hom-space between finite dimensional vector spaces. ($X,Y$ are assumed finite.)

abenthy

@LeakyNun I mean it in the sense of embedded submanifold, i.e. $A$ is a manifold (secound countable, locally euclidean, Hausdorff space) and $A \hookrightarrow \mathbb{R}$ is a topological embedding.

Mike Miller

The point is that this dips above and below the unit circle infinitely many times. The places between intersections with the unit circle below the unit circle are your intersection. If you did the above so that there are infinitely many crossings, you will get infinitely many components.

loch

@Arrow i think you can find a basis indexed by $G$-equivariant maps $X\rightarrow Y$ ?

Arrow

@loch the clue about the orbits of the product action make that seem a bit unlikely since the product action is symmetric w.r.t $X,Y$.

Mike Miller

If you're worried about smoothness, I will remind you that one can come up with smooth maps with arbitrary closed sets as zeroes, and the graph of any smooth function is an embedding. The 'second Jordan curve' can be thought of as the graph of a map $S^1 \to (1-\epsilon, 1+\epsilon)$, the codomain being radius.

Leaky Nun

1:31 PM

I think a $k[G]$-linear map $k[X] \to k[Y]$ is determined by the image of $1$

abenthy

Wow, so you mean its even false if I require that $M$, $A$ and $B$ are smooth Mike?

Mike Miller

Yeah

abenthy

Thats a nice example with the two hemispheres of $S^2$!

loch

@LeakyNun that doesn't seem true

Mike Miller

If the boundaries don't intersect (or intersect transversely) I think you get finitely many path components

Arrow

1:32 PM

@LeakyNun what do you mean by $1$ in $\Bbbk[X]$?

Mike Miller

The argument being that the intersection is a compact manifold with boundary or corners or something

abenthy

What do you mean by boundary? (I have implicitely assumed that $M$, $A$ and $B$ are manifolds without boundary.)

loch

@Arrow ah that's true - so by your hint it's also suggesting that the dimension is the same as $Hom_{k[G]}(k[Y],k[X])$..

but now if you take $X,Y$ to be finite sets of different sizes with the trivial $G$-action, then i dont think the claim is true as stated

Arrow

@loch Even disregarding the hint, I don't see how to index a basis via $G$-equivariant maps as you suggested.

Mike Miller

@abenthy I must misunderstand your question. I thought these were codimemsion 0 submanifolds, which must have boundary if they are closed but not the whole connected component.

Arrow

1:44 PM

@loch For what it's worth, the next part of the exercise is to prove an action of $G$ on a finite set $X$ is 2-transitive iff $\mathbb C[X]$ is the direct sum of two simple representations. Clueless here as well.

Mike Miller

But I guess you can take my example about graphs of functions from the circle to get two circles in the plane that intersect at countably many points.

quallenjäger

What is an arc again? an injective curve?

abenthy

The question I had in mind is that if $M$ is a smooth manifold (without boundary) and $A,B \subset M$ are embedded smooth submanifolds (also without boundary) and $A$ is compact and $B$ closed as a subspace. Then can the topological space $A \cap B$ have infinitely many path connected components?

Mike Miller

Yes, I just didn't have the right picture for your submanifolds. The circle example works.

abenthy

Sorry, I just added the "compact" and "closed as subspace" conditions

So I take $M = S^1$ ?

Mike Miller

1:47 PM

The point being that any closed subset of $M$ can arise as the zero set of some function $M \to \Bbb R$.

abenthy

But will it also be an embedded submanifold of $M$?

These seem more well behaved to me.

Mike Miller

No, you want $M = S^1 \times \Bbb R$.

loch

@Arrow actually i dont think what i said there about your hint makes sense - of course if $G$ acts trivially then those two vector spaces ($Hom_k(k^n,k^m)$ and $Hom_k(k^m,k^n)$ have the same dimension lol

Mike Miller

$A = S^1 \times \{0\}$

Arrow

@loch meh.

Mike Miller

1:49 PM

$B = \{x, f(x) \mid f: S^1 \to \Bbb R\}$

$B$ is an embedded submanifold (its definition gives you an injective immersion from the circle to it)

The intersection $A \cap B$ is the set of zeroes of $f$

Now just cook up a smooth function with countable zero set (or worse! You could write down something smooth with zero set a fat Cantor set or something).

abenthy

Oh yes, It's incredible how bad it can get.

loch

@Arrow at least for this part i think i can say something not entirely dumb - try see that the action of $G$ on $X$ is $2-$transitive iff the product action on $G\times G$ decomposes into the union of two orbits, then apply the first part of your question along with Schur's lemma

Leaky Nun

do we really need Schur?

loch

2:05 PM

in my head i think so - maybe you can think of another way to do it

Leaky Nun

I think if $y'$ is a function of $y$, and $y(0) = y(1) = 0$, then $y = 0$, but I can't prove it

should I think a bit more?

let's just say $y=0$ for $0 < x < 1$

I can only think of local things, and I don't think that is enough to establish a global result

loch

@LeakyNun if you take $y=\sin(x)$ on $(0,\pi)$, then $y' = \cos(x) = \sqrt{1-y^2}$, so $y'$ is a function of $y$ and satisfies your requirements?

Leaky Nun

no, $\cos(x) \ne \sqrt{1-y^2}$

it's only valid before $\pi/2$

loch

oh lmao

in any case i think you need to be more specific than just 'a function of $y$'

Leaky Nun

let's say the function is $C^1(\Bbb R)$

$y' = g(y)$ where $g \in C^1(\Bbb R)$

well $y$ is bounded so the domain of $g$ can be compact

loch

2:21 PM

i think if you suppose $g(y)$ is non-constant, $g$ continuous, then for some value of $y$ you necessarily have two different derivatives which is bad

Leaky Nun

I can't make that argument rigorous

oh and I think it is equivalent to the claim that if $y'$ is a function of $y$ then $y$ is monotonic

maybe not

Arrow

@loch thanks for the hint about 2-transitivity. For the first question, I'm thinking along the following lines: an equivariant $\Bbbk$-linear map $\Bbbk[X]\to \Bbbk[Y]$ probably amounts to an equivariant set-function $X\to U\Bbbk[Y]$. For starters such a set-function is constant over the orbits of $X$. However, its fibers might be finer. Perhaps specifying the fibers precisely amounts to specifying a connected component of the product action? Edit - it has no reason to be constant over X-orbits.

Leaky Nun

2:46 PM

@loch I feel like this is an easy question but I can't think of anything, literally

but I don't want any hints either...

really, the only global condition I know is that it has a global maximum, and I should prove that it must be 0

loch

@LeakyNun it seems like a good exercise to see if you can make your intuition is rigorous :p

@Arrow what's U?

Arrow

Underlying set

Rudi_Birnbaum

@user2236 1:44:30 ???

Leaky Nun

every time I try to prove it, I somehow run into situations that can be avoided locally

but I don't have enough skills to deal with global situations

WLOG $g(0.5) = 0.5$ is the global maximum

WLOG $g(0.25) = 0.25$

my idea failed again, as I constructed a partial example where $g'(0.25) = 0$

partial, as in, the condition fails in other places

oh I have been using $g$ where I meant $y$

I think I can establish, with more work, a contradiction if $g'(0.25) > 0$ and another contradiction if $g'(0.25) < 0$

and then generalize to obtain $g'(x) = 0$ for all $x$

this is harder than i thought

abenthy

3:08 PM

@MikeMiller By the way, I now think the statement becomes true if one adds the requirement that $M$ is a Riemannian manifold and the submanifolds $A,B$ are totally geodesic submanifolds.

Mike Miller

3:24 PM

@abenthy Does this imply they intersect more or less transversely?

Leaky Nun

I think the answer is "uniqueness of solution to first order homogeneous ODE"

but that requires continuity of the auxiliary function (or even differentiability, idk)

which I'm not very eager to assume

Rudi_Birnbaum

4:27 PM

Who can give a short abstract of the primary achievements of the other three Fields medal winners (PS one I have a rough idea about)?

Mike Miller

Quanta has articles about them

Rudi_Birnbaum

@MikeMiller great!

Alessandro Codenotti

here

loch

4:44 PM

i know way too little to say anything about what birkar did, but i can ramble a tiny bit about birational geometry..

roughly speaking the goal of birational geometry (birkar's field) is to classify varieties birationally.

as a very basic example of a classification result in geometry/topology is the classification of closed orientable surfaces - where such surfaces are determined by the number of holes they have (genus).

varieties are zero loci of polynomials. e.g. take your favourite polynomial (e.g. $y^2=x(x^2-1)$, and look at its zero locus in $\mathbb{C}^2$).

3

Alessandro Codenotti

5:03 PM

In your last paragraph is everything equipped with the Zariski topology?

mercio

how did you know my favourite polynomial

Mike Miller

A good guess

@AlessandroCodenotti Yup

Ted Shifrin

5:15 PM

hi demonic @Alessandro, mercio, MikeM

oops, and loch

hi @TedShifrin

@TedShifrin Hi

hi Tobias

interesting and seems difficult: math.stackexchange.com/questions/2867261/…

notice that the measure is of a subset of the domain, not codomain

my gut feeling is that the answer is no but i have made no good progress

Alex Provost

5:36 PM

I've been trying to solve the geodesic equations for the Poincaré disk model in polar coordinates unsuccessfully -- does anyone know if explicit parametric solutions can be found? (It's well-known that the geodesics are circular arcs orthogonal to the boundary of the disk)

I had some hopes because the process can be carried out quite nicely in the upper half plane

Antonios-Alexandros Robotis

hi @TedShifrin

Mike Miller

Yes, certainly. You just need to understand the set of circles which intersect the unit circle perpendicularly, @AlexProvost. As it turns out these are the same as those circles which are fixed under inversion. Then there's a not-horrible formula to parameterize every circle on the list.

If I had my inversive geometry book still on me I would be able to quote it at you

Alex Provost

@MikeMiller Nice! This gives me hopes, but at the same time, also makes me wonder whether I messed up somewhere when solving the ODEs. I arrive at some particularly nasty thing to integrate

(which W|A cannot find explicit solutions for)

Mike Miller

@AlexProvost I can't say, and I definitely don't want to get my hands dirty :D

Alex Provost

I don't blame you ^^

Mike Miller

5:43 PM

Stupid trick: once you check that all of these are geodesics, you know all of the geodesics, because a geodesic is determined by a starting point and a unit tangent vector at that point, and you have a circular arc or bisecting line passing through each one of those ;)

Alex Provost

The best my google-fu could find for a geodesic is the implicit formula $r^2 - r(h cos(\theta) + k sin(\theta)) + 1 = 0, where h,k are the cartesian coordinates of the center of the circle the geodesic is part of

Which is not very satisfying

Mike Miller

Maybe I will try to find the book

Alex Provost

(For one, this represents a whole circle, and just an arc, and also yields no parametrization, so I can't even check it satisfies the ODEs)

Thanks, that would be great!

and not *just an arc

mercio

My intuition is that vertical half lines are geodesics, and the rest are their conjugates, which happen to be half circles orthogonal to the real line

Alex Provost

I think the usual smart way of obtaining the geodesics is by pulling back to the half plane, where they are more easily understood, and then mapping back to the disk

But I really would like to obtain them directly somehow

Mike Miller

5:52 PM

@AlexProvost the circle perp to the unit circle, centered at $(a,b)$ where $a^2 + b^2 > 1$, is given by $x^2 + y^2 - 2ax - 2by + 1 = 0$. We can complete the square to find out the radius of this circle

$(x-a)^2 + (y-b)^2 = a^2 + b^2 - 1$

Alex Provost

@MikeMiller Yeah, this agrees with the polar version I wrote down above

Mike Miller

I dunno how much better you're liable to get than this, since you can parameterize a circle

Alex Provost

You can, but you don't want to parametrize the whole circle, just an arc. So it's not obvious to me when to start and when to end. And also, the parametrization should have constant speed wrt to the metric

Mike Miller

"The circle at $(a,b)$ of radius $\sqrt{a^2+b^2-1}$" is then $\sqrt{a^2+b^2-1}(\cos \theta - a, \sin \theta - b)$

oh, i see your point, you want to unit speed reparameterize

determining where to start and where to end is the same as determining when that intersects the unit circle, which is just algebra - but the reparameterization is a more serious problem

the half-plane trick is probably the best thing to do (assuming you know the parameterizations for the circular arcs in the half plane model)

mercio

the reparametrizatoin should give you the endpoints

since you can never reach them

Mike Miller

5:56 PM

was also thinking that

mercio

is the parametrisatoin of the vertical half line simply $i \exp(t)$ ?

Mike Miller

prolly

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