Mathematics - 2021-02-06

dc3rd

00:17

So I did have a question about looking at things from a different perspective @TedShifrin. In the example of looking at the hyperboloid (let's take 2-sheet), you observe what the shape/ value that $x$ and $y$ terms are going to give. So can I look at this as you are decomposing things as "functions inside of functions"?

specifically: $x^{2} + y^{2} - z^{2} = -1 \to x^{2} + y^{2} + 1 = z^{2}$

then: $g(x,y) + 1 = z^{2}$. so you are going to see what happens with $g(x,y)$ and then build on top of that what happens to the $z$.

Ted Shifrin

00:34

Well, you see $|z|\ge 1$,

and you get a union of two graphs. But I'm not sure what your point is.

dc3rd

I'm just curious in terms of an approach, so expanding on that example say I had some object of the form $x^{2} + y^{2} + z^{2} - w^{2} = 1$. I could rewrite it as: $x^{2} + y^{2} + z^{2} + 1 = w^{2}$, so I could look at what the sphere does in $\mathbb{R}^{4}$ and use that behaviour to get a feel for what is happening in $\mathbb{R}^{5}$.

anakhro

00:51

Is there an error here: $$\frac{\cot{\pi (-z-ia)}}{2i} - \frac{\cot{\pi (-z+i a)}}{2i}= \Im\left(\cot{\pi(-z-i a)}\right)?$$

They seem to be forgetting the conjugate on the z when they are doing the imaginary part?

Astyx

hi chat

anakhro

Like, wouldn't it be $$\frac{1}{2i}(\cot \pi(-z-ia) - \cot\pi(-\bar z + ia))?$$

Instead of the LHS above.

Ted Shifrin

Hi, Astyx.

Oh, I see. Yes, reasoning by analogy with 2-D and 3-D can be useful. You can let a circle in $\Bbb R^2$ represent a sphere in $\Bbb R^k$.

Thorgott

the general way of reasoning with R^k is pretending that k=3

Leaky Nun

yep

dc3rd

01:05

@TedShifrin, so take what I read in Flatland and put it on steroids for the higher dimensions...............

Ted Shifrin

Yeah, basically.

anakhro

01:22

most certainly hellish

EM4

01:55

For a friend, is panicking like crazy. In all of his math classes he got A's and one C-. He thinks at Real Analysis he will get C/D will that hurt him for grad school.

Ted Shifrin

Not everyone belongs in math grad school.

Blunt, but truth.

EM4

true true.

will that hurt him to apply.

he got A- (Abstract Algebra 1) and A (Abstract Algebra II)

got A on PDEs, A on Complex Analysis etc.

Ted Shifrin

02:11

Depends on level of school, GRE scores, letters of rec ....

EM4

do you know good website of level of school.

Ted Shifrin

In the US the AMS has rankings of Group 1, Group 2, Group 3. I assume it will show up on Google.

There's also a big dependence on quality/rigor of undergraduate school.

This is why students should talk to faculty advisers. They should give sound advice based on experience and knowledge of the individual student.

EM4

this is what i told him.

Ted Shifrin

You may quote me.

36 years of advising students.

robjohn

03:38

@TedShifrin "me"

dc3rd

@robjohn you may be able to answer this as much as @TedShifrin. How is quality/rigor of the schools measured? I was looking at the site Ted suggested and the information is about number of PhDs in the dept, employment, number of PhDs granted, etc.

so how do you measure "quality" and "rigor"?.............I looked at the groupings and you see the usual "big name" schools, but what is the metric?

Mike Miller

Assessment by other mathematicians on gut feel, mostly

anakhro

@robjohn If we have something like $\sum_{n=1}^\infty \frac{2z}{n^2 - z^2}$, then can we still apply the same residue calculations somehow (navigating the fact it is not a Laurent series)?

Mike Miller

Considering that's also the metric used by mathematicians to choose grad students or new hires it's not a bad choice

anakhro

That's also how Mike chooses to grade final exams.

"gut feel"

On occasion, he even tries a RNG.

dc3rd

03:52

Gut feel?........hmmmm interesting......so are you saying there are "cliques" in the world of higher math @MikeMiller?

anakhro

Of course there are. They are called fields.

dc3rd

Lol.

Ted Shifrin

They also look at research reputation of faculty, how many PhDs are granted, etc, but Mike's right — a lot is peer rating.

dc3rd

04:08

I guess ratings from peers will depend on if a certain faculty is able to do research in an "interesting" field that may not be possible at their own institution.

anakhro

Okay, Conway seems to think it follows that: $$\frac{1}{a} + \sum_{n=1}^\infty \frac{2a}{a^2 - n^2} = \pi\cot\pi a.$$
This might be handy for the alternating case.

But I don't see how to handle the non-Laurent series.

He says to use similar methods to the calculation of Laurent series via residues.

Is it just because it is symmetric?

I guess that's really all. The n=0 case is the 1/a

Wow I am i d i o t

dc3rd

04:52

I had a question about part c) of this question....I've done all three parts no problem, but I was trying to understand what they were getting at with regards to part c).

So i established that the matrix is not diagonalizable because it has only one vector in its null space.

I'm wondering if it is implying that just because it is a square matrix that could be a converted to a scalar matrix, that it is not necessarily diagonalizable.

If that is the case I was trying to figure out what would be the invertible matrix $Q$ that would allow me to turn the given matrix into a scalar matrix.

uhoh

05:07

6

Q: How can orbital resonance sometimes have a stabilising effect, whilst other times, it has a destabilising effect?

I have just started learning about orbital resonance. I understand how bodies in orbital resonance will line up according to the orbital ratio number, and there will be increased gravitational effects upon alignment. However, I do not understand how resonance can often have a stabilising effect, ...

Asked in Astronomy SE

copper.hat

05:23

@dc3rd it is not diagonalisable because it has only one eigenvalue (1) and if it was diagonalisable then (b) & (a) would imply that it is the identity which is it not.

dc3rd

@copper.hat, what is the catch about this matrix chosen then?

123

Hello World.. Of Mathematics

copper.hat

Not sure there is one, just showing the Jordan form cannot be diagonalised, I supppose?

123

Hi @copper.hat @TedShifrin

copper.hat

Or maybe that tha geometric & algebraic multiplicities are different?

@123 Hi

dc3rd

05:30

Well I haven't learned about the Jordan form yet, don't even know what a geometric multiplicity would be in comparison to an algebraic one..........

I'll just circle around back to this when I get more knowledge.

copper.hat

The algebraic multiplicity of $A$ is 2 (the characteristic poly, is $p(x) = (x-1)^2$)( and the geometric multiplicity is 1 ($\dim \ker (A -\lambda I))$.

dc3rd

Oh....that's what the terms mean.....nice

copper.hat

The Jordan form is a useful theoretical tool, it is sort of a canonical (modulo ordering) representation.

dc3rd

the thing about multiplicities definitely has the smell that it is going to be determining when it comes to evaluating if a transformation is diagonalizable.

as in if the geometric and algebraic multiplicities are not equal then something is going to break

robjohn

05:46

@anakhro That sum is $\frac1z-\pi\cot(\pi z)$

it has residue $-1$ at all integers except $0$

@anakhro I see that Conway agrees ;-)

Ted Shifrin

06:03

@dc3rd That's all in chapter 9 of my book.

dc3rd

@TedShifrin, wew!......as you know from my progress in your book right now I got a fair bit of work to do before I get there. Going to be very fun to merge them together.

user 147593

09:30

for those not interested in group study

perhaps, the pandemic will create such a generation

Balarka Sen

10:49

@User873110 No. Take two parallel geodesics in $\Bbb H^2$, let $A$ be the region bounded by them, and then paste them by a translation.

user586228

12:20

What should be the thinking process for such questions.

?

Astyx

13:09

Do you know about series expansions ?

user586228

13:33

Slightly.

Astyx

Do you see why there is a unique a for which this works?

user586228

13:59

Yes I do otherwise things would tend to infinity...

You need to have the coefficient of x to be 0

Astyx

That's not a very concise argument, but ok

Astyx

14:21

Now you could find that the only value that works by asking that $\sin 2x - a\sin x\over x$ goes to zero

Assuming you know how to compute that

This doesn't show that this specific $a$ works, it just shows that if there is one that works, it has to be this one

@user586228

And following on this idea, you can realise that this is asking for successive derivatives (up to order 3)

You want $(\sin 2x)' = (a\sin x)'$

user586228

@Astyx ok

TheReal__Mike

15:00

Hola

Astyx

hi

user 85795

16:02

👋

Alessandro Codenotti

16:47

Let $G=\Bbb Z^2$, what does $[G,G]$ look like? Some AT notes claim it's generated by $4$ elements, but that seems very wrong to me

Astyx

The commutator subgroup?

Alessandro Codenotti

yeah

I think it's not even finitely generated

Astyx

Isn't $\Bbb Z^2$ commutative ? Shouldn't that be zero?

Alessandro Codenotti

Oh I can't write, sorry, I wanted $G=\Bbb Z^2\ast\Bbb Z^2$

Astyx

Ok, that expalins my confusion :p

Alessandro Codenotti

16:50

yeah my bad

Astyx

I agree with you it seems unlikely

Alessandro Codenotti

I mean the commutator subgroup of $F_2$ is not finitely generated, and this thing surely contains it

Astyx

Is that sufficient?

Thorgott

no

Alessandro Codenotti

No, after all $[F_2,F_2]$ (just like any countable group) is contained a 2-generated group

Astyx

17:04

There's a surjective map to $[F_2,F_2]$, which is not finetly generated however

By projecting $\Bbb Z^2*\Bbb Z^2\to F_2$, $(n,m)_1\mapsto s_1^n$ and $(n,m)_2\mapsto s_2^n$

Thorgott

true

Alessandro Codenotti

Nice

So what's the proper way of telling apart the fundamental groups of the wedge of two torus vs that of their connected sum?

Thorgott

I would suspect there's no surjection from $G$ to $\Sigma_2$

are there non-trivial homomorphisms $\mathbb{Z}^2\rightarrow\Sigma_2$ at all?

I mean, yeah, of course

are there any such non-trivial maps where both basis elements get mapped non-trivially?

ah no, there aren't

Alessandro Codenotti

Why not?

Thorgott

It would have to come from two elements in $\Sigma_2$ that commute. These come from two words in $F_2$ whose commutator is in $\langle[a,b][c,d]\rangle$, but that doesn't exist.

i.e. $[a,b][c,d]$ is not itself a commutator

so the image of a homomorphism $\mathbb{Z}^2\rightarrow\Sigma_2$ is cyclic, whence the image of a homomorphism $G\rightarrow\Sigma_2$ is $2$-generated, but $\Sigma_2$ isn't $2$-generated, because that would imply $\mathbb{Z}^2$ surjects onto $\mathbb{Z}^4$ after abelianizing

user586228

17:40

There is a very important step involved where they are taking logarithm outside and limit inside.A slight mentorship to the above technique will be highly welocme.

welcome.*

Thorgott

logarithm is a continuous function

user586228

ok

@Thorgott But how do you know that?

Thorgott

it's the inverse of the exponential function, which is continuous, and the inverse of a continuous function between intervals is continuous

user586228

@Thorgott I am getting e as the answer however the solution manual reads 1..

Where did I go wrong..It might also be the situation that the given asnwer is wrong.I so please help.

If*

Thorgott

1 is the correct answer

I can tell you where you went wrong if you tell me what you have done

user586228

17:55

ok

Thorgott

you forgot to differentiate the denominator when apply L'Hopital's rule

Balarka Sen

18:25

@AlessandroCodenotti They're both K(G, 1) spaces, so if they had isomorphic fundamental groups they would be homotopy equivalent. Contradiction in H_2.

Alessandro Codenotti

I was hoping for an approach that makes sense to a student who only knows $\pi_1$

Balarka Sen

Can't help you sorry

Thorgott

my argument makes sense to someone who only knows $\pi_1$, no?

Alessandro Codenotti

unrelated but Cult of Luna released a new album yesterday

Thorgott

or are you not convinced by it?

Mike Miller

18:28

One has Z^2 subgroup the other doesn't.

Thorgott

oh right, that's even shorter

Balarka Sen

You need hyperbolic geometry to argue that.

More than knowledge of pi_1

Thorgott

what's wrong with my argument?

Balarka Sen

Didn't read

Mike Miller

Maybe consider homomorphisms to 4x4 Heisenberg group.

It's a group theory problem find a group theory answer

Thorgott

18:36

No two non-trivial elements of $\Sigma_2$ commute, because two such commuting elements would come from two words in $F_4$ whose commutator lies in the subgroup generated by $[a,b][c,d]$, but that doesn't contain any commutators.

Balarka Sen

a and a^2 commute

Mike Miller

@Thorgott Prove this.

Also you mean F_4.

Balarka Sen

The centralizers are cyclic. I don't know how to prove that without hyperbolic geometry

Thorgott

that's a convincing point Balarka

Mike Miller

You can @BSen, this was known to Nielsen I believe. But I don't like to think about words.

Thorgott

18:38

but I don't see the error in t he argument

Balarka Sen

"You can" No, I cannot.

Someone can maybe

Thorgott

is some power of $[a,b][c,d]$ magically a commutator? surely not

Alessandro Codenotti

@Balarka if we go the GGT route how many ends does $\Sigma_2$ have?

Balarka Sen

Infinitely many.

Thorgott

ok, I see why my argument was stupid

Balarka Sen

18:39

It's quasi isometric to H^2

I mean one end lol

robjohn

@user586228 That shouldn't be a problem since $\log$ is continuous at $1$.

Balarka Sen

Yeah so maybe that's a good argument

Mike Miller

Your argument is not wrong but it has a giant gap.

Balarka Sen

One has one end the other has infinitely many

Thorgott

you have to alter it: two elements of $\Sigma_2$ that commute come from two words in $F_4$ whose commutator is contained in the subgroup generated by $[a,b][c,d]$. the only commutator in this subgroup is the identity, so the words must already commute in $F_4$. this only happens if they're in the same cyclic subgroup, so the same must hold in $\Sigma_2$.

Mike Miller

18:40

Since you have to explain why that product of commutators is not a commutator.

Alessandro Codenotti

Stallings's theorem now tells us that $\Sigma_2$ is not a free product, done

Thorgott

@MikeMiller I'm not gonna do it, but that can be done by just looking at words

Alessandro Codenotti

That's an argument using way too much machinery though, but a nice one nontheless

Mike Miller

You have yet to do this

@Thorgott That's the kind of thing someone who can't produce a proof says

"Sure I can calculate number of generators in an index k subgroup of F_m. Just look at words"

Balarka Sen

@Thorgott These kinds of things are usually not clear. It's not clear for example why you cannot produce a word which is 0 in every finite group.

(This is residual finiteness of free groups)

Alessandro Codenotti

18:45

@BalarkaSen Wait can taking a quotient ever increase the number of ends of a group?

Balarka Sen

Z+Z to Z?

One to two

Ted Shifrin

Why doesn't Z+Z have four?

:)

Alessandro Codenotti

because it looks like $\Bbb R^2$

Balarka Sen

No matter how large a ball you take in the integer lattice the boundary is connected in the subspace metric

Ted Shifrin

I was yolking.

Balarka Sen

18:48

So one end, all the way to infinity

Michael Albanese

19:13

Hi @TedShifrin

Ted Shifrin

Heya @Michael

Michael Albanese

Have you seen this question?

robjohn

@TedShifrin such a hard boiled attitude

Ted Shifrin

Nah, @robjohn, I'm a softie.

No, @Michael. I don't know this stuff, anyhow. :)

robjohn

@MichaelAlbanese what is the topological generalization of conformality?

Michael Albanese

19:18

Sorry, I keep thinking you did Riemannian geometry. My mistake.

Simone

evening

Ted Shifrin

LOL, silly mistake, @Michael :P

Michael Albanese

@robjohn: I'm not sure what that would mean. If you're referring to the question I linked, I'm not sure the OP is asking the question they intend to ask.

Ted Shifrin

Morning here, @Simone :P

Michael Albanese

Sometimes I conflate differential geometry and Riemannian geometry.

Simone

19:20

@TedShifrin good saturday morning Ted.

robjohn

@MichaelAlbanese That might be. I was asking about the question

Ted Shifrin

You'll find zero Riemannian stuff in my publications, I'm relieved to say. More algebraic geometry and projective differential geometry.

robjohn

@TedShifrin just barely

Ted Shifrin

Thank you, @Simone :) Did you ever finish that compactness argument?

Balarka Sen

@MichaelAlbanese Weird question. Maybe they mean to ask about conformally flat metrics. In which case it is true.

Simone

19:22

I did, but I had to abandon my idea... I was going nowhere. And even if I could continue it would have been a bit redundant.
https://math.stackexchange.com/questions/4011170/do-you-find-this-proof-convincing

Balarka Sen

In fact, compact simply connected conformally flat manifolds are conformal to spheres.

Simone

@TedShifrin The proof comes after the EDIT part

Ted Shifrin

You do understand my complaint. Any such proof must use the nature of $f$ in an essential and transparent way, @Simone.

Michael Albanese

@BalarkaSen: Their working seems to indicate they are starting with a metric induced from $\mathbb{R}^4$, so maybe they are only interested in knowing whether such metrics are conformal to the round metric.

Simone

Indeed, but I had a plan for that ;)

Anyway I think I successfully proved the proposition

Ted Shifrin

19:24

My standard proof, @Simone, is to take a ball of radius $R$ (centered at the origin, say) containing all the $\mathbf a_j$ and then it's an easy estimate to see what happens on the boundary and outside a ball of radius $2R$.

Simone

I did it with a ball of radius $f(\mathbf a_1)$ centered at $\mathbf a_1$

Balarka Sen

@MichaelAlbanese Of course not, right? Conformal automorphisms of R^n for n >= 3 are so restrictive!

Simone

sorry radius $\sqrt{f(\mathbf a_1)}$

Balarka Sen

A generic smoothly embedded sphere in R^4 should have Weyl tensor nonzero.

Michael Albanese

@BalarkaSen: Why must the conformal factor come from a conformal automorphism of Euclidean space? I was thinking that $g$ is a metric on $S^n$ obtained by restricting the Euclidean metric on $\mathbb{R}^{n+1}$ to some embedded $S^n$.

Ted Shifrin

19:29

Interesting idea, @Simone. I was trying to worry about distance to all the points, and you're showing that you only need to do that once.

You're back to conformal and the Weyl tensor yet again, a @Balarka :D

Balarka Sen

To check for conformal flatness one just needs to check vanishing of Weyl tensor. So one should try to compute Weyl tensor of a hypersurface in R4, this is now a problem in exterior geometry much to the happiness of @Ted

Simone

@TedShifrin :thumbs up:

Ted Shifrin

I've never computed a Weyl tensor in my life :D

@Simone: Sometimes the estimates on the "right" compact sets gets pretty tricky. I made sure all my exercises were doable, but some are challenging.

Simone

The last exercises of every sections of your book tend to be the hardest; I like the way you increase the difficulty gradually.

Your textbook is very good Ted. I'm really enjoying it

Ted Shifrin

I'm glad you are. Most of my students did, too, but they liked to suffer :) Some took even more classes from me :D

Simone

19:35

What do you teach apart from Multivariable calculus/linear algebra?

Mike Miller

Many people teach most things

You just usually don't write books on them

Ted Shifrin

Nothing now. I retired almost 6 years ago. I taught lots of things. Undergraduate and graduate differential geometry. Undergraduate algebra. Topology, differential topology. Graduate complex analysis. Even some applied stuff years ago (which I loved teaching).

Simone

@MikeMiller not one person in existence teaches most things ;)

Ted Shifrin

@MikeM: It's interesting, though. I had several algebraist colleagues who basically would teach nothing but calculus and various algebra courses. And some topologist colleagues who (out of laziness?) taught only calculus and undergraduate and graduate topology.

Well, @Simone, one can aspire to teach most every course at the undergraduate level, but some choose to stay very narrow.

Mike Miller

Most [lower-division classes in the math department].

Sometimes even many [upper-division courses that the department does not have a specialist in].

Ted Shifrin

19:39

Yeah, one of my friends in San Diego is a combinatorist who teaches the upper-division topology class because the [few] topologists have no interest in doing it.

Mike Miller

Boo.

Ted Shifrin

Just reporting the fact.

Mike Miller

Sure, I'm not a point set guy. But you can make it fun.

Ted Shifrin

Well, he does a lot of what you did at the end in the second quarter ... fundamental group, covering spaces, etc.

Simone

I was beeing facetious :)

Ted Shifrin

19:41

Anyhow, @Simone, when I came to UGA essentially no one was teaching the undergraduate curves and surfaces course, so I started teaching it a lot and then wrote a book on it because I was unhappy with the five or six books I tried using.

Simone

Which other books were you using? SPivak 2?

Ted Shifrin

No, no. Undergraduate.

Simone

ah ok

Balarka Sen

@Ted @MichaelAlbanese I looked it up, apparently a conformally flat hypersurface in a conformally flat manifold has either 2nd fundamental form a multiple of the identity or 2nd fundamental form has two distinct eigenvalues, one of multiplicity 1 and the other n-1.

Seems like an exercise for me. But surely you can make embeddings of S^3 in R^4 which do not satisfy these. Make it curve in all three directions differently, at a point.

Ted Shifrin

DoCarmo, Millman/Parker, Oprea, McCleary, and at least one other.

Balarka Sen

19:44

I'll try to prove that fact later.

Simone

DoCarmo is a cheapo

I though about buying that instead of yours but then you sold me on yours XD

Balarka Sen

To carry it out at a point consider graph of $w = \lambda_1 x^2 + \lambda_2 y^2 + \lambda_3 z^2$ in $\Bbb R^4$

$\lambda_i$ all different.

Throw away everything outside a large ball containing the origin, and cap it off by a whatever disk

This is an embedding of S^3 in R^4 which has $\Bbb{II}$ with three distinct eigenvalues at the origin.

Ted Shifrin

That seems like an interesting exercise, a @Balarka. Sort of a conformal version of the famous exercise/theorem I've mentioned before. Schur's lemma.

Mike Miller

@TedShifrin Oh, I have no problem with him doing it. I'm glad. I just feel sad the topologists abandon it.

Ted Shifrin

There seems to be a lack of enthusiasm, in general, for undergraduate teaching at UCSD.

Michael Albanese

19:51

I think a closed hypersurface will have to have a point of positive scalar curvature (or maybe that's only for surfaces). So just choose a metric with scalar curvature non-positive. Such metrics exist on all closed manifolds of dimension three or more.

Balarka Sen

That seems much harder than my explicit construction :)

But I didn't know of this scalar curvature fact.

Michael Albanese

Actually, there are always metrics of negative Ricci curvature.

Balarka Sen

@TedShifrin Yeah, I'll work it out later. It's interesting, because the larger eigenspace seems to be giving a natural distribution on the manifold. Is it integrable? Is it a foliation by S^(n-1)'s, dare I say?

Dragging round S^2's along loops in R^4 might be ways to create conformally flat hypersurfaces (with the same topology, S^2 x S^1...)

Ted Shifrin

Seems interesting, @Balarka, but I have no intuition.

Balarka Sen

Actually by topology if those were integrable all the conformally flat (compact) hypersurfaces in R^4 would have been S^3, S^2 x S^1

Topologically

Ted Shifrin

19:59

I was going to say that Cartan classified isoparametric hypersurfaces (constant eigenvalues of II). You can have only certain combinatorial situations in terms of the multiplicities.

Balarka Sen

@MichaelAlbanese Is this Lokhamp's h-principle

@TedShifrin Oh, ok

Ted Shifrin

20:13

a @Balarka: I have no idea how the Gauss equations play with the Weyl tensor (and whatever replaces it in dimension 3).

Balarka Sen

Yeah, I meant the Cotton tensor. Sorry about that.

Ted Shifrin

I don't know Mme Cotton.

It feels like a Schur's Lemma type of computation (with or without moving frames). I would do it with flat ambient space first and then figure out how to deal with conformal flatness later.

Michael Albanese

@BalarkaSen: Yes, this is Lohkamp's result.

Balarka Sen

@TedShifrin Yah, I'll try that.

@MichaelAlbanese Cool.

robjohn

20:31

@MaryStar Here is the shape with a constant of $24$. With a constant of $25$ or greater, there is no surface.

Alessandro Codenotti

Does $\Bbb R^2$ retract on an embedded Cayley graph (the usual embedding with segments parallel to the axis getting shorter and shorter) of $\Bbb F_2$? I guess not but I don't see an argument

Thorgott

Ok, I claim $([a,b][c,d])^k$ is a commutator iff $k=0$. Assume $k\ge1$ WLOG and say $([a,b][c,d])^k=[v,w]$ for two reduced words $v,w$. If $w$ is one letter, we obtain an immediate contradiction, since then $[v,w]=vwv^{-1}w^{-1}$ ends with $a^{-1}w^{-1}$. Thus, $w^{-1}$ has to end with $c^{-1}d^{-1}$, i.e. $w$ has to start with $dc$, say $w=dcw^{\prime}$. Then $[v,w]=vdcw^{\prime}v^{-1}(w^{\prime})^{-1}c^{-1}d^{-1}$. Now $dc$ doesn't appear anywhere in $([a,b][c,d])^k$, so part of this has to cancel. There are two possibilities: 1. $w^{\prime}$ is the empty word and $c$ cancels with $v^{-1}…

Balarka Sen

Lol

@AlessandroCodenotti Are you sure? R^2 retracts to four teardrops wedged at a common point. Iterate?

Thorgott

pure algebra Balarka

Alessandro Codenotti

@BalarkaSen what's a teardrop

Balarka Sen

20:40

That is, R^2 retracts to closed unit disk D. Consider union of x and y axis in D. D retracts to D/(union of x and y axis) which is 4 teardrops wedged at a point. Iterate the same construction on each teardrop.

In the limit, you'd have retracted D to the Cayley graph of F_2. Am I wrong?

You see it right? It's the natural candidate for the retract disk -> F_2

Alessandro Codenotti

Hm that seems believable

Yeah I got the picture

Balarka Sen

I bet you can write a (quasi-)geodesic retract as well

In the hyperbolic disk

Think of collapsing these boxes bounded by the grey lines to the crosses they contain

By an almost-geodesic retract. You just push the four corners to the center of the cross, and the four chambers given by the diagonals joining opposite corners to the respective edges they contain

I think this is continuous, but one should check carefully.

These quasi-stuff are in general not continuous so that must be the subtle point

Thorgott

fun fact: every countable group can be embedded in a six-generator simple group

Balarka Sen

Yeah, not continuous.

@Thorgott Is this Higman-Thomson

Thorgott

20:59

I don't think so

it's by Hall, apparently

Balarka Sen

mmk

Ah no so the result I am thinking supersedes your quoted thing

In 1949 Graham Higman, Bernhard Neumann and Hanna Neumann proved that every countable group can be embedded in a two-generator group

(note the initials: HNN... the group is actually an HNN extension iirc)

Thorgott

yeah, right

but that construction doesn't yield something simple necessarily

Balarka Sen

got it

Thorgott

apparently Goryushkin and Schupp showed that in fact you can embed every countable group in a 2-generator simple group

Alessandro Codenotti

@BalarkaSen yes I think the construction was introduced in that paper

chat.stackexchange.com/transcript/message/52979611#52979611 that's the reference

user10478

21:14

Can I do an exponential shift if my linear operator has variable coefficients, i.e., $(xD^2 + (1 - x)D - 1)[ln|x| e^x] \implies e^x (x(D + 1)^2 + (1 - x)(D + 1) - 1)[ln|x|]$, or is that only for constant coefficient problems?

Ted Shifrin

Did you work out both sides to see if they agree, @user10478? The point is that $D(e^x f(x)) = e^x(D+1)f(x)$ and a coefficient in front of $D$ is irrelevant. However, what is the case is that $cD(f(x)) = D(cf(x))$, but $xD(f(x)) \ne D(xf(x))$.

user10478

Okay, that makes sense. I'm doing Frobenius on ODEs with variable coefficients so my powers of $x$ are already outside of the powers of $D$.

robjohn

21:35

$[D,x]=1$

Thorgott

22:27

how does one spot repeated linear factors in a multivariate polynomial?

Ted Shifrin

One doesn't. Or one computes the resultant, I suppose.

user76284

0

Q: Functional differential equation for the Fabius function

Is the Fabius function the unique solution (up to scaling) of the following functional differential equation? $$ f'(x) = 2 f(2 x) $$ If so, how can this be proven? As I understand it, the Picard–Lindelöf theorem cannot be applied here, since the equation is not of the correct form. If not, what m...

real-analysis ordinary-differential-equations derivatives functional-equations

Anyone have an idea?

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$Mathematics$ Mathematics