Mathematics - 2019-10-26 (page 1 of 2)

Tanner Swett

00:45

Hey everyone. I've got a really vague control theory question. :D

I'm implementing a fly-by-wire system for an aircraft in a video game. My goal is to control the angle of attack $a$ of the aircraft by manipulating the elevator deflection $e$.

The system is modeled as a damped harmonic oscillator whose center point is at $e$

So if the system changes $e$ and then holds it constant, then $a$ will oscillate about, and approach, $e$.

The vague question is, what sort of control strategy would work here?

Leaky Nun

01:10

hej @MatheinBoulomenos

TREE vs Graham's Number - Numberphile

►

MatheinBoulomenos

@LeakyNun hi

Leaky Nun

@MatheinBoulomenos on a problem sheet I submitted I used an argument that I don't quite believe myself

so I'm claiming that if $p$ is a prime and $K$ doesn't contain the $n$-th roots of unity, where $n$ is prime to $p$, then $K(\zeta_n, p^{1/n})/K$ is not abelian

MatheinBoulomenos

why is this unbelievable?

I think it's obvious

Leaky Nun

I don't know

yeah maybe it's obvious

MatheinBoulomenos

01:26

Is $K(p^{1/n})/K$ a normal extension?

Leaky Nun

no

oh

ah

ok that's obvious

lmao

MatheinBoulomenos

01:37

@LeakyNun I think I maybe have a proof that $\kappa^2=\kappa$ that doesn't use ordinals

Leaky Nun

what is it?

MatheinBoulomenos

Let $\kappa$ be an infinite cardinal. Let $k$ be a finite field. By Skolem-Löwenheim, there exists a vector space $V$ with $|V|=\kappa$. Choose a basis $B$ for $V$. $B$ is necessarily infinite, so we can find a bijection $B \cong B \sqcup B$, this extends to a linear isomorphism $V \cong V^2$

So this reduces $\kappa^2=\kappa$ to $\kappa+\kappa=\kappa$

Leaky Nun

ha...

MatheinBoulomenos

It suffices to find an injection $B \sqcup B \to B$.
Let $\mathcal{X} =\{(A,f) \mid A \subset B \sqcup B, f:A \hookrightarrow B\}$ with the partial order $(A,f) \leq (A',f')$ if $A \subset A'$ and $f'|_A=f$. By an easy Zorn argument, we get a maximal element $(X,f)$ with an injection $X \hookrightarrow B \sqcup B$. If $X \subsetneq B \sqcup B$,we can just use the fact that $B$ is Dedekind-infinite, so $B$ admits a non-surjective injection into itself to extend $f$ to $X \cup \{x\}$ for some $x \notin X$

done

@LeakyNun I think this is easier than this stuff with order types ...

Leaky Nun

01:53

let's hope Loewenheim--Skolem doesn't use $\kappa^2=\kappa$

MatheinBoulomenos

you can completely avoid Löwenheim-Skolem

Leaky Nun

@MatheinBoulomenos also your argument gives an injection $\Bbb R \to \Bbb N$

MatheinBoulomenos

hmm

okay here's a correct proof:
we prove that every infinite set can be written as a disjoint union of infiinite countable sets.
Let $X$ be a set and let $\mathcal{X}$ be the set of all disjoint families of infinite countable subsets. Apply Zorn's lemma to this. Get a maximal disjoint family of infinite countable subsets $(X_i)_{i \in I}$, then $X \setminus (\bigcup{i \in I} X_i)$ must be finite, so we can just add the remaining elements to any of the $X_i$

Now let $X$ be an infintie set, then write $X=\coprod_i X_i$ where all the $X_i$ are countably infinite. For $\Bbb N$, we can easily construct a bijection $\Bbb N\cong \Bbb N \sqcup \Bbb N$, so we can do this for all the $X_i$, this extends to a bijection $X \cong X \sqcup X$

Leaky Nun

looks good to me

MatheinBoulomenos

To avoid Skolem-Löwenheim, just to be safe, we just have to prove that if $k$ is some fixed finite field, say $k=\Bbb F_2$, then the cardinality of a vector space with dimension $\kappa$ is $\kappa$.

I think there should be an easy argument for that, too

Leaky Nun

02:09

@MatheinBoulomenos I don't think so

the one I know uses $\kappa^2 = \kappa$

Eric Wofsey

02:32

You can also prove $\kappa^2=\kappa$ by a more direct Zorn's lemma argument, from $\kappa+\kappa=\kappa$.

Leaky Nun

@EricWofsey could you enlighten me?

Eric Wofsey

Let $X$ have cardinality $\kappa$, and by Zorn's lemma get a maximal subset $Y$ of $X$ equipped with a bijection $Y\to Y\times Y$.

If $Y$ has cardinality $\kappa$ we're done.

Otherwise, the cardinality of $X\setminus Y$ must be $\kappa$, since if both $X\setminus Y$ and $Y$ were smaller than $\kappa$ then their union would be too.

(OK, so here I'm also using the fact that cardinals are totally ordered, but that can also easily be proved with Zorn's lemma.)

This means that we can find a subset $Z$ of $X\setminus Y$ of the same cardinality as $Y$.

But now we can extend our bijection $Y\to Y\times Y$ to $(Y\cup Z)\to (Y\cup Z)\times (Y\cup Z)$.

MatheinBoulomenos

nice, thanks @EricWofsey

Eric Wofsey

(What we need to add is a bijection that witnesses $|Y|=|Y|^2=3|Y|^2$.)

Note that the original application of Zorn's lemma requires you to know that $X$ has a countably infinite subset and that $\aleph_0^2=\aleph_0$, to prove the poset in question is nonempty.

Or I guess if you allow $Y$ to be finite, to prove that a singleton $Y$ would not be maximal.

Alessandro Codenotti

07:46

@MatheinBoulomenos this argument uses choice though

Well you need to choice to show that $|X|=|X^2|$ for all $X$ (this should actually be equivalent to AC), but you need no choice to show that for well ordered cardinals, which is the case you were talking about originally

Rudi_Birnbaum

11:29

Hi all, i have a complex vector field on R^3 and the real part is divergence-free and the imaginary part is curl free. I wonder if there is a name for such a structure or if it rings otherwise a bell with anyone here?

Jack Don

Why should $[x,[x,y]]$ be zero in the universal enveloping algebra?
$$[x,[x,y]]=x\otimes[x,y]-[x,y]\otimes x = xxy-xyx -xyx+yxx$$

(where I'm lazy so don't wanna always denote the tensors

I can keep using relations, but never seem to be able to show that this is zero

Mike Miller

11:45

It's not zero on every Lie algebra, so why should it be zero in the universal enveloping algebra

Jack Don

Okay right, what I say is false. The actual claim is that $ad(E_i)^{1-A_{ij}}E_j$ is zero in the UEA

Where $E_i$ is the operator corresponding to $\alpha_i$, and $A_{ij}=\alpha_j(\alpha_i^\vee)$ is the Cartan matrix entry

So I'm taking $x=E_i$ and $y=E_j$

And I assume say that $A_{ij}=-1$

So the claim is that $ad(x)^{2}y=0$ i.e. $[x,[x,y]]=0$ in the UEA

I'm not quite sure how to check this, since I don't know how to utilise $A_{ij}$ in the UEA

FuzzyPixelz

12:17

Can someone give an example of a metric subspace where the the boundary is not exactly the set of limit points?

Alessandro Codenotti

@FuzzyPixelz $[0,1]$ as a subspace of $\Bbb R$

FuzzyPixelz

What about one where the set of limit points is not the closure?

Alessandro Codenotti

$[0,1]\cup\{2\}$

FuzzyPixelz

That clears my confusions, I didn't think in terms of exterior points

Alessandro Codenotti

I think you mean isolated points?

ÍgjøgnumMeg

12:38

Ello everyone

Alessandro Codenotti

Hi @ÍgjøgnumMeg

And hi @Mathei @BalarkaSen

MatheinBoulomenos

Hi @ÍgjøgnumMeg @Alessandro

ÍgjøgnumMeg

Hiya :)

Balarka Sen

Hi @Mathein, @ÍgjøgnumMeg, @Alessandro

MatheinBoulomenos

Hi @Balarka

ÍgjøgnumMeg

12:42

waddup

I just shaved my head

I look like a bowling ball

Balarka Sen

I shaved but kept my mustache

ÍgjøgnumMeg

hahahaha

nice

Balarka Sen

I look retarded which I am

ÍgjøgnumMeg

rofl savage

Mike Miller

I had a phase like that

It was ill advised

ÍgjøgnumMeg

12:43

hahahaha

My hair was too thin and I had a massive bald patch in the middle which was turning into a bit of a comb-over

skullpetrol

a phase like what?

Balarka Sen

Bald patches are a-ok. I wouldn't mind if I get one

Alessandro Codenotti

I keep my hair very long and I'm terrified of going bald lol

ÍgjøgnumMeg

hahah, I used to have very long hair, went shorter because the weight of the hair was causing a rift in the middle of my head

Balarka Sen

I wish I had long hair so that I can uh headbang

but instead I have a bird's nest

good for nothing piece of crap

ÍgjøgnumMeg

12:45

hahaha

skullpetrol

hair is overrated

ÍgjøgnumMeg

yeah, at least I don't have to pay for the barber anymore

skullpetrol

you can headbang without hair

ÍgjøgnumMeg

yeah

Jens Kidman

Balarka Sen

Pitbull

Mike Miller

12:50

I imagine Balarka dresses like pitbull

skullpetrol

but does he have the chicks :P

Balarka Sen

you can come to my hotel room biatch

ÍgjøgnumMeg

dang

google.com/…:

I look like this now

skullpetrol

lol

sauve

rico

Alessandro Codenotti

So do you want to hear about weird compactifications? @Balarka

Balarka Sen

13:00

Yup

Alessandro Codenotti

Nice so you about the correspondence between subrings of $C_b(X)$ (real valued functions) and compactifications of $X$?

(There's a similar construction with sub $C^\ast$-algebras and complex valued functions, but I think this one is nicer to think about)

Balarka Sen

Hm, no. I know you can embed $X$ in $C_b(X)$ in various way and take the closure to get various comactifications.

Alessandro Codenotti

There's some niceness assumption on $X$, for the Higson stuff we can assume that it is proper metric and not compact, in general locally compact plus some separation axiom is enough but I don't know

Balarka Sen

@MatheinBoulomenos $M$ is an $A$-module. Are these slogans correct?

Alessandro Codenotti

Hm ok, tbh I'm also kind of blackboxing this result but let me state it precisely at least

Balarka Sen

13:05

Aight

Alessandro Codenotti

Put the usual sup norm on $C_b(X)$ and look at a closed subring $F$ containing the constant functions and generating the topology of $X$. There's a few way to get a compactification of $X$ from this ring, one is to look at the set of maximal ideals of $F$ with the Hull-kernel topology and $x\in X$ is identified with $\{f\in F\mid f(x)=0\}$

Balarka Sen

What do you mean $F$ generates the topology of $X$?

Preimage of open sets by elements of $F$ generates the topology on $X$?

Alessandro Codenotti

A nicer one is, denoting with $I_f$ the smallest closed interval of $\Bbb R$ containing the image of $f\in F$, to embed $X$ into $\prod_{f\in F}I_f$ via $x\mapsto \prod_{f\in F}f(x)$ and then take the closure of $X$ in this product

Balarka Sen

Is Hull-kernel topology just the Zariski topology?

Alessandro Codenotti

It's a noncommutative Zariski topology

Balarka Sen

13:10

@AlessandroCodenotti Alright, sounds reasonable

Alessandro Codenotti

@BalarkaSen I think so

Mike Miller

So it's actually an algebra?

What is the algebra

Balarka Sen

@Alessandro In this case of $C_b(X)$ you have a commutative setup, right?

Alessandro Codenotti

That's true

Mike Miller

I guess a noncommutative compactification is an algebra map $A \to C_b(X)$ whose image contains $C_0(X)$

If $A$ is injective (injective isometry maybe) then this is a commutative compactification given by some space, a quotient space of $\beta X$

Alessandro Codenotti

13:13

If you prefer the language of $C^\ast$-algebra you can look at nice unital sub $C^\ast$-algebras of $C_b(X)$ and get a compact space out of them by looking at their spectra, but I like real valued functions

Mike Miller

You said the hull kernel topology is a noncommutative topology

I'm just trying to parse what that means

Alessandro Codenotti

I just mean that it works in noncommutative rings too, but that actually doesn't matter here

Mike Miller

Lame

Alessandro Codenotti

Anyway the point is that if you fix such a closed subring $F$ then the compactification $X^F$ associated with it has the property that every function $X\to\Bbb R$ in $F$ extends to $X^F$

Balarka Sen

That makes sense of course

In the Busemann compactification the Busemann functions extend

Alessandro Codenotti

13:19

Easy examples for $X=\Bbb R$ are $F=\{f\mid\lim_{x\to\pm\infty}f(x)\text{ exists}\}$, giving $[0,1]$ or you can ask the two limits to be equal and get $S^1$

Balarka Sen

Right

Alessandro Codenotti

And of course if you use $F=C_b(X)$ you get $\beta X$

So now Higson came up with an $F$ that somehow should capture the asymptotic behaviour of the space $X$, the intuition is not super clear to me, but he took the subring of all slowly oscillating functions

That is $f\colon X\to\Bbb R$ is slowly oscillating iff for every $r>0$ and every $\varepsilon>0$ there is a compact set $K$ such that for every $x\not\in K$, $\mathrm{diam}(f(B_r(x)))<\varepsilon$

More intuitively this is saying that for every $r>0$ the diameter of $f(B_r(x))$ goes to zero as $x$ goes to infinity

Balarka Sen

Hmm

Interesting!

Alessandro Codenotti

Turns out that the Higson compactification can be characterized by a nice universal property. Suppose that $X$ is a proper noncompact metric space. The Higson compactification $hX$ of $X$ is the unique compactification of $X$ such that if $Y$ is any compact metric space and $f:X\to Y$ is a continuous map, then $f$ extends to $hX$ iff it is slowly oscillating

(I defined slowly oscillating for functions $X\to\Bbb R$ above but the same definition works with any metric space as codomain)

Mike Miller

sin(ln(|x|)) seems to be a good slowly oscillating function

@AlessandroCodenotti This universal property has nothing to do with the Higson compactification, right? Just to do with algebras of bounded functions

Alessandro Codenotti

13:36

It looks like a sine curve where the waves keep getting less steep right?

@MikeMiller Hmm yes I think so

Mike Miller

It's that the waves are longer / fatter

Which I guess is partly about being less steep yeah

Alessandro Codenotti

The proof is based on the fact that $Y$ embeds into the Hilbert cube to compose everything with the projections to $\Bbb R$ and fall back on the fact that a function $X\to\Bbb R$ extends to $hX$ iff slowly oscillating, but it seems like it should work with any other compactification

Mike Miller

How do you prove Y embeds in the Hilbert cube

Alessandro Codenotti

Every compact metrizable space does

Mike Miller

Yeah this will surely be unrelated to Higson

Alessandro Codenotti

13:40

Even every Polish space embeds into the Hilbert cube

The only bit that worries me is where the proof says that a function $f\colon X\to Y$ which is sowly oscillating is still slowly oscillating after embedding $Y$ into the Hilbert cube and looking at $\pi_n\circ f\colon X\to [0,1/n]$, I don't think I can do that regardless of the subring $F$

Also the fact that when $Y$ is compact being slowly oscillating doesn't depend on the metric on $Y$ so we don't care if the embedding is not an isometry

Well there's also other issues, like if $F$ is the subring to get $S^1$ out of $\Bbb R$ it doesn't really make sense to talk about the limit at $\pm\infty$ unless the codomain is $\Bbb R$ again

But for example the same exact proof should work to show that not only a function $X\to\Bbb R$ extends to $\beta X$ iff bounded, but also that the same holds for any compact metric space instead of $\Bbb R$

I need to leave for a while but I'll think about this

skullpatrol

cya

Rithaniel

Morning, chat

skullpetrol

hi

Rithaniel

13:55

So, I've shown that the Sylow $2-$subgroups of $\text{GL}_3(\mathbb{F}_2)$ are isomorphic to the dihedral group on the square by simply finding one (upper triangular matrices with 1 along the diagonal) and writing out the Cayley table. Is there a more abstract or quick way of doing this?

Balarka Sen

14:37

$GL(3, 2)$ is isomorphic to $PSL(3, 2)$ which acts on $\Bbb P^2_{\Bbb F_2}$. That has size $(2^3 - 1)/(2 - 1) = 7$. So $GL(3, 2)$ embeds in $S_7$

The largest power of $2$ dividing $|GL(3, 2)| = 168$ is $8$. So the Sylow 2-subgroups are either $D_8$ or $Q_8$

But $Q_8$ does not embed in $S_7$

Well, the Sylow 2-subgroups can be abelian as well. I don't know lol

Someone can argue it's nonabelian. @Mathein?

Rithaniel

14:50

Well, the center of $S_7$ is trivial, but that's tantamount to a nonsequitur

Balarka Sen

Yeah I don't know a super slick argument for showing they are nonabelian

genescuba

15:04

Hi, I'm having some trouble understanding math.stackexchange.com/questions/186697/…

$\max _{x} \min _{y} f(x, y) \leq \min _{y} \max _{x} f(x, y)$

Let $f(x, y)=\sin (x+y)$

I guess firstly Im having a bit of trouble visualizing this function. So I am thinking about just $sin(x)$ so $x \in [0, 2 \pi]$ and we know that $sin(x)$ will be minimum at 0 and at $\pi$

Should I use a trig identity with $sin(x+y)$ to think about this or is there a more intuitive way to think about $\min _{y} f(x, y)$

Thorgott

That's what $sin(x+y)$ looks like

genescuba

Yup

Leaky Nun

@BalarkaSen it's the upper triangular matrices

genescuba

So basically I am trying to find a value $y$ which minimize $sin(x+y)$

Balarka Sen

@LeakyNun Yeah I know

Leaky Nun

15:13

so they're nonabelian

Balarka Sen

It also directly tells you it's D_8 :P

He wanted an abstract argument

Leaky Nun

a b s t r a c t

Balarka Sen

There's clearly two order 4 and order 2 sheers that generate D_8

(It's worth drawing the 8 point lattice and figuring this out, btw, @Rithaniel)

Thorgott

Yeah, but for a fixed $x$ this is reduces to a one-dimensional minimization problem.

Leaky Nun

@BalarkaSen is it triply or doubly transitive?

genescuba

15:16

yup that makes sense so maybe it is easier to see if I look at it like this

Rithaniel

point lattice? That's something I'm not familiar with

Leaky Nun

I don't remember my mobius transforms

Balarka Sen

Triply transitive, right?

Leaky Nun

does this tell us something

genescuba

the lhs becomes $f = sin(x + c)$ in which I want to find a value of x that maximizes $f$ and c is a constant

Balarka Sen

15:16

@Rithaniel GL(3, 2) acts on F_2^3, right? That's a 3D space but with 8 points, sorta

@LeakyNun Hm, maybe. Not sure

genescuba

To solve a maximization problem I can take the derivative of $f$ and set it to 0

but maybe just thinking about $f$ will be better here

Rithaniel

Ah, yeah, it's the automorphism group of $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2$

Leaky Nun

or just $C_2^3$

saves you typing time :P

Rithaniel

Okay, so the upper triangular matrices might be able to be thought of as a rotation of this "cube," perhaps?

This is true, Leaky

Thorgott

$f$ is just a sine function shifted $c$ to the left on the $x$-axis. Can you maximize a sine function?

genescuba

15:20

Yup, so basically the x which maximizes my $f$ is $\frac{\pi}{2} - c$

Leaky Nun

I really wonder how a group of 8 permutations can be triply transitive on 7 objects

oh wait I'm dumb

... the 168 permutations are triply transitive, not the 8 permutations

Thorgott

Yes, but the value we are interested is not the $x$ which maximizes the function, but what the maximum itself is.

genescuba

Oh right because this is not argmax but just max

the max of a sin function is 1

Thorgott

The maximum of $\sin(x)$ is $1$ and the maximum of $\sin(x+c)$ is $1$, because it's just a sine shifted along the $x$-axis (informally speaking). However, the max of, say, $\sin(\sin(x))$ is smaller than $1$, so you need to be careful when you talk about "a sin function".

genescuba

yes you are right, ok I understand math.stackexchange.com/questions/186697/… for the $sin(x+y)$

But my next question is why is this inequality $\max _{x} \min _{y} f(x, y) \leq \min _{y} \max _{x} f(x, y)$ true in general

To me, it seems like we got lucky with our choice of $f(x,y)$

Thorgott

15:27

Have you looked at the proofs in the answers?

That said, it should be pointed out that this inequality is only true if those minima and maxima actually exist, which is not necessarily the case.

genescuba

Yes, I didn't understand in the proof:

From all this equation you get the following inequalities:

$f\left(x, y_{0}\right) \leq f\left(x_{0}, y_{0}\right) \leq f(x, y) \leq f\left(x_{1}, y_{1}\right) \leq f\left(x_{1}, y\right)$

how do they get $f\left(x_{0}, y_{0}\right) \leq f(x, y)$ and $f(x, y) \leq f\left(x_{1}, y_{1}\right)$

Thorgott

I think that answer is faulty. Look at the one from Did instead

genescuba

In that answer how come The assertion is equivalent to the fact that, for every 𝑥 and 𝑦, $\min _{s} f(x, s) \leqslant \max _{t} f(t, y)$

ÍgjøgnumMeg

@Mathein Vogel wants us to practice Sütterlinschrift for AZT 1

I always just write some form of underline for prime ideals hahaha

genescuba

$\max _{x}\left(\min _{s} f(x, s)\right) \leqslant \min _{y}\left(\max _{t} f(t, y)\right)$

so essentially we can we are saying $\max_x$ of some function of $x$ is less than or equal to the $\min_y$ of some function of $y$

or $\max_x \min_s g(x, s) \leq \min_y \max_t h(y,s)$

Thorgott

15:44

Generally, if $p(x)\le q$ for all $x$, then $\max_xp(x)\le q$

The LHS depends on $x$, the RHS does not, so if you maximize the LHS over $x$, you achieve $\max_x\min_sf(x,s)\le\max_tf(t,y)$. Then minimize the RHS over $y$ to achieve the desired inequality.

genescuba

Ah I see, thank you for your help @Thorgott

Thorgott

np

anakhro

17:06

Punks.

ÍgjøgnumMeg

you wanna take this outside ?

anakhro

@ÍgjøgnumMeg what math have you been up to?

ps i l u

ÍgjøgnumMeg

rofl

I'm in courses on modular forms, ANT, and construction of L-Functions

so mostly just been trying to survive in those

how about you?

Rithaniel

asdfmovie7

►

anakhro

Just doing some standard algebra stuff right now.

Like solvability.

Review stuff, really.

Balarka Sen

17:10

I have an extremely dumb question.

ÍgjøgnumMeg

Nice :)

anakhro

@BalarkaSen it's about time you get your share.

Balarka Sen

lool

Given a closed manifold $M$, it should be true that it always admits a nontrivial vector bundle. How do you construct one?

anakhro

Mike is here to save the day.

Mike Miller

Why should that be true

Balarka Sen

17:13

I have a nuke idea

First, assume $M$ is orientable

Mike Miller

It's false but I won't tell you a c/e until I get your logic

anakhro

Does {*} have a non-trivial vector bundle?

Balarka Sen

OK, something which is not a point.

anakhro

;)

Leaky Nun

some cohomology should classify vector bundles right

Mike Miller

17:14

Tell me the idea

Leaky Nun

Chern class

In mathematics, in particular in algebraic topology, differential geometry and algebraic geometry, the Chern classes are characteristic classes associated with complex vector bundles. They have since found applications in physics, Calabi-Yau manifolds, string theory, Chern-Simons theory, knot theory, Gromov-Witten invariants, topological quantum field theory, the Chern theorem etc. Chern classes were introduced by Shiing-Shen Chern (1946). == Geometric approach == === Basic idea and motivation === Chern classes are characteristic classes. They are topological invariants associated with ...

Mike Miller

We will see what can be salvaged

Lol

Balarka Sen

OK, I was thinking of orientable vector bundles over $M$. Rank $n$ ones are classified by $[M, BSO(n)]$, yes?

Mike Miller

Yes, keep going

Balarka Sen

There's a fibration $S^{n-1} \to BSO(n-1) \to BSO(n)$

I dualize that to $[M, S^{n-1}] \to [M, BSO(n-1)] \to [M, SO(n)]$

Mike Miller

17:17

Yes, so you have the exact sequence of pointed sets $\pi^{n-1}(M) \to \text{Vec}_{n-1}(N) \to \text{Vec}_n(M)$

Balarka Sen

If the two terms are trivial, this forces $[M, S^{n-1}] = 0$.

Maybe not

It's not short exact

Leaky Nun

hey @Ted

Ted Shifrin

Hi @Leaky, @MikeM, a @Balarka

Leaky Nun

Balarka is presenting his PhD thesis here

Balarka Sen

$[M, SO(n-1)] \to [M, SO(n)] \to [M, S^{n-1}] \to [M, BSO(n-1)] \to [M, SO(n)]$ is the correct thing.

All I get is $[M, SO(n)] \to [M, S^{n-1}]$ is surjective, I suppose?

Mike Miller

17:19

Yeah the issue is you can never say anything about the last map (which should be to [M, BSO(n)]) being surjective

Balarka Sen

(For some reason I was thinking all cohomotopy groups of $M$ are trivial, which is not possible since $[M, S^{\dim M}] = H^{\dim M}(M) = \Bbb Z$)

Yeah I am stupid

anakhro

waves at Ted from afar.

Balarka Sen

@MikeMiller Is the counterexample extremely easy?

Ted Shifrin

Oh, hi, @anakhro.

Mike Miller

Just to be clear there are no principal G-bundles over $S^3$ for any compact Lie group $G$.

ÍgjøgnumMeg

17:20

Henlo @Ted

Ted Shifrin

Hi @ÍgjøgnumMeg

Mike Miller

This is a theorem of Bott from (equivariant) Morse theory.

He proves $\pi_1 \Omega G = 0$, which is to say $\pi_2 G = 0$, which is to say $\pi_3 BG = 0$.

Balarka Sen

Oh. $\pi_2(O(n)) = 0$ for all $n$?

anakhro

Is there a good modern textbook on Morse theory? That is, not Milnor's book.

Balarka Sen

Oh

Fuck I should have known this

Ted Shifrin

17:21

$\pi_2$ of any L.G. is $0$, @Balarka.

Mike Miller

That's what I said!

Ted Shifrin

(Although I've forgotten the proof.)

Balarka Sen

I know the result but forgot it completely

Ted Shifrin

I can't read to keep up with you, @MikeM.

Balarka Sen

Anyway this proves how much topology I have forgotten

Ted Shifrin

17:23

Or how much I never knew.

anakhro

or how not dumb of a question it was, you let me down mr sen

>: (

anakhro

17:37

I showed this to my business calculus students:

I thought it was pretty cute.

Ted Shifrin

Business calc students are being told about uniform convergence? Um ...

I guess the point is that triangles would work just as well ...

anakhro

@TedShifrin I never phrased it as "uniform convergence".

s.harp

lets calculate some business

Ted Shifrin

So this is a baby version of what goes wrong with approximating surfaces to get surface area.

anakhro

I just asked them what they thought the red family of curves was "limiting" or "converging" or "getting closer" to.

"the line" was the consensus. :P

Ted Shifrin

17:42

In the $C^0$ norm, I assume you stressed :D

anakhro

I just kept it at a visual thing. ;)

Ted Shifrin

But, gee, the lengths don't converge, daddy!

I mean, they do, of course, but not to the right number.

Leaky Nun

2 = pi confirmed

Balarka Sen

The infamous reason why h-principles are true

Poline Sandra

hello, please what is equal to $\sum_{k=1}^n \frac{1}{2^k}$

Ted Shifrin

17:44

@Poline: You have to know that yourself.

Leaky Nun

well it's always equal to itself, we can be sure of that

anakhro

Try a few examples by picking your favourite $n$, @PolineSandra

Balarka Sen

Any curve in R^3 can be uniformly approximated by curves of constant curvature... in C^0 norm

Ted Shifrin

Hint: $1/2^k = (1/2)^k$.

I don't like the $C^0$ norm, a @Balarka.

I demand $C^1$.

Poline Sandra

it is a geometric series with reason 1/2 , but the first term is 1/2

Ted Shifrin

17:45

So what? You can make it start with $1$ by doing what?

Balarka Sen

@TedShifrin OK, it's also true for $C^1$ norm

Ted Shifrin

It is? Agh.

Poline Sandra

@TedShifrin I don't know

anakhro

The curves can easily be changed to be smooth in that problem. They don't have to be those semi-circles.

Balarka Sen

Curvature is a serious obstruction only in $C^2$, of course. But yeah, this continues to hold.

Ted Shifrin

17:47

Think back to basic algebra principles, @Poline.

Balarka Sen

@anakhro High frequency low amplitude waves, yeah

Ted Shifrin

Yeah, we can smooth triangles easily enough.

Poline Sandra

we start by 0 and finish by n-1?

anakhro

@TedShifrin you can tell you are a geometer.

Ted Shifrin

Well, sorry, @anakhro. I don't think that was hidden from anyone in here :P

Yes, @Poline, but what do you do to arrive at that?

anakhro

17:51

@TedShifrin you can just use $f_n(x) := (1/n)\sin(nx)$ on $[0,\pi]$ for a nice smooth result.

I guess the more interesting part of the question is if the length function isn't continuous, then what is the strongest "continuity"-like result that holds?

Ted Shifrin

Sure, @anakhro, I know :P

You need $C^1$ convergence.

It is continuous if you're working in $C^1$ with the appropriate norm, of course.

Maybe Lipschitz convergence is good enough. Exercise for you.

anakhro

Well I was thinking more like semicontinuity. :P

Ted Shifrin

That's immediate from the definition of length.

I'm not sure that's interesting.

Leaky Nun

I mean length isn't really defined for any continuous curve

or maybe it's variation

Ted Shifrin

No, but if you have a rectifiable curve, ... what's the definition?

anakhro

17:54

@TedShifrin well I think lower semicontinuity is as strong as it gets. That's the interesting part.

I don't think you can guarantee any better unless you change the notion of convergence.

But I haven't thought about it too much more than that.

Balarka Sen

18:08

@MikeMiller So in general nothing can be said about surjectivity of [M, SO(n)] -> [M, S^(n-1)]? I was wondering for which manifolds it is true that every vector bundle is trivial

anakhro

{*}

puts his arms up in victory.

Ted Shifrin

I thought that meant you gave up.

anakhro

Putting one's arms up?

A quick google images search for "victory" yields people with arms up in the air.

Leaky Nun

try searching for "surrender"

anakhro

A lot of christian imagery, @LeakyNun, and white flags

Poline Sandra

18:28

@TedShifrin we do a change of variable ?

@TedShifrin we get $\sum_{k=0}^{n-1} (1/2)^{k-1}$

schn

To find the negative index of inertia of a quadratic form, one can count the sign changes in the sequence $D_n$ of the determinant of the $n$ by $n$ matrix representing the quadratic form. Say $D_i$ equals 0. Is this a sign change?

anakhro

@PolineSandra I think the nicest way of thinking about it is that you have some sum $S$ which you know is the geometric sum $G$ without the first term. So $S = G - \text{$1^{st}$ term}$.

Mike Miller

@BalarkaSen I can't really see why one would be able to say something about that, no.

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$Mathematics$ Mathematics