Mathematics - 2019-06-13 (page 1 of 2)

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12:08 AM

@user193319 the orbit of x

@Leaky: I would write that as $G\cdot x$.

or $\mathscr O_x$.

but that seems like the most likely explanation

I'm used to notation like that for $G$-invariants, but it didn't make sense with just $x$. So I hushed.

oh right there's also that

@Ted btw maybe your president finally did something correct

That's what I figured. Thanks!

12:10 AM

Doubtful.

@TedShifrin scmp.com/news/hong-kong/politics/article/3014267/…

I interpret that as kissing Xi's ass.

I'm still trying to find a primary source

you really need to fact check everything these days

I can't find it here whitehouse.gov/briefings-statements

You won't find facts there!

> Hong Kong’s chief executive, Carrie Lam, has said her government will push ahead with the bill after the historic march on Sunday and fresh protests this week.

this is bad

oh and there's this petition petitions.whitehouse.gov/petition/…

8 hours ago there were 56,576 signatures

now it has 91,930 signatures

it's growing really fast

@TedShifrin will you sign the petition?

12:19 AM

I just did.

thanks

scmp.com/video/hong-kong/3014269/…

Well, like no other site I've been to, it wants to validate my email before it will accept my signature. And I've gotten no email to which to reply.

wait you've never signed a petition on that website?

I've signed zillions of petitions. I don't recall.

check the spam folder

12:22 AM

I did.

that's strange

The WH may already have me listed as an evil person.

ok I finally found it

> Q (Inaudible) your reaction to the demonstrations in Hong Kong? Is China overplaying its hand here?

PRESIDENT TRUMP: Well, they’re massive demonstrations. I looked today, and that really is a million people. A lot of times, people talk about they had 2,000 people but it was really 1,000 or it was 200. I see it all the time. I see it all the time. But when you look at this demonstration, they said it was a million people, and that was a million people. That was as big a demonstration as I’ve ever seen. So, I hope it all works out for China and for Hong Kong.

whitehouse.gov/briefings-statements/…

that's the vaguest speech I've ever read

You haven't been following this illiterate president.

12:35 AM

user image

$g$ is holomorphic, since $g$ is analytic in the puctured disk, and continuous at $0$.

$|g(z)|\leq 1$ by using maximum modulus principle.

(c) I tried to use Schwarz pick lemma $|f'(z)|\leq \frac{1-|f(z)|^2}{1-|z|^2}$. But without the knowledge of $f(z)$ I can't.

yay Schwarz lemma related stuff

just be aware that they never told you that $f(0)=0$

yes

so you can pick say f(z)=0.5

then g is immediately not holomorphic

okay

How do we find the bound for the derivative?

mathworld.wolfram.com/Schwarz-PickLemma.html

but for Schwarz pick lemma we don't need $f(0)=0$

I think you can prove (4) using generalized Cauchy integral formula

(1) and (2) are disproved using my example

12:46 AM

yes.

Thank you very much for example.

I have no idea about (3)

okay. let me ask in main.

1:04 AM

0

Q: Question for finding bound for $f"(z)$

My attempt:- (1)Taking $f(z)=.5$, So, $g(z)= \begin{cases} \frac{.5}{z} & z\neq 0 \\ 0 & z=0 \end{cases} $ So, I can eliminate (1) and (2) I am trying to apply Schwarz pick lemma for (c), But I am not able to make $|f'(z)|\leq \frac{1-|f(z)|^2}{1-|z|^2}\leq 1$ Please help me.

complex-analysis

2 hours later…

2:52 AM

To be discussed later in Star Wars Room: Triangular noncommutative lattice as a model of tridirectional spaces

On another note: Trump response is very disappointing, but then the past experience of USA meddling with other countries often do not end well, so whatever...

3:37 AM

Hello

Simplify $$d/dx(\int^{x^3}_{t^2}\frac{dx}{\sqrt{x^2+t^4}})$$

Ignore the above

Its actually

Simplify $$d/dx\int^{x^3}_{t^2}\frac{dt}{\sqrt{x^2+t^4}}$$

The answer given is

${\frac{3x^2}{\sqrt{x^2+x^12}}$$-\int^{x^2}_{t^2}{\frac{xdx}{(x^2+t^4)^{3/2}}}$

Ignore above.The answer given is
${3x^2}/{\sqrt{x^2+x^{12}}}$$-\int^{x^2}_{t^2}{\frac{xdx}{(x^2+t^4)^{3/2}}}$

Please someone help

4:13 AM

@Jasmine Are you there?

@AjayMishra yes

If yes, you have probably used $x^2$ instead of $x^3$ in the limits

4:52 AM

Gallian, in his book Contemporary algebra says,

Consider $a^2-5b^2=\pm7$. (We are trying to find integral solutions $a,b$.) Viewing this equation modulo 5, and trying all possible cases for $a$ reveals that the only solution is $a=0$. But this means that a is divisible
by 5, and this implies that $|a^2-5b^2|=7$ is divisible by 5, which is false.

So, we look at $a^2=2$ or $a^2=3$, right? and then plugging $a=0,1,2,3,4$ does not solve this equation. Isn't this enough? Why do we need extra line observing '$|a^2-5b^2|=7$ is divisible by 5, which is false'?

Akiva Weinberger

5:41 AM

@Silent Where did you get $a^2=2,3$ from?

@AkivaWeinberger $a^2-5b^2=7\pmod5$ is $a^2\equiv2\pmod5$ and $a^2-5b^2=-7\pmod5$ is $a^2\equiv3\pmod5$.

Am i right?

Subhasis Biswas

Let ${A}$ be a countable subset of $\mathbb{R}$ which is well-ordered with respect
to the usual ordering on $\mathbb{R}$ (where ‘well-ordered’ means that every
nonempty subset has a minimum element in it). Does ${A}$ have an order
preserving bijection with a subset of $\mathbb{N}$?

Now, one doubtful case is when $A=(2,1,3,5,6,8,10,9,...)$

Akiva Weinberger

@Silent Oh yeah that looks right

I'm now confused by the proof from the book

Thank you!

huh :)

Subhasis Biswas

$A$ just has to contain a minimal element, but does not have to maintain the order?

Akiva Weinberger

5:54 AM

@SubhasisBiswas $A$ is a set, not a sequence

Subhasis Biswas

@AkivaWeinberger any countable set can be arranged into a sequence

define $f(n)=x_n$

Akiva Weinberger

$A=\Bbb N$, which clearly has an order-preserving bijection with $\Bbb N$

(the identity map)

You want a subset of $\Bbb R$ where no such bijection exists

Subhasis Biswas

@AkivaWeinberger does it hold for ALL $A$?

yes.

Akiva Weinberger

The one you just gave

Subhasis Biswas

@Silent hello

@AkivaWeinberger works, right?

5:56 AM

hello !

Subhasis Biswas

@Silent do you have any interesting questions?

How is it going?

@SubhasisBiswas not right now! but reading about UFD, PID, ED. So I am sure that i will be having questions every hour.

what are u studying these days?

Subhasis Biswas

@Silent I have not yet read about those. I will start in a while

Akiva Weinberger

@SubhasisBiswas ?

Subhasis Biswas

@AkivaWeinberger my example.

5:58 AM

alright

Akiva Weinberger

No, the $A$ you just gave has an order-preserving bijection with $\Bbb N$

Note that the order we're using is the usual order on $\Bbb R$

not the order of the sequence you put it in

Subhasis Biswas

the answer provided is false. I don't know how that is

Akiva Weinberger

You said $A=\{2,1,4,3,6,5,\dots\}$

but that equals $\{1,2,3,4,5,6,\dots\}$

which clearly doesn't satisfy the hypothesis

Subhasis Biswas

I agree

Akiva Weinberger

You need to use nonintegers

Subhasis Biswas

6:08 AM

I still don't know how that is going to work either

the set will still remain in bijection with N. And ordering can be preserved

Akiva Weinberger

I don't know how to give a good hint without just giving away the answer :/

Note that the fact that it's well-ordered is important

Subhasis Biswas

@AkivaWeinberger give away the answer

wait

Akiva Weinberger

Otherwise you could do things like $\Bbb Z$ or $\{1/n:n\in\Bbb N\}$ or $\Bbb Q$

Subhasis Biswas

@AkivaWeinberger indeed

is the result true?

Akiva Weinberger

Also note that $\{1-1/n:n\in\Bbb N\}$ is well-ordered, but it has an order-preserving bijection with $\Bbb N$

Subhasis Biswas

6:14 AM

or, I could use this

Akiva Weinberger

(and $\{-1/n:n\in\Bbb N\}$ which I guess is a bit simpler)

Subhasis Biswas

$\{n, 1/n\}$

Akiva Weinberger

That's not well-ordered. What's its smallest element?

You mean $\{n,1/n:n\in\Bbb N\}$?

Subhasis Biswas

@AkivaWeinberger yeah

If I have a metric and a connection on my manifold such that they are compatible with each other, if I lift them to the tangent bundle (assuming such a lift can be defined) will the compatibility condition still remain even on the tangent bundle?

Akiva Weinberger

6:16 AM

@Albas I think so?

Subhasis Biswas

i don't think it is going to work. The set is not going to be well ordered

it has to "contain" a minimal element

or we could just adjoin $\{0\}$ to it

Akiva Weinberger

OK, hint: the "order type" of $\Bbb N$ and $\{-1/n:n\in\Bbb N\}$ is called $\omega$. The set we're looking for has an order type called $\omega+1$

@SubhasisBiswas Sure but then what's the smallest element greater than zero

Let $A$ be a nont necessarily symmetric matrix. If for all non zero vector $x$, $x^{T}Ax>0$, how is $A$ non-singular. I am asking this with reference to the linked question, where the first answer concludes that $A$ is non-singular and then says that determinant is hence non-zero.

12

Q: Does a positive definite matrix have positive determinant

Let $A$ be a positive-definite real matrix - in other words, $x^T A x > 0$ for every real vector $x$. I don't require $A$ to be symmetric. Does it follow that $\mathrm{det}(A) > 0$?

linear-algebra matrices

@AkivaWeinberger Hmm okay. So what I am kind of trying to do is take a killing vector field on the manifold and look at it at the tangent bundle and check if it gives me anything new over there ?

Akiva Weinberger

@AnjaniGupta If $A$ were singular there would be some $v$ such that $Av=0$

meaning $v^\top Av=0$

@Albas I don't actually know enough about this, sorry

6:20 AM

ohhh! thanks a lot!

Subhasis Biswas

@AkivaWeinberger we are looking for an order that is not isomorphic with $\omega$, the order on $N$?

Akiva Weinberger

Yes

Two sets are said to have the same order type if there is an order-preserving bijection between them

OK I'll just say it

$\{-1/n:n\in\Bbb N\}\cup\{0\}$

$\{-1,-\frac12,-\frac13,-\frac14,\dots,0\}$

Any time you have $\{a_0,a_1,a_2,\dots\}\cup\{b\}$ where $a_0<a_1<a_2<\dotsb$ and $\forall n,a_n<b$ you can do this

Subhasis Biswas

got it

Akiva Weinberger

Oh, you know what, a good hint would've been "Give me a well-ordered countable set with a maximum element"

Subhasis Biswas

but, what is the next element after $0$?

Akiva Weinberger

6:26 AM

It's the maximum

Well-ordered means every nonempty subset has a minimum element

As a consequence, every element other than the maximum has a successor

but the maximum doesn't need a successor

If you want one with no maximum you can do $\{-1/n:n\in\Bbb N\}\cup\{1-1/n:n\in\Bbb N\}$

(This has order type $\omega+\omega$)

or $\{-1/n:n\in\Bbb N\}\cup\Bbb N$, which has the same order type

Subhasis Biswas

@AkivaWeinberger I have to read about ordering first

Akiva Weinberger

This stuff, by the way, is very drawable

user image

That's an illustration of something called $\omega^2$

(a.k.a. $\omega\cdot\omega$ a.k.a. $\omega+\omega+\omega+\dotsb$)

The vertical bits are just to make it easier to see, it's really just the set of $x$-coordinates

{1/n + 1/m : n, m in N} I believe

It's a good example of a subset of R with the set of limits points itself having a limit point

Akiva Weinberger

@BalarkaSen The negative of that

{-1/n-1/m}

Otherwise it's backwards

Ah yeah sure

Akiva Weinberger

6:36 AM

math.ucr.edu/home/baez/week236.html

^Good introduction to ordinals

They're not subsets of $\Bbb R$ but they're well-ordered

and in fact every well-ordered set has an order-preserving bijection with ("is order-isomorphic to") some ordinal

Say I start with some subset of the powerset of a set. The topology generated by that subset would be the intersection of all topologies containing this subset. What if I inductively want to generate this smallest topology by taking their intersections and unions and adding them to the subset if it was not there already. Would this process terminate eventually to give the same topology as I would get from intersections?

Akiva Weinberger

That's a weekly blog so there's some stuff on that page that is unrelated to the ordinal stuff

madore.org/~david/math/drawordinals.html#?v=eee

I even doubt if it will terminate

Subhasis Biswas

@BalarkaSen hey man. One question

For any closed set $A$, there exists a continuous function $f$ on $R$ that vanishes exactly on $A$

okay

Akiva Weinberger

6:39 AM

@AnjaniGupta What about the discrete topology on an uncountable set like $\Bbb R$?

That's generated by the set of singletons

@SubhasisBiswas I don't see a question :P

Akiva Weinberger

($\big\{\{x\}:x\in\Bbb R\big\}$)

but any union of those will only get you finite sets

Subhasis Biswas

i will answer it later. But, would the result hold true if the set $A$ were open?

I believe not

Akiva Weinberger

when in fact the topology is all sets

Of course not

Subhasis Biswas

6:40 AM

since $Z(f)$ is a closed set

right?

Akiva Weinberger

and even if you take the "limit", whatever that means, you'll only get at most countable stuff I think?

@SubhasisBiswas You shouldn't be asking for verification on this

Subhasis Biswas

ok ok

now, I will try to "create" a function to show that the result is indeed true. Check that out

Akiva Weinberger

"Exactly" on $A$ meaning it vanishes nowhere else?

Subhasis Biswas

the problem is, what about the cantor set?

Akiva Weinberger

6:42 AM

So for example a continuous function that's only zero on $(0,1)$?

Subhasis Biswas

how would I do that?

@Albas Yeah, given a metric and a metric connection on $M$ there's a canonical metric and a metric connection on $TM$ which are compatible, sure.

Subhasis Biswas

@AkivaWeinberger it won't hold true

Akiva Weinberger

@SubhasisBiswas $f(x)={}$the distance from $x$ to the Cantor set

I think the question went unclear. Start with some 'arbitrary' space $X$ and take a subset of its powerset (which may not yet be a topology). Let that not be a topology because the intesection of some two elements of the subset is not within the subset. Add it now to the subset. But still to see it makes(or not) a topology, one has to recheck all the axioms with this new element.

Akiva Weinberger

6:43 AM

The graph of that function looks like a lot of "mountains"

"/\"

Just take a limit of continuous functions vanishing on the middle-third process at each step, scaled appropriately so it converges uniformly, if you want a picture.

Akiva Weinberger

One big one, two smaller ones, four even smaller ones between these, etc

and a big \ on the left and a / on the right

Subhasis Biswas

What I started with is this:

Firstly, consider a closed interval $[a,b]$. Define $f(x) = (x-a)$ on $x \in (-\infty, a)$, $f(x)=0$ on $[a,b]$ and $f(x)=(x-b)$ on $x \in (b, \infty)$

Or, I have another idea

@BalarkaSen I see. So then like if I have a killing vector field in my manifold then I can also in a certain sense lift this Killing vector field to another Killing vector field because I can lift the metric and the metric connection?

The above question might be a little vague. I can try to formalise it a little. The basic idea is to try to understand the tangent bundle by using killing vector fields of the underlying manifold.

@Albas Let $\pi : TM \to M$ be the bundle projection of the tangent bundle. Given a metric $g$ and a metric connection $\nabla$, there's a canonical decomposition $TTM \cong V \oplus H$ of the double tangent bundle into a "vertical" and "horizontal" factor, where $V = \ker d\pi$ is what the differential of the projection kills and $H$ is a complementary distribution that guides the $\nabla$-parallel lifts of vector fields on $M$.

$g$ induces a fiberwise metric $g_v$ on the bundle $V = \ker d\pi$ (identify the fiberwise tangent spaces with the tangent spaces by trivializing $T T_p M \cong T_p M \times T_p M$) and a fiberwise metric $g_h$ on $H$ by $g_h = d\pi^* g$.

Akiva Weinberger

6:52 AM

Fun fact: the population of the New York City Metropolitan Area is larger than the population of New York State

Then just define the total bundle metric on $TTM$ by $g_h + g_v$. That's the canonical Riemannian metric on $TM$.

Is there something like en.wikipedia.org/wiki/… but for set theory?

$\nabla$ pulls back similarly, because you can set the connection to be flat along $V$ and just $d\pi^* \nabla$ on $H$.

Okay

Akiva Weinberger

(New York State, about the size of Greece, is the state that New York City is in)

Subhasis Biswas

6:54 AM

$\chi_{A}(x) = 0$.

@Albas Should be right, sure, why not?

Subhasis Biswas

now...

define $f(x)=min{|x-p|}$, $p\in A$

$x \notin A$

$f(x)=0$ on $x \in A$

is it going to work? @BalarkaSen

Yeah but have to formally write it down though. Also I am just guessing but this kind of a canonical construction for lifting the metric and the connection will not necessarily work for any kind of a vector bundle.

Subhasis Biswas

or rather $f(x) = \inf |x-p|$, $p \in A$, $x \notin A$.

@Albas Well you need the additional data of a bundle metric for a general vector bundle to start off. A Riemannian metric on $M$ gives a bundle metric on $TM \to M$ automatically so this extra piece of information is redundant in that setup.

But if you have a metric on the base, a fiberwise metric connection and a fiberwise metric on the bundle, sure, there's a canonical metric on the total space by an identical construction.

I don't quite understand what it means to "lift the connection for a general vector bundle" though. Lift what to what?

7:08 AM

So I mean that if you have a connection on the underlying manifold then lifting that connection from the underlying manifold to a vector bundle defined on the manifold.

I don't know what that means, is the thing. A connection on a vector bundle is a very different beast; a connection on $M$ is an operator $\Gamma(TM) \times \Gamma(TM) \to \Gamma(TM)$. A connection on a vb $E$ over $M$ is an operator $\Gamma(TM) \times \Gamma(E) \to \Gamma(E)$

Ah okay. I should not talk about this. I clearly don't know it properly.

Subhasis Biswas

7:54 AM

is this going to be a valid series : $ \lim_{n \to \infty}\sum_{m=0}^\infty(1/2)^{n-m}$

?

It came up while I was trying to create a nonnegative continuous function on $[0, \infty)$ such that $\int_0^\infty {f(t) dt} = \mathbb{a \ finite \ value} $

The function I came up with is

$y(x)=\lim_{n \to \infty}[1-\{x-(2m+1)\}^2](\frac{1}{2})^{n-m}$, $x \in [2m, 2m+2]$

$m \in \{0, 1, 2, ...\}$

@AkivaWeinberger , can you please check this out

the graph of $y$ without the factor $\frac{1}{2}^{n-m}$ would look like infinite number of small bumps all over the real line from $0$ to $\infty$

The actual question asked, if such a function exists, whether is it always true that $\lim_{x \to \infty}f(x)=0$?

Subhasis Biswas

8:27 AM

3

A: finding examples for a non negative and continuous function for which the infinite integral is finite but the limit at infinity doesn't exist

For (b), and therefore (a), let $f(x)=0$ with the following exceptions. For every positive integer $n$, $f(x)$ climbs linearly from $f(x)=0$ at $x=n-2^{-2n}$ to $f(x)=2^n$ at $x=n$, then falls linearly to $0$ at $x=n+2^{-2n}$. The area of the triangle "at" $n$ is $(2^{-2n})(2^n)$, that is, $2^...

wow

2 hours later…

10:19 AM

Isnt $$ \lim_{(x,y) \to (\alpha,0)} \biggl(1 + \cfrac{x}{y}\biggr )^y = 1$$ ?

1 hour later…

11:42 AM

Does anyone know how to use wolframalpha to multiple two permutations (in the sense of $S_n$)?

Figured it out: blog.wolframalpha.com/2011/11/29/…

What does the presence of a baby manbrot set indicates about the dynamics of the system?

2 hours later…

1:27 PM

Just as a sanity check $f:\Bbb R^3\to \Bbb R^2$ given by $f(x,y,z)=(x+y^2,y+z^2)$ is submersive right? I was asked to sketch the set of critical points, which seems to be empty

1 hour later…

2:45 PM

@AkivaWeinberger well, sagemath isn't going to be much help: it only lets you do polyhedra computations over the rationals, whereas rotation matrices will in general introduce square roots

so rip to that

3:21 PM

@AkivaWeinberger Conjecture: for the case of (rotational) tetrahedral symmetry, the possible polyhedra include the tetrahedron and the tetrakis hexahedron

Akiva Weinberger

Interesting

pretty close to testing it out, though i dunno how quick it'll run

Akiva Weinberger

That has 24 faces, the symmetry group of the tetrahedron only has 12 elements

How do you get more faces than things in the group

Maybe the triakis tetrahedron?

hmm. simplest example I'm getting right away is just the cube

not entirely surprising: tetrahedral symmetry = A4 is a subgroup of cubic symmetry = S4

Akiva Weinberger

Yup

Two-color the vertices of the cube, the vertices of one color form a tetrahedron

so you get tetrahedral symmetry by looking at symmetries that preserve that coloring

3:36 PM

In fact there are two such embeddings, which are the two conjugate A_4's in S_4

looks like the rhombic dodecahedron shows up too

which has the same symmetry group as the cube, so no surprise there

@AkivaWeinberger looks like you were right about this

found the tetrahedron now as well

Akiva Weinberger

@BalarkaSen Two embeddings but only one subgroup, yeah?

Ya

Well leaky, maybe you are not brave enough to ask the really pertinent questions at hand there, maybe we need to start asking why any federal laws should matter to any of us, maybe it's time we start seeing international law as more than a diplomatic farce

Akiva Weinberger

Like we have $\iota_1,\iota_2:A_4\to S_4$ but they have the same image

3:44 PM

no sign of the triakis hexahedron rho

Akiva Weinberger

@Semiclassical Oh, that's interesting

And it has the right number of sides so that's good

ya

Akiva Weinberger

Hm. Bipyramid:trapezohedron::triakis tetrahedron:rhombic dodecahedron?

I never thought I'd be looking at a question where all these shapes come up naturally

dunno about that yet

Akiva Weinberger

You can partition the 3akis tetrahedron into four groups of three sides that each correspond to a side of the tetrahedron

(In fact you can do that it two ways 'cause it's self-dual)

The same is true of the rhombic 12hedron

3:47 PM

test:

user image

dangit, i always forget how to get wikipedia images

user image

there we go

that's the fundamental domain of tetrahedral symmetry

going back to Mike's original formulation (rather than mine) each polyhedron should amount to a choice of where the cutting plane axis falls within the fundamental domain

i.e. one polyhedron for each point in the yellow portion

I think the diamond yields the cube, and the triangles yield regular tetrahedrons

Akiva Weinberger

@Semiclassical It took me way too much staring to understand what I'm looking at

@Semiclassical There are six edges ("diamonds") so that makes sense

(though I think the orientation you get for the tetrahedra at the 'red' points is opposite that for the 'purple' points)

Akiva Weinberger

Equivalently the diamonds are 2-symmetries and 12/2=6

so the 12 faces that we usually get collapse into 6

bravery is a very underrated virtue actually

Akiva Weinberger

Found this GIF again

Stared at it for a bit and now the world looks like it's warping

I think it's called the waterfall effect

3:55 PM

oof

yeah

Akiva Weinberger

'cause apparently if you state at a waterfall for too long, things you stare at start looking like they're moving upwards

okay, halfway between the diamonds and the triangles seems to give triaksis tetrahedra

Akiva Weinberger

Motion aftereffect

The motion aftereffect (MAE) is a visual illusion experienced after viewing a moving visual stimulus for a time (tens of milliseconds to minutes) with stationary eyes, and then fixating a stationary stimulus. The stationary stimulus appears to move in the opposite direction to the original (physically moving) stimulus. The motion aftereffect is believed to be the result of motion adaptation. For example, if one looks at a waterfall for about a minute and then looks at the stationary rocks at the side of the waterfall, these rocks appear to be moving upwards slightly. The illusory upwards movement...

halfway between the triangles seems to give the rhombic dodecehedron

whoa

Akiva Weinberger

Whoa re: motion aftereffect or re: shapes

3:59 PM

the midpoint between all three of the points seems to give the regular dodecahedron

wait no

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