Mathematics - 2019-05-16 (page 1 of 3)

schn

12:01 AM

@Semiclassical Okay. The actual function was $\frac{x+2}{\sqrt{x^3+x^5}}$. And this was simplified to $\frac{2+0}{\sqrt{x^3}+0}$. When I draw this in Desmos, they are almost similar, and so I can see why the definite integral would be the same.

Right, cause this function behaves very much like $1/x$, so it's the small x that probably contribute most to the integral. Thanks a lot!

Ted Shifrin

12:20 AM

@schn: To be a bit more precise. When $x\approx 0$, $x^3+x^5$ looks like $x^3$. So the function looks like $2/\sqrt{x^3} = 2x^{-3/2}$.

This sort of reasoning is very important for looking at infinite series and improper integrals.

Jacksoja

@TedShifrin Hi uncle Ted

Keshav Srinivasan

12:57 AM

Is anyone here familiar with LyX (the LaTeX editor)?

Ted Shifrin

LOL, hi @Jacksoja.

Jacksoja

@TedShifrin hehe

I was meaning to ask you about a smal problem

Ted Shifrin

What's that?

Jacksoja

in trying to prove that there cannot be a surjective map between S and S* ( the subsets of S)

we are not assuming finetness of S

Ted Shifrin

Ah, this is a classic problem.

Jacksoja

1:05 AM

am not sure how to tackle it

the idea is , one element subsets would exaust all of S

hence there are no room for 2 or more element sets

but this arguement does not work for infinity right?

Ted Shifrin

Nope, it doesn't work.

Jacksoja

i thought so

can you drop me a hint?

Ted Shifrin

There are fancy ways with infinite arithmetic, but I think you can do a Bertrand's paradox type proof. You suppose you have a surjection, and you invent a subset that cannot possibly be in the image of it.

Jacksoja

okay

i have to google this Bertnand

Ted Shifrin

well, that gives it away.

Jacksoja

1:12 AM

okay then later I shall

ÍgjøgnumMeg

I'm desperate for a pun but that would give it away too

Ted Shifrin

LOL, @ÍgjøgnumMeg. Bad boy.

Jacksoja

@ÍgjøgnumMeg Hi , go ahead do that pun

:D

ÍgjøgnumMeg

I shan't!

I cannot, nor will I

Jacksoja

What Ted said holds true then

Ted Shifrin

1:13 AM

@Jacksoja: Suppose you have a map $f\colon S\to S^*$. Can you think about whether $s$ and $f(s)$ are related?

That's a generous hint.

Jacksoja

well i do want that s to be in f(s)

so f(s) is a proper name of it

the subset of S that contains our element s

but what else does it contain ?

Ted Shifrin

Maybe. That's not quite the idea. You want to make a subset of $s$'s with some property.

Jacksoja

if just a singelton, then i cannot do much with it

Ted Shifrin

Make a subset $X = \{s: \text{something holds}\}$. Can $X=f(t)$ for some $t\in S$?

Jacksoja

okay I see

so the idea I want to make a subset of S , that is not reached by my funtion

Ted Shifrin

1:18 AM

Right.

Jacksoja

no matter what function is

hmm that is quite evolved

Ted Shifrin

@ÍgjøgnumMeg: Have I misled?

ÍgjøgnumMeg

No that's what the standard method is I think?

Ted Shifrin

I'm just trying not to give it away entirely. Yes, I know only the standard method.

ÍgjøgnumMeg

It's just a nice choice of something holds I guess

Ted Shifrin

1:21 AM

That's for @Jacksoja to contemplate.

BBIAB

Jacksoja

would this be good, if i consider a subset of T of S such that, for each a in S, f(a) and T differ in at least one element

@TedShifrin what is BBIAB?

darn it ,I assume you had to go, thanks anyway all

ÍgjøgnumMeg

@Jacksoja I assume "Be back in a bit"

Jacksoja

@ÍgjøgnumMeg such a short sentence why not spell it out as it is

old generations are driving me crazy with these abbriviations

ÍgjøgnumMeg

1:36 AM

Let $M$ be an $R$-module and $\varphi : R \to S$ a ring hom. Then $S$ is an $R$-module with $R$-action given by $rm := \varphi(r)m$. Thus we can form $M^S := M \otimes_R S$.. this is an $S$-module with $S$-action given by $(m \otimes s) \cdot s^\prime = m\otimes ss^\prime$

so I wanna check that this $S$-action is well defined

Ted Shifrin

@Jacksoja LOL ... Mostly they come from your generation, guy.

ÍgjøgnumMeg

and I guess $\varphi$ lets me do that?

Ted Shifrin

Well, of course you need to use $\varphi$ :P

@Jacksoja: My hint was that your condition should involve relating $s$ and $f(s)$.

What you wrote is a good idea, but how would you define such a $T$?

Oh, @Rithaniel snuck in.

ÍgjøgnumMeg

So I have $m \in M$ and $r_1, r_2 \in R$, $s_1, s_2 \in S$ I have $m \otimes (r_1s_1 + r_2s_2)s^\prime$.. and the expression $r_1s_1 + r_2s_2 = \varphi(r_1)s_1 + \varphi(r_2)s_2$ and you get $m \otimes (r_1s_1s^\prime + r_2s_2s^\prime)$

and then by $R$-bilinearity $r_1(m \otimes s_1s^\prime) + r_2(m\otimes s_2s^\prime)$

Spam end

Balarka Sen

My sleep schedule is fucked

ÍgjøgnumMeg

1:49 AM

S a m e

N. Maneesh

Let $A$ be a $4\times 4$ matrix over $\mathbb C$ such that $\operatorname{rank} A=2$ and $A^3=A^2\neq 0$.Suppose $A$ is not diagonalizable.
$\exists$ a vector $v$ such that $Av\neq 0$ but $A^2v= 0$. Is the statement true or false?

Balarka Sen

It's holidays for me so I also have no "real" reason to fix it which makes it worse because I won't even though I want to

ÍgjøgnumMeg

I'm off work atm so I'm slowly ruining my life

N. Maneesh

I know that rank (A)=2. so Geometric multiplicity of ($\lambda=0$)=2

Ted Shifrin

LOL, a @Balarka. How unusual.

Balarka Sen

1:53 AM

@ÍgjøgnumMeg I wonder if this corresponds to pushforward of the coherent sheaf defined by $M$ on $\text{Spec}\, R$ to $\text{Spec}\, S$.

The jay lower star you know

(This is extension of scalars, of course)

ÍgjøgnumMeg

lol I'm not there yet, this is just me justifying that $k[X]/(f) \otimes_k \bar{k} \cong \bar{k}^{\deg f}$

N. Maneesh

So Eigen space corresponding to $\lambda=0$ consists of two linearly independent vectors. $E_{0}=span\{x_1,x_2\}$. I can rule out the cases and show that characteristic polynomial is $x^3(x-1)$.

Ted Shifrin

@N.Maneesh: I vote false. What do you vote?

No, that isn't the right characteristic polynomial.

N. Maneesh

Then obviously it must be false.

Ted Shifrin

Hmm.

What are you suggesting the Jordan normal form will be?

Balarka Sen

2:02 AM

@ÍgjøgnumMeg Oh, that's because the spectrum of the left hand side is the fibered product given by limit of the pullback diagram $\text{Spec}\, k[X]/f \stackrel{\pi}{\to} \text{Spec}\, k \leftarrow \text{Spec}\, \overline{k}$, which is nothing but the fiber over the basepoint $\text{Spec} \, \overline{k} \to \text{Spec}\, k$, i.e., $\deg f$ many copies of $\text{Spec}\, \overline{k}$ which is $\text{Spec}\, \overline{k}^{\text{deg}(f)}$!

Ted Shifrin

What's the algebraic multiplicity of $0$ and of $1$, @N.Maneesh?

Balarka Sen

Covering spaces ftw

ÍgjøgnumMeg

$$\operatorname{Spec}^{-\deg f}! = \operatorname{Spec}^{-\deg f}(\operatorname{Spec}^{-\deg f} - 1)(\operatorname{Spec}^{-\deg f} - 2) \cdots$$

uh oh

missed a $k$

N. Maneesh

$$
J=
\left[\begin{array}{rrrrr}
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1\\

\end{array}\right]
$$

Balarka Sen

@ÍgjøgnumMeg ripriprip

ÍgjøgnumMeg

2:04 AM

@Balarka sad

Ted Shifrin

So that has rank $3$, @N.Maneesh.

ÍgjøgnumMeg

I guess this book is supposed to be giving me an explanation for the obvious similarity between the galois correspondence for fields and for covering spaces, but i'm only on chapter 1

Balarka Sen

I don't know the explanation to be honest. Also I don't know any field theory; I can pretend to because I know covering spaces and I know the statement of the correspondence >:)

It's good for party tricks if you are a topology enthusiast want to surprise people with your quick Galois theory skills

N. Maneesh

$$
J=
\left[\begin{array}{rrrrr}
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1\\ \end{array}\right]
$$ sorry

Ted Shifrin

Use \begin{matrix}

ÍgjøgnumMeg

2:08 AM

Alas, I don't really know any topology "properly", only things I've gotten from reading number theory books

lol

Ted Shifrin

Yes, that is the only possibility.

Balarka Sen

You're on the other side. Learn the correspondence and come to topologists' parties to surprise people with your quick covering space theory skills

Ted Shifrin

So then the statement must be correct, @N.Maneesh.

You have to show that that is the only possible Jordan form.

OK, I'm disappearing.

N. Maneesh

@TedShifrin How can be from the unique jordan form statement correct?

ÍgjøgnumMeg

I like whole numbers too much

Despite my poor numeracy skills

Jackozee Hakkiuz

2:12 AM

Hey guys.

Take $n$ vectors $\{v_i\}$ in an $F$-vector space $V$.

Does the existence of linear maps $\phi_i:V\to F$ with $\phi_i(v_j) = \delta_{ij}$ imply linear independence of the $v_i$?

I mean, for any $\alpha_i\in F$ we have
$$\sum \alpha_iv_i = 0 \implies \sum \alpha_i\phi_j(v_i) = 0 \implies \alpha_j = 0$$

Is that the proof? haha.

I it's so simple I feel I'm missing something important hahaha.

N. Maneesh

2:29 AM

$$
J^2=
\left[\begin{array}{rrrrr}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1\\ \end{array}\right]
$$. For $J^2$, I can prove that for $J^2$ there is $v=(a,b,c,0)$. For suitable $a,b,c$, I can prove that $Jv \neq 0$ and $J^2v=0$. I don't know about $A$

BAYMAX

@N.Maneesh math.stackexchange.com/questions/1295229/…

N. Maneesh

2:47 AM

@BAYMAX We know that there is a matrix $P$ such that $J=P^{-1}AP$. So, $J^2=P^{-1}A^2 P$. Similar matrices have rank same. So, Rank of $A^2$ must be 1. So, Nullity($A^2)=3$. Nullity(A)=2. So, there is a vector $v$ such that $Av\neq 0$ and $A^2v=0$.

Is my argument correct?

BAYMAX

I am thinking about

> Nullity($A^2)=3$. Nullity(A)=2. So, there is a vector $v$ such that $Av\neq 0$ and $A^2v=0$.

@N.Maneesh

Rithaniel

3:52 AM

Ah yeah. I just leave this tab open on my computer. So I imagine that I just randomly pop up whenever I connect my internet. :] @TedShifrin

Jackozee Hakkiuz

4:15 AM

Hey.

I have vector spaces $U$ and $V$, and I'm trying to show that the usual construction $F(U\times V)/S$ (where $F$ is the free functor from Set to k-Vec) together with the map $\xi:U\times V\to F(U\times V)/S$ given by $(u,v)\mapsto [(u,v)]$ satisfies the universal property of the tensor product.

So I take an arbitrary bilinear map $f:U\times V\to W$, and I want to show it factors through $\xi$ via a unique $\tilde{f}$.

Haha is anybody following?

Rithaniel

This sounds like Category Theory, and, unfortunately, I do not personally know category theory well enough to be of any assistance.

Jackozee Hakkiuz

It's linear algebra. I just use categorical jargon because it's easier haha.

But thanks :)

Secret

5:10 AM

[Random]

On the problem of the Category of Mathematics

Consider the following diagram:

Top -> Ord

..........|

.........V

Rudi_Birnbaum

Hi all, where is the playground?

Secret

..........Alg

The morphisms that links Top to Ord and Ord to Alg may not be associative, as the morphisms Top -> Alg may be in general very different in structure

Thus mathematics is something much larger than a category

May 11 at 16:52, by Tobias Kildetoft

If you lecture long into the void, the void starts lecturing back.

@Rudi_Birnbaum what playground, there is no chat room call that

Rudi_Birnbaum

@Secret where I can test mathjax code (regularly).

I do it usually in the answer windows in MSE

that would be irregular I guess

Secret

Sandbox

Where you can play with chat features (except flagging) and ch...

5

Rudi_Birnbaum

@Secret Sandbox, that was the term! Thanks!!

Axoren

5:45 AM

I'm about to go to bed, but I don't even know how to ask this question in a front-page appropriate way. Take a hyperbolic plane ($H^2$), mapping it to a disk, and compactifying it (including the circumference). Adding a Euclidean dimension $X$, to turn it into a cylinder, what type of geometry is along the surface of the $\text{Circumference}\times X$?

I'm probably even talking nonsense at this point, in regards to "what type of geometry", but I'm just having trouble grasping a geometric concept discussed in a physic discussion with someone

And this shape came up

I can't even begin to reason about the Circumference of a compactified Hyperbolic Disk, let alone see how extending it with a euclidean dimension affects it.

What should I ask to accomplish learning anything?

N. Maneesh

7:02 AM

Let $f:[0,\infty)\to \mathbb R$ be a real valued function such that $f(1)=1$ and forall $x\in \mathbb R$ $f'(x)=\frac{1}{x^2+f(x)^2}$. Then $\lim_{x\to \infty}f(x)$

I know that $f'(x)\ge 0$. So, Monotonically increasing

$f''(x)\leq 0 \implies f$ is concave down

Silent

7:47 AM

Let $f:[0,1]\to \Bbb R$ be a continuous function, define $g:[0,1]\to\Bbb R$ and $g(x)=(f(x))^2$, then $\int_0^1g\,dx=0\implies\int_0^1f\,dx=0$. How to show this? I first thought that it is because $\int fg\,dx=\int f\,dx\int g\,dx$ kind of rule, but there is none such rule.

Subhasis Biswas

8:10 AM

can someone help me with this

1

Q: Proving that the set function $m$ is regular

I want to finish this proof. OP has considered the interval differently. I choose here, in $\mathbb{R}^p$, the interval $a_i - (\epsilon/2)^{(1/p)} ≤ x_i ≤ b_i + (\epsilon/2)^{(1/p)}$ (the closed set $F$). I am trying to show that $m(A)≤m(F)+ \epsilon$. Clearly, $F \subset A$. We break up the...

Leaky Nun

@Silent that's because $g$ is always nonnegative

so if its integral is zero, it must be identically zero

Subhasis Biswas

@Sil

@Silent, suppose $f(x)=k$ somewhere. Then, in some neighbourhood of $x$, $f(x) \neq 0$. Therefore, $g(x)>0$ in there. Let the nbd be $(e,f)$. Now, split the integral of $g(x)$ into three parts: from $0$ to $e$. $e$ to $f$. $f$ to $1$. middle term is positive and the other parts are nonnegative, thereby yielding $\int_0^1g(x)>0$. A contradiction

this is basically what @LeakyNun said, I guess

Rithaniel

Suppose we have a function $f:B\rightarrow\ddot{B}$ where $B$ and $\ddot{B}$ are algebraic structures with the same underlying set and $f$ satisfies the property that for $a,b\in B$ then $f(a)f(b)=\phi(f(ab))$ where $\phi :\ddot{B}\rightarrow\ddot{B}$ is a permutation on $\ddot{B}$. Is there a term for this sort of relationship?

Subhasis Biswas

8:34 AM

can anybody on earth say why this is wrong?

math.stackexchange.com/questions/3228074/…

$p$$L$$E$$a$$s$E$

Rithaniel

8:45 AM

I don't see anything necessarily wrong with your work, Subhasis, though I always doubt myself when I see expressions like this.

Subhasis Biswas

for one last a time, this is my request to go through this

maybe you can find some errors here

part 1 ibb.co/BPC3h2q; part 2 ibb.co/kSVt9R2

very sorry for the handwriting

Rajath Krishna R

Hello everyone. I have a small query. Is writing a matrix in its Smith normal form a basis change?

Leaky Nun

10:05 AM

@ÉricoMeloSilva there is one such function for the upper half plane though: G(r;r0) = 1/2π [log|r-r0| + log|r-r0'|]

the "divergence theorem" proof doesn't work because the test function 1 isn't $C_0(V)$

Silent

12:42 PM

@LeakyNun @SubhasisBiswas, thank you.

Silent

1:35 PM

Let $f:\Bbb R\to\Bbb R$ be a polynomial such that $f(0)>0$ and $f(f(x))=4x+1$ for all $x\in \Bbb R$, then what is $f(0)$? The options given are: 1/4, 1/3, 1/2, 1

I have no idea what to do

Secret

1:49 PM

6

Q: What eigenvector-like tools are there for analyzing tensors of rank three and higher?

If I have a rank-two tensor that I want to analyze ─ say, an electric quadrupole moment, or a moment of inertia ─ it can often be very easy to analyze by moving to its principal-axes frame: one rotates to a reference frame where the tensor is diagonal, and this simplifies all sorts of understandi...

tensor-calculus mathematics linear-algebra

$f(0) > 0$

o great, f is not necessarily order preserving

trying again...

$ff(0) = 1$

Earth Cracks

2:01 PM

@Silent $f(x)=\frac13 + 2x$

Silent

Wow! how did you figure that out?

Earth Cracks

Just take an arbitrary polynomial $f(x)=a_0+a_1x+\dots +a_nx^n$ and you get $$f(f(x))=a_0+a_1(a_0+\dots+a_nx^n)^1+\dots + (a_0+a_1x^1+\dots +a_nx^n)^n=4x+1,$$
and over an integral domain you can deduce that $a_2=a_3=\dots=a_n=0$

Silent

hmm

Earth Cracks

So $f(x) = a_0+a_1x$ and $f(f(x))=a_0+a_1(a_0+a_1x)=a_0+a_0a_1 + a_1^2x=1+4x$

so $a_0+a_0a_1 =1 $ and $a_1^2=4$ which tells you that $a_1=\pm 2$

And thus you have two cases, either $a_1=2$ in which case $a_0=\frac13$ and you have $f(x)=\frac13 + 2x$ which satisfies $f(0)>0$

Or else you have $a_1=-2$ and thus $a_0=-1$ which gives you $f(x) = -1-2x$ which does not give you $f(0)>0$

Hence you can conclude that $f(x)=\frac13 + 2x$ yielding $f(0)=\frac13$

Silent

Thank you so much, @EarthCracks!

Earth Cracks

2:04 PM

Not a problem

ÍgjøgnumMeg

2:18 PM

Afternoon

schn

How does one estimate the behaviour of a function? Say I have the function $\frac{x+2}{\sqrt{x^3+x^5}}$, how come one can simplify this to $\frac{1}{\sqrt{x^3}}$. I'm asking cause I'm determining the convergence of an integral from 1 to infinity with this function, and this approximation is made. But according to my drawing, the approximated function is less than the original, isn't that a problem?

I would simplify the function to $\frac{2+x}{\sqrt{x^3}}$ cause for large x, this is GREATER than the original function, and so, if the integral with this simplified function convergence, the original has to as well, no?

But I can't see how one can go even further and approximate $\frac{2+x}{\sqrt{x^3}}$ to $\frac{1}{\sqrt{x^3}}$. This is making compromises on both numerator and denominator, which gets itchy.

Axoren

2:36 PM

Can someone help me figure out how to turn this into a front-page appropriate question? I was thinking about this last night and I honestly, don't know what I'd ask if I posted it on the front page

9 hours ago, by Axoren

I'm about to go to bed, but I don't even know how to ask this question in a front-page appropriate way. Take a hyperbolic plane ($H^2$), mapping it to a disk, and compactifying it (including the circumference). Adding a Euclidean dimension $X$, to turn it into a cylinder, what type of geometry is along the surface of the $\text{Circumference}\times X$?

ÍgjøgnumMeg

@schn Note that $\frac{x + 2}{\sqrt{x^3 + x^5}} \sim \frac{x}{\sqrt{x^3 + x^5}}$. Now $x^3 + x^5 = x^2(x + x^3)$ so you have $\frac{x}{\sqrt{x^3 + x^5}} = \frac{1}{\sqrt{x + x^3}}$

Mathphile

is 2, 3 the only prime of the form $(p-1)!+1$ where $p$ is prime?

if yes then why so?

ÍgjøgnumMeg

probably smth to do with Wilson's Theorem

schn

@ÍgjøgnumMeg Okay. But when approximating from $\frac{x + 2}{\sqrt{x^3 + x^5}}$ to $\frac{x}{\sqrt{x^3 + x^5}}$, aren't we approximating to a lesser function, and so a lesser integral over 1 to infinity? And how would one arrive at $\frac{1}{\sqrt{x^3}}$ from your last expression $\frac{1}{\sqrt{x + x^3}}$. It seems like here one would have to make the function greater again, no?

My intention is to determine the convergence of the definite integral over the function $\frac{x+2}{\sqrt{x^3+x^5}}$ from 1 to infinity, and the approximation $\frac{1}{\sqrt{x^3}}$ is made.

ÍgjøgnumMeg

@schn the denominator can be further factored to give $\sqrt{x^3}\sqrt{\frac{1}{x^2} + 1}$

Now in the limit $\frac{1}{x^2} \to 0$

Also, in the limit the $2$ in the numerator becomes irrelevant (think $x \to \infty$, so $x + 2 \to \infty$)

Could've done this all in one go by writing $x^3 + x^5 = x^5\left(\frac{1}{x^2} + 1\right)$

schn

3:02 PM

@ÍgjøgnumMeg True. Would you say then that $\frac{x + 2}{\sqrt{x^3 + x^5}}\geq \frac{1}{\sqrt{x^3}}$ or the other way around? Would this matter in determining the definite integral of $\frac{x + 2}{\sqrt{x^3 + x^5}}$ from 1 to infinity?

I'm thinking, wouldn't one want to find a function that behaves very similar to the one in question but that is greater than it, so if the integral with the approximated or simplified function, $\frac{1}{\sqrt{x^3}}$ in this case, is converging, then the integral with the original function has to as well

ÍgjøgnumMeg

Well you can think of $\int_{1}^\infty$ as $\lim_{c \to \infty} \int_1^c$

so that probably helps idk

I can't do analysis

topologicalmagician

4:02 PM

@TedShifrin on a final exam I just took, there was another typo :\

Mathphile

6:46 PM

is infinitation of x equal to the infinite tetration of x?
Lets introduce a new notation for this question:
let $x^{/1/}=x+x$ (addition)
$x^{/2/}=x.x$ (multiplication)
$x^{/3/}=x^x$ (exponentiation)
$x^{/4/}=^xx$ (tetration)
so infinitation according to my notation is $x^{/\infty/}$
infinite tetration of x is $x^{x^{x^{x.....}}}$
so my question is whether $x^{/\infty/}=x^{x^{x^{x.....}}}$

randomgirl

When people talk about the tails on box plots, are they talking about just the whiskers or are they talking about anything after or before the median?

ÍgjøgnumMeg

7:00 PM

Why does anything I read involving a topological group specify a basis of open neighbourhoods of the identity? What's useful about having that information?

Balarka Sen

Context? Having a local basis at identity gives a local basis at every point which might come in handy

ÍgjøgnumMeg

I see, context is varied, I've just seen it a lot of times and wasn't sure what the use of it was

The kernels of the connecting maps in a profinite group give a basis of open neighbourhoods of 1

No not the connecting maps

The projections down from the inverse limit

Balarka Sen

So in particular if it's a countable inverse system, the inverse limit is a first countable profinite group. That's always a plus.

ÍgjøgnumMeg

ergh how far can I get without actively learning any topology and just picking up the relevant definitions as and when I need them

lol

Balarka Sen

Very far

I do that all the time with point set topology

Alessandro Codenotti

7:06 PM

@ÍgjøgnumMeg because you can just translate it around and describe a basis around every point, hence the whole topology

ÍgjøgnumMeg

So having a local basis at every point lets you piece together the whole topology?

@alessandro ah sniped

Balarka Sen

ye

ÍgjøgnumMeg

That's nice

Alessandro Codenotti

The idea is that you only need to know what happens near a point to talk about convergence

(random remark: if you're convinced of this fact it's obvious that the metrics $d(x,y)$ and $d'(x,y)=\min\{1,d(x,y)\}$ induce the same topology)

Balarka Sen

Throw away all the big balls and you're still fine

ÍgjøgnumMeg

7:12 PM

lol

Alessandro Codenotti

It is sometimes useful to know that every metric is equivalent to a bounded one

Proving that a countable product of Polish spaces is Polish is the first example I could think of but there's surely more

ÍgjøgnumMeg

sounds way further from the surface than I care to venture

Balarka Sen

Even more fundamentally you need that to metrize the product topology, if you have a countable family of spaces

Alessandro Codenotti

Oh right, of course, that's how it's used in my example too

Balarka Sen

Right

ÍgjøgnumMeg

7:15 PM

might go ahead and learn p-adic analysis before I learn real analysis properly

Leaky Nun

@ÍgjøgnumMeg why not both

write a program to calculate exp in 3-adic :P really cool

ÍgjøgnumMeg

and then just claim that the p-adic metric is more natural and that real numbers don't make sense to me

F.White

if i want to consider the antipodal map on the n-sphere, i can consider it as being restricted from $-Id_{n\times n}$ on $R^{n+1}$. can I make rigorous through this embedding taking the map on tangent spaces on the sphere as just coming from the Jacobian $-Id_{n+1\times n+1}$

or should I instead take the stereographic projection charts on the n-sphere, and determine the Jacobians in local coordinates

ÍgjøgnumMeg

whenever I get stuck on something I go back in time and remember when I didn't know the difference between $\cap$ and $\cup$

Leaky Nun

What's the subgroup of PGL(2,C) consisting of transformations leaving the unit circle invariant?

Alessandro Codenotti

7:18 PM

We had a lecture on Lean yesterday, but I slept in... @Leaky

Leaky Nun

@AlessandroCodenotti quel dommage

F.White

@LeakyNun {1}?

Leaky Nun

definitely not

F.White

you mean taking the unit circle into itself?

Leaky Nun

they arise in the context of classifying Aut(D)

F.White

7:19 PM

or fixing the unit circles elements

Leaky Nun

maps like [z-a]/[1-conjugate(a)z]

$\dfrac{z-a}{1-\overline{a}z}$

@ÉricoMeloSilva I'm wondering whether we can use these maps to generate a Green's function for the Laplacian in the unit disc

s.harp

Does anybody have a nice sounding soundbite as to why a geometric structure on a manifold should always have a finite dimensional symmetry group?

Balarka Sen

Remind me what the Green's function is again? Solution to Laplacian f = delta-mass at the boundary?

s.harp

ie if the symmetry group were infinite dimensional "its topology and not geometry"

F.White

@s.harp what's a geometric structure?

Leaky Nun

7:23 PM

@BalarkaSen $G(r;r_0)$, $\nabla^2 G = \delta(r-r_0)$, $G|_{\partial M} = 0$

s.harp

@F.White here just a structure that is about geometry, so the question is vague and I'm trying to motivate why it should be so.

there is a notion by Gromov called geometric structure, but thats not what I mean

(or it is what I mean, but I don't want to say it)

(cf people.mpim-bonn.mpg.de/hwbllmnn/archiv/geostr00.pdf )

F.White

0

A: References for actions of infinite-dimensional Banach-Lie groups on infinite-dimensional Banach manifolds

Here are some partial answers: It is known that Banach Lie group actions on finite-dimensional manifolds are quite restricted. What I mean by this is: Due to a theorem by Omori, see Hideki Omori, On Banach-Lie groups acting on finite dimensional manifolds, Tohoku Math. J. (2) 30 (2), 223-25...

Balarka Sen

Myers-Steenrod theorem states for example that Isom(M) for any Riemannian manifold M is actually a finite-dimensional Lie group. I don't know a good argument for it to be finite-dimensional; you might want to argue that the tangent space to Isom(M) at the identity is finite dimensional - which I suppose is equivalent to claiming that the space of Killing fields is finite dimensional.

That should be doable but I haven't tried it

F.White

@s.harp

"if a Banach-Lie group acts smoothly, effectively and transitively on a finite-dimensional manifold, then it automatically is finite-dimensional. Hence for many manifolds which come up, one can only consider actions which do not satisfy these requirements or one is forced outside of the class of Banach Lie groups (e.g. if one considers diffeomorphism groups of finite-dimensional manifolds).

Quotient theorems, or smooth structures on orbits for infinite-dimensional group actions are in general much harder to establish than in the finite-dimensional case. Something along the lines y…

s.harp

@F.White that is very interesting, thank you for that link. Just a remark: Diffeo(M) is infinite dimensional modelled on a Frechet space and acts on M, for me to motivate with this remark would require me to distinguish between Frechet spaces and Banach spaces, which is too subtle I believe

Leaky Nun

7:31 PM

@BalarkaSen rip me I got topology exam tomorrow

year 2

Alessandro Codenotti

What kind of topology?

ÍgjøgnumMeg

Viel Glück :)

Balarka Sen

Good luck!

Leaky Nun

@AlessandroCodenotti "metric space and topology"

Balarka Sen

Oh, way easier.

s.harp

7:33 PM

@BalarkaSen I believe that that theorem follows from you being able to see that an isometry is uniquely determined by its derivative at a point (if manifold is connected). So the idea that geometric structures have to be rigid (that is symmetries are uniquely determined by a fixed amount of derivatives at a point) would be enough to motivate finite dimensional

loch

@LeakyNun I remember when I took the exam one question asked us to state and prove Baire's theorem and I was afraid I remembered the statement wrong :p

Leaky Nun

yeah I'm just afraid of missing some things that are considered basic :P

oh no

Balarka Sen

@s.harp Yeah, that's it, I figured it out right before you mentioned.

Leaky Nun

@loch that's hardcore

lemme dig out your year's paper...

loch

2015 I think

Leaky Nun

7:35 PM

yeah found it

this is not good

Balarka Sen

Fixing a point $x \in M$, you can write down a map $\text{Isom}(M) \to M \times GL_n(T_x M)$

This is going to be injective.

Leaky Nun

@loch what if they ask us to prove sequential compactness => compactness lol

Balarka Sen

That's actually surprisingly complicated if you think about it ^

Leaky Nun

yeah I remember the whole proof I think

and I think I also remember the skeleton of Baire

and also the filter proof for Tychonoff

but I'm kinda feeling that they're pointless

like who cares about proofs of technical result

Balarka Sen

Oh remembering them are, yes. Cramming to keep them in your head for an exam the next day is very pointless.

I hate doing it.

Alessandro Codenotti

7:37 PM

@LeakyNun review the Hausdorff iff closed diagonal thing, a lot of people like to put it in exams apparently

Leaky Nun

@AlessandroCodenotti that's trivial

seriously

in Lean you just "follow your nose"

Alessandro Codenotti

So is the filters proof for Tychonoff, so what?

loch

it's true I don't remember most of these things since I don't use them

Leaky Nun

you're probably right

ok FIP characterisation for compact is also follow-your-nose

eh, what more technical results do you have in mind

@BalarkaSen et al

Balarka Sen

Baire is pretty simple it's just that some applications are weird.

s.harp

7:39 PM

prove that isometry of a compact space is a compact group

Alessandro Codenotti

@LeakyNun well I sure hope you have a metric space if they ask that

Leaky Nun

@AlessandroCodenotti yeah, metric space

Alessandro Codenotti

I suppose that Baire theorem means "complete metric spaces are Baire" in this context then?

Leaky Nun

yeah

Balarka Sen

What's your proof that a complete metric space with no point isolated is uncountable?

@Leaky

Leaky Nun

7:41 PM

@BalarkaSen "BCT1" :P

Balarka Sen

Ah, good.

I didn't think of applying Baire when this came up in my exam. Just did the Rudin-style proof. Essentially proving Baire but still.

Leaky Nun

but isn't Z a complete metric space with every point isolated

Alessandro Codenotti

@BalarkaSen without isolated points*

Balarka Sen

Thanks

Alessandro Codenotti

With a bit more work you can prove it actually has cardinality $\geq |\Bbb R|$

s.harp

7:44 PM

always assume continuum hypothesis is true to save yourself as much effort as possible

Alessandro Codenotti

Con(ZFC) implies Con(ZFC+GCH) so I guess you can even assume GCH if you really want to :P

s.harp

I dont know the contents of GCH so I'll assume that too why not

Balarka Sen

Also, $\Bbb R^n$ cannot be written as a countable union of disjoint closed balls. That's actually notoriously complicated.

Alessandro Codenotti

$2^\kappa=\kappa^+$ for every infinite cardinal $\kappa$. $\kappa=\aleph_0$ is CH

Balarka Sen

It's "obvious" but you need Baire to prove this.

s.harp

7:47 PM

can't $\Bbb R^1$ be done like that?

Balarka Sen

Thanks again :P

skillpatrol

@BalarkaSen 0celo is back.

(as Ryan Unger ;)

GFauxPas

I'm having trouble visualizing this,

if $\dot{\mathbf x} = \mathbf f(t,\mathbf x(t))$, what's the graphical interpretation of $\langle \mathbf x, \mathbf f \rangle < 0$?

Leaky Nun

Any metric space $X$ embeds into $C(X,\Bbb R)$

$X$ is complete iff the image is closed

Mathphile

could i get some help for this question:

1 hour ago, by Mathphile

is infinitation of x equal to the infinite tetration of x?
Lets introduce a new notation for this question:
let $x^{/1/}=x+x$ (addition)
$x^{/2/}=x.x$ (multiplication)
$x^{/3/}=x^x$ (exponentiation)
$x^{/4/}=^xx$ (tetration)
so infinitation according to my notation is $x^{/\infty/}$
infinite tetration of x is $x^{x^{x^{x.....}}}$
so my question is whether $x^{/\infty/}=x^{x^{x^{x.....}}}$

Balarka Sen

7:56 PM

@LeakyNun Busemann embedding, very important

I wouldn't say technical though

Leaky Nun

I'm just wondering if passing to the space $C(X,\Bbb R)$ would help me prove that any complete perfect metric space is at least continuum

Balarka Sen

@s.harp So actually $\dim \text{Isom}(M^n) \leq n(n+1)/2$, because that's the dimension of $M \times SO(T_x M)$. Which is interesting: this bound is tight, it's attained for eg $S^n$.

Is it true that it's attained iff $M$ is isotropic?

s.harp

@Balarka its is also attained by euclidean space (with symmetry group $O(n)\ltimes \Bbb R^n$)

Sandbox

Transcript for

$Mathematics$ Mathematics