This is the best place to on earth to discuss math.

No one is making fun of how awkward mistake I make lmao.

@BalarkaSen I'm back. I'm pretty sure that $g_1$ and $g_2$ can be assumed to be nondecreasing without issues here

That follows from my observation I think yeah

@BalarkaSen iirc there's a conjecture about closed geodesics on spheres (always at least 3? maybe) which is false in Finsler land

Now if you time reverse geodesics in Finsler manifolds they aren't geodesics anymore

Oh dang. Also isn't that a theorem for Riemannian spheres?

I'm not sure why you wanted them to be increasing though

So somehow this time reversal property must be essential

Yeah I got confused with that

The conjecture is infinitely many

Ahh

Katok constructed Finsler metrics on $S^n$ with only finitely many

@Alessandro Seemed like the right thing to do if I wanted g_i and sup/inf to commute

But I haven't done any computation

Ah I see makes sense

@MikeMiller Ah because Finsler metrics are not quite norms fiberwise, your time reversibility thing makes sense now

They are positive homogeneous

OK

Bizarre

Katok is such a beast, these Russians are mad

Uhm maybe the definition $d_H(X,Y)=\inf\{r\geq 0\mid X\subseteq B_r^d(Y)\land Y\subseteq B_r^d(X)\}$ is actually good here. If $X\subseteq B_r^d(Y)$ then $X\subseteq B_{g_2(r)}^{d'}(Y)$ or something like that

I don't understand, though. Once you get $g_1$ and $g_2$ increasing, isn't it direct?

Is it? What am I missing?

@AlessandroCodenotti This looks good to me now, also if $X\subseteq B_r^{d'}(Y)$, then $X\subseteq B_{g_1(r)}^d(Y)$, so $g_1$ and $g_2$ are also the functions witness that $d_H$ and $d'_H$ are coarsely equivalent

Hi guys, I have a doubt

about continous spectra of an operator. If I have an operator on l2(Z;C) like Af(x) = 1+1/(1+e^{-|k|})

times f

here, the continous spectrum is 2 because it s an accumulation point, right?

@BalarkaSen Lol for about a minute I forgot norms have |v| = |-v|

@Alessandro {d(x, y) : y in B} is continuous image of a compact set so is a compact set in [0, infty). The inf of g_2 over this compact set is attained at the minimum of this compact set because g_2 is increasing, so everything is commuting

Similarly for sup of g_2 and maximum of {d(x, B) : x in A}

Hmm makes sense

I don't see why it's so complicated

@BalarkaSen It shouldn't be

My argument with balls doesn't need $g_1$ and $g_2$ to be increasing (not that it makes any difference)

so passing in zero right? for k=0

that is the continous spectra

3/2

no matter, was another alessandro, sorry

I ve misunderstood

thanks all

how many Alessandri are there...?

a finite amount

At most 7, to be specific

If $$z_1 + z_2 + z_3 = 1$$ then is it a valid to write $$ \big| z_1 +z_2 + z_3 \big|= 1$$

@AlessandroCodenotti this is a lie, I'm actually using it

yes

ok so no one can explain me the continous spectra there?

Leaky was that “yes” for me?

yes

does a homeomorphism $h:X \to X$ on some compactified lorentzian submanifold imply nonzero curvature on $X$?

or any compactified manifold for that matter.

maybe if you assume $X$ is quasiprojective left-invariant under $h$

@MikeMiller Even I can remember that.

@Jasper "Even", get it? ;-)

@robjohn Other than even and odd, I don't get anything. But I guess that is what you mean!

@Jasper Even you can remember that norms are even...

@robjohn Oh I see. I never used that terminology with norms...

@robjohn That made me chuckle.

I think I am going to put up some nice colour for my square.

@Jasper you have a non-uniform gravatar! I almost didn't recognize you.

@robjohn Well, nobody else talks like I do. So that is the litmus test.

for a valuation can $v(0)=-\infty$?

or does it have to be $\infty$

@robjohn Tada! Refresh your browser to see my brand new square.

@geocalc33 why?

@Edward nevermind I was just misreading the definition

Okie dokie

are you familiar with the basics of valuations? @Edward I'm learning the definition today and practicing

Hello, i could use some help with my problem, i have a square root of x-1 and i am trying to rewrite it as (something)^2. Is it even possible, i need 2ab to be equal to zero, but that can only be done if my b or a is 0, is that even the right way to look at this. I don't need an answer, a hint would do just fine.

@geocalc33 it's easier if you just ask questions, I know what a valuation is at least :P

Let $V$ be a subspace in some Hilbert space. I am trying to show that $V^{\perp \perp} = \overline{V}$. I already showed that $\overline{V} \subseteq V^{\perp \perp}$ (this was the easy direction). However, I am having trouble with the other inclusion. I could use some help.

@user193319 Did you prove that, on a prehilbert space $H$, for any non-empty subset M, we have that $M^\perp = [\overline{\text{span}(M)}]^\perp$?

@user193319 Using that, and the fact that on a Hilbert-space, the biorthogonal complementer of a closed, linear subset is the set itself, you can prove the statement.

@Edward can you define $v(x)=\log(x)$ to be a valuation? And is this is what is called a logarithmic valuation?

and how does this fit in with the p-adic norm? $|x|_p=p^{-v_p(x)}.$ I'm not sure what $v_p(x)$ is explicitly...

@geocalc33 $\nu_p(x)$ is the number of times $p$ divides $x$

@tobias oh okay thanks :)

any discrete valuation on a field gives rise to a norm on it in the same way

so you can define log x to be a valuation?

is $\log(0) "=" \infty$ ?

it equals negative infinity

so..

so no?

Guess not

$\log(x)$ is not defined for negative $x$ to begin with

so what's a really simple example of v(x) that's not the trivial valuation?

$v_p(x)$

what's a really simple example of v(x) that's not discrete?

uhh

are they worth considering

dunno

isn't there a whole field called valuation theory

probably multiple

can you assign each real number a magnitude? so for example, $.245 \mapsto .3546$

Is this a valid proof that the knot sum of two nontrivial knots is nontrivial?

Suppose it's trivial, so A#B=0. Draw a sphere around the A portion so that it intersects A#B in two points. Perform an ambient homotopy taking A#B to the unknot.

The sphere is mapped to a new configuration, but it can be homotoped to a standard sphere, containing a simple arc of the unknot. Looking at the contents of the sphere on its own, then, you get an ambient homotopy of A to the unknot.

(Think of collapsing the boundary of the sphere to a point)

Suppose I am given a polynomial $f(x) = ax^2 + bx + 1$, how do I factorize it? can i still write it in $f(x) = (x - y_1)(x - y_2)$ where $y_1, y_2$ are the roots to the $f(x)$

@quallenjäger Almost

@AkivaWeinberger This seems circular. Why is the arc in the end simple?

I have to normalize it right?

right

because it is +1

@MikeMiller It's just a subset of the unknot, no?

(of course, you need to work somewhere where the roots exist)

@quallenjäger No, you need to normalize because the highest degree coefficient might not be 1

So like a geometric arc

A geometric arc with boundary points on some wacky embedding of the sphere

Hmmm

Maybe this actually does work

The interior of the sphere is a ball

So this is actually fine

Nice argument, why have I not seen this before

what letter should I use for a chart after $\varphi,\psi$?

the form $f(x) = (x-y_1)(x-y_2)$ applies only if a = 1 right?

Why not $\varphi_i$

@MikeMiller No idea but I'm worried there's a subtlety

It's just one of the multiple proofs which work in the PL or smooth category

what is the standard procedure to write a general polynomial $f(x) = ax^2 + bx +c$ in root form?

$a(x-r)(x-s)$ where the roots are $r$ and $s$

The true proof is the swindle

@AkivaWeinberger You have to formalize the following: given a smoothly embedded ball in R^3, and a straight line going from one boundary point to another, pulling back by the embedding gives you a trivial knot / tangle

Yeah but that doesn't work if you want to stay in tame knot land

But this strikes me as straightforward.

I guess there's no reason not to use indices

Yeah when there's more than 2 things I usually start using indices

@Thorgott Use $\phi$ and $\varphi$

$[\varepsilon,\epsilon]$

@AkivaWeinberger is there any proof to that formula? I googled it but I always found some high school maths stuff

@AkivaWeinberger lol the interval between two infinitesimals

There's a theorem - I forget what it's called - that if $r$ is a root of a polynomial then $x-r$ is a factor of the polynomial

There's a lot of cases where I have neat conventions for three objects, like $p,q,r$ for primes, $r,s,t$ for ring elements, $x,y,z$ for reals, $f,g,h$ for functions, but I'm not sure if there's something that goes nicely with $\varphi,\psi$

@AkivaWeinberger I am not really sure that has a name

I guess $\phi$ works

@Akiva it's called factor theorem lol

Once you factor out one root from a quadratic, you're left with a linear polynomial

and you can factor out the coefficient of x from that

how does it give me the other root?

Do you know how to do polynomial long division?

@Thorgott Wait don't

yes

$\varphi$ $\chi$ $\psi$ maybe

I know how to do it but I am not sure why this give me the other root.

@quallenjäger Let's do an example. Say your quadratic is $x^2-4x+3$, and say we know $x=1$ is a root

(which you can check)

The theorem I mentioned says $x^2-4x+3$ is a multiple of $x-1$

So we find $(x^2-4x+3)/(x-1)$

Through long division, we get $x-3$

That would be alphabetically appropriate, but it's not pleasing aesthetically

$\chi$ is reserved for Euler characteristic

So $x^2-4x+3=(x-1)(x-3)$

that means I have a root at 3 as well right?

Ah sigma could be a good chart

@quallenjäger Yes

$(3-1)(3-3)=(2)(0)=0$

$3^2-4(3)+3=9-12+3=0$

Why does this long division $(x^2 - 4x + 3) / (x-1)$ leads to the other root

I mean it can't be coincidence right

@Jasper the same as it ever was...

letting the days go by

Tu uses $\sigma$, it seems

carrying the drops of time

@quallenjäger Suppose $r$ is a root of $x^2-4x+3$ other than $1$. Then I claim $r$ is also a root of $(x^2-4x+3)/(x-1)$

In fact, if $r$ is a root, then $r^2-4r+3=0$

so $(r^2-4r+3)/(r-1)=0/(r-1)$

Told you it would be a good chart

We need to be careful we're not dividing by zero

but I said $r\ne1$ so that's fine

God what the fuck are you two talking about

Call it $\Sigma$

You guys have been talking about notation for chart for like the last 18 hours

so $(r^2-4r+3)/(r-1)=0$, and therefore $r$ is a root of $(x^2-4x+3)/(x-1)$

I sent four messages

You've sent more messages about hating notation than I have about notation

All I saw you say was charts

You're obsessed

@quallenjäger It's easier to find the root of a linear polynomial than a quadratic one

The root of $x-3$ is $3$

because when $x=3$, $x-3=0$

Another example. Let's find the roots of $2x^2-x-1$

@BalarkaSen what's your choice of notation for a third chart

the first two being $\varphi,\psi$

$1$ is a root (because $2-1-1=0$, and because I chose the example so I could make it nice)

I just don't know why you're looking at three charts

@Thorgott I'm not falling for this one

The other root is also a root of $(2x^2-x-1)/(x-1)$

which is $2x+1$

Interesting, my blue looks approximately exactly like the blue of @MikeMiller

What's the root of $2x+1$? @quallenjäger

I couldn't say, Jasper

Three charts are natural

Most compact 2-manifolds have Lusternik-Schnirelmann category 2

I know that there is exactly and approximately, and one day I figured I would give a new meaning to approximately exactly.

If I want to see the whole manifold I'll just look at it

I don't need to cover it with charts bro

Lmfao

I've spent like ten minutes thinking about how to call this chart that I will need for a calculation that's probably only going to take one minute, because none of these choices pleases me

Three charts: $\alpha$ $\beta$ $\gamma$

Yeah that's silly

You're doing manifolds stuff

Nobody has good notation

how about $\varphi$

And everyone disagrees

Make the symbol for the third chart a picture of your closest childhood friend

It is strange that one can lmao so many times, because after one time, there would be no more a to l o.

How about "😂"?

@AkivaWeinberger 1/2 then

got cha

That's a good suggestion

thanks akiva

ok, I'm gonna try doing that

Suppose we have three charts, 😂, 🤩, 😴: R^n -> M

@AkivaWeinberger that was a nice intuitive way, Thanks alot

Three charts: 地, 図, 帳

OK fine you need 3 charts if you want to check that the transition functions of a vector bundle form a Cech 1-cocycle

Three charts: $\phi$ $\psi$ $p$

Use $\varpi$

$\overline \Xi$

Three charts: $p$ $\rho$ $\wp$

Fail

Consider the three charts 1, 2, 3: R^n -> M

@AkivaWeinberger Do you usually use traditional characters?

12*23*31 = 1

@Jasper Japanese usually uses what's called traditional in China

That's Cech theory summarized

@AkivaWeinberger Oh I see. No wonder that seemed Chinese but not Chinese.

$[n]$ is honestly a great notation for simplices. Whoever came up with that is a genius and will be remembered

but 8555 =/= 0 so Cech theory so it's only consistent mod 5, 29, and 59

chart1 notchart1 notchartnotchart1ornotchart1

Wait

Three charts: $\varphi$, $\psi$, $\varphi\!\psi$

Hey @CaptainAmerica16 what math book are you studying now? I am back!

$\circle \!\!|!\!\!\! \wedge$: a chart

How do I do this

$\imath\jmath$

I forgot

There are many silly messages being starred.

How do I draw a circle in latex

O

This is ridiculous

the Prince symbol isn't available in Unicode

Ƭ̵̬̊ is the best I can get

@BalarkaSen I think you can use the picture environment, or use a package like pgf.

ж

$\bigcirc \!\!\!\!\,\! \wedge$

$\theta$

Alright I'm going to do something better than this

Well y is i grec

Well that was rather insane

I'll try to have panic attacks elsewhere from now on

@Jasper Hey!

I'm doing intro real analysis and linear algebra right now.

@CaptainAmerica16 Did you finish Spivak's Calculus or the set theory book I recommended you?

@Jasper I'm working through a textbook recommended by the university I'll be attending next year. I use Spivak as a reference or for extra practice. I didn't get through the set theory book :/ I had some set backs preparing for college within the last school year, but now that I'm in I've been devoting my free time to math.

@CaptainAmerica16 Oh I see. What is the textbook?

@Jasper Introductory Real Analysis by Bartle and Sherbert

@CaptainAmerica16 Oh I see. A book commonly used to teach the very basics of real analysis in a simply way and no more.

@Akiva @Mike Suppose $M$ and $N$ are compact $n$-manifolds and suppose $M \# N \cong S^n$. Under this homeomorphism, the separating $S^{n-1}$ goes to an embedded locall flat sphere in $S^n$, which separates $S^n$ in two balls by Jordan-Schoenflies. Connected components go to connected components, so $M$ minus ball must be homeomorphic to $D^n$ and $N$ minus ball must be homeomorphic to $D^n$. Adding the ball back and using $D^n \cup_{\varphi} D^n \cong S^n$, you get $M \cong N \cong S^n$

Should be fine right? Same idea

Exactly, I was just going to say that it's not the most challenging text. Definitely less so than Spivak, I seek out more in-depth explanations and problems where it seems necessary

@MikeMiller the deed has been done