@EricSilva oh nooooooo

Doesn't Federer just call outer measures, measures?

lmao does he

You have A <= B+ C, the cycles of that, and A+B+C<=2pi @BalarkaSen

And that’s all

oh A, B, C were my angles?

Ya

If $\arctan x + \arctan y + \arctan z = \pi$, prove that $x+y+z= xyz$. Please let me know if this is solvable through complex numbers?

Ahh

@EricSilva page 53

this is going to be confusing

heck

Why's this relevant to the problem I wrote?

I don't understand

I tried to define three complex numbers ... then use the properties of arguments..

But it didn't work.I don't want the solution btw..

Eh, I only said it was an example of a geometric inequality

Four of them, in fact

... Ok

@EricSilva Oh I also forgot that he doesn't use the sigma algebra framework

he just has a measure and then those things satisfying the Carathedory criterion are the measurable sets

It’s a simple enough result. I just like that it ends up being a tetrahedron in [0,pi]^3

that's not sooo bad

idk

this is a garbage book

I have properly labeled Federer

lmaooo perfect

@EricSilva time for... Hitman mean curvature flow

oh no actually I volunteered to teach calculus of variations on thursday

I should probably figure out what to do

If you were a physicist I’d say something something Hamilton’s principle or geometric optics

But you’re not so...good luck?

@Semiclassical I actually volunteered to teach the physics

As much as I hate physics, I still love you guys

I'll do Noether's theorem and applications to wave equations

Lol

problem is, the best wave equations (wave maps, Yang-Mills) require too much geometry to be appropriate here

pde schmidiees

@Semiclassical hmm

why do physicists even care about Noether's theorem

like, it's cool and all...but so what

Because conservation laws woo

TBH I don’t do that stuff much

for wave-type equations you can get energy estimates, which are useful for analysis

Klainerman gave talks at my school 20 years ago on Noether's theorem in hyperbolic PDE and I have the notes

I mean, it is neat to be able to say “laws of physics are time invariant” => conservation of energy

@Semiclassical the professor is being anti-physics to annoy me

And similarly for translation-invariance => conservation of linear momentum

yeah yeah I know that stuff

but it's hard to motivate that to a mathematician

I’d also say it’s useful insofar as it gives you a recipe for deriving conserved quantities from symmetries

But yeah

Ah. Noether's theorem for 1D system leads to a first integral

it helps to solve Newton's equations

Eh, I guess

I know it’s important in QFT

but why?

Because you write down theories with higher gauge groups and you want to figure out what charge conservation means there? TBH I dunno

Particle physics is a strange realm

Is there a name for a dynamical system whose entire set is pre-periodic?

@Semiclassical we've really only been talking about very mathematical examples so I'll do stuff like the free particle and SHM in the Lagrangian formalism

mmkay

maybe mention electromagnetism

Lagrangian is $\propto E^2-B^2$ right?

Hamiltonian should be $E^2+B^2$

sounds right

that's one place where I do like having conservation laws tbh: You have $E^2-B^2$ and $\mathbf{E}\cdot\mathbf{B}$ as invariants under Lorentz transformations iirc

@Semiclassical I'll mention path integrals and stationary phase to really freak them out.

which in particular means that if there's a reference frame where there's only electric fields, then there's no reference frame where there's only magnetic fields

and vice versa

nice

first one is due to $F_{\mu\nu}F^{\mu\nu}$ being a scalar product and therefore invariant under Lorentz transformations

right

second is $G_{\mu\nu}F^{\mu\nu}$ with $G$ being the Hodge dual of $F$

ugh, 2-forms

do not want to deal with that

yeah, no argument there

@Semiclassical oh is that the "unexpected" one?

right

there's some duality hidden

that's the E.B one

for reference: en.wikipedia.org/wiki/Electromagnetic_tensor#Properties

maths.ed.ac.uk/~jmf/Teaching/Lectures/EDC.pdf

lol

there's a simpler version of electric-magnetic duality

the title is pretty awful

considering the text is like algebraic geometry

something like $(E,B)\mapsto(B,-E)$ in the absence of charge

well, physics algebraic geometry

idk

@Semiclassical that leaves the Lagrangian invariant

why $-E$

oh, this is the duality they discuss at the start of the paper

equation 1.2

oh right

you need to leave the equations invariant

right

wow, they use the Bourbaki symbol

sophisticated!

hmm?

page 13

the road sign

Bourbaki came up with it in their books

oh huh

i didn't know that

This is for math.stackexchange right?

@Semiclassical reading shankar

@ManishKundu yes but we're taking it over with physics

Thats sad

@Semiclassical such a good book

@ManishKundu no need for that

lol

Does any algebraic number have a unique decomposition as a product of (rational) prime powers?

Example: Does $1+\sqrt{2} = \prod_{i=1}^\infty p_i^{a_i}$ where $a_i \in \mathbb{Q}$ and $p_i$ is the $i$th prime number?

Definitely false in general, since there are polynomials with integer coefficients which can't be solved by radicals

At least, not if you only allow finitely many such powers

What if there were infinite radicals, i.e. if there are infinitely many nonzero $a_i$?

Yeah

If you allow infinitely many, I think you'll lose uniqueness

I thought so too, but I'm not sure. Perhaps the fact that the $a_i$ must be rational guarantees uniqueness?

I'm dubious

No, I'm pretty sure it's not. Here's my idea

Start with $1+\sqrt{2}$, and pick the smallest $n$ such that $1<2^{1/n}<1+\sqrt{2}$

You mean $n=1$?

I guess that's true here. I have in mind the next few steps

then form $\frac{1+\sqrt{2}}{2^1}$ and find the smallest $n$ such that $3^{1/n}$ is smaller than this

And repeat ad infinitum

The number you get each time will always be bounded below by 1, but is also getting smaller

I'd conjecture that it converges to 1

If so, that'd give such a decomposition. However, I think you could just as well have skipped a prime--say, gone from 3 directly to 5---and this would still work

In which case the decomposition is certainly not unique

There would, however, be a unique decomposition for which the powers are all nonzero unit fractions

So this is saying there are multiple such "prime decompositions" of 1?

yeah

wait, no

not of 1. my approach won't work for that, since 2^(1/n)>1 for all n

...oh, but then you just take the ratio of the two decompositions I gave

nm, i see what you mean

so yeah, I agree in that case

I see, thanks!

Please someone look at this:

@0celo7 I think you can use the energy to solve some non-linear PDEs

It's used in topological defects

Sine-Gordon also has that weird thing where it has infinitely many symmetries that allows it to get solved exactly

@LeakyNun hey leaky

@LeakyNun Wake up ._.

I would be very happy if someone can help me

@jublikon remember that $1 - y^2 = (1 - y)(1 + y)$

@KasmirKhaan ?

@TobiasKildetoft Thanks! I understand it now :)

@LeakyNun did you study rep theory ?

@TobiasKildetoft morning!

@KasmirKhaan Hi

Tobias ! i kinda need help with last question

I solved 5 now :D

cool

serre book , question number 4

page 12

if you have time id be glad :D

otherwise ill keep fighting it =p

@KasmirKhaan Someone else asked that yesterday I think. There is no trick there, just writing up what everything means

Cool ：D

when they say linear representation with character

X

chi

what does that mean ?

@KasmirKhaan It means that $\chi$ is the character of the representation

its not something unique no

I know that, but hmm

yes, the character is uniquely determined by the representation

(and vice versa, but that is more complicated)

how to calculate it?

i thought each matrix has a character

a trace

no, each matrix has a trace

am not sure if am making lots of sense

the character is a map from the group to the complex numbers

hmm

can you clarify that Point more? :D

Not really. You need to look back at the definitions

hmm

they defined it by the X_p(s) = trace (p_s)

but we could have many values in C

the map from G--->C

is unique in a sense that, we get some complex numbers associated with the trace of matrices?

So anyone familiar with the use of quaternions and rotation matrices in CG? I want to improve a wiki page about it but I'm lacking in knowledge. It now reads "Compared to rotation matrices they are more compact, more numerically stable, and may be more efficient." The last bit is seriously jarring. Are they or are they not? Or does it depend and if yes, on what? Or is it unknown or disputed?

You can represent quaternion rotations with only 4 values

Compared with the 6 values of rotation matrices

I'm not sure of the number of operations to apply for each but I'm pretty sure it's in the quaternion's favor

How to extract the maxima and minima of $2\dfrac{(x^2+2)}{x^2+1} \tag{precalculus}$?

It can be written as $2(1+\dfrac{1}{x^2+1})$

Clearly, maxima occurs at $x=0$,$\therefore max= 4$

Minima occurs at $x= \infty$, $\therefore min = 2$

How to formally write the proof of minima? I am sure writing x= \infty is not allowed.

@Abcd do you know what derivative is

it can simplify findinf max min

@Jacksoja yes yes but calculus will be overkill for that question.

It doesn't have a global minimum. It's asymptotic to the line $y = 2$.

do we call that a local minimum?

No we don't call it anything.

fine.

Do you say $x = \infty$ is a maximum of $f(x) = x^2$?

you mean $y= \infty$, I think.

No, we don't say that.

Well, rather "maximum of $f(x) = x^2$ occurs at $x = \infty$". Point vs value

@Abcd So you shouldn't call it so in the case of the above either

The right notion is what I wrote; it's asymptotic to the line $y = 2$

@BalarkaSen yeah sorry, I was a little confused because in many physics problems that I has solved before we took $\dfrac{1}{\infty}= 0 $...

why is $\dfrac{1}{\infty}=0$? Can it be proven..? (though its somewhat intuitive...)

It's not correct to write that in precalculus level mathematics, indeed. You can construct a rigorous formalism in which writing it makes sense

You can just crowbar infinity in the real line but then operations lose a lot of important properties

For instance you no longer have $x + 1 > x$ since $\infty + x = \infty$

@BalarkaSen You might be able to, but usually the formalisms which include infinity do not allow for it either

@Tobias I'm thinking of the Riemann sphere, really.

Not the extended real line or whatever

One point compactification

On the Riemann sphere there is a holomorphic involution which switches infinity and 0. That involution is precisely $1/z$.

@BalarkaSen Sure

Riemann sphere cheats by only having one infinity

Thats the right notion in complex analysis

but in $\Bbb C^2$ eg you want to have a Riemann sphere's worth of infinities

for each complex subspace passing through origin

which, upon addition, makes $\Bbb{CP}^2$

but then again you can also do the one point compactification of $\Bbb R$, too

The Riemann circle, if you will

true

$\infty/\infty$ is still undefined on the riemannian sphere however

but i am not sure how analytically important that notion is @Slereah

$\infty / \infty$ is defined by the graph $\infty = a \times \infty$

which has no unique solution

Isn't the one point compactification of the reals just the real projective line $\mathbb{RP}$?

well it's a circle

So yes

Wow, for once an email from a publisher which is not spam. How unusual

When is the one point compactification of a manifold a manifold? Not very often I'd say

@Alessandro Only if the endspace of the manifold is a sphere, I think. But I at least need the generalized topological Poincare conjecture to prove this.

Let's try to write down an argument

Firstly, if $M$ is a noncompact manifold, I claim that a neighborhood $U_\infty$ of infinity inside the one-point compactification $M^\infty$ is homeomorphic to the cone on $E$, the space of ends of $M$

Intuitively if the end result is not a sphere it looks like a manifold where some circles were quotiented to a point I think

This should be a definition-check.

What happens if you do the one point compactification of the long line

Is it just a circle

I know there's no long circle

It's not a manifold

Don't you just add one point at the right end?

A neighborhood of the infinity is homeomorphic to the long line again

Which is not homeomorphic to R

Because it has a minimum already

Long line, not long ray

Ah, wait, the one I'm thinking about is called the long ray apparently

fun fact : the manifold $S \times \mathbb L^+$ is called the long drink

You can drink your $\omega$ miles island in it

@Alessandro Actually you don't need to meddle with the endspace. If $U_\infty$ is a neighborhood of infinity in $M^\infty$, it has to be homeomorphic to a ball $B$. Therefore $U_\infty - \{\infty\}$, which is complement of a compact subset $K \subset M$, has to be homeomorphic to the deleted ball $B-\{0\}$.

Taking closure says there must be a subset $U \subset M$ with compact closure such that $M - U$ is homeomorphic to a deleted ball.

That's, like, a good restriction

I am sure it also means that the space of ends is a sphere.

Yeah it does.

OK, I'm happy. We didn't need any cone or Poincare conjecture business after all

Does the tubular neighbourhood theorem say that the bundle of the neighbourhood is trivial?

Or is it not

Huh?

(and is it the case for hypersurfaces, hopefully)

The normal bundle need not be trivial, no

Consider $\Bbb{RP}^1 \subset \Bbb{RP}^2$

It has a Moebius strip as normal bundle

if you have orientable it is trivial

Ah good

Any orientable line bundle on a manifold is trivial

@BalarkaSen cool, thanks!

Diff geo ain't easy

@Alessandro The reason I was thinking Poincare was the following question: If $M$ is a closed manifold, when is the cone $CM := M \times [0, 1]/M \times \{1\}$ a manifold (with boundary)?

This reminds of poncaire hopf: en.m.wikipedia.org/wiki/Phase_rule

The answer is only when $M$ is homeomorphic to $S^n$. The easiest proof I know is by taking a neighborhood of the cone point, which is homeomorphic to $CM$. It must also be homeomorphic to the $n$-ball $B^n$ because it's a manifold by hypothesis

Delete the cone point to get a homeomorphism $M \times \Bbb R \cong B^n - p \cong S^n \times \Bbb R$

I mean in a sense: it is number of edges = number of (something) - number of faces + 1

This gives a homotopy equivalence between $M$ and $S^n$. Any homotopy sphere is homeomorphic to the sphere

That's precisely the Poincare conjecture

@AlessandroCodenotti It can be more complicated. What about a surface of infinite genus?

You'll get a sphere with smaller and smaller handles accumulating to a point as one point compactification

Given the tubular embedding $f$, do we have an associated level-set function $\phi$ by $\phi = \pi_1(f^{-1})$?

That is, the function $f^{-1}$ maps points of the manifold to $\Bbb R \times S$ and then $\pi_1(f^{-1}) : M \to \mathbb R$

Only when the normal bundle is trivial.

If and only if

That's alright

I don't plan to glue projective planes together

@BalarkaSen you're right, looks like I was too optimistic

@BalarkaSen Is there some inverse transformation for this?

If we have some level-set function $\phi$ for the (orientable) hypersurface $S$, can we define the bundle from this

I'm guessing we can have $f(p) = (\phi(p), ???)$

Not sure if we can map to the surface just from this

@Slereah Yes.

You can pull $T_0 \Bbb R$ back by $d\phi$ to get a line bundle on $S$ that is orthogonal to $TS$

(i.e., a normal bundle)

Also the orientability hypothesis is redundant. Any level set is orientable.

what's the projection of $p$ on $S$, is it related by the flow by $d\phi$?

It's just pullback of the projection $T_0 \Bbb R \to \{0\}$.

Alright

Thanks my dude

np

@Alessandro If you want to know more about the space of ends, I can refer to a few pages in a book that we could read togather.

It's like 3 pages

@BalarkaSen ever read Tondeur's book on foliation?

It's quite comical

I haven't heard of it

It's a regular book on manifold foliations

140 pages on foliations

"Foliations on Riemannian Manifolds"?

And then

140 pages of bibliography

lol

He did the bibliography of every paper on manifold foliation

From the early 20th century to 1995

For a total of 2500 papers

I kinda wonder if he did it as a joke

Sounds like something that could be useful for me

There's too many fucking things about foliations out there

Well if you want a good biblio reference on the topic

@BalarkaSen I'm busy with exams atm but I'd be happy to read it at the end of February

probably a good book

That's a lot. Most I have seen is about 450 in a book by Humphreys. But I think he actually references all of those in the book (which is only just over 200 pages)

Foliation geometry, contact geometry and symplectic geometry come bundled as a package

@Alessandro Okie-dokie

Does bundle geometry come bundled

@ypercubeᵀᴹ are you there ?

@Tuki yes

are you good with space integrals ?

nope!

if i have this $$\int_{2}^{1}\int_{0}^{z}\int_{0}^{y+z} \frac{1}{(x+y+z)^3}dxdydz$$

oh ok

i think the easiest way in this would be to substitute $u=(x+y+z)$ and the work the chainrule backwards and adjust the ingration bounds

I'll just create a post about this and see if someone is able to answer

The almighty sun

easy to make these on the rising sun japan flag