Mathematics

12:19 PM

0

Q: Algorithm to count irreducibles in a non-UFD?

Consider a non-UFD that only has 2 units ( $-1,1$ ) and the min difference between 2 elements is $1$. Also there are only a finite amount of elements for any given fixed norm. ( Maybe that follows from the other 2 conditions ? ) I wonder about counting the irreducible elements bounded by a lower...

Any ideas ??

Tobias Kildetoft

@mick what does the min difference between elements mean?

quallenjäger

12:53 PM

What does the notation $\times_{G}$ mean?

Narcissus jewel

@quallenjäger Depends on context. Here it likely means the orbit space under $G$-action.

Tobias Kildetoft

@quallenjäger probably fibered product over $G$

quallenjäger

I am studying the principal bundle, they write the associate bundle from a principal bundle $P$ as $E\times_{G}F$

Mike Miller

@TobiasKildetoft For group actions it's usually orbit space of a product of two G-spaces, in my experience

Leaky Nun

Mike Miller

12:59 PM

In this case it's definitely that though

Leaky Nun

If a and b are coprime, does that mean that b and a are prime?

Mike Miller

observe that because the fibers of $P$ are identified with $G$ as $G$-spaces, the fibers of $P \times_G F$ are identified with $G \times_G F \cong F$, given by sending $(g,f) \mapsto gf$; recall $G$ acts on the fibers on the right

I think I have my right/left handedness all correct there but I occasionally get those wrong

Tobias Kildetoft

@LeakyNun No, of course not

Mike Miller

@Narcissusjewel I sort of didn't respond to your question the other day but I can try now

Ehhhh

Narcissus jewel

@TobiasKildetoft I think he was joking haha

Mike Miller

1:03 PM

I'm too lazy to give a broad overview, nevermind

But maybe John Ma will :D

Narcissus jewel

@MikeMiller Haha sure, let me know if you're ever not feeling lazy :p.

Yep I will prod him about it shortly. Have to make coffee or obtain an energy drink.

quallenjäger

@MikeMiller What is $P \times_G F$? I am building the cross product over a principal bundle?

Mike Miller

@quallenjäger $G$ acts on $P$ on the right and $F$ on the left. We set $$P \times_G F = (P \times F)/(pg,f) \sim (p,gf).$$

Aka, the quotient/orbit space of the product.

this retains a left G-action by acting on the left by G, and this action on each fiber can be identified with the given G-action on F

quallenjäger

@MikeMiller I see, thanks

Mike Miller

@Narcissusjewel The two-line catchphrase, I'd say, is that it's a type of geometry in which there's a lot more local flexibility than you're used to (all symplectic manifolds are locally isomorphic to a "standard" symplectic manifold, unlike Riemannian manifolds, where curvature is a local invariant)

So you can't possibly tell them apart via simple invariants. Then the features of symplectic situations tend to be global in character, and often feature counts of certain types of (global) submanifolds of symplectic manifolds, or symplectic versions of e.g. the Riemannian volume

Semiclassical

2:13 PM

morning chat

some notes for myself: in spherical trig, one has the half-side angle identities $$\sin\frac{A}{2} = \sqrt{\frac{\sin s\sin (s-a)}{\sin b \sin c}},\; \cos\frac{A}{2} = \sqrt{\frac{\sin (s-b)\sin (s-c)}{\sin b \sin c}},$$

where $a,b,c$ are the side lengths of a spherical triangle, $A$ is the angle opposite to side $a$, and $2s=a+b+c$

hence $$\sin A=2\sin \frac{A}{2}\cos\frac{A}{2}=\frac{2 \sqrt{\sin s\sin(s-a)\sin(s-b)\sin(s-c)}}{\sin b \sin c}$$

and hence $4\sin s\sin (s-a)\sin(s-b)\sin(s-c)=\sin A\sin b\sin c$. cyclic permutation gives two other such formulas.

Secret

This part $$\sqrt{\sin s\sin(s-a)\sin(s-b)\sin(s-c)}$$ so reminds of heron's formula

Semiclassical

this last conclusion also implies the sine rules $\frac{\sin A}{\sin a}=\frac{\sin B}{\sin b}=\frac{\sin C}{\sin c}$

@Secret agreed.

Secret

is that really heron's formula analogue in spherical trig?

Semiclassical

I don't know.

Semiclassical

2:34 PM

@Secret what I'm really after is a geometric proof / interpretation of the following identity:

$$4\sin s\sin(s-a)\sin(s-b)\sin (s-c)=\det\begin{pmatrix} 1 & \cos a & \cos b \\ \cos a & 1 & \cos c \\ \cos b & \cos c & 1\end{pmatrix}$$

Anant Saxena

2:46 PM

hey all

Semiclassical

oh. error in my above: should be $\sin^2 A \sin^2 b\sin^2 c$

Master Yushi

3:03 PM

A cycloid is the locus of a point on the circumference of a rolling circle, and it's arc length for one revolution of the circle is 8R, does this result change if the circle is not moving with constant velocity?

that is the angle doesn't vary with time linearly

Antonios-Alexandros Robotis

@TedShifrin I have the solution.

Leaky Nun

@MasterYushi it doesn’t as we only consider it rolling

anon

3:35 PM

welp

Kirill

Hey guys, for a function that is $1$ for positive values of $x$ in $\mathbb{R}$ and $0$ for non-positive $x$, is $f((x_n))_{n \in \mathbb{N}}$ with $x_n = (-1)^n\frac{1}{n}$ divergent or convegent? It alternates between $1$ and $0$, so for every $\epsilon$ there is an $n$ such that for all $N>n$ there is an $N_1$ such that $f(x_{N_1}) - f(x) \ge \epsilon$, so yes it deverges. But $\lim_{n \to \infty}f(x_n) = 0$, so the limit exists, so it conveges. What is right?

Pritt Balagopal

3:53 PM

Hello is anyone on?

I have a question thats really bothering me, I hope someone can help me out

Karl Kronenfeld

@PrittBalagopal Is it along the lines of, what is the meaning of life?

Or, is it more directly related to math?

Balarka Sen

lol

Pritt Balagopal

Math related lol @KarlKronenfeld

Kirill

@PrittBalagopal yes there some here, but they will answer you only if the question is enough good and they will like your question

Karl Kronenfeld

Ah, now your question is bothering me, simply because it is shrouded in mystery.

Pritt Balagopal

4:01 PM

OhhKayy, so basically, how do you find minimum distance of a point from a surface defined by 3d cartesian equation?

Semiclassical

usual method is Lagrangian multiplers

Karl Kronenfeld

mathworld.wolfram.com/LagrangeMultiplier.html

Semiclassical

with the distance squared being the objective function and the surface being the constraint.

Karl Kronenfeld

Darn it, @Semiclassical beat me to it.

Semiclassical

(you do distance squared rather than distance b/c that makes the calculation easier without changing the result)

Balarka Sen

4:04 PM

You do distance square because you can do the calculations in the first place

The distance function is not very differentiable

Semiclassical

yeah

Karl Kronenfeld

@BalarkaSen smoothness is not the issue, because the point is presumably not on the surface

Semiclassical

yeah, the distance function will be a smooth function of position so long as the distance never goes to zeor

that said, it's waaay more convenient to minimize distance squared

Karl Kronenfeld

Oh, man, I got distracted. I was looking something up in a mse question, and then I found my way into the chat!

Semiclassical

lol

Balarka Sen

4:06 PM

@KarlKronenfeld Ah, fair enough.

Pritt Balagopal

Okay, I currently don't know what Lagrangian Multipliers are, trying to read a Wikipedia article and see if I can understand anything. Btw, I am just a college first year student, so I know very little math.

Richard

4:18 PM

If a series converges uniformly on $[a+\varepsilon,b) $ for all $\varepsilon>0$ and pointwise on $[a,b)$, is then the convergence uniform on $[a,b) $?

Semiclassical

@PrittBalagopal the geometric idea goes like this. suppose you've got a surface and a point not on that surface.

if you put a small sphere at that point and expand it slowly, eventually said sphere will touch the surface

Pritt Balagopal

Oh, I see

Semiclassical

at the moment this happens, the sphere is tangent to the surface and the radius of said sphere is the minimum distsance to the surface

Pritt Balagopal

Oh never mind

Yes, and we have to find the radius right?

Semiclassical

right

geometrically, what Lagrangian multipliers enforces is that condition that the sphere (i.e. a level set of the objective function) be tangent to the constraint surface

Pritt Balagopal

4:22 PM

Oh I see.

Semiclassical

algebraically, you do that by demanding that $\nabla F=\lambda \nabla G$ where $F=F(x,y,z)$ is your objective function, $G=G(x,y,z)$ is the constraint function, and $\lambda$ is some nonzero scalar

Pritt Balagopal

I looked at the wikipedia page, it says they define a new function L(x,y,z,k), such that L(x,y,z,k)=f(x,y,z)-kg(x,y,z)

Semiclassical

the gradients there determine the normal vectors for the tangent planes at the point of mutual tangency, and the above requirement enforces that the two normal vectors are proportional

yeah, same thing. in that case you demand $\nabla L=\nabla f-k \nabla g=0$

I don't think the geometry is nearly so obvious when written like that, though

Pritt Balagopal

Oh wait, I don't know what gradients are

Is it like a slope analogy for 3D?

Semiclassical

i wondered if you would. it's a vector function consisting of partial derivatives: $\nabla F = (\partial F/\partial x,\partial F/\partial y,\partial F/\partial z)$

it's not quite the slope analogy; that'd be more like assuming a surface of the form $z=f(x,y)$ and taking derivatives $\partial f/\partial x,\partial f/\partial y$

what it does have in common with the slope analogy, though, is that gradients are related to tangents to surfaces

user8663905

4:27 PM

Question: suppose you had a curve represented by a pair of parametric equations. This curve goes to the same point as t goes to either positive or negative infinity. How would you go about finding a point that is equidistant from all points on that curve in the xy plane?

Semiclassical

that's a topic for multivariable calc, though

where this leads to is that, when you use a lagrangian multiplier to optimize an objective function

Maneesh Narayanan

0

Q: Doubt on the equation 8.11.9 ( Green's function's construction ). How do they take limit of the integral?

Reference: Linear Partial Differential Equations for Scientists and Engineers(Authors:-Tyn Myint-U, Lokenath Debnath) I understood till the red box. I have doubt on the equation in the red box. How do they take limits of the integral? I don't understand. Please help me. What is the motivatio...

differential-equations proof-explanation eigenfunctions greens-function

Can you please help me on the equation 8.11.9

Semiclassical

you end up with the following equations: $\frac{\partial F}{\partial x} = \lambda \frac{\partial G}{\partial x}$, $\frac{\partial F}{\partial y} = \lambda \frac{\partial G}{\partial y}$, $\frac{\partial F}{\partial z} = \lambda \frac{\partial G}{\partial z}$, and the constraint equation written as $G(x,y,z)=0$.

as an example, take the surface to be $z=0$ and the point of interest to be at $(0,0,h)$ (i.e. $h$ above the origin.)

the surface constraint is that $G(x,y,z)=z=0$ and the objective function is $F(x,y,z)=(x-0)^2+(y-0)^2+(z-h)^2=x^2+y^2+(z-h)^2$ (i.e. the distance squared between $(0,0,h)$ and a particular point on the surface).

Pritt Balagopal

@Semiclassical Is there any way to render the LaTeX commands? I'm not really able to translate all those equations you wrote.

Semiclassical

Use the LaTeX in chat link in the room description. that'll give instructions

the above equations then become: $2x=\lambda\cdot 0,$ $2y=\lambda\cdot 0$, $2(z-h)=\lambda\cdot 1$, $z=0$.

so $x=y=z=0$ in which case $F(0,0,0)=h^2$. So the minimum distance from a point at height $h$ to the horizontal plane $z=0$ is just $h$ (as it has to be).

This example is trivial, but the same idea can work for $F,G$ which are decidedly nontrivial.

Any book on multivariable calculus should discuss Lagrangian multipliers, so I'd look there for further examples.

Pritt Balagopal

4:39 PM

Wow thanks @Semiclassical ! You told me alot of stuff, I'll take quite a while to fully understand. Thanks for your help, really appreciate it!

Semiclassical

np

Secret

So one thing I found intriguing about that positive semidefinite symmetric bilinear form is as follows:
1. Why the diagonal of ones

I always felt like the ones has to have some meaning besides to make the det work

Maneesh Narayanan

Please help me.

2

Secret

In completely unrelated news, given how fast AlphaZero master all chess it seems that soon there will be AI which can learn how to explicitly count to $\omega_1$

Integrate Sin ² x by ExamSolutions

►

Maneesh Narayanan

I don't understand.

Semiclassical

4:58 PM

mathematica...wtf is wrong with you

Kevin Driscoll

Stephen Wolfram created me and Im a monster

Semiclassical

basically: I've been cutting and pasting a certain displayed MatrixForm object repeatedly

but somehow mathematica's interpretation of that internally isn't the same as what it's showing visually

Balarka Sen

@KevinDriscoll did you see a new kind of science in the night sky lately?

Kevin Driscoll

@Semiclassical Ya Ive seen lots of screwey things happen with MatrixForm

@BalarkaSen uuuummmmm now that I know of?

Balarka Sen

I see you're not familiar with the meme

Oh well

Semiclassical

5:05 PM

ANKS

Balarka Sen

this man here knows his science

Kevin Driscoll

OH BECAUSE OF STEPHEN WOLFRAM

I forgot all about that

Semiclassical

hmm

suppose I've got a positive semidefinite matrix $I+S$ where $S$ is symmetric and zero on the diagonal

then $e_2^T(I+S)e_1=S_{12}$.

Is there an easy way to argue from that that there exist unit vectors $v_1,v_2$ such that $v_2^Tv_1 = S_{12}$?

brot

5:24 PM

Hi

Is the automorphism group of a topological group always topological?

Semiclassical

okay, got it: since $I+S$ is positive semidefinite with $S$ symmetric (and real---forgot to say that) there's a cholesky factorization $I+S=V^T V$. Hence $S_{12}=e_2^T (I+S)e_1=e_2^T V^T V e_1=(Ve_2)^T (Ve_1)=v_2^T v_1$.

Secret

I felt like there's a whole world of matrix factorisation I did not knew exists until now...

I was only aware of things like QR, Jordan, SVD, LU

Semiclassical

yeah

MatheinBoulomenos

@brot every group is a topological group with the discrete topology. A more interesting question to ask is if there is some topology on the automorphim group which captures some of the topology of the space. If the space is locally compact (or just compactly generated), you can use the compact-open topology

Secret

@LeakyNun The unit ball is the same as that in the 1-norm. Thus the metric space might be $L^1(\Bbb{R}^2)$

MatheinBoulomenos

5:37 PM

@Secret Being a topological vector space, $L^1(\Bbb R^2)$ is homogenous. But $\Bbb R$ with this topology is not homogenous, because, as Leaky observed $\{0\}$ is not open, but the other singletons are

Secret

hmm...

Alessandro Codenotti

topology.jdabbs.com/spaces/S000133

Secret

7

Q: Plotting the open ball for the post office metric space

The post office metric space, $P$ has the distance function defined as follows: $$ d_P (\mathbf{x},\mathbf{y}) := \begin{cases} 0 & \mathbf{x} = \mathbf{y}\\ \Vert \mathbf{x}\Vert_2+\Vert \mathbf{y}\Vert_2 & \mathbf{x}\neq \mathbf{y} \end{cases} $$ where $\Vert \mathbf{x}\Vert_2 = \sqrt{x_...

plotting topology

hmm, very strange...

Alessandro Codenotti

Wait why are the singletons open? That doesn't sound right

MatheinBoulomenos

Choose any $\varepsilon < |x|$

Then the ball around $x$ is $\{x\}$

Alessandro Codenotti

5:45 PM

Hmm so the topology is almost discrete

Since singletons are closed

brot

@MatheinBoulomenos Ok, I see. I guess I thought there was some kind of induced topology.

MatheinBoulomenos

@brot the compact-open topology is always defined. But in general composition is not continuous. (It is if you assume locally compact or compactly generated)

Liad

i need to show that the decomposition of harmonic and analytic function is harmonic, someone can help?

i thought that would follow from cauchy riemann equations but it does not seem to work

MatheinBoulomenos

analytic as in complex analytic? I think you can use that every harmonic function is the real part of some holomorphic function and that the composition of analytic functions is analytic

Liad

so why would it be harmonic?

and yes complex analytic

MatheinBoulomenos

5:57 PM

the real part of a holomorphic function is again harmonic (which does indeed follow from Cauchy-Riemann)

Liad

yea what i just wanted to write :P

but if $f$ is the analytic function , and $u$ is the harmonic one , we have $u = Re(g)$ for some analytic function $g$ , so you say that $u $ composition $f$ is the real part of $g$ composition $f$ , right ?@MatheinBoulomenos

MatheinBoulomenos

right

Liad

it seems a bit strange to me, you say the $Im(g)$ composition $f$ is $im(g $ composition $f ) $ (sorry im not sure what is the sign for the composition)

user193319

0

Q: Rudin: Sequence of Complex Measurable Functions

Here is theorem 1.14 from Rudin's RCA: If $f_n : X \to [-\infty, \infty]$ is measurable for $n=1,2,3,...$, and $g = \sup_{n \ge 1} f_n$ and $h = \lim_{n \to \infty} \sup f_n$, then $g$ and $h$ are measurable. Immediately following this is the corollary (a): The limit of every pointwis...