Mathematics - 2019-01-03 (page 2 of 2)

Leaky Nun

21:00

man I feel like an idiot for missing that trick lol

Akiva Weinberger

Hm actually, what are the conditions for $f(\sqrt x)f(-\sqrt x)$ to be irreducible? Clearly $f$ needs to be irreducible, and not an even function. Are those enough?

Leaky Nun

so if $\alpha$ is a root of $X^3+cX+d$ then $\alpha^2$ satisfies $(X^{1.5}+cX^{0.5}+d)(X^{1.5}+cX^{0.5}-d) = X^3 + 2cX^2+c^2X-d^2$

you're brilliant

if $\alpha$ is a root of $X^3+bX^2+cX+d$ then $\alpha^2$ satisfies $(X^{1.5}+cX^{0.5}+bX+d)(X^{1.5}+cX^{0.5}-bX-d) = X^3 + 2cX^2+c^2X-b^2X^2-2bdX-d^2 = \\ X^3 + (2c-b^2) X^2 + (c^2-2bd) X - d^2$

Akiva Weinberger

The coefficient of $x^{n-1}$ gives you the sum of the roots, right?

Leaky Nun

the negative of

Akiva Weinberger

Right yeah

So the coefficient of $x^{n-1}$ should give you (minus) the sum of their squares

Leaky Nun

21:05

no you would need $x^{n-1}$ and $x^{n-2}$ together

maths researcher

If I want to show p, q, and r equivalent, is it enough to show p implies q and q implies p and q implies r?

Leaky Nun

in fact if $a$ and $b$ are the coefficient respectively then the sum of squares is $a^2-2b$

Akiva Weinberger

I meant the coefficient of $x^{n-1}$ of $f(\sqrt x)f(-\sqrt x)$, sorry

Leaky Nun

which is $2b-a^2$

assuming monic for sanity

Akiva Weinberger

Neat

@LeakyNun Yeah

So we've basically just derived that relation

Leaky Nun

21:08

@AkivaWeinberger you're looking at how the prime $f$ splits in the extension $k(\sqrt X)/k(X)$ (i.e. ramification theory)

hi @MatheinBoulomenos

MatheinBoulomenos

hi @LeakyNun

Leaky Nun

the ring of integers are $k[\sqrt{X}]$ vs $k[X]$

yeah I see why function fields are also global fields, lol

$k[\sqrt{X}]$ is still a P.I.D.

MatheinBoulomenos

$k(X)$ is usually only considered a global field if $k$ is finite

Leaky Nun

yeah sure but you get what I mean

Akiva Weinberger

What's a global field

MatheinBoulomenos

21:11

there's an inherent ambiguity, though, you could use $k[X]$ or $k[\frac{1}{X}]$ to have the same role as the ring of integers

Akiva Weinberger

Is that when the world is finally turned into a corn farm

3

Leaky Nun

given $f \in K[X]$ monic prime, either $f \in K[\sqrt{X}]$ is still prime (e=1) or $f=g(\sqrt X)g(-\sqrt X)$ for $g \in K[X]$ monic prime (e=2)

MatheinBoulomenos

a global field is either a finite extension of $\Bbb Q$ or a finite extension of $\Bbb F_p(x)$ for a prime $p$

Leaky Nun

the former is a number field; the latter is a function field

Akiva Weinberger

I imagine it's the opposite of something called a "local field"?

Leaky Nun

21:13

a local field is a global field completed w.r.t. a place (i.e. a valuation / a norm)

it is either $\Bbb R$ or $\Bbb C$ or a fin. ext. of $\Bbb Q_p$ or a fin. ext. of $\Bbb F_p((t))$

@MatheinBoulomenos but they're conjugates... kinda

MatheinBoulomenos

still different from the number field case

Akiva Weinberger

Fun fact: the earliest spy satellites took their pictures on film and dropped them onto earth; they were caught by planes midair

At least, the US ones did that

(I dunno how you deorbit a film canister)

Leaky Nun

maybe just throw the film canister backward

Akiva Weinberger

Re: summing squares of roots

The annoying thing is you can't do the reverse; you can't sum the square roots of the roots

Leaky Nun

right

Akiva Weinberger

21:19

I mean, you technically can - it's just that you have to include both square roots, the positive one and the negative one

Leaky Nun

and then you get 0

Akiva Weinberger

Ya

Same reason $\zeta(3)$ is so hard to find, in a sense

We can find $\sum_{n\in\Bbb Z\setminus\{0\}}1/n^s$ for any integer $s$

When $s$ is even we get $2\zeta(s)$ and when $s$ is odd the positive and negative bits cancel and we get $0$

Leaky Nun

If $f(x) = \prod (x-\alpha_i)$ then $f(\sqrt x)f(-\sqrt x) = \prod (x-\alpha_i^2)$. We know that the action of the Galois group on $\{\alpha_i\}$ is transitive, and therefore so is the action on $\{\alpha_i^2\}$... so if $\prod (x-\alpha_i^2)$ is reducible then there must be a duplicate, i.e. some $\alpha_i^2 = \alpha_j^2$ with $i \ne j$, and after some thoughts one can conclude that $f(x) = f(-x)$, so $f$ is even

@AkivaWeinberger that's the proof

Akiva Weinberger

I see

Leaky Nun

basically pass it to the splitting field

Akiva Weinberger

21:22

Basically either $\alpha_i$ is conjugate with $-\alpha_i$ or it isn't

Leaky Nun

and if one is, then every is

Akiva Weinberger

Right, by transitivity

Leaky Nun

right

though I have not given enough caution to the char-2 case

I shall not devote any attention to that little prime

Akiva Weinberger

Oh lord

In char 2, $f(\sqrt x)f(-\sqrt x)=f(x)$, actually

because $f(x)^2=f(x^2)$ and because minus signs don't exist

Leaky Nun

really?

aha

Akiva Weinberger

21:25

$(a+b)^2=a^2+b^2$

So I guess, if $\alpha$ is a root of $f$ then so is $\alpha^2$

MatheinBoulomenos

$f(x)^2=f(x^2)$ only works when $f$ has coefficients in $\Bbb F_2$

Akiva Weinberger

Oh crap you're right

Leaky Nun

ok i'm way too slow on these things

Akiva Weinberger

So I guess the deal is, square the coefficients to get the minimal polynomial of $\alpha^2$

Leaky Nun

sure

Akiva Weinberger

21:27

unless $f$ is ev—I mean, unless $f$ has no terms of odd power

Also, if $f$ does have coefficients in $F_2$, then the squaring map must be a permutation on the roots

Weird

Leaky Nun

the squaring map is the frobenius homomorphism afterall

all finite fields are perfect, i.e. the frobenius homomorphism is an automorphism

Akiva Weinberger

Hm, are all polynomials in $\Bbb F_2$ factors of $x^n+x$ for some $n$?

Leaky Nun

all irreducible ones

MatheinBoulomenos

if $f$ is irreducible of degree $n$ over $\Bbb F_p$, then $f$ is a factor of $x^{p^n}-x$

Leaky Nun

$x^2$ for example isn't

Akiva Weinberger

21:29

Ah

@MatheinBoulomenos Is that polynomial squarefree

MatheinBoulomenos

in fact $x^{p^n}-x$ is the product of all irreducible polynomials over $\Bbb F_p$ of degree dividing $n$

Leaky Nun

yes

@AkivaWeinberger it's even separable

Akiva Weinberger

Oh wait I see

MatheinBoulomenos

@AkivaWeinberger yes

Akiva Weinberger

Its derivative is $-1$

Leaky Nun

21:30

right

Akiva Weinberger

@LeakyNun Remind me what separable means

Leaky Nun

no double roots

so in particular it can't have a double factor

Akiva Weinberger

@MatheinBoulomenos Oh cool

I guess that's just a consequence of the other things you said

and the reason the first thing is true (on how if $f$ is irreducible of degree $n$ over $\Bbb F_p$ then it divides $x^{p^n}-x$) is basically, look at the multiplicative group $\Bbb F_p[x]/\langle f\rangle^\times$, yeah?

Leaky Nun

yes

Akiva Weinberger

and what's the name of the thing

orders divide cardinality

Leaky Nun

21:33

lagrange

Akiva Weinberger

Ah, thanks

Of the Lagrange points

MatheinBoulomenos

you also want that $x^{p^n}-x$ has no factors of degree larger than $n$

Akiva Weinberger

Oh. How do you prove that

MatheinBoulomenos

if $f$ is irreducible and divides $x^{p^n}-x$, then $\Bbb{F}_p[x]/(f)$ is contained in the splitting field of $x^{p^n}-x$ which is $\Bbb{F}_{p^n}$

ramose

hi, can someone help me understand why i can1t

MatheinBoulomenos

21:36

now $[\Bbb{F}_p[x]/(f):\Bbb{F}_p]=\mathrm{deg}(f)$ and by the tower rule $\mathrm{deg}(f)$ divides $[\Bbb{F}_{p^n}:\Bbb{F}_p]=n$

Akiva Weinberger

@MatheinBoulomenos How do we know that the splitting field of that is $\Bbb F_{p^n}$?

Leaky Nun

If $\alpha$ has minpoly $X^3+cX+d$ then $\alpha+p$ has minpoly $(X-p)^3+c(X-p)+d$, and then $(\alpha+p)^2$ has minpoly $X^3+(2c-3p^2)X^2+(c^2+6pd+3p^4)X+(-p^2c^2+(2pd-2p^4)c+(-d^2+2p^3d-p^6))$ (thanks PARI)

MatheinBoulomenos

@AkivaWeinberger by Lagrange every element of $\Bbb{F}_{p^n}$ is a root of $x^{p^n}-x$ and there are $p^n$ of those

Akiva Weinberger

Ah

That makes sense

ramose

can someone help me understand why i cant use the discriminant to work out a constant in a paramtric equation?

Akiva Weinberger

21:38

These are all such nice proofs

@ramose Could you give an example?

@LeakyNun What's PARI

Leaky Nun

PARI/GP

PARI/GP is a computer algebra system with the main aim of facilitating number theory computations. Versions 2.1.0 and higher are distributed under the GNU General Public License. It runs on most common operating systems. == System overview == The PARI/GP system is a package that is capable of doing formal computations on recursive types at high speed; it is primarily aimed at number theorists. Its three main strengths are its speed, the possibility of directly using data types that are familiar to mathematicians, and its extensive algebraic number theory module. The PARI/GP system consists of the...

Akiva Weinberger

Ah

Google says it's also a Bollywood film

maths researcher

guys does this type of relation have a name?: if pRq and rRq then pRr

Akiva Weinberger

Transitive relation

Oh wait

Leaky Nun

euclidean relation

maths researcher

21:41

do implications have a euclidean relation?

Leaky Nun

no

maths researcher

what type of relations do implications have?

Akiva Weinberger

If A implies B and B implies C then A implies C

and that's transitive

Leaky Nun

implication forms the morphisms of a category

it's closed under finite product and coproduct

maths researcher

I have no idea what that means @LeakyNun

ramose

21:43

@ Akiva Weinburger Thankyou. postimg.cc/kRPTCX0c I am assuming there is only one collison, but I was wrong, I was told even if there is one collision im still wrong though. I equated the y values, simplfioied and set the equation to 0.

I then took the discriminant and set it to 0 to try and find the value of k. if there is just one collision, then there is one root etc. i was told the collision isnt related to there being a root. its what i did in ages ago when learning the discriminant, I have no idea why that fails here. Thankyou

Akiva Weinberger

@ramose Did you ever use the equations for the x-coordinates?

The particles collide if the $x$ coordinates and the $y$ coordinates are equal

We already have enough information to find when the $x$-coordinates of A and B are equal

ramose

akiva: i didnt, ah so what i did would have worked out, if the y and t values were equal, but ignored the x?

Akiva Weinberger

Yeah, but that doesn't give you a collision, that gives you two particles whose y-coordinates are equal briefly

ramose

ok, that makes perfect sense thanks.

Akiva Weinberger

Lol

(For context, オ is the sound "o" and レ is the sound "re")

Alessandro Codenotti

21:51

@LeakyNun Is that a fancy way to say Lindenbaum algebra?

Akiva Weinberger

(and と is "and")

Leaky Nun

@AlessandroCodenotti yes

Akiva Weinberger

English version

Leaky Nun

lol

Akiva Weinberger

Tys I'm orererererererereo

Leaky Nun

21:53

rereo puts me off ok

Akiva Weinberger

Source

(No, I can't read that - I originally saw the English version and then spent way too much time trying to track down the original)

Leaky Nun

hi @CaptainAmerica16

CaptainAmerica16

@LeakyNun Hey

Akiva Weinberger

You made that impressively fast

Leaky Nun

@AkivaWeinberger no it's from the comments

Akiva Weinberger

21:56

Oh

♫ Olé, olé olé olé ♫

ramose

that looks like an oreo alexa

er, echo

Perturbative

Let's say I have a topological space $X$ that's a polyhedron, then there exists a simplicial complex $K$ and a homeomorphism $h : |K| \to X$. If we have a subspace $A \subseteq X$, is there a subcomplex $L$ of $K$ such that $h[|L|] = A$?

Akiva Weinberger

That's so sad, Alexa order $500 worth of pool noodles

and play Despacito

@Perturbative What's $|K|$?

Is $K$ a combinatorial thingy and $|K|$ the topological version of it?

In any case can't you choose $A$ to be a really nasty subspace

Like it doesn't need to be homeomorphic to a polyhedron or anything

It could be the Cantor set or the topologist's sine curve or things like that

Perturbative

Yeah @AkivaWeinberger, $|K|$ is the topological version of it

Hmm yeah that's a good idea

Akiva Weinberger

Also even if it's "nice" and piecewise linear it could have more edges than $K$

Like imagine if $X$ is a cube and $A$ is a cube with lots of square holes punched out of it

Perturbative

22:09

Yeah so $A$ would have more edges and we'd need a 'bigger' simplicial complex than $K$ to yield $A$ I guess?

Akiva Weinberger

I feel like you could just do the barycentric subdivision of $K$ a billion times until it's big enough, in that situation

Leaky Nun

I haven't heard the word barycentric subdivision for such a long time

Akiva Weinberger

Repeated barycentric subdivision looks really weird and creepy

Perturbative

@AkivaWeinberger Those are really good ideas, thanks!

Akiva Weinberger

(Basically what it is is, to barycentrically subdivide an edge you just divide it in half, and to barycentrically subdivide a polygon you subdivide the edges, add a point in the center of the polygon, and connect that point to the subdivisions of the edges)

(and you can define it inductively for arbitrary dimension)

(It's a theorem that, if you repeat it, the diameters of the pieces go to 0)

If you do it to a cube or octahedron you get this:

which I think is related to polyhedral models of the projective plane?

Leaky Nun

22:18

hi @MatheinBoulomenos

MatheinBoulomenos

hi @LeakyNun

Perturbative

@AkivaWeinberger Yeah I think I've seen that simplicial complex used in simplicial homology calculations of the projective plane but I could be wrong

Akiva Weinberger

Hm actually there's a simpler polyhedral model of the projective plane I think

self-intersecting

where you take an octahedron, delete every other face

(which you can do because the faces are two-colorable, you can color it like a checkerboard and get rid of the faces of a certain color)

Ben Beri

interesting problem: you have an unsorted array of N positive numbers. You have to find out atleast one number that is > the median number in the array with complexity better than O(1/2n) . e.g array of 1-10 numbers, median number is 5, so acceptable number is 6+

Akiva Weinberger

and then you add in the squares that go through the center

('cause any pair of opposite edges on the octahedron determines a square, so you get three of them)

Leaky Nun

22:21

Given rational A, parameterize d such that A+d^2 is a square of a rational...

Akiva Weinberger

(mutually perpendicular and intersecting)

and the resulting thing, combinatorially at least, is a projective plane I think

I've just described the polyhedral version of this thing I think

Oh wait it has a name

Tetrahemihexahedron

In geometry, the tetrahemihexahedron or hemicuboctahedron is a uniform star polyhedron, indexed as U4. It has six vertices, 12 edges, and seven faces: four triangular and three square. Its vertex figure is a crossed quadrilateral. Its Coxeter–Dynkin diagram is (although this is a double covering of the tetrahemihexahedron). It is the only non-prismatic uniform polyhedron with an odd number of faces. Its Wythoff symbol is 3/2 3 | 2, but that represents a double covering of the tetrahemihexahedron with eight triangles and six squares, paired and coinciding in space. (It can more intuitively be seen...

Tetra hemi hexahedron

It's double-covered by the cuboctahedron

Leaky Nun

d=(A-t^2)/(2t)

Akiva Weinberger

I think I remember learning that a polyhedral version of sphere eversion can be done on a cuboctahedron, I think this is why

Leaky Nun

A+d^2=[(A+t^2)/(2t)]^2

Akiva Weinberger

22:43

These are neat-looking trophies

Eurographics Award

Dunno what that is

They made it out of snow

"Whirled White Web"

Nooo

Annoying website but cool slides:

slideplayer.com/slide/5337982

Fewer pictures, more words:

people.eecs.berkeley.edu/~sequin/PAPERS/Isama03_WWW.pdf

Akiva Weinberger

23:10

Fargle

23:30

A manifest is something that locally looks like a party

Akiva Weinberger

commons.m.wikimedia.org/wiki/User:JasonHise

720 degree symmetry

Two full rotations of the cube bring the ribbons back to the start

Fargle

23:45

I think that's the gif on the Wiki page for spinors because it exhibits that characteristic symmetry.

Akiva Weinberger

This can be extended to an ambient isotopy of space. It would take a 720 degree rotation to bring the ambient isotopy back to the identity.

@Fargle What's a spinor

(In 2D, the ribbons would never return to the start)

maths researcher

guys

i have a question

is my proof correct? : The closure of a set A is a closed set: Proof: Assume p is a limit point of the closure A and that p $\notin$ the closure so since p is a limit point then every neighborhood of p contains a point q in the closure such that p $\neq$ q, but since p$\in$ $\mathbb{R}^n - closure of A$ then there exists a neighborhood of p completeley contained in $\mathbb{R}^n - closure of A$ which is a contradiction, because p is a limit point of the closure

@AkivaWeinberger may you please help me?

Fargle

@AkivaWeinberger I, uh, don't rightly know very well, but off the top of my head it's a thing that takes 2 full rotations to return to original position

Akiva Weinberger

That looks right but I'm too tired to know for sure @mathsresearcher

I need to go to bed

Bye

Transcript for

$Mathematics$ Mathematics