Mathematics - 2020-10-19 (page 1 of 3)

12:23 AM

TIL

Define a grid diagram of a knot to be a knot diagram made of horizontal and vertical edges, such that every horizontal edge is beneath every vertical edge

Call the number of vertical segments the "complexity" of the grid diagram

Then, for any diagram of the unknot, you can do a sequence of elementary moves such that the complexity of the diagram monotonically decreases to zero

That is, to unknot it, you never have to increase the complexity

This isn't true for usual knot diagrams, the crossing number, and Reidemeister moves

NeverEnoughTime

What's the definition of the complexity of a knot?

Akiva Weinberger

3 mins ago, by Akiva Weinberger

Call the number of vertical segments the "complexity" of the grid diagram

Don't know why the image isn't showing up but here's an example of a grid diagram: katlas.org/wiki/File:3_1_AP.gif

Note that the horizontal segments are undercrossings and the vertical segments are overcrossings

That has complexity 5

All that's really important is all the overcrossings are oriented the same way, and we're counting all segments of a certain orientation

I haven't defined the elementary moves

Some are called "exchange moves", which don't change the complexity, and some are called "stabilization (destabilization) moves", which increase (decrease) the complexity

You can get from any grid diagram of an unknot to the minimal one through exchange moves and destabilization moves

The exchange moves are essentially swapping the y-coordinates of a pair of horizontal segments in cases where that doesn't change the knot type

(Sometimes those increase the number of crossings)

(On the top figure, destabilization moves are the ones that go to the center diagram; stabilization moves go the other way)

("Arc presentations" are equivalent to grid diagrams)

$user image$

Image source: homepages.math.uic.edu/~culler/gridlink

Hawk

1:00 AM

If F is a field, is it true that F \approx End(F)?

F might not be finite dim.

Thorgott

as?

what does approx even mean? isomorphism?

yes isomorphism.

isomorphism as?

module isomoprhism

Can you name a natural map $F\rightarrow\operatorname{End}(F)$?

Hawk

1:10 AM

f \to \alpha (f)...?

Thorgott

i dont know what alpha is

Hawk

alpha is an element of End(F)

Thorgott

yeah, but which

Akiva Weinberger

@AkivaWeinberger The number of horizontal and vertical edges is equal of course

Hawk

any map

Thorgott

1:25 AM

I'm not sure where you're going with that

Hawk

u asked me to name a natural map F \to End(F), i just thought f \to alpha(f) would be one

Thorgott

that's not a specific map

I'm saying there's a specific map that's fairly natural to write down and perhaps it turns out to be an isomorphism

if I given you an element of $F$, do you have an idea how to turn it into an endomorphism of $F$?

Hawk

my first guess would be to associate it with identity map

Thorgott

that won't give you a linear map $F\rightarrow\operatorname{End}(F)$

Hawk

the only map i thought of is the one i told u about. i can't think what specific map u mean

Thorgott

1:37 AM

think about a concrete case like $F=\mathbb{R}$ and recall that every linear map can be represented by a matrix, if that helps

Hawk

this is sounding ike exactly the map i was going for before.

Thorgott

there are endomorphisms of $\mathbb{R}$ other than the identity

can you name some

Hawk

the zero map..

Thorgott

yeah, that's one example

can you name more

Hawk

technically projection map is one but there is nothin to project here

Ted Shifrin

2:54 AM

@Hawk, if $\alpha$ is an endomorphism, then $\alpha(f)$ is an element of $F$, not an endomorphism.

Hawk

4:05 AM

i should've used the notation \alpha_f instead

Snapdragon-X

4:35 AM

Any idea why Recurrence relation using polynomial ($\lambda^n$) is always done when the relation is linear?

why cant it be applied here oeis.org/A248049

Hawk

actually i just found the answer in my book, the required endomorphism is by multiplication

Ted Shifrin

5:19 AM

@Hawk Yup!

Hawk

6:04 AM

If R is a division ring, then it is simple. Now the proof of this looks is almost identical to field being simple. Now the subtlety is that it is only for one-side ideals in division rings. But that if I \subset R (R is a div ring), then i^{-1}i \in i^{-1}I \subset I \subset R is the same proof for field, so how od I extract the one-sideness in this?

Leaky Nun

7:04 AM

@Hawk what do you mean by extract the one-sidedness?

Hawk

7:35 AM

@LeakyNun i mean how to demonstrate division rings have no two sided ideals (simple) in the proof

Leaky Nun

7:56 AM

@Hawk well a two-sided ideal is also a one-sided ideal

Unknown x

I could prove tha $\Nu$ is $\mathbb R^3$

$\mathbb R^3 +\{\vec{a}\}$ is $\mathbb R^3$

So, B and E are correct.right?

C is also correct.

D is also correct.

Is A and F a proper justification. I am feeling that it is incomplete.

Since I can not conclude anything from A and F.

Leaky Nun

8:21 AM

@Unknownx C is not correct, because containing 0 doesn't make it a subspace

you should think about the other options this way

Unknown x

okay. Thank you

Faust

too early in the morning

Unknown x

@LeakyNun F is true. right? I am not able to judge A.

Leaky Nun

@Unknownx F is not true.

just think about whether the explanations will work for other subspaces

we're talking about the logic of the explanation

arriving at the right conclusion with the wrong reasoning is still wrong

Faust

@Unknownx A is clearly in contradiction of B

you defintly cant have both be true

Same with C and F

and well F is in contradiction of the very statement of the question...

Unknown x

8:43 AM

@Faust let me think bro.

Faust

About what?

What does linearly dependent mean?

Unknown x

one vector can be written as the linear combination of other vectors..

Faust

ok so in A its states that V is linearly dependent show me how to write one of those vectors as a linear combination of the other 2

You still there?

Unknown x

yes

Faust

are you still trying to do what i asked then?

Unknown x

8:54 AM

The vectors given in in question B={(1,0,0),(0,1,0),(1,0,1)} is a basis. V is the span of B, hence any vector in span(B) is the linear combination of vectors in B

Faust

So are you saying that there is no way to write one of the vectors as a combination of the other two then?

Unknown x

I don't understand your question.

Faust

Frank is claiming that V is already linearly dependent

but you just wrote down above that V is linearly independent

i wanted you to say V is linearly independent so A must be false!

Unknown x

V is linearly dependent. I said B={(1,0,0),(0,1,0),(1,0,1)} is linearly independent.

Faust

V is not linearly dependent

if it was then you could write one of its basis vectors as the sum of the other 2!

Unknown x

9:01 AM

V=R^3. right?

Faust

yes...

Astyx

hi chat

Unknown x

it has uncountable numbber of elemrnts. right?

Astyx

@TedShifrin I put myself to sleep out of shame

Faust

yes, though that is irrelavent...

Unknown x

9:02 AM

any colloection of vectors with more than 3vectors, is linearly dependent. right?

Faust

no

$C= \{ (1,0,0),(0,1,0)(1,1,0)\}$

C is an example of 3 vectors that are not linearly independent

that is to say $(1,1,0) = (1,0,0)+(0,1,0)$

Unknown x

mmm

Faust

so C is not a basis for $\Bbb R^3$

Unknown x

but given vectors is a basis

Faust

but since B as your wrote above has 3 linearly independent vectors B is a basis for $\Bbb R^3 $

Unknown x

9:05 AM

Faust

oh thats a z

what you proved it that V is made up of 3 linearly independent vectors

so V is not linearly dependent!

it is linearly independent!

Unknown x

yes. it is not linearly independent

Faust

are you trolling me?

NOT linearly independent = linearly dependent they are opposites

Unknown x

I mean to say any set with more than cardinality 3 is linearly dependnt.

Faust

that's nonsense

Unknown x

9:11 AM

@Faust no bro. I am really confused with the options.

Faust

you have confused the meaning of linearly independent and linearly dependent

Unknown x

No, the options

Faust

A states that V is linearly dependent so it has to wrong because V is linearly independent!

Unknown x

' V is linearly independent'. V=span(B).right?

Faust

YES!

Unknown x

9:17 AM

How V is going to linearly independent. I can understand B is linearly independent.

Faust

Lets say $w = (17,12,6) $

We can see that $w\in V$ because $w = 11v_1 + 12v_2 +6v_3$

this is true for any vector in $\Bbb R^3 $

since any vector in $ \Bbb R^3 $ is in V we have that $ V= \Bbb R^3$

In order for a vector space to span $\Bbb R^3 $ it must contain 3 linearly independent vectors

Leaky Nun

@Astyx salut !

Astyx

hello

Faust

since V has 3 linearly independent vectors in its basis V is linearly independent

Leaky Nun

on met espace avant de ? et ! seulement ? @Astyx

Astyx

9:23 AM

Les deux !

Et :

Leaky Nun

seulement ces deux?

Astyx

Mais pas;

Faust

@Astyx welcome! hows your day going?

Leaky Nun

seulement ces 3?

Astyx

Je crois

Unknown x

9:24 AM

@Faust okay bro, I think I have to revise the definitions. Thank you very much for the help.

Astyx

Ces trois parmis ? ! , . ; :

Leaky Nun

@Astyx y a-t-il une raison historique?

Faust

@Unknownx i strongly agree =p

Astyx

@Faust Hi, I'm good, wby ?

@LeakyNun Pas que je sache

Leaky Nun

ok

Faust

9:24 AM

Not bad at all ^^

Astyx

Je vais chercher

Faust

@Unknownx if you need more help after you review the definitions feel free to ask ^^

Unknown x

okay bro. shall i ping? if I have doubts on this question?

Faust

sure, if nothing else someone still here will help you though

the question is mostly just an exercise in understanding the definitions ^^

Unknown x

okay

Anush

10:11 AM

does this theorem have a name? "Over any field, a nonzero polynomial of degree d has at most d roots."

Faust

common sense?

Anush

:)

Faust

i mean you could prove it in about a line so i doubt it has a name, i doubt its even a theorem.

its probally a corollary

The fundemental theorem of algebra might be what your looking for but its usually stated over C

If the $n+1$ distinct roots are $\alpha_i$, then we have that $x = [a_0, a_1, \dots, a_n]^{T}$ is a solution of $Ax = 0$ where $A$ is invertable.

Since the A is invertible for distinct $\alpha_i$, it follows that $x = [0, 0, \dots, 0]$.

Thus if $a_j \neq 0$ for some $j$, then your polynomial can have at most $n$ different roots.

@Anush

Thorgott

11:03 AM

It's a direct consequence of the Factor Theorem

the argument works over all integral domains as well

Leaky Nun

so it's called "a corollary of factor theorem"

Thorgott

it's called "well-known fact"

Leaky Nun

"trivial"

Astyx

"small enough to fit in the margin"

Thorgott

Let $x$ be the fact that a degree $d$ polynomial over an integral domain has at most $d$ roots

Charlie

11:48 AM

Regarding a bounded linear operator $T$ on a normed vector space, it is correct to interpret "bounded" as "the codomain of the operator $T$ contains only elements with finite norm", right?

Seems deceptively simple

Astyx

No

Charlie

damn

Astyx

All elements have finite norm

Charlie

But for boundedness to fail there must not exist a $c\in \Bbb R$ such that $c||x||\geq ||Tx||\forall x$

the only way I can imagine this fails is if $||Tx||=\infty$

Thorgott

not for all $x$

Astyx

11:51 AM

That's a flawed reasoning

Thorgott

what you're describing isn't what not being bounded means

Charlie

oh

Thorgott

it's called bounded because the operator itself has finite operator norm

Astyx

If I look at C[X] (polynomials) and I assume the $X^i$ form an orthonormal basis, then the derivation operator sends $X^i$ to $iX^{i-1}$

Charlie

the "operator norm" is just the minimum value of $c$ right

is that an example of an unbounded operator @astyx?

Thorgott

11:54 AM

infimum, yeah

Astyx

This operator (let's call it D) is not bounded, because for every $c>0$, there exists $n>c$ such that $||DX^n|| =n||X^{n-1}||=n$

Thorgott

what's your norm on $\mathbb{C}[X]$?

Astyx

Defined on the orthonormal basis $X^i$

Thorgott

ah nvm

Astyx

You can do that right ?

Charlie

11:57 AM

I don't follow how that related to boundedness

I'm not a mathematician I have had limited experience with polynomial spaces :c

Thorgott

yeah

Astyx

The derivation operator is a linear operator

Charlie

My interpretation of boundedness basically follows from what I thought it meant on say $f:\Bbb R\rightarrow \Bbb R$

could you say in words why that derivation operator fails to be bounded?

Astyx

Sure

For it to be bounded you ask that there exists a $c$ such that for all $x$, $c||x||\ge ||Dx||$

The opposite of that statement is : for all $c$, there exists $x$, such that $c||x||<||Dx||$

Charlie

ok with it so far

Astyx

12:01 PM

It turns out that, with the norm I defined above, $||DX^n|| = n$ and $||X^n|| = 1$ for all $n$

Thorgott

this notion of boundedness has nothing to do with the other notion of boundedness you're familiar with

the only linear map that is "bounded" in the sense of "having bounded image" is the zero map

Astyx

(It kind of does: you're asking for the restriction of the linear operator on the unit sphere to be bounded)

So for all $c$, there exists $n>c$, such that $||DX^n|| =n >c = c||X^n||$

Charlie

@Astyx is $||X^n||=1$ here meant to be $||X^{n-1}||=1$?

Thorgott

(yeah, true, I was gonna bring that up)

Astyx

@Charlie No, it's $||X^n||=1$ for all $n$

Charlie

12:06 PM

@Astyx what's happening up here at the end in that case?

Astyx

Well, the image of $X^0 = 1$ through $D$ is 0

Charlie

ah

Thorgott

another way to look at it is that a bounded operator is characterized by taking bounded sets to bounded sets

Astyx

Do you know fintie linear algebra @Charlie ? Eigenvalues and eigenvectors ?

Charlie

I'm fairly happy with that yeah

i'm just writing down what you've written so i can see it all in one place :P

Astyx

12:12 PM

For a diagonalizable operator, bounded means that the eigenvalues are bounded

Charlie

bounded in the $<\infty$ sense?

Astyx

Yes

Charlie

does this imply that unbounded operators can't be written as matrices?

Astyx

Well, all finite dimensional operators are bounded

Thorgott

what does diagonalizable mean between infinite-dimensional spaces

existence of an Eigenbasis?

Astyx

12:14 PM

Yes

So you're left with inifinite dimensional spaces, for which the operators cannot really be written as a matrix anyway

If you take $XD$ (derivation, then multiplication by $X$), you find that $XDX^n = nX^n$, so $X^n$ is an eigenvector with eigenvalue $n$

(you check by hand that the $n=0$ case works as well)

Charlie

Would it be correct to state boundedness instead as "if an operator $T$ is unbounded, there exist elements in its codomain with arbitrarily large norm"?

Astyx

So if you were to write $XD$ as a matrix in the canonical basis (ie $(X^n)$), it would be a diagonal matrix with entries $(0,1,2,\dots)$

No, because as Thorgott stated there always are such elements as long as $T\ne0$

Charlie

ok I can see that

oh yeah

Thorgott

If $T\neq0$, there is a $v$ such that $Tv\neq0$, hence $\lVert Tv\rVert\neq0$. Then $\lVert T(\lambda v)\rVert=\lVert\lambda Tv\rVert=\lvert\lambda\rvert\lVert Tv\rVert$ gets arbitrarily large as $|\lambda|$ gets arbitrarily large.

Astyx

For the simple reason that if $Tx$ is in the codomain, $T\lambda x = \lambda Tx$ is in the codomain and has norm $|\lambda||Tx||$, which can be arbitraily large as long as $||Tx||\ne 0$, ie $Tx\ne0$

Thorgott

12:20 PM

sniped

Astyx

:(

Charlie

lol

Astyx

A correct statement would be : if an operator is unbounded, there exist elements in the codomain of its restriction to the unit ball (or sphere) with arbitraily large norm

You can even restrict to $||x||=1$ (this is a small check which you can do)

Charlie

If I could rephrase my question then, why does what you've written above not imply that every operator is unbounded, if I can make $\lambda$ arbitrarily large, then I can always fix it so that $c<\lambda$

sorry if we are going in circles

what you've said about the unit ball is what I would have expected

Astyx

The $\lambda$ affects both sides, both $||\lambda x||$ and $||T\lambda x||$ have the same dependency on $\lambda$

Charlie

12:24 PM

oh yeah, ok that's fair

Astyx

That's pretty much the check I mentionned above

For all nonzero $\lambda$, for all $x$,$$c||\lambda x||\ge ||T\lambda x||\iff c||x||\ge ||Tx||$$

So you can always take $\lambda = ||x||^{-1}$ and take elements with norm 1 only

Charlie

and the restriction to the unit sphere is necessarily because if we allow $\lambda$ to be arbitrarily large it would make every operator unbounded

I think I get it

It's strange to me that the definition of boundedness doesn't place such restrictions on the domain

I've gtg, thanks for your time guys

porridgemathematics

2:21 PM

why is it that the set of points in $C \cap D$, where $C$ is the zeros of $y^2 - x^3 - x$ in $\mathbb{R}^2$ and $D$ the set of zeros of some polynomial $P(x,y) \in \mathbb{R}[x,y]$ is exactly the set of solutions to $y^2-x^3-x = 0 = Q(x,y)$ where $Q(x,y)$ is the polynomial obtained by replacing $y^{2m} $ in $P(x,y)$ with $(x^3-x)^m$ for all $m \geq 1$? I can see that $C \ cap D$ is a subset of that solution set, but i dont know why the other side inclusion holds

sorry,* $C \cap D$ is a subset of that solution set,...

Thorgott

Cause $y^2=x^3+x$ for any $(x,y)\in C$

porridgemathematics

but why does $Q(x,y) = 0$ imply $P(x,y) = 0$?

Astyx

2:37 PM

It doesn't

porridgemathematics

so without more information we can only say $C \cap D \subset$ that solution set?

Astyx

$Q(x,y)=0$ and $y^2 -x^3-x=0$ implies $P(x,y)=0$

porridgemathematics

oh, because we can just replace each $(x^3+x)^m$ with $y^{2m}$ again..

which gives $P$ back

Astyx

yup

geocalc33

What is $f_A(B)=AB^{-1}$ where $A,B$ are matrices?

Thorgott

2:46 PM

not well-defined

anakhro

@geocalc33

Note that B needs to be invertible and if A is mxn then B has to be nxn.

Hence Thorgott's suggestion.

@Thorgott what are you up to?

geocalc33

@anakhro $A,B$ are square matrices and $A$ has only scalars in it, while $B$ has functions in it. That's why I wrote $f(B)$

Edward Evans

yo @Anakhro @Thorgott

anakhro

3:02 PM

@geocalc33 you still need B to be invertible.

@EdwardEvans what's up? Anything new lately?

Edward Evans

just a lot of number theory as usual hahaha

you good?

anakhro

Pretty good. If you had to teach some high school students about some number theory topics, what would you teach them? (assuming they didn't have calculus under their belt)

Edward Evans

Maybe about Pythagorean triples?

they'll know about Pythagoras' theorem presumably

anakhro

Yeah.

Edward Evans

so you can talk about there being infinitely many primitive pythagorean triples

anakhro

3:08 PM

Do you think pythagorean triples is the way to infatuate them with number theory?

Edward Evans

Idk maybe, it's an interesting piece of elementary number theory that you can connect with some geometry they'll have already seen

you can also talk about the congruent number problem as an "easy to state but open" problem

Thorgott

I just learned that the course "algebraic geometry" this semester is actually "algebraic geometry III"

so no AG for me :/

Edward Evans

ha glhf

anakhro

heh @Thorgott

Edward Evans

@anakhro in fact there's a talk by Richard Taylor on youtube where he talks about some elementary problems

anakhro

3:11 PM

@EdwardEvans I have not heard of the congruent number problem. I guess number theory has the benefit of having a lot of hard "simply stated" problems.

Edward Evans

Right, so you can talk about Pythagoras' theorem, talk about generating primitive triples, and then ask "can you find any integer triples that satisfy $x^3 + y^3 = z^3$" and then laugh while they struggle

nOn-TrIvIaL

before anyone says it

or talk about the congruent number problem, which also has a link to geometry

Thorgott

exponent 3 isn't that hard, I think?

Edward Evans

you'll have trouble finding non-trivial integer triples satisfying that

I think you read wrong lol

Thorgott

I mean, showing there aren't any isn't that hard

At least I feel like I've seen a proof of that before

Or am I tripping

Edward Evans

Nah you're right, it's elementary fuckery

anakhro

3:18 PM

I think we did the exponent 3 case in my elementary number theory course I took.

Thorgott

some infinite descent magic

Edward Evans

Yeah even Kummer's proof for regular primes was by descent

I think even Wiles' proof was basically descent

happy that there is an opportunity to waffle about FLT

anakhro

@EdwardEvans at this stage in your mathematical career, to what degree do you understand Wiles' proof?

Edward Evans

I know what a modular form is, what a Galois representation is, and what an elliptic curve is

fin

I think after my master's I could probably read Wiles' proof at a rate of 1 page per month but I'd get there eventually hahaha

anakhro

Is it one of your goals to understand Wiles' proof?

Edward Evans

3:23 PM

I think my main motivation to start studying maths seriously was to understand Wiles' proof, but I wasn't aware of what I was letting myself in for when I started

now I'm in too far and I can't get out

anakhro

Sum 41 - In Too Deep (Official Music Video)

►

Edward Evans

now I'm gonna start studying p-adic Galois representations and something Langlands-y

Thorgott

since I now don't have an AG lecture this semester, I might commit the cardinal sin and take a physics lecture

Edward Evans

ew

anakhro

@Thorgott that would be possibly good.

Get to see the stuff that spurred along the development of the math.

Thorgott

3:26 PM

gotta minor in something and I respect myself enough to not do any more programming

Edward Evans

lool

@Thorgott here you can register your "Anwendungsgebiet" as pure mathematics hahaha

Charlie

I have a newb question about "proof", if I need to show that A is true if and only if B is true it is sufficient to show that A implies B and B implies A?

anakhro

@Charlie yes.

Edward Evans

that's what iff means

Charlie

this demonstrates that there is a 1-to-1 correspondance

merci

Thorgott

3:27 PM

I wish, man, I wish

Edward Evans

@Thorgott I still have to take one module in applied math though

but the other two that would be part of my Anwendungsgebiet will just be more number theory

Thorgott

the dream

Edward Evans

literally the dream

Thorgott

numerics was already too applied for me

anakhro

@Charlie "1-to-1 correspondence" does not have to do with "if and only if".

Edward Evans

3:28 PM

I need to wait and see what p-adic Hodge theory will be like before I can be sure that I take it

Charlie

yeah I guess not

Edward Evans

cuz it might be some comparison between étale cohomology and some other cohomology theory and I have no idea what étale cohomology is

Thorgott

it sounds like something where there's no going back once you've seen it

Edward Evans

well it looks like one of the main results is a categorical equivalence between p-adic Galois reps and some category of weird modules

but idk

the dude only ever holds really advanced courses

anakhro

@Charlie "1-to-1 correspondence" is a term we use to describe functions which are both injective and surjective. "if and only if" however is a logical connective (something which connects two propositions, like "and", "or", "if...then...", etc.)

Charlie

3:33 PM

ty @anakhro

Azmuth

3:53 PM

How such shots are captured?

I mean, are they randomly thrown up in the air and a guy with slo-mo cam captures it?

seems very unrealistic...

anakhro

I assume that it is the same process for each item, photoshopped into the same image one bit at a time.

Azmuth

@anakhro Ahhhh!!! That's great idea!

But, that pose looks lit... How are these done?

anakhro

so probably just hanging things in front of a green screen, then cutting the image out, then putting them all on the same bg and correcting lighting and stuff

The person's pose could have just been done while they are on the ground, or you can just take a shot while they are in the air I guess. But his hair looks like he is stationary.

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