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

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

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)

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

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

F might not be finite dim.

as?

what does approx even mean? isomorphism?

yes isomorphism.

isomorphism as?

module isomoprhism

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

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

i dont know what alpha is

alpha is an element of End(F)

yeah, but which

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

any map

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

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

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 $F$?

my first guess would be to associate it with identity map

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

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

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

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

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

can you name some

the zero map..

yeah, that's one example

can you name more

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

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

i should've used the notation \alpha_f instead

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

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

@Hawk Yup!

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?

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

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

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

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.

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

you should think about the other options this way

okay. Thank you

too early in the morning

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

@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

@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...

@Faust let me think bro.

About what?

What does linearly dependent mean?

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

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?

yes

are you still trying to do what i asked then?

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

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

I don't understand your question.

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!

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

V is not linearly dependent

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

V=R^3. right?

yes...

hi chat

it has uncountable numbber of elemrnts. right?

@TedShifrin I put myself to sleep out of shame

yes, though that is irrelavent...

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

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 $(1,1,0) = (1,0,0)+(0,1,0)$

mmm

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

but given vectors is a basis

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

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!

yes. it is not linearly independent

are you trolling me?

NOT linearly independent = linearly dependent they are opposites

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

that's nonsense

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

you have confused the meaning of linearly independent and linearly dependent

No, the options

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

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

YES!

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

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

@Astyx salut !

hello

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

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

Les deux !

Et :

seulement ces deux?

Mais pas;

@Astyx welcome! hows your day going?

seulement ces 3?

Je crois

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

Ces trois parmis ? ! , . ; :

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

@Unknownx i strongly agree =p

@Faust Hi, I'm good, wby ?

@LeakyNun Pas que je sache

ok

Not bad at all ^^

Je vais chercher

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

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

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

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

okay

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

common sense?

:)

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

@Anush

It's a direct consequence of the Factor Theorem

the argument works over all integral domains as well

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

it's called "well-known fact"

"trivial"

"small enough to fit in the margin"

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

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

No

damn

All elements have finite norm

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 $||Tx||=\infty$

not for all $x$

That's a flawed reasoning

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

oh

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

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 $iX^{i-1}$

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

is that an example of an unbounded operator @astyx?

infimum, yeah

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

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

Defined on the orthonormal basis $X^i$

ah nvm

You can do that right ?

I don't follow how that related to boundedness

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

yeah

The derivation operator is a linear operator

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?

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 $c||x||<||Dx||$

ok with it so far

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

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

(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 $||DX^n|| =n >c = c||X^n||$

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

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

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

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

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

ah

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

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

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

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

bounded in the $<\infty$ sense?

Yes

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

Well, all finite dimensional operators are bounded

existence of an Eigenbasis?

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)

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"?

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 $T\ne0$

ok I can see that

oh yeah

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.

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 $Tx\ne0$

sniped

:(

lol

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)

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

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

oh yeah, ok that's fair

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

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

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,...

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

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

It doesn't

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

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

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

which gives $P$ back

yup

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

not well-defined

@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?

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

yo @Anakhro @Thorgott

@geocalc33 you still need B to be invertible.

@EdwardEvans what's up? Anything new lately?

just a lot of number theory as usual hahaha

you good?

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)

Maybe about Pythagorean triples?

they'll know about Pythagoras' theorem presumably

Yeah.

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

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

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

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

so no AG for me :/

ha glhf

heh @Thorgott

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

@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.

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

exponent 3 isn't that hard, I think?

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

I think you read wrong lol

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

Nah you're right, it's elementary fuckery

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

some infinite descent magic

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

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

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

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

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

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

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

ew

@Thorgott that would be possibly good.

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

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

lool

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

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?

@Charlie yes.

that's what iff means

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

merci

I wish, man, I wish

@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

the dream

literally the dream

numerics was already too applied for me

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

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

yeah I guess not

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

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

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

@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.)

ty @anakhro

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...

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

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

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

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.