Mathematics - 2020-03-08

Thorgott

00:00

Well, $I^2\subseteq I$, but they needn't be equal, as your counter-example shows

Rithaniel

Also, note that $aI+bI+I^2\subseteq I$

Lucas Henrique

I'm really stuck

EnjoysMath

(a) + (b) = (gcd(a,b))

in $\Bbb{Z}$ ring

$(a + I) (b + I) := ab + I$ is defined that way so that $R/I$ is indeed a ring, but it doesn't workout elementwise in general

You're allowed to define stuff so that stuff works out

Lucas Henrique

what do you mean by "is defined that way"?

EnjoysMath

$R/I$ is another ring right?

Lucas Henrique

00:14

yup

EnjoysMath

But only if you define multiplication carefully, it doesn't work out if you do it purely elementwise

and derive a definition from an elementwise expansion

Rithaniel

Yeah, I actually started googling things because I wasn't sure, myself. The multiplication is simply defined to be $(a+I)(b+I)=ab+I$, it seems.

EnjoysMath

As your expansion shows, you can't in general get $aI + bI + I^2 = I$.

Thorgott

also note that $bI$ is not necessarily the same as $Ib$

Rithaniel

Non-commutative rings? Terrifying

Thorgott

00:16

indeed, but I assume that's the setting as otherwise specifying two-sided ideal would be superfluous

EnjoysMath

@JackOhara need help with your algorithm still?

Lucas Henrique

I thought that $II^\prime = \{xx^\prime\colon x \in I \land x^\prime \in I^\prime\}$

EnjoysMath

You'll need to choose whether you're handling formal polynomials or polynomials as functions. There's a huge difference. Usually in research they use formal polynomials

@JackOhara, but I think if formal then there are infinitely many monomials in $K[X,Y]$ even when $K$ is finite, right?

Therefore you can't enumerate them all, but you could return larger and larger monomials by degree, and then your program terminates on some other unrelated criterion

Thorgott

yes, that's how you define the product of two ideals

multiplying subsets of a ring element-wise is generally not the same as multiplying cosets in the quotient ring

geocalc33

how does one conceptualize a lift in geometry

EnjoysMath

00:22

@LucasHenrique that works in $\Bbb{Z}$ but in general you need $IJ = \{ \sum_{i=1}^n r_i s_i : r_i \in I, s_i \in J, n \in \Bbb{N}\}$ i.e. finite sums

So it's defined that way

Which of course restricts to your simpler definition in $\Bbb{Z}$

because it's a PID i think

@geocalc33 what's a lift?

I'm new to alg geom

geocalc33

@EnjoysMath en.wikipedia.org/wiki/Lift_(mathematics)

I suppose there are many different definitions depending on what field ones in

EnjoysMath

@geocalc33 it's just a refactorization of a morphism

anakhro

@geocalc33 how goes the geometry?

EnjoysMath

If you have $f = g_1 \circ \cdots \circ g_k$ then you have "lifting" going on and vise versa.

It's like reducibility for category morphisms

@geocalc33 my trading system failed, that I emailed you about. Trying some different data, then finally giving up on it. Requires game theory and stochastic processes to analyze, which I don't know yet

I think another money maker would be formally verified learning software answers. Like Brilliant.org but using a formal verifier. So learning out of Pierce's programming language book

@geocalc33 what geometry book are you using?

anakhro

00:40

OH THAT IS BANANACATS

HI

EnjoysMath

hi

:D

Yeah, that bananacats stuff requires some advanced coding skills in type theory so trying to learn some

Otherwise it's just an error-prone diagram editor with limited math features

anakhro

Have you ever worked with Coq?

EnjoysMath

A few tutorials

I was going to support Lean, but learning the language is just as hard as writing my own algorithm

anakhro

It's great.

EnjoysMath

for type checking

@anakhro it uses CoC and Thierry Coquand's original thesis algorithm

I thought of doing every statement in lean visually, which of course leads to all arrows being represented by a graphic arrow

The logic of a type system isn't so bad, it's the parsing that's hard, but making my way through Benjamin Pierce's book here:

ropas.snu.ac.kr/~kwang/520/pierce_book.pdf

I want the system defined in the visual language itself

so a user could ideally create their own type system by dragging some shapes around

and entering in LaTeX

anakhro

00:47

I liked Selinger's notes on lambda calculus.

EnjoysMath

link?

anakhro

mscs.dal.ca/~selinger/papers/lambdanotes.pdf

EnjoysMath

Nice, I bookmarked it

geocalc33

@anakhro ive been doing a lot of GR and looking at convolution metrics and warped products lately. And of course branched manifolds

anakhro

Wild.

Did you end up figuring out that one thing about the pushforward metric, @geocalc33 ?

EnjoysMath

00:49

Be careful of branching manifolds, they could lead you to another universe :D

geocalc33

@anakhro almost

I think I'm a few years out from completely figuring it out

@EnjoysMath what?

anakhro

@EnjoysMath there is also this gem: paultaylor.eu/stable/prot.pdf

geocalc33

@anakhro this is basically what I've been working on math.stackexchange.com/questions/3571855/…

but I'm gonna take a little break now

Jack Ohara

@EnjoysMath Hi ! yeah sorry was not close to my pc

EnjoysMath

no worries

Thanks @anakhro

Jack Ohara

01:04

Might need to postpone it for later

might be not what I want to do , let me check what is it really I need to compute and come back to you ^^

EnjoysMath

@JackOhara what are you using to code in?

SymPy lib / Python has polynomials

Jack Ohara

mathematica

I have another question for you

EnjoysMath

Oh, cool, I could look at it, but I don't usually code in that IDE

Jack Ohara

Do you know about elliptic curves over finite fields?

EnjoysMath

nope

I guess, elliptic curve equation but coefficient field is finite?

Jack Ohara

01:07

no worries !

no they are equations of this form : Y^2 = X^3 +AX +B

the name is missleading haha

I have been trying to write a program that calculates the order of these groups

EnjoysMath

I guess the points on an elliptic curve form a group

that's all I know though

Jack Ohara

and what possible order are to be achieved

Yes they do !:)

that is the intresting part

EnjoysMath

Is this an open problem?

Jack Ohara

there is something called Hasse's bound

EnjoysMath

Link to article?

Jack Ohara

01:09

not sure but I assume yes

there are tons of unsolved problems on elliptic curves

I don't have any link for the moment but any googling will give you some useful results

EnjoysMath

I see, so you're trying to compute the exact number of points in the group

Surely there's a mathematica function already for that

Jack Ohara

Really ?

do you know the name

I would like both addition of elements

and group order

7

Q: What are some open problems regarding elliptic curves in finite fields?

I accept that my question seems so vague and broad, and I already looked into some similar questions in MO. But I would like to learn specifically about some open problems and conjectures regarding elliptic curves in finite fields. Also if there is something about their isogeny in particular it i...

btw this is fun and your kind of questions!

EnjoysMath

@JackOhara library.wolfram.com/infocenter/Conferences/6911

They present an algo

somewhere

Jack Ohara

@EnjoysMath Oh thanks!

that is one thing I dont have to worry about then :)

EnjoysMath

Well you should make sure you're able to run their impl

I would test it

What do you mean by addition of elementse?

You'd like to know the group law table?

Jack Ohara

01:16

Yes the addition law is somewhat involved

( this is why they use it for most new security communication )

RSA and others in comparasion need maybe 2048 bit for the same security provided by Elliptic with 512 i think

or even less

well not so sure about they are the next best thing after symmetric encryptions schemes

in terms of security and practicality

EnjoysMath

Mathematica should also be able to enumerate the points

if it can count the points

Then you just call the GroupLaw(x,y) function on each two points

which it also should have

Jack Ohara

let me check if this is feasible

EnjoysMath

It's $O(|G|^2)$ and is optimal since you have to visit each entry in the table to view it

Jack Ohara

the file is just a slide

made by someone to show in class i think ^^

EnjoysMath

I would be looking into role elliptic curves played in FLT (Wiles' proof)

but that's just me

Jack Ohara

01:24

That is way above what I am confortable with

EnjoysMath

The reason is that would guarantee a steady supply of elegant math to study

Jack Ohara

that proof is the hardest i think is all history of proofs lol

geocalc33

When did Hume do his best philosophizing?

EnjoysMath

yes, true

Jack Ohara

Who is hume?

geocalc33

01:24

a german phisolophser

Thorgott

uh, no

Jack Ohara

:)

geocalc33

He did his best philosophising when he was vac(hume)ing

Jack Ohara

Well I sorta saw that coming

geocalc33

he was Scottish

my bad

Jack Ohara

01:26

that is hardly the point of what you are trying to do so no worries

geocalc33

okay

EnjoysMath

mathworld.wolfram.com/Taniyama-ShimuraConjecture.html

Taniyama-Shimura-Weil theorem

Abwatts

Hey guys, can a function map an element in its domain to the same element in the co-domain three times?

EnjoysMath

Yes

any number of times

Jack Ohara

Yes

take the constant function f : x --> 5

EnjoysMath

01:28

It can actually do it a continuum of times

geocalc33

well slow down there

I think I solved 10% of the Novikov conjecture

all it took was a little homotopy!

Abwatts

that's interesting :] so my reasoning for this question "How many functions $f:\{1,2,...,2019\} \to \{0,1\}$ are there that are not surjective" is wrong

EnjoysMath

The last 10% is the hardest part

geocalc33

90% ^

yeah

EnjoysMath

80% is routine for writing a program the last 20% is hard, by curry-howard isomorphism program = proof

Abwatts

01:34

would the answer be $2$ for the question I posted above?

Thorgott

yes

geocalc33

@EnjoysMath you know anything about a fiber?

other than it being part of a healthy diet

Jack Ohara

it is not commonly used term, preimage is better

geocalc33

oh

Jack Ohara

not better but more elementary i think

geocalc33

01:41

I didn't know that

Abwatts

@Thorgott would it because first if we have $f: \{1\} \to \{0,1\}$, then the function wouldn't be surjective because nothing maps to both elements in co-domain, and secondly, if we have $f: \{1, 2\} \to \{0,1\}$, then it can be the case that we map $1$ to $0$ twice?

Jack Ohara

in order for it to be not surjective you have to map all elements of the domain to one element of the codomain

in this case that you have only 2 elements in the codomain

Thorgott

The question is about functions $f\colon\{1,...,2019\}\rightarrow\{0,1\}$. I don't see how what you're saying is related.

Abwatts

Sorry, I am not sure I quite understand what this question is asking. I know the definition of a surjective function, but I don't know how the answer is $2$..

Jack Ohara

@Abwatts you have to send all elements in the domain to either 0 or all of them to 1

Thorgott

01:44

Well, what is the definition?

Jack Ohara

you cannot do anything else , or else it would be surjective, not wht you want

Abwatts

We need it to be the case that the image of $f$ is equal to the co-domain of $f$ for it to be surjective?

Jack Ohara

and by definition of function, it has to act on every single element of its domain

Thorgott

That's wrong

Abwatts

corrected it, my bad.

@JackOhara I see, so if we send $1, ... , 2019$ to $0$, it would be a different function of if it was the case we sent $1, ... , 2019$ to $0$ to $1$?

Jack Ohara

01:48

@Abwatts yes!

Abwatts

@JackOhara Got it, thank you! :)

Jack Ohara

no problem

geocalc33

what if a set of fibers is the open unit interval

or equivalently, what if a set of preimages is the open unit interval

Thorgott

Ok, so what can the image of such a function look like if it's not surjective?

Abwatts

Just one of the elements in the co-domain of f

EnjoysMath

01:55

math.wisc.edu/~boston/869.pdf

Thorgott

Right, and for each of these two options, there is necessarily only one function that does that.

eigenvalue

Hi there!
If the determinat-function (i.e. det(g_ij)) has a non-zero differential, what does it actually mean? What are the consequences/implications of the differential being non-zero/zero?

Jack Ohara

04:33

@LeakyNun Hello

Rithaniel

04:57

Hmmm, how to show that $40+11\sqrt{-41}$ is irreducible in $\mathbb{Z}[\sqrt{-41}]$

Orb

05:09

hi

Leaky Nun

05:20

$\begin{array}{rcl} \Bbb Z[\sqrt{-41}]/(40+11\sqrt{-41}) &=& \Bbb Z[X]/(X^2+41,40+11X) \\&=& (\Bbb Z/3^8\Bbb Z)[X]/(X^2+1112_3,1111_3+102_3X) \\&=& (\Bbb Z/3^8\Bbb Z)[X]/(X^2+1112_3,2110102_3+X) \\&=& (\Bbb Z/3^8\Bbb Z)[X]/(2110102_3+X) \\&=& \Bbb Z/3^8\Bbb Z \end{array}$

@Rithaniel are you sure it is irreducible?

oh or does irreducible not imply prime again

$3^8 = (40+11\sqrt{-41})(40-11\sqrt{-41})$

Rithaniel

Yeah, in this ring, irreducible does not imply prime. Though, the class group is isomorphic to the cyclic group of 8 elements, so that actually will help

So, the norm of $40+11\sqrt{-41}$ equals $6561=3^8$ and ideals with norm equal to $3$ are in the class which generates the class group (which I've shown elsewhere). So no power less than $8$ is principal. All I need to show is that $3$ does not divide $40+11\sqrt{-41}$

(Unless I'm mistaken on something)

user714630

05:51

Why do you need the last step? Disclaimer: I'm not a number theorist

Rithaniel

06:08

Because if $3$ divides that element, then it can be reduced to a product of $3$ and some other element, thereby making it reducible. See, showing that only 3 divides the norm of the element only gives us that it's corresponding ideal decomposes into primes of norm 3, but not necessarily which primes. There are two possibiliies for which prime ideals can appear. If one of each type shows up, they cancel and we get a factor of 3.

user714630

I see thanks.

Secret

07:21

What is reality?
I am really starting to think that reality is the part of totality that is self consistent
The fact that regardless of what philosophy we use, and what we observed (and to some lesser extent, what other animals and life have reasoned about their surroundings) and the absence of abrupt discontinuous events, suggests that at least the subjective reality that interacts with us humans (which if the objective reality is really out there, is the interface where humans and it interacts) has a strong bias to be self consistent

This perspective also raised interesting metaphysical questions, such as self consistency does not require a lot of criteria, yet most things in real life can be modelled mathematically, which is a stronger version than mere self consistency. It thus lead to the question on why is our human sense of mathematics evolutionary significant enough to become a cognitive bias in our thinking, and only be interrupted by our disposition to create beliefs

adesh mishra

07:37

@TedShifrin Sir, I'm getting no answers on that symmetry question of vector fields. What should I do?

pregunton

13:20

Quick stupid question

In a Steiner triple system, is there a name for the property that when [abx], [cdx], [adx] are blocks then [cbx] is a block? For example, the Fano plane STS(7) has the property but STS(9) doesn't:

^sorry, I meant [abx], [cdx], [ady] are blocks -> [cby] is a block

Secret

13:55

Huh unbelievable, looks like that extended absence from SE and then return from this have completely removed all weirdness channels

now only the weird respond. All response probabilities of the non weird (not just antiweird) have decayed to zero

That will explain why there are not even joke responses to my comments and questions

Looks like yet another character inversion phenomena. Because it seems my reputation have restored a bit in the physics community, at the cost of the maths one

Last time a character inversion happened in 2014, between chemistry and philosophy

Whatever, that does not really matter anymore, considering 3 new channels for maths have been opened to keep the flows going

I should stay focused, do the final proofread of these 3 articles on hand, and submit the lot to the reviewers so that it will finally be activated...

geocalc33

14:34

If you have a branched manifold embedded in $(0,1)^2$ with "intersection" points as $x^s-(1-x)^t$ for some $s,t \in \Bbb R^+ $ can you put a group structure on it? It does not seem likely because, if $s,t$ are in the positive reals then the formula I gave is not a polynomial

Is there a branched manifold notion in algebraic geometry?

I think it would be in the scope of "algebraic manifolds" and algebraic varieties

adesh mishra

14:50

@geocalc33 Hello!

alan2here

How well are arbitrary_paths and/or random_walks aproximated in higher dimensional flat space?

better or worse that with few dimensions?

Edward Evans

@Lukas I have a somewhat computational question for you when you're around :P

Astyx

16:25

I have a bounded self adjoint operator (A, D(A)) on a separable Hilbert space H. I have a subspace $V \subset H$ such that $f(A)V\subset V$ for every bounded function f. Set (B, D(B)) the operator on the Hilbert V with the induced scalar product of H, such that Bf = Af on $D(B) = D(A)\cap V$ (assume it's dense in $V$). I want to show that B has values in $V$

anakhro

16:48

What's D(A)?

Well... $D(\cdot)$

Astyx

The space on which the operator is defined

anakhro

Oh so it's not even on all of $H$. Okay.

anakhro

17:51

@TedShifrin got any good cast iron recipes?

The Terrible Puddle

19:31

How would you compute $\dim_\mathbb{C}(\mathbb{C}[x,y]/(x,y)^n)$ as a function of $n$?

an integer $n>0$

johnny09

20:17

0

Q: Can one show a relation between two Lagrange multipliers of similar convex optimization problems?

Suppose we have the following two convex optimization problems: $$\max_{x_i} \; \sum_{i \in I} w_i \cdot f_i(x_i) \\ \text{subject to:} \quad 0 \leq x_i \leq 1, \quad \forall i \in I \\ \sum_{i \in I} x_i \leq c$$ and given that we reserve $x_i$ beforehand: $$\max_{x_{i '}} \; \sum_{i ' \neq i...

calculus optimization convex-optimization lagrange-multiplier

any help would be much appreciated!

The Terrible Puddle

How do you link to questions?

Simple