Mathematics - 2020-05-26 (page 2 of 3)

Knight

11:00

@Thorgott I meant, what operation is that $\ast$ doing

Leaky Nun

@AlessandroCodenotti how does italian look

Knight

Relation is not defined in that question

Thorgott

yes, it's arbitrary subject to the given identities

Leaky Nun

@AlessandroCodenotti let's say we're following the homophone relation instead

Thorgott

the question is whether any such relation satisfying the given identities is necessarily associative

Leaky Nun

11:01

so the paper Balarka linked

two words are the same if they are homomophones in Italian

Category:Italian terms with homophones

Fundamental » All languages » Italian » Terms by lexical property » Terms with homophones Italian terms that have one or more homophones: other terms that are pronounced in the same way but spelled differently....

wow this category is much smaller

only 47 entries

Alessandro Codenotti

Homophones are rare in Italian

q is read the same as the hard c, and h is silent, but apart from that I cannot think of other groups of letters that are read the same

Leaky Nun

anno = hanno
cappa = kappa
ceco = cieco

abhas_RewCie

heya

sup?

Alessandro Codenotti

We have homophones in the sense of whole words having different meanings but the same pronunciation, like costa which can be both coast and (she) costs

@LeakyNun Ah right, there's a few with cie/gie vs ce/ge

feynhat

she costs? umm... what?

Leaky Nun

11:04

Thorgott

how much?

Knight

$$x*(x*y)=y$$
$$(y*x)*x=y$$
Is it given or is this just OP's attempt?

Alessandro Codenotti

@feynhat Objects are gendered in Italian

Balarka Sen

@feynhat asking the right questions

Alessandro Codenotti

@LeakyNun Aren't some of those homographs rather than homophones? Like Monaco and monaco

Leaky Nun

11:06

I suppose that's just automatically generated

like words with different capitalizations treated as different words

Alessandro Codenotti

costa is the third female singular person of the present tense, indicative mood, of the verb costare (to cost)

abhas_RewCie

@Knight What's it? logic?

Alessandro Codenotti

The third person of the verb is declined according to the gender of the subject

Knight

@abhas_RewCie This question pal

[math.stackexchange.com/questions/3692195/is-associative](this)

abhas_RewCie

What is $*$ operation?

Knight

11:08

OP didn't define the operation

abhas_RewCie

Okay...

Knight

I couldn't understand the question

abhas_RewCie

Replace, $y$ with $(y*x)*x$

in the first equation.

Now, it looks kinda associative

Leaky Nun

@AlessandroCodenotti what's the masculine form?

feynhat

@AlessandroCodenotti You probably should have written (feminine it) instead of (she). lol.

Knight

11:10

@abhas_RewCie Is that relation given or did he make that attempt?

Alessandro Codenotti

@LeakyNun Uhm well, also costa

abhas_RewCie

@Knight given

Alessandro Codenotti

The example sentence I had in mind had a chair, which is female

Leaky Nun

@AlessandroCodenotti the masculine and feminine forms for 3rd sing pres ind are always the same

Alessandro Codenotti

Yeah I realized

I don't know grammar lol I just speak Italian

Leaky Nun

11:11

fair

Alessandro Codenotti

Hmm maybe only the past participle has a gender in the verb then

Balarka Sen

ok chomsky

Leaky Nun

Costa is also a brand of coffee in London

I remember going to Costa stores in year 1

they're like Starbucks stores

(do those have a generic name?)

Alessandro Codenotti

I know, founded by Mr. Costa

Leaky Nun

cafe?

Balarka Sen

11:12

everyone has an inbuilt deep grammar anyway so theres no reason to learn language

thats just surface structure

nothing compared to this DEEP STRUCTURE BRO

towc

@robjohn oh, right, I should have floored all the xs

$\int_{2}^{n-1} \left \lfloor{\left \lfloor{\frac{n}{\left \lfloor{x}\right \rfloor}}\right \rfloor - \frac{n}{\left \lfloor{x}\right \rfloor} + 1}\right \rfloor dx$

abhas_RewCie

sxpt

towc

you can also assume $n \in \mathbb{N}, n > 2$, for the sake of that problem

I guess at that point it's more like $\sum_{2}^{n-1} \left \lfloor{\left \lfloor{\frac{n}{x}}\right \rfloor - \frac{n}{x} + 1}\right \rfloor$

4 internet points to whoever figures out what I'm trying to achieve with that

5 if you also send a 4 cheeses pizza to my address

geocalc33

11:54

can a kernel be anything?

in terms of an integral transform

for the laplace transform the kernel is $K(s,t)=e^{-st}$

there are other kernels I know of, such as the kernel for the fourier transform

I'm under the impression that a "good" integral transform is invertible for a large class of functions. Is this on point?

Sha Vuklia

guys, why do we need to take a quotient when dealing with the push-out in the category of sets?

that is, if we have $f\colon Z\to X$, $g\colon Z\to Y$, and we want the push-out of $f$ and $g$, why do we need to put an equivalence relation on $X\sqcup Y$ given by $f(z)\sim g(z)$?

Mike Miller

What is a pushout?

Alessandro Codenotti

Because it doesn't have the right universal property otherwise? Not sure what's your doubt

feynhat

Hi Mike.

Sha Vuklia

this is the general definition (thought it might be better to share this screenshot, since the commutative diagram is easier to read than what I could have written down)

Mike Miller

12:10

so does X cup Y satisfy that definition?

Sha Vuklia

I feel like it does

Mike Miller

That's not an argument

Hi feynhat

Sha Vuklia

oh oops

I now see where it goes wrong

Yuvraj

@Semiclassical hi

Sha Vuklia

I was checking the wrong part of the diagram

Mike Miller

12:10

Where does it go wrong?

Ah

Sha Vuklia

in the left part of the diagram

$\iota_X f=\iota_Y g$

Yuvraj

how are you?

feynhat

The other day we worked out when pullback gives an embedded submanifold (of the product). Is there a similar result for pushouts?

Balarka Sen

Pushout of manifolds is mostly not gonna be a manifold

Mike Miller

Yeah there you go

feynhat

12:13

sad

Balarka Sen

It exists if you pushout along open submanifolds, but you stop being Hausdorff (take R <- R\0 -> R, pushout is line with two origins). In schemes it's a bit better, pushout exists along open immersions and closed immersions both

Mike Miller

You should think of examples more @feynhat --- any example here would have convinced you the answer is "no" :)

Balarka Sen

feynhat should think about schemes more am I right

feynhat

Bruhlarka

Balarka Sen

Pushouts must exist in the stacky sense

geocalc33

12:23

Balarka

Thorgott

why do people even care about pushouts and pullbacks

Balarka Sen

arxiv.org/pdf/1011.5484v1.pdf

AHA

theorem A.1

pushouts exist in the $\infty$-category of derived stacks

Oh man no they also demand closed immersions

What the fuck is the point of this then

Mike Miller

@Thorgott Pushouts are union along a subspace; van Kampen says that pi_1 preserves a certain kind of pushout; these sorts of things are usually comprehensible to algebraic topology; the universal property shows up here and there

Pullbacks are fiber products

or if you like transverse intersections

those also show up

Alessandro Codenotti

They are products and coproducts in the slice category over an object, which AG people will swear is the right category to work in because of some relative viewpoint nonsense

(might be comma category, I can never remember them)

Balarka Sen

Mike prove that pushout of a general diagram in Sch/S exists as a derived stack

Thorgott

12:28

I shall learn algebraic topology some day

Balarka Sen

algebraic geometry should be banned if this is not known

Mike Miller

Sorry to hear Thorgott

geocalc33

has anyone ever wondered about the equal sign

feynhat

umm... sorry because he hasn't learnt it yet or because he shall someday?

geocalc33

do some people go insanse from trying to learn AG

Balarka Sen

12:32

the people who try to learn AG are kind of fucked in the head from the beginning

geocalc33

oh

well I have a question about equivalence

in algebra 1 we learned about things like how to graph $f(x)=x^2$

the key point is that f(x) is exactly equal to x^2

but mathematicians on the historical perspective when given time, developed several things like looser equivalence

why?

like jacob lurie. He developed certain category theories in which the power of the equal sign is diminished. It's almost as if the equal sign is becoming less and less of what it used to be

Mike Miller

Like the equals sign no longer equals itself

geocalc33

yeah

kind of

but my main concern is, if we build all of this elaborate structure of mathematics based on all these loose forms o equivalence what do you gain from it?

Yuvraj

is it an inequality then?

geocalc33

maybe a more positive viewmpoint is that okay now mathematicians have maybe in the future have understood the entire spectrum of equivalence

Mike Miller

12:48

I think the point is that even in the fancy new stuff you still have f(x) equal to x^2, let's say --- but you also have a way to compare x^2 to y^2

The new stuff doesn't displace equalities so much as enrich them; there are now not just two things being equal but ways of going back and forth between different things

Yuvraj

@geocalc33 I always have an impression that sign of equivalence has equal has a story

geocalc33

that's a good point

it's like having two languages and figuring out how to translate between them or something

this relates to integral transforms because apparently (integral transforms) map a problem into a different domain that's easier to work in

Yuvraj

sorry if I interrupted your conversation!

Mike Miller

13:06

Not at all I just didn't have much else to say

geocalc33

13:32

Why does $x^{\frac{1}{2}}=(1-x)^2$ have the same real roots as $x^3=(1-x)^{12}?$

Question no.1

.

let's define a polynomial $p=x_1^2x_2+x_2^2x_3+x_3^2x_4+x_4^2x_1$

How would you solve stabilizer and orbit of this polynomial given variable $x_1,x_2,x_3,x_4$

there are 4! ways you can arrange

If you have intutive approach let me know. Tag me.

May be this is not a good place to ask pathetic math question. I quit.

Semiclassical

15:25

@geocalc33 If you raise each side of thee first equation to the sixth power, then you get the second equation.

that said, this only shows that a real root of the first equation is also a real root of the second equation

to see the issue, note that (for instancne) you could apply the same trick to $x^{1/2}=-(1-x)^2$. Luckily, this equation has no real roots.

feynhat

16:04

I gave up and decided to read Hatcher's proof. I don't understand a few things.

How does he lift the homeomorphism $p_i : X_{i} \to X_{i-1}$ to bundle isomorphisms $E_i \to E_{i-1}$?

Alessandro Codenotti

What are you trying to prove again?

feynhat

pullbacks of bundles over homotopic maps are isomorphic

He defines $h_i(x, \psi_i(x), v) = \begin{cases} (x, \psi_{i-1}(x), v) & \text{ on } p^{-1}(U_i \times I) \\ (x, \psi_{i}(x), v) & \text{ elsewhere } \end{cases}$.

topologicalorientablesurface

If $\sigma,\tau \in S_k$ and $f$ is a k-tensor on a real vector space $V$.

Then $\tau(\sigma f)=(\tau \sigma)f$

Is he composing or taking product?

Tobias Kildetoft

@topologicalorientablesurface I only see one way to interpret that

feynhat

I don't get why this should even be continuous.

topologicalorientablesurface

16:16

composing is the way i interpret it

but $\sigma \circ f$ makes no sense

Tobias Kildetoft

@topologicalorientablesurface I am not sure what you mean by composing here

topologicalorientablesurface

sigma is a bijection on $\{$ $1,2....,k$ $\} $

Tobias Kildetoft

you have an action of $S_k$ on the space. That is the action being considered

feynhat

$\sigma f$ means you permute the arguments of $f$.

Tobias Kildetoft

The equality just states that this is an action

topologicalorientablesurface

16:17

ohhh $\sigma f$ means you permute the arguments

okay

oh haha it was written that in the page before

that its just rearranging the arguments

thanks

feynhat

oooh... $\psi_i|_{X\setminus U_i} = \psi_{i-1}|_{X\setminus U_i}$

Thorgott

$\tau(\sigma f)=(\tau\sigma)f$ is actually evil

I always get the order wrong

Tobias Kildetoft

@Thorgott Do you prefer to act on the right but write it on the left?

Thorgott

16:35

nah, but every time I rethink this, I need to remind myself again why writing $\tau(\sigma f)(v_1,....,v_k)=\sigma f(v_{\tau(1)},...,v_{\tau(k)})=f(v_{\sigma(\tau(1))},...,v_{\sigma(\tau(k))})=(\sigma\tau)f(v_1,...,v_k)$ is garbage

Tobias Kildetoft

@Thorgott Yeah, you act via the inverse on the arguments

feynhat

Munkres writes $f^\sigma$. So for him $(f^\sigma)^\tau = f^{\tau\sigma}$. Now that's confusing.

Thorgott

acting via inverses is even worse

Balarka Sen

This extension by zero sheaf is nuts

feynhat

Okay. That was neat. But how does one look at vector bundle over $X \times I$ and go "I should look at a finite cover $U_i$ such that restriction to $U_i \times I$ is trivial, consider the partition of unity $\phi_i$ over $U_i$, and consider the graphs of $\psi_n = \sum_{i=1}^n \phi_i$...." umm... hello?

Thorgott

16:48

why not

Balarka Sen

I don't know what else to do feynhat. You have trivializations on stuff of the form $U_i \times [t_0 - \epsilon, t_0 + \epsilon]$, and I know how to push this trivialization along in the forward direction in time

The way to push it forward is by taking graphs

It's the natural idea because there's nothing else to do!

To be 100% honest with you this is actually just the homotopy lifting property idea in covering space theory (indeed, a variant of the proof gives you that fiber bundles have the homotopy lifting property)

Mike Miller

@BalarkaSen Why is Hom_{AlgGp/k}(k*, k*) = Z for alg closed k?

Tobias Kildetoft

@MikeMiller I don't think you need alg. closed for that

Balarka Sen

Isn't it clear that if f : A^1\0 -> A^1\0 is a regular function such that f(zw) = f(z)f(w) then f(z) = z^n

Mike Miller

@TobiasKildetoft You're right

Probably that's clear

Tobias Kildetoft

16:55

Basically, being anything but a single monomial breaks the comultiplication on the coordinate algebra

as does any non-idempotent coefficient

Balarka Sen

That's a strange way to say monomials are the only multiplicative polynomials, chief

Tobias Kildetoft

@BalarkaSen being multiplicative means respecting the comultiplication, not the usual multiplication

Mike Miller

But regular functions aren't polynomials, which is why this isn't obvious to me.

Balarka Sen

@Tobias Fair point

@MikeMiller Oh ok you want elements of k[z, z^-1]

Still seems like a calculation

Tobias Kildetoft

right, you need the negative powers since you are to get all the integers

Mike Miller

16:59

Oh of course. f/g where f is nonzero except at 0 and g is nonzero except at 0.

You don't even need multiplicativity to see that all regular maps are cz^n I think

That's probably sloppy

Tobias Kildetoft

@MikeMiller Well, there you do need alg. closed if you don't want to use multiplicative

Mike Miller

Got it

Balarka Sen

They have some atrocious notation for the circle

Tobias Kildetoft

I am not sure I ever went through all the details of this actually. I just saw it mentioned as being obvious and quickly convinced myself that it seemed reasonable and then moved on with using it

Balarka Sen

$\mathbb{G}_m = \text{Spec} \;k[z, 1/z]$

Dude, it's a circle

Tobias Kildetoft

17:07

@BalarkaSen it is also the $m$ultiplicative group

Balarka Sen

$m$ultiplicative $\Bbb{G}$roup?

Tobias Kildetoft

Hence the similar notation for the additive group

Balarka Sen

Nuts

Alessandro Codenotti

@BalarkaSen You should knot theorists, they have hundreds of objects with different names, but they are all circles...

Balarka Sen

@Alessandro Yeah say that it's "just a circle" when I tangle your headphone into a Perko pair

These sassing set theorists

Alessandro Codenotti

17:09

lol

I'm actually not going to do (pure) set theory in my PhD most likely! Checkmate

Tobias Kildetoft

@AlessandroCodenotti Does that mean you will be doing applied set theory?

Balarka Sen

Well at thee rate I will do my PhD in Grothendieck universes

Going to have to understand the upper shriek before it's too late

Gonna grab dinner, cya

Alessandro Codenotti

@BalarkaSen $\infty$-stacks you mean, that's what all the cool kids do

@TobiasKildetoft Well I'm still going to do some set theory, so I guess?

feynhat

17:28

@BalarkaSen Not sure what you mean by pushing along in the forward direction in time.

Don't we already have that restricted to $U_i \times I$, the bundle is trivial. What we did in the proof was extend this $U_i$ to $X$ using the graphs.

Right?

Thorgott

18:19

goddamn, why are modules so complicated

anakhro

18:37

Hey cool cats.

@Thorgott what seems to be the problem?

Alessandro Codenotti

Modules are the one place in algebra where literally everything that can go wrong will go wrong

geocalc33

19:01

anyone want to connect on LinkedIn

anakhro

No, but how are you?

geocalc33

I'm pretty good. I am getting a birthday present called "Semi-Riemannian Geometry" by O'Neill from my friend

anakhro

Nice!

He had a nice Elementary Differential Geometry book.

geocalc33

also read an non rigorous article about repulsion principles

"Things with number-theoretic significance tend to push each other apart. The name for this is a repulsion principle,” said Jacob Tsimerman of the University of Toronto

Also played tennis today and my serve is 65-80 mph

Semiclassical

was looking at this blog post of Terry Tao's: terrytao.wordpress.com/2012/09/05/…

and trying to figure out what his 'extensions of the base ring' would look like in the case of the ring being $\mathbb{C}$

Ted Shifrin

19:20

@AlessandroCodenotti Well, if all you knew was abelian groups, you might be upset at how much goes wrong with non-abelian groups. :P

Alessandro Codenotti

That's fair

Semiclassical

he only requires that the ring extension be commutative and contain $\mathbb{C}$ as a subring

Ted Shifrin

I remember (when I took algebra second year of university) being so pleased at how module theory unifies f.g.a.b. and canonical forms of linear maps.

Semiclassical

he does talk about how one could construct certain extensions of $\mathbb{R}$ to his own taste, so I guess I should stare at that for a while

Thorgott

non-abelian groups behave much nicer than modules over non-PIDs

Ted Shifrin

19:32

Yes, and non-finitely-generated modules are a pain, too. I was just offering an analogy, not a formal comment :P

Semiclassical

poorly-founded question about divisors incoming

over an elliptic curve, it seems as if we've got div(f(z)dz) = div(f)+div(dz) for all f

first off, is that right

and second, if it is right, does it break down when we go to higher-genus curves?

(i am pretty well certain that $\text{div}(f\omega)=\text{div}(f)$ where $\omega$ is the invariant differential on the elliptic curve)

Ted Shifrin

It's true in total generality, @Semiclassic.

But $dz$ is nowhere-zero on an elliptic curve.

Semiclassical

hmm

Ted Shifrin

In higher genus, you don't have such convenient formulas for the holomorphic differentials. If it's a plane curve or a hyperelliptic curve, you can write things down explicitly, but not so easy in general.

Semiclassical

what I've got in mind as far as what's confusing me: suppose we now consider the case of a hyperelliptic curve (say, degree-5)

Eduardo Maza

19:46

Good day. My algebra teacher asked me to do a mini project, I'm looking for a topic that covers all (or most) of the following topics:
1. Free modules. Matrix bases and finitely generated modules in main domains,
structure and classification
2. Finely generated abelian groups, structure and classification.
3. Similarity of matrices over fields, rational and canonical forms of Jordan, diagonalization of
matrices, Cayley-Hamilton theorem, Jordan-Chevalley decomposition
4. Quadratic forms, Sylvester's inertia theorem, definite positive and negative forms, bases

Ted Shifrin

Representation theory will invoke a lot of that, but I can't think of anything that hits all the topics. But I'm no algebraist.

Semiclassical

brief computer problem---back now. let me simplify to the case y^2=p(x) where p(x) is quintic with distinct roots

Ted Shifrin

So you're doing only hyperelliptic curves presented as plane curves.

Semiclassical

right, with as small of degree as possible for convenience

in that case, I thought I understood that there are now two holomorphic differentials: one is still $\omega=dx/y$ and the other is $x \omega$

Ted Shifrin

So a smooth plane quintic has genus 6, or something.

Semiclassical

19:53

wait, what

i thought smooth plane quintic was genus 2

Ted Shifrin

Um, $(d-1)(d-2)/2 = g$, best I remember. Maybe I'm rusty.

Semiclassical

hrm. i see what you mean

the wiki page on hyperelliptic curves gives the degree as either n=2g+2 or n=2g+1

so g=2 gives n=5,6

but the degree-genus formula seems at loggerheads with that

Ted Shifrin

Well, one has to think about what singularity there is at infinity.

Semiclassical

ah

Ted Shifrin

I'm going to eat lunch now, so I won't work it out now.

Semiclassical

20:03

fair

Ted Shifrin

I never did anything much with hyperelliptic curves, I confess, other than assigning a few homework exercises.

Semiclassical

fair

Ted Shifrin

I still say you should look at Griffiths's little book I suggested.

Semiclassical

which one was that?

@TedShifrin i think i was taking for granted that hyperelliptic curves wouldn't have singularities, which I see now was an incorrect assumptionn

Semiclassical

20:59

@TedShifrin the statement on y^2=p(x) with degree >3 seems to be that one shouldn't embed it in the projective plane, but rather view the projective curve as a double cover of the projective line and use Riemann-Hurwitz

which ...neat?

geocalc33

yo

does anyone think "momentum" exists in sports games?

Semiclassical

Hot hand

The "hot hand" (also known as the "hot hand phenomenon" or "hot hand fallacy") is the purported phenomenon that a person who experiences a successful outcome has a greater chance of success in further attempts. The concept is often applied to sports and skill-based tasks in general and originates from basketball, whereas a shooter is allegedly more likely to score if their previous attempts were successful, i.e. while having "hot hands.” While previous success at a task can indeed change the psychological attitude and subsequent success rate of a player, researchers for many years did not find...

is essentially what you're asking about

(with the present status being a firm ".maybe"?)

Ted Shifrin

21:18

@Semiclassical This is the definition of a hyperelliptic curve, in fact. The canonical mapping (which is for most curves an embedding) given by taking a basis of the holomorphic differentials maps you to $\Bbb P^{g-1}$. When the curve is hyperelliptic, this gives a branched double-cover of (a curve isomorphic to) $\Bbb P^1$.

The book is Griffiths's Introduction to Algebraic Curves.

Semiclassical

21:41

mmkay

Transcript for

$Mathematics$ Mathematics