Mathematics - 2018-06-03 (page 1 of 5)

MatheinBoulomenos

00:00

@TedShifrin I will deal with Bezout in classical projective geometry this semester in a seminar talk. Does this count as concrete and smooth case?

Leaky Nun

hi @EricSilva

Eric Silva

sup

Balarka Sen

\o @Eric

Leaky Nun

@MatheinBoulomenos in lemma 2.2 (P.4) of this, what is $\widehat E$ ($E$ is a field)?

loch

i like the fact that with enough machinery developed, then bezout's theorem is a consequence of $d$ lines intersect $e$ lines at $de$ many points :p

Leaky Nun

00:02

parallel lines

Daminark

@Eric did you go to beer skits? Also hey Eric, Ted, loch

MatheinBoulomenos

@LeakyNun haven't looked at the lemma, but it's completion

Eric Silva

no i was busy

Ted Shifrin

hi Eric and Demonark

MatheinBoulomenos

Sup @EricSilva @Daminark

Balarka Sen

00:02

I wrote down a summary here.

Eric Silva

i never go to funny things either

Balarka Sen

Once upon a long time ago

Ted Shifrin

@Mathein: Not really.

Eric Silva

hi @Ted and @Mathei

Leaky Nun

@MatheinBoulomenos is there infinite algebraic extension that is not algebraically closed?

Ted Shifrin

00:03

@Balarka: That's good. You'll think more about spectral sequences and the Thom class, among other things.

MatheinBoulomenos

@LeakyNun sure

I'm sure you can find one yourself. I believe in you

Daminark

@Mathein lol, I mean it prob wouldn't be too hard for you to pick up some difftop

MatheinBoulomenos

I don't want to learn diff top, I didn't really enjoyed diff geo

Leaky Nun

lol, maximal abelian extension, maximal unramified extension, separable closure, maximal tamely ramified extension @MatheinBoulomenos

Daminark

It's basically just Sarding everything really hard

Leaky Nun

00:04

> Sarding everything really hard

Balarka Sen

An interesting verb

@TedShifrin I am starting to think in terms of integration over fibers a lot.

Ted Shifrin

Differential topology has nothing to do with connections, @Mathein. Transversality and intersection numbers are fundamental.

Daminark

I have seen some amount of diffgeo and wasn't too fond, but I kinda liked difftop. At least the way we did it had a really different feel

MatheinBoulomenos

I want to learn more alg top and some complex geometry

Ted Shifrin

@Balarka: Good. Next you'll be ready for my Ph.D. thesis. It was all invariant forms and integration over fibers.

MatheinBoulomenos

00:05

that's enough topology/non-algebraic geometry for me

and maybe Lie groups, if that counts

Daminark

@Balarka everything is a verb if you verb it

MatheinBoulomenos

4

Daminark

:O

Balarka Sen

Lol

MatheinBoulomenos

Okay maybe I'll give diff top a try

I suppose Bott-Tu doesn't count just because they do smooth manifolds?

Ted Shifrin

00:09

Diff top is about smooth manifolds.

Having tangent spaces gives you linear algebra.

Bott-Tu doesn't have all the beautiful transversality stuff.

Balarka Sen

Ok, I'm making this chat

Ted Shifrin

I mean, they use the notion.

Balarka Sen

"cohomological garbage" is going to be the name of the room

2

Daminark

Actually as of right now difftop is the only way I know how to compute any homotopy groups. That, covering spaces, and the Hopf fibration

Balarka Sen

better suggestion?

Ted Shifrin

00:10

Thom-Pontryagin construction, Demonark?

Leaky Nun

@BalarkaSen nice name

Eric Silva

garbological cohomology

Balarka Sen

@TedShifrin I still don't know how to do $\pi_4(S^2)$ using T-P

Ted Shifrin

Eric, don't forget we're supposed to work on Bryant this summer.

Daminark

I don't think I've seen that yet

Leaky Nun

00:11

@BalarkaSen obviously there aren't any non-null-homotopic map $S^4 \to S^2$... :P

Balarka Sen

I can T-P your house if you want, @Daminark

Eric Silva

@TedShifrin i have not forgotten!

Ted Shifrin

@Balarka: I sure wish I hadn't shredded my notes from the time I taught diff top and did that.

MatheinBoulomenos

@BalarkaSen "cohomological semi-locally quasi-garbage for derived spectral nerds"

Leaky Nun

@MatheinBoulomenos nooooo

Balarka Sen

00:12

@LeakyNun There is one.

Leaky Nun

you meme-infected ...

@BalarkaSen :o

Balarka Sen

$\pi_4(S^2) \cong \Bbb Z_2$

I think it's generated by $h \circ \Sigma h$ where $h : S^3 \to S^2$ is the Hopf map

0 idea why double of that is nullhomotopic tho

Ted Shifrin

Framings on normal bundles would suggest $\Bbb Z$ rather than $\Bbb Z_2$. Hmm ...

Balarka Sen

I did get $\Bbb Z$ by doing a wrong computation once until Mike pointed out there's a weird invariant which classifies framed cobordisms of surfaces

Hi @Mike

Ted Shifrin

Oh, not Arf or something ...

Balarka Sen

00:15

Yeah Arf I think

Mike Miller

Yes, exactly Arf

Ted Shifrin

Well, getting $\pi_1(SO(2))$ was just the first approximation :P

Mike Miller

I was just trying to think of the form we actually plug in

I agree

MatheinBoulomenos

Is $\pi_4(S^3)$ easier to compute (it's the same by Hopf fibration)? maybe Freudenthal if the first stable homotopy group is somehow easier

idk any homotopy theory

just guessing

Leaky Nun

@TedShifrin isn't SO(2) just S^1

Mike Miller

00:17

I think Balarka wants to see the Z/2, not prove it

Ted Shifrin

Yes, Leaky.

Mike Miller

Proving it is easier

Ted Shifrin

LOL

Balarka Sen

Yeah

Mary Star

Ok! Some derivatives of $f(x)=\cos (x)$ at the point $x_0$ are:
$$f^{(0)}\left (\frac{3\pi}{2}\right )=0, \ \ f^{(1)}\left (\frac{3\pi}{2}\right )=1, \ \ f^{(2)}\left (\frac{3\pi}{2}\right )=0, \ \ f^{(3)}\left (\frac{3\pi}{2}\right )=-1 \\ f^{(4)}\left (\frac{3\pi}{2}\right )=0 , \ \ \ldots$$

So we get
\begin{align*}T_{x_0;m}(x)&=\sum_{k=0}^m\frac{f^{(k)}\left (\frac{3\pi}{2}\right )\cdot \left (x-\frac{3\pi}{2}\right )^k}{k!}\\ & = \frac{f^{(0)}\left (\frac{3\pi}{2}\right )\cdot \left (x-\frac{3\pi}{2}\right )^0}{0!}+\frac{f^{(1)}\left (\frac{3\pi}{2}\right )\cdot \left (x-\frac{3\pi}{2…

Balarka Sen

00:17

It's just running the Hopf fibration

MatheinBoulomenos

Hopf fibrations gives you an isomorphism to $\pi_4(S^3)$, but how do you compute that?

Balarka Sen

Serre spectral sequence

Mike Miller

Framed bordism or Serre SS or Adams SS

MatheinBoulomenos

oh sorry

Balarka Sen

Freudenthal says it eventually stabilizes

Doesn't say when

MatheinBoulomenos

00:19

misrembered the condition

Ted Shifrin

Now you put $x-\frac{3\pi}2 = \big((x_1-\pi) + (x_2-\frac{\pi}2)\big)$ (or whatever is right), and use the binomial theorem, @MaryStar.

Mike Miller

It also says when

It just didn't do it yet

Balarka Sen

Oh. I do not know that part of the theorem then.

When does it?

Mike Miller

@MatheinBoulomenos that was a really aggressive 'no' on my part

MatheinBoulomenos

@MikeMiller it's alright

Mike Miller

00:21

@Balarka I do not remember. A constant away from pi_{2n} S^n

Balarka Sen

Huh

Mike Miller

pi_{2n+2} S^{n+2}?

I dunno you translate

en.m.wikipedia.org/wiki/Freudenthal_suspension_theorem

Yes I think that's right

Balarka Sen

Looks right. Weird

MatheinBoulomenos

so is any higher $\pi_{n+1}(S_n)$ easier to compute?

Leaky Nun

@MatheinBoulomenos So if you are maximally unramified, then your residue is algebraically closed, and going up the chain doesn't change your residue?

MatheinBoulomenos

00:26

yes

totally ramified extensions never change the residue field

Leaky Nun

what is the definition of tamely ramified?

MatheinBoulomenos

for finite extensions, the residue characteristic doesn't divide the ramification index

Ted Shifrin

holy s***, batman ... @anon is here!

hugs @anon

Leaky Nun

holy mother Ted just swore

Mary Star

@TedShifrin So, we get $$T_{x_0;m} = \left (x_1-\pi\right ) + \left (x_2-\frac{\pi}2\right )-\frac{ \left (x_1-\pi\right )^3}{6}-\frac{\left (x_1-\pi\right )^2\left (x_2-\frac{\pi}2\right )}{2}-\frac{\left (x_1-\pi\right )\left (x_2-\frac{\pi}2\right )^2}{2}-\frac{\left (x_2-\frac{\pi}2\right )^3}{6}+\ldots$$ right? Or didn't you mean that? Do we have to find now a specific pattern?

anon

00:29

yeah

hello

hugs back awkwardly

MatheinBoulomenos

@LeakyNun there is also a definition in terms of higher ramification groups

Ted Shifrin

I don't know what your professor wants for an answer, @MaryStar. It's easier to substitute $u=x_1-\pi$ and $v=x_2-\pi/2$ and you can write the general formula with a sum inside a sum, using the binomial theorem.

LOL, @anon ... you didn't need to be awkward :P

Benja

Hi

MatheinBoulomenos

@LeakyNun tamely ramified extensions are really easy

Balarka Sen

@Mathein @loch @Leaky @Daminark @MikeMiller chat.stackexchange.com/rooms/78363/…

Ted Shifrin

00:31

hi Benja

Benja

I wonder if you could help me with a question

I'm currently doing physics

Ted Shifrin

depends on the question

MatheinBoulomenos

The Galois group of the maximally tamely ramified extension with residue characteristic $p$ (both in the p-adic and function field case) is just $\widehat{\Bbb Z} \rtimes \prod_{q \neq p} (\Bbb Z_q)^\times$, where the topological generator over $\widehat{\Bbb Z}$ acts on that other group by the map $x \mapsto x^p$

Benja

I find it much easier and more intuitive to see physics from a mathematical point of view

Ted Shifrin

the physicists may not appreciate that

Benja

00:34

It makes sense for me to change careers if I want to be in physics

?

Leaky Nun

is the subgroup lattice a complete lattice?

Benja

Yes jajaja

Ted Shifrin

I taught plenty of intended physics majors who switched to math or at least double-majored.

Mary Star

@TedShifrin Does it hold that $$T_{x_0;m}=\frac{\left (u + v\right )}{1}-\frac{ \left (u + v\right )^3}{6}+\ldots=\sum_{k=0}^m(-1)^{k}\frac{(u+v)^{2k+1}}{(2k+1)!}$$ ?

MatheinBoulomenos

@LeakyNun yes

Leaky Nun

00:35

but AB isn't a subgroup?

MatheinBoulomenos

take the subgroup generated by AB

Leaky Nun

oh

Ted Shifrin

Well, that's $T_{x_0,2m+1}$? @MaryStar

MatheinBoulomenos

@LeakyNun all the interesting parts of the Galois group of a local field correspond to wild ramification, the tamely ramified extensions are easily classified

Leaky Nun

@MatheinBoulomenos I see

Mary Star

00:38

@TedShifrin Ah. So we take the sum from $0$ to $\frac{\lfloor{m-1}\rfloor}{2}$, or not?

Ted Shifrin

Oh, I suppose.

MatheinBoulomenos

@LeakyNun actually, a prof at my uni wrote down generators and relations for the absolute Galois group of a p-adic field

Leaky Nun

@MatheinBoulomenos :o

Mary Star

Oh, it should be $\lfloor{\frac{m-1}{2}}\rfloor$. Or do we not the floor? @TedShifrin

Leaky Nun

I thought that's very complicated

MatheinBoulomenos

00:39

it's a lot easier than in the global case

Ted Shifrin

Oh, yeah, one of those. Figure it out.

Leaky Nun

how

MatheinBoulomenos

well, actually $\operatorname{Gal}(\widehat{K}^{sep}/\widehat{K})$ is a subgroup of $\operatorname{Gal}(K^{sep}/K)$

and it's simpler in many ways: One can show that at least in the p-adic case, there exist only finitely many extensions of fixed degree $n$, so this means the Galois group has only finitely many open subgroups for every index

that's false already for $n=2$ obviously in the global case

and all finite quotients are solvable

you can even write down a series of normal subgroups with abelian factors that have arithmetic meaning

Leaky Nun

do you have a link?

MatheinBoulomenos

for every finite Galois extension

Leaky Nun

00:43

did he write a paper?

MatheinBoulomenos

it's in a book

Ted Shifrin

What's that?

MatheinBoulomenos

Neukirch-Schmidt-Wingberg "Cohomology of Number Fields"

Leaky Nun

which one of them is your professor?

MatheinBoulomenos

Both Schmidt and Wingberg are professors here, but the theorem is due to Wingberg

Mary Star

00:44

@TedShifrin Does this mean that the m-th Taylor polynomial of $\cos(x_1+x_2)$ at $x_0=\left (\pi, \frac{\pi}{2}\right )$ is $$T_{x_0;m}=\sum_{k=0}^{\lfloor{\frac{m-1}{2}}\rfloor}(-1)^{k}\frac{\left (\left (x_1-\pi\right )+\left (x_2-\pi/2\right )\right )^{2k+1}}{(2k+1)!}$$ ? Or do we have to simply that expression further?

MatheinBoulomenos

it's freely (and legally) available here: mathi.uni-heidelberg.de/~schmidt/publ_en.html

Ted Shifrin

@MaryStar: That's not a question for me. As I said, you can use the binomial theorem to write an inner sum. I don't know what your professor wants for an answer.

Leaky Nun

@MatheinBoulomenos which one?

MatheinBoulomenos

@LeakyNun cohomology of number fields

as I wrote above

Leaky Nun

ich kann download nicht

MatheinBoulomenos

00:46

why?

Leaky Nun

es keep loading

MatheinBoulomenos

works for me

try another browser?

Leaky Nun

nvm done

840 pages

no wonder

MatheinBoulomenos

yeah, there's so much stuff in this

Leaky Nun

welche Folie?

MatheinBoulomenos

00:48

Folie?

Leaky Nun

page

Ted Shifrin

Seite?

Leaky Nun

hmm

Mary Star

@TedShifrin Ok, but if we leave it like that it is also correct, or not?

At the next question we have to justify that for all $x\in \mathbb{R}^2$ it holds that $f(x)=\lim_{x\rightarrow \infty}T_{x_0;m}$.

For this question is it easier to have the expression of $T_{x_0;m}$ as it is now or using the binomial theorem?

And to calculate this limit: We get an alternating series, right? Isnt it then the Taylor expansion of $\sin(x)$ ?

MatheinBoulomenos

it's the section VII.5 "Explicit Determination of Local Galois Groups", I think you can find the exact page as well as I do

Ted Shifrin

00:49

@MaryStar: You're annoying me. I've told you two or three times I cannot tell you what your professor wants for an answer.

MatheinBoulomenos

the section starts at p 409

Ted Shifrin

You need to use Taylor's Theorem with Remainder to answer the convergence question.

Leaky Nun

@MatheinBoulomenos 7.5.14?

(P.419 printed, P.433 in pdf)

MatheinBoulomenos

@LeakyNun yes, that's it

Mike Miller

@MatheinBoulomenos The framed bordism approach proves that pi_m S^n is equal to the group of "bordism classes of submanifolds of S^n of codimension n-m with a trivialization of their normal bundle"

Sorry, submanifolds of S^m of codimemsion n

To compute pi_m S^{m-1} you want to compute bordism classes of framed links in S^m.

If you have the trivial bundle of rank k over S^1, it has [S^1, SO(k)] automorphisms (and therefore that many different possible framings on the same vector bundle). So there are Z = pi_1 SO(2) many framings on a knot in S^3, and Z/2 = pi_1 SO(k) for a knot in S^{k+1}, k> 2

Links are trivial up to isotopy in S^k, k>3, so you're asking about framed trivial links up to bordism, where each knot is labeled with +- (the framing in Z/2). Using a bunch of pairs of pants, you can add these framings into a single framing on a single knot

Leaky Nun

01:14

@MatheinBoulomenos if I do the Witt vectors thing on $\overline{k}/k$, what do I get for the local field?

Mike Miller

So you get that the homotopy group pi_{k+1} S^k is either Z/2 or 0, depending on whether or not both framings on the unknot extend to a framing over a framing of the normal bundle in a surface in B^{k+1}

I totally don't remember how to do that tho

Abdullah UYU

How do we denote the length of a set? For example $\{1,2\}$ has 2 elements, so has length 2.

I am trying to reformulate my last question.

Mary Star

01:48

@TedShifrin Thank you!!

Akiva Weinberger

02:07

@AbdullahUYU The size (or cardinality) of a set is written $|A|$ or $\#A$

Those are the two notations I've seen most often

If the set is finite, "cardinality" might be too fancy a word to use, 'cause it's usually used with infinite sets

Ted Shifrin

Nah, cardinality is used with finite, too.

And hi, DogAteMy. You been transported across the ocean again?

Akiva Weinberger

Yeah, I'm back home

Stateside

Ted Shifrin

Oh, I thought you'd been back and gone back again.

So, how does it feel to be an alumnus, DogAteMy?

Akiva Weinberger

Not quite graduated, that's June 14

but pretty good

Ted Shifrin

Oh.

I guess they'll pass you somehow.

Akiva Weinberger

02:18

Apparently, if you don't graduate, during the ceremony where people go onstage and grab their diplomas one-by-one, they just hand you something with a smiley face on it

Ted Shifrin

If I were in charge, it would be a grim face :P

Well, nevertheless, glad you're home safely and almost congratulations ... and good wishes on the next adventure.

Abdullah UYU

If we're trying to find a function that is defined $\mathbb{R} \rightarrow \mathbb{R}$ and has exactly two $x$ correspondences for all $y$'s, is it correct to reformulate the question as "Find a function $\mathbb{R} \rightarrow \mathbb{R}$ that validates $\forall y\in\mathbb{R}, |\{x: f(x)=y\}| = 2$"?

Ted Shifrin

@AbdullahUYU: I would prefer to say a function that has the property ... as opposed to validates, but yes.

Akiva Weinberger

Yeah, that works

Ted Shifrin

If you know the notation $f^{-1}(\{y\})$, this would be a good place to use that.

Abdullah UYU

02:24

The inverse image, I suppose?

Ted Shifrin

Right.

Akiva Weinberger

I'd probably write it in words though tbh. Like, "Find a function from $\Bbb R$ to $\Bbb R$ such that every number has exactly two preimages" or something

You don't want continuous, by the way, right?

Ted Shifrin

tee hee

MatheinBoulomenos

@TedShifrin How does the smooth completion of the hyperbola $V(xy-1)$ look like?

Abdullah UYU

Hmm, I am trying to remember the details of the question @AkivaWeinberger

Akiva Weinberger

02:27

I remember doing that problem a while ago, actually, the continuous version

(Proving that there is no such function)

I don't remember what it was

but it was probably needlessly complicated

Ted Shifrin

You mean the variety $xy-z^2=0$ in $\Bbb P^2$?

It's a smooth conic.

MatheinBoulomenos

how many point do we need to add?

Ted Shifrin

One.

MatheinBoulomenos

thanks!

Ted Shifrin

DogAteMy: Spivak has a wonderful question like that, contrasting even and odd ...

Oh, sorry, @Mathein. Two.

Abdullah UYU

02:29

Well, i don't remember a constate about the continuity. @AkivaWeinberger

Ted Shifrin

The points $[1,0,0]$ and $[0,1,0]$ both get added.

In my head I was visualizing $V(y-x^2)$.

Akiva Weinberger

If you don't want continuity, you're essentially asking for a bijection between $\Bbb R\times2$ and $\Bbb R$ (where by $2$ I mean any two-element set, such as $\{0,1\}$)

which is also not obvious

Doable, but not obvious. I think I see it.

Ted Shifrin

Do we accept that $\Bbb R$ is in bijective correpondence with $\{0,1\}^\omega$, DogAteMy?

Akiva Weinberger

Oh, you know what, what I was just thinking just then was too complicated^

I see an easier way

@TedShifrin Ah, that would do it easily

but it would be harder to translate it into a function $\Bbb R\to\Bbb R$

Ted Shifrin

Sure would.

Abdullah UYU

02:34

Anyway, the thing is, my lecturer adviced me to consider the function $a\sin(x)+bx$, examining it for a while, getting no where.

Ted Shifrin

For certain $a$ and $b$, @AbdullahUYU, presumably? You can get $2$ for most preimages, but still just one for an infinite number.

Akiva Weinberger

That would be continuous and I don't think that's doable

I found a simple closed-form solution

Ted Shifrin

Well, don't give everything away, DogAteMy.

Abdullah UYU

Yes, he want me to adjust the $a,b$ @TedShifrin

Akiva Weinberger

@TedShifrin Not if $b$ is near zero

Ted Shifrin

02:36

Of course. I said certain.

Jacksoja

I was asked to see what a matrix A^n converge to

using mathematica

does anyone know how to do that ?

Akiva Weinberger

@AbdullahUYU That's odd, which means $0$ will have an odd number of preimages

($0$ and some positive-negative pairs)

so it can't possibly be a solution anyway

Ted Shifrin

So you aren't using mathematical knowledge like eigenvalues, @Jacksoja? You're just asking about how to input into Mathematica?

Jacksoja

@TedShifrin Yes , the thing is it does no converge when I checked

Ted Shifrin

What's the matrix $A$? Huge?

Jacksoja

02:38

Let me send the matrix am looking at to make it easy

Abdullah UYU

@AkivaWeinberger Let me read the things we read.

Jacksoja

3x3

Akiva Weinberger

Is it determinant 1?

Jacksoja

{{-1,-8.4,-8.4} ,{-1,-4.4,-5.4},{1.2,6,7}}

It does not converge what I found

Ted Shifrin

You're correct.

Jacksoja

02:40

does any entry converge?

Ted Shifrin

Oh wait.

Jacksoja

sorry about that

Now it is correct

Ted Shifrin

Yeah, two of the eigenvalues have modulus way bigger than $1$, so it's going to diverge.

Jacksoja

but entrywise ?

does anything entry converge to something?

Ted Shifrin

Do you know about diagonalization?

Jacksoja

02:42

from my cal , none does, entry 1,1 and 2,1 swings +- 1

Am getting to that soon but not atm

Ted Shifrin

One eigenvalue is positive, real and less than $1$, so in the limit it goes to $0$. But when you translate back to the original coordinates, that won't mean that any particular entry converges, no.

Jacksoja

The question asks if the matrix sequence converge to something

Ted Shifrin

Depends on the actual eigenvectors.

No, for it to converge, you'd need every entry to converge. And that can't happen.

Jacksoja

okay thanks , was a mistake from the teacher then

this is a course on mathematica not actual math yet

Ted Shifrin

Oh.

Jacksoja

02:45

ie no advanced theory by hand yet

Thanks alot @TedShifrin

Ted Shifrin

He should have given you guys a stochastic matrix instead.

Jacksoja

We have in other problems

Ted Shifrin

Ah.

Abdullah UYU

So @AkivaWeinberger, what is the conclusion about the function I gave?

Akiva Weinberger

What function? $a\sin(x)+bx$?

Won't work

You need something discontinuous

Try replacing sine with some discontinuous function

Abdullah UYU

02:51

Well, it doesn't work with $2$, but I think it works with $3$.

For $4<a<5$ and $b=1$.

I couldn't determine the $a$ exactly but it is in the very interval.

i.sstatic.net/shyJU.png @AkivaWeinberger

Akiva Weinberger

Ah, for 3

That looks like it works

Ted Shifrin

It's a great question to investigate.

Jacksoja

How to find a forth degree polynomail that best fits the function e^x in given points in mathematica ? @TedShifrin

say from -2,...,1 integers

Akiva Weinberger

03:07

@AbdullahUYU Or, $a=1$ and $b\approx0.2$ (by dividing by your value of $a$)

@AbdullahUYU Graph $\dfrac{\sin(x)}x$

It has a local minimum of value $-0.21723\dots$

I don't think that has a closed form

$\sin(x)+0.21723\dots x$ works as a function where every value has exactly three preimages.

Fun fact: The local extrema of $\dfrac{\sin x}x$ are exactly where its graph intersects the graph of $\cos x$

Abdullah UYU

Wait :) Too much input.

Those are so interesting by the way.

Silent

04:37

I saw somewhere that $G$ abelian group of odd order then $x\to x^2$ is automorphism of $G$. But if $G=Z/3Z$ then $0\to 0, 1\to 1, 2\to 1$ @LeakyNun

anon

$x\mapsto x^2$ is for a multiplicatively written group

if it's additively written then the map is $x\mapsto 2x$

Silent

oh! thank you very much!

Silent

05:45

How many elements of order $2$ in group $S_5$? I thought this way: order $2$ element iff either it is a transposition or product of two disjoint transpositions. So, in first case, pick any two elements out of $\{1,2,3,4,5\}$ and interchange them

In second case, pick two elements out of $5$ and then from remaining three, so 5 C 2+3C2

So, is the answer 40?

anon

06:55

@Silent 5C2+3C2 counts the ways to pick 2 out of 5 things or pick 2 out of 3 other things. To count the ways to pick 2 out of 5 things and pick 2 out of the remaining 3, you must do (5C2)x(3C2).

Silent

@anon yes, and then i thought that we need to divide (5C2)x(3C2) by 2 so that we don't count $(1\,2)(3\,4)$ and $(3\,4)(1\,2)$ twice, right?

anon

yes

Silent

so total is 25

thank you very much

@anon, how can we show that there is no polynomial $P(x)$ with integer coefficients such that $P(5)=5$ and $P(9)=7$?

anon

Consider Q(x)=P(x)-5, which is divisible by (x-5) since Q(5)=0

Silent

ok

anon

07:03

What is Q(9)? What happens when you plug x=9 into (x-5)?

Silent

Q(9)=2 and $x-5=4$ when $x=9$

anon

do you get why that doesn't work?

Silent

Since $Q(x)=(x-5)F(x)$

so, Q(9)=2F(9)

or 1/2=F(9)

If all coefficients were integers, then F(9) would be integer too. @anon

am i right?

anon

yep

Silent

thank u very much!

Leaky Nun

09:19

@MatheinBoulomenos I'm basically overwhelmed by notation right now

in this thing you sent me

Secret

10:08

[Random]

Order theory

Order theory is a branch of mathematics which investigates the intuitive notion of order using binary relations. It provides a formal framework for describing statements such as "this is less than that" or "this precedes that". This article introduces the field and provides basic definitions. A list of order-theoretic terms can be found in the order theory glossary. == Background and motivation == Orders are everywhere in mathematics and related fields like computer science. The first order often discussed in primary school is the standard order on the natural numbers e.g. "2 is less than 3", ...

The realm of spacelessness that is structural

Leaky Nun

10:20

@MatheinBoulomenos if $A$ is a local ring that is PID but not a field, and $\pi$ generates the maximal ideal, must every element of $A$ be in the form of $u\pi^b$ with $u \in A^\times$ and $b \in \Bbb N$?

MatheinBoulomenos

@LeakyNun yes

Leaky Nun

why?

MatheinBoulomenos

that's just unique factorizatino in a PID

you have one maximal ideal, so one prime element up to units

Leaky Nun

aha

Silent

11:18

Is $\Bbb Z[x]/(x^2-3)$ integral domain?

Leaky Nun

11:29

@Silent sure

@MatheinBoulomenos @Daminark Is there a sense in which the first isomorphism theorem and the orbit-stabilizer theorem can be derived from each other?

Leaky Nun

11:58

Let G act on S. Pick x in S. Pick h in G.
Then, Stab(hx) := { g | ghx = hx } = { g | h^-1ghx = x } = { hgh^-1 | gx = x } = h Stab(x) h^-1.

Transcript for

$Mathematics$ Mathematics