Mathematics - 2017-12-27 (page 2 of 5)

Kasmir Khaan

09:00

but we can rewrite that as H is contained in gHg'

so H=gHg'

Leaky Nun

that isn't a particularly good proof

by good I mean rigorous

Kasmir Khaan

we have the condition for all g in G that saves us

but let me do it Another way

Typhon

@LeakyNun that made no sense whatsoever to me

Leaky Nun

@Typhon exactly

Typhon

Exactly what?

What does Q[sqrt5] mean?

is it a function?

Leaky Nun

09:02

the smallest field extension of Q containing sqrt(5)

MatheinBoulomenos

An example of a theorem in number theory that you're probably familiar with is the fact that every natural number may be written uniquely (up to rearrangement) as a product of prime numbers

Typhon

Thats a theorem?

MatheinBoulomenos

yes

Leaky Nun

@Typhon that's the fundamental theorem of arithmetic

Typhon

oh

Leaky Nun

09:03

and yes, you need to prove it

MatheinBoulomenos

it's not obvious

Leaky Nun

by "you" I mean the generic you

Typhon

i assumed it was true but i never had that actually assserted

Kasmir Khaan

let h_1 be in H, and we consider gh_1g' = h_2

Leaky Nun

by generic I mean just about everyone

Kasmir Khaan

09:03

we can do this because gHg' is contained in H for all g in G

Leaky Nun

by everyone I mean each individual in this room and not

Typhon

I thought there were numbers with non unique primes

Leaky Nun

by room I mean this particular setting where we discuss mathematics or other things

Typhon

:(

Kasmir Khaan

so h_1 = g'h_2g , we took element in H , and we showed that is an element of gHg'

Leaky Nun

09:04

by mathematics I mean the study of numbers and shapes and abstract notions

Kasmir Khaan

@LeakyNun that what u wanted?

Leaky Nun

by number I mean the real numbers, complex numbers, integers, or quaternions

Typhon

What about divergent limits?

Leaky Nun

by real number I mean the cauchy sequences of rational numbers modulo some relation

Typhon

does math study those?

Leaky Nun

09:04

no, divergent limits are not numbers, and hopefully we don't have to go there again

or else I can invite someone to this room

Typhon

I wasnt meaning that

i meant the expression exists so its not just numbers we study

Kasmir Khaan

leaky

i want to understand the second part

Typhon

Surely we also study statements that are and are not true or whatever

Kasmir Khaan

that gHg' is contained in H without being equal to H

Typhon

amd also pepperoni

Lllilillilil

Leaky Nun

09:07

@KasmirKhaan To prove that "gHg' contained in H for all g" implies "gHg'=H for all g":
Assume that gHg' contained in H for all g. Now, we want to prove that gHg'=H for all g. Fix an arbitrary g. We know that gHg' is contained in H by our assumption, so we need to prove that H is contained in gHg'. By multiplying g' on the left and g on the right, we only need to prove that g'Hg is contained in H. However, that is true by assumption, since gHg' is contained in H for all g, and in particular for g'.

Kasmir Khaan

I did that

the exact arguemnt

Leaky Nun

yes, but I don't like your rigour

or the lack thereof

14 mins ago, by Kasmir Khaan

@LeakyNun but I dont see the distinction why would that fail to be normal, ( did not do the example yet )

if you understood the argument, you would know which element to choose as a witness

of the failure of the normality of the subgroup in question

Typhon

Sorry about the lilli post

i nodded off

Im awake now

Balarka Sen

$\nexists$ crisis intensifies

3

Typhon

Sorreee

ill stop falling asleep

Kasmir Khaan

09:11

@LeakyNun hmm by saying that gHg' is contained in H, for all g in G, what are we excluding here?

the two definitons look exactly the same

MatheinBoulomenos

@LeakyNun there's no finite-dimensional representation of that

Typhon

@LeakyNun for a surface made up of trianglws

Leaky Nun

@MatheinBoulomenos why not?

Typhon

if you have a line connecting two points

Kasmir Khaan

@MatheinBoulomenos can you help clear my comfusion ? :D

Alessandro Codenotti

09:13

@LeakyNun nitpicking: that's the smallest ring containing Q and sqrt5 (which also happens to be a field)

Leaky Nun

@KasmirKhaan we are excluding H is contained gHg'

Typhon

can you state that locally the circle is the same as a regular circle?

Leaky Nun

@AlessandroCodenotti you're right

MatheinBoulomenos

You're asking for a matrix which is conjugate to its own square. Look at the Jordan normal form to see that this is impossible (two matrices are conjugate iff they have the same JNF)

Typhon

Im trying to figure out how to construct obe

Leaky Nun

09:14

@MatheinBoulomenos ah, the JNF has to have all 1

MatheinBoulomenos

yeah, so we get only the trivial representation

Leaky Nun

the JNF has to be the identity

thanks

Kasmir Khaan

but leaky, we just said that gHg' is contained in H for all g in G, is equivalent to gHg' = H

now we saying that it can be that gHg' is contained in H for all g in G, without being equal

that is a contradiction to me

Typhon

@LeakyNun anything interesting that would be worth doing over the winter break that yo

you can think of

ive ran out of interesting things to look at

Leaky Nun

no idea

MatheinBoulomenos

09:16

@KasmirKhaan the thing is: the statement $\forall g \in G: gHg' \subset H$ is equivalent to $\forall g \in G:gHg'=H$. But for one single $g$, $gHg' \subset H$ is not equivalent to $gHg'=H$

Leaky Nun

29 mins ago, by Leaky Nun

but you can't pull out the "forall g in G" and claim that I have said that "gHg' = H and gHg' is contained in H" is equivalent for all g

@KasmirKhaan did you even read this?

Kasmir Khaan

aha so the idea about "for all part"

fixing g in G, it can happen that gHg' is contained in H, without being equal

but to be equal, we need that gHg' in contained in H for all g in G

I see it now

thanks for help guys :)

Leaky Nun

I think logic is really helpful to maths

Kasmir Khaan

by saying that g In G

i did not Think of that g being fixed

i thought we could pick any G

and that gHg' is contained in H

Leaky Nun

that's why you should learn logic

Kasmir Khaan

09:19

that way the defintions are identical

Leaky Nun

then you will stop saying nonsense like "g is fixed" and that stuff

Kasmir Khaan

the course starts in marsh i will take it

Leaky Nun

and start using "for all" to make your statements more precise

@KasmirKhaan I would be very expectant for a change in you

Kasmir Khaan

what does that mean?

:D

Typhon

For all means something is true with every element in a set

Leaky Nun

09:21

that means

I expect you to write more rigorous and precise proofs after your course in march

so I can stop being frustrated over your imprecision

Kasmir Khaan

aha thanks :)

Typhon

there exists means there is some element in a set for which a statement is true

Kasmir Khaan

well it was very delicate point

Typhon

@LeakyNun me?

Kasmir Khaan

first we say that g in G, an elemnet of G, it is abstract

Leaky Nun

09:22

@KasmirKhaan which statement implies the other? 1. "for all x, there exists y such that P(x,y)." 2. "there exists y, such that, for all x we have P(x,y)."

Kasmir Khaan

so I assumed it was not fixed

2 ==>1

Leaky Nun

@KasmirKhaan this is true:
"gHg'⊆H for all g" and "gHg'=H for all g" are equivalent
this is false:
"gHg'⊆H" and "gHg'=H" are equivalent for all g

now I don't want to see any more "fixed"

Kasmir Khaan

:D

Leaky Nun

@KasmirKhaan does 1 imply 2?

Kasmir Khaan

well putting it that way it was good

Typhon

09:24

NO

it doesnt

Leaky Nun

@Typhon I'm not asking you

Typhon

Srry

I know i just realized

Leaky Nun

@KasmirKhaan is it any different from what I said at the beginning?

38 mins ago, by Leaky Nun

"gHg' =H for g in G" and "gHg' is contained in H for all g in G" are equivalent

37 mins ago, by Leaky Nun

but you can't pull out the "forall g in G" and claim that I have said that "gHg' = H and gHg' is contained in H" is equivalent for all g

Kasmir Khaan

well it is not i know

Typhon

@LeakyNun i thought basic logic was how to make number sets from the empty set?

Kasmir Khaan

09:25

but it was a Communication problem

first i thought they were identical because i assumed for all g in G

which was wrong assumption

Leaky Nun

@Typhon I wouldn't think so

Typhon

Oh

my bad

Then what is basic logic?

Leaky Nun

@Typhon well, statements and rules about $\neg, \lor, \land, \implies, \iff$

that's zeroth order logic, or predicate logic

Typhon

@LeakyNun oh so common sense?

Leaky Nun

and after that you have $\forall$ and $\exists$, aka first order logic

Typhon

09:28

XD

Leaky Nun

@Typhon you'll realize that it isn't that common of a sense

Typhon

Eh?

doesnt everyone learn that?

Daminark

Well, stuff like the basics of logical implication and interpreting a simple statement probably is common enough

Leaky Nun

as demonstrated above

Daminark

But studying logic takes things to the next level

Leaky Nun

09:30

("as demonstrated above" didn't send out)

it was supposed to follow "you'll realize that it isn't that common of a sense"

Typhon

@LeakyNun if that is all there is to logic then why do people claim i dont know logiv and use me not being able to come up with a formal function for some abstract concept as evidence? Isnt thst more of an engineering failure?

Leaky Nun

@Typhon well as demonstrated above, someone confused "$(\forall g \phi) \iff (\forall g \psi)$" with "$\forall g(\phi \iff \psi)$" for like a week

Typhon

Me no latex

Leaky Nun

"(∀gϕ)⟺(∀gψ)" with "∀g(ϕ⟺ψ)"

and that comment follows "as demonstrated above", not your rant

Typhon

Ah

Yeah i know

Leaky Nun

09:33

I hardly care about your rant

Typhon

I know

Hoe

is that a rant though?

its a question

a serious question

@LeakyNun what is real analysis anyways?

Leaky Nun

@Typhon is this a typo?

Typhon

Yes

Alessandro Codenotti

@Typhon that isn't all there is to logic, try browsing the questions tagged with logic on mse

Typhon

No thanks

im more interested in geometry and graphics

Ill probably study that if i go to college

which is unlikely

Im not interested in being a logician

Cant study too much or i wont have room for programming stuff

Leaky Nun

09:44

good luck programming without any logic

curry howard correspondence, anyone?

Typhon

Well theyll teach m e what logic i need to knlw

surely i need no

not lesr.

*learn second order logic though

Ok now my head actually hurts

is that a sign of sleep deprivation

?

Leaky Nun

10:14

Why Socrates Hated Democracy

►

Arrow

12:20

Hello. Where can I find proofs of Ehresmann's theorem that a proper surjective submersion is a fiber bundle? I would like to see as many different proofs as possible.

ÍgjøgnumMeg

13:06

Do authors usually exclude fields from the definition of local rings by requiring that the unique maximal ideal be non-zero?

Arrow

No, fields are not usually excluded from being local rings.

ÍgjøgnumMeg

Alrighty cheers

philmcole

13:41

Hi guys. Can somebody tell me what the notation $(x,\to)$ means?

Akiva Weinberger

13:56

Is ;; a full colon

ÍgjøgnumMeg

@Akiva lol

Balarka Sen

14:31

@Arrow I don't know a reference, but I think it's a Morse theory argument.

GFauxPas

@philmcole it means $(x,+\infty)$

Leaky Nun

@GFauxPas interesting

Semiclassical

Hi chat

Leaky Nun

high

GFauxPas

hai

Leaky Nun

14:34

if you look up the chat by searching p-adic

you’ll find it being used in physics

@Semiclassical any idea?

Mike Miller

Locally you have a proper smooth submersion to R^n. You wan o lift yeah be n basis vector fields downstairs to fields upstairs. Flow along them to get the desired local triv respecting the projection.

@Arrow

Play this argument out globally when the codomain is R to see that every smooth fiber bundle over R is trivial

Of course one provavly already knows this but whatever

Balarka Sen

That flow argument is where Morse theory comes in right? You have a map M --> R with no critical points

Maybe it's easier than Morse theory :p

Leaky Nun

btw this is an exact sequence: $1 \longrightarrow 1+J(R) \longrightarrow R^* \longrightarrow (R/J(R))^* \longrightarrow 1$

where $J$ is Jacobson radical

MatheinBoulomenos

@LeakyNun interesting. I think I saw special cases of this for $R$ p-adic integers

Balarka Sen

something something Nakayama's lemma

Semiclassical

14:40

There may be a few mathematical physicists who use p-adics but they’re a minority within a minority

An experimental physicist won’t have a clue about them, and even among theorists they’re an obscure subject

MatheinBoulomenos

it's not hard to prove, but I never thought about it

Balarka Sen

@Daminark: New sensation of today's meeting was that Anosov diffeomorphisms are structurally stable. If $f : M \to M$ is Anosov, any small $C^1$-perturbation $g : M \to M$ is $C^0$-conjugate to $f$ by a homeomorphism $h$ of $M$ such that $h$ and $h^{-1}$ are both close to the identity.

Leaky Nun

right, coz the kernel is just things mapped to the unity of R/J(R), ie 1+J(R)

MatheinBoulomenos

as Balarka mentioned, you need Nakayama's lemma to prove that $R^\times \to (R/J(R))^\times$ is surjective

Leaky Nun

ie if x+J(R) is a unit then so is x?

Balarka Sen

14:46

Ya

Leaky Nun

ie if xy in 1+J(R) then xy is a unit

well you also need that to prove that 1+J(R) is a group

the proof is that otherwise there would be a maximal ideal containing xy

but xy-1 is also in that ideal

so 1 is in that ideal, contradiction

do 3 lines get named after someone?

Balarka Sen

Yes, because it's extremely important, but easy to prove.

It's like an inverse function theorem.

Akiva Weinberger

@BalarkaSen You saw the exercise I pinged at you that I couldn't get?

Balarka Sen

@Akiva I pinged you a reply :)

Leaky Nun

btw is there any reason (maybe from graph theory) that 0+1+2+4 in V(2^7) after cycling through the 7 generators and adding 0 forms a vector space?

Akiva Weinberger

14:50

Ah, I didn't see.

It didn't look closed at first glance…

Balarka Sen

@LeakyNun The most general form is if M is a f.g. R-module and IM = M for some ideal I of R, then there exists an r that annihilates M such that r = 1 (mod I).

Akiva Weinberger

(Closed is the $d\omega=0$ one, right? Or is that exact?)

(The names make no sense)

Leaky Nun

ie (0+1+2+4)+(1+2+3+5)=3+4+5+1

Balarka Sen

@Akiva Yeah that's closed. I think it's exact, actually.

Leaky Nun

is there some graph theoretical reason why it is closed under addition?

@BalarkaSen do you get me?

Akiva Weinberger

14:51

Balarka Sen

@LeakyNun I'm not paying attention.

Leaky Nun

i should ask @MatheinBoulomenos

Balarka Sen

@Akiva Okay so maybe it's worth checking if it's closed.

That's the first obstruction to being exact :p

Akiva Weinberger

$$(3x+y^2+2xz)dx+\\ (2xy+ze^{yz}+y)dt+\\ (x^2+ye^{yz}+ze^{z^2})dz$$

So hold on, if I integrate that first part wrt $x$ I get

$\frac32x^2+xy^2+x^2z+C_{yz}$

Balarka Sen

yup

Akiva Weinberger

14:55

where $C_{yz}$ is something that only involves $y$ and $z$

Balarka Sen

Similarly, for the second term $xy^2 + e^{yz} + y^2/2 + C'_{xz}$ should happen.

Akiva Weinberger

Now differentiating that with respect to $y$ is $2xy+C'_{yz}$, so $C'_{yz}=ze^{yz}+y$, or $C_{yz}=e^{yz}+\frac12y^2+C_z$

So so far I have $\frac32x^2+xy^2+x^2z+e^{yz}+\frac12y^2+C_z$

Mike Miller

@BalarkaSen I would say the other way around: Morse theory runs on this flow argument

Balarka Sen

@MikeMiller That's true.

Akiva Weinberger

$C_z'$ should be $ze^{z^2}$

So $C_z$ is $\frac12e^{z^2}$

So in the end I have that this is all$$\boxed{d\left(\frac32x^2+xy^2+ x^2z+e^{yz}+\frac12y^2+\frac12e^{z^2}\right)}$$

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$Mathematics$ Mathematics