Mathematics

Kasmir Khaan

12:00 AM

this is from an old one =p

did not figure that out yet

z= 4-4x^2-y^2 , z>0

its a paraboloid

Maks

@TedShifrin So I do cross product on my vectors, I get the normal vector to the plane

Ted Shifrin

Yes @Maks

Maks

And I apply it on the equation <x-p,n> = 0

Ted Shifrin

Yup.

Maks

That's the normal equation right ?

thz

Kasmir Khaan

12:06 AM

@TedShifrin can you please tell me what is the surface ?

i know its a paraboloid but from my picture nothing exists for z>0

i got it 0<z<4

Aksel'sRose

1:39 AM

If I am asked to prove that $g\in Bil(VxV)$ do I just prove bilinear?

Adeek

1:50 AM

hey @KajHansen

Kaj Hansen

Hey there

Adeek

just want to verify something with you.

So I am computing the galois group of $x^8 - 2$

I computed the splitting field to be $\mathbb{Q}((2)^{1/8},i)$ and this is degree 16 extension ok ?

Kaj Hansen

That sounds reasonable

I suspect the Galois group will come out to be $D_8$. Confirm that for yourself though; I could be wrong

Adeek

let us denote it by $\mathbb{K}$. Consider $\tau((2)^{1/8} \mapsto (2)^{1/8}\omega$ and fixes i, where $\omega$ is 8th root of unity, and $\sigma_1$ complex conjugation. I verified that $H = <\tau_1>$ and $K = <\sigma_1>$ generate this group.

Ted Shifrin

I don't think so @Kaj, but I don't know for sure.

Let's see. What's the splitting field?

Adeek

1:56 AM

Since the index of H is normal. Then let $G = H \rtimes K$

Kaj Hansen

I actually was wrong. I Googled it out of laziness @Ted :P

Adeek

now we have to see how is our semi direct product behave.

$\sigma_1 \tau_1 \sigma_1^{-1} = \tau_1^7$

Ted Shifrin

I count on you to be my algebra czar, @Kaj.

Adeek

So $G = <a ,b : a^8 = 1 = b^2, bab^{-1} = a^7>$ right ?

Ted Shifrin

Karim: "Since the index of $H$ is normal"?

Adeek

1:58 AM

Since the index of H is 2

so it is normal

Kaj Hansen

Definitely the first part @Adeek, let me think about bab^{-1} = a^7

Adeek

$\sigma_1 \tau_1 \sigma_1^{-1}(2^{1/8}) = \sigma_1 \tau_1 ( (2)^{1/8}) = (2)^{1/8}\sigma_1( \sqrt{2}/2 + i * \sqrt{2}/2) = (2)^{1/8}*(\sqrt{2}/2 - i*\sqrt{2}/2) = (2)^{1/8}\omega^7$.

$\sigma_1 \tau_1 \sigma_1^{-1}(i) = i$ so we get it is $\tau^7$ right ?

want to verify my calculation

Kaj Hansen

Let's see, $\sigma^{-1}$ fixes $2^{1/8}$

Then $\tau$ sends it to $\omega 2^{1/8}$

Then $\sigma$ sends it to $\omega^7 2^{1/8}$

Yeah

Adeek

okay cool

Ted Shifrin

Hmm, so it actually is dihedral?

Kaj Hansen

2:03 AM

This is why I thought it was $D_8$ @TedShifrin. It looks like $\tau$ is the rotation subgroup and $\sigma$ is acting as a reflection

But an MSE post says otherwise :S

Ted Shifrin

Don't make me work.

Adeek

but this isn't the presentation for $D_8$ @KajHansen

Kaj Hansen

Aren't presentations not unique though?

Adeek

yeah I guess

Ted Shifrin

Huh, slow down. I'm slow.

Kaj Hansen

2:05 AM

Also, that looks like the dihedral presentation :S

Ted Shifrin

Yeah, doesn't it?

Kaj Hansen

$bab^{-1} = a^{-1} \implies ba = a^{-1}b$

Adeek

I just saw that mse post

the guy who posted it made a mistake

Ted Shifrin

goes back to sleep

Adeek

the guy who posted the answer made a mistake

Kaj Hansen

2:05 AM

Which is right. Dihedrals have that sort of "anti-commutivity" where you pick up an inverse

Ayup.

also $a^{-1} = a^7$

mhm

so that makes sense

Another source is saying it's the semidirect of $\mathbb{Z}_8$ with $\mathbb{Z}_2$

Can't open it though

Adeek

2:07 AM

2

Q: Galois group for $x^8 - 2$

My textbook asked me to find the Galois group $G$ for $x^8 - 2$. Ok, so the roots of $x^8 - 2$ are $e^{2\pi ik/8} * 2^{1/8}:0 \le k < 8$, by my calculations, so the splitting field is $\Bbb{Q}(2^{1/8}, i)$ and the Galois group is generated by $f: 2^{1/8} \to e^{i\pi/4} * 2^{1/8}, i \to i$ and $g:...

Ted Shifrin

This is semidirect.

Adeek

$D_8$ is a semi direct product @KajHansen

the guy who posted solution made a mistake

it is not $f^3$ it is $f^7$.

Kaj Hansen

I've actually never been introduced to the semidirect in classes and such. I always have to look it up on wiki to remind myself. :/

Adeek

It is very useful.

Ted Shifrin

OK, @Kaj. mr. algebra, you need to learn that :P

Kaj Hansen

2:09 AM

On it tonight @Ted

Adeek

all we do in our class is semidirect product this semidirect product that

Ted Shifrin

LOL, hugs @Kaj

Kaj Hansen

hahaha

Ted Shifrin

First step to representation theory :)

Yikes.

Adeek

ok interesting $Aut(\mathbb{K} / Q(\sqrt{-2})$ is $\mathbb{Q}_8$

I want to prove this

Kaj Hansen

2:10 AM

Does this generalize easy to the Galois group of $x^n - a$?

I mean, complex conjugation is always going to be available

Adeek

where K is the splitting field that we began with.

Ted Shifrin

Now that you remind me, Karim, I think this is an exercise in my book.

Adeek

really ?

Kaj Hansen

And it seems like $\sqrt[n]{a} \mapsto \omega \sqrt[n]{a}$ will be an automorphism

Adeek

should I approach this directly ?

or should I use something else ?

I mean compute the order of $\mathbb{K} / Q(\sqrt{-2})$ etc

Ted Shifrin

2:12 AM

You could look at intermediate fields/subgroups and see they're all normal.

Adeek

I see

I have to probably look @ the lattice of subgroups for $D_8$ ?

Ted Shifrin

Only the part corresponding to the extension you're talking about, Karim.

Adeek

I think we can achieve $Q_8$ by some semi direct product let me remember which one.

Fargle

Hi chat.

Ted Shifrin

@Fargle!! You owe me a re-email.

Fargle

2:14 AM

@Ted, just know that wasn't the first time I forgot an attachment. I sent it to you.

Ted Shifrin

Oh.

Fargle

:D

Adeek

I see @TedShifrin so I have to see that extension and the subgroup generated by that extension and see what it is

Fargle

I feel like problem 1 didn't really bear inclusion, but, hey.

Kaj Hansen

What is $K$?

Ted Shifrin

2:15 AM

No, but maybe it's bared inclusion.

Adeek

$\mathbb{K} = \mathbb{Q}( (2)^{1/8},i)$

Kaj Hansen

Ah

Adeek

so $\mathbb{Q}(\sqrt{-2}) = \mathbb{Q}( i * \sqrt{2})$

Fargle

@TedShifrin This is a most appreciable pun.

Ted Shifrin

appreciable or appreciated? :D

Kaj Hansen

2:16 AM

Find a polynomial over $\mathbb{Q}$ with quaternions Galois group

Adeek

that is brute force way to do it I think I can do it constructively

Fargle

Both, but only because I like puns.

Ted Shifrin

@Fargle: Re #23, usually you would just choose balls of radius $1/k$. But "unwieldy"? You need to use triangle inequality to check details, but yes, you have it

Aksel'sRose

Hello all, doing some studying and not really sure how to prove bilinear for something such as $x_\alpha: (v,v')\mapsto x(\alpha(v), \alpha(v'))$.

Adeek

what are the normal subgroups of $D_8$ is it just the cyclic group ?

Fargle

2:20 AM

@TedShifrin That's an easy check by my construction, as the $|q_k - x| \leq \sum_{i=1}^k r_i < r$.

Adeek

there are many :S

Fargle

No reason not to write it in.

Ted Shifrin

Maybe I missed details, @Fargle. There weren't any.

Fargle

@TedShifrin Good to know there weren't hiccups. I don't know why the 1/k construction didn't occur to me--this one just felt right, diagrammatically.

I certainly did a lot of literal hand-waving and drawing of imaginary open balls in the air while proving it...

Ted Shifrin

BTW, @Fargle: I'll look more carefully tomorrow. But if you're gonna LaTeX, you need to know that you do `` and then ''.

Fargle

2:25 AM

@TedShifrin Wait, where?

Ted Shifrin

wherever.

Fargle

Oh, I see where you meant. Thanks, I never knew that one.

Ted Shifrin

I'm a regular fount of trivia.

Fargle

I felt proud of the brevity of my proof for the last problem I did. But I think that's just the topology course helping me out.

Ted Shifrin

Of course it is.

Let's get on to analysis :P

Fargle

2:27 AM

Oh boy, sequence time.

Kaj Hansen

(You can study sequences in greater generality with topology @Fargle) ;)

Ted Shifrin

He already done did that ... but not the estimates part of series :)

Fargle

@TedShifrin Somehow that ellipsis makes it more intimidating.

Ted Shifrin

Good :D

Heya DogAteMy

Akiva Weinberger

Hi

Ted Shifrin

2:30 AM

Heya @PVAL

DogAteMy: sorry if it upsets you that @Balarka is now copying your name :P

user228700

Hi, everyone :-)

Akiva Weinberger

Hi!

DHMO

@Kaumudi hi

user228700

I've a very quick homework-tsy question.

user228700

How am I to solve $e^{-\lambda t} = 0$ ?

Akiva Weinberger

2:32 AM

$e^{\rm anything}$ is never zero.

Ted Shifrin

That'll do it

Akiva Weinberger

Well, $e^{-\infty}$ is zero in the limit (where by that I really mean $\lim_{x\to-\infty}e^x$)

but that doesn't really count

(It's a bit interesting how the $e^x=\lim_{N\to\infty}(1+x/N)^N$ definition kind of implies that there's an infinite-degree zero at $-\infty$.)

DHMO

Can $\exp(x)$ have a root in the octonians?

user228700

Akiva Weinberger

I guess that means $f'$ is never zero?

Or you made a mistake somewhere

@DHMO No idea; doubt it.

(I don't think it does in the quaternions, at least.)

user228700

2:37 AM

Actually, no, $R$ may be equal to $\lambda N_\circ$ but that still doesn't give me the correct answer :-|

Akiva Weinberger

Idk

user228700

Alright, thank you :-)

DHMO

@AkivaWeinberger but the octonions have zero divisor

by the way, $\exp(x)$ does not make sense in a non-commutative non-associative algebra like octonion, so what am I asking?

user228700

You know, substituting $e^ {-\lambda t} = 0$ in the function directly does give me the correct answer!

user228700

(I don't got any terms of $t$ that aren't associated with $e$ as $e^{-\lambda t}$. So really, I don't need to solve for $t$ in the first place but e^{-\lambda t}$ does equal 0 :-/)

meow-mix

2:43 AM

@AkivaWeinberger i found a solution to the puzzle thing

Akiva Weinberger

I saw above

I didn't analyze the solution or anything

@DHMO Really? What about the power series?

meow-mix

all solutions have an odd sequence length

obviously

DHMO

@AkivaWeinberger Well, $z^3$ is not even consistent, since $(zz)z \ne z(zz)$

Akiva Weinberger

Oh right

But I thought that the octonions satisfied that anyway? @DHMO

DHMO

@AkivaWeinberger satisfied what?

Akiva Weinberger

2:48 AM

Even though they're non-associative in general, I think that they satisfy $z(zz)=(zz)z$

meow-mix

im confused about the dual projective plane

whats the point of defining a copy of $\mathbb{P}^2$ when you can just use $\mathbb{P}^2$ itself?

DHMO

@AkivaWeinberger no idea

Sophie

3:08 AM

I have a conjecture. If a number $n$ can be written as a sum of two squares and is not a multiple of 4, then the equation $x^2-ny^2=-1$ has integer solutions.

meow-mix

@Sophie just found a counterexample

$n=34=5^2+3^2$

Sophie

yes wrong statement

meow-mix

?

Sophie

I'm trying to prove if $n$ is a prime congruent to 1 mod 4 then $x^2-ny^2=-1$ has solutions, and then I generalised it wrong

mathoverflow is so weird. Often you'll find answers by Gowers or Terence Tao

Fargle

3:29 AM

@Ted: wait, do I really need the triangle inequality? I've constructed a (finite) descending chain of neighborhoods $N_r(x) \supset N_{r_2}(q_1) \supset N_{r_3}(q_2) \supset \cdots \supset N_{r_{k-2}}(q_{k-1}) \ni q_k$.

@AkivaWeinberger They in fact satisfy $(xx)y = x(xy)$ and $(yx)x = y(xx)$. The sedenions do not, but they do satisfy the property you stated.

Secret

4:13 AM

Hey guys have a topology conceptual question here:
Suppose I have a 3 element set $\{a,b,c\}$ wit the following topologies:
$$\tau_1=\{\emptyset,\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},\{a,b,c\}\}$$
$$\tau_2=\{\emptyset,\{b\},\{a,b\},\{b,c\},\{a,b,c\}\}$$
$$\tau_3=\{\emptyset,\{a,b,c\}\}$$

Is the following interpretation correct (omitting the description about the emptyset):
For $\tau_1$ all elements form an open set, and no two elements are considered near to each other in this topology
For $\tau_2$, b form an open set, while a,c and a,b form two other open sets. Therefore a and c are c…

Martin Sleziak

4:30 AM

@Secret I am not sure how you would formally define "close to each other". But there exists notions of topologically distinguishable points and separated points. See Wikipedia article on separation axioms.

I have also reposted your message in general topology chat room. (Although it is not sure that it improve chances of getting a response too much. The room seems to be rather inactive.)

Fargle

@MartinSleziak I think Secret is taking the convention that "close" = "topologically indistinguishable".

@Secret For $\tau_2$, as a minor nitpick, it'd be a and c as "close", but no other pair.

Martin Sleziak

@Fargle Well, a and c are topologically distinguishable in $\tau_2$, but Secret calls them close.

Fargle

@MartinSleziak Separated then? I'm not strong on these definitions myself.

arctic tern

@DHMO no. purely imaginary unit octonions are square roots of -1, so de Moivre's formula works for them. that is, any octonion x is of the form r+su where r and s are real and u is a pure imaginary unit octonion, in which case exp(x)=exp(r)(cos(s)+sin(s)u) just like with complex numbers.

Martin Sleziak

Not being strong on definitions is the reason why I added a link. (I had to look there myself.)

arctic tern

4:37 AM

@DHMO octonions do not have zero divisors

Fargle

I mean, yeah. >_>

DHMO

@arctictern except that not every octonion x is of the form r+su where r and s are real and u is a pure imaginary unit octonion

@arctictern they do have zero divisors

arctic tern

@DHMO no, octonions have no zero divisors. and yes every octonion is of that form.

Fargle

@arctictern $1 + e_1 + e_2$? Or are you calling $\frac{e_1 + e_2}{\sqrt{2}}$ a unit octonion for these purposes?

arctic tern

@Fargle yes, $(e_1+e_2)/\sqrt{2}$ is a unit octonion

not just "for these purposes" though. it clearly satisfies the definition of being a unit octonion.

Secret

4:38 AM

@MartinSleziak O sorry I made a typo, I mean a,b are close together and b, c are close together but not a, c

Fargle

@arctictern I just wasn't sure what you meant by "unit"--whether you meant literally some $e_i$, or just an octonion $x$ with $|x| = 1$

Martin Sleziak

"Two points x and y are separated if each of them has a neighbourhood that is not a neighbourhood of the other; that is, neither belongs to the other's closure." In $\tau_2$ closure of b is the whole set. So no point is separated from b.

a and c are separated

So Fargle might be correct in guessing that what you mean by close is the same as not separated.

arctic tern

now, sedenions have zero divisors, but even they are power-associative so powers make sense, and the thing I said about octonions having the form r+su and u being a square root of -1 still applies to sedenions IIRC so once again exp has no roots

DHMO

@arctictern you're right

@Fargle I think he means $x^2 = -1$, right

Fargle

@DHMO He's not requiring that. Just that it have no real part and norm 1.

arctic tern

4:44 AM

by "unit octonion" I mean $|x|=1$, and by "is a square root of $-1$" I mean $x^2=-1$

an octonion is a square root of -1 if and only if it is purely imaginary and has unit norm

Fargle

Don't drink and MSE, kids

bows out

Martin Sleziak

Since I am trying to get the bounty room going, perhaps a reasonable thing could be to promote it here. So here is link to the relevant meta post and here is the room.

Secret

5:05 AM

if S={a,b} is an open set, does it mean the set actually does not contain the elements a and b, i.e. $a,b \not\in S$? , analogous to how for reals in the usual topology (0,1) does not actually contain 0 and 1?

arctic tern

@Secret it is not possible for a set to not contain its elements

dunno why you would think that

best guess for why you would think that is that no isolated point a in a subspace can be an interior point

in Euclidean space specifically

but in topology, "open" doesn't have to mean anything other than "one of the sets we have in this collection of subsets that we decided to call open" (where the collection satisfies some axioms)

Martin Sleziak

@Secret The notation $S=\{0,1\}$ and $S'=(0,1)$$ is different. They are two completely different sets.

Secret

Well, the interval (0,1) in reals in the usual topology has elements x between 0 and 1, but not 0,1. Knowing that topology is a generalisation of the concept, I initially thought {a,b} being open will mean something similar

(slow typing, pretty much what I said above is what you have guessed)
I see

Martin Sleziak

However, if you are looking for somtheing similar to set $S=(0,1)$ and point $p=0$, you can define in any topological space the notion of boundary of a set, defined as $\partial S=\overline S\setminus S$. In this case $0\notin S$ but $0\in\partial S$.

Wikipedia: Boundary (topology)

Secret

ok

arctic tern

5:14 AM

@Secret what does "between" mean in an arbitrary topological space? intervals generalize to partially ordered sets, but not really topological spaces

Martin Sleziak

BTW it is your first encounter with general topology Secret? What subject are you currently studying - analysis, multivariable calculus, real analysis, metric spaces, functional analysis, ... ?

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$Mathematics$ Mathematics