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03:00
(123)(132)(124)(e)(142)
mm thats not right
i have no idea what it looks like =\
I told you to list the 12 elements of $A_4$
(234)(243)
thats all i can think of
there are 12 elements
[or else you can go list out all 24 elements of $S_4$ first]
[and this isn't even related to Galois theory]
can you just tell me what anthor one looks like?
you should separate distinct permutations with commas, Faust
03:03
everything i can think of isnt even
then think of more
@Faust what is a permutation?
I like your golden hair @anon lol
hey jasps
03:04
@anon do you do Galois theory?
I know some basics
hmm
@Faust we can come back to this later and continue on my explanation if you want
@LeakyNun we havent defined those in terms of permutations
your choice
@Faust you haven't defined what?
@Faust
sorry dinner is here
03:08
:c I haven't explained to you the use of Galois theory
and its fucked up
ok go ahead
do you know quotient groups?
yeah
ok, so in the subgroup chain $S_4 > A_4 > V_4 > \Bbb Z_2 > \{e\}$, every subgroup is normal in the previous
03:09
and the quotient groups (i.e. $S_4/A_4 = \Bbb Z_2$, $A_4/V_4 = \Bbb Z_3$, ...) are all abelian
ok
a group for which such a subgroup chain exists is called a solvable group
so $S_4$ is solvable
let's say we have a quintic polynomial
it has 5 roots, a,b,c,d,e
we consider the field extension $\Bbb Q(a,b,c,d,e)$
and its Galois group $\Gal(\Bbb Q(a,b,c,d,e):\Bbb Q)$
If the Galois group is solvable, then $a,b,c,d,e$ can be expressed using radicals
actually, $a,b,c,d,e$ can be expressed using radicals if and only if the Galois group is solvable
whatcha mean by radicals\\
03:11
@Faust it means you can express them like $\sqrt{2-\dfrac12\sqrt[3]4}$
it was a big problem to find solutions to arbitrary quintic polynomials
Galois proved its impossibility
that there are polynomials whose roots cannot be expressed with radicals
intresting
$A_5$ has no non-trivial normal subgroups, so it isn't solvable.
so if we can prove that the Galois group is $A_5$, then the polynomial isn't solvable by radicals
and this was done when Galois was 19 years old.
and he didn't have any groups (he is one of the co-founders of modern abstract algebra)
go have you dinner :P
why are you awake
EATINGN IT WHILE LISTENING
ups caps
@0celo7 I had a nap and I'm reading Galois
@anon let's do Galois together lol
I mean, I had a nap in the afternoon so I'm relatively awake
03:13
k
@anon what should we talk about
@LeakyNun ehhh. co-founder is a bit of a weird word for someone who died that young.
was your idea to talk
@Semiclassical founder?
no, i mean
03:14
@anon how much do you know?
if he had lived longer, he'd have been in a position to shape the discipline etc. etc.
@Semiclassical I see
well afaik three guys independently came up with groups
Modern abstract algebra?
maybe forerunner?
@Semiclassical I see
03:15
@LeakyNun ¯_(ツ)_/¯
What abstract algebra was there before
@0celo7 eh, lol
200 year old math is hardly modern
I think I'd parse it as: Galois was a forerunner of modern algebra
@anon $\newcommand{Gal}{\operatorname{Gal}}\Gal(\sqrt[4]2,i)$?
@Semiclassical I see
03:16
but, eh. words
you mean ${\rm Gal}(\Bbb Q(\sqrt[4]{2},i)/\Bbb Q)$?
Hallo
funny how the famous algebraists die young/unexectedly
@anon yes, pebkac
Abel and Galois, we hardly knew ye
03:17
Galois, Abel, Fermat
@0celo7 solution: be an algebraist but don't be famous
"Along the way, [Galois] introduced some amazingly original and powerful concepts, which form the framework of much algebraic thinking to this day. Although Galois did not work explicitly in axiomatic algebra (which was unknown in his day), the abstract notion of algebraic structure is clearly prefigured in his work." (Pinter)
@Brody go study your quasigroups :P
@Daminark or just be an applied physicist and be...useful
aka engineer, although that's probably a taboo word here
I mean I'd consider being more applications of number theory to compsci but I've had an attempted run with physics and it seriously didn't blow over well
03:18
or go into finance
apparently the Brits call it soluble
and get real world experience in faustian bargains :P
@Semiclassical finance people are hardly useful :P
me be like: what, salts?
They're useful for making money for certain people
03:19
@0celo7 I mean if we're talking about being useful as an end in itself versus usefulness with the end of having a good career, I don't think I'm particularly interested in the former
@LeakyNun ok. call it G. there's a map from G to S_2, permutations of {i,-i}. It's surjective because of conjugation. Now we must find the kernel K. Then K permutes the four roots of 2, the only question is how.
Fermat died age 58, so I don't think you can count him as dying young
though compared to Euler, most mathematicians died young :P
high school algebra people be like what, four roots of 2? [roll eyes]
@Semiclassical I said young or unexpectedly
he was about to write his proof and then kicked the bucket
@anon can you know what one root maps to if you know what another root maps to?
03:22
Like, in principle I'd rather spend time doing what I want to do. If I am gonna sacrifice that, it won't be for abstract usefulness so much as being broke. So finance would be an open option, though admittedly I'd rather try to see if something like data science or being an actuary is an option ahead of like, Wall Street style stuff. The latter just seems to be a bit... dodgy
@Semiclassical whom is that directed at
Fermat wrote that remark in 1637 and died in 1665
oh, wtf
@LeakyNun Say $\alpha$ is an automorphism that fixes $i$ and sends $\sqrt[4]{2}$ to (wlog) $i\sqrt[4]{2}$. If it exists, it is fully determined, has order four, so must generate the kernel $K$.
03:23
so he was a giant troll
@Semiclassical which remark?
"Proof is too long to fit in the margins"
@anon good
Or he thought he had a general approach and was wrong
Poll: do you think that Fermat was trolling?
03:24
Nah
@Semiclassical otoh, Yamabe wrote his paper than kicked the bucket. The error wasn't found for 8 years
He probably had what he thought was a solution
he died thinking he had solved a great problem :,(
or he did and took it to his death
a relevant point: "In 1847, Gabriel Lamé outlined a proof of Fermat's Last Theorem based on factoring the equation $x^p + y^p = z^p$ in complex numbers, specifically the cyclotomic field based on the roots of the number 1. His proof failed, however, because it assumed incorrectly that such complex numbers can be
factored uniquely into primes, similar to integers. This gap was pointed out immediately by Joseph Liouville, who later read a paper that demonstrated this failure of unique factorisation, written by Ernst Kummer.
03:27
@Semiclassical what was the failure of unique factorization?
@anon so what is the group?
at least Galois had the decency to write down his work the night before the duel
@Brody how did he keep all his work in his head :o
he tried several times to share and publish his results but people either lost them or were uninterested
03:29
@LeakyNun see how $\alpha$ interacts with conjugation $\beta$
@anon what is $\alpha$?
already said
1/137
this issue is specifically stated on the top of page 8 in that link
@Brody sad
@anon ok, then how?
03:31
namely that the ring $Z[\zeta_p]$ (p-cyclotomic integers), where $p$ is an odd prime, isn't necessarily a UFD
compute e.g. $\beta\circ\alpha\circ\beta^{-1}$
since $K=\langle\alpha\rangle$ is normal (it's a kernel)
this is of course assuming $\alpha$ even makes sense as defined
(it should)
@Semiclassical thanks
an example is on P.9: in the ring $\Bbb Z[-\sqrt5]$, $6 = 2 \times 3 = (1+\delta)(1-\delta)$
@anon there's an easier way... you know the order of the group right
the extension has degree 8
the following remark shows up on Wikipedia: "For a square-free positive integer n, the quadratic integer ring Z[sqrt(-n)] will fail to be a UFD unless n is a Heegner number." (these are 1, 2, 3, 7, 11, 19, 43, 67, 163. )
03:33
@anon and you know the degree of the polynomial
Chickens murdered
wings feasted upon
@LeakyNun wait, are you asking for my help, or just felt like chatting about something you already know?
@Semiclassical ever heard of the "Noether transform"?
@anon sorry :P
03:34
me neither
and I have no idea what these notes are about
I mean, Noether's theorem on conserved quantities, sure
@anon that's why I asked you what you know
and noether charge in that context
@Semiclassical this is something more sinister
anyway, yes, if you know the groups of order 8, and this one has a normal cyclic subgroup of order 4, you know what it is
03:35
in The h Bar, 5 hours ago, by 0ßelö7
@ACuriousMind it's this awful thing $$m(\xi,X)(\dot v_1,\dot v_2)=(\xi\cdot X)h(\dot v_1,\dot v_2)-h(\xi\otimes \dot v_1\cdot X,\dot v_2)-h(\dot v_1,\xi\otimes \dot v_2\cdot X)$$
wb @Faust
@anon hmm, my pathway is a little bit different: the degree of the polynomial is $4$, so the group must be a subgroup of $S_4$, which means it is $D_8$
@LeakyNun know a good way to show this? Prove that φ : Sn → GLn(R). is a homomorphism
@0celo7 madre dios
@Semiclassical it's not much better in index notation
03:36
@Faust what is the definition of $\varphi$?
oh
one sec
I can't really figure out what Christodoulou is trying to do with it
@anon so why don't you test me lol
googling it doesn't reveal much
one of my favorite things I did in Galois theory was this with chebotarev density
03:37
so i guess that's the no-idea-either transform
@Semiclassical this looks like Spanish
@Semiclassical I think the point is that it makes the definition of hyperbolicity for a Lagrangian system easier
but you're missing a de in the middle
Fix a positive integer n. For each σ in Sn, define an n×n matrix, φ(σ)
by
$φ(σ)_{i,j} =\{1$ if $i = σ(j), 0 $i \neq σ(j) \}$
03:38
33 mins ago, by anon
I know some basics
@anon I'm never trusting anyone who claims to know "some basics" again :P
@0celo7 oook
well, it's true, I do know some basics
@anon more like, a lot
fuck im out of it
@Faust no, I don't know a good way to solve that
03:39
cause its shit!
@anon asking for help: could you run me through the proof that a polynomial is solvable iff its Galois group is solvable?
@Semiclassical What for...I don't really know. I might have to get his book on Lagrangian PDE
@Semiclassical I'm reading some lecture notes because in the middle there's a proof I really need to learn
out of perverse curiousity
03:40
the first part isn't great
what does it look like in index notation?
@LeakyNun ok, can't say I remember what the proof is
@Semiclassical puts equation on second monitor
let's see...
lol
it reminds me of something, but I can't remember what
@anon lol never mind then
03:42
well, don't remember why cyclic galois group implies radical extension
I guess the proof itself isn't that important
I was tortured for an hour by this fun exercise: prove that if $\Gal(K:\Bbb Q)=\Bbb Z_4$ then $i \notin K$
where "$i \notin K$" is defined as "$\forall k \in K: k^2+1 \ne 0$"
@anon do you know the answer?
I was going to say $\Bbb Q(i,\sqrt{2})$ until you edited
@anon sorry :P
complex conjugation would have order 2, which means it'd have a sqrt in the galois group, and where would that element send the imaginary unit i?
@anon oh I am not asking for help this time :P
03:46
I know, that was a sketch of the proof idea.
but that's an approach I haven't heard of
$$m(\xi,X)^{\mu\nu}_{ab}\dot v_{1\mu}^a\dot v_{2\nu}^b=\xi_\rho X^\rho h^{\mu\nu}_{ab}\dot v_{1\mu}^a\dot v_{2\nu}^b-h^{\mu\nu}_{ab} \xi_\mu \dot v_{1\rho}^a X^\rho \dot v_{2\nu}^b-h^{\mu\nu}_{ab}\dot v_{1\mu}^a \xi_\nu \dot v_{2\rho}^bX^\rho$$
calm down, satan physicist
@Semiclassical
@0celo7 oh fur die Gottesliebe
03:47
@0celo7 if I reorganize it to $$(\xi\cdot X)h(\dot v_1,\dot v_2)=m(\xi,X)(\dot v_1,\dot v_2)+h(\xi\otimes \dot v_1\cdot X,\dot v_2)+h(\dot v_1,\xi\otimes \dot v_2\cdot X)$$ it sorta reminds me of the Leibniz rule
@anon it would send it to $a$, and then?
what's $a$?
(are you thinking of a proof or is that the whole proof already?)
@anon I prefer the term "analyst"
well, that's a ...thing
03:47
@anon I don't care
$a$ is a number that gets mapped to $-i$
@Semiclassical the fact that I wrote that by inspection and without TeX errors probably means I've done too much GR in my life
probably
@0celo7 what
Let's call $\sigma$ a generator of the Galois group. Since $\sigma^2$ is the unique element of order $2$, and complex conjugation has order $2$, we find that $\sigma^2$ must be complex conjugation. So $\sigma^2(i)=-i$. There are two possibilities for $\sigma(i)$, it's either $i$ or $-i$, but both cases lead to $\sigma^2(i)=i$, a contradiction.
@LeakyNun what what
03:49
@anon that's a nice proof
@Semiclassical huh
do you want part 2, which I adjoined by myself to the field?
that was a bad pun
show me why your argument fails if I replace $i$ by $\sqrt5$
@Semiclassical if you want more of this crap, see amazon.com/…
chapter 2 is a debauchery of indices and tensor crap
03:52
"noether transform" does show up on google a few times, but not very helpfully
the guy might have made it up himself
he's famous enough to do that
it shows up here‌​, for instance
that's exactly what is here!
...so it is
@Semiclassical Christodoulou wrote that article
03:54
yeah, i don't know how i missed that
lol
@LeakyNun not as clear to me right now
out of all the google hits, one of the few that isn't DC's article is this thesis: stogin.com/images/thesis.pdf
@anon hmm
do you want to work with an actual example?
" The technique in this paper is
an extension of the Christodoulou–Klainerman vectorfield method"
03:56
ok
...
why can't i read tonight
@anon $\Bbb Q \subset \Bbb Q(\sqrt5) \subset \Bbb Q(\zeta_5)$
@Semiclassical all roads lead to rome
all roads lead to the guy who dreamed up that bloody name
where $\Gal(\Bbb Q(\zeta_5):\Bbb Q) = \Bbb Z_5^* = \Bbb Z_4$
epic subtitle
15.2 An Insoluble Quintic
Watch carefully, there is nothing up my sleeve...
03:58
@Semiclassical Oh, this is actually interesting stuff. He's defining an abstract causality theory for Lagrangian systems
LEMMA 15.10 [...]
the subtitle is too brilliant lol
the thesis?
No, the lecture notes
it's fairly light on details, however
not that I really want details right now, but it would be nice to know that they're there...
04:05
@anon another problem for you if you want: prove that the Galois group of $f(x) = ax^4+bx^3+cx^2+dx+e$ is a subgroup of $\Bbb D_8$.
bye everyone
04:27
[Random] As I see more and more textbooks and attempted explanations of math, the more I realize just how good the MIT open courseware materials I studied from really are:
This material breaks the "Wikipedia barrier" between an interested high school student or high school graduate and someone who can actually read math and understand what is meant.
@Semiclassical Interesting Remark: Nother's theorem assumes the Lagrangian is first order in derivatives, but the GR Lagrangian isn't. Moreover, the GR Lagrangian is essentially unique so this is a fundamental issue.
@Semiclassical If you try to kill off the high order derivatives you have to break coordinate freedom and you end up with pseudotensors.
ew
Is there an obvious physical consequence to that?
@Semiclassical Yeah, it's the attempt to isolate the "gravitational energy" terms.
Einstein did it more or less by gut feeling, but Chris is explaining it a bit better
04:40
is this to do with how energy conservation works in GR?
Yeah
my understanding was that it basically doesn't work due to the lack of symmetry
@Semiclassical Even if you have symmetry it's nontrivial.
You still have to say what the energy is.
Nother's theorem doesn't give anything because the Lagrangian has second derivatives.
So the goal is to roughly throw out the second derivatives and look at the remaining Lagrangian
04:43
ah
@Semiclassical One has to put on the physics glasses because these things aren't tensors, but are supposed to be defined everywhere
for all I know there might be some theory for these objects...
"The Lagrangian is invariant under translations" ...on a manifold. @Semiclassical is there a recommended liquor for such things?
Alcoholic root beer. (I don't know why I'm picking that.)
@Semiclassical Alright. I guess the motivation should take place on a curved $\Bbb R^4$, and then in the end you pull a vast generalization and define the stuff like that in the general setting
And the integrals of these things turn out to be well-defined
it's a mess
04:53
friends don't let friends touch pseudotensors?
friends advise friends to do something other than theoretical physics or math
too late for that :/
Theoretical compsci
 
1 hour later…
06:08
[Random]
Max
Max
my book gives a description for an algo that finds the MST of a graph: "Grow a tree one edge at a time by adding the minimum weight edge possible to the tree, making sure that you have a tree at each step." to my understanding, this is known as Prim's algorithm?
06:33
Hai
Hi @Daminark how's the group theory going?
It's going alright, I'm still having a bit of trouble with the problem I mentioned earlier
It's actually a bit tough
What was it? Writing an n-cycle as a product of 2-cycles?
But honestly the last pset was a bit lame so I'm happy
So the question asks which elements in S_n can be written as a product of 2 elements of order 2
The idea is that it should be all of S_n, since you should be able to do it for a cycle, and then you use disjoint cycles and push things around
But I dunno how to execute it on a cycle yet
06:40
I got $(1234) = (12)(34)(24)$ and $(123456) = (12)(34)(56)(24)(64)$
But I dunno how to pull it off for 5
And I've yet to see a general pattern
Oh I think I know what to do for 5
@Daminark Your $4$-cycle does not seem to be a product of two elements of order $2$
sorry, your $6$-cycle I mean
Oh shit, yeah I have too many 4s
Actually even my 5 cycle case has too many 2s
(it would have been easier to spot if you had written the last one as $(46)$ instead)
06:46
Yeah
::thonk::
Crap I actually have no idea how to make this work
Or maybe I've got the 5 cycle
$(12345) = (12)(34)(54)(25)$ might work?
Oh wait no I don't think I can shift things around the way I want
Goddammit

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