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02:00
sorry, what do you mean by there are none?
there is no solution to the DE which is everywhere nonzero if that's what you are asking
best one can do is worry about $x>0$ and $x<0$ separately
Well I wasn't thinking everywhere, that's why I said $C^{\infty}(whatever)$
Anyway I'm done with this differential equation
@Daminark ah ok
my bad
yeah, that seems pretty done to death
02:02
I'm done with differential equations in general tbh
the ODE which always makes my head hurt is Riccati: $y''=a(x)+b(x)y+c(x)y^2$
nonlinear odes either have a trick or they're just kinda miserable
I propose we talk about... I dunno algebra or smth?
Oh actually
50 mins ago, by Leaky Nun
@Daminark $[a^2,b^2] = (a[a,b]a')(ab[a,b]b'a')(aba'[a,b]ab'a')([a,b])$
shorten this :P
it's important, alas, because it can be shown equivalent to a linear second order ode $u''+A(x)u'+B(x)u=0$
hah, that problem is probably up my alley
I've been meaning to find out how to prove the structure theorem for finite(-ly generated) abelian groups
02:05
@Daminark just find it online :P
though tbh when I think commutators I usually have in mind $[A,B]=AB-BA$ not $[A,B]=ABA^{-1}B^{-1}$
I think you use Jordan canonical decomposition over Z-modules
@Semiclassical hang him
Oh yeah that's a thing I never got around to
02:06
there is absolutely nothing wrong with the way SemiC is thinking about commutators
that is precisely how you write it for vector fields
using smith normal form is the way i last saw the proof of the structure theorem for whatever modules over whatever
@BalarkaSen /s
Its nothing but (AB)(BA)^-1
eh, give me a moment. $[\frac{d}{dx},x]=\frac{d}{dx}x-x\frac{d}{dx}=1$
@Eric yeah that's the same thing as Jordan form right
02:08
Ye $(x\cdot f)'-x\cdot f'=f$
which in quantum mechanics leads to $[\hat{x},\hat{p}]=i\hbar$ since the canonical momentum operator in the coordinate representation is $\hat{p}=-i\hbar \frac{d}{dx}$
Oh this stuff about commutators reminds me
@Balarka they're not quite the same
So in the REU last year Laci had us prove that Semi's commutator (he referred to it as such :P) cannot equal the identity after rambling about quantum mechanics for a bit
@Daminark Where things get weird is the anticommutator $\{A,B\}=AB+BA$
then you're dealing with fermions
02:09
@Semiclassical yeah
@EricSilva Oh hm
Can someone explain to me the usefulness of defining infinite integral domains? I know all finite integral domains are just fields...
you need a bit more structure to do the Jordan form
which is why SNF is kind of preferable sometimes
And he was like "And thus we disprove quantum mechanics... Just kidding, they do operators on Hilbert spaces
@gian the integers is one of them
@Daminark Here's a question on similar lines when people talk about the Dirac equation in quantum electrodynamics
02:11
Right, but why study the integers under the context of an integral domain?
@gian to generalize the concept of integers
I mean so the integers have a lot of structure and you're trying to isolate how much of it falls out of less
Oh I see, because the integers do not form a field.
it's all fun and games until you start calling $\Bbb R$ an integral domain
@gian right
the set of holomorphic functions on some domain is an integral domain
02:12
Gotcha, thanks.
Dirac wanted to have a set of four matrices which all squared to the identity matrix and which all mutually anticommute.
So properties that hold over all integral domains don't use, say, unique factorization even if perhaps it can be proven though it.
@Semiclassical quaternions?
@Semiclassical finite dimensions?
@Daminark ...
hi chat
02:14
I mean matrices in infinite dimensions are sketchy
@LeakyNun Well, he wanted to use hermitian matrices.
@Daminark and yeah, he wanted finite dimensional matrices
The three quaternion units don't square to the identity matrix
Given that $a,b,c,d,e,f,g$ are elements of a group, and that $ab=c,bc=d,cd=e,de=f,ef=g,fg=a,ga=b$. Prove that $a^{11}=1$.
where $1$ is the identity.
My understanding is that one can do it with quaternions, but usually we ask for it to be done with complex hermitian n-by-n matrices.
The question is: How small can $n$ be if you want to have four such matrices, each of which square to the relevant identity matrix and which mutually anticommute?
In Gallian's Algebra book, he considers powers of a single element $a \neq 1$ that is an element of a finite integral domain to prove that it must be a field. Is this just invoking Cayley's Theorem?
02:18
@gian how is this related to Cayley's Theorem?
Not really, the point is that a^n are all in the group, but there are only finitely many of them
pigeonhole principle :P
Oh never mind, I guess you don't need Cayley's Theorem as long as you are referring to powers of an arbitrary $a$.
@gian note: that can only get you to skew fields. it's more difficult to prove that it is a field.
Integral domains are defined as commutative, no?
02:21
What do you mean? Don't we just need to show that an arbitrary element is a unit?
oops, :P
(Skew fields = division rings = noncommutative fields)
(Like the quaternions)
To say a bit more about Dirac's problem: If each matrix $\gamma_\mu$ has $\gamma_\mu^2=I_n$, then the only possible eigenvalues of $\gamma_\mu$ are $\pm 1$ (act on a possible eigenvector to see why).
Actually wait wait there is a Cayley-esque proof of this fact
It doesn't involve taking powers
@Daminark of course there is, but it's overkill
02:23
It turns out that finite skew fields are fields (in other words, finite implies commutative), though I have no idea how to show this
I don't think so
@Daminark why Cayley when you can pigeonhole
I mean it's not invoking Cayley, but it has a similar feel
$R = \{a_i\}_{i\in [m]}$
> 24.10 Theorem (Wedderburn's Theorem) Every finite division ring is a field.
Proof See Artin, Nesbitt, and Thrall [24] for proof of Wedderburn's theorem.
@Daminark, my confusion about Cayley's arose out of interpreting the proof incorrectly. At first I thought Gallian just took a single element $a$ and all of its powers to represent all elements of the finite integral domain by mapping each original element to a power of $a$. I don't know what I was thinking lol...
02:25
Choose $a_j\ne 0$, then $a_jb = a_jc \implies b=c$, $a_j(b-c) = 0$
So $a_jR = R$
Oh, I see @gian
so the abelian requirement is extraneous in a ring with unity?
a+a+b+b (1+1)a+(1+1)b = (1+1)(a+b) = 1(a+b)+1(a+b) = a+b+a+b
You could take an $a$ and its powers. Then show that there must be $a^i = a^j$ such that $i \neq j$. From there, the inverse of $a$ is $a^{i-j-1}$.
@LeakyNun, what do you mean the abelian requirement? Multiplicative commutativity?
@gian additive commutativity
Oh wow, nice result :P
@Daminark blah blah blah, one finds that the matrices have to be at least 4-by-4
02:32
What is $\langle a,b | a^2,b^2 \rangle$ isomorphic to? @Semiclassical @Daminark
this is coming in the context of trying to describe electrons in a relativistically proper way, btw
gut feeling tells me that it's something like $\Bbb Z$
I see
What is that angle notation @LeakyNun?
group presentation
02:33
In mathematics, one method of defining a group is by a presentation. One specifies a set S of generators so that every element of the group can be written as a product of powers of some of these generators, and a set R of relations among those generators. We then say G has presentation ⟨ S ∣ R ⟩ . {\displaystyle \langle S\mid R\rangle .} Informally, G has the above presentation if it is the "freest group" generated by S subject only to the relations R. Formally, the group G is said to have the above presentation...
sniped
The relations are $a^2$ and $b^2$?
@Daminark yes
That they're what?
I think the elements would just be powers of $ab$
so yeah, isomorphic to Z
@gian informally: a free group of $a$ and $b$ such that $a^2=b^2=1$
no, it's isomorphic to $V_4$.
02:34
Oh I see.
hmm
oh, yeah
I mean, it has to contain at least a,b by themselves
$abba = 1$
and (ba)(ab)=b(1)b =b^2=1
Lol sniped
so it's isomorphic to $\Bbb Z$?
02:37
still not sure now
BALEETED
DATA ESPUNG
But yeah we have $ba = (ab)^{-1}$
If it's going to be Z in the way I said above, I'd need to have $a=(ab)^n=ab(ab)^{n-1}$ for some $n$
and that seems sketchy
RIP I start school tomorrow
Hi @AlexClark
02:40
let's ask it on main :P
Not yet, let's try to reason through it a bit more first
So are we assuming this presentation isn't redundant?
I think every element has to reduce to some string like abab...ab, but possibly starting with b or ending with a
@Daminark what does redundant mean?
if you had any repetitions, you'd delete them
let $c=ab$
02:41
yeah
then $a$ and $c$ generate the group
a?c*a?
regex :P
user147690
Hey @BalarkaSen, have you read much Hartshorne II.5?
b=ac
@Daminark I don't think it's Z anymore, no
02:42
@Daminark right, it isn't cyclic so it isn't $\Bbb Z$
@AlexClark Nope, I know 0 Hartshorne beyond ch I
(Given that $\mathbb{Z}$ is only a group under multiplication, but you don't have $2x = 1$ for $x\in \mathbb{Z}$
Are you all in school yet
user147690
@BalarkaSen Have you thought much about vector bundle and locally free sheaf correspondence?
@AlexClark A little, I guess?
02:43
It should be the Klein group
user147690
I'm trying to do an exercise from Hartshorne haha
$(1,0) + (1,0) = (0,0)$ and $(0,1) + (0,1) = (0,0)$
user147690
I wrote something about it in the geometry & topology chat, but not sure if it's interesting to you atm
@Daminark but it's infinite
we're trying to identify <a, b|a^2 = 1, b^2 = 1>?
That's Z/2 * Z/2
infinite dihedral group iirc
02:44
how is it finite
which product is that?
oh, free product
free product
Like the dihedral group satisfies that presentation
right
and then it makes sense, yeah
02:45
welp, that's obvious.
@AlexClark Let me look at it
In mathematics, the infinite dihedral group Dih∞ is an infinite group with properties analogous to those of the finite dihedral groups. In two dimensional geometry, the infinite dihedral group represents the frieze group symmetry, p1m1, seen as an infinite set of parallel reflections along an axis. == Definition == Every dihedral group is generated by a rotation r and a reflection; if the rotation is a rational multiple of a full rotation, then there is some integer n such that rn is the identity, and we have a finite dihedral group of order 2n. If the rotation is not a rational multiple of a full...
wikipedia confirms :P
unfortunately my mental toolkit of infinite groups is pretty small
What about finite simple groups
Monster, Tits
true, they're pretty weird
02:46
uh
pls dont
I don't actually think I know any more
This is most of my group theory knowledge
Mathieu group(s)
Other than, like, $\Bbb Z_n$, $D_n$, $Q_n$, small stuff
02:48
In the area of modern algebra known as group theory, the Mathieu groups are the five sporadic simple groups M11, M12, M22, M23 and M24 introduced by Mathieu (1861, 1873). They are multiply transitive permutation groups on 11, 12, 22, 23 or 24 objects. They were the first sporadic groups to be discovered. Sometimes the notation M9, M10, M20 and M21 is used for related groups (which act on sets of 9, 10, 20, and 21 points, respectively), namely the stabilizers of points in the larger groups. While these are not sporadic simple groups, they are subgroups of the larger groups and can be used to construct...
@AlexClark actually I don't really know about locally free sheaves well enough to help you with this
Sorry
I don't actually remember how $Q_n$ works, never mind
The multiplicative group of $\pm1,\pm i,\pm j,\pm k$ is a thing though
I know every group of order 15
@AkivaWeinberger it's GAP(8,4), the quaternion group
just Z_15 right
02:49
Yeah
Baez has some weird stuff on the Mathieu group $M_{12}$ here
What is the general approach to proving that a certain morphism cannot exist between objects?
Every group of order 9 is commutative
Z_9 and Z_3xZ_3
9 hours ago, by Leaky Nun
:o TIL quarternian group (the group above) isn't isomorphic to other groups I know
@AkivaWeinberger GAP is a programming language about groups
02:50
and the quaternion groups have a label
Proof is here
@Balarka
user147690
@BalarkaSen All good, I'll keep banging my head against it periodically :P
8 is the order, 4 is the id
@Alex wai
Hartshorne = suicide
user147690
02:51
@BalarkaSen It feels that way
@AlexClark By Bott Periodicity...
user147690
@Daminark Oh my
user147690
Apparently the exercises are lemmas, propositions and theorems from the EGA
yeah you need 8-periodic banging
> I've been looking for a good excuse for that pun for years. I could stop here and be completely satisfied!
02:51
lolyes
Is the 24 rotations of the cube embedded in $S_6$?
@Balarka Or 2, since this is some complex material
should be, since each rotation permutes the 6 faces
I think the group of isometries of the cube are in fact S4
@Daminark true
$|S_4| = 24$ so...
02:53
I think the isomorphism is given by action on the diagonals
@LeakyNun It's isomorphic to $S_4$
Yes^
it should act freely and transitively on the four diagonals
what the hell
SNIPED
GET SNIPED
REKT4LIFE
is schnapped
02:54
why am i doing this with mah life
3
wow, that's more amazing than I thought
That was a pretty long snipe too
I was just... um like just in case Leaky doesn't trust Balarka
I have verifications
1 min ago, by Balarka Sen
I think the isomorphism is given by action on the diagonals
simple proof :P
Like point-blank sniping
02:55
It is cool
it seems we need to confirm the kill here
user147690
@AkivaWeinberger That is amazing.
Akiva can't be sniped and not recognize it
now the rotations of the icosahedron... I'm just kidding, the rotations of the tetrahedron.
02:56
Wait how do we exchange two diagonals
Akiva has raised a valid point of contention!
by 360 noscoping
2
@BalarkaSen what are you doing
user147690
lmao @BalarkaSen
Does he even know @LeakyNun
user147690
02:57
Did @BalarkaSen become a memelord in my absence?
How can I show that there is no ring homomorphism $\mathbb{Z}_3 \rightarrow \mathbb{Z}_4$?
of course it isn't isomorphic to $S_4$
I have thought about it before
No but seriously I think you rotate 180 degrees about an edge that intersects the two diagonals
@gian of course there is
@LeakyNun It is
02:57
New millennium problem to replace Poincare's conjecture: uw0tm8 Balarka?
2
It isn't isomorphic because the rotations have 3 elements of order 4 with different orbits
and the elements of order 4 in $S_4$ is like, you know, bye
so let's just agree that one can only embed it in $S_6$
(1234), (1324), (1423), (1432)
#resistshitposting
@LeakyNun $(1234)$, $(1324)$, $(1432)$?
GodFUCKINGok I'm calm
@AkivaWeinberger, the problem reads: Prove that there can be no morphism $\mathbb{Z}_3 \rightarrow \mathbb{Z}_4$ of rings.
02:59
Meh, I'm done with this shitz
though, wouldn't that be four elements of order 4?
@gian Well we know $1+1+1=0$ in the image
user147690
Many of my good math friends are memelords, so I am not criticizing :P
Applying our hypothetical function $f$ we get $f(1)+f(1)+f(1)=0$ in $\Bbb Z_4$ @gian

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