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2:58 AM
Last night dream: A strange number
$\cdots 11 \cdots 11 \cdots 11 \cdots 11$
it is in some way, bigger than $11\cdots 11 \cdots 11 \cdots$
The set is not well ordered obviously due to the infinitely decreasing chain, but it is linearly ordered
 
 
1 hour later…
4:18 AM
@MatheinBoulomenos I thought I know Galois theory... until I read his original paper
it's definitely worth reading if you haven't
without Artin's embellishment
 
"Injective linear maps between finite-dimensional fields of the same dimension are bijective" is basically linear algebra's version of the pigeonhole principle
(and very useful eg in Galois theory)
 
cool
 
Random thought, no context
 
[More random]
A set with the elements {1,11,111,1111,11111$\cdots$,} is smaller than a set with the elements {1,11,111,1111,$\cdots$11111,}
Their union will have elements of the form $\cdots 111 \cdots$
and thus ceased to be well ordered in the usual ordering
The resulting set is a not dense subset of supernatural numbers
 
 
2 hours later…
6:57 AM
The principle of left no stones unturned. If a proposal make sense, however minority it is, it is included into curation
 
 
2 hours later…
8:39 AM
Hi all, what is $l^2(\Bbb Z)$?
 
8:52 AM
Servus @Rudi! That I believe is the set of square summable sequences of integers
 
It is
If you're familiar with the definition of $L^2(X,\Sigma,\mu)$ for a generic measure space this is the same as taking $X=\Bbb Z$, $\Sigma$ the full powerset and $\mu$ the counting measure
 
Hi @Alessandro
 
9:16 AM
@ÍgjøgnumMeg Servus! OK. Makes sense!
Hi @AlessandroCodenotti, unfort. I am not that familiar with it, but I can guess and look it up.
I am interested in probability currents for quite a while and now I came across that en.wikipedia.org/wiki/Probability_current#Discrete_definition which is facinating, as it seems to suggest that there is quantum mechanical world on discrete spaces...
 
9:31 AM
Can someone tell me what is difference between elementary number theory and algebraic number theory?
I know algebraic nt uses group theory, but elementary does not. Is this the only difference?
 
elementary number theory isn't characterised by its use of algebraic methods
 
@ÍgjøgnumMeg okay. Algebraic method means group theory?
 
well yeah
among other things
but groups usually show up in texts on elementary number theory too
 
@AlessandroCodenotti, @ÍgjøgnumMeg I try to imagine what a Hilbertspace vector (on that space?) looks like? Is it an infinte sequence of integers, or rather a sequence of field elements, like $\in\Bbb C$?
 
@ÍgjøgnumMeg Okay. Thanks for telling me!
 
9:39 AM
more specifically is that $\Bbb Z$ the domain or the codmain? I suppose the domain, but whats then the codomain?
 
beyond me lol
 
Its a bit physical seems
 
It's functions $\Bbb Z\to\Bbb C$
 
10:05 AM
@AlessandroCodenotti Grazie Alessandro e buona domenica!
then differentiation I suppose corresponds to formation of the difference series?
 
 
3 hours later…
12:40 PM
What hypothesis do I need to get $X/G$ metrizable, where $G$ is a group acting on a metric space $X$?
 
12:52 PM
I'm pretty sure that having $G$ act by isometries works, but can it be weakened somehow?
 
1:27 PM
Whaddup
 
Hi @ÍgjøgnumMeg
 
Hey @Ultradark
 
1:50 PM
Hi @ÍgjøgnumMeg
Already done with the ANT sheet?
 
Hi @Mathei @ÍgjøgnumMeg
 
Hi @Alessandro
 
How was the first week?
 
It was cool
 
1:58 PM
The L-function lecture is nice, adic spaces as well
And for you? I think it was your second week, right?
 
Yep
So far I'm really enjoying the set theory course, I'm getting lost in higher categories and the two analysis courses are alright
Do you happen to know an answer for my group actions question a few messages back?
 
@Mathein yeah lol
except
I have a stray factor of $[L:K]$ in the trace
not sure if it's supposed to be there or not lol
 
@Alessandro no idea about the group actions question
@ÍgjøgnumMeg there is no [L:K] factor in the trace
 
Alright I have a factor of $n$ in there so I'm not sure where that came from, but otherwise I have what I want
hehe
but what I did was, as you said, to look at the trace of the multiplication by $\alpha$ matrix (where $L = K(\alpha)$) and ended up with a companion matrix, and that guy has minimal polynomial equal to the minimal polynomial of $\alpha$, factored that guy over $\overline{K}$, and thus have that the multiplication by $\alpha$ matrix is similar to a matrix with $\sigma_i(\alpha)$ along the main diagonal
 
2:14 PM
Yeah, that's correct
So where did you get the [L:K] factor from?
 
Well when I then consider the trace of an element $x$ I use the linearity of Tr and oh yeah I don't have a factor of $[L:K]$ ignore me
$K$-linearity lol
so it should fix elements of $K$
duh
 
The trace does not fix K
 
ffs
that's right it should give me back $[L:K]k$ for an element of $K$
which is why I'm confused I guess
$\Tr(x) = \Tr(k_0 + k_1\alpha + \cdots + k_{n-1}\alpha^{n-1})$, split this up by $K$-linearity
 
Okay so if you consider a general element of x of L, you want to choose a K basis for K(x) (just the standard one by powers of x) and a K(x) basis for L, then multiply those to get a nice basis for L over K
You get a block diagonal matrix with a bunch of companion matrices on the diagonal
@ÍgjøgnumMeg I don't think that's all that useful, since you may know the trace of $\alpha$, but not of powers of $\alpha$
 
yeah I see my error now
woopsie
hahaha
Although can't I easily work out the trace of the powers of $\alpha$ by using the fact that they share the same minimal polynomial as $\alpha$?
 
2:31 PM
They don't
 
Right I'm being naive then
I'm so rusty :(
 
You get $\mathrm{Tr}_{L/K} (x) = [L:K(x)] \mathrm{Tr}_{K(x)/K} $ if you construct a basis as I described above
Now you have to consider the question "for every K-algebra homomorphism $K(x) \to \overline{K} $ in how many different ways can I extend it to $L \to \overline{K} $?"
 
okey :) Thanks for the help
otherwise the ENT questions were not hard lol
 
3:32 PM
@MatheinBoulomenos hier?
 
3:43 PM
How can a non-abelian group be associated to a non-zero curvature geometry
Can cayley graphs be extended into 3 dimensional versions
Why can one think of nilpotent groups as slight perturbations of abelian groups?
 
@Ultradark Abelian groups are all amenable, while hyperbolic groups (which are the generalization of non positive curvature) are amenable iff they are virtually cyclic, so an Abelian group needs to be essentially $\Bbb Z$ (to be precise it needs to be quasi-isometric to $\Bbb Z$) to also be non positively curved
@Ultradark I'm not sure why you consider the Cayley graph of a group to be two dimensional here
 
3:59 PM
@AlessandroCodenotti yeah, nevermind about the dimensional thing. Do you know if there is a possible Cayley graph with two generators whose arrows only point in one direction.
 
I don't know what that means, how do you decide whether two arrows are pointing in the same direction
 
I mean can a Cayley graph be legit if there are no edges that admit arrows pointing towards each node
nvm I answered it
This one works
A generated group $(G,S)$ is abelian iff the generators in $S$ pairwise commute with each other.
Can someone please help me with this statement.
I'm mostly stuck on "pairwise commute"
 
4:15 PM
What's the issue exactly?
 
do any of the generators above pairwise commute with each other ?
it's a cyclic group so I'm assuming the answer is yes
 
@LeakyNun I am now
 
That's a Cayley graph for Z/6Z with respect to a generating set with a single element
 
@MatheinBoulomenos have you ever read Galois' original paper?
 
4:20 PM
it's worth a read
 
@AlessandroCodenotti yeah, I still don't understand what pairwise commute means though
 
well galois didn't know about fields and groups
 
does it have to do with arrows?
 
to me it's the core of the theory without Artin's embellishments
 
4:21 PM
You fix a generating set $S$, its elements pairwise commute iff for all $s_1,s_2\in S$, $s_1s_2=s_2s_1$
 
I think I like Artin's embellishments
 
you're too abstract
 
And also Grothendieck's embellishments
 
I like things to be concrete
 
$(\infty,1)$-categories are clearly the correct setting for Galois theory
 
4:22 PM
oh no
 
I mean, they are obviously the correct setting for all math, which just happens to include Galois theory
 
The main theorem of Galois theory is that the étale site of Spec(K) is equivalent to the site of finite discrete G-sets where G is the absolute Galois group of k
 
...
 
Which is of course a special case of the main theorem on Galois categories
 
@AlessandroCodenotti I guess I'm asking if you can simply look at a Cayley graph and see whether the elements pairwise commute, or is it not that simple
 
4:25 PM
this is not maths
what Galois wrote is maths
what you wrote is just abstract nonsense
 
Well then what Galois did is a special case of abstract nonsense
 
triggered
 
Start from the vertex corresponding to $e$, follow the arrow label $s_1$, then the one labeled $s_2$. Go back to the vertex corresponding to $e$, follow the arrow labeled $s_2$, then the one labeled $s_1$. If you end up in the same place then $s_1$ and $s_2$ commute, otherwise they don't
 
"The introduction of the digit 0 or the group concept was general nonsense too, and mathematics was more or less stagnating for thousands of years because nobody was around to take such childish steps..." - our Lord and saviour
 
@MatheinBoulomenos ok the text I'm reading defines a real structure on a C[G] module V to be a involutive conjugate-linear map V->V
now I swear this is just Galois descent
 
4:30 PM
@LeakyNun yes, it is
 
and I'm just like "you're restating Galois descent the 100th time now"
it's for a undergrad seminar course
 
Why is a real structure on a C[G]-module galois descent?
 
and I swear if I drink too much I will say in my presentation that it is Galois descent
"what a waste of time, this is just Galois descent yall, trivial"
 
Do you mean in the sense that you can recover an $\Bbb R$-form of $\Bbb{C}[G]$ by taking fixed points of that conjugate-linear map, which is acting as $Gal(\Bbb C/\Bbb R)$?
 
4:35 PM
@MatheinBoulomenos brilliant
@GaloisintheField sure
 
@GaloisintheField the equivalence of complex representations with a equivariant real structure and real representations follows immediately from Galois descent
 
That makes sense
 
@MatheinBoulomenos I feel like I'll immediately fail the presentation if I ever mention the two words "Galois descent"
 
why though?
you should also mention "monadic adjunction"
 
it's forbidden knowledge
 
4:37 PM
Galois descent follows from Beck's monadicity theorem, soo
 
Is there a stacky/fibered category interpretation of galois descent?
 
@MatheinBoulomenos wie war Oktoberfest?
 
I never attended the Oktoberfest
 
why not?
 
4:45 PM
idk, the beer there is overpriced and the music is bad (imo)
@GaloisintheField Galois descent can be seen as a special case of descent along a torsor
 
@MatheinBoulomenos So descent for etale torsors on a point?
 
@MatheinBoulomenos do you know that the Galois group is actually a torsor
 
@GaloisintheField yes
@LeakyNun yeah
 
The galois group isn't itself a torsor Leaky, Spec(E)-->Spec(F) with Gal(E/F) acting on Spec(E) is a torsor
 
Proposition II of Galois' paper says that if E is an intermediate field for L/K Galois, then Gal(L/K) decomposes into p isomorphic groups
one for each conjugate of E
 
4:53 PM
(or maybe your definition of torsor is different to mine, mine are geometric torsors)
 
now a person who knows group theory would be like "you can't decompose a group into smaller groups!"
but it actually makes sense if you interpret "group" as torsor
 
what
Can you explain the not being able to decompose thing?
 
well you only have one identity
 
For a distance to define a norm it is sufficient to have $\text{dist}(\alpha x, \alpha y)=\alpha \text{dist}(x, y)$ but is this the only way we can define a norm using a distance?
 
the Galois group of (x^2-2)(x^2-3)=0 is {abcd, abdc, bacd, badc}
over the intermediate field $\Bbb Q(\sqrt2)$ it decomposes into {{abcd, abdc}, {bacd, badc}}
 
4:58 PM
@Mathein I can extend $K$-algebra homs $K(x) \to \overline{K}$ to $K$-algebra homs $L \to \overline{K}$ exactly $[K(x):K]$ different ways right?
 
@ÍgjøgnumMeg no
 
what if $K(x)=L$? Then you should get $1$ as an answer
 
Then I guess $[L:K(x)]$ different ways
 
right
and that's exactly what you need
 
5:00 PM
I see
 
surely it's $[L:K(x)]_s$ different ways
 
we're in a Galois setting, but you're right of course
 
@Mathein thanks for your help :)
I think I'm gonna struggle for a few weeks lol
 
howdy @ÍgjøgnumMeg and @Mathein (and @Leaky and @Galois)
 
Hey @TedShifrin :D
 
5:08 PM
hi @Ted
 
(dies inside)
 
@TedShifrin Hi
 
hi @Mats and dead @Fuzzy
 
@FuzzyPixelz You want to see what extra conditions you have to give for a metric to define a norm via $\|a\|:=d(a,0)$?
 
My soul is now clean
Yes, and I want to know if you can go back and forth between them in other ways (I can't think of anything other than $\text{dist}(x, y)=||x-y||$)
 
5:12 PM
Heya @Ted
 
@Fuzzy: So you're on a vector space to start with, of course. The formula you just wrote down is the standard way to get a metric out of the norm, but of course you could take any positive multiple of that, so there's nothing unique.
 
@Mathein tbf I still have 5 days to hand in my solutions lol, maybe I'm just rushing
 
Okay, so we have multiples, what else?
 
Hmm, maybe a convex function would work — have to worry about triangle inequality.
 
I mean if you want a norm then it should be scale-invariant. It seems to me the only interesting question is the following. Say a subset of R^n is a unit sphere if it intersects every ray from the origin exactly once. Under what conditions is this the unit sphere of a norm?
As Ted says the corresponding unit ball must be convex; and the unit sphere must be a closed set.
 
5:35 PM
@MikeMiller Could you please rephrase that..? I'm really not familiar with those concepts
 
I'm just saying I don't understand how to extract a question from what you asked beyond something unrelated and then I got kind of interested in that other question
 
Yes and I got interested in it too but couldn't get what you saying
 
6:00 PM
@LeakyNun Can you explain the counterexample you sent me earlier?
 
6:18 PM
Hey! I am having quite a hard time trying to figure out what I am supposed to do in this question:
Am I supposed to show that this relation is reflexive, symmetrical and transitive?
 
@HarryBattersby Show that it satisfies the three conditions for being an equivalence relation
right
 
I see. So, do I need to assume the first part (P) and conclude the second (Q) using these three properties?
 
There are no P or Q that I can see
 
But isn't it supposed to be in the form P implies Q and Q implies P since it states that this statement only holds true if and only if a-b is divisible by 4?
 
No, the if and only if is in the definition of the given relation. There is nothing to prove there
And please remember to introduce variables before referring to them. Otherwise we will just end up confused
 
6:24 PM
@MikeMiller

let $U$ be an open convex neighborhood of $0$ in $V=\Bbb R^n$ such that $v \in U \Rightarrow -v \in U$ and such that $\partial U$ has the property that every ray from the origin interesects $\partial U$ exactly once.
Define $\|0\|=0$ and for any $v \in \Bbb R^n \setminus 0$, let $\|v\|=1/\alpha$ where $\alpha \in \Bbb R_{>0}$ is the unique constant such that $\alpha v \in \partial U$. Then clearly $\| \lambda v \|=\lambda \|v \|$ and since we have $-v \in U \Rightarrow v \in U$, we also have $v \in \partial U \Rightarrow -v \in \partial U$ which implies $\|-v\|=\|v\|$.
 
Okay, sorry about that. So we're pretty much given in that question that: $a (mod 4) = b (mod 4)$ with which we can prove that this relation is indeed an equivalence relation?
 
@HarryBattersby No, that is not what we are given. We are given that the relation is defined in the given way
Once we start writing up what we need to prove for each of the properties, we will probably assume it at some point, but it is not given yet.
 
@TobiasKildetoft Okay. I understand. So, to prove that this relation is reflexive, would it be valid to say "For any m ∈ Z, m ≡ m (mod 4), as m − m = 0 is divisible by 4 as given." Therefore, this relation is reflexive?
 
and for symmetrical would that also be valid: "If a ≡ b (mod 4), then a−b is divisible by 4. But then b−a = −(a−b) is also divisible by 4, which implies that b ≡ a (mod 4)", as needed to show symmetry?
 
6:33 PM
yes, perfect
(well, you don't actually need the mod stuff, as that is not mentioned in the definition given, but it means the same)
 
I see. So basically by discarding the part where I mention that a ≡ b (mod 4) and b ≡ a (mod 4), the proof would stay valid?
 
yes, since the definition you were given just talk about the difference being divisible by 4
 
Oh, okay. I understand. That makes sense. And for the last part (proving that this relation is transitive), would this be acceptable? "Suppose that a-b and b-c are both divisible by 4 by definition. Consequently, their sum, (a − b) + (b − c) = a − c is also divisible by 4".
 
yep, looks good
Just add the conclusion ("and hence...")
 
Will do. Great, thank you!
 
7:20 PM
Hey! Are the following conclusions correct? For part 1: the number of elements of the largest transversal subset of that list is $7$. For part 2: the number of elements of the largest equivalence class that is a subset of that list is $3$ since the equivalence class subsets are ${1/2, 5/10}, {0/1, 0/2}, {2/1, -10/-5, 10/5}$.
Sorry for the second part, the sublists are: {$1/2, 5/10$}, {$0/1, 0/2$}, {$2/1, -10/-5, 10/5$}. Therefore, the last list contains the most elements ($3$).
 
7:52 PM
Would that be true?
 
8:08 PM
but where x = 0 is not allowed
 
@MikeMiller I calculated the +1 surgery on the trefoil thing
 
hey @BalarkaSen
 
If I draw a naive parallel curve to the trefoil I get an embedded curve which hits the Seifert surface of the trefoil at 3 points, so I put three twists in the opposite direction to make it the "true parallel" - intersection of the unit normal bundle of the trefoil with the Seifert surface.
If $a, b$ are the Wirtinger generators, then the knot group is $\langle a, b | aba = bab \rangle$ and the parallel comes out to be $ba^2ba^{-4}$ I think.
Which does have zero abelianization, as it should (the Seifert surface has that as boundary)
Because when I abelianize, $a = b$
 
But why is it the Brieakorn sphere
 
Yeah I am not sure I quite see it. But see the fundamental group: The meridian is $a$ and the longitude is $ba^2ba^{-4}$ (hoping I got the signs right). The +1 curve is $ba^2ba^{-4}a = ba^2ba^{-3}$
 
8:22 PM
Hey there, a @Balarka!
 
I am making this bound a disk, so $\pi_1 = \langle a, b | aba = bab, a^3 = ba^2b \rangle$
 
I think the key thing you want to see is the Seifert fibering over the 2-sphere with 3 singular fibers
Singularity type 2,3,5
Maybe this comes from the open book decomp from the trefoil being fibered
 
Let $x = aba$, $y = ab$. Then $x^2 = y^3 = ababab = aba^2ba = a^5$
 
Nerd
 
Hm, I messed up signs. $a = y^{-1} x$. It's ok, I got $\pi_1 = \langle x, y | x^2 = y^3 = (y^{-1} x)^5 \rangle$.
 
8:25 PM
So many letters bruh
 
Sorry
 
Don't be
 
@MikeMiller Ah OK
I don't understand the +1 curves though. If I just do a 0-surgery on the trefoil I glue a disk to the boundary of the Seifert surface, which gives me a torus bundle over the circle?
Is it just T^3?
Hi @Ted
 
0-surgery is a homology S^2 x S^1
 
Yeah I seem to get fundamental group $\langle x, y | x^2 = y^3 = (xy)^4 \rangle$ (is that correct?)
Hm maybe not
$(xy)^6$
Meh there's some sign mistake in my calculations
Wish I could do these things fast but also correctly
OK, it has to be $\langle a, b | aba = bab, a^4 = ba^2b \rangle = \langle x, y | x^2 = y^3 = (y^{-1} x)^6 \rangle$. The abelianization matches at least, given what you said
 
8:39 PM
@MatheinBoulomenos what's the generalization of the statement that an H-module is the same thing as a C-module with a negative-involutive conjugate-linear map?
 
@MikeMiller I don't understand where the $5$-fold symmetry is coming from when making the Poincare homology sphere, though. Do you have a guess?
 
I understand your comment about the Brieskorn sphere is that if I look at $x^2 + y^3 + z^5 = 1$ in $\Bbb C^3$ and a link of the $(0, 0, 0)$ singularity, that's a $3$-manifold inside $S^5$. What is the fibration over $S^2$?
 
I always forget. Milnors paper about these explains it
 
I vaguely recall Mahan told me about this way to see it once. I'll look at Milnor's book
Borrowed it from the library a few days ago hah
 
8:45 PM
@BalarkaSen Oh
S^1 acts on sphere in obvious way
 
Oh right
 
Does it preserve that eqn
It does not
Lol
 
$A_5$ acts on the $3$-manifold right (pi rotation on first component, 2pi/3 rotation on second, 2pi/5 rotation on the last)? It preserves the equation.
Not sure where to go with that
 
It shouldn't act right? It's a fundamental group
 
You're right I'm dumb. $A_5$ leaves the equation fixed lol
 
8:49 PM
It def does not act trivially
Very confused
Imo look at Milnor's paper on brieskorn spheres it's very short
 
Alright checking out
 
@LeakyNun uhm, there should be one, but I'm not sure what it is
 
what is H/C even
 
Hi @Balarka
 
Hi @Alessandro
 
8:51 PM
what's the analogue of C=R[x]/(x^2+1) for H/C
 
@LeakyNun C is a maximal subfield of H
 
what's the correct polynomial ring to form H from C
 
I don't think that's the correct question
 
a C-module is the same as an R-module with a negative-involution
 
$H^2(\Bbb C/\Bbb R,\Bbb C^\times)$ is cyclic of order 2, the nontrivial element corresponds to $\Bbb H$
 
8:53 PM
I guess I wasn't making sense above; I just produced $\Bbb Z/2 \times \Bbb Z/3 \times \Bbb Z/5 \leq S^1 \times S^1 \times S^1$ which acts on the hypersurface $x^2 + y^3 + z^5 = 1$ by $(x, y, z) \mapsto (-x, y, z)$, $(x, y, z) \mapsto (x, \zeta_3 y, z)$, $(x, y, z) \mapsto (x, y, \zeta_5 z)$ respectively.
 
@MatheinBoulomenos and the trivial element?
 
@LeakyNun $\Bbb R$
 
there's definitely some general phenomenon happening here
we can go from C to R and H
this must correspond to the Brauer group generally
 
well, yeah
 
there's something general behind this that I don't know
this is so unsatisfying
@MatheinBoulomenos any idea?
 
8:58 PM
@LeakyNun let $L/K$ a Galois extension with $G=\mathrm{Gal}(L/K)$ and let $f \in H^2(G,L^\times)$ a 2-cocycle, then we can construct a central simple algebra over $K$ like this. Take as a set the group algebra $L[G]$ and define the multiplication based on the relations $\lambda \cdot \sigma = \sigma \cdot \sigma(\lambda)$ and $\sigma \cdot \tau = (\sigma \circ \tau) f(\sigma,\tau)$
 
R isn't a central simple algebra over C
 
but H is
 
but the trivial element corresponds to R
 
so how does this work
 
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