1:33 AM
I was wondering - would the power set of the set {{$1,2,3$}} contain $2^1$, $2^1∗2^3$ or just $2^3$ elements?

None of the above? What's the power set of {a}? Of {{a}}?
I mean, it's been a long time for me, but it's just the power set of a one-element set, right? That the one element is, itself, a set of three elements doesn't matter.

The power set of {$a$} would be {{}, {a}}, therefore contain $2^1$ elements, I reckon?

Me, too.

I see that makes sense. So, it would be essentially equivalent to finding the power set of the set {$a$}?

Yup.

1:38 AM
No problem ;D thanks!

o/

2 hours later…
3:15 AM
I am fairly fresh in set theory, and I was wondering why is [$0.5, 7.5$] $∩ N$ ≠ $[1, 7]$?
Is it due to the reason that $0.5$ and $7.5$ are $\notin$ the set $[1, 7]$ and for [$0.5, 7.5$] $∩ N$ = $[1, 7]$ to be true, both sets must be subsets of each other which isn't the case here? Or is there a different reason?

3:52 AM
@Abwatts Can you enumerate the elements of $[0.5, 7.5] \cap \mathbb{N}$? (I think so.) Can you enumerate the elements of $[1,7]$? (I think not.)
(Assuming you're using the square brackets for continuous intervals in $\mathbb{R}$, at least.)

That makes sense. So, basically, because the interval [$1, 7$] is not restricted to only natural numbers, we will have some rationals in that interval too, which are obviously not natural numbers?

3 hours later…
7:02 AM
Mooornin'

7:21 AM
0

yes you are correct. P(EDGE)=0, so you cannot condition on that. This is just a trick question

@EdwardEvans wanna play?

2 hours later…
9:27 AM
Composing music is very hard when you have a anti-motif bias
You end up with numerous chunks of completely distinct melodies and you struggle to glue them together into a song
Basically, most western type and video game music are ABA, ABCA, or some variation of this to ensure loopability
and so it usually involve 1-2 motifs and the whole song will be just variation of these motifs
My issue is that I often end up with music like the figure shown above
where there are like 10 pieces of melodies, none of them pairwise equal
This means it is usually hard for me to compose a full music piece since it tends to birfrucate in the middle and produce a new melody
and hence you end up with motif-less music
Maths chat people only: This makes me wonder, is there a mathematical formula that detect motifs

10:02 AM
@Leaky I was learning a generalisation of the Dirichlet unit theorem lol
a very small generalisation but cool nonetheless
in particular there's an exact sequence

$$0 \to \mathcal{O}_K^\times \to \mathcal{O}_{K,S}^\times \to \bigoplus_{\mathfrak{p} \in S} K^\times/(\mathcal{O}_K)_{\mathfrak{p}}^\times \to \operatorname{Cl}(K) \to \operatorname{Cl}_S(K) \to 0$$

where $S$ is a finite set of primes of $K$ and this guy gives you $\mathcal{O}_{K,S}^{\times} \cong \mu(K) \oplus \Bbb Z^{\lvert S \rvert + r_1 + r_2 -1 }$
where the usual notations are the usual notations
rofl

2 hours later…
12:20 PM
Does anyone here study non-commutative algebra?

12:53 PM
@SolvingProblems I do

Hello Lukas, I posted a question on main.
0

Let $A$ be a commutative ring, $B$ be a non-commutative $A$-algebra, $P$ be a finitely generated projective left $B$-module. Let $E=\text{End}_B(P)$ as a left $A$-module (by restricting from $B$). If I were to know that $E$ is in fact a free $A$-module (since $A$, and the map $f:A\to B$ defining...

Perhaps to rephrase the question:

Let $A$ be a commutative ring, $B$ be a non-commutative $A$-algebra, and $P$ be a finitely generated $B$-module, $P\subset B^n$ with known generators. I want to take $\text{End}_B(P)$ and restrict it to be an $A$-algebra, which I'll denote by $E$, where in my case it is now free as an $A$-module. I want to find the rank of $E$.

2 hours later…
3:25 PM
I'm interested in the simplest case of the edge-of-the-wedge theorem: If $f\colon\mathbb{C}\rightarrow\mathbb{C}$ is continuous and holomorphic on both half-planes, it is holomorphic everywhere.
It is claimed that this follows from Morera's theorem by splitting any contour up along the real axis into parts in the upper and lower half-plane, where they vanish by hypothesis. However, I don't understand how that works. because one still necessarily has parts of the contour along the real axis and I don't think the Cauchy Integral theorem necessarily holds on the boundary. What am I missing?

3:36 PM
> > We asked [composer Gyorgy Ligeti] about his musical "influences." We had read—in the New York Times—that he had been influenced by the new field of fractal geometry "whereby physical objects are conceived as consisting of smaller and smaller identical shapes." He said:
> "Yes, fractals are what I want to find in my music. They are the most complex of ornaments in the arts, like small sea horses, like the Alhambra where the walls are decorated with geometric ornaments of great minuteness and intricacy, or like the Irish Book of Kells, those marvelously decorated borders and capitals.
> The most complicated ornaments—perhaps not art, perhaps geometry. It is a very complex music, difficult to describe. I only want to give a metaphysic for my music. After all, music is not a science."
@Secret Thought you might find that interesting
I think this is the quote from the New York Times they're referring to:
> ''I do not use direct mathematical translation into my music, like Xenakis,'' Mr. Ligeti said. ''The influence is poetic: fractals are the most complex ornaments ever, in all the arts, like the Book of Kells or the Alhambra. They provide exactly what I want to discover in my own music, a kind of organic development.''

Hello, I'm learning pde and I'm trying to find an example for a function u that holds: for every parallelogram ABCD with parallel lines to the lines y=+-x, u(A)-u(B)+u(C)-u(D)=0 and u does not hold u_tt=4u_xx

@Secret I can relate to the problem of having lots of chunks of music and no coherent large song
When I'm entertaining myself on the piano I often just play them in a random order with strange transitions
(some of which are more natural than others)
(Well, random-ish, influenced by how natural the transition would be)
@Eran Is u a function of x and y, or of x and t?

@AkivaWeinberger: A function of x and t

4:00 PM
I will try posting this code here in chat to get myself motivated:
Clear[a, b, nn];
nn = 60;
a[n_] := Total[MoebiusMu[Divisors[n]]*Divisors[n]];
Monitor[a1 =
Table[Sum[Sum[a[GCD[m, r]], {m, r, n}]/r, {r, 2, n}], {n, 1,
nn}];, n]
g1 = ListLinePlot[a1, PlotStyle -> {Red, Thick}];
Monitor[a2 =
Table[Sum[
If[Sum[-Abs[Sum[a[GCD[m, k]], {m, k, n}]], {k, 2,
r}] >= -(n - 1), -Abs[Sum[a[GCD[m, r]], {m, r, n}]]/r,
0], {r, 2, n}], {n, 1, nn}];, n]
g2 = ListLinePlot[a2, PlotStyle -> {Thick}];
Show[g2, g1]
Marilyn vos Savant said that learning is boring. Therefore don't multitask. Multitasking is just a way of easing the feeling of boredom to make it feel like you are progressing although you actually are not.
Prove that the blue curve times a constant is greater than the red curve.
1,2,3,4,5,...
In other words/formulas:
Let:
$$a(n)=\sum\limits_{d \mid n} \mu(d)d$$
Show that:
$$\sum\limits_{r=2}^{n} \frac{\sum\limits_{m=r}^{n} a(\gcd (m,r))}{r} \geq c\underset{\sum\limits_{k=2}^{r} -\left|\sum\limits_{m=k}^{n} a(\gcd (m,k))\right|\geq -(n-1)}{\sum _{r=2}^n} -\frac{\left|\sum\limits_{m=r}^{n} a(\gcd (m,r))\right|}{r}$$
where $c$ is the constant.
The following is not well formulated, at least not yet. Under what conditions/variable bounds is a sequence that sums to 1 and consists of positive and negative terms bounded by c*Floor(Sqrt(Length(sequence)))?

4:18 PM
@Thorgott Take part(s) of your path parallel to, and arbitrarily close to, the real axis. You do know continuity (hence uniform continuity on a compact subset).

Solve that and then determine into what category, greater than or less than c*Floor(Sqrt(Length(sequence))), the right hand side falls into.
That is what I am going to do next, I am going to find those mentioned conditions/variable bounds.

I'm trying to prove that u:R^n-->R that holds the spherical mean value property also holds the ball mean value property. I understand the intuition but I'm having trouble proving it formally.
I thought about change of variables comparing the integral on the sphere to the integral on the ball

4:42 PM
So, say we are (wlog) looking a triangle $T$ with vertices $x,y,w$, the former two lying on the real axis and the latter lying in the upper half-plane. Then we look at the triangles $T_n$ with vertices $x+\frac{i}{n},y+\frac{i}{n},w$.
We have $\int_{[x+i/n,w]}f(z)dz=\int_0^1f(w+t(x+\frac{i}{n}-w))(x+\frac{i}{n}-w)dt\rightarrow\int_0^1f(w+t(x-w))(x-w)dt=\int_{[x,w]}f(z)dz$ by uniform continuity on some compact rectangle containing all these triangles.
But $\int_{[x+i/n,w]}f+\int_{[w,y+i/n]}f+\int_{[y+i/n,x+i/n]}f=\int_{T_n}f=0$ by Cauchy, hence $\int_Tf=\int_{[x,w]}f+\int_{[w,y]}f+\int_{[y,x Ok, that works. Thanks. 5:00 PM @Eran Are there such functions which are not harmonic? 5:57 PM Polar coordinates work for that problem, but I feel like there should be a more direct argument for why integrating over the ball is the same as integrating over all the spheres of smaller radius and then the radius. Can't think of one right now though. 6:37 PM I like your polar coordinates+Fubini approach @AlessandroCodenotti @Thorgott Thanks! 6:53 PM It feels like the relationship between integration on balls and integration on spheres (geometrically speaking) should be more intrinsic than an explicit parametrization, idk Ted will probably say something about Stokes soon Probably, but I want there to be a direct way with CoV/Fubini Isn't Stokes's theorem such a way? Buonasera @Alessandro Ciao Lukas 7:01 PM The only thing about Stoke's theorem is that it generalizes the FTC, so I have no clue, really Stokes' tss Stokesens oh, right see, that's how little I know about it Ritter von Stokes 7:04 PM hey @Edward still working through cyclotomic fields? Hey @Lukas, I finished the chapter on cyclotomic fields and on localisations, gotta go back over it tho :P There was a slightly generalised form of Dirichlet's Unit Theorem at the end which I liked and posted above lol ah, the S-unit theorem Right, where the rank of the S-units is |S| + r_1 + r_2 - 1 Z-rank a friend of mone wrote his bachelor thesis on something with S-units orly 7:06 PM yeah, in the S-units there are only finitely many solutions to the equation$u+v=1$ha wtf and you can reduce some other diophantine finiteness results to that Nice :) also, did the cyclotomic proof of quadratic reciprocity ah, my favorite proof and the 2nd Ergänzungssatz was the Übungsaufgabe but yeah it's very cool 7:08 PM you need to work with$\Bbb Q(\zeta_8)$for that iirc Hmm maybe I did it wrong but I worked with$\Bbb Q(\zeta_p)$as in, for (2/p) and look at the decomposition of$2$in that guy there are probably many ways to do it Yeah right You look at the field$K := \Bbb Q\left(\sqrt{(-1)^{\frac{p-1}{2}}p} \right)$and then the splitting number of$2$is even iff$K \subset \Bbb Q(\zeta_p)^{Z_2}$which is equivalent to$2$being split completely in$K$, which happens when$d_K \equiv 1 \bmod 8$and then that's equivalent to$p \equiv 1$or$7 \bmod 8$and that happens to be equivalent to$(2/p) = 1$@Thorgott How would you prove it with polar coordinates? the proof with$K=\Bbb Q(\zeta_8)$goes like this. Every odd prime$p$is unramified in$K$, so we have a Frobenius element in the Galois group which is just given by$\zeta_8 \mapsto \zeta_8^p$. We have a subgroup in the Galois group$\mathrm{Gal}(K/\Bbb Q) \cong (\Bbb Z/8)^\times$given by$\overline{\pm 1}$, the fixed field$L$of that is generated by$\zeta_8+\zeta_8^{-1} =\sqrt{2}$. Now the Frobenius element at$p$fixes$L$iff$p$splits completely in$L=\Bbb Q(\sqrt{2})$but$p$splits splits completely iff the minimal polynomial of$\sqrt{2}$, i.e.$x^2-2$splits over$\Bbb F_p$thus$x^2 \equiv 2 \pmod{p}$has a solution iff$p \equiv \pm 1 \pmod{8}$I like both solutions :) @Edward 7:18 PM Brilliant :D and then our other exercise was to prove FLT for n=4 lol which was just a couple of successive pythagorean triples and coprimality arguments +descent @EdwardEvans in L-functions we defined Artin L-functions today so we're doing the more Galois/ANT-heavy stuff right now bleh nice Should've stayed in it but ah well Define them for me if you want can't do everything lol Pretty much just write it down. E.g. in two dimensions you have$\int_{|x|<r}f=\int_0^rs\int_{-\pi}^{\pi}f(r\cos\varphi,r\sin\varphi)d\varphi ds=\int_0^r\int_{|x|=s}f$It feels backwards though, I'm not satisfied with it 7:26 PM okay so Rösner did motivate them really well. Start with the cyclotomic field$\Bbb Q(\zeta_n)/\Bbb Q$. The Galois group of that is isomorphic to$(\Bbb Z/n\Bbb Z)^\times$, so a one-dimensional representation of the Galois group corresponds in this case simply to a Dirichlet character. We should expect that the Artin L-function for such a representation is a Dirichlet L-function we can start by looking at Dirichlet L-functions and reinterpreting everything in ANT/Galois terms to try and generalize it to a more general setting if$\chi:(\Bbb Z/n\Bbb Z)^\times$is a Dirichlet character, then the L function$L(\chi,s)$has an Euler product $$L(\chi,s)=\prod_{p \not \mid n}(1-\chi(p)p^s)$$ now for the reinterpretation: viewing$(\Bbb Z/n\Bbb Z)^\times$as the Galois group of$\Bbb Q(\zeta_n)/\Bbb Q$, we can interpret the primes in$\Bbb Z$as prime ideals$(p)$, the condition$p \not \mid n$is of course the fact that$(p)$is unramified. For these unramified prime ideals,$\chi(p)$is the evaluation of the representation at the Frobenius element of the prime ideal$(p)$Now suppose that$L/K$is a Galois extension of number fields and$\rho:G=\mathrm{Gal}(L/K) \to \mathrm{GL}_n(\Bbb C)$is a representation. Then for all unramified prime ideals$\mathfrak{p}$of$K$we can do the following: choose any prime ideal$\mathfrak{P}$above$\mathfrak{p}$and evaluate$\rho$at the Frobenius element$\mathrm{Frob}_{\mathfrak{P}/\mathfrak{p}}$, then the factor at$\mathfrak{p}$is$\det(\mathrm{Id}-\rho(\mathrm{Frob}_{\mathfrak{P}/\mathfrak{p}})N(\mathfrak{p}))$, i.e. the characteristic polynomial of$\rho(\mathrm{Frob}_{\mathfrak{P}/\mathfrak{p}})$evaluated at th Niiice If$\mathfrak{p}$is ramified, then there's a slight modification we can make: we can still take a lift of the Frobenius in the residue Galois group, but that will only be well-defined up to an element in the inertia group. That's no problem if the inertia group acts trivially. We can now force the inertia group to act trivially by simply looking only at the invariants$V^{I_{\mathfrak{P}}$. As the inertia group is normal in the decomposition group, this subspace carries an action by the decomposition group, so we take the characteistic polynomial of the action of the Frobenius on$V^{I_\ma
and then the product over all those local factors is the Artin L-function

7:41 PM
Nice man
Sounds super interesting
T_T
I can just read it all in the back of Neukirch instead I guess

we're proving Cebotarev next week

Nice, I only briefly glanced over that in Milne's notes
just says the density of split primes is $1/[L:K]$ or smth right?
or completely split primes?

completely split

Cool :P

the more general statement is that for any conjugacy class of size $n$ in a Galois group of an extension of number fields $L/K$, the density of those prime ideals whose Frobenius is in that conjugacy class is $\frac{n}{[L:K]}$
and you can recover the statement about completely split primes by looking at the conjugacy class of the idenitity