Mathematics - 2019-01-02

MacroGuy

00:21

hi

is there any Literature/Information on a "theory of equations" for the algebra of sets

A, B, C, D, ..., and so on are fixed subsets of a universal set, U. X is a subset of U and is restrained only by its particular equation. Find the condition(s) under which a certain equation has a solution and obtain all the solutions.

Ex) X union C = D only has a solution if C + D (symmetric difference of C and D) is a subset of X, and X is a subset of D.

Leaky Nun

@MacroGuy boolean algebra

user193319

00:49

Let $X = [0,1]/\sim$ where $0 \sim 1$. I've been asked to show that $X$ is homeomorphic to $S^1$. I figured that $f : X \to S^1$ given by $f([r]) = e^{2 \pi i r}$ would be the homeorphism, but even proving it continuous is hard. How should I proceed. I tried showing that $f^{-1}(S^1 \cap B(e^{2 \pi i x}, \epsilon)$ is open in $X$ for $x \in X$ arbitrary, but it got messy pretty quickly.

Leaky Nun

@user193319 universal property of quotient

and then (compact -> hausdorff) ~~> (bijection => homeomorphism)

Mike Miller

@user193319 Think about continuity in the metric-space sense, and that part of the question starts looking more like a real analysis exercise.

Bob

I am hoping that somebody who is good in Calculus can look at my post: math.stackexchange.com/questions/3058165/…

I suspect I made a computational error.

famesyasd

01:30

Let's say I'm on a real number line and I want to move to the number 1 and I'm on 0 and the way that I can move is like: if I'm given a sequence {1-1/n} n in N I can move on the sequnce points (so on like 0, 1/2, 2/3, 3/4, 4/5, ... and I can also move onto the limit of this sequence so I can get to 1 immediately).
But the game may not be so generous and after one sequence I could only be at 1/2 for example, and after the second sequence only at the 2/3, after the third sequnce on 3/4 etc so after a sequence of such sequences I can get to 1 eventually. But the game may not be so generous an…

Leaky Nun

VTC very unclear

famesyasd

what's vtc?

Leaky Nun

vote to close

famesyasd

01:44

okay I'll ask that later

Rithaniel

02:11

So, I had an idea for a topology, but I'm unsure if it's a particularly useful one. Take any set $X$. The topology on $X\times P(X)$ is defined such that a set $U\subseteq X\times P(X)$ is open iff $U=X\times P(X)$ or if for every $(a,A)\in U, a\in A$.

It's a fairly simple concept, so I imagine that, if it's a "useful" topology, it probably already has a name and has been studied.

BAYMAX

Hi chat

I was thining about the terminologies -

Along the curve

and

across the curve

what is the difference?

Rithaniel

(I was about to ask about that "square cup" symbol you used, Leaky. I've not seen it before.)

Leaky Nun

that's disjoint union

If $Y = \{ (a,A) \in X \times P(X) \mid a \in A \}$ then we have a map $Y \to X \times P(X)$; the topology on the latter is the coinduced topology along the map, with the topology on the former being discrete

Rithaniel

Ah, so $(A\bigcup B)\setminus (A\bigcap B)$?

Leaky Nun

no

$A \sqcup B = A \cup B$

with the emphasis that $A \cap B = \varnothing$

Rithaniel

02:22

Ah, okay. It's the same, but with an extra emphasis.

Leaky Nun

yes

Rithaniel

Also, yes, I believe your description is accurate, but I need to read a definition of the coinduced topology to be certain.

Leaky Nun

my point is just that the topology is useless =)

Rithaniel

Okay, thought it might be.

So, the discrete topologies tend to be useless?

Or, rather, it's useless because it's just another way of describing a discrete topology?

Leaky Nun

you don't get much from the topology, is what I'm saying

the discrete topology is basically the "free" topology

you can, however, put a product topology on $P(X) = 2^X$ which makes it compact (by Tychonoff)

or the compact open topology on $P(X) = \{f : X \to 2\}$

(they might be the same topology, I don't know)

Leaky Nun

02:50

conjecture: cyclic extensions of $\Bbb Q$ of degree $p$ correspond to elements of $K^\times/K^{\times p}$ where $K = \Bbb Q(\zeta_p)$

Leaky Nun

03:45

$\Bbb C[G]^{op} \cong \Bbb C[G]$

Leaky Nun

03:59

$\Bbb C[G]$ with conjugation is isomorphic to $\Bbb C[G]$ with left-regular action?

denote by $\Bbb C[G]^c$ and $\Bbb C[G]^l$ I guess

we want $\varphi : \Bbb C[G] \to \Bbb C[G]$ with $\varphi(ghg^{-1}) = g\varphi(h)$

Leaky Nun

04:28

was it compact <-> discrete again

man I’m so unfamiliar

1. C[G] is semisimple

2. C[G] = V1 + ... + Vn

3. End(C[G]) = sum End(Vi)

4. C[G]^op = sum End(Vi)

this is so confusing

Leaky Nun

04:46

2 is as C[G]-modules, 3 is as C-algebras?

what role does the conjugation action play here then

Buddhini Angelika

06:36

Wish a very Happy New year to all!

Brown Ninja

08:44

Happy new year everyone

Let $\hat{X} = \left( x^0,x^1,\cdots , x^{N-1}\right)$ be the sequence of $N$ i.i.d realizations of a random variable $X$ with pmf $P_1, P_2, \cdots , P_{d}$, where $P_i$ is the probability of appearance of symbol $i$. Can we show that
\begin{align}
\hat{P}_{i} &= \frac{N_i}{N}\\
&= \frac{1}{N}\sum_{k=0}^{N-1} I\left( x^k = i \right)
\end{align}
is a consistent estimator of $P_i$, where $I\left( x^k = i \right)$ is the indicator function (=1 when $x^k = i$, and 0 otherwise)?

ÍgjøgnumMeg

10:28

@BuddhiniAngelika the presentation you gave is a semi-direct product of $\Bbb Z_q$ with $\Bbb Z_{p^2}$ so it has cardinality $qp^2$, while the semi-direct product you wrote down has cardinality $qp$

Leaky Nun

12:00

So apparently End(C[G]) = C[G]^op as C-algebras

$\newcommand{End}{\operatorname{End}}$$\varphi : \End(\Bbb C[G]) \to \Bbb C[G]^{op} : \psi \mapsto \psi(1)$

$\varphi(\sigma \circ \tau) = \sigma (\tau (1)) = \tau(1) \sigma(1) = \varphi(\tau) \varphi(\sigma)$

but $\End(\Bbb C[G])$ is also a $\Bbb C[G]$ module by $(g \cdot \varphi)(x) = g \varphi(g^{-1} x)$?

wait that's nonsense

hi @MatheinBoulomenos!

MatheinBoulomenos

12:15

hi

Leaky Nun

maybe I should figure this out on my own

MatheinBoulomenos

the step $\sigma(\tau(1))=\tau(1)\sigma(1)$ doesn't work

Leaky Nun

what do you mean by doesn't work?

MatheinBoulomenos

in which category are you taking the endomorphisms?

Leaky Nun

C[G]-modules

MatheinBoulomenos

12:17

no, it's okay nevermind

Leaky Nun

the idea of group rep is that C[G] and C[G] are isomorphic, right?

@MatheinBoulomenos

MatheinBoulomenos

every object is isomorphic to itself

Leaky Nun

yeah no

C[G] with the left-regular action and C[G] with the conjugation action

MatheinBoulomenos

consider the case G abelian

then conjugation is trivial, but the left-regular action isn't

Leaky Nun

oh man

Rick

12:21

Hey guys, newbie here. I wanna ask why Excel use \y=ax^{b} as power function regression model, instead of \y=ax^{b} +c which I think is more appropicate?

Leaky Nun

why do you keep crushing my dreams

happy new year btw where have you been

MatheinBoulomenos

happy new year

Leaky Nun

as in, you haven't been here for a long time

not asking where you have been

user 1357113

@AlexanderGruber haha, yeap, nice part! :-) Thank you for the feedback.

MatheinBoulomenos

let $G$ act on itself by conjugation and let $V$ be the associated representation over $\Bbb C$ and $\chi$ its character. Then for $g \in G$, we get that $\chi(g)$ is the number of elements that commute with $g$

not sure how more there is to say

the dimension of $V^G$ is equal to the number of conjugacy classes

Leaky Nun

12:39

@MatheinBoulomenos when is Hom(A,B) an R-module?

MatheinBoulomenos

are A and B R-modules?

Leaky Nun

yes

MatheinBoulomenos

okay say A and B are left R-modules. Then if B is an (R,S)-bimodule, Hom(A,B) is a right S-module. Similiarly if A is a (R,S)-bimodule, then Hom(A,B) is a left S-module

Leaky Nun

what about just R-module?

ok replace S by R got it

MatheinBoulomenos

you can take S=R

Leaky Nun

12:43

what is Hom(R,M) isomorphic to as left R-module?

MatheinBoulomenos

M

Leaky Nun

yes M

ok great

MatheinBoulomenos

but if both cases apply, then you have two natural R-module structures

Leaky Nun

but somehow End[R-Mod](R) = R^op as R-algebras

this is very confusing

MatheinBoulomenos

if you take Homs in the category of right-modules you get Hom(R,R)=R

Leaky Nun

12:45

maths is confusing

MatheinBoulomenos

left-right stuff (I guess chemists would call this chirality) is confusing

Leaky Nun

what is group rep in one sentence?

MatheinBoulomenos

I don't think you can summarize it so concisely (at least I can't)

Leaky Nun

like LCFT in one sentence is $W_{F}^{ab} \cong F^\times$...

MatheinBoulomenos

for the case $k=\Bbb C$ and $G$ finite everything is basically a consequence of Maschke's theoerem and Schur's lemma

well, at least a lot of basic results

Leaky Nun

12:49

isn't group rep also about two things being isomorphic

welp group rep is about $\operatorname{Fun}(G,\Bbb C)^G$ having two bases

yes?

MatheinBoulomenos

isn't $\mathrm{Fun}(G,\Bbb C)^G$ one-dimensional?

$Z(\Bbb C[G])$ has two bases one coming from conjugacy classes, the other from Artin-Wedderburn that gives you that the number of irreps= number of conjugacy classes

Leaky Nun

13:49

great

@MatheinBoulomenos if $\Bbb C[G] = \bigoplus_i V_i$ then $V_i \ncong V_j$?

MatheinBoulomenos

@LeakyNun no

Leaky Nun

ah

MatheinBoulomenos

each irreducible representation $V_i$ is contained in $\Bbb C[G]$ with multiplicity $\mathrm{dim} V_i$

Leaky Nun

yeah I know where my notes went wrong now

oh, the only purpose of taking End is to count?

MatheinBoulomenos

the endomorphisms will just be a $\Bbb C$-vector space so we only care about the dimension, yeah

Leaky Nun

13:52

hey it's a C-algebra also

MatheinBoulomenos

ah true

Leaky Nun

and it tells us that C[G] as C-algebra is isomorphic to $\bigoplus_i \End(V_i)$ also

MatheinBoulomenos

you need to account for multiplicities

$\bigoplus_i \mathrm{End}(V_i^{n_i})$

Leaky Nun

right

MatheinBoulomenos

if you use $\mathrm{End}(V_i^{n_i}) \cong M_{n_i}(\Bbb C)$ you've got the Wedderburn decomposition

Leaky Nun

13:56

and [group rep is] also a bunch of tricks that I cannot remember

MatheinBoulomenos

you might enjoy this answer that I wrote some time ago:

1

A: Can the averaging of a linear arrow of modules over a group algebra be described functorially?

Here is a higher-level description of this operation, mostly from the perspective of Hopf algebras. You'll have to decide yourself how "functorial" this is. Let $R$ be commutative ring and $G$ be a group. (Usually we have $R=\Bbb Z$ in group cohomology and $R$ a field in representation theory.) ...

Leaky Nun

@MatheinBoulomenos I like measure theory / harmonic analysis more :P

MatheinBoulomenos

harmonic analysts probably think differently about rep theory

Leaky Nun

it's just taking orthogonal complement

or integration, idk

probably $l^2$ of a locally compact group

idk I would really love to learn that

connecting pontryagin duality and group rep

MatheinBoulomenos

I only know harmonic analysis for things like ideles or $\mathrm{GL}_2(\Bbb Q_p)$

Leaky Nun

14:05

@MatheinBoulomenos can you teach me :P

MatheinBoulomenos

not right now

Leaky Nun

do you have any reference?

MatheinBoulomenos

14:33

@LeakyNun Tate's thesis is included in Cassels-Fröhlich. For $\mathrm{GL}_2$ over a local field, there's Bushnell-Henniart. Also Bump "Automorphic Forms and Representations" covers both

Leaky Nun

great

@MatheinBoulomenos do you know pontryagin duality?

MatheinBoulomenos

yeah

Leaky Nun

how to prove it?

like briefly

also is there biquadratic reciprocity...

MatheinBoulomenos

haven't looked at the proof in detail

you use Fourier transforms

yes, there's biquadratic reciprocity

Leaky Nun

really

I'm trying to show that if p%20=11 then x^4+3x^2+1=0 has no roots mod p

I've reduced it to showing that if d^2=5 then 2d-6 is not a QR

MatheinBoulomenos

14:53

Okay we have Galois group $\Bbb Z/2\Bbb Z \times \Bbb Z/2 \Bbb Z$, so this the composite of two quadratic extensions. If the polynomials has a root mod p, then it splits completely in the splitting field, hence also in every intermediate subfield

splitting in quadratic fields is well understood

Leaky Nun

actually we can't have that as Galois group

I don't follow at all

MatheinBoulomenos

Magma says that this the Galois group

Leaky Nun

mod p...

MatheinBoulomenos

no I mean over $\Bbb Q$

Leaky Nun

are you using Dedekind-something theorem?

I don't follow what you said at all

sorry

MatheinBoulomenos

15:00

the idea is this: let $f=x^4+3x^2+1$. And $K$ be a splitting field of $f$ over $\Bbb Q$.
then $K$ is the composite of two quadratic extensions.
If for some prime $p$, $f$ has a root mod $p$, then $(p)$ splits completely in $K$

(well actually in $\mathcal O_K$, but that's standard abuse of language)

Leaky Nun

lol I've reduced that case to this case and now you're reducing this case to that case

but ok

MatheinBoulomenos

and if $(p)$ spilts completely in $K$, then it also splits completely in those intermediate quadratic extensions which I'm too lazy to determine right now

Leaky Nun

I know the splitting field

MatheinBoulomenos

what is it?

Leaky Nun

Q(i,sqrt(5))

MatheinBoulomenos

15:02

but that spitting may be described in terms of quadratic residues

Leaky Nun

I think I got it, thanks

MatheinBoulomenos

great, so if $p$ splits complety in $K$; then we have $\left( \frac{-1}{p} \right) = \left( \frac{5}{p} \right) =1$

Leaky Nun

no, I don't got it

I don't understand why splits in K implies splits in quadratic extensions

MatheinBoulomenos

you know ramification index and intertia degree?

Leaky Nun

sure

MatheinBoulomenos

15:04

these are multiplicative in towers and splitting completely is equivalent to both being $1$

Leaky Nun

ah

I'm an idiot

@MatheinBoulomenos where can I learn about hilbert class field?

btw it was fun determining the integers of that field :P

MatheinBoulomenos

your favorite book on CFT

Leaky Nun

what, milne?

MatheinBoulomenos

for example

Leaky Nun

ok

MatheinBoulomenos

15:12

in Neukirch it's chapter VI §7

Leaky Nun

@MatheinBoulomenos why is Q the only unramified extension of Q?

Captain Bohemian

@MikeMiller Is your anti-involution just antiautomorphic involution?

Leaky Nun

15:25

hi @AlessandroCodenotti

Alessandro Codenotti

Hi

Leaky Nun

@AlessandroCodenotti do you want a number theory exercise?

Mike Miller

@CaptainBohemian Yes.

Alessandro Codenotti

No

Leaky Nun

ok

Captain Bohemian

16:00

when it comes to number theory, I just read something about it. Particularly I read [Affinely Extended Real Numbers] (mathworld.wolfram.com/…), which tells me some arithmetic operations involving +∞ and -∞ and 0 are undefined. I have sometimes encountered that kind of arithmetic operation and wonder how to calculate them. But I don't know why this extension is called affinely extended.

MatheinBoulomenos

16:17

@LeakyNun you can use Minkowski's bound to derive a lower bound for the discriminant of any proper extension of Q

the absolute value of the discriminant, that is

and then use that primes dividing the discriminant are ramified

note that this implies via CFT that $\Bbb Z$ is a PID :P

user616128

16:42

This definition isn't the proper definition is it:

https://imgur.com/a/ItqJrEb

(as it is phrased I mean)

Leaky Nun

what's wrong with it?

user616128

Nvm it's fine actually

I was ignoring the word algebraic in algebraic extension like a dum dum

Akiva Weinberger

17:32

Good video if you have an hour

The mathematics of machine learning and deep learning – Sanjeev Arora – ICM2018

►

Leaky Nun

hi @AkivaWeinberger

Akiva Weinberger

19:01

Did you know that a toenail can die, and a new toenail can grow underneath it while the first one is still attached?

Help

Thorgott

19:34

The categories $\mathbf{Mat}$ and $\mathbf{FinVect}$ (each over a field $\mathbb{F}$) are equivalent. There is a contravariant duality functor $(-)^\ast\colon\mathbf{FinVect}\rightarrow\mathbf{FinVect}$ sending vector spaces and linear maps to their duals each and there is a contravariant transpose functor $(-)^t\colon\mathbf{Mat}\rightarrow\mathbf{Mat}$ sending natural numbers to themselves and matrices to their transpose. These two contravariant functors on equivalent categories each are related by the following fact:

user193319

20:13

If $X$ is a connected space, why is the suspension $\Sum X$ simply connected? I've only found proofs which assume that $X$ is path connected, and it makes me wonder whether it is false.

Leaky Nun

@user193319 isn't $\Sigma X$ simply connected no matter what $X$ is?

wait no

Mike Miller

20:43

I doubt that's true for arbitrary connected spaces but I don't have a counterexample in mind

When people talk about simple connectedness, 90% of the time they're assuming the space is decent

Locally contractible eg

Leaky Nun

path-connected, locally path-connected, semi-locally simply connected

Mike Miller

sure, for covering spaces. To do algebraic topology more generally you really want more than just that.

Mike Miller

21:01

I googled a little bit and did not find a counterexample to the claim "if $X$ is connected, then $\Sigma X$ is simply connected"

I still don't think it's true but it probably means people have not thought very hard about this

Alessandro Codenotti

21:31

In general $\kappa\leq 2^{<\kappa}$ but there can an equality right? For example it should be equal for $\kappa=\aleph_\omega$ under GCH

MatheinBoulomenos

let $X$ be the topologists sine curve. This has $\tilde{H}_0(X)=\Bbb Z$, as $X$ has two path components. so by the suspension isomorphism $H_1(\Sigma X)=\tilde{H}_1(\Sigma X)=\tilde{H}_0(X)=\Bbb Z$ which implies that $\Sigma X$ is not simply connected by Hurewicz

note that the suspension isomorphism just follows from Eilenberg-Steenrod, so we don't need any assumptions on the space $X$

Leaky Nun

and you've also given us a stronger theorem!

MatheinBoulomenos

@user193319

Leaky Nun

I've just realized that I know nothing about global fields

MatheinBoulomenos

GCFT is harder than LCFT

user193319

21:42

@MatheinBoulomenos Oh, very nice. Thank you!

Leaky Nun

@MatheinBoulomenos hast du ted gesehen?

MatheinBoulomenos

not this year

Leaky Nun

nein, ich meine her

Mike Miller

@MatheinBoulomenos I feel like a fool for not noticing this

Good call...

Alessandro Codenotti

Is anyone familiar with the concept of "Samuel compactification with respect to a proximity"?

Mike Miller

21:59

noooope

Fargle

Samuel might be

Alessandro Codenotti

It's mentioned on the second page of this thing I was reading

Leaky Nun

Let $M$ be a left $R$-module and $T : M \to M$ linear such that $T^2 = T$

then $M \cong \ker T \oplus \operatorname{im} T$

When is $\operatorname{Hom}_R(M,N) \cong M^\ast \otimes_R N$?

Ultradark

22:41

I'm doing an exercise in a textbook and I need to calculate the effective dimension of a possibly non-Euclidean network lattice. I have $r^d=2r^2+2r+1$ where $d$ is the dimension of the network and I need to count the number of nodes up to $r$ connections away from a given node. I'm trying to figure out if $d$ converges as $r$ goes to infinity.

MatheinBoulomenos

@LeakyNun if $M$ is finitely presented and flat (or equivalently finitely generated projective), then it works

proof sketch: there's always a natural map $M^* \otimes_R N \to \mathrm{Hom}_R(M,N)$ given by $\xi \otimes n \mapsto (m \mapsto \xi(m)n)$. One checks that this is an isomorphism when $M=R^n$. In general, take a finite presentation $R^n \to R^m \to M \to 0$, then apply the two functors $\mathrm{Hom}_R(-,N)$ and $(-)^* \otimes_R N$ and use the result for $R^n$ and the five-lemma

Leaky Nun

22:57

@MatheinBoulomenos what does class field mean?

MatheinBoulomenos

the name is due to this theorem I think:

Principal ideal theorem

In mathematics, the principal ideal theorem of class field theory, a branch of algebraic number theory says that extending ideals gives a mapping on the class group of an algebraic number field to the class group of its Hilbert class field, which sends all ideal classes to the class of a principal ideal. The phenomenon has also been called principalization, or sometimes capitulation. == Formal statement == For any algebraic number field K and any ideal I of the ring of integers of K, if L is the Hilbert class field of K, then I O ...

Ultradark

Would the down voter care to explain their point of view?

Daminark

23:22

Principal ideal theorem? More like, principal nerd theorem!

How's everything going?

MatheinBoulomenos

pretty well, had some nice holidays. and yourself?

Daminark

Same here, modulo some bumps. But yeah kinda ready to get back into things, classes are gonna start Monday

You know what you're gonna be doing next term?

Leaky Nun

man when can I learn L/GCFT properly

TeaFor2

If we have two polynomials f and g with roots a_i and b_i, we can construct a polynomial who's roots are the union of a_i and b_i with f*g or f+g. Is there any easy way to construct a polynomial whose roots are the intersection of a_i and b_i?

Fun problem I heard a bit back, thought you guys might enjoy although I haven't come up with a solution myself yet

MatheinBoulomenos

23:49

just take the gcd

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