Mathematics - 2020-12-03 (page 1 of 2)

Akiva Weinberger

00:15

Consider a set in the plane such that every intersection with a vertical line is at most countable. Can such a set have positive measure?

The set is not required to be closed

Leaky Nun

00:35

@AkivaWeinberger do you have a proof if the set is closed?

porque podemos reducir a este caso gracias al hecho que la medida es regular

Akiva Weinberger

Remind me what a regular measure is

"every measurable set can be approximated from above by open measurable sets and from below by compact measurable sets" - Wiki

I do not know a proof where it's closed

or compact

Leaky Nun

es decir la medida de un conjunto es el supremo de los subconjuntos compactos (que es la condicion relevante a esta situacion)

(estaba buscando la palabra supremo jaja)

pienso que hay una clasficiacion para conjuntos cerrados de R, que no se si se aplique a R^2 tambien

Ted Shifrin

DogAteMy: Sounds like Fubini's Thm, assuming the set is measurable.

Leaky Nun

es que cada conjunto cerrado es la union de un conjunto numerable y un conjunto perfecto

y cada conjunto perfecto es innumerable

Akiva Weinberger

Fubini does sound like the right tool to use

Leaky Nun

00:41

hmm

Akiva Weinberger

I'll need to double-check how nice the sets need to be for that

Leaky Nun

it just needs to be R^2-measurable right

Akiva Weinberger

Arright so lemme tell you why I was thinking about this

> "If you associate to each real x a countable set A(x) ⊆ ℝ then there must be nonequal reals x,y so that x isn't in A(y) and y isn't in A(x)." This is independent of ZFC.

Called "Freiling's axiom of symmetry" apparently (but I haven't looked that up to check)

So thinking geometrically about this

graph A (i.e. $\{(x,y)|y\in A(x)\}$)

Flip $A$ across the diagonal, call that $A^\top$ I guess

$\{(y,x)\mid y\in A(x)\}$

Then this is saying $A\cup A^\top$ never equals $\Bbb R^2$

And I suppose that only can happen if $A$ is nonmeasurable, by Fubini

Oh, $A$ intersect any vertical line is countable

Wikipedia says this is equivalent to the negation of the Continuum Hypothesis (assuming the rest of ZFC)

So Freiling XOR CH

So I guess, metamathematically, CH is a bad axiom iff Freiling is a good one

Ted Shifrin

01:31

Ugh.

copper.hat

01:45

@robjohn Thanks! I have found over the years that many of my attributions are incorrect, so appropriate, pithy quotes are now surrounded by legal disclaimers :-).

Leonhard Euler

02:58

Henlo, anyone here?

Unknown x

03:11

1

Q: Show that $G:[c,d]\to \mathbb R$ defined by $G(x)=\int_0^{f(x)}F(t)dt, x\in [c,d] $ is differentiable and $G'(x)=F\circ f(x) \cdot f'(x).$

Let $F$ be continuous on $[a,b].$ let $f:[c,d]\to \mathbb R$ be differential function satisfy $f([c,d])\subseteq [a,b].$ Show that $G:[c,d]\to \mathbb R$ defined by $G(x)=\int_0^{f(x)}F(t)dt, x\in [c,d] $ is differentiable and $G'(x)=F\circ f(x) \cdot f'(x).$ My attempt $\lim_{h\to 0} \frac{ G(x...

copper.hat

03:33

why are you posting the question here?

@robjohn is the mean green for Christmas?

Mike

03:49

Hello, can anyone take a look at my post regarding orthogonal probability measures?

0

Q: Show that two orthogonal probability measures have the following representations using Vitali covering

Problem: $\alpha$ and $\beta$ are two mutually orthogonal probability measures in the sense that for some Borel subset $E\subset[0,1]$, $\alpha(E)=0$ and $\beta(E)=1$. If $F(x)=\alpha\{[0,x]\}$ and $G(x)=\beta\{[0,x]\}$ for $0<x\leq 1$, show that for any $\varepsilon>0$, there is finite collectio...

real-analysis probability-theory measure-theory solution-verification

123

04:26

Hi All.

In group $G = \{ 1 , -1 , -\iota , -\iota \}$

$O(1) = 1$

$O(-1) = 2$

$O(\iota) = 4$

$O(-\iota) = 4$

Question is that $n$ is the order of $a \in G$ where $n$ is the least positive integer.

How do we apply $a^k = e$ Where $k$ is some integer.

$k = nq + r , 0 \le r < n$ Where $q$ & $r$ integers.

(1) What is $q$ and $r$ as in my example???

(2) Also what is the meaning of $0 \le r < n$. Why we take $r$ value less than $n$ and greater than equal to $0$???

Pls explain these questions with my example.

love_sodam

04:44

I know that if $R$ is a PID, then any submodule of a finitely generated free R-module is free

Can we drop the condition 'finitely generated?'

Leonhard Euler

04:59

beautiful

does someone know anything about the number of ways of representing a number as sums of cubes?

123

Hi @TedShifrin Good Morning.

Leonhard Euler

hello @TedShifrin

hello @123

you new to this room?

123

Hello @LeonhardEuler . Yes i am new in this room.

Leonhard Euler

okay

which field of math you study?

maybe algebra

123

University Math. But now i am focusing on math more deeply rather than computation.

Leonhard Euler

05:09

I study NT

Analytic NT

Want to know about the number of ways of representing a number as sums of squares

123

@LeonhardEuler Great

yes sure definitely.

Leonhard Euler

But I can't find much information

they are not studied very much

123

@LeonhardEuler Pls see my question. If you can answer.

Leonhard Euler

which question?

123

Pls see few comments above.

Leonhard Euler

05:13

@123 this one?

123

Yes this one.

Leonhard Euler

okay let me read it

123

Sure Thanks.

Leonhard Euler

$k=nq+r,\,0\le r<n$ reminds me of division algorithm

it's like we are dividing $k$ by $n$

you can think $q$ to be the quotient

and $r$ to be the remainder

this is what division algorithm says

a very trivial answer is $$q=\lfloor\frac{k}{n}\rfloor$$

123

@LeonhardEuler Yes it is Euclid division algorithm.

Leonhard Euler

05:22

I dunno much about group theory

123

Yes but in group theory, to find an order of element what $q$ and $r$ in my example?

@LeonhardEuler Okay.. no prob.

Leonhard Euler

please ping me if someone gives a proper answer

123

@LeonhardEuler Sure. Yesterday i have discussed with @TedShifrin may today someone gives me answer.

copper.hat

05:37

@123 Your question is very unclear.

123

05:47

Hello @copper.hat

I want to understand theorem for finding the order of element of group with example.

In this theorem we used division algorithm. If $n$ is order of element $^n = e$. Why we use $a^k = e$ Where $k = nq + r$ and $0 \le r < n$.

What is $q$ and $r$ here as in my example for elements in group???

copper.hat

06:12

@123 i don't know what theorem you are referring to or how the division theorem is used.

123

@copper.hat I sahred yesterday. let me tag.

copper.hat

ok, just fyi i'm leaving in a few mins.

123

@TedShifrin Pls the theorem above this comment.

robjohn

@copper.hat See here and here.

copper.hat

:-)

@123 sry, how do i find the theorem you are talking to. i don't have the energy to trawl through the comments looking

Leonhard Euler

06:23

what would be the most important consequences of a formula for primes?

just wondering

Leaky Nun

36

Q: Submodule of free module over a p.i.d. is free even when the module is not finitely generated?

I have heard that any submodule of a free module over a p.i.d. is free. I can prove this for finitely generated modules over a p.i.d. But the proof involves induction on the number of generators, so it does not apply to modules that are not finitely generated. Does the result still hold? What...

abstract-algebra ring-theory modules principal-ideal-domains

Leonhard Euler

ah found it:

10

Q: What would be the immediate implications of a formula for prime numbers?

What would be the immediate implications for Math (or sciences as a general) if someone developed a formula capable of generating every prime number progressively and perfectly, also able to prove (or disprove) the primality of every N-th number. I know this is a very large and subjective answer,...

number-theory prime-numbers philosophy number-systems prime-twins

does someone want to see an exact fromula for the nth prime?

crazy

123

@Measureme Theorem is below this comment.

feynhat

07:13

I have a nonsense question. When we say 'a contravariant functor F is left exact' we mean that given an ES A -> B -> C -> 0, then 0 -> F(C) -> F(B) -> F(A) is exact. Is 3 a special number here? Will this not imply A->B->C->D->0 is ES then 0 -> F(D) -> F(C) -> F(B) -> F(A) is ES.

ES = exact sequence

Alex D

10:46

Hey all, I was wondering whether anyone took Apostol's calculus

Leonhard Euler

11:05

@AlexD I have that book on my laptop but didn't read it

Alex D

Could you check whether he covers infinite series, please?

If so, would you mind sharing on what chapter it is?

Leonhard Euler

Okay it would take some time lemme check

Alex D

Thank you

Leonhard Euler

Tom M. Apostol calculus volume 1 One-Variable Calculus, with an
Introduction to Linear Algebra has a chapter on infinite series

Chapter 10 @AlexD

Alex D

Thank you

Leonhard Euler

11:16

You're welcome

Simone

Hello

Leonhard Euler

Henlo

Yeah I greet everyone

Simone

I like that :-)

I'm trying to solve a problem on implicit differentiation using partial derivatives but I keep getting the same wrong answer, obviously something escapes me

Can't I use the simple division rule we use in single variable calculus with partial derivatives?

Leonhard Euler

What is the problem you are trying to solve?

Simone

I'll Latex it. It'll take me a few minutes

Suppose $f: \mathbb{R}^3 \to \mathbb{R}$ is $\mathcal{C}^2$ and $\frac{\partial f}{\partial z} (\Bbb{a}) \neq 0$, so that $f=0$ defines $z$ implicitly as a $\mathcal{C}^2$ function $\phi$ of $x$ and $y$ near $\Bbb{\bar{a}}$.

$\Bbb{\bar{a}}$ is $\Bbb{a}$ without the third component.

Show that:

Rithaniel

11:39

How about that for a fractal?

Simone

$$ \frac{\partial^2 \phi }{\partial {x}^2 } (\Bbb{\bar{a}}) = - \frac{\frac{\partial^2 f}{\partial z^2} \left(\frac{\partial f}{\partial x}\right)^2 - 2 \frac{\partial^2 f}{\partial x\partial z} \frac{\partial f}{\partial x} \frac{\partial f}{\partial z} +\frac{\partial^2 f}{\partial x^2} \left(\frac{\partial f}{\partial z}\right)^2}{\left(\frac{\partial f}{\partial z}\right)^3} $$

I can't believe I did it (more or less XD)

What I did is use implicit differentiation with partial derivatives on $\phi$ to get:

Oh the partial derivatives on the right are all evaluated at $\Bbb{a}$

with implicit differentiation I get:

$$\frac{\partial \phi}{\partial x} = - \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}}$$

But if I just differentiate a second time using the rule for fractions I get the wrong result. so that's clearly the wrong way to go about it. what is it that escapes me?

Astyx

12:06

If I have a linear operator $T : Z^r \to Z^r$ such that $T^3 = Id$ and 1 is not an eigenvalue of T. How do I show 3 divides the index of the image of I-T ? So far I've shown that 3 divides the norm of all the eigenvalues of I-T and that I-T is diagonalisable if we extend the map to $\Bbb Q[w]^r$ (where w is a primitive third root of unity).

Also it's clear that the eigenvalues come in pairs: the dimension of the eigenspace for w is the same as the one for w^2

Measure me

@123 @123 I don't really see a Theorem.

Thorgott

12:24

@Astyx the index of the image of $I-T$ is precisely $|\det(I-T)|$ as long as that is nonzero (which is the case since $1$ is not an eigenvalue of $T$)

Astyx

Oh yeah

Thanks!

robjohn

12:35

@LeonhardEuler That looks to be based on $$\pi(n)=\sum_{k=2}^n\left\lfloor\cos^2\left(\pi\frac{(k-1)!+1}k\right)\right\rfloor$$

Astyx

12:45

@EdwardEvans Re this: I'm not using the fact that B is a DVR am I?

Edward Evans

I'm just thinking about how it works with the valuation

err

Astyx

Maybe it's a superfluous hypothesis

Edward Evans

Yeah I don't think you need that B is a DVR, it's just any old ring between A and Frac A I think

Astyx

It's just that $Frac(A) = A[p^{-1}]$ where $v(p) =1$

Edward Evans

13:02

what about smth stupid like Frac(A) = A[X]/(pX - 1) lmao

I gotta go listen to someone talk about L-functions now

Thorgott

what's stupid about that, that's how you localize at an element smh

Edward Evans

No Frac(A) is formed by writing A x 1/A

even for 0

Thorgott

@EdwardEvans direct product, right

Edward Evans

16:35

@LeonhardEuler giphy.com/gifs/cauliflower-xasZeFOhWS5cQ this is smth for you

uh

Astyx

How do I show that there are an inifnite number of primes of Z that totally split in $Q[\sqrt{d}]$ ?

I've shown that for every nonconstant polynomial P, there's an infinite number of primes p such that there exists an n such that p|P(n)

Edward Evans

uhh

I can't remember

Astyx

In other words I want to show that the decomposition group is trivial

supinf

@Simone I haven't read everything, but since you already did some work yourself, this might be a good question for MSE.

Simone

I did precious little I'm afraid

I'm only confused as to how I should go about to do a second order partial derivative of phi again

Why can't I use the calculus 1 division rule (converted to partial derivatives)?

Even if I use the more rigorous approach of the Jacobian matrix and the differentiation rules on those I get the same (wrong) result.

I think I haven't yet digested partial differentiation very well

supinf

16:48

If you include your attempt at calculating the second derivative, then it might be a good question on the site, just saying.

Simone

I just tried the division rule, getting:
$$\frac{\partial^2 \phi}{\partial x^2} = (-1) \left( \frac{\frac{\partial f}{\partial z} \frac{\partial^2 f}{\partial x^2} - \frac{\partial f}{\partial x} \frac{\partial^2 f}{\partial x \partial z}}{ \left( \frac{\partial f}{\partial z} \right)^2} \right)$$

Which is wrong.

thanks for the consideration btw, you guys always help.

I also tried some reverse engineering on the solution, and I saw that what I got is a part of the solution, but that didn't give me an epiphany.

supinf

17:07

@Simone I didn't do anything to help you :D

Simone

then at least give some moral support damn you!! XD

jk

Rithaniel

What about amoral support?

supinf

Fine I'll have a look at it

Simone

sure, or immoral support... if you meet me at the back of the alley...

@supinf thank you.

Thorgott

@Simone What you're missing is the points at which you actually take those derivatives. The complete formula reads $\frac{\partial\phi}{\partial x}(x,y)=-\frac{\frac{\partial f}{\partial x}(x,y,\phi(x,y))}{\frac{\partial f}{\partial z}(x,y,\phi(x,y))}$, so if you differentiate again, you don't only have to use the quotient rule, you also have to use the chain rule and the Jacobian of $(x,y)\mapsto(x,y,\phi(x,y))$ will come into play

Simone

17:15

riiiiiiight.

I didn't think of that

that's why I didn't ask this on MSE, I knew I was being stupid

Thank you Thorgott.

supinf

ok someone was faster than me :)

Thorgott

np

Sayan Chattopadhyay

18:48

Hello @MikeMiller

How's it going?

user2103480

btw @SayanChattopadhyay

we will introduce the poisson bracket and liouville operator

so I guess symplectic stuff isn't that far after all

Sayan Chattopadhyay

Oo awesome! What was the course specifically on again?

user193319

$M$ is a semisimple means it is a direct sum of simple submodules. I am trying to prove that the direct sum decomposition is unique up to isomorphism and permutation. That is, if $N_1 \oplus ... \oplus N_{t} = M = M_1 \oplus ... \oplus M_{s}$, I want to show that $t=s$ and there exists a permutation $\tau \in S_t$ such that $N_i \cong M_{\tau (i)}$.

I'm not sure where to begin.

user2103480

@SayanChattopadhyay stochastic processes in fluids

so the liouville operator will determine the time-evolution of a stochastic process, if I understand it correctly

Sayan Chattopadhyay

Oh nice, maybe a naive question. But in such cases is the formalism more Lagrangian dependent or Hamiltonian? I have a feeling Lagrangian because there's more of a hands on approach, dealing with E-L equations and so on?

user2103480

19:03

hamiltonian it is

I think

Thorgott

@user193319 try induction

user2103480

We're always dealing with the hamiltonian, at least

user193319

@Thorgott On either $s$ or $t$?

Thorgott

ye

user2103480

we'll then be interested in "slow" observables, which apparently correspond to conservation laws, and "fast" observables, which are hard to measure and classified as noise

Sayan Chattopadhyay

19:05

@user2103480 Yeah you would get $iL = \{\cdot,H\}$, where the right hand guy is the hamiltonian vector field

@user2103480 Oh I have never heard of this

user2103480

so we calculate around with correlation functions, determine eigenvalues and resolvent of the liouville operator

and then we go on to this stuff:

Mori-Zwanzig formalism

The Mori-Zwanzig formalism, named after the physicists Hajime Mori and Robert Zwanzig, is a method of Statistical Physics. It allows to split the dynamics of a system in a relevant and an irrelevant part using projection operators, which allows to find closed equations of motion for the relevant part. It is used e.g. in fluid mechanics or condensed matter physics. == Idea == Macroscopic systems with a large number of microscopic degrees of freedom are usually well described by a small number of relevant variables, e.g. the magnetization in a system of spins. The Mori-Zwanzig formalism allows to...

"Mori-Zwanzig projection operator formalism"

Sayan Chattopadhyay

Dude this stuff is so fascinating, it's like you're doing some kind of decomposition on the algebra of observables

user2103480

Yeah I hope I will be able to appreciate that and not get lost in the details

at quite a few points I get lost since I don't know the underlying physics, but atm it's all about correlation functions so that's OK

Sayan Chattopadhyay

So is like stat mech is a prerequisite to this course?

user2103480

Yes and No

I was told it's fine if I either took advanced stat mech or a course on stochastic analysis

(technically I took neither but I'm fine on the stochastic analysis since we don't need anything like girsanov)

But the course is, at the moment, still based on a previous run of a similar course

Meaning that the sheets and the lecture notes are identical

Sayan Chattopadhyay

19:15

I see

user2103480

And this time we're only mathematicians. He mentioned that it's unfortunate, since in the last run, there were many physicists, and the exercises/material/notation are accordingly more physics-ish and with more assumptions implicit than I can follow

Sayan Chattopadhyay

Oh that's bad. I always wanted to take a course on stat mech, but the major system in my university puts so much workload on you, I could never sit do the prerequisites and go do a course in stat mech from the physics department

And we do not have anyone who takes a "mathematical" course on stat mech so that it can be offered as an elective or something

user2103480

I still hope that the physics leaves a trace of intuition for what's happening in these and those dynamical systems at the end of the semester. I've noticed that uni and studying for exams, then forgetting it all, made me compartmentalize a bit too much of the knowledge

And that needs to be broken up; I think "applied" courses can really help there

@SayanChattopadhyay And the intersection with other courses isn't large enough to learn stat mech while studying something else?

@SayanChattopadhyay How would a mathematical stat mech course look like?

Sayan Chattopadhyay

@user2103480 Yeah for instance let me tell you what I have to take this upcoming semester. I have Enumerative Geometry, Riemannian Geometry, Geometric Group theory, Lie Groups and its representations, Knots and Braids, Homology and Cohomology, and maybe Algebraic Groups. With all this and a master thesis to think of and write (which is going to be something "applied" as of now, I think), I do not know where I can find the time to do some physics

user2103480

wtf

that's too much to learn properly in one semester for a normal person

Sayan Chattopadhyay

19:30

Exactly, I am thinking I will drop one course, that is the max that I can drop, or they change something in the next semester because of all these online shenanigans

Thorgott

??????

user2103480

smartest to drop would probably be either enumerative geometry or riemannian geometry?

I'd think these are the least interconnected with the rest

Sayan Chattopadhyay

Nah those are the ones I badly want to take, the enumerative course is very rarely offered, and what I maybe want to do for my thesis uses some sort of enumerative geometry, I believe. I am thinking of dropping either Algebraic groups, or GGT.

user2103480

sad @AlessandroCodenotti noises

anyways that's a horrifying workload

Sayan Chattopadhyay

I am expecting the Lie groups course to be an easy one, Homology Cohomology should be manageable since I have seen some of it before. I have no clue what Knots and Braids is going to be like, and the Enumerative Geometry course has a weird ass course structure

user2103480

19:36

I'm at my limit with 3 courses, each 1 lecture per week, and one 2-lecture course

although 2 of the courses with 1 lecture move at a pretty fast pace

Tobias Kildetoft

@SayanChattopadhyay Don't drop algebraic groups. I never get to talk about algebraic groups in here otherwise :)

Sayan Chattopadhyay

If someone can tell me what exactly is happening with that course, it would be great. Here is what they plan to do:
Examples of counting problems in geometry: Grassmanians and intersections. Chasles problems on conics. Fixed points of transformations. Bezout’s theorem and 9-point circles.
How to “count properly”: Transversal intersections. Fundamental theorem of algebra and multiplicity.
Grassmanians and Projective space: Cell decompositions. Schubert cells. Schubert calculus.
Moving and blowing-up: Resultants and plane curve intersections. Singular intersections by moving. Singular interse…

user2103480

ouch

Sayan Chattopadhyay

@TobiasKildetoft Lol, well I am pissing someone off in the chat anyway, its either Balarka, Alessandro or you :p

Thorgott

ah yes, the formal grothendieck-riemann-roch theorem

smells like dried tears in a corner

user2103480

19:40

@Thorgott lmao yeh

Sayan Chattopadhyay

lmao

The "references" are, Milnor's Characteristic Classes, and Fulton's Intersection Theory, with some random K-theory book. These are hardly references, they are damn mountains

user2103480

Thank god I don't do all that stuff. Yesterday I needed to show convergence of an infinite product by first completing squares, then taking the logarithm and doing the full log(1-x) taylor expansion, and afterwards bounding by a geometric series. Felt nice not to do some abstract shite

Tobias Kildetoft

Probably a course on algebraic groups will never get to the stuff I am mainly interested in anyway (i.e. their representation theory in positive characteristic)

Ted Shifrin

@Sayan Beautiful course!

Classic stuff motivating modern technology.

Sayan Chattopadhyay

@Ted, help me understand, what does the course plan to do? I couldn't find any direct references and things seem very scattered

user2103480

19:44

A friend of mine is interested in algebraic combinatorics and schubert calculus. She doesn't like topology and geometry, but her advisor makes her do equivariant cohomology of the grassmannian and k-theory

You can go into the dried-tears-corner with her

Sayan Chattopadhyay

Lol, after this course, I hope to read Sheldon Katz's Enumerative Geometry and String Theory, maybe I can already die.

Tobias Kildetoft

Schubert calculus: arxiv.org/abs/1309.5055

Sayan Chattopadhyay

@TobiasKildetoft About this, yeah I do not think there is any representation theory. The course ends at Borel's fixed point theorem

Tobias Kildetoft

Yeah, that is pretty inevitable. It takes a lot of prerequisites to really get anywhere

Ted Shifrin

My favorite exercise for you is: How many lines meet four lines in general position in $\Bbb P^3$?

Sayan Chattopadhyay

19:50

By general position, do you mean no three of them meet at the same point?

Ted Shifrin

You can do this completely classically, so that a high school student could follow, or you can learn Schubert calculus and tge cohomology ring of the Grassmannian.

No, pairwise skew.

Sayan Chattopadhyay

Oh okay

Ted Shifrin

Side exercise: Classify the (smooth] doubly-ruled surfaces in $\Bbb P^3$.

Sayan Chattopadhyay

Nice, let me give them a shot

Alessandro Codenotti

@SayanChattopadhyay but GGT is so good

Sayan Chattopadhyay

19:57

You guys are going to mess my entire thing up lol, maybe now I will drop Lie groups and take GGT.

Alessandro Codenotti

I went to two different seminars today and they were both about hyperbolic group even though one was supposed to be a model theory seminar (I'm not complaining though)

user193319

@Thorgott For the inductive step, assuming that I have $N_1 \oplus ... \oplus N_{t-1} \oplus N_t = M_1 \oplus ... \oplus M_s$, should I mod out by $N_t$ or something like that? I'm not sure to use the inductive hypothesis.

user2103480

@SayanChattopadhyay take a course on spde as well, they're useful in quantum field theory!

take an entire master's workload in one semester

Sayan Chattopadhyay

Yup, anyway the world's going to shit, so why not finish a master's in a semester

Thorgott

20:15

yeah, I'd try something like that

Alessandro Codenotti

Anyone want to think about some ~~atrocious~~ interesting point set topology?

user193319

Hmm...The LHS will be isomorphic to $N_1 \oplus ... \oplus N_{t-1}$, but I don't know what the RHS is...

The $N_t$ can't be fully contained in any of the $M_i$'s, since they are simple modules...

Thorgott

@AlessandroCodenotti I'll say yes and you better make me regret it

Alessandro Codenotti

Do you know what the Vietoris topology on the space of closed sets of a topological space is?

Thorgott

no, but I'm pleasantly surprised to hear that there is something else named after Vietoris

Alessandro Codenotti

20:23

lol

So you take some space $X$ and $F(X)$ is the set of closed subsets of $X$. We put a topology on $F(X)$ with basis $\langle U_1,\ldots,U_n\rangle=\{F\in F(X)\mid F\subseteq\bigcup U_i\land F\cap U_j\neq\varnothing \text{ for all $j$ }\}$ where $U_i$ range over open sets in $X$

Thorgott

did you know that Vietoris was one of, if not the, oldest mathematician ever

Alessandro Codenotti

yep, oldest Austrian ever

Almost 111 years

user193319

@Thorgott Is it possible that there is some $i$ and $j$ such that $N_i$ and $M_j$ are equal? Then could I mod out by this common factor?

Edward Evans

@AlessandroCodenotti eleventy one*

user2103480

^

@EdwardEvans or one-hundred-onety-one

Edward Evans

20:28

onety onety one

user2103480

ah yeah

Edward Evans

$\ell\ell\ell$

Juan Jandrad y Juanti Juan

user2103480

wait

Edward Evans

idk i'm bored

user2103480

isnt it onety-onety onety-one

Edward Evans

20:29

It depends

Thorgott

@AlessandroCodenotti Ok, if we have two collections $\{U_i\}$ and $\{V_j\}$, then every closed set in $\langle U_i\rangle\cap\langle V_j\rangle$ also belongs to $\langle U_i\cap V_j\rangle$ for some subcollection of indices (maybe omitting index range was a bad notational choice), so they form a basis, alright

Alessandro Codenotti

yeah

So this thing I'm reading claims that if you take a chain $Y$ of closed subsets of $X$, meaning that for all $E,F\in Y$ either $E\subseteq F$ or $F\subseteq E$, then the closure of $Y$ in $F(X)$ is still a chain

Which seems very reasonable, but I'm not seeing a formal argument

Oh here $X$ is even compact hausdorff just to be safe

random general facts: $K(X)$, which is the subspace of $F(X)$ of compact sets, tends to inherit many properties of $X$, metrizability (if $X$ is compact metrizable then the Hausdorff metric induces the vietoris topology), connectness, being Polish, zero dimensional, separation axioms etc.

user193319

20:52

Easier way: Jordan-Holder theorem.

Question: What does it mean to say $k^n$ is the unique simple module of $M_n(k)$?

Thorgott

21:03

@AlessandroCodenotti I think every ascending sequence of closed subsets converges to its union. Without further contemplation, I'd like to claim the same thing holds for nets. Adding unions of elements of a chain to it doesn't make it not a chain, clearly.

If $C_n$ is a decreasing sequence of closed sets and $\bigcap C_n\subseteq U$, is $C_n\subseteq U$ for large enough $n$?

smells like some compact Hausdorff separation trick

supinf

@AlessandroCodenotti what is $F(X)$?

Thorgott

closed subsets of $X$

Alessandro Codenotti

@Thorgott But the union is not necessarily closed

supinf

thanks

Alessandro Codenotti

Since $X$ is compact Hausdorff convergence of sequences here is the same as convergence in the topological liminf=topological limsup sense (points whose neighbourhoods meet cofinitely many/infinitely many points of the sequence)

Thorgott

21:11

oops, that's a fair point

that makes it more complicated

@AlessandroCodenotti are those closed sets necessarily?

Alessandro Codenotti

yes

Lukas Heger

@user193319 it just means what is say: $k^n$ is a simple module over $M_n(k)$ and it's the only one

but more is actually true: $M_n(k)$ is semisimple, so because $k^n$ is the unique simple module, every module over $M_n(k)$ is just $\bigoplus_{I} k^n$ for some index set $I$

Thorgott

ok, then all left to do is convince yourself that adding those to the chain doesn't destroy chain-ness, no?

(and pray that replacing "sequence" with "net" doesn't change anything)

Alessandro Codenotti

yeah I can trust that nets won't mess this up :P

Being in the closure of $Y$, for a closed $F$, should mean something like "if you cover $F$ with finitely many open sets, then you necessarily also cover an element of $Y$"

(this is not accurate, but that's my possibly misleading intuition)

Thorgott

why is the topological limsup=liminf of an ascending sequence of closed sets not just the closure of their union?

meeting infinitely many of them <=> meeting cofinitely many of them <=> meeting any of them <=> meeting their union

and clearly adding closures of union would keep it being a chain

@user193319 That works too, of course. For the record, what I had in mind was observe that the $N_t$ on the LHS corresponds to some submodule on the RHS, which necessarily has to look like (up to isomorphism) some partial diagonal. Then quotienting also drops only one factor on the RHS and induction does the rest.

user193319

21:31

Ah, I see. That's what I was thinking would happen! Thanks for that!

Amin Idelhaj

21:48

Eyy @lukas what's up

Lukas Heger

Hey @Amin

Ted Shifrin

Howdy, Demonark.

user193319

How does one show that any simple module over the nxn upper triangular matrices (over a field) is one dimensional?

Amin Idelhaj

Hey Ted!

How are you guys doing?

Ted Shifrin

Hassling with semi-incompetence which may or may have nothing to do with the covid mess.

user193319

21:50

I don't know much about the nxn upper triangular matrices. Jacobson radical (all strictly upper triangular matrices), I know it has a nice direct sum decomposition...hmmm...I think that's about it.

@LukasHeger I'm guessing 1 dimensional as a vector space.

Lukas Heger

oops I didn't read the upper triangular thing

user193319

That's okay. It happens (to me all the time).

Amin Idelhaj

Rip, that's no good Ted

Lukas Heger

@user193319 for any ring $A$, simple modules over $A$ are the same as simple modules over $A/J(A)$, where $J(A)$ is the Jacobson radical

user193319

Wait, is $A/J(A)$ always guaranteed to be a simple module?

Lukas Heger

21:53

no

$A/J(A)$ is another ring

user193319

But it has an $A$-module structure.

Lukas Heger

I'm saying that simple modules over $A$ are the same as simple modules over $A/J(A)$

user193319

Oh, I see. Hmm...I wonder if I have that theorem available. Let me check my notes.

Amin Idelhaj

Also that's a nice starboard we've got going

Lukas Heger

for $A$ the ring of $n \times n$ upper tringular matrices, then $A/J(A) \cong k^n$ (as a ring)

for $k^n$ it's easy to show that all simple modules are one-dimensional

Amin Idelhaj

21:55

@Ted I think you might appreciate this, in my most recent meeting with my advisor some hyperbolic stuff came up that I wasn't totally sure about, and he's like yeah your hyperbolic geometry is rusty

Lukas Heger

@user193319 it just says that $J(A)$ acts trivially on each simple module

Amin Idelhaj

And then I'm like :0 that's giving a bit too much credit to me lol

Also AMS had a sale a couple days ago and I bought Helgason!

Ted Shifrin

Helgason is rather dense, but well-written

I no longer possess him, of course

user193319

@LukasHeger When you say simple modules over $A$ are the same as simple modules over $A/J(A)$, what exactly do you mean by "same as"?

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