Mathematics - 2024-01-10

Tanner Swett

00:28

Had to brag a little bit: after being on the math site for about 12 years, I've finally hit 10k reputation 🎉

Ted Shifrin

Congratulations!

@copper will send you a trophy. I’m still waiting for mine at 100K.

Tanner Swett

Thanks 😁

copper.hat

:-). I am still working towards my orange mse jumpsuit

Ted Shifrin

00:51

How’s the recovery, copper?

leslie townes

$ellouts.

Thorgott

01:14

@Jakobian they know what it means, but they lack the formalism to actually express it

Jakobian

01:33

@Thorgott probably not by name

Obliv

How do I calculate the probability of an event occurring if I'm given $P(X) = 1/30000$ and the number of "rolls" as 5? they're separate events

If I flipped a coin $n$ times I can simply get the number of outcomes $2^n$ and subtract by the outcomes I don't want

If for example I flipped a coin 3 times and wanted to calculate how probable just getting one heads outcome would be, I take $3^2 - 1 = 8$

Oh wait I meant to write $n^2$ not $2^n$ before

Obliv

01:59

nvm i got it, it's around 1/28k

one potato two potato

02:27

If $X$ is a (smooth) manifold then its interior $X^\circ$ and $X$ are homotopy equivalent?

Thorgott

yes, use a collar neighborhood

one potato two potato

ah right. thanks

there is a certain statement that some family of manifolds is always homeomorphic to the interior of some compact manifold so I thought then that that family of manifolds always has f.g. $\pi_1$ and was surprised but I re-read the statement and found that the statement actually assumes "some family of manifolds with f.g. $\pi_1$"...

copper.hat

03:40

@TedShifrin good progress. can 'walk' (think Frankenstein) without a cane now. some strange twinges from time to time, but par for the course, I gather. Thanks for asking.

Ted Shifrin

@copper.hat Great to hear.

copper.hat

During my PT session yesterday, she said "The notes here say you are impatient and stubborn"...

leslie townes

sir, these 'strange twinges' as you call them are what your kind might know better as 'the shakes,' and we can't treat that here

go across the street to the starry plough, they have a therapist on site from 6am onward

copper.hat

i have 26 percocet remaining. unfortunately (or fortunately) they did nothing for me.

thankfully i can consume alcohol now.

Ted Shifrin

@copper.hat Leslie and I ratted you out!

copper.hat

03:53

:-)

Ted Shifrin

I shudder to think what my PT’s notes say. I am not sxercising much these days — first the cataract surgery, now yesterday’s sinus lift.

Akiva Weinberger

0

Q: Why must $n-\lfloor\frac n2\rfloor+\lfloor\frac n3\rfloor-\dotsb$ grow like $n\ln2$?

Let $$a(n):=n-\left\lfloor\frac n2\right\rfloor+\left\lfloor\frac n3\right\rfloor-\left\lfloor\frac n4\right\rfloor+\left\lfloor\frac n5\right\rfloor-\dotsb.$$ Note that this is a finite sum. Naïvely, $$a(n)\approx n-\frac n2+\frac n3-\frac n4+\frac n5-\dotsb=n\ln2,$$ and so in particular \begin{...

@copper.hat Ooh, I had a cane for a bit in high school

Ted Shifrin

HNY, DogAteMy!

Akiva Weinberger

You hurt your leg? I hope you feel better!

@TedShifrin Happy 2024!

(Excitement or factorial? You decide)

copper.hat

i was caned in high school (well, secondary school)

Akiva Weinberger

03:56

Very different!

copper.hat

@AkivaWeinberger i had a hip replaced so i can do Brasilian Samba

ok, the latter part is not quite true

Akiva Weinberger

The hip was inside you all along

- moral of a movie about hip-hop

copper.hat

not sure where my left hip is now. hopefully not dog food...

Akiva Weinberger

So you got a replacement that is "following the latest fashion, especially in popular music and clothes"

copper.hat

a new titanium model apparently

with some sort of tracking device in case i lose it

Akiva Weinberger

04:01

@copper.hat Stylish… That is a hip replacement

copper.hat

a hip hip replacement maybe

Ted Shifrin

04:22

Hip hip hooray?

copper.hat

04:40

my sister sent me a birthday card with the front matter: hip hip [1] hooray or [2] replacement

what does this notation mean $f_n \overset{\ast}{\rightharpoonup} \bar f$ ? in the context of, say, $L^\infty$ ?

pointwise? weak?

or Alaoglu

ignore, i found it, weak *.

why the funky arrow? what is wrong with $\to$ with an asterisk above?

Ted Shifrin

What else could it be? 🤷‍♂️

copper.hat

that was my question, i suppose, i was wondering if it was different to $f_n \overset{\ast}{\to} \bar f$

@robjohn Why the calendar on your avatar?

robjohn

05:08

@copper.hat It is the new year

I did the same thing last year with Jan 2023

copper.hat

:-) as long as its not measles

robjohn

@copper.hat monthly measles

copper.hat

:-)

Ted Shifrin

05:23

Ideas?, @robjohn?

nickbros123

06:07

theres a little similarity between "span" of a collection of vectors belonging to a vector space V, and the "smallest" $\sigma$-field that contains a collection of subsets of $\Omega$, $A \subset P(\Omega)$, well the similarity is the smallest "something" that contains a given collection. There is the equivalence between span and set of all Linear combinations (for countable dimensioned v-spaces), is there a similar analog for $\sigma$ fields?

leslie townes

06:21

the span of a subset of a vector space (without any other structure) is usually defined in terms of finite linear combinations, irrespective of the dimension of the vector space or the size of the subset. the generalizations of 'span' that you often see in infinite dimensional spaces often involve structure beyond vector space structure (e.g. "infinite linear combinations" are usually interpreted in terms of limits of finite linear combinations in some topology).

you can give a 'bottom-up' description of the sigma algebra generated by a set of subsets, but it isn't as nice, because the sigma-algebra operations (unlike the formation of a linear algebraic span) involve 'genuinely infinite' operations (using quotes here to indicate that i intend no precise definition of this term, i just mean to distinguish from taking finite linear combinations). the usual bottom-up description is via a transfinite inductive process.

5

Q: Constructively generating a sigma algebra

We have a collection $\mathcal{C}$ of sets (includes $\Omega)$ and would like to constructively generate the sigma algebra $\sigma(\mathcal{C})$. Would the following process work? Let $\mathcal{S}=\mathcal{C}$ 1.Take the complement of each set in $\mathcal{S}$ and add it to $\mathcal{S}$ 2.Take...

measure-theory set-theory

nickbros123

ah, i see

thanks for the link

leslie townes

you can already see this complexity when trying to describe the sigma algebra generated by the open intervals in R. starting with intervals, each step in the process of "take countable unions of intervals," "take complements of sets like those just mentioned," "then take countable unions of sets like those just mentioned," etc. adds more sets. it doesn't 'stop at one step,' as "take finite linear combinations of things that are finite linear combinations of elements of S" does

Borel hierarchy

In mathematical logic, the Borel hierarchy is a stratification of the Borel algebra generated by the open subsets of a Polish space; elements of this algebra are called Borel sets. Each Borel set is assigned a unique countable ordinal number called the rank of the Borel set. The Borel hierarchy is of particular interest in descriptive set theory. One common use of the Borel hierarchy is to prove facts about the Borel sets using transfinite induction on rank. Properties of sets of small finite ranks are important in measure theory and analysis. == Borel sets == The Borel algebra in an arbitrary...

but, very similar idea, maybe at some level the same idea, just with a different level of complexity due to what you require of a sigma-algebra vs. what you require of a vector space.

a book on 'general algebra' or 'universal algebra' might include galaxy brain definitions of which both things are a speical case

one potato two potato

11:10

If $X$ is a topological manifold such that $\pi_1(X)\simeq G\times H$ for some nontrivial groups $G$ and $H$, then can I always find topological manifolds $Y$ and $Z$ such that $X\cong Y\times Z$ and $\pi_1(Y)\simeq G$ and $\pi_1(Z)\simeq H$?

Jakobian

13:56

@leslietownes its true that generating subalgebras requires only finiteary amount of operations, but that doesn't imply this case is equally as simple as vector spaces in general.

Change finiteary to, each element requires finite amount of operations

Furthermore, even for vector spaces, its required that we apply $n$th step for each $n$

So the case of vector spaces just has an easy description, where we can proceed with iterating some process first, and then another one

Instead of applying them all at the same time

If we think of it as a tree, then vector spaces have a subtree which works equally well

But sigma-algebras don't have this luxury

Mad

14:15

in order to prove lies third theorem, which states a fintie dimensioanl real lie algebra g there iexists a simply connected lie group G whos lie algebra is isomorphic to g, one usually quotes Ados theorem, which states there exists a group G with isomorphic lie algebra, and then replaces G with the universal covering, however, the universal covering is only simply connected if G is connected, does Ado theorem guarantee G connectedness

Jakobian

If you look at the process of generating sub-magmas, you'll see that this generating process looks equally as complicated, or maybe rather, hard to work with, as the case of sigma-algebras

Jakobian

14:48

actually I'm not sure if I even agree with my statement since the tree for magmas is very simple, and the one for sigma-algebras is simple as well

Ted Shifrin

14:59

@Mad if not, take the identity component. That’s all that the Lie algebra determines, anyhow.

leslie townes

17:14

jakobian: mentally insert my qualifier "using quotes here to indicate that i intend no precise definition of this term" everywhere i use quotes in the discussion above, and not just where i said that, and you'll have the spirit i was talking in :)

certainly wasn't expressing any formal idea about magmas, anyway

Ted Shifrin

Volcanos, anyone?

leslie townes

i might prefer being thrown into a volcano to studying magmas

Mats Granvik

(*start*)
nn = 15;
p = 2;
f = x^p + y^p - z^p;
g[n_] := DivisorSum[n, MoebiusMu[#] # &]; Monitor[
Table[Sum[
Sum[Sum[Sum[If[GCD[f, n] == k, 1, 0]*g[k], {x, 1, n}], {y, 1,
n}], {z, 1, n}], {k, 1, n}], {n, 1, nn}], n]
(*end*)
Output:
{1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
g[n_] is the Dirichlet inverse of the Euler totient function.
I suspect that p=1 and p=2 are the only values for which the Fermat last theorem equation outputs the charachteristic sequence of 1.

Sine of the Time

youtube.com/shorts/r3abfhlvIw8?si=x7Vim2AHdVUYUnEu

leslie

Mats Granvik

oeis.org/A368197

Jakobian

17:32

@leslietownes Surely, there's no such thing as studying magmas since its too general of a concept

maybe you could study certain class of magmas

Soumik Mukherjee

@SineoftheTime do you play chess?

Jakobian

sas.rochester.edu/mth/undergraduate/honorspaperspdfs/…

hmm interesting

I guess its sort of like geometric group theory?

leslie townes

dunno, i'm on my way to the volcano now

Jakobian

must be cozy and warm

here its -7 C

leslie townes

goodbye, chilly world

Sine of the Time

17:37

@SoumikMukherjee yes

you?

Soumik Mukherjee

Yes

Sine of the Time

what's your online rating?

Thorgott

@onepotatotwopotato this should be very far from true

for example, splitting as a product also implies that any higher homotopy groups split as products

Soumik Mukherjee

@SineoftheTime Around 2200 on lichess

Sine of the Time

wow nice

Soumik Mukherjee

17:39

Your?

Sine of the Time

between 2300 and 2400 on lichess

Thorgott

if $X$ is a top. space and $f\colon X\rightarrow X$ a homeomorphism, you can form the mapping torus $T(f)$ and obtain a fiber bundle $X\rightarrow T(f)\rightarrow S^1$

Soumik Mukherjee

@SineoftheTime Nice!

Thorgott

assume $X$ connected, then the LES implies that $\pi_2(T(f))\cong\pi_2(X)$ and there is a SES $0\rightarrow\pi_1(X)\rightarrow\pi_1(T(f))\rightarrow\pi_1(S^1)\rightarrow0$

Soumik Mukherjee

In blitz or bullet?

Sine of the Time

17:40

bullet

I'm addicted and I play only one minute games

I used to play otb, FIDE rating 1900

Soumik Mukherjee

Oh, my highest rating in bullet was 2277

@SineoftheTime nice

robjohn

18:20

@TedShifrin The first thing that came to mind was Vandermonde. Considering $\frac{\mathrm{d}^n}{\mathrm{d}x^n}x^n(1+x)^n$ is not simpler.

unless I'm missing something

Ted Shifrin

18:38

@robjohn No, it's not simpler, but I don't see how to make it work. :(

@Thorgott An immediate counterexample comes to mind for homology, but not for $\pi_1$.

Thorgott

I got distracted from what I was writing earlier. I believe the idea works, but I now realize there's some subtleties I overlooked.

Ted Shifrin

18:56

This question makes me realize I don't know nearly as much about fundamental groups as I do about homology/cohomology. Since I'm far from a topologist, I suppose this isn't too surprising.

Unfortunately, I expect that when $G$ is a product, a $K(G,1)$ will naturally be a product.

Thorgott

yeah, Eilenberg-MacLane spaces are compatible with products (also for higher $n$)

Alessandro Codenotti

@Thorgott @Ted given a group $G$ if we do the usual construction of a $2$-dimensional complex with $\pi_1=G$, embed in $\Bbb R^5$ and take a regular nbhd to have a manifold, there's no way this thing is a product in general. But how to produce a specific counterexample is unclear

Ted Shifrin

Let's start with the homotopy version of the Klein bottle :) What is a non-product space with $\pi_1 = \Bbb Z\oplus \Bbb Z_2$?

Or the Klein 4-group, if that's any easier ...

Alessandro Codenotti

19:20

@SineoftheTime @SoumikMukherjee damn you guys are way better than me haha

Ted Shifrin

Maybe you're not quite demonic enough, Demonic Alessandro.

Thorgott

19:40

ok, I'm tripping over silly stuff

if I have two indecomposable groups $G,H$ is the way of writing $G\times H$ as a product unique?

Soumik Mukherjee

@AlessandroCodenotti I feel the same way about you (and most people here in chat) in math:)

Thorgott

ah wait yeah, this is Krull-Schmidt

Thorgott

20:04

god this problem is a lot more annoying than I want it to be

Thorgott

20:26

ok, I think I got it

Alessandro Codenotti

@SoumikMukherjee chess is harder than maths

leslie townes

there should be blitz and lightning variants of math where you have to balance accuracy against speed

Thorgott

nevermind, I don't got it

Mad

i am having an issue understanding this, for starters, it seems like the exponential map is being differentiated like it is a function with codomain of Reals. I just read, that the differential of it is the identity, should not the differential here give i * t *s `

And in the end, it is not clear what the author means, plugging in the one, and what is exactly trying to be said?

Context: One parameter group is the exponential function.

Thorgott

20:44

the exponential map is defined so that if $\varphi$ is the unique one-parameter subgroup satisfying $\varphi^{\prime}(0)=v$, then $\varphi(1)=\exp(v)$, that's where the 1 comes from

and for the differentiation, you identify $T_1S^1$ as the subspace $i\mathbb{R}\subseteq\mathbb{C}$ via the inclusion $S^1\subseteq\mathbb{C}$

leslie townes

i'm not 100% sure of the confusion here, it might be in the definitions, do you understand "we can identify g with T_1 S^1 = iR"

and under that identification can you write out whatever your definition of differentiation would give you for that map

it might or might not help to draw a picture in which tangent vectors to the circle at 1 are literally arrows whose tails are placed at 1 and point vertically up or down

Mad

so basically, because we identify this, we can use our standard derivation defintion.

In this course, this has been extremly confusing to me, this whole "identification" of spaces, and jumping between definitions and so on and so forth. I am not a mathematician by a long shot, this might be the reason for my suffering.

But what is the author is trying to say with this whole deal?

Thorgott

the author is trying to say that the exponential map of $S^1$ in the sense of Lie groups is "the same thing" as the usual exponential map $t\mapsto e^t$

Mad

Ok, i see.
i understand the derivation, i can perfectly derive an exponential map on R or C. But just a passage above, i had a theorem stating, that the Exponential (as a map to Lie groups) is differentiable, and its differential given by the identity. shouldnt then we have i * t *s

Thorgott

where?

Mad

21:01

Hmm, i see, they wrote dexp_o is the id

Thorgott

@onepotatotwopotato though I think my Ansatz is sound, I have not succeeded in producing a counter-example. I'm still beyond certain the answer is negative, but at least it's non-trivial enough that I think it's worth asking this question on main

Ted Shifrin

I concur.

Mad

21:18

@thorgott can you explain again, why the number one in this case significant?, does it have to do with T_1 S

In another example, the Author again evaluate at one.

leslie townes

how does your text generally define the exponential map on the lie algebra of a lie group

Mad

At first, it was defined on Matrix Space GL(n,C) as a Series.

leslie townes

there's a 1 that appears in many formulations of that definition, that plays a different role from the subscript 1 in T_1 S^1 (the significance of this subscript 1 is that it is the identity element of the lie group S^1, and not so much that it is the real number 1)

but it will all likely turn on what your text's definition is, or what theorems they've proved about it

Mad

In this section, it is not defined at all.

Its only being stated that this parametergroup which one has shown to be unique is $\phi_x(t) := exp(tx)$

leslie townes

the excerpt you quoted says "to understand the exponential map of G, we need to find the unique one-parameter group in G in direction . . . ," is that ringing any bells about what the general definition of the exponential map may have been

thorgott's commentary above was with reference to what en.wikipedia.org/wiki/Exponential_map_(Lie_theory) refers to as "the typical modern definition," but under that definition your question almost collapses into a statement of the definition, so i wonder if it's something else

Mad

21:30

I see.

Thanks for the link, that makes sense.

Ted Shifrin

@Mad You always work with the tangent space at the identity element.

Mad

Identity at what, we are considering these " Parameter groups" which go from R + as an abelian lie group, here the identity is 0.

I think in this particular case, this evaluation is rather not to be interprtated in a complex manner

Author wants to show that the "exp" they defined using these parameter groups coencidide with known "exps"

And plugging in 1 results into that

Ted Shifrin

A group has an identity element, right?

Mad

@TedShifrin That is true, but i am not sure what you want to say with this.

Ted Shifrin

The Lie algebra of a Lie group is defined to be the tangent space of the Lie group at the identity element. That is the reason for the $1$ you kept asking about.

Mad

21:45

I understand that Mr Shifrin, but i was refering to the one being plugged into the function, not the one in the index of the tangent space. i do not think they corruoulate

leslie townes

it's 1 of the 1s he was asking about.

there's also, in the graphic above, the 1 in "phi_it(1)," which relates either to the definition of the exponential map, or hopefully at least some theorem about it, in the text.

Ted Shifrin

No, sorry. The other $1$ is conventional to make things simpler.

Mad

I think it is just to make things simpler, i agree. i have found no theorem or such that is pointed to "plug in one here.."

leslie townes

mad, well, FYI, in some texts, to define what exp(v) is, they will literally declare, OK, first find the one parameter subgroup whose tangent at the identity is v, and just plug 1 into that. with background theorems supporting why that recipe makes sense as a definition.

it seems like maybe your text doesn't use that definition, but is (in the image above) somehow assuming familiarity with that property of the exponential map, wherever they got it from.

Thorgott

@leslietownes I was assuming this, for the reocrd

Ted Shifrin

22:00

I also. That’s the only sane and prevalent defn.

Sha Vuklia

Let $X$ be a compact Hausdorff space. Let $\mu$ be a positive Borel measure on $X$. Then
$$
\pi_\mu:C(X)\to \mathscr B(L^2(X,\mu)),\phi\mapsto(f\mapsto \phi\cdot f)
$$
defines a cyclic representation. One can show that any cyclic representation on $C(X)$ is of this form. If $\nu$ is another positive Borel measure on $X$, then $\mu\perp\nu$ implies that $\pi_\mu\oplus\pi_\nu\cong \pi_{\mu+\nu}$ is cyclic. I want to show that it goes the other way around too. So if $\pi_\mu\oplus\pi_v\cong\pi_{\tau}$ where $\tau$ is a pos. Borel measure on $X$, then $\mu\perp\nu$.

ehh, I was about to edit the last part, but ran out of time

so the last paragraph can be ignored

psie

22:41

I need to prove the following (here $\mathbb T$ is the unit circle):

> Prove that if $f \in C^{k}(\mathbb{T})$ and $c_n$ are the Fourier coefficients of $f$, then $$\lim_{n\to \pm \infty} n^k c_n = 0$$

The Fourier coefficients of $f^{(k)}$, denoted $b_n$, are given by, recursively applying partial integration, \begin{align} b_n &= \dfrac{1}{2\pi}\int_{\mathbb T}f^{(k)}(t)e^{-int}dt \\ &= \dfrac{1}{2\pi}[f^{(k-1)}(t)e^{-int}]^{\pi}_{-\pi}+in\int_{\mathbb T}f^{(k-1)}(t)e^{-int} dt \\ &= \ldots \\ &=(in)^k\int_{\mathbb T}f(t)e^{-int}dt = (in)^kc_n.\end{align} I'm stuck here. Maybe, by the Riemann Lebesgue lemma, the result should somehow follow. I'm unsure.

A version of the Riemann Lebesgue lemma states that $$\lim_{n\to\pm\infty}\int_{\mathbb T}f(t)e^{int}dt =0,$$if $f$ is (absolutely) integrable in the Riemann sense over $\mathbb T$. But I do not see yet how this is applicable here.

Well, maybe I'm overthinking this. Just apply the lemma on $f^{(k)}$...

Ted Shifrin

23:27

Correct @psie

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