Mathematics - 2024-01-08

mick

00:10

0

Q: Properties of a 3D Julia set from squaring a 3D number?

Consider a commutative 3D algebra $T$ where the nonreal units $x,y$ satisfy $$x^2 = A_1 + A_3 y $$ $$xy = 1 + B_2 x + B_3 y$$ $$y^2 = C_1 + C_3 y$$ where all the parameters $A_1,A_3,B_2,..$ are real. Also we want this algebra to be a non-associative algebra. Squaring a number in this algebra is d...

3d question :)

Xander Henderson

00:54

@mick Please don't use comments to complain about downvotes.

Jakobian

@XanderHenderson why?

Xander Henderson

@Jakobian Voting is intentionally anonymous. If someone wants to leave a comment explaining their downvote, that is their prerogative. But complaining about downvotes is probably not going to cause the downvoter to leave a comment (indeed, the downvoter has probably already moved on), and is not likely to lead to productive discussion.

It is just noise.

Jakobian

Is it forbidden by the policies of math.se?

Xander Henderson

@Jakobian The actual policy on comments is pretty restrictive (see the help center). Comments are only supposed to be used to offer suggestions for improving a post, or for asking questions which will help to clarify a question.

"Why did I get downvotes?" does neither.

Ted Shifrin

01:35

Many of my comments are socratic or pushing the OP to figure more out. I’m considered inappropriate for MSE.

leslie townes

i dunno, a lot of that could be seen as broadly directed at improving the question. maybe not the exact way you do it.

Jakobian

how would you suggest for the author to improve on this proof?

Xander Henderson

@TedShifrin I typically consider those kinds of comments to be a little borderline. I tend not to delete them until either (a) the question is resolved or (b) they get flagged.

Jakobian

errors start after definition of $U$

no, sorry, they are before that, but never mind that

Xander Henderson

Is the author you?

That being said, without knowing what is being proved, it is a bit hard to say what can be improved.

Jakobian

01:41

No, those are online notes of Stefan Friedl that I'm trying to improve on

Xander Henderson

I will again note that I hate the mathematical "we".

Jakobian

That $\rho_A(x) = \sup\{r\|x\| : rx\in A\}$ is continuous, where $A$ is compact, convex, $0\in\text{int}(A)$

Xander Henderson

It is a weird verbal tic that I would like to see die a fiery, fiery death. Kill it dead. With fire.

Maybe nukes from orbit.

Dead.

With no chance of resurrection when the rapture cometh.

Jakobian

The error is in that the author wrongly assumed that $U$ is a neighbourhood of $x$

Xander Henderson

@Jakobian I have no idea. I don't know what statement is meant to be proved, and there appears to be notation in that proof which isn't defined in the proof (likely it is in the statement of the result).

Jakobian

01:45

@Jakobian this

Xander Henderson

What is $f_0$?

And why not just reproduce the statement of the result from whatever paper you are reading it from?

Jakobian

@XanderHenderson That part is wrong anyway, replace it with $x\cdot \frac{\rho_A(x)}{\|x\|}$

Xander Henderson

X(

Jakobian

this is a goldmine of errors

Xander Henderson

Yeah, I am not going to invest more time in this. There is too much notation which seems to be defined elsewhere, and the whole thing feels very out-of-context to me, which makes it very hard for me to process it. And if it is full of errors (as you claim), it is going to be a game of what-a-mole. I'm out.

Jakobian

01:52

I've provided all the notation. I'm just wondering how do you usually handle the case of $x\in\partial A$ i.e. $\rho_A(x) = \|x\|$ because I'm struggling to find appropriate $U$ in that case

Ted Shifrin

@XanderHenderson I disagree. I do not want to write “I” and “you” is not right.

Xander Henderson

@TedShifrin Oh, those are even worse.

I prefer a writing style which uses no pronouns.

Write everything in the imperative.

Ted Shifrin

Ugh.

leslie townes

how about "They" as the pronoun and everything in the past tense

Jakobian

6

A: Minkowski functional $p_E$ is continuous if and only if $0\in E^0=\text{int}E$

continuity at $0$: Let $\epsilon>0$. Since $0 \in E^0$, the set $$U := \epsilon E := \{ \epsilon \cdot x; x \in E\}$$ is a neighborhood of $0$. Moreover, by the sublinearity of $p_E$, $$|p_E(x)| = p_E(x) = \epsilon p_E \underbrace{\left( \frac{x}{\epsilon} \right)}_{\in E} \leq \epsilon \cdot 1$...

I guess I'll just use this

Xander Henderson

01:59

@leslietownes Yuck.

Jakobian

02:09

alright, I fixed it

Ted Shifrin

@leslietownes Poifekt.

Thorgott

02:23

@XanderHenderson what about the royal "we"

Xander Henderson

@Thorgott Nope. Even worse than the mathematical "we".

Ted Shifrin

@Thorgott I am always regal — if not royal — with my we.

leslie townes

you may is, but we'm ain't

Thorgott

@TedShifrin we have that in common

Juliamisto

Hello. I want to ask a soft question about the English language. Why is it the case that one often capitalizes "Euclidean" but does not often capitalize "abelian"?

leslie townes

02:35

39

Q: Why is "abelian" infrequently capitalized?

Posted with input from meta for improvement. I usually read, e.g. "Gaussian integers" and "Riemannian metrics", and occasionally "euclidean" or "cartesian" or even "lorentzian space", but the latter examples are relatively uncommon. I have also seen, e.g. "artinian" (and it has been commented tha...

ho.history-overview

the short answer is that nobody knows, but it is definitely a convention, and maybe a convention that is pretty specific to abel and not an instance of more general phenomenon

Juliamisto

What a convention, which really confuses me, it is.

leslie townes

helpfully, it seems pretty infrequent 'these days' that people's names get turned into adjectives. it seems to be something that, if we haven't abandoned the practice entirely, we tend to primarily do with names from 'long enough ago'

so you won't have to worry about offending someone in your department by not capitalizing their name, or by capitalizing it, whichever they would regard as more offensive

Juliamisto

I see.

By the way, does "positive" sometimes mean "nonnegative"?

leslie townes

yes, in some subfields this is even conventionally what it means

Xander Henderson

@Juliamisto Yes.

Jakobian

02:43

its the French way

Juliamisto

Hence there is "strictly positive".

English is really challenging for me, a native speaker of Chinese.

Xander Henderson

03:18

In English, "positive" almost always means "strictly larger than zero". If you want to indicate nonnegative, you should say "nonnegative". However, there are some authors (Barry Simon, for example) who adopt the convention that "positive" means $\ge 0$ and "strictly positive" means $> 0$.

There may also be certain fields where this convention is adopted (I don't know of any off the top of my head, but that doesn't mean such fields don't exist).

leslie townes

03:56

operator algebras (and maybe other algebras) and at least some corners of functional analysis are examples of such fields. in these areas, the gap between x being positive and strictly positive is also about slightly more than just whether or not x is allowed to be zero.

that may or may not be a coincidence, as is the fact that these folks seem to do what barry simon does :)

of course, the proper term for "x is nonnegative, not necessarily strictly positive" is "simonian," lowercase s.

one potato two potato

07:31

@Thorgott I think I figured out what it means. It follows from two facts: 1. Any nonsingular closed 1-form $\omega$ on $M$ whose periods have rational ratios induces a fibration $M\to S^1 = \Bbb R/\mathrm{periods}(\omega)$ by $x\mapsto \int_{x_0}^x\omega$.

2. If $M$ is a smooth compact oriented $n$-manifold then the pair $\langle e(\tau_M),\mu\rangle = \chi(M)$ where $\tau_M$ is a tangent bundle of $M$, $e(\tau_M)\in H^n(M;\Bbb Z)$ is an Euler class of $\tau_M$ and $\mu$ is the fundamental class of $M$.

and I think fibration here means fiber bundle

Sahaj

07:50

If I know $\bmod p: f(x) \equiv 0$, how can I work out $f(x)$ mod $p^2$? Is there any relevant theorem?

leslie townes

you'd need some assumption on what f is to be able to say much

Hensel's lemma

In mathematics, Hensel's lemma, also known as Hensel's lifting lemma, named after Kurt Hensel, is a result in modular arithmetic, stating that if a univariate polynomial has a simple root modulo a prime number p, then this root can be lifted to a unique root modulo any higher power of p. More generally, if a polynomial factors modulo p into two coprime polynomials, this factorization can be lifted to a factorization modulo any higher power of p (the case of roots corresponds to the case of degree 1 for one of the factors). By passing to the "limit" (in fact this is an inverse limit) when the power...

may be of interest

one potato two potato

ah, Hensel's lemma

first reason I give up studying number theory

Sahaj

Oh right

I actually forgot about that

I learnt about that a couple years back and I can't even recall the statement

one potato two potato

08:10

Jan 4 at 14:09, by one potato two potato

Let $M$ be a closed orientable 3-manifold and suppose there is a fibration $K\to M\xrightarrow{f}S^1$ and consider a closed 1-form $u = [f^*(d\theta)]\in H^1(M;\Bbb Z)\simeq Hom(H_1(M),\Bbb Z)$ such that $u(\pi_1(M)) = \pi_1(S^1)$. If $S_0$ is a component of $S\subset M$ in which represents $u$, then $\pi_1(S_0)\in\ker u$ @Thorgott ?

I think Fact 1 also solves this because $f$ is a fibration induced by $u$ and $K$ can be thought as a leaf of a codimension 1 foliation induced by $u$

PM 2Ring

08:39

I see Ted's answered a great circle path question that was migrated from Physics.SE. I made a Sage thing a year or so ago for plotting such paths in 3D. It also prints the GC distance (in degrees).

sagecell.sagemath.org/…

And if the 2 points have the same latitude it also prints the small circle distance.

Traditionally, the GC distance is computed using the spherical trig cos law, probably in haversine form. But on a computer it's easier to convert to Cartesian & do the dot product. ;)

12

Q: What is the difference between Hensel lifting and the Newton-Raphson method?

So in the Newton-Raphson method to iteratively approximate a root of a real polynomial, we start with a crude approximation $x_0 \in \mathbb{R}$ for $f(x)=0$ where $f(x) \in \mathbb{R}[x]$. For the next iterate $x_1$, we put $x_1 = x_0 + \epsilon$, and we want to determine $\epsilon$ to get a bet...

elementary-number-theory numerical-methods congruences

Xander Henderson

11:25

@leslietownes This. :)

user 85795

11:45

+1

User13114

14:04

Does a statement like all directional derivatives exist for a function just mean they are not undefined

Jakobian

14:17

Directional derivative of $f$ at $x$ in direction $v$ is the limit $\lim_{t\to 0} \frac{f(x+tv)-f(x)}{t}$

existence of all directional derivatives at a point $x$ just means such limit exists for all $v$

psie

16:11

I'm trying to show the statement "(a continuous, periodic function) $f$ is real-valued $\iff$ the complex Fourier coefficients satisfy $\overline{c_n}=c_{-n}$". Consider the $\impliedby$ direction. To this end, I define $g(x)=f(x)-\overline{f(x)}$ and my goal is to show all coefficients $b_n$ of $g(x)$ are zero.

Let's compute them, $$b_n=\frac1{2\pi}\int_{-\pi}^\pi g(x)e^{-inx}dx=\underbrace{\frac1{2\pi}\int_{-\pi}^\pi f(x)e^{-inx}dx}_{=c_n}-\frac1{2\pi}\int_{-\pi}^\pi \overline{f(x)}e^{-inx}dx.$$ Just how do I deal with the last integral and show it equals $\overline{c_{-n}}$? I'm stuck.

Xander Henderson

@psie Why don't you look at the integral of $(f(x) - \overline{f(x)}) \mathrm{e}^{-int}$?

$f$ is real valued, right? What does that say about $\overline{f}$?

psie

@XanderHenderson we have $f=\bar{f}$, right?

Xander Henderson

Is that a question? Be more assertive!

You should already know the answer to that question!

User13114

@Jakobian Does that include $v$ = 0?

Xander Henderson

@User13114 Is that a direction?

Though note that it is trivial...

User13114

16:21

Yes for the directional derivative

Xander Henderson

@User13114 Are you sure?

User13114

Well I guess your not really going anywhere

Xander Henderson

How do you define a "direction"?

But, like I said, it is trivial, in in any case (where "trivial" is being used in the sense of "it evaluates to zero", not "it is easy").

$$\lim_{t\to 0} \frac{f(x+t\cdot 0) - f(x)}{t} = \lim_{t\to 0} \frac{f(x) - f(x)}{t} = 0.$$

User13114

Ok my book doesn't specify what the definition of direction is

Xander Henderson

Well, like I said, it doesn't end up mattering, because it is trivial. But a lot of texts will define directional derivatives only in the direction of a non-zero vector.

Or, equivalently, they will assume that $\|v\| = 1$.

Since nothing interesting happens when $v = 0$.

Jakobian

16:33

@User13114 yes

any vector $v$

User13114

Can you get something nonlinear when you compute a directional derivative?

Jakobian

More intuitively, they take only vectors $v$ with $\|v\| = 1$

since directions can only be made with respect to unit vectors, if you think about it

but what really matters here is the map $df_x(v) = \lim_{t\to 0} \frac{f(x+tv)-f(x)}{t}$

If $f$ is differentiable at the point $x$, then $df_x$ exists for every $v$ and is a linear map

User13114

If f is not differentiable at a point then you can get something nonlinear right?

Jakobian

In general, even if $df_x(v)$ exists for all $v$, it doesn't imply linearity

The map $df_x$ is called a Frechet derivative of $f$

some authors assume that $df_x$ needs to be linear for it to be called Frechet derivative

it depends

User13114

Ok I was doing an exercise where you have to show that the all directional derivatives for a function $g$ exist but $g$ is not continuous

I was getting a nonlinear directional derivative so I wanted to make sure I didn't make a mistake

Jakobian

16:43

@Jakobian I'm sorry, I've said rubbish here

its Gateaux derivative, not Frechet one

@Jakobian and here by differentiable I mean Frechet differentiable, which means that $f$ can be approximated by a linear map near $x$

User13114

17:01

How would you go about proving that $f(x,y)$ =
\begin{cases}
\frac{x^2y}{x^4+y^2}, & \text{if $(x,y) \neq (0,0)$} \\
0, & \text{if $(x,y)$ is $(0,0)$}
\end{cases} is bounded at the origin?

Xander Henderson

It is zero at the origin. Zero seems to be bounded...

Or do you mean on a neighborhood of the origin?

In any event, I might make the change of variables $t = x^2$, then try to think about it in polar coordinates.

User13114

I should say there exists a neighborhood around the origin where the function is bounded

Xander Henderson

17:17

@User13114 Again, make the change of variables, make the change of coordinates, and you should be fine.

nickbros123

17:28

my book on statistics lists one of the axioms of probability set function, for the infinite series case: if {$A_i$} is a sequence of subsets, pairwise disjoint, then $P(\cup_{i=1}^{\infty} A_i)=\sum_{i=1}^{\infty} P(A_i) $

why is this the case

doesnt the finite case for arbitrary $n$ allow us to write this?

or is there a problem to do with taking limits on either side, and commuting P and the limit?

ok, i think i got the issue: the infinite union (or intersection) has nothing to do with a limiting process?

Jakobian

@nickbros123 wdym

Xander Henderson

Jeebus, enrollment in math is looking terrible this semester. Across all five full-time faculty, it looks like we have maybe 200 students. Total.

nickbros123

@Jakobian

Xander Henderson

That is a definition...

Jakobian

I know the definition

I just really don't understand what you're confused about

Xander Henderson

17:41

@nickbros123 The function $$P : \mathcal{B} \to \mathbb{R} : A \mapsto 1$$ satisfies the first two properties, but not the third...

Indeed, we can cook up all kinds of arbitrary set functions which satisfy the first two properties, but not the third. There are many set functions which are not probability set functions.

nickbros123

well I was confused as to why the finite case, as in {$A_1,A_2...A_n$} pair wise disjoint means $P(\cup_{i=1}^n A_i =\sum_{i=1}^n P(A_i)$ wouldnt imply this. I think I got the problem here, is it that infinite unions are not a limiting process like real sequences and hence I cant apply limits both sides to the finite case?

Jakobian

The condition for finite unions does not imply it

The is a limit process here, $\sum_{n=1}^\infty P(A_n)$ is a limit

Xander Henderson

@nickbros123 Your definition doesn't say anything about "finite" or "infinite" cases...

Jakobian

the condition for infinite sums tells you, more or less, that $P$ is "continuous" in a sense

nickbros123

I see

Jakobian

17:46

$\lim_{n\to\infty} P(\bigcup_{k=1}^n A_k) = P(\bigcup_{k=1}^\infty A_k)$

nickbros123

I guess a quick google search of the definition of $\bigcup_{i=1}^{\infty} A_i$ couldve avoided this question

Xander Henderson

@Jakobian I'm not entirely sure I like that phrasing... it is more just an assertion that this function has one of the nice properties of a measure, e.g. additivity.

Though I do see how you are using "continuity" here.

Jakobian

It can be formalized if you prefer

See "Measure theory I" by Bogachev

Xander Henderson

No no... I'm just expressing a distaste for that view of things. It isn't the intuition I like for dealing with these things (and I am familiar with Bogachev).

nickbros123

@Jakobian to prove that you can commute the P and limit here, do we use the 3rd condition in the definition in the way?

Jakobian

17:49

@nickbros123 yes

nickbros123

i see

Jakobian

finite additivity + continuity is equivalent to countable additivity

Xander Henderson

@nickbros123 I really don't think that is the right way to think about it. My guess is that your book has defined $\bigcup_{n=1}^{\infty} A_n$ to be the set of all all $x$ such that $x \in A_n$ for some $n$. While this can be phrased as a limit in a broader mathematical framework, this is not the way that analysts / probabalists generally think about it.

nickbros123

@XanderHenderson I agree, Even i dont really think of it like that.. after googling how the infinite union is defined

Xander Henderson

@Jakobian Yes, indeed. But the definition here skips straight to (countable) additivity---it doesn't seem to much about with finite additivity at all (unless there is context missing), so invoking that theorem is kind of orthogonal to the given definition.

Jakobian

17:54

Nickbros was asking why can't we just assume finite additivity

Xander Henderson

@Jakobian I mean, you can just assume finite additivity, as finite additivity is a special case of additivity (as presented in that definition).

Jakobian

then they mentioned limiting processes

I've explained in which ways does this definition deal with limiting processes

the series $\sum_{n=1}^\infty P(A_n)$, and the definition of countable additivity itself includes a hidden continuity condition

as for exposition to general user, I wasn't concerned much about that, rather I tried to answer Nickbros' concerns

nickbros123

well you pointed out that commuting the limit and P there is not possible without 3, also I dont know how, in an analysis context, we can apply a limiting process given the definition of infinite union. I sort of realised this after understanding how the infinite union is defined

so i guess my concerns are cleared

Xander Henderson

@nickbros123 The way that analysts define infinite unions is pragmatic for the purposes of analysis. But there are other valid approach, e.g. via category theory.

But this is usually not required for analysis, and is not the way that analysts / probabilists tend to think about things.

nickbros123

@XanderHenderson I personally dont know how to do it in any other way than epsilons and Ns, but even if i knew, as jakobian pointed out, we cant interchange the limits..

Xander Henderson

18:06

Chat is borken...

The point that I am making is that most analysts don't regard $\bigcup_{j=1}^{\infty} A_n$ as a limit (it can be regarded as such, but that isn't the usual analyst's intuition). The sum is certainly a limit, but the set is not generally treated as a limit.

Thorgott

@onepotatotwopotato ah, that was not at all obvious

Jakobian

I also don't regard it as a limit

Thorgott

no, increasing unions are definitely limits

Jakobian

intuition-wise

leslie townes

yeah. huge difference between saying "X is Y" with the meaning of X is an instance of mathematical notion Y, and saying "X is Y" with the meaning of "Y is the best way to think about X"

one is a head thing, the other is a heart thing

2

Thorgott

18:12

well, it's pretty literal

you just need to grant that subsets are the same thing as their characteristic functions

Xander Henderson

@Thorgott This sounds like category theory talk.

Jakobian

no it doesn't, lol

Thorgott

no, I think this is an analytical point of view

category theory talk would be if I told you it's a colimit (which it is, but I don't think that's relevant right now)

when doing integrals and $A\subseteq X$, you always use facts like $\int_Af=\int_X1_Af$

Xander Henderson

Mostly I was responding to the word "just".

Thorgott

fair, that's a bad word

Xander Henderson

18:15

I can't tell you how many times I've heard people with a love of category theory say "Well, [X] is just a [Y] in the category of [Z]."

Thorgott

I love saying that too, but that's not what I'm doing right now

leslie townes

that's their mantra

Xander Henderson

@Thorgott You are one of those people, eh?

Jakobian

Unions are just colimits in the category of sets

Thorgott

only filtered unions!

@XanderHenderson I have no shame left

Xander Henderson

18:16

Your mom is just a... something something... in the category of... uh... something something...

Thorgott

lol

Xander Henderson

(There's a joke in the somewhere, but I'm too lazy to figure it out.)

Jakobian

a VERY NICE LADY

leslie townes

your mom is not an object in Cat

Xander Henderson

Yes, but is Cat an object in YourMom?

(I'm saying that your mother barbecues cats.)

leslie townes

18:20

in looking for a comic example of how far "X is just Y" can go, i discovered that ncatlab has a page on the schrodinger equation

sadly its not an example of what i wanted

Xander Henderson

Heh.

Soumik Mukherjee

18:31

@XanderHenderson "A monad is just a monoid in the category of endofunctors"- Sun Tzu

3

leslie townes

the entirety of sun tzu's written work is just the yoneda lemma

Xander Henderson

@SoumikMukherjee I can't tell you how much that hurts my soul.

Sine of the Time

@SoumikMukherjee "Why the people are using me for random quotes?" - Sun Tzu

Xander Henderson

"Sun Tzu said what?" ---Oscar Wilde

Soumik Mukherjee

@XanderHenderson "I don't need to fight. My quotes are enough to hurt you" -Sun Tzu, The Art of War

Sine of the Time

18:40

"Note to self: publish the book only after you've won the war"- Sun Tzu, the Art of War, Second Edition

Soumik Mukherjee

@SineoftheTime Statistically it's either Sun Tzu or Einstein.

Jakobian

@leslietownes Lets adapt Hegel's work into category theory

ah, no, that already exists

Math Horse

19:15

Hi, how can I prove this?

I tried:
Putting x=0 and x=1 and the inequality was right

Xander Henderson

Induction?

What tools do you have?

In what context are you being asked to prove this statement?

Math Horse

Calculus I course, we didn't study induction

I think I need to use derivatives

I got the f'(x) and did the same process with the x=0,1, it worked

Xander Henderson

Yes, you could look at the the derivative of $1 - nx - (1-x)^n$.

Math Horse

but I don't know what to do next

Xander Henderson

What is $f$?

Math Horse

19:18

f'(x) = -n * [1 + (1-x)^n-1]

f'(x) =< 0

n is at least 1 so for x=0,1 the statement is true

Xander Henderson

I asked what $f$ was... but sure. What does that tell you?

Math Horse

f(x) = 1 - nx - (1-x)^n

@XanderHenderson That's where I'm stuck at

Xander Henderson

Good. $f'(x) \le 0$ for all $x \in [0,1]$. What does that mean?

Math Horse

That $f$ is a constant?

Xander Henderson

No. If $f$ were constant, then its derivative would be zero, yes?

What does it mean if a function's derivative is negative?

Math Horse

19:23

The graph of f(x) will always decrease

Negative slope

Xander Henderson

@MathHorse Okay, sure. The function is decreasing. Good. What does that tell you?

Math Horse

Maybe

if f(o) = 0

Than f(0+) will always be < 0?

Xander Henderson

Not always, but certainly as long as the derivative is negative, yes.

Math Horse

Hm.. I still can't really see what I'm trying to find here
Calculated all values and it looks correct, but I am missing a certain purpose

Xander Henderson

The argument should go something like the following: $1-nx \le (1-x)^n$ if and only if $1- nx - (1-x)^n \le 0$. When $x=0$, $1-nx-(1-x)^n = 0$. Moreover, $\frac{\mathrm{d}}{\mathrm{d}x} (1-nx-(1-x)^n) = -n - n(1-x)^{n-1} = -n(1-(1-x)^{n-1}) \le 0$ for $x \in [0,1]$. In particular, $f(x) \le f(0) = 0$ for all such $x$.

Basically, the function is $0$ at $x=0$, and decreasing.

So it stays below zero until at least $x=1$.

Math Horse

19:42

I see.. makes more sense now
I'll try to phrase it somehow based on my findings

Thanks

Daniel Donnelly

@Shaun Congrats on your Tits Alternative... paper. I don't know that math! Looks good though.

Xander Henderson

@DanielDonnelly Watch your mouth!

Daniel Donnelly

@XanderHenderson it's not pr0n. There was a guy named Tits

@Shaun this is the UI of the future for CD's:

That's what it will look like on your phone!

Xander Henderson

@DanielDonnelly I'm aware. He did work with Bruhat. I was making a joke. :(

Daniel Donnelly

In desktop mode though everything is always more spacious. I happen to have a touch-screen laptop ($400 getac)

@XanderHenderson I know, I was joking back

That search button in the lower left is what I'm hooking up rn. I have had search working before and still have the code, so... But it's only the start of many features that enable actual proofs, but not as rigorous as PA's like Lean, so everything still has to be community verified. It's cool 😎 though.

The CoREACT (Nicolas Behr et al) team that is actually developing a Diagrammatic Editor in conjunction with Coq proofs, wants me to present for 30 minutes my own independent work (namely regarding my nice GUI / UX) on March 12. I am trying to get more of it done so I have some cool features to show off by then

There tool will probably be much better than mine, but mine will have its own advantages too

MathDOPE = Database of Proof Explanations

Not drugs!

Though we don't care if you're on drugs editing your diagrams, just be safe

I'm talking to you Erdos

Xander Henderson

19:54

@DanielDonnelly I've got some bad news for you...

Daniel Donnelly

What's up?

Xander Henderson

@DanielDonnelly nytimes.com/1996/09/24/us/…

Daniel Donnelly

I know, but I meant modern day Erdos' and or the spirit of Erdos. I don't know many math people, but I do know that Erdos took pills

Xander Henderson

Again... it was joke...

I give up. :/

Daniel Donnelly

Dude, I'm joking too. Drugs are bad

Ted Shifrin

19:58

puts in eyedrops and takes an antibiotic pill

Daniel Donnelly

Well smoking is bad at least 🚭

Don't smoke, don't joke :|

😐

"A monad is just a monoid in the categoroy of endofunctors" in the sidebar

I wonder how that works with all those commutative diagrams in the definition of monad (usual)

So endo functors as objects would mean natural map arrows

Interesting! Will have to explore that later

I'm still wondering whether my site should support Arrow-to-Arrow, Arrow-to-Object arrows

the database doesn't do that naturally, so you'd typically need to fake it with a node and two database arrows (relationships), the node for connections

Quiver definitely does support it

(Quiver is 100% frontend btw)

database => backend

Jakobian

20:26

drugs are wonderful

Sha Vuklia

@leslietownes do you know if any $a\in A$ where $A$ is a $C^\ast$-algebra can be written as $a=x+iy$, where $x$ and $y$ are Hermitian? If $a$ is normal, then the answer is yes by functional calculus. However, what if $a$ is not normal? (I'm trying to argue that a certain element is hermitian, and I could use the above result if it were true)

Daniel Donnelly

@Jakobian ikr

You have to be a Hermit to solve that problem, @ShaVuklia

Sha Vuklia

hm, I think I can show my element is normal

in which case my question is not relevant anymore (as the normal case is already covered)

Jakobian

We're Not Candy "Singing Pills" PSA (1980's) | Marc's Misc Shorts

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leslie townes

x = (1/2) (a + a-star), y = (i/2) (a-star - a)? is it important for some reason that x and y commute?

Sha Vuklia

20:34

oh I was stupid, I wrote $x=1/2(x+x^\ast)+i/2(x-x^\ast)$, which is false of course

Daniel Donnelly

Kara Jackson - no fun/party (Live on KEXP)

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Sha Vuklia

and because it was wrong, I wrongfully concluded that it must be because of commutativity issues or sth (since the result holds in $\mathbb C$)

sorry for sloppiness, but thanks for your remark!

leslie townes

you are definitely right to care the about the commutativity issues, there are a lot of results around expressing elements of arbitrary cstar algebras as combinations of 'nice' ones, that sound super cool and potentially useful, then when you realize that the 'nice' things generally won't commute, the coolness and the potential usefulness vanish

Ted Shifrin

I don't want to be bothered with mathematics in which linear operators fail to commute. That's just passé.

Sha Vuklia

@TedShifrin so matrices are passé?

Ted Shifrin

20:47

Nah, in my world all matrices commute.

I'm just acting like leslie to see if it's fun.

leslie townes

ted, i'm not entirely sure i see where you're coming from, i don't see an attack on general non commutativity in what i wrote above, not even as a joke

Ted Shifrin

Oh, it had nothing to do with this particular discussion. I was just referring to your general behavior and so I wanted to be absurd to see if it's fun.

leslie townes

it's a little weird that the snake in the garden of eden gave eve two things that didn't commute, but that's what the good book tells me, so it's true

when jesus said "the last shall be first, and the first last" he was talking about commutativity in the afterlife

Ted Shifrin

I only read the atheists' bible.

leslie townes

what's that? lam's noncommutative rings?

Jakobian

21:19

@TedShifrin I wouldn't want to study an empty theory either. Why do you bother with such trivialities when all rings are commutative?

Xander Henderson

21:31

@TedShifrin Is it? It looks like you are having fun. :D

Ted Shifrin

@Jakobian Indeed.

robjohn

@TedShifrin watch out, you might be ex-commuted.

Xander Henderson

21:50

So I just learned that, based on the way compensation is handled here, if I were to quit my job and then get rehired, I would get a $5k per year pay raise...

(I would have to go back to being probationary faculty, but... double you tee eff?!)

robjohn

@XanderHenderson When I retire, I will be getting more than I am making now.

Xander Henderson

@robjohn Had my father managed to retire, that would have worked for him, too.

But his retirement would have been calculated based on his salary as vice president at the college, while in his last two years, he returned to teaching.

Sha Vuklia

23:23

@leslietownes Today I learned about the Spectral Mapping Theorem for normal operators on a Hilbert space. I was already impressed by continuous functional calculus (and how it's just a simple application of the Gelfand transform!), but I would have never guessed there is even a Borel functional calculus xD

leslie townes

haha, i love it

Xander Henderson

23:40

@ShaVuklia Oh, that is some lovely mathematics. I really enjoyed funky anal in grad school.

3

Sha Vuklia

:D

Jakobian

@XanderHenderson I'm going to cite this out of context

Transcript for

$Mathematics$ Mathematics