Mathematics - 2024-10-26

Sine of the Time

00:01

can't be over $\Bbb C$?

Ben Steffan

I guess you can have the norm be valued in a different field than your vector space is over

then you can have something over $\mathbb{C}$ valued in $\mathbb{R}$

you can't have something over $\mathbb{C}$ valued in $\mathbb{C}$

Sine of the Time

triangle inequality would not make that sense

Ben Steffan

yeah

Thorgott

00:32

norm makes sense in any vector space over a valued field

Xander Henderson

01:37

@Thorgott Norm only makes sense when he walks into Cheers. After a couple of beers, he gets a bit sloppy.

@Thorgott Hrm... I think that this is not necessarily a generally accepted use of the terminology. In my world, the codomain of a norm is always $\mathbb{R}$.

I would generally use "absolute value" to refer to a function which eats things in a field and spits out things in an ordered ring. I would probably also use the term absolute value for a thing which vectors and spits out elements of an ordered ring.

Or maybe modify the term "norm" somehow, e.g. perhaps an "$R$-norm" is a function from a vector space to an ordered ring $R$.

Though that's gross, too.

Thorgott

01:53

@XanderHenderson gotta love Norm

@XanderHenderson I didn't intend otherwise

the way I know things (which is an algebraist's way, I suppose), a valued field is a field $K$ together with an absolute value $K\rightarrow\mathbb{R}_{\ge0}$ satisfying definiteness, multiplicativity and the triangle inequality and a normed vector space $V$ over $K$ is a vector space equipped with a norm $V\rightarrow\mathbb{R}_{\ge0}$ satisfying definiteness, homogeneity and the triangle inequality

Xander Henderson

@Thorgott No, I'm not trying to argue with you. I'm really just thinking out loud.

Thorgott

yeah, I'm just clarifying

Xander Henderson

@Thorgott Yeah, that makes sense. I don't like it (as an analyst). I don't think that you are abusing the language, I just think that you are wrong, and should be the first up against the wall when the revolution comes.

I mean... uh... whatever floats your boat.

I mean... really! What kind of monster wants a "norm" defined over some stupidly abstract field $K$?!

:P

In other news, I have insufficient ingredients to make any cocktail. :(

I need to order more vermouth.

think_meaning_builds

02:31

Anyone watching the world series?

⚾

Xander Henderson

03:00

@think_meaning_builds Haven't had time. I don't even know who's playing this year. :/

Oh, two teams I hate. Bah.

Go Dodgers, I guess...

Do you know why they are called the Dodgers?

think_meaning_builds

Coastal heavyweights slugging it out.

I do now, thanks 🙏

Xander Henderson

@think_meaning_builds Aw... you cheated and looked it up!

:P

It is about dodging street cars, which is something you just don't do in LA, since there are no street cars in LA. But they kept the name.

Like the Utah Jazz.

I mean, if there is any place in the world where jazz isn't, it's Utah!

But the team was originally from NoLa, so...

think_meaning_builds

My love for baseball has been reignited after the disappointment in the cricket world cup :-)

The winning history of the Yankees is second to none in all pro sports world wide.

think_meaning_builds

03:39

GRAND SLAM!!!

🎇🎆🎇🎆🎇🎆

Game #1 poetry.

leslie townes

05:53

@psie you may be sort of 'misapplying' intution from the case of function spaces, e.g. L^p spaces familiar from measure theory, where it is not uncommon to have "the same" function either sit, or not sit, in various L^p spaces depending on whether certain integrals involving the function (which a priori might take the value +oo) are finite or not.

in an abstract setting (as you seem to have cleared up with ben above) there is no notion of the norm being finite or maybe not (and, similarly unlike many function spaces, no "formula" for an abstract norm whose finiteness gives a test for membership in the space). the finiteness is built into the definition. the closest thing to an idea of 'the same thing maybe sitting in different normed spaces' might be notions of there being (or not being) a bounded linear map from one space into another...

.. perhaps with various properties that make it a lot like an inclusion map in a function space setting.

when that matters you sometimes do see people identifying an element of one normed space with its image under some map into another normed space, as you might use the same letter f to denote an element of L^oo or L^1. but then you usually have to be very punctilious about keeping mental track of what "the norm of f" means, as what that number is, or even whether it makes sense at all, will vary in a way that you are visually suppressing when you do that.

at a minimum a good practice in that setting would be to decorate the notation for the norm in some way so there is no question what it is, e.g. || ||_Y for the norm in the space Y. there is no standardized way of doing this, you just have to pay attention to what definitions authors are using and be careful on your own.

it goes without saying that it doesn't help to speed through this material. :)

leslie townes

06:14

labeling the norms yourself might be a helpful way of making sense of these basic arguments, e.g. in the inequalities in chat.stackexchange.com/transcript/message/66512637#66512637 the same symbol || || is used to denote three different norms (the norm in Y, the norm in X, and the norm in "bounded linear maps from X to R," which is an operator norm arising out of the norm of X and the norm in R), with context alone distinguishing them.

it might be helpful/instructive to add that kind of notational context in until you get sick of seeing it. most books do not notationally belabor points like this because of pedagogical choices (likely a belief that on balance it would be more distracting, rather than less distracting, to do this every time)

Gian's Pizzeria

09:49

$\int_{\gamma} \frac{1}{z(z^3-1)}dz$

$\gamma:|z-1|=\frac{1}{2}$

Sine of the Time

10:23

@Gian'sPizzeria do you know how to proceed?

Gian's Pizzeria

@SineoftheTime I have to find the singular points first

Jakobian

What I could prove is that if every compactification of $X$ is zero-dimensional then $X$ is pseudocompact. @AlessandroCodenotti

Sine of the Time

@Binky the singular points inside the curve

Binky

10:39

Are the singular points z=0 and z=1?

Sine of the Time

how many roots does $z^3-1$ have?

Binky

No I was wrong

@SineoftheTime 2

Sine of the Time

?

Pizza

Hi

Sine of the Time

@Pizza hi

Binky

10:42

$-\frac{1}{2}±\frac{i\sqrt{3}}{2}$

These are the complex ones

Pizza

@SineoftheTime I'm still waiting for the results, I today or tomorrow

Sine of the Time

what will you study next? Methods?

@Binky what about $z=1$?

Binky

Also this

Pizza

I need algebra and geometry, but the November appeal got cancel :(

Sine of the Time

so which singularities are inside the curve?

Binky

10:44

So the singular points are 0,1,$-\frac{1}{2}±\frac{i\sqrt{3}}{2}$

Sine of the Time

@Pizza that sucks.

Pizza

@SineoftheTime yes, I don't know why

Sine of the Time

you can email the professor asking why that happened

and if it's only postponed

Pizza

said the next one will be in January

Sine of the Time

:(

Pizza

10:52

Well at this point I'll have to prepare some other subject

Sine of the Time

yeah

Pizza

Are you studying something?

Sine of the Time

yes

Binky

@SineoftheTime i dont know

Can you give me a Hint?

Sine of the Time

then you should revise complex numbers

Binky

11:04

Okay!

I'm going to do a review

Sine of the Time

do you know how to compute the modulus of a complex number?

psie

@leslietownes yup, I'm taking it easy :) baby steps is the key :)

Soham Saha

Hi everyone! Quick question: This answer explains a given question in just a few words, while this one goes about the same question in a convoluted way. Is the first answer deficient in logic in any way compared to the other?

I personally did it by the first way, because I thought that the ‘prime’ condition doesn’t affect the probability in any way, and given that the sum is a perfect square, I can focus only on those particular cases where we do get a perfect square as my ‘reduced sample space’

Is my thinking correct?

Binky

11:20

@SineoftheTime yes

Sine of the Time

then you should have no problems finding which singularities are inside the curve

Binky

$|z|=|a+bj|=\sqrt{a^2+b^2}$

Sine of the Time

are you able to continue?

Binky

$|z-1|=|a+bj-1|=\sqrt{(a-1)^2+b^2}$

$\sqrt{(a-1)^2+b^2}≤\frac{1}{2}$

now I think I have to replace the z's I found in here

Sine of the Time

correct

I suggest you taking a look at the book Schaum's outlines Complex varibles

psie

11:34

> Theorem Let $\{T_n\}_{n=1}^\infty$ be a sequence of continuous linear maps from a normed space $X$ to a Banach space $Y$. Suppose that $\sup_n \|T_n\|<\infty$ and $\lim_{n\to\infty} T_nx$ exists for $x$ belonging to a dense subset of $X$. Then there exists a continuous linear map $T$ from $X$ to $Y$ such that $$Tx=\lim_{n\to\infty}T_n x,\quad x\in X.$$

My book claims, after showing that $Tx=\lim_{n\to\infty}T_n x$ indeed holds for all $x\in X$, that $T$ is linear because addition and scalar multiplication are continuous on $X$. I don't see what this has to do with continuity of those operations on $X$. Isn't it just because of the sum law for limits, i.e. $\lim (a_n+b_n)=\lim a_n+\lim b_n$?

$$Tx+Ty=\lim T_nx+\lim T_ny=\lim(T_nx+T_ny)=\lim T_n(x+y)=T(x+y).$$ I only used the linearity of $T_n$ and the sum law of limits.

Binky

@SineoftheTime okay!

Thanks for the information

Sine of the Time

The operation of limit is a linear operator

Binky

$z=0\to a=b=0$

1≤1/2

so it is not internal

Jakobian

@SineoftheTime uh... sure but you need to possibly justify that

Binky

$z=1\to a=1, b=0$

0≤1/2 , so It Is not internal

Jakobian

11:41

@psie the "sum law for limits" is for real numbers

as how its usually stated

here you're dealing with vector spaces

psie

oh, ok

Jakobian

its true that $\lim_{n\to\infty} (x_n+y_n) = \lim_{n\to\infty} x_n + \lim_{n\to\infty} y_n$ still but its because of continuity

name the continuity of the map $(x, y)\mapsto x+y$

the property used is that if $f$ is continuous and $x_n$ is a sequence convergent to $x$ then $f(x_n)$ converges to $f(x)$

here take the sequence $(x_n, y_n)$

this is the way in which continuity is used

Sine of the Time

@Binky A part from the fact that $z=1$ is inside the curve, no one here will check if you're computing the modulus correctly

this is basic, you should have zero doubts

Jakobian

in fact the "sum law for limits" can be thought of as continuity of the operation of addition $+$ for $\mathbb{R}$, at least if we aren't taking limits where one of the sequences converges to $\pm \infty$

its sequential continuity, equivalent to continuity because we are in a metric space

Binky

@SineoftheTime I was wrong

Jakobian

11:45

well, continuity always implies sequential continuity but yeah

Binky

I also wrote 0≤1/2....

psie

ah ok, I think I understand, thanks. That they say it is continuity on $X$ also makes sense. First I thought it was continuity on $Y$.

Jakobian

yeah, note that continuity on $X$ refers to those operations being on $X$ but they are actually on $X\times X$ as functions

Binky

@SineoftheTime Only z= 1 is internal

Sine of the Time

I'm not going to check this claim

Jakobian

11:46

just a minor thing

psie

👍

Binky

@SineoftheTime okay!

Thorgott

@XanderHenderson :D

Jakobian

oh. Actually they refer to "on $X$" in order to distinguish them from operations on $Y$ and so on

its actually wrong because you are using continuity of them as operations on $Y$

@psie all of the things here are vectors in $Y$

psie

yeah, upon further thought continuity on $X$ doesn't sound right

Jakobian

11:51

@Jakobian on $X\times X$ and on $\mathbb{K}\times X$ where $\mathbb{K}$ is your scalar field, actually, if we include multiplication by scalars

Jakobian

14:40

@AlessandroCodenotti Let $\alpha$ be a large enough limit ordinal and $C$ the Cantor set. Can $C\times \omega_\alpha$ have compactification which isn't zero-dimensional?

Alessandro Codenotti

No clue

Xander Henderson

My day today: drive for three hours, watch 2.5 hours of ballet, get dinner, drive for three hours. And it is totally worth it. :D

Jakobian

@AlessandroCodenotti let me ask another question. Can there exist compactification $Z$ of $X\times \omega_\alpha$ where $\alpha$ is big enough and $X$ is compact, such that there exists continuous surjection $f:X\times (\omega_\alpha+1)\to Z$ and $Z$ contains a connected set of size $\geq 2$?

what interests me is when $X$ is zero-dimensional and there can't be such connected set

so the remainder of the compactification $Z$ is like a "limit" of $X\times \{\alpha\}$

can this be set up so that it never results in some connected set

oh and $X$ needs to be at least continuum in size

my suspicion is that this function $f$ might be "rigid" enough as to not change it in significant way, but then again, lots of things are quotient of the Cantor set

and if $C\times \omega_\alpha$ has one-point compactification then why couldn't the remainder be, say, $[0, 1]$

Thorgott

15:18

@Ben do you happen to have ever come across an example of a simplicial map that is anodyne and a bijection on vertices yet not in the saturated class generated generated by the horn inclusions in dimension $\ge2$?

Ben Steffan

15:28

no :(

I've never really needed to work with explicit maps that much, at least so far

Thorgott

understandable, I don't really need this example either

it would just calm me down morally

Ben Steffan

If you ask on the main site I'm sure Daniël will tell you an example :)

Thorgott

15:47

yeah, I'll just do that

Binky

Hi

Ryder Rude

hi

Sine of the Time

hi

Binky

$\int^{\infty}_0 e^{-x^2} \cos \begin{pmatrix} 0 & x & x \\ 0 & 0 & x \\ 0 & 0 & 0 \end{pmatrix} dx$

How do you calculate the integral of a matrix?

How can I write the integral proportionate to the text?

I mean in LaTeX

Sine of the Time

first find the cosine of the matrix

Binky

15:58

em

Ben Steffan

@Binky try \displaystyle

I don't know if mathjax has the appropriate packages for proper scalable integral signs

$\bigint$ <- yeah that doesn't work

Binky

$\cos(A) = \begin{pmatrix} 1 & 0 & -\frac{x^2}{2} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

Found it

@BenSteffan :(

Now I have to multiply the exponential for each element of the Matrix

Vladimir Lysikov

@Binky Should be $-x^2$, not $-\frac{x^2}{2}$

Binky

are you sure?

Sine of the Time

the integral of a matrix is the matrix of the integrals

Binky

16:07

@VladimirLysikov You have to multiply by -1/2

Vladimir Lysikov

@Binky Yes, but $A^2 = \begin{bmatrix}0 & 0 & 2x^2 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$ has a multiple $2$ here

Binky

Oh got it

However I had to compute $A^4$

$\cos(A) = I - \frac{A^2}{2!} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} - \frac{1}{2} \begin{pmatrix} 0 & 0 & x^2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$

Sine of the Time

this appeared on my yt home a couple of days ago

Binky

Simplifying I get what I wrote above

@SineoftheTime can you send the link pls

Sine of the Time

Come si calcola l'integrale di una matrice? Esercizi integrali difficili

►

Vladimir Lysikov

16:13

@Binky It's not hard to compute $A^4$ here, but one can also invoke the fact that strictly upper triangular matrix is nilpotent, and so $A^3 = A^4 = \dots = 0$

Sine of the Time

yeah, you just need $A^2$

Vladimir Lysikov

@Binky Oops, sorry. I'm wrong about $A^2$

Binky

I checked the video and it turns out to be like mine

@VladimirLysikov ah ok dw

Sine of the Time

@Binky where did you find it?

since you're italian and it was posted by an italian channel, I assume you saw that video

Binky

they sent it to me

Sine of the Time

16:24

the feds?

Binky

?

Wdym

Sine of the Time

nothing

Binky

Oppon Gangnam style

Sahaj

19:06

Perelman is a legend.

Martin Brandenburg

any idea what the fitting subgroup of $F_2$ (free group on two gens) is? (by def the fitting subgroup is generated by all normal nilpotent subgroups.)

leslie townes

19:32

huh. subgroups of free groups are free, right? i guess any cyclic subgroup of F_2 is nilpotent (because abelian), but is any such subgroup normal? can a non abelian free group be nilpotent?

maybe its the trivial subgroup

Thorgott

19:50

good point, a non-cyclic free group isn't nilpotent and a cyclic subgroup isn't normal

psie

20:11

Let $\{T_n\}$ be a Cauchy sequence of bounded linear maps. I'm reading a proof of the fact that the space of bounded linear maps from $X$ to $Y$ is Banach if $Y$ is. There's this statement that I'm getting hung up on. The authors have shown $\{T_nx\}$ is also Cauchy and converges to $Tx$. Then the authors estimate $\|T-T_m\|$ from an earlier theorem by $$\|T-T_m\|\leq\liminf_{n\to\infty}\|T_n-T_m\|,$$and say $T_m\to T$ in norm as $m\to\infty$ because $\{T_m\}$ is Cauchy.

Intuitively this makes sense, but I want to show this formally. I'm a bit uncertain how to use the definition of the $\liminf$ here to show $\liminf_{n\to\infty}\|T_n-T_m\|\to0$ as $m\to\infty$. How would you deal with the above statement?

Thorgott

the $\liminf$ is unnecessary, $(\lVert T_n-T_m\rVert)_n$ is a convergent sequence

psie

20:29

Ok, I have to think about this, so there's no circularity in the logic of the proof.

Thorgott

20:41

$\lvert\lVert T_n-T_m\rVert-\lVert T_{n^{\prime}}-T_m\rVert\rvert\le\lVert T_n-T_{n^{\prime}}\rVert$

so $(\lVert T_n-T_m\rVert)_n$ is a Cauchy sequence of real numbers, thus convergent

psie

21:00

Ok, it makes sense now. We can replace the $\liminf_n$ with $\lim_n$, and then as we take the limit with respect to $m$, we just get a double limit and $\{T_n\}$ is Cauchy. Bingo. Thank you Thor! :)

psie

22:29

Has anyone tried ChatGPT for converting images to Latex? I don't know if it's possible, if it accepts images?

Ben Steffan

good luck lol

you might be interested in mathpix, however

psie

@BenSteffan well, I know of mathpix, but it's not all free

Ben Steffan

no, but chatgpt is useless

also mathpix doesn't produce great output

psie

ok :) mathpix will probably do then

yeah mathpix sometimes gives you \backslash instead of \setminus, that's annoying

And there are other commands which it kind of gives you some alternative which isn't really what you wanted

Ben Steffan

the last time I used it mathpix couldn't even spit out proper enumerates

rather embarassing for a paid service

but I personally wouldn't use it either way because it does not fit in with my personal macros, environments etc.

psie

22:39

I see.

copper.hat

23:00

@PM2Ring For some reason I thought of you when I hear this youtu.be/3coSlNZX_nI?feature=shared

I tried CrapGPT once. That was sufficient. It was as if all the useless service bots I have ever interacted with read Ulysses, had a few drinks and then let loose.

leslie townes

we live in a golden age

copper.hat

Just imagine the Irish version of CrapGPT.

BlarneyGPT

Very quite here since our friend took a break.

leslie townes

23:20

tumbleweed rolls by

think_meaning_builds

Did anyone watch the only walk-off grand slam?

copper.hat

jewel lake is full again

i had to look up what a grand slam is

other than when i was young and and i talked back to my mother

think_meaning_builds

The first ever in the world series.

copper.hat

indeed, i had to look up what the world part of world series meant as well

Ben Steffan

why is AG terminology so messed up

who hurt them

flabby sheaf, soft sheaf, perverse sheaf

copper.hat

23:25

jk. i was introduced to the world of baseball when my son was asked to play in aaa

indeed, mathematicians spend a lot of time discussing balls

there was the time when i asked my mathematics professor about the delta durex function.

Ben Steffan

the sheaf is flesh. it is warm to the touch. you can feel it pulsate under your hand, as if somewhere amidst its stalks a heart was beating, slowly.

copper.hat

BibleGPT

Thorgott

@BenSteffan im okay with flabby and soft, but the perverse ones really earned their name

spent an entire afternoon with a friend once trying to parse the definition, it wasn't very successful

copper.hat

just the name alone AG makes you wonder

like mathematical physics

i haven't even started drinking yet

think_meaning_builds

what about physical mathematics

::pours the drinks::

psie

23:31

@copper.hat not only balls, there is the Tits group.

Ben Steffan

and also the tits alternative, and also tits buildings

Sine of the Time

our favourite group

Ben Steffan

god bless Jacques Tits

copper.hat

i have always had an interest in the tits group, i must say

think_meaning_builds

🍷🍸🍹🍺🍻🚰🥛🥃

copper.hat

23:32

related to Jacques Eauff

Ben Steffan

pff

copper.hat

the mind is full of possibilities

Ben Steffan

the mind is full of boggles

psie

another one; Cuntz algebra

copper.hat

when my daughter was in her teens she used to whup me in boggle, so much so that we decided that in order for he to win, she had to get 2x mu score.

there, i self censored

Thorgott

23:40

my favorite is the Cox-Zucker machine

Sine of the Time

@psie Wiener Measure

Ben Steffan

cracking up

copper.hat

i have a feeling i will be getting another suspension

psie

boy I'm laughing

Ben Steffan

isn't that a good thing? your homotopical properties will improve :)

Sine of the Time

23:42

The important problems in life

copper.hat

how come there are no heterotopies?

lol to the happy ending

Thorgott

@BenSteffan he will lose all his cup product information

and his Steenrod squares!

Ben Steffan

he will not lose his Steenrod squares

Thorgott

wait, uh, the other thing

Ben Steffan

but clearly he was rational to begin with so losing the cup product information is a big deal

Sine of the Time

23:44

I'm crying: en.wikipedia.org/wiki/Slutsky%27s_theorem

Thorgott

I remember this one from probability class

copper.hat

group theory suddenly became more interesting

Ben Steffan

don't look up what notation algebraists use for associated primes

or how topologists abbreviate the Adams spectral sequence

copper.hat

would that be an asinine remark?

Ben Steffan

asine? isn't that the inverse to sine?

copper.hat

23:47

i was thinking more of 7 of 9

Sine of the Time

yes, it's my inverse

copper.hat

Seven of Nine

Seven of Nine (born Annika Hansen) is a fictional character introduced in the American science fiction television series Star Trek: Voyager. Portrayed by Jeri Ryan, she is a former Borg drone who joins the crew of the Federation starship Voyager. Her full Borg designation was Seven of Nine, Tertiary Adjunct of Unimatrix Zero One. While her birth name became known to her crewmates, after joining the Voyager crew she chose to continue to be called Seven of Nine, though she allowed "Seven" to be used informally. Seven of Nine was introduced in the fourth-season premiere, "Scorpion, Part II". T...

ok, i need to do something useful. i am going to buy eggs.

think_meaning_builds

cya

Transcript for

$Mathematics$ Mathematics