A question was recently migrated to Puzzling. It is blatantly off-topic there. We (Puzzling) do not accept subjective or open-ended questions. If something's looking for a big list (as the original tags indicated) then it is not welcome at Puzzling. </rant>

filthy mathematicians posting on Puzzling.stackexchange

they should be ashamed of themselves

that's actually unpleasant. it's a question that probably doesn't belong anywhere and maybe somebody thought instead of flushing, off to some other SE

I mean, we accept good math puzzles. This isn't it.

i'm not responsible, but my apologies. puzzling is puzzling, not... whatever that is.

it is a conspiracy to destroy the site

To think they were that close to being successful!

wedohomework.com

the question reads like the beginning of a conversation where maybe after an hour there's a good math question, or puzzle.

Coinbase presents Wedohomework.com. convert your SE points to Lesliecoin at very generous rates.

the problem with language is that you can say or write any string you want. there is no validation.

I don't think that's a good math question either

not sure where it belongs

it probably doesn't belong anywhere. it's half-formed. i have notions too.

move it to stackoverflow, those **** will kill it in a flash

we have a saying around the office when someone is just spouting. "you can have a thought." meaning a mental picture of something that is just yours. no sounds or images accompany it. it is just in your head. you can have those. entirely permissible.

i keep getting asked to spam check amwhy messages. why is that?

is that like amway?

someone should have spam checked them a long time ago. i guess now it's your problem.

yep :-) my spulling aoint so gud

oh i see it's a SE user. i don't know. you get too much reputation and that's what happens.

my father in law gave me a proof of the existence of god which i keep in my little spivak as a reminder.

words and meaning live in separate universes

little spivak isn't wide enough to contain what i'd need. (i'm thinking an encyclopedia set with holes carved through them and maybe 6 glasses of porter inside)

i think in my entire 60yrs i have not cumulatively consumed a single pint of guinness

i'm probably around 10 or 20.

when you're in a foreign land and you don't understand the local stuff, a pub with shamrocks and stuff in its logo will reliably serve that. that accounts for half of them.

there's usually something better but who cares.

that could be their slogan.

at least one of my brothers has consumed 24 in a similar number of consecutive hours.

gosh, i hope he was hydrating.

that wasn't high on his priority list

no california mum to make sure he drinks every 10 seconds

my daughter has internalized california culture. after going to bed last night, she woke up about 2 hours later demanding more agua. and again four hours later. she had drained the bottle.

maybe she is trekking across a desert in her dreams.

we went through a tiring bpa removal process during elementary school. god know how we survived growing up with mercury, asbestos, drinking out of cow ponds etc.

i don't have such a good memory since i drank my thermometer, but i whittles what i sees

i can still do the mercury thermometer flick that nurses used to learn

my mom learned that flick

i think we even had mercury thermometers. i suppose they are banned now

a liability opportunity waiting to happen

liability 'opportunity.' music to my ears

@copper.hat I was in line waiting to get into the Farmers Market. My appearance was but fleeting.

@Rover $x$ and $y$ change with each other. You can either consider $x$ as the independent variable that you control and watch what happens to $y$ or consider $y$ as the independent variable and watch what happens to $x$.

@leslietownes depends on how many people liked to crunch them in their teeth.

The glass would be bad enough, then there is the Hg poisoning.

oh, yeah, they made an oral version too. forgot about that.

@leslietownes I know that we still have mercury thermos in our bathroom cabinet.

i do remember the day we threw it out. we probably shouldn't have put it in the trash.

and where was this? I want to make sure I get the address correct on the report...

123 Green Apple Way, Anytown, CA, 90210

Wait... that's my address.

sometimes i use Granny Smith Apple way. if i'm feeling facetious, Honeycrisp Apple way. i don't know why but i tend to use apple names in fake addresses.

Thank goodness I don't live around the corner!

it's a very bad neighborhood.

it is a very open neighborhood, however

it's a boundary point, and also a frontier point.

Wild West, huh?

Untamed Occident

shootout at the $\mathcal{O}(k)$ corral.

The herds grow, but not too quickly

Good one!

yes, that wins.

No, my comment was for you. Robjohn's comment interloped (maybe even anteloped)!

we're way out west right now.

I'm sorta proud of my quip 🤷‍♂️

@TedShifrin I ruin the chat again.

Nah, you set me up! 😍

@leslietownes in the valley, there is a lot of distance to the west of us; Oxnard, etc. Not so much to the south of us. The Pacific is much closer to the south, here.

same. the ocean is a straight shot south. owing to the length of the beach.

Yeah, the coastline goes surprisingly almost due west heading toward SB.

When driving home from UCLA, I take Sunset to the coast, then a few miles to Topanga Canyon, then north to my house (which is just a few blocks from Topanga Cyn)

It's a drive that I enjoy immensely

i go up and down PCH between long beach and newport beach. it's nice except when i turn on the news and see some kind of pro-coronavirus protest in huntington. "you're blocking traffic!" i think. "what would i do if the office were open and i were trying to come home!"

lots of good birding on the drive, and i can always see when the surf's up. love it.

@leslietownes they have pro-coronavirus protests?

that's my term for them. and white power rallies. huntington really is a cesspit. nice beaches though.

Yes, the drive along the coast and then in the canyon has such wonderful scenery and there are lots of birds to be heard in the canyon, especially in the morning.

Ah, non-masked, less than 6 foot crowds.

i talked to a guy at the ADL once and apparently an enormous amount of time they spend keeping an eye on stuff in california is spent in huntington beach. it's something in the water.

protest was the wrong word. rallies, i guess.

hah, i finally got my old computer to talk to my new computer. it is now populated with ebooks and notes and homework problems from way back. if i seem cleverer from now on it because i am cut-pasting from the archives.

Why is the unit disk not orientable? :x

I expected it to be orientable (I have no argument tho, since I feel unsure about orientations atm), but then one would expect it to have a genus, since it's a compact 2-manifold, and that's not the case...

er, isn't it orientable?

Um, simply connected mumble mumble mumble ...

@ShaVuklia what makes you say it isn't?

what's the definition of 'orientable'? full disclosure, i never understood orientation, and i'm left handed, which means i'm going to get it wrong even if i do understand it.

Compact manifold WITH BOUNDARY.

stops yelling

I meant closed unit disk x')

oh yeah, wow. that's not what i normally use 'disk' to mean. still feels pretty orientable.

That has a boundary, no?

i'm walking around it. i've been in the first quadrant, the second, third, and fourth. i've been all over. i'm still on one side of it.

can someone please tell me the class equation of $Q_{8} \times S_{3}$. I am trying to learn the class equation of external direct product of non-abelian group, and this is the only example I have for which I have a partial answer (trying to see pattern). wolfram alpha does not understand the query, and google also dosen't have something related.

I know how to write it for abelian times non abelian groups, so trying to see pattern.

Ok, so... if it is orientable, then why doesn't it have a genus?

isn't it just genus zero?

where are these assertions coming from? genuinely curious.

this closed v. open thing seems to matter.

Sha, you missed the point of my yelling.

i'm thinking. on a sunday. i hope you're happy

@TedShifrin Ye sry, I didn't understand what you were saying.

Classification you're thinking of is for compact manifolds with no boundary.

Ahh

I had missed that

Thanks, that explains it

That's why I yelled!

ok, so I was thinking of the unit disk with boundary, which is not a manifold. got it.

@TedShifrin I didn't miss the yelling x'D I just couldn't guess what you meant x'D but thanks for the clarification x'd

There is a modified version with capping off boundary components with disks. Work it out ....

@leslietownes Using the simplest triangulation on the unit disk, I found that its Euler characteristic would be odd, and that would contradict a result which relates the Euler number to the genus (which would have to been even). However, I had missed that the result only deals with manifolds without boundary

Contractible things have the homotopy type of a point.

i see it too now. it's a little confusing.

math.stackexchange.com/questions/219290/… raises similar issues.

palio's answer.

I detest people who are ambiguous/lazy and say manifolds for manifolds with boundary.

@TedShifrin Ye, I might do that another time. I'm actually busy reading about Riemann surfaces, and the book only briefly mentions some results about triangulations/Euler characteristic, etc. I will find some time in the future to go through these concepts a bit more formally.

you know ted i was about 2/3 through my first topology class before i realized the distinction. the prof and book never brought it up.

it really frightened me. it had been swept under the rug.

@TedShifrin or ignorant of the distinction as I was a couple of days ago.

@leslietownes Ya, indeed!

@TedShifrin I've don that

granted, I always make it clear from the start whether I'm working with boundaries or not

i had a similar time embarrassingly late in my thesis research, when i realized some of the sketches i had made to explain things to myself were real-scalar pictures, and because i was working over complex scalars, everything i was actually dealing with was twice the dimension.

frightening stuff.

a few years ago somebody emailed me out of the blue asking for fully worked out solutions for a textbook my phd advisor had authored. i had a set of those, but i don't know how she knew i had them. it would be a weird thing to ask for if you didn't know it existed.

i still think about that from time to time.

Is the class equation of $S_{3} \times Q_{8} : 48 = 1+1+2+2+2+2+2+4+4+4+3+3+6+6+6$ ?

Who the hell knows.

if $\lim\limits_{x \to \infty} f(x) = 0$ is it possible for $\lim\limits_{x \to \infty} f'(x) \neq 0$?

yeah. consider something like /\/\/\/\/\/\/ except you fiddle with the height of the peaks.

fiddling with the height of the peaks does not affect the slopes of the lines.

if i'm wrong someone stomp on me now. i've been addled by reading legal documents all morning.

Leslie is correct, for once.

I'm not sure I follow, might have to consider the example in a bit more detail.

is that like a saw tooth?

try something like (sin(x^2))/x. it's gonna go to zero because |sin| <= 1. but when you take the derivative you get stuff that isn't going to zero.

I'm not gonna work out the example for you, but the conjugacy classes in a product are just products of conjugacy classes of the factors

i'm trying to keep it differentiable.

ok

I will play witht he sin example, thanks

so you just have to work it out for the factors, which will just be a handful computations

and the key property there is that the function can be as crazy as it wants, slope-wise, as long as it's getting closer to zero, height-wise. going to zero is no restriction on how crazy the slope can get. that's how i fit it together in my mind.

Take $1/x$ times $\sin(x^2)$, for example.

ted has ingeniously stolen my idea and concealed his tracks by time traveling into the future with it. the past would have been a more devious option.

we're agreeing so much it's beginning to bother me.

Hey, I said you were correct!

I just provided the technicalities .

well you did put it in dollar signs, which i should be better about.

I'd better go before you and ted become the same person

thanks again

i remember 20 years ago everybody was like 'we need some way of seeing latex on irc.' now it's actually here. i should use it.

LOL, bye, @Andrew.

:)

cheers andrew.

so I can just write the class equation of $S_{3} \times Q_{8}$ by multiplying each element in the class equation of $S_{3}$ to each element in the class equation of $Q_{8}$? @Thorgott

might this more profitably be phrased in terms of conjugacy classes?

the class equation has this dusty, 19th century feel to it. we can do better. or maybe we can't.

class equation is classy

it's official, i'm downloading a distribution of latex to my newer computer. look out, universe. arxiv math.GM is about to see a bunch of trash posts.

pun intended

the miktex distribution site is giving me like 100k/s. i might as well be using xmodem.

yes

Dial up on your cellphone.

though phrasing this in terms of the class equation seems awkward

what you care about is the size of the conjugacy classes

the miktex default is A4. i'm sorry, did we lose a war? letter. always letter. i don't know what A4 is and i don't care to.

Isn't it a highway in England?

probably. i'll take the A4 from Bishops Womble down to Potty Knob. that's a normal thing for an englishman to say. but not an american.

thanks :) @Thorgott

i can now read DVIs on my computer. look out, universe. i'm reading my math history term paper from 2001.

@leslietownes I disapprove of the racism

as do i. are we back to huntington beach?

Hi! I have $A\in\mathcal{M}_{3\times 3}(\mathbb{R})$ and $A^\dagger A=k\operatorname{Id}$ where $k\in\mathbb{R}_{>0}$, also we assume that $A$ has an eigenvalue $\lambda$ with multiplicity $3$ (denote the eigenvectors by $a_1,a_2,a_3$. Consider the following operator: $M=k(A^t-A)$ I showed that $Ma_1=k(\frac{k}{\lambda}a_1-\lambda a_1)$, now I suspect that $\lambda^2=k$ but I can't show it from $A^\dagger A=k\operatorname{Id}$ alone, can anyone shine light here with some hint?

hi all

sorry to spam but it might be an interesting edit that might end in an answer

@user153330 if you have three linearly independent eigenvectors with the same eigenvalue, the matrix is diagonal.

@TedShifrin thanks, yes i forgot that fact but i figured it out after :)