The h Bar

I can understand why the map might not be bijective if $k$ were not relatively prime to $n$ but for it to not be surjective means either one of the elements in the domain cannot be mapped to $x^{ak}$ or the map isn't a map

ACuriousMind

@Obliv Look at easy examples. For instance, consider $\mathbb{Z}/4\mathbb{Z}$ and the map $1\mapsto 2$.

Obliv

4:13 PM

additive?

ACuriousMind

Yes, written additively.

(it's not a group under multiplication)

Obliv

$\{0,1,2,3,\} \to \{0,2\}$

right?

ACuriousMind

What is "e", and what is that supposed to mean?

The neutral element in $\mathbb{Z}$ is 0!

Obliv

oh right

oh wait it should be multiplication when raised to an exponent

that's surjective!

0 maps to 0, 2 maps to 0

1 maps to 2, 3 maps to 2

ACuriousMind

How on earth is that surjective?

Obliv

4:17 PM

all elements in domain get mapped to the codomain?

it's not injective

ACuriousMind

...name one function where that isn't the case

Obliv

isn't that the definition of surjectivity though

ACuriousMind

No.

Obliv

like $y = x^2$ is surjective

but not injective

ACuriousMind

$f(x)=x^2$ is neither surjective nor injective as a function $\mathbb{R}\to\mathbb{R}$.

Obliv

4:19 PM

wtf

0celo7

@Obliv I'm jealous, you get all of Bajoran's attention these days

Obliv

imgur.com/NMlQyvy this is from my textbook

ACuriousMind

@0celo7 Well, I have no idea what your problem is. Milnor does not claim there that such functions always exist, so what is the issue?

Obliv

@0celo7 math chat is just really off topic otherwise i'd ask them and not bother mother duck

ACuriousMind

@Obliv And it is correct. You have to learn to read more carefully.

Obliv

4:24 PM

there is some element in $\{0,1,2,3\}$ that gets mapped to $\{0,2\}$ for all $\{0,2\}$

there's something crucial that i'm missing

ACuriousMind

@Obliv Wait, when did you edit that to $\to\{0,2\}$.

Of course the function becomes surjective if you restrict to the elements it maps to!

Obliv

after i said it should be multiplication when raised to an exponent

ACuriousMind

The exercise is about the map being surjective as a map from the group to itself.

Obliv

well isn't the codomain the image of the domain?

OH

did it specify that? I'm so mad if it did and I didn't notice that

ACuriousMind

@Obliv It did not but that is supposed to be obvious.

Because you do a priori not know what the image of that map is like, you can only define it as a map $G\to G,x\mapsto x^k$.

Of course you could write $G\to f(G),x\mapsto x^k$ but that would just be silly.

Obliv

4:28 PM

it just said prove $x \to x^k$ is surjective. I wouldn't read that as prove that $x\to x^k$ is a surjection to the group

nvm

Emilio Pisanty

btw

-2

A: How do I ask homework questions on Physics Stack Exchange?

We can ask question with some conceptual doubt or the students can show their try about the question .

Say, @ACuriousMind, what's your take on that one ↑?

ACuriousMind

@EmilioPisanty I agree with your comments and voted to delete

I'm also not sure what exactly the answer was supposed to say

Emilio Pisanty

@ACuriousMind Yeah, that too

Physics Meta

4:46 PM

0

Q: How is it possible that a user casts more than 40 vote in a day?

I wondered when I saw this: How is it possible?

bug

lucas

4:58 PM

Should I delete this post and ask it in Meta Stack Exchange?

Isn't it better to migrate it rather than deletion?

TheQuantumMan

Hello, could I ask you guys and gals about your opinions on some physics textbooks?

Bernard Meurer

@EmilioPisanty It should be renamed DopeHammer

0celo7

@ACuriousMind ??? Of course it is

Surjective means precisely that f(X)=Y

ACuriousMind

@0celo7 "All elements in the domain get mapped to the codomain" is not the same as "All elements in the codomain are images of an element in the domain". The former just says that $f(X)\subset Y$.

0celo7

5:15 PM

@Obliv please explain what you meant

Obliv

for what @0celo7

0celo7

To me those are the same statement

Obliv

well actually @acuriousmind the next question poses this : $\sigma_a: Z_n \to Z_n$ given by $\sigma_a (x) = x^a$ for all $x \in Z_n$. Prove that $\sigma_a$ is an automorphism if $a$ and $n$ are relatively prime to each other.

as part $a)$

but isn't that essentially what the last question was asking

eh maybe not. for the map to be surjective to the group (assuming the group is all of the codomain) perhaps injectivity wasn't necessary

ACuriousMind

@Obliv Being an automorphism is stronger than being surjective, though in this case not by much.

It might be stronger though if you haven't proven that any bijective homomorphism is an isomorphism.

Obliv

well isn't that a matter of definition

what's there to prove

ACuriousMind

5:21 PM

Usually one would define an isomorphism as a homomorphism that has a two-sided inverse.

Since e.g. a continuous map between two topological spaces can be bijective without being a homeomorphism.

Bijective only means a map has an inverse as a map on sets, but it is not a priori guaranteed that the inverse is a homomorphism, too. (Although this is guaranteed for groups)

Obliv

A bijective map's inverse is guaranteed to be a homomorphism for groups?

ACuriousMind

I should have been more precise...

0celo7

@ACuriousMind Proof?

Obliv

hmm do maps between groups have to be homomorphisms/isomorphisms?

ACuriousMind

An isomorphism is a homomorphism that has a two-sided inverse homomorphism. For group homomorphisms, it is true that a bijective homomorphism is always an isomorphism.

0celo7

5:26 PM

Oh

ACuriousMind

@0celo7 Left as an exercise for the reader.

0celo7

@ACuriousMind I misunderstood you.

Is that true for all groups or just finite ones?

ACuriousMind

All groups

Obliv

@acuriousmind if we assume the codomain to be the entire domain (this is the case of automorphisms in a cyclic group), does surjectivity assume bijectivity in this case?

rather does one lead to the other

being able to map all domain elements of a cyclic group to itself without being injective seems fishy but I haven't really investigated that

ACuriousMind

@Obliv Well, that's a question for you! When are surjective maps from a set to itself already injective?

0celo7

5:32 PM

@ACuriousMind Let $\phi:G\to K$ be a bijective group homomorphism, that is, $\phi(gh)=\phi(g)\phi(h)$. Let $\psi=\phi^{-1}$. We will show that $\psi$ is a homomorphism. Let $u,v\in K$. Then $\psi(uv)=\psi (\phi(g)\phi(h))=\psi(\phi(gh))=gh=\psi(u)\psi(v)$.

Obliv

wait duh if the domain and codomain are required to be same size, as well as surjectivity, the only way is to have the map be injective as well. Otherwise some elements would have two images?

ACuriousMind

@Obliv That is wrong. The map $\mathbb{Z}\to\mathbb{Z}, n\mapsto\lfloor n/2 \rfloor$ where $\lfloor n/2\rfloor$ is the number rounded down is surjective but not injective although the two sets are the same size (it's the same set after all).

Obliv

oh is this wrong in the case of infinite groups then

I should specify then that this applies for finite cyclic groups

ACuriousMind

@0celo7 Yep

Obliv

I tried $Z_4$ and $Z_5$ and noticed injectivity. Then I realized if it were not injective, the codomain would not be same size as the domain

ACuriousMind

5:37 PM

@Obliv Well, but how do you know it is true for finite sets? Prove it, don't wave your hands!

0celo7

I am noticing similarities between @Obliv and me a year ago.

@ACuriousMind Do you see them too?

Obliv

let $\varphi: G \to G$ be defined as $x \to x^k$ and let $|x| = n$. If $(k,n) \ne 1$ we have codomain < domain in that $\varphi(x^a)$ where $1 \leq a < n$ produces less elements than the domain. if $(k,n) = 1$ then $\varphi(x^a)$ produces $\langle x \rangle$. If the map is to be surjective, this requires $\varphi(x^a) = x^b$ to be unique otherwise we have $\varphi(x^a) = \varphi(x^c) = x^b$ so that the codomain is again smaller

@0celo7 except I'm not learning any geometry yet :p what similarities do you see though

0celo7

@Obliv You're dominating the chat with involved questions directed at ACM

You're going to become too reliant on him

I learned that it's sometimes good to stew on things

Also he's told me to prove things without handwaving many, many times.

Because he likes algebra so much he'll tolerate your questions for longer.

But I've since moved on to math that he doesn't know or find interesting so I'm SoL most of the time.

Obliv

@0celo7 seeing as how I have never taken a formal class on proofs/abstract algebra or even linear algebra I don't see how I can go through this textbook alone and be able to do every exercise unless I asked other people. I am not reliant on just ACM either, I ask people in math chat or just ask questions on math.se

It's not like if ACM decided not to answer me or wasn't in the h-bar I wouldn't do any studying.

ACuriousMind

@0celo7 Some, yes

@0celo7 "SoL"?

Emilio Pisanty

5:52 PM

Any thoughts on a cosmic-neutrino-background tag?

Bernard Meurer

@EmilioPisanty Yes, potato

Emilio Pisanty

gah

one character too long

ACuriousMind

@EmilioPisanty Just one: ...there's a cosmic neutrino background?

Emilio Pisanty

@ACuriousMind You've not seen the recent question spree?

There's all this cool stuff

44

Q: Where are all the slow neutrinos?

The conventional way physicists describe neutrinos is that they have a very small amount of mass which entails they are traveling close to the speed of light. Here's a Wikipedia quote which is also reflected in many textbooks: It was assumed for a long time in the framework of the standard mo...

special-relativity particle-physics neutrinos

Slereah

Well obviously

Emilio Pisanty

5:53 PM

6

Q: Is it possible that all "spontaneous nuclear decay" is actually "slow neutrino" induced?

This thought was inspired by a comment from the current leading answer, by @Sentry, to the question Where are all the slow neutrinos? This [slow-neutrino induced nuclear decay] will still be an extremely rare process and the big problem is to distinguish it from normal spontaneous nuclear dec...

special-relativity particle-physics dark-matter neutrinos

Slereah

Lots of neutrino production during baryogenesis

Emilio Pisanty

11

Q: What does the cosmic neutrino background look like today, given that neutrinos possess mass?

This question is inspired by (or a follow-up to) the threads Where are all the slow neutrinos? and Is it possible that all “spontaneous nuclear decay” is actually “slow neutrino” induced? The cosmic neutrino background (CνB) consists of the "primordial" neutrinos from the time when the universe ...

cosmology big-bang neutrinos

a few older ones

13

Q: Do primordial background neutrinos orbit in dark matter halos?

According to Wikipedia, neutrinos separated from other matter seconds after the Big Bang and formed a separate background radiation field which now fills space at a temperature ~2 K. Supposing neutrinos have a rest mass of 2 eV/c2, they should be traveling at an average speed $$v_\nu = c \sqrt{...

cosmology big-bang neutrinos

8

Q: If we could build a telescope to view the cosmic neutrino background, what would we see?

If we could build a neutrino telescope capable of viewing relic neutrinos that decoupled after the big bang, with a similar angular and spectral resolution that is possible now for the CMB (e.g. with Planck), what would we see How would the C$\nu$B differ because of the finite neutrino mass and ...

cosmology big-bang neutrinos

7

Q: Seeing beyond the CMBR with neutrinos?

As I understand it, you can't see beyond the Cosmic Microwave Background Radiation because the plasma of the early universe was opaque to electromagnetic radiation. What if you had a "neutrino telescope" with sufficient resolution? Would you be able to observe the primordial universe from before ...

big-bang neutrinos cmb

John Rennie

@0celo7 you're a student. They are regarded as being slightly less valuable that stationery. Graduate students are worth about the same as a low end calculator.

Emilio Pisanty

My own humble contribution

0

Q: How would cold neutrinos get trapped by stars?

Continuing on from the cool physics Q&A'd on the threads Where are all the slow neutrinos?, Is it possible that all "spontaneous nuclear decay" is actually "slow neutrino" induced?, and What does the cosmic neutrino background look like today, given that neutrinos possess mass?, I have a follow-u...

particle-physics cosmology astrophysics big-bang neutrinos

Bernard Meurer

@JohnRennie My calculator can run Linux, can a graduate student? No.

Useless

ACuriousMind

5:55 PM

@EmilioPisanty I didn't notice anything unusual about questions with neutrinos...then again, I only glance over the cosmology questions, usually

John Rennie

Correction: graduate students are worth slightly less than Bernard's calculator.

Though having said that, in a tight spot you couldn't eat a calculator.

2

Emilio Pisanty

@JohnRennie Oi, there's grad students around

ACuriousMind

@BernardMeurer You have not yet installed Linux on a human?

Bernard Meurer

@JohnRennie That's a good point, I wonder whether graduate students taste like pizza

Emilio Pisanty

Anyway, seven cool question makes a tag, right?

Bernard Meurer

5:57 PM

@ACuriousMind It's on the works, I need to overcome my fear of needles

ACuriousMind

And here I thought you went around asking people "May I install Linux on you?" who are then slightly disappointed that it is not a weird pickup line

@JohnRennie In a tight spot, the grad student might also eat you, though.

3

John Rennie

@ACuriousMind I'm too old and tough - you'd have to stew me for days. A nice young tender student though ...

Bernard Meurer

@ACuriousMind I have tried that line, she didn't know what a Linux was

@JohnRennie gasp

ACuriousMind

@BernardMeurer And you didn't offer to show her your Linux?

Bernard Meurer

@ACuriousMind I couldn't help it, I gave her an overview on the kernel

yes, that is an innuendo

Emilio Pisanty

5:59 PM

@BernardMeurer Terrible strategy

Better start with the Barry White collection

Bernard Meurer

@EmilioPisanty Hahaha

John Rennie

@BernardMeurer isn't that a type of supository?

Bernard Meurer

@JohnRennie Not that kernel either

The FreeBSD Kernel!

It's easier to explain quickly :p

John Rennie

Oh well, that joke fell a bit flat. Perhaps I should put it behind me.

Bernard Meurer

@JohnRennie Hahahaha

Good one, good one

John Rennie

6:02 PM

Childish jokes are always the best ones.

Scatalogical childish jokes are even better.

Bernard Meurer

@ACuriousMind I'll visit you in Heidelberg, make sure you buy potatoes

John Rennie

Bernard Meurer

@JohnRennie AAAAAAAAAAAAAAAAAAAAAAAH

John Rennie

I just saw this. Isn't it a beauty?

Bernard Meurer

::flees::

John Rennie

6:04 PM

It's a hawk moth - quite harmless.

Bernard Meurer

It's HUUUUUUGE

(that's what she said)

John Rennie

It was on the windowsill at my local supermarket. It is big - about the size of my palm.

Bernard Meurer

I'd walk with a flamethrower if I lived there, holy crap

John Rennie

Those pictures were taken with my phone. It's amazing just how good phone cameras are these days. A Linux derived OS too, of course.

Bernard Meurer

Android is p. cool

I wish ubuntu mobile was good

Emilio Pisanty

6:14 PM

@JohnRennie yeah, about that

twitter.com/swiftonsecurity/status/…

Bernard Meurer

@EmilioPisanty Oh please Linux has had malware for ages

It's not that hard really

I can write a malware for linux with more ease than for windows

make a rootkit for the Linux kernel is pretty easy too

what makes Linux safeis that users know what they are doing more often

Emilio Pisanty

Just jumping at the chance to quote Swift on Security

Bernard Meurer

and don't just go around running everything as root

ACuriousMind

@EmilioPisanty I was going to ask if you have twitter messages from two years ago bookmarked, but I guess that answers that... ;)

Emilio Pisanty

@ACuriousMind Nah, I just googled for the quote (precise search google.co.uk/…)

John Rennie

6:18 PM

@EmilioPisanty well I'm kind of an old time nerd whose preferred user interface is a command line. But the OnePlus 3 phone I've recently bought is just wonderful. And yes, you can open a command line if you want to :-)

Bernard Meurer

I hate the whole "OSX/Linux doesn't have viruses" talk, it takes the blame away from the user; your Windows PC gets e-HIV because you don't know what you're doing, don't just blame everything on the OS

@JohnRennie Ah, you got the OP3, cool! Could you give me you're thoughts/review on it

John Rennie

@BernardMeurer It's perfect.

Oh, did you want more detail? :-)

lucas

@EmilioPisanty Hi! Do you have some time? Less than 10 min!

Bernard Meurer

@JohnRennie If possible :p

I imagine the whooping 6GB of ram must be cool

John Rennie

@BernardMeurer Basically it's like a Nexus but beautifully made and with an excellent camera. It has the current fastest CPU (Qualcomm 820) and while 6GB RAM is overkill I guess it's nice to have. The user interface is streamlines like a Nexus, and the few additions OP have made are all good.

Bernard Meurer

6:25 PM

@JohnRennie Sounds awesome, I've been wanting to drop the iPhone 5 for a while

0celo7

@JohnRennie cool

John Rennie

Basically I cannot recommend this phone enough. A Galaxy 7 or (spit) iPhone will be twice the price, and they are maybe 10% better, if that.

Bernard Meurer

it's battery barely works

0celo7

@ACuriousMind Shit outta luck.

Bernard Meurer

@JohnRennie How is it with waterproofing? I spill a lot of beer

John Rennie

6:26 PM

I don't keep the WiFi and Bluetooth on, and my phone uses about 8% of the battery per day.

Bernard Meurer

Le Tombeau de Couperin III.Menuet

►

John Rennie

If you turn everything on you're looking at about a day and a half of battery.

Bernard Meurer

Ravel was ambulance driver in the first world war, lost 6 friends, and wrote this after the war

for our french mates out there

John Rennie

It isn't waterproof. Don't spill beer on it.

Bernard Meurer

@JohnRennie A DAY AND A HALF WITH EVERYTHING ON?!

Holy cow that's a lot!

How does that speed charge thing work? Well?

John Rennie

6:28 PM

To be fair that's pretty standard for the top end Android phones these days.

Bernard Meurer

My iPhone 5 needs to be fully recharged 3 times a day

it's horrible

John Rennie

The charging speed is astonishing. I've never had to charge it froma very low level, but from say 70% it's fully charged in about 20 minutes.

ACuriousMind

I killed my OP1 a while ago. I miss it :(

2

Bernard Meurer

@ACuriousMind What do you run now?

John Rennie

The iPhone 4 and 5 battery life are rubbish. To be fair Apple have improved things in the 6.

ACuriousMind

6:29 PM

@BernardMeurer Uh...a Galaxy Mini whose number I have forgot

John Rennie

@ACuriousMind the OP1 was nice, but the 3 is on a whole new level. I genuinely feel it's perfect i.e. I can't think of anything about it I'd change.

Bernard Meurer

@JohnRennie I know, a microSD slot

John Rennie

And of course in Euros its price hasn't changed. In GBP the price just went up 10%. Now why might that be I wonder ... :-)

ACuriousMind

I mainly bought that because it came with a deal that they will replace it in the next three years free of charge no questions asked. Considering I killed the OP1 in an unfortunate drunken accident that seemed like a rather good deal.

John Rennie

@BernardMeurer it's got 64GB built in. Why do you need a microSD card?

Bernard Meurer

6:32 PM

@JohnRennie If I compress all my music down to 128kbits AAC it's still ~50GB

ACuriousMind

@JohnRennie To use the data on that SD card without having to transfer it wirelessly, perhaps?

Bernard Meurer

I use a 128GB card on my music player

John Rennie

Fine, then your entire collection will fit on the built in memory

Bernard Meurer

@JohnRennie Yeah but then I won't have space for all cool apps and pictures

I don't like cloud storage

Emilio Pisanty

@lucas I've got two minutes if you can make it snappy.

John Rennie

6:34 PM

Well, it doesn't have an SD card slot. But apart from that it's perfect :-)

Bernard Meurer

@JohnRennie Thanks for your thoughts :)

John Rennie

Seriously, it is an awesome phone. And if you fancy developing for a phone it's a perfect platform.

vzn

@EmilioPisanty hi cant recall exactly but did you say once you were open to guest speaker?

John Rennie

I bet someone will port Linux to it.

Bernard Meurer

@JohnRennie Is this a challange?

John Rennie

6:37 PM

In fact:

linux.softpedia.com/blog/…

Bernard Meurer

@JohnRennie Lame, I thought you meant real Linux

like compiling a kernel for it, working on a bootloader

you know, art

vzn

@JohnRennie quite impressive thx for sharing... looks like something from a south american jungle... thought it was BM posting it at 1st lol

John Rennie

@BernardMeurer well, there's a challenge for you :-)

Bernard Meurer

@vzn I don't photograph insects the size of my hand, I either flame'em or run

Cause down here everything is poisonous

vzn

reminds me some of a crazy viral picture of a large bat from s.america from awhile back... huge wingspan... several ft...

Emilio Pisanty

6:42 PM

@vzn In principle, sure

At the moment, that's pretty hard

John Rennie

@vzn I don't recall seeing many hawk moths in past years, but I've seen two this year. I wonder if global warming is increasing their range.

vzn

@EmilioPisanty re this have been thinking about it

yesterday, by Emilio Pisanty

@vzn You're "open to handing that off"? I imagined you'd be saying "I'm perfectly happy to take charge of it"

John Rennie

I grew up in Khartoum in The Sudan (my father was working there) and we used to get lots of hawk moths there. But it's a bit hotter in Sudan than in the UK - as in Sahara desert hot :-)

vzn

right dont really see a (feasible) way to mod the chat room/ spker session much better than what DZ et al are doing (& its natural to work with them on that). think/ agree fmt is not ideal right now, but dont see alternative either. would like to see it scale further but its very early/ experimental right now.

Bernard Meurer

We should hand modding to AI overlords

Emilio Pisanty

6:45 PM

@vzn The point was that you often seem awfully hands-off on a project that requires a lot of dedication. If you really want to push this through no matter what, then if there's something that needs doing and no one volunteering then you need to be up to catch any slack. Or the project just slogs down to a halt.

vzn

@EmilioPisanty what do you mean "pretty hard"

Emilio Pisanty

@vzn I mean I'm busy and don't think I'll be able to find much time to commit to something like that.

I've got a thesis to write and a hard deadline to write it by.

vzn

@EmilioPisanty need cooperation/ contribution of other volunteers. impossible to do alone. seems DZ/ DS are both "pretty hands off" also. they dont consistently respond to my queries wrt it. DS seems very tied up lately also. etc

@EmilioPisanty yeah fair enough thx for keeping an open mind will keep you in mind longterm

Emilio Pisanty

I'm just saying, you're free to take it or leave it.

@ACuriousMind You killed the OP?

Bernard Meurer

"Life During Wartime" Talking Heads LIVE

►

Emilio Pisanty

6:48 PM

That's definitely a Stack Exchange suspension looming up, if nothing else

Bernard Meurer

How did they do such good performances?

It's just amazing

vzn

@EmilioPisanty re "slogging down to a halt" it has had quite respectable/ substantial energy so far (esp considering how groundbreaking it is) but yeah, wondering myself if it is not exactly "mounting" either...

Emilio Pisanty

@vzn Seriously, I've said that piece and don't want to re-discuss it endlessly

ACuriousMind

@EmilioPisanty lol, don't worry I've got sockpuppets to protect me

(idontreallyhavesockpuppetsplzdontbanme)

2

Bernard Meurer

Who's getting suspended?

Emilio Pisanty

6:51 PM

@BernardMeurer @ACuriousMind killed the OP.

Comment discussion got really out of hand.

Bernard Meurer

@EmilioPisanty These bloody germans

John Rennie

It's a new high rep privilege. You can vote to kill the OP.

vzn

@EmilioPisanty fine & also plz consider contributing something other than (incisive) criticism too :P

Bernard Meurer

I hope that's not true, otherwise I might as well just start writing my will

ACuriousMind

@BernardMeurer Why the plural? I am but one

Emilio Pisanty

6:52 PM

@JohnRennie Coming to you at a Stack Exchange site near you, early next April.

Bernard Meurer

@ACuriousMind Sure you are

ACuriousMind

@EmilioPisanty Shouldn't that be "in six to eight weeks"?

Emilio Pisanty

@vzn You mean apart from precise diagnosis of the moderation tools available for the different formats?

John Rennie

What's happening in six to eight weeks?

ACuriousMind

And I think that April Fools joke probably would bring chaos and mayhem

vzn

6:54 PM

@EmilioPisanty you know what am saying, am sure you could think of something constructive

Emilio Pisanty

@ACuriousMind precisely six weeks after February 19

i.e. sometime in mid-December two years later =P

ACuriousMind

@JohnRennie There's a running gag on meta.SE that feature requests will be implemented "in six to eight weeks" regardless of their actual status of completion

vzn

@EmilioPisanty JR had a very nice writeup in meta of the 1st event, he captured all the questions, maybe just a post-writeup by someone would be helpful, it wouldnt have to be long/ detailed, maybe just highlights, and could edit out all the noise from the transcript etc

John Rennie

Ah :-)

Emilio Pisanty

437

A: The Many Memes of Meta

Meme: 6 to 8 Weeks Originator: Jeff Atwood First Heard: May 13th, 2008 Cultural Height: In about 6 to 8 weeks Definition: The time estimate given "off the top of my head" when the Stack Overflow team has only a vague idea of how long a task will take because they have little-to-no formal sche...

John Rennie

6:56 PM

@vzn I didn't do it this time mainly because I didn't understand the questions ...

vzn

@JohnRennie yeah but theoretically the transcription could be done by someone with basic physics bkg, ie not necessary to understand it all to transcribe it halfway decently, ie mostly copy/paste job

Emilio Pisanty

@vzn To be clear, I wish the project all the best, and I will contribute as and when I can and find time to (including constructive criticism and possibly limited to that when circumstances so dictate), but I'm not appreciative of pressures to do so. Sorry to be blunt, but that's the way it is.

vzn

@EmilioPisanty lol 2nd time youve said sorry for being blunt re prj. anyway whatever you can think of would be helpful. maybe types of guests youre interested in etc... agree it all looks pretty flimsy/ feeble right now but thats what we get in cyberspace with unpaid volunteers/ "enthusiasts" & chat rooms etc... think focus should be on speakers/ topics and not the management/ coordination etc

thx for voicing your high stds/ expectations & lets continue to figure out what can turn everyone into ~~accomplices~~ allies :)

0celo7

7:14 PM

High stds?

@vzn if you're sleeping with something with high stds, get out of the relationship

Obliv

something

classy

0celo7

@Obliv I don't judge.

Obliv

@0celo7 how's the morse theory going

Bernard Meurer

@Obliv .. - .----. ... / --. --- .. -. --. / .-- . .-.. .-..

Obliv

lol @bernard that's impressive

did you learn that in milnor

0celo7

7:29 PM

@Obliv busy with other things

and @ACuriousMind didn't explain my doubt properly

Bernard Meurer

.. / .-.. . .- .-. -. . -.. / .. - / .-- .. - .... / .-- .... .. - -. . -.-- / .... --- ..- ... - --- -.

0celo7

bob knows morse code

Obliv

whitney houston huh

didn't know she was such a good teacher

0celo7

@Obliv what?

Secret

7:54 PM

@0celo7

Excerpt from Sakurai

https://scontent-lhr3-1.xx.fbcdn.net/v/t1.0-9/13658962_1219084281500640_4238653674961260481_n.jpg?oh=d42a97cca1b35483f523e71dd2f479d3&oe=57F82FCA

I wonder if I am overthinking this a bit, cause I found one case as follows:

Start with 1.3.3.
(a'-a"*)<a'|a">=0 ------1.3.3.

If we take (a'-a"*)=0, we obtain a'=a"*-----(2), thus this tell us that a' and a" are complex conjugates of each other

Now <a'|a"> becomes <a"*|a">=(a"1*)* a"1+(a"2*)* a"2 + (a"3*)* a"3 = a"1 a"1+a"2 a"2 + a"3 a"3 which is not only nonzero in general, it is in fact a complex number

vzn

8:08 PM

@0celo7 maybe a topic/ theme for a rap song? :P

bolbteppa

8:26 PM

Jesus, if you want to build a 2-dimensional representation of a rotation in $\mathbb{R}^3$, you can assume $\mathbb{R}^3 = \mathbb{R} \times \mathbb{C}$, so that $(x,y,z) \mapsto (z,x+iy)$. Then, just as a rotation in space breaks down into separate planar rotations $\begin{bmatrix} \cos(\theta) & - \sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}$, where the columns are the images of basis vectors, we can form the spinor representation of this rotation

$\begin{bmatrix} z & - (x - iy) \\ x + iy & z \end{bmatrix}$. Since a reflection matrix is of the form $\begin{bmatrix} \cos(\theta) & \sin(\theta) \\ \sin(\theta) & - \cos(\theta) \end{bmatrix}$ it's spinor representation is $\begin{bmatrix} z & x - iy \\ x + iy & - z \end{bmatrix}$. Plugging in $\hat{i}$, $\hat{j}$, $\hat{k}$ gives the Pauli matrices, wtf :o

0celo7

8:38 PM

@Secret What are you confused by?

I cannot parse that block...

@Secret Do you understand why $(a-b^*)\langle a|b\rangle=0$ is true?

If $|a\rangle=|b\rangle$, then $b=a$. But $\langle a|a\rangle\ne0$, so $a-a^*=0\implies a=a^*\implies a\in\Bbb R$.

Suppose now that $a\ne b$.

Then $a\ne b^*=b$, because we just showed that $b$ is real.

Thus $a-b^*\ne 0\implies \langle a|b\rangle=0$.

Emilio Pisanty

117

A: If there are $74$ heads and $196$ legs, how many horses and humans are there?

A hypercentaur is a creature with two heads and six legs; an anticentaur is a creature with no head and two negative legs. Since $74$ heads make for $37$ hypercentaurs, with $74\cdot 6/2=222$ legs, you have $(222-196)/2=13$ anticentaurs. Since a hypercentaur is the same as a human on a horse, a...

↑ lolz

0celo7

That's a spiffy arrow

Emilio Pisanty

@0celo7 what, ↑?

Yeah, it's pretty spiffy

0celo7

Yes.

Emilio Pisanty

←↓→↑ for the full set

0celo7

8:52 PM

Oh lordie

Emilio Pisanty

Sitting right next to ⟨ and ⟩ on my software keyboard

0celo7

What is that??

>

Emilio Pisanty

But those display better inside $⟨a|b⟩$ → $⟨a|b⟩$.

0celo7

Whaaaaaaat

I'm blown away

Emilio Pisanty

@0celo7 They're all unicode

0celo7

8:54 PM

I'm sure you'd not be surprised to know I don't know what that is.

Secret

Basically according to Sakurai, for hermitian operators, you obtain this expression by combining brackets
$$(a'-a"^*)\langle a"\lvert a' \rangle =0$$

If we set $a'-a"^* =0$ we get $a'=a"^*$ which suggests a' and a" are complex conjugates of each other. Then $\langle a"\lvert a' \rangle$ becomes $\langle a"\lvert a"^* \rangle$. But $\langle a"\lvert a"^* \rangle$ is nonzero in general (e.g. consider $\lvert a"^* \rangle = 1+i$ and $\lvert a" \rangle$ = 1-i$

@0celo7 For the line a=a*, we only showed a is real if the eigenvalues are the same. But for $a\neq b$ we cannot say a or b is real because the condition a=a* only applies if the eigenvalues are equal. Thus that left open a case $a=b^*$ which I am not sure how to rule it out (because $<a|b^*> \neq 0$)

Emilio Pisanty

depending on your OS, it's relatively easy to make your keyboard do whatever unicode you feel you're going to remember consistently

Unicode

Unicode is a computing industry standard for the consistent encoding, representation, and handling of text expressed in most of the world's writing systems. Developed in conjunction with the Universal Coded Character Set (UCS) standard and published as The Unicode Standard, the latest version of Unicode contains a repertoire of more than 128,000 characters covering 135 modern and historic scripts, as well as multiple symbol sets. The standard consists of a set of code charts for visual reference, an encoding method and set of standard character encodings, a set of reference data files, and a number...

0celo7

@Secret What?

If you show that $a$ is real, $a$ is any eigenvalue.

This applies to $b$ as well

Or $c$

Once you show that all eigenvalues are real (which you just did), then you may assume all eigenvalues are real.

The equation $(a-a^*)\langle a|a\rangle=0$ applies to all eigenvectors.

So you can conclude that all eigenvalues are real.

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