@Rithaniel Yup. And $q \overline q = \|q\|^2$ for any quaternion. So for imaginary quaternions, that says $q^2 = -\|q\|^2$.

The best part of that is that you get a 2-sphere's worth of $\sqrt{-1}$s!

Dunno why Lovecraft hated non-Euclidean geometry, it's so pretty

(Yes, those are right-angled pentagons)

@Rithaniel Also, that means, any quaternion can be written as $a+bu$ where $a$ and $b$ are real and $u^2=-1$

(Specifically, $a+bi+cj+dk=a+\sqrt{b^2+c^2+d^2}u$ with $u=\dfrac{bi+cj+dk}{\sqrt{b^2+c^2+d^2}}$)

So every quaternion is contained inside a plane isomorphic to $\Bbb C$

Is there a standard way to write something like $a_1 i+a_2 j+a_3 k$ for quaternions?

e.g. $\vec{a}\cdot \vec{i}$

except probably not that

$\sum a_ie_i$?

hmm

I know that, in the octonions, instead of come up with seven letters, they just use $e_1$ through $e_7$

for comparison, in physics you deal a lot with the Pauli matrices $\sigma_1,\sigma_2,\sigma_3$

which satisfy $\sigma_i\sigma_j = \epsilon_{ijk}\sigma_k$ (with implied summation)

$\epsilon_{ijk}=\pm1$ depending on the order the subscripts are in, right?

right

Like $\epsilon_{123}=1$ and $\epsilon_{132}=-1$

these are complex hermitian matrices, and if consider the set $i\sigma_1,i\sigma_2,i\sigma_3$ then you get the quaternion algebra

In coordinate geometry it's not unusual to write $x_1=x$, $x_2=y$, and $x_3=z$

ya

so I'd imagine writing $e_1:=i$, $e_2:=j$, and $e_3:=k$ wouldn't be too radical

anyways. you often write $\vec{\sigma}\cdot \vec{a}=a_1 \sigma_1+a_2\sigma_2+a_3\sigma_3$

Ah

(Physicists.)

Another fun fact: conjugation of complex number can't be written as a polynomial, but conjugation of quaternions can

and then a useful result is $$(\vec{\sigma}\cdot \vec{a})(\vec{\sigma}\cdot \vec{b})=(\vec{a}\cdot\vec{b})I_2+i\vec{\sigma}\cdot (\vec{a}\times \vec{b})$$

@AlessandroCodenotti cool, how so?

@AlessandroCodenotti excuse me?

@ÉricoMeloSilva there's certainly no complex poly with p(z) = z bar

@AlessandroCodenotti That's related to this

@MikeMiller ofc

$\bar{q}=-\frac12(q+iqi+jqj+kqk)$

the quaternião thing kind of blows my mind

If you write $\vec{e}=i\vec{\sigma}$, then that becomes $(\vec{e}\cdot \vec{a})(\vec{e}\cdot\vec{b})=-(\vec{a}\cdot \vec{b})I_2+\vec{e}\cdot(\vec{a}\times \vec{b})$

quaternion

Should have guessed

Nice

which is then basically the quaternion version

Ah, yeah, looked at that one too. I came to the conclusion that no, that is never true. The number is always purely imaginary, right? @AkivaWeinberger

You changed your username? Nice

@ÉricoMeloSilva

@Rithaniel Yeah. It always has only $j$ and $k$ components, specifically

yeah

There's a great discussion somewhere about how there are a few ways to make sense of "quaternion holomorphic" functions, and at least one of those definitions only gives you linear functions

Is it OK if I pronounce it like it's Spanish

'cause I have no idea how Portuguese orthography works

Yeah, I had a professor during my bachelor who mostly works in real algebraic geometry and "trying to make sense of analysis on quaternions"

I think that you're trying to get the pronunciation right at all is a good start

@MikeMiller And Möbius functions, should be

They're conformal (and they're the only things that are conformal in dimension${}>2$)

Uh, are you sure about that? Do you know the reference I'm thinking of?

No

I also vaguely recall a definition that seems natural but turns out to only allow linear functions

Wait, which "linear" functions? Is $iqi$ linear?

Is $q+iqi$?

$i(q+p)i=iqi+ipi$

@AkivaWeinberger the r is tapped and the o is like oo

In that case you get anything that's linear in the $\Bbb R^4$ sense

@Semiclassical Ah, all right. So, linear over $\Bbb R$

$(p+q)+i(p+q)i=(p+ipi)+(q+iqi)$

and the e is like eh

I gotta go

(As opposed to complex linear functions, which are linear over $\Bbb C$ and also they're more restricted)

(and I guess you could try asking for things that are rotations of the space, but unlike in 2D, those aren't closed under addition)

I proved that <K> is closure and hence forms subgroup but i am unsuccessful in proving that ( or disproving ) that if last collection is a subgroup ( for proving closure we need to take too many cases any short or intuitive proof ? ) First answer in math.stackexchange.com/questions/157876/normal-subgroups-of-s-4

Thanks for any help

Any idea ?

@WillHunting

here is a paper on quaternionic analysis. theorem 1 (old and well-known) is the statement that if you try to define quaternion analytic by saying that the derivative from the left - ie $\lim_{h \to 0} h^{-1}(f(x+h) - f(x))$ - then quaternion differentiable functions are precisely the same as maps $x \mapsto a + xb$

the version that includes everything that satisfy the Cauchy-Riemann equations is probably the right one and is developed there

@MikeMiller You have any idea

no and i'm going to put you on ignore for pinging me about something i never asked about.

@MikeMiller What i did ?

@MikeMiller can you please tell me .Any one can tell what i did wrong ?

i never pinged you @MikeMiller before

@AkivaWeinberger can you clear my doubt

@nitsua60

This chatroom has a bit of a problem with people pinging people at random to solicit them for help.

@Hansie

Sorry

no worries =)

So please can you just see at my doubts

@Hansie You're STILL going. It's rude to just randomly ping someone to ask them to solve their problem. You can post your question and see if people are interested.

@Hansie (who are you talking to?)

There is something of a stigma against pinging people who didn't willingly join the conversation, at this point.

The more you do this, the less likely it is someone will be interested.

@MikeMiller Ah, I see

"Just ask, don't ask to ask" is a policy for the room, not for specific people

@MikeMiller @AkivaWeinberger sorry but this is my first doubt

never asked before

Probably you people are taking me wrong

You came in here and randomly pinged someone asking for help because they were visibly present

Ok

i am sorry

i did'nt thought people will mind such small things

*did not

The nice way to do it would be something like "hey everyone--I have a question about [link] which is about <insert intriguing summary here>. Anyone who could help me understand <tricky part> please ping me."

It happens often enough with random people that regular denizens of the chat simply don't have the patience for it anymore, but it's understandable that you wouldn't know that on your first visit.

Sorry i was beginner

@Hansie (there you go =D )

Can any one help me in this

(Now i am not tagging any one )

So if any one wants can help me .Thanks

perfect. now just wait patiently for an answer to come, if any does.

Thanks .Sorry to every one .

That was my first and last mistake

Yeah, you're good. Just give it some time.

Thanks.

waiting ........

Sorry, I had to listen to an announcement

I'm back

(briefly)

And gone again

@Hansie (patiently? no one here works for you--you have no reasonable expectation that someone's going to address this, just a reasonable hope.)

If anyone replies, it might take a while, man. (Best to lurk and do some other stuff.)

ok thanks

every one taught me to wait but no one even gave a suggestion in my doubt .sad

@rockdock 4 star comment is right .Every one just gossips

I started to about 10 minutes ago because I was impressed you changed behavior but then you immediately whined about how you were waiting. Get over yourself. You are not entitled to anybody's time.

@Hansie True. This isn't a help room, it's a room where a lot of people who like mathematics tend to congregate and will often help others with math.

@Ted I suggest that at some point, regular users of this room decide on a specific set of rules, and then you add more people to the position of room admin / moderator / whatever to actively moderate based on those rules.

Well i think we should help him

@Hansie wait .If i come with some solution i will tell you

Eh, I don't think insistence should be rewarded with a response

But I guess you can answer if you'd like

Thanks @neraj

@MikeMiller if you got any solution to @hansie doubt share.Let us see his right part and help him

Also please see his yesterday chat .Then judge him

Also i am still waiting patiently .Please share your help .( Not tagging any one )

I respect you people time

Do realize that part of being patient is also not repeating/bumping. Just wait, maybe if in a few hours you don't get a response you can ask again

"Am I patient yet?"

@Hansie I don't understand. What do you mean by "last collection"?

@AkivaWeinberger oh, they came out with the answers for that coin flip question: fivethirtyeight.com/features/…. they picked a bit different route than I expected

@AkivaWeinberger Probably @hansie is referring to last subgroup (8+3+1)

example: "To achieve a 1 in 3 chance for Anna, have the players alternate flips, with Barry going first. The first person to flip a head wins. In each pair of flips, Barry is twice as likely as Anna to win, because Barry went first. Somebody will win — therefore Anna will win one time in three."

I think all four are subgroups .

Ah

In last subgroup , closure property is holding

But we need to check case by case

"And, finally, to achieve a 1 in 5 chance for Anna, have the players alternate again. But this time, each gets two flips, so that Barry flips twice, then Anna flips twice, then Barry, and so on. The first to flip a head wins."

It's the alternating group

$A_4$

Set of even permutations

@hansie yes that is lengthy process

Now I'm curious how their procedure stacks up against the ones we came up with in terms of "expected # of flips" @AkivaWeinberger

@Semiclassical Ahh. H+TH+TTH+…=1/2+1/4+1/8+…=2/3

Clever

yeah

Expected number would be $\sum n/2^n$

Wouldn't it be H+TTH+TTTTH+...?

${}=2$

@Semiclassical For the 1 in 3 thing

yeah

unless you meant 1/2+1/4+1/8+... = 1

One has H+TTH+TTTTH+... = 2/3 and TH+TTTH+TTTTTH+... = 1/3

Well, this is embarrassing

lol, happens

What's not obvious to me is how to generalize it

Still expected 2 flips though, I think

The "flip twice, discard TT" strategy has how much expected

Yes @AkivaWeinberger is right . Since even permutation group has (1/2) * n ! elements and all the three groups in last has even permutations . @Hansie You are done

See just post you doubt and wait from next time .People will help you

the "flip n times and discard enough outcomes" approach has the virtue of being obvious in its generality

Out of curiosity, what's your native language?

(@neraj and @Hansie)

French?

sup nerds

i mean, there's some obvious things one can do. for instance, to get 1/6 chances, have Barry flip once initially to see if he wins; if not, then do the game with 1/3 chances

but what about 1/7?

@AkivaWeinberger what is problem in my language

also, the case of 1/4 is a bit of odd one out since there's no need to keep repeating flips

@ÉricoMeloSilva Making fun ?

what

what is sup nerds

i was saying hi

it's the supremum over all nerds

pls mathein

communication

dont do me dirty like this

is hard

@MatheinBoulomenos and you exhibit a maximum!

lmao @Mike u got 'im

Thanks

nah, he just exhibits a maximal

You people have time to JOKE on my language and not to help a person

we're joking on Erico's language. but yes---this room is not a job

Please i take it very personal .Don't make fun of some one's language .It feels offensive

@MikeMiller I never really tried not to be nerd tbh

I wasn't joking about your language

I was just curious

Sorry if it offended you

@AkivaWeinberger yes it offended me .

@Semiclassical I think nerdery is a scalar, so linearly ordered

none of us are here under a sense of obligation regarding helping people.

@MikeMiller eh. i think nerdery with regards to a certain subject can be ordered

do not help him but don't make fun of others language

i'm not so sure about cross-subject nerdery

lets not get

this technical

about how much y'all are nerds

Yeah, my bad anyway

Clearly Semiclassical is strictly more of a nerd than Mathein

i guess we're just being pos(et)uers

They just make fun and nothing.Did not have time to help a wounded man but time to make joke on his language

@Mike ur right and he just proved it

+eu, -ue :)

"Sup nerds" had nothing to do with you

what's more nerdy: going to an anime convention, doing a reading group on infinity categories or playing scrabble in Latin?

"Sup" means "What's up"

which means "Hello"

I am proud of my language

@MatheinBoulomenos I think the Latin Scrabble is the least on the list

No one said anything about your language except for me @neraj

latin scrabble is like nerdy but cool

Hi does anyone know whether we can say a vertex transitive graph is locally connected?

id have a beer w u if u do that

But bigger than both of those us going to an infinity categories workshop at an anime convention

if u go to an anime convention i aint gonna have a beer w u

@MikeMiller lmao

Thanks @AkivaWeinberger and @neraj

@ÉricoMeloSilva pffff

Don't make fun of our language

@MatheinBoulomenos reading groups on infinity categories... sounds pretty cool to me

Hey @lush!

@MatheinBoulomenos more badass than nerdy

heyho ^^

Anyone else remember the 2006 (?) ad for an anime convention that had a bunch of people suddenly getting excited about random stuff and announcing it loudly before the chef tells them to leave

@Semiclassical anime was a mistake but i unironically think so

riiiiight

@ÉricoMeloSilva I proudly tell people I play Latin scrabble lol, seems like the right choice

imo if it's a nerdy thing that is also a skill it's cooler

like when people do introduction rounds where you have to say some random fact about yourself

knowing latin is a useful skill

it is?

it helps u know more word good my man

what a ringing endorsement

@MatheinBoulomenos Hold up a second do you do all three of those

no comment.

Semiclassical has been unseated!

ah nuts

we need a nerd off

i'm trying to decide what my most nerdy feature is beyond academic pursuits

Out of curiosity, is chat sockpuppetry accepted?

probably the fact that there's an online text MUD that I've had a character on since spring 2003, and I still play that character occasionally

I think it's banned on main for point reasons or whatever but I would guess it's just considered weird in chat

@Semiclassical less nerdy than liking anime imo

lol

dont get me wrong it's still hella nerdy

due to the way that the timeline works in that game, my character is 468 years old lol

(i think the oldest active character is 639, lol. though neither he nor I have been active the entire time)

Though I probably don't stack up to the people who have invested a lot of RL money into the game over the years.

that rabbit hole goes deep

what is RL money

oh

real-life

oh

vs. IG = in-game

RL has more syllables than real

but fewer characters

yeah

none of us speak aloud anymore my man

we all just beam text brain to brain

i guess u got me

it's a knee-jerk reaction, i suppose, since when talking to other players about it you're as likely to be talking about IG currencies

@MikeMiller do you know about arithmetic hyperbolic 3-manifolds? it kinda surprised that there's a class of manifolds for which number theory seems to be important

I don't really know about them, but I do know that the arithmetic governs a lot of important stuff there

@AkivaWeinberger oh, the easier riddle from this week is kinda neat too

>To make this concrete, assume that landlines and cell numbers in my area code are assigned randomly, such that a person is equally likely to get any of the 10,000,000 numbers between and including 000-0000 to 999-9999. Given that assumption, what is the probability that the last seven digits of the cell number are an exact scramble of the last seven digits of our landline?

i heard about them from Benson Farb and im convinced he's just lying to me

There was a beautiful recent paper that studied Seiberg-Witten Floer homology for a class of hyperbolic 3-manifolds. To do that, you need to have an explicit understanding of the set of "monopoles on $\Bbb R \times Y$". This is quite difficult even for very simple manifolds, unless you can prove that you get zero; that approach usually only works for things with positive scalar curvature

(Which in 3D is $S^2 \times S^1$, spherical space forms, and connected sums of those)

Oh, I forgot to go on

Well, these authors studied monopoles on $\Bbb R \times Y$ for some small-volume hyperbolic 3-manifolds, and used the Selberg Trace Formula to relate arithmetic properties to the first eigenvalue of the Ricci tensor on 1-forms, which was then used to obstruct the existence of monopoles in these cases

So I think they proved maybe 20 examples of hyperbolic manifolds with no differential in the SWFH chain complex, and thus computed the answers

What's interesting is what happens as they get to the "boundary" of where this technique will work, and I think they are thinking about that

arxiv.org/pdf/1810.06346.pdf

I just really liked the appearance of the explicit Selberg formula in a field that usually calculates in a very different fashion

The relationship between hyperbolic geometry and floer stuff is very mysterious

RIP Stan Lee

oof. not that surprising, but that sucks

jeeze, 95?

he had a good run, at least

One day we'll all be dead so there's that

@MikeMiller that sounds interesting though I didn't understand even what it is about. Can you make precise what you mean by arithmetic properties? I don't really understand the arithmetic part from skimming the paper for it

I am pretty ignorant

The relevant line seems to be

@MikeMiller we were dead before we set sail dude

we’re on a sinking ship w no land in sight

The Selberg trace formula is a very powerful tool as it allows to extract seemingly inaccessible information regarding spectral geometry of X via the understanding of the lengths of its geodesics; and the latter quantities are directly computable from the traces of the elements π1(X) in PSL2(R).

in particular it's about the subgroup of PSL_2(R), but i guess that's still not arithmetic

@MatheinBoulomenos Maybe this is more up your alley

@MikeMiller wow this is really cool! section 1.6 on motivation from NT is intriguing

For what it's worth I really really don't know about this stuff

I mentioned the first only because I knew part of it

What does $\Bbb R^4 \times \Bbb C^2$ mean?

In what context? Do you know the phrase "Cartesian product"?

@MikeMiller hello ^^

Hi @s.harp

I guess I said hello to you another day and this is a response to that

@MikeMiller I was reading about mathematical spaces and $\Bbb R^4 \times \Bbb C^2$ was defined as the upper bound of the space. In other words, $\Bbb R^4 \times \Bbb C^2$ can hold all the elements of the space inside it but if you decrease the bound to $\Bbb R^3 \times \Bbb C^2$ you lose information and can't completely catalogue the space so to speak. I'm still reading and not understanding too much but I just thought I'd ask about that cause it confused me when I was reading

I think yes it's talking about the cartesian product too

@Mike yes, I'm often here for a short while

I can't wait for my textbook to come so I can ask some more concrete questions

I have a question

If you take a finite periodic distribution of points, and reflect it about it's centre, you get the exact same distribution back right?

I have absolutely no idea what anything in your first response meant, so I will not comment.

If you take a finite non-periodic distribution of points along a straight line, and take a reflected copy of that distribution, superimpose the two together, will the resulting distribution still be non-periodic. Categorize the degree of randomness of the resulting distribution

can't tell if that's a good question to ask or not

I think the answer is that yes, the distribution will still be non periodic

How is an arrangement of points periodic? If they are of the form $x_n = c + b\,n$?

if they have the same spacing between them

ok, that means they are of the form I described for some constants $c$ and $b$

if you reflect a point $x$ about another point $y$, what is the new point?

well if $x$ is greater than $y$ I guess you'd substract $y$ from $x$ and then

I don't know

well, in the case $y=0$ the result of the reflection is $-x$

Any of the usual algebraists on?

so if you try to draw a line, y----x the reflection should go to r(x)----y----x

basically you are moving "the distance between y and x in the other direction from y"

if you think about it the formula is $r(x) = y + (y-x)$

oh okay

right, so if your arrangement of points is given by $\{c+ b\, n\mid n\in\Bbb Z\}$, the reflection along some point $y$ will be of the form $\{y+ (y-( c + b\,n)) = 2y-c -b\, n\mid n\in \Bbb Z\}$

the tricky question is now, when is it so that these two lattices together form a lattice (ie the arrangement is periodic using your vocabulary)

how hard of a question is that?

its not very hard

if you draw the right picture you will see the answer

okay good

I'll try it

sure^

Hi everyone

@s.harp having a little trouble, but I'll come back to it later, thanks for walking me through the process though

@TobiasKildetoft not an usual algebraist, but I really like algebra as well. @MatheinBoulomenos is online tho

@lush I am looking for a good introduction to Galois theory which does not have too many prerequisites and also covers some of the basic results for finite fields.

I have two students that would like to do project on Galois theory and I need a good reference to give them for a start

mhm

@MikeMiller Fair enough... Are all classifying spaces such infinite-dimensional manifolds?

@MikeMiller sad life

@TobiasKildetoft I think I learned with Dummit&Foote the first time

I liked that one

@MikeMiller I was actually not trying to use curvature (I want classes in $H^*(X;\Z)$), but Cech cohomology for the definition for line bundles and then the splitting principle

@TobiasKildetoft John mackintosh is also good reference although I don't know whether they discuss about finite fields as deep as Dummit and Foote.

@MatheinBoulomenos Perhaps it's not so strange, given that they must all be quotients of the hyperbolic 3-ball by some subgroup of isometries, so their classification is purely group-theoretic, and I find it very plausible that number theory plays a role there! (I don't know anything about them, though)

@ÉricoMeloSilva We call it a "macronym." =)

hi everyone

@TobiasKildetoft how is this

finite fields isn't really that hard I think

@LeakyNun I am looking for something with some more detailed explanations

I will take a look at Dummit-Foote

@TobiasKildetoft I could write more explanations... given enough time lol

(btw I wrote that thing in case you haven't noticed)

Is Morandi too advanced?

im trying to show $\{(0,0,z) : z\in K\}$ is prime ideal in $k[x,y,z]$. so if $fg$ in the set, i want to show that either $f(0,0,z)= 0 $ for all z or $g(0,0,z)= 0$ for all z. not sure how. any hints?

@AlessandroCodenotti Not familiar with it

What you wrote is not a subset of $k[x,y,z]$ @Liad

since you guys are talking about Galois theory, there is a formula for the roots of a polynomial of degree 2 and it involves the number $\frac12$, so it should be valid for any complete field of char $\neq2$

what about char $=2$? is there a sort of abel ruffini theorem that here you have no closed formula?

i have absolutely no idea about anything related to galois theory (so maybe the question is dumb)

@AlessandroCodenotti you are right, the corrret set is $\{f : f(0,0,z) = 0 $ for all $z\in K\} $ . sorry.

@Danu Any countable CW complex should have such a model I think.

If you're asking if they have to be infinite: BG for G compact Lie foes, but not for G infinite discrete

Liad: Suppose fg ist in your set and there is some z s.t. f(0,0,z) \neq 0. Does that help?

Speaking of CW-complexes do you know where should I look for a proof that CW-complexes are metrizable and locally contractible? (I checked Hatcher but it's not proved there)

@Alessandro finite CW-complexes?

Not necessarily

Liad: nevermind, I confused sth

@lush its "is"? (you wrote ist which can mean either is or isnt ^^) if it is then if $f(0,0,z) \ne 0$ then $g(0,0,z) = 0$

@lush ah ok.

@Alessandor that makes me realise that the Hawaiian earring is not a CW-complex, although when you draw it it definitely does look like one

@AlessandroCodenotti any ideas? :-)

@AlessandroCodenotti I guess it is hard to read morandi for first time; Dummit Foote is perfect as it contains lots of examples

Liad: is K algebraically closed? If so you can use the Nullstellensatz if you know it

I don't want to think about algebraic geometry right now @Liad, I'm still salty I messed up some easy T/F on today's AG midterm :P

@AlessandroCodenotti ah , got ya. take it easy ^^

@TobiasKildetoft I sometimes get confused between the left-multiplication action and the conjugation action...

I mean, it seems like both are used

and it just confuses me

@lush it is algebraically closed, but we didn;t study Null.. yet

@lush what is it saying?

@liad can't we can show k[x,y,z] / (0,0,z) is integral domain to prove (0,0,z) is prime ideal

@Liad $f(0,0,z)$ and $g(0,0,z)$ are polynomials in $z$, i.e. elements of $K[z]$