the cosets are of the form ti +R

fbar (ti+R) = f(ti)

@Ted I guess so

@AlessandroCodenotti it's not a big leap from there to see what happens for arbitrary f

@EricSilva I know that $A\mapsto\int_Af\mathrm{d}x$ is a signed measure for arbitrary $f$ and that is also all I know about signed measures

Maybe try proving a decomposition theorem for signed measures

In a positive and negative part, I suppose

Mhmmm

@Ted I was thinking about this concrete example; say I have the holomorophic function $\sqrt{z}$ away from $0$. I take the germ $g$ at some point $p \neq 0$ and try to analytically continue it; that's done by taking the component $X$ of $|\mathscr{O}(\Bbb C - 0)|$ that contains the germ $g$ so $p : X \to \Bbb C - \{0\}$ is a covering map and the function $f : X \to \Bbb C$ is given by $f(\xi) = \xi(p(\xi))$.

$p$ should be the double cover of $\Bbb C - \{0\}$, right?

It's not immediately clear to me why

I guess the point is if I take a circular loop on the base from $p$ that winds once around the origin, then the germ returns back to the germ of $-\sqrt{z}$? So that should guarantee $p$ is a two-fold cover?

This makes me wonder what $f$ is after I identify $X$ with $\Bbb C - 0$. Should just be the coordinate function $z$, shouldn't it?

(I guess a better way to parse the two-fold thing is that monodromy of $p$ is $\Bbb Z/2\Bbb Z$-valued)

@EricSilva Well in this case where the measure is induced integrating a function, I can just divide the space into $A=f^{-1}([-\infty,0))$ and $B=f^{-1}([0,\infty])$, all subsets of the first have negative measure and all subsets of the second have positive measure

Right, @Balarka. You can make it explicit by putting in the $e^{i\theta}$ and seeing that you get $e^{i\theta/2}$ so that you get $e^{i\pi} = -1$ when you make one circuit around.

ooo, monodromy

@Ted True. Thanks!

@Balarka: No, $f$ is the squaring map, right?

Hm.

trying to decide whether or not to deduct a point on HW for using $A^2$ instead of $|A|^2$.

The European style is to write $AB$ for $A\cdot B$ and $A^2$ for $\|A\|^2$, but I won't allow it, @Semiclassic. It leads to serious errors. I would take off.

well, here $A$ is just a scalar. but it's conceivably a complex scalar.

Just to clarify, $f$ is the analytic continuation of the germ of $\sqrt{z}$ to the cover $p : \Bbb C - 0 \to \Bbb C - 0, p(z) = z^2$. You're claiming $f$ is the squaring map?

Actually, I first saw the $A^2$ in Chern's differential geometry notes, I think.

That sounds off

Oh, I thought $f$ was the covering map.

I thought so.

So $f$ is the tautological function on the Riemann surface?

it's the normalization constant in a plane wave state $\Psi=Ae^{i (kx-\omega t)}$.

If we think of the R.S. as a "multigraph" $\{(z,w)\in\Bbb C^2\}$, then $f(z,w)=w$, right?

Thats not a real number

the tricky thing is that one could take $A$ to be positive real without any loss of generality; the phase of $A$ has no physical consequence.

oh right duh

@TedShifrin Yeah I think so.

$$\int_0^1 [(1-x^{1/17})^{23} + (1-x^{1/23})^{17}] \ \mathrm dx$$

(I should say, one could take it to be positive real in this case. there's cases where it wouldn't be legit.)

It's just projection on the range of the "function," Balarka, hence tautological :)

@Semiclassical unless they mention that they are assuming $A$ to be a positive number, that is $A=1/\|\Psi\|$ I would for sure deduct points

I'm packing my stuff as I saw this written on one of my rough worksheets

hmm.

Ohhhh, @Semiclassic, your $A$ is a complex number, rather than a vector? Even worse, IMHO.

just the question is there somehow

being clear and explicit about what you are doing is necessary

@TedShifrin Right, @ted

the corrector is not supposed to have to find an interpretation of the text in order to make what the student is writing correct

eh, I would argue with that a touch. I think one does have to take the point of view of "what was the student trying to accomplish"

not when giving out points :)

otherwise you're not providing meaningful feedback.

Homework is the place to learn what's right and what's wrong, so that by exam time one doesn't write garbage.

True.

Ted what is C/R isomorphic to ?

You can write on the sheet that you think they mean this, because that would be the only thing that makes sense

but you can still deduct points, the whip is a tool of learning

I cant get how to do it

@KasmirKhaan come on

the cosets are of the form ai +R

what is the map ?

[earlier Kasmir said C/R is isomorphic to R^2 and I said he's wrong. On second thought, C/R is indeed isomorphic to R^2]

So isn't that isomorphic to $\Bbb R$?

[but not naturally isomorphic]

@Ted: Right, OK, I am not confused anymore. All we have to check is that $\text{germ}_{\sqrt{p}}(f)$ pushforwards to the germ of $\sqrt{z}$ at $p$ by the (inverse of the) isomorphism of stalks $\mathcal{O}_{\Bbb C\setminus 0, p} \to \mathcal{O}_{\Bbb C\setminus 0, \sqrt{p}}$

No, Leaky. $\Bbb R^2/\Bbb R \cong \Bbb R$.

@TedShifrin but $\Bbb R$ also $\cong \Bbb R^2$

@s.harp what you see on their pages is for instance $\Psi=Ae^{i (k x-\omega t)}\implies \Psi^\star =A e^{-i(kx-\omega t)}.$

how to think about it without a map ?

Oh, good grief.

@TedShifrin they are.

Clearly the identity function $f(z) = z$ is the desired thing.

the annoying thing is that one really could assume $A>0$ without loss of generality.

C/kerf is isomorphic to im (f)

I dont see how you can say anything without a map

all depends on the map

The right structure here is $\Bbb R$ vector space. This is not good for Kasmir to be thinking about.

(And, I should stress, this is a physics course not a math one. so i'm leery of expecting that phrase.)

@Semiclassical in my opinion, this is wrong and if it is too trivial to deduct points I would make a frowny face writing why it is wrong. If they had written more things, like wlog assume $A$ is real, then that would be okay

@TedShifrin I know

@Kasmir: How about the homomorphism $f(a+bi) = b$?

@TedShifrin what is the right thing to think about then ?

well, I've already marked on their sheets that they should include the * and absolute values

f : C--> ?

R?

Yes.

Who in the god's own nickname thinks of the group structure of $\Bbb R$?

to its kernel ?

that just looks wierd

No, it just so happens that the kernel and image are each $\Bbb R$.

The image is coming from the imaginary part.

why would we need to go to quotient group if we have a map between a group and its kernel

@KasmirKhaan that's because $\Bbb C$ is naturally isomorphic to $\Bbb R^2$

NOOOOO.

@TedShifrin That's it. "naturally isomorphic" it is.

@s.harp Part of the reason I'm debating this is because I don't want to treat it harsher than it actually is

@TedShifrin Hmm let me see that map and come back

@Semiclassic: They need to know that complex numbers work differently from real numbers.

well, tbf, they are correctly doing $(e^{i k x})^*=e^{-i k x}$.

@Semiclassical I can understand that sentiment, I did physics tutorials for a while before I gave up and stuck to math ones. There I can make all the frowny faces I want without feeling guilty :)

define $\phi (gxg^{-1})=g C_{G}(x)$, where $g$ and $x$ belong in $G$ and $C_{G}(x)$ is the conjugacy class of $x$. I was able to show that this map is well defined and 1-1, but i am stuck at proving onto. what should i write exactly?

but $z^2 \ne |z|^2$ !!

yeah...

@TedShifrin Ted on your book you did rings before groups, we are doing groups first , should I read from chapter 1 ?

Can it be hashed out as a typo or do they write $A^2$ all over the place

@NV-US: Please define the map with domain and range first.

@TedShifrin I really like how you explain things and good examples but I dont want to do something not as the class

Is that a map from a group to the group quotiented by a centraliser?

They write $\Psi = A e^{i k x}$ and then $\Psi^* = A e^{-i k x}$. Not a lot of ambiguity there.

No, @Kasmir, I already told you that you should be fine reading groups. It will occasionally refer to stuff you haven't done, but don't worry about it.

alright im off for today

Moreover, this is not a matter of just one paper.

good bye all ;)

@Semiclassical Oh dear.

OK, @Semiclassic. That's good.

$\phi | C_{G}(x)->\frac{G}{C_{G}(x)} $ @TedShifrin

Provided we know $A\in\Bbb R$, that is.

$A$ is a complex number here isn't it?

Or am I misreading

eh, there's the thing. typically in the book they don't make that assumption

and so they would have $|A|^2$.

@NV-US: I'm confused. You're fixing $x$?

@TedShifrin yes i noticed that , you said stuff like , where quotion in rings allways work, we need extra condtion on the subgroups , in case of groups

@TedShifrin so Ill start from chapter 6

@Kasmir: Still you should use D&F as your primary source, I would say.

it's about 15 papers out of 75, so a minority but not insignificant

yes, $C_{G}(x)$ = {$gxg^{-1}$ | $g$ is in $G$} @TedShifrin

@TedShifrin okay thanks :)

whoa, that's very dangerous notation o.O

@NV-US: And the thing on the right is just a set, not a group?

o/

hey @danu

So, well-definedness is: If $gxg^{-1} = hxh^{-1}$, then is $gC_G(x) = hC_G(x)$?

Heya @Danu.

Tomorrow's my defense!

I haven't looked carefully at your paper yet, but it looks snazzy.

@TastyRomeo very dangerous

huzzah

yes @TedShifrin that is how i proved it is well defined

Snazzy... Hmmm... Not sure what to take away from that ;D

kek

@Danu QM grading question for you. how annoying is $\Psi = Ae^{i k x}\implies \Psi^*=Ae^{-i k x}$ to you?

@BalarkaSen nice. how'd u do that?

@Danu good luck

OK, @NV-US, and what are we trying to show?

@BalarkaSen ctan.org/pkg/halloweenmath go

@Semiclassical What's the problem?

i will write the whole question

@Semiclassic: I think you really have to say -1 pt and insert "We take $A$ to be real."

hrm hrm hrm

For $A$ being real? Naw don't subtract points

unless you guys are super strict

it's out of 5 points, so I have to be judicious about what I dock for

Don't dock, man

This is going to get them in trouble later.

put a remark and deduct no point

if they think for themselves they will read the remarks

Unless the course always takes $A$ real.

@TastyRomeo beautiful

The way these problems are set up the coefficients are always chosen to be real in physics; You always write $\Psi=Ae^{ikx}+B e^{-ikx}$ and use boundary conditions to eliminate $B$

This is such standard notation in QM textbooks

Q : The number of elements in the conjugacy class of $x$ in $G$ is $[G : C_{G}(x)]$. for this i defined the map $\phi$ as is. @TedShifrin is my thinking correct?

@NV-US C_G(x) is NOT the conjugacy class of x

I'd note it but definitely not subtract points

quoting from a footnote in the text: "Incidentally, normalization only fixes the modulus of $A$; the phase remains undetermined. However, as we shall see, the latter carries no physical significance anyways."

It is the centraliser of x

@NV-US: You are confused.

@Semiclassical Yeah, because you can always choose $A$ real.

pls dont dock

@TastyRomeo it is the conjugacy class of x

So you have a group action by conjugation. One thing is the stabilizer. Another thing is the orbit.

think of all those poor students with their anime sad faces

@Danu right.

No, @NV-US. You're wrong.

your heart will instantly melt

@Balarka: My heart never melts.

It's funny how Balarka has turned into a meme-obsessed teenager lately :D

eyes of Argus, heart of ice

see this @TedShifrin

$C_G(x)$ is the centralizer. It's $\{g: gxg^{-1}=x\}$.

then why have they written conjugacy class

?

And then it's an immediate application, as I said above, of orbit/stabilizer.

@Danu plus, there are worse errors one can make on the problem (find the probability current of a free particle)

Where have they written $C_G(x)$ is called the conjugacy class?

exactly @Semi

@Danu I shall send you a meme worthy of thy memelord along with good wishes and luck for thy thesis defense.

@Semiclassic: OK, I'm swayed. Don't take off. But warn them in class or in writing what they're assuming and to pay attention.

For instance, some people made the problem harder than intended and took $$\Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \phi(k)e^{i (k x-\omega t)}\,dk$$

ok @TedShifrin i'll read the text again, more carefully this time. ty.

now, if they did this and wrote out the current $$J(x,t)=\frac{i\hbar}{2m}\left(\Psi \frac{\partial \Psi^*}{\partial x}-\Psi^* \frac{\partial \Psi}{\partial x}\right)$$

in a correct if tedious way, then I'm giving full credit.

I agree; you don't dock for tedium.

on the other hand, some of those people are then writing $\frac{\partial \Psi}{\partial x}=i k \Psi$

and thaaaaaaats bad

complete garbage

@Balarka: Are we done with our R.S. question?

@Ted Yeah I'm totally good with it now

Credit goes to @SemiC for getting me to think about this

yep. as math it's nonsense, since $k$ is an integration variable. and as physics it's no better, since it leads to the final answer containing a $k$ which has no determinate meaning (the wavefunction in the above case is a superposition of waves of different $k$)

He loves monodromy. And then you get to learn Gauss-Manin connnections and Variation of Hodge Structure. :)

I'm giving people 3/5 for that.

It'll give me a good excuse to reread Griffiths's seminal stuff on that.

I have never actually rigorously spent time thinking about analytic continuation of germs of algebraic functions.

before this

@Ted lol

Picard-Fuchs equations are neat.

oh there he goes

lol

I keep meaning to actually try to read up on Hodge theory

I've got Voisin's book on the subject

I've never seen it, but she's great.

but ugh. I really shouldn't be trying to learn new things now :(

Right.

@Danu: Glückliche Wünsche, aber natürlich.

@Semiclassical Because you're out of time, or because you're leaving academia?

the former, really

Please stick with us even after you leave acacacacademia; we learn a lot from you.

I'd like to think I"ll still try to read/learn stuff once I"m out in "the real world"

kk

@TedShifrin that's the most ü in a sentence

Despite his protestations to the contrary, @Semiclassic loves teaching and problem-solving.

@Leaky: I'm sure we can do better.

Not sure I've ever denied that.

@TedShifrin Wüe?

Problem is, I like research as exploration

I really don't care for it as exploitation (to borrow a usage I saw recently)

loool

@TedShifrin Worst. Speaker. Ever.

e.g. indigosim.com/tutorials/exploration/t0s1.htm

@Danu oof, really?

@Danu: I only know of her research reputation. I've never seen her speak.

I did, and it was really exceptionally bad

sucks

Perhaps it's her superstar reputation that allows her to get away with it :D

thinks it's a good thing Danu's never seen me speak

ghehehe

presentations are tricky, though

At least she is not Gromov

but, come to think of it, I think I prefer presenting stuff to writing it up

@TedShifrin $C_{G}(x)$ is the centralizer, sorry abt that.Then too, how to prove that the mapping is onto? i am stuck

the former doesn't give me as much of an opportunity to get stuck in my own head

@NV-US: This is just the powerful $\mathscr O_x \cong G/G_x$, where $\mathscr O_x$ is the orbit of $x$ and $G_x$ is the stabilizer. Review.

^this is the best thing ever

Why thanks @Danu

"Go forth and use likewise."

@TedShifrin i dont get it

The Ted hath spoken.

@BalarkaSen I don't think his talks are that terrible.

@Danu Really? Wow

They're sorta confusing but in some sense pretty interesting, usually (what I've seen on YouTube at least)

Look at the basic orbit/stabilizer discussion in the book, @NV-US.

Voisin must be pretty bad then

ok

Gromov is just like a magician, in the way he speaks

but usually there's like 1-2 tidbits that you can pick out and understand

uhhhh uhh i uhhh agree

At least that kind of talk sounds interesting if over my head.

I went to an applied math talk on monday because it happened to be in the physics building

it started out interesting: 3D mathematical modeling of glioblastoma

you know who looks like a wizard

(the kind of brain cancer which McCain has been diagnosed with and which has really poor outcomes)

the perelman

loook at that man

The problem was interesting, but it was just this overwhelming stream

got it. we just changed the orbit, so $G_{x}$ is replaced by $C_{G}(x)$, yes? @TedShifrin

he looks like a super twisted wizard

lots of details on how they'd modeled it

Yup, @NV-US.

thank u

as if he's gonna trick someone in the woods

@NV-US: Group actions is one of the most important things you'll learn in the whole course. Work on mastering them.

barry simon sorta looks like gandalf

oh hey

indeed he does

roger that @TedShifrin

:)

I guess I'd say that Simon looks like a wizard and Perelman looks like a sorcerer

yeah thats pretty close

by contrast, Alan Moore (not a math person) looks like Rasputin:

Imagining that kind of hair on my head/face makes me very uncomfortable.

oh damn

thats raskolnikov bruh

can't agree more

straight out of ze dostoyevsky

pretty much

who is this guy

oh man

how close am I

I bet this is photoshopped, but I so want this to be real:

hi @PVAL

hahaha

He wrote Watchmen right?

yeah

@PVAL Ah

also V for Vendetta

hi @Ted

tvtropes page for him here: tvtropes.org/pmwiki/pmwiki.php/Creator/AlanMoore

Ok, that has a pretty twisted story line so everything fits

Ah ok the Killing Joke is by him

right

I always thought people who wrote like mainstream comics wrote like 100's of stories.

Like I figured there was probably one superman writer for like 10-15 years or whatever who just wrote tons of stuff.

I think that's right

once in a while some isolated guys come up with sudden groundbreaking stuff and takes over though

eg that's what happened with Grant Morrison

@BalarkaSen Well this guy wrote like 3 superman stories.

There is a lock company in Watchmen called Gordian knot.

@PVAL-inactive aha, strange

hey guys, question about vectors, what exactly happens when two vectors are multipled together by using the dot product?

nvm stupid question, sorry for bothering

@SylentNyte it becomes a scalar! :D

a vector is projected onto the other vector and multiplied by the length of the other vector

[in order to make the operation commutative]

@LeakyNun oh okay, im gonna see if i can find a visual representation on youtube (what im looking for), thank you

@SylentNyte try 3blue1brown

@LeakyNun tho this does help a lot actually!

@SylentNyte best math channel so far

unbelievably and unprecedentedly good

Hello @LeakyNun !! Could you take a look at my question: math.stackexchange.com/questions/2448179/… ? Do you have an idea?

@MaryStar entschuldigung ich weiss nichts ueber exterior algebra

@LeakyNun Ah ok. Kein Problem.

is there a site where i can graphically play around with vectors?

geogebra

@SylentNyte desmos

@BalarkaSen That pic is so cool, lol.

@LeakyNun Still awake?

@Jasper about to fly

I never slept

Ugh

Apparently I have been banned from asking questions in stackoverflow

I have a positive vote ratio on my questions and yet I am banned!

I have almost 70k rep on my combined accounts on SE and I find out Im banned the one time I need to ask something.

SO is really full of jerks, MSE is like the exact opposite of SO

I finally managed to find the solution after an hour of looking

So I dont care if SO is full of jerks anymore

How can we show that the sign of the permutation $\pi: (1, \ldots , k, k+1, \ldots , k+\ell) \mapsto (k+1, \ldots , k+\ell, 1, \ldots , k)$ is $(-1)^{k\ell}$ ?

@MaryStar it's presumably a product of "simpler" ones, then use induction

At what do we have to apply the induction? @0ßelö7

@MaryStar Well, you can move the first $k+1$ to the left past the first $k$ things, right

then you move $k+2$ back until it gets to $k+1$

etc.

A fully formal proof would use induction, but it's really quite clear that you pick up a sign of $(-1)^k$ $\ell$ times