did i hear electric potential? :) @BalarkaSen

ah, the qualitative bit: Gauss's law is $\nabla\cdot \mathbf{E}=-\rho/\epsilon_0$ i.e. divergence of electric field is sourced by charge density

and the electric potential is given by $\mathbf{E}=-\nabla \phi$ where $\phi$ is the electric potential. so $\nabla^2\phi = -\rho/\epsilon_0$

if $\rho=0$ in some region, you get laplace's equation. so the electric potential is harmonic so long as there's no charge around @wowdavers

Hi @Semiclassic.

hi @ted

You making progress with Griffiths?

not today. had other stuff for most of it

Me neither.

i have been looking at my riemann surface, and trying to boil it down to a simple characteristic case whose branching i can display nicely

Heya @MikeM. Butting in as usual, I see.

Always.

like i did with the $(y^2-1)^2=x^4-1$ case i considered

Well, I was helping little kids with addition and subtraction of small integers yesterday, so don't mind me :)

@Semiclassic, but that one's very non-generic.

I unfortunately don't have the patience for that.

eh, is it? obviously the roots being nicely positioned isn't generic (nor is the symmetry)

@TedShifrin me too. Though these little kids seemed to believe they were college students.

ROFL, @PVAL. Sadly, I know of what you speak. :(

Mine were 2nd and 3rd graders.

but all i care about is the structure, not the symmetry of it

@Semiclassic: Repeated roots is special, not generic.

But, damn, some of these old logic puzzles (five people with first names, last names, and occupations ... with clues) that they give these kids take me a half-hour to figure out. Do they really expect these little kids to get those?

anyways, the case i've got is $\frac14 (x^2+y^2)^2+\frac12 (x^2+y^2)+\mu(x^2-y^2)+c=0$

But there's a sweet little kid who comes to me every week, and he's actually very smart and doesn't get discouraged. But seriously ...

The university is stalling on meeting with us about fixing healthcare stuff, of course.

where $\mu,c$ are parameters

@TedShifrin Most of the teachers probably do not look at the work they assign, or they are so incompetent that looking doesn't help.

Damn, @MikeM.

Hi guys! I'm new to the chat. Nice to meet you all.

welcome

Hi @user3925758 ... what a lovely name you have.

Thank you, @TedShifrin I designed it myself.

@Semiclassic: I already mentioned making the base change $(x,y)\to (x^2,y^2)$. You shouldn't ignore that.

@user3925758: Yes, I know.

Hi @Ted

yeah, i know

Are you up early again, @Balarka?

i was planning to do $u=x^2$, $v=x^2+y^2$

That's too complicated, IMHO, @Semiclassic.

to take advantage of the form of the first two terms

Sounds like it.

Well, OK, try it.

It should be very easy for an organized graduate student body to bring a school to its knees. Taing or raing isn't a job that should be very easy to find scabs for.

In GA we had nothing resembling unions. So I'm ignunt.

@PVAL-inactive You'd be surprised. A lot of people are raised with scab attitudes.

i mean, the differential I'm interested in $y(x)\,dx$. so that'd become $\frac12 \sqrt{\frac{v-u}{u}}\,du$

LOL @inactive.

In any case, we can't strike during a contract period or for a few months after negotiations fail.

So I am assuming the situation is something other than grad students are not getting healthcare.

Because if it was at that extreme, I wouldn't be working and I'd be looking to find any other source of funding.

Oh, I see what you mean. That happened to somewhere between 30 and 60 (we're not sure) grad students due to university negligence, and lasted over a week. They had no access to healthcare, some had to cancel appointments, etc. There is a nontrivial "Well, I have mine" attitude out there.

On the other hand we got plenty fo support while that was happening. We learned in the meantime it happens every year, we just haven't known about it before. So we're meeting with administration and demanding a number of things to make right what happened and ensure it doesn't happen in the future. They are stalling meetings and the meeting we actually have is likely to be unhelpful.

I'd think if that happened you would have that group file a large lawsuit against the university. I don't know if I see the point of negotiation.

Lawsuits take time.

"In any case, we can't strike during a contract period or for a few months after negotiations fail." This doesn't make any sense. You cannot strike so the university has some time to find scabs?

We legally cannot strike. It's part of the contract and part of the law.

Everything seems broken sometimes.

American union laws, dontcha love it

It's extremely frustrating. Usually strikes don't happen anyway, though, only threats. That's what the postdoc union threatened, and I think they got through with negotiation today.

uaw5810.org/category/bargaining-2016

@MikeMiller Do you know how people prove words in the MCG?

No idea.

marking is soo boring

I wish I could code a program that automatically marks everything for me

$I sent my advisor a reasonably long email and he cut it up into four paragraphs, and responded with "no, no, yes, not sure how you'd do that".$

Mike

@MikeMiller

This is what happens when I email someone in my field who is at another university

I expect a thoughtful reply, but I don't get it :(

@KeithYong wikipedia has a constructive algorithm in its chinese remainder theorem article. in effect, first solve the system with {a_1,a_2}={0,1} (two possible systems), then linearly combine the results

@arctictern thank you i will check it out

):

:(

@arctictern professor gave me a tip, that x = a1*m1*n1 + a2*m2*n2 where m1 and m2 are mult. inverse from extended euclid algorithm. I played around with it and couldn't figure out what that equation means

you know what multiplication means. you know what addition means. that equation has nothing but multiplication and addition. so are you sure it's the equation that you don't understand?

I'm pretty new to mod/divisibility so most of this stuff is flying over my head >_<

I came up with $x \equiv a_2m_2n_2 \mod a_1m_1n_1$ and $x \equiv a_1m_1n_1 \mod a_2m_2n_2$

why are you taking $x$ mod the numbers $a_1m_1n_1$ and mod $a_2m_2n_2$? are you trying to check that $x$ is a solution to the system? use your words!

Is there a shorthand for \left[\ begin{matrix}BULK\ end{matrix}\right]?

Yeah honestly I'm gonna go to office hours for this. There's probably something fundamental about mods and divisiblity that I missed

Reading up on chinese remainder theorem and gonna do what you suggested

@KeithYong do you have any questions about the tip your professor gave?

I'm sure your questions could be answered here, but you have to actually form the questions with words.

@Axoren huh?

Like, \begin{bracmatrix}1 & 2 \\ 3 & 4\end{bracmatrix} to make $$\left[\begin{matrix}1 & 2 \\ 3 & 4\end{matrix}\right]$$

Instead of that really long stuff.

\begin{bmatrix}\end{bmatrix} for brackets, \begin{pmatrix}\end{pmatrix} for parentheses

@arctictern You're a godsend.

Thank you.

@arctictern So for $x = a_1m_1n_1 + a_2m_2n_2$, I'm at a loss at how that turns into $x \equiv a_1 \mod n_1$ and $x \equiv a_2 \mod n_2$.

@Axoren see help: displaying a formula on wikipedia. also note if you ever see any latex in the wild on SE you can right click and view the tex commands

@arctictern I've only seen it done the hard way.

@KeithYong okay, let's reduce $a_1m_1n_1+a_2m_2n_2$ mod $n_1$.

@Axoren seriously? that sounds crazy.

I didn't believe it existed until now. It was all blind hope that led me to this moment.

what happens to the first term, $a_1m_1\color{Red}{n_1}$ when you reduce it mod $n_1$ @Keith ?

I'm genuinely considering reading the entire latex documentation for the base package, amsmath, amssymb, and tikz

There have to be better ways to do everything.

@arctictern it's the remainder of $a_1m_1n_1$ divided by $n_1$

@Axoren honestly the wiki page I linked should have everything you need, at least until you write actual documents

@KeithYong which is what?

there's a reason I used red color, Keith

should it be zero because $a_1m_1n_1 / n_1 = a_1m_1$

right

since $a_1m_1n_1$ is a multiple of $n_1$, it is congruent to $0$ mod $n_1$

therefore $x$ becomes just $0+a_2m_2n_2$ mod $n_1$

now, how is $m_2$ defined again?

ok ok I'm starting to get it. Let me work with this for now

$m_2$ was from the extended euclid algorithm

tell me more

how is it defined?

@arctictern I didnt study extended euclid deep enough, I just programmed a script for it. But I'll look into how $m_1$ was made from it

@KeithYong who cares about the algorithm. how is $m_2$ defined? not how is it calculated in practice. how is it defined?

But from what I know so far its made from the quotients and remainders of the regular euclid algoirthm

you even told me what m1 and m2 were back up in the beginning

so $n_1m_1 + n_2m_2 = 1$ if $gcd(n_1, n_2) = 1$

you said $m_1$ and $m_2$ were the multiplicative inverses of $n_1$ and $n_2$

yes

that means $m_1n_1\equiv 1\bmod n_2$ and $m_2n_2\equiv 1\bmod n_1$

oh ok

Yeah I get it now. So by using definition of multiplicative inverse I should be able to solve it

so then, $\bmod n_1$ we have $$ x ~\equiv~ a_1m_1n_1+a_2m_2n_2 ~\equiv~ a_2(\color{Red}{m_2n_2}) $$

what does the red term become, since we're working mod $n_1$?

actually I think I got them backwards hold on

$m_2n_2$ becomes $1$ while working in $\mod n_1$ space, by definition of multiplicative inverse

yeah, if $m_1,n_1$ are inverses mod $n_2$ and $m_2,n_2$ are inverses mod $n_1$, it should be $x=a_1m_2n_2+a_2m_1n_1$

the things are backwards

I think I get it now to work it out on my own. Thanks a lot for the help

since mod $n_1$ we want $x=a_1m_2n_2+a_2m_1n_1\equiv a_1\cdot 1+0=a_1$

aight

Thanks again, screenshotting this. Super helpful

Superposition using Laplace on a 2D plate, blah blah, method of separation

That's a physics question, isn't it? Wouldn't it be better off asked at the Physics SE?

It's a PDE question

(Application) - it's just a rectangle with 4 boundary conditions.

My question is: how do you know whether your G(x) or H(y) turns into an exponential/hyperbolic or a sin function?

Because you'll have 2 boundary conditions for either G or H, but only 1 condition (not classified boundary condition anymore) for the other.

Bad wording - rather, what's the significance of 2 boundary conditions v.s one (boundary) condition?

might be good to include how you separated your variables, since you reference G and H

presumably you've got G''(x)/G(x) = -H''(y)/H(y) = separation constant

Yup

well, the fact that it has to vanish identicallyat x=0 and y=0 means that the only possible solutions are of the form sinh(lambda x)sin(lambda y)

and it doesn't really make a difference if you say it's instead sin and sinh, since that's just the same as making \lambda into i\lambda

so the solution should be built out of that

@wowdavers one way to proceed, after thinking about this for a bit: Suppose you replace the boundary condition for u(2,y) with u(2,y)=0, and get the resulting solution as a sum of sines/sinhs. by construction, this will vanish on every edge but y=1

now suppose you instead replaced the boundary condition for u(x,1) with u(x,1)=0. the resulting solution vanishes on every edge but x=2.

if you now superimpose these two solutions, you get correct boundary conditions on all edges

so you solve two using two different boundary conditions, and then combine them

@ForeverMozart what the heck is the motivation for separability

why would I want a space to be separable

hmm

old geometry texts define manifolds to be separable

something like second countability, metrizability, or paracompactness is more natural imo

i don't know about motivation but usually you can prove more properties if the space is separable

even sigma compactness works

but why separable?

Separable metric spaces embed into the Hilbert cube

And if the dimension is small enough it embeds into a Euclidean space

We define manifolds to be (a) Locally Euclidean (b) Hausdorff and then one of 5 equivalent properties

separability is one of them

So you can think of manifolds as subsets of Euclidean space

but why would you state that as a definition

@ForeverMozart you have to go through and prove it's metrizable, which is hard if you start with separability

or not, there might be some metrization theorem that works with just separability and locally compact Hausdorff

maybe

I wrote down proofs that all the various definitions are equivalent

the only useful ones are second countability and paracompactness

so I don't know why old books use separability

because when they want partitions of unity they end up proving paracompactness and second countability anyway

my guess would be that by a standard theorem it embeds into the Hilbert space and you can prove more from there

but what

the point set topology of manifolds is like 3 pages

it's not very interesting

a lot of old books use "separable" for second countable

why would they do that?

@MikeMiller Famous yet?

Nope

but you proved something I bet

Balarka the hurricane

@BalarkaSen

Hurricane Balarka

25ft waves did I read that correctly?

Georgia is going to be destroyed

Yeah, Matthew's gonna wreck. He needs to chill out.

Hurricane?

yes we are gonna get wiped out

There's a Hurricane named Matthew. Yeah, that

this is why you don't live next to the ocean

It's a compromise. You trade hurricanes for tornadoes.

And depending on where near the ocean, you also avoid earthquakes.

lol

We had a tornado here about 5 years ago

In Georgia?

no

Where on the coast are you?

oh i am not on the coast

That's your problem.

I hope none of you are on that side of the coast.

Massachusetts gets the remnants of whatever they're getting right now.

It looks like a boat sinking until you realize it's supposed to be a road.

Or maybe I'm seeing it wrong and that's just beach under there

^ Surf and wind from Hurricane Matthew crash on the waterfront in Baracoa, Cuba, Tuesday, Oct. 4, 2016. (Photo: RAMON ESPINOSA/AP)

Hi @Brody

Used to live in Miami, riding out hurricanes never seemed too bad. At least more in-land it's usually just wind and precipitation, never 25 ft waves thankfully

Aloha! @BalarkaSen

what's up?

Having fun reviewing some basic set theory / logic stuff whilst further depriving myself of sleep

hbu?

Set theory is nice. It's about the same with me: topology.

How's that?

sorry i.e. how's that going?

More or less good.

Good to hear. Topology seems neat and useful, so I'm glad you're faring well

It's pretty cool.

I'm actually considering transferring unis bcuz my current uni only has one undergrad course for it

for @BalarkaSen and any lurkers, every injection has a left-inverse and every surjection a right

in either case, the inverse is necessarily unique?

@Brody Suppose $f : \{1, 2\} \to \{1, 2, 3\}$ is just the inclusion map. Let $g_1, g_2: \{1, 2, 3\} \to \{1, 2\}$ be given by $1 \mapsto 1, 2 \mapsto 2$ each and $3 \mapsto 1$ for $g_1$, $3\mapsto 2$ for $g_2$ respectively.

Both $g_1, g_2$ are left inverse of $f$.

sorry I'm still here

Okay, perfect. That makes sense @BalarkaSen

I missed the last part that $g_2$ maps 3 to 2 so I was confused

$f$ is injective non-surjective. there are "leftover" elements in Y, which allows one to define more than one left-inverse for the mapping

Right

whereas if $f$ were bijective the one-to-one correspondence would not allow such, hence the complete inverse is necessarily unique

That's it.

and similar argument from before for surjective non-injective. thanks!

@Brody Here's an exercise for you. Say you have a function $f : A \to B$, and call it a monomorphism if $f \circ g = f \circ h$ for all functions $g, h : X \to A$ imply $g = h$.

Prove that the class of monomorphisms are precisely the class of injections (so a monomorphism $A \to B$ is an injection, and an injection $A \to B$ is a monomorphism).

@BalarkaSen I think I intuitively understand the reasoning but am wondering how to write it up

$\circ$ (test)

($f\circ g=f\circ h \Longrightarrow g=h$) implies ($f$ has a left-inverse iff $f$ surjective)

more strongly, former if and only if latter

very hand-wavey I suppose. guess I could use more detail

sorry, "surjective" should say injective lol

Sorry, I had to go for a while.

@Brody I don't entirely understand how to construct a left-inverse from the property of a monomorphism.

Neither do I, lol. Let me try something else then @BalarkaSen

Sure, think about it.

Sorry for being long. Still here? @BalarkaSen

Yup

So I'll go the other way then and obtain the left-inverse by letting $f:A\to B$ be injective to begin with. Suppose $g,h:X\to A$. Assuming $f\circ g=f\circ h$ we apply the left-inverse and see that $g=h$, satisfying the monomorphism.

Not sure if it's substantial enough to demonstrate anything but that's Take #2

@Brody Right, but note that you didn't actually need a left inverse, just the definition of injection. $f \circ g = f \circ h$ means $f(g(x)) = f(h(x))$ for all $x \in X$, hence $g(x) = h(x)$ - that's exactly what injectivity means.

(aka $f(a) = f(b)$ implies $a = b$)

Hmm I noticed the two definitions had essentially the same exact format

but I suspected the application was too simple

This direction is not too hard. But, you have proved that injections are monomorphisms. Can you prove monomorphisms are injections?

That's the harder direction.

The other direction follows in the same vein, I'd guess

I'm confused. Did we not previously prove that monomorphisms are injections? @BalarkaSen

Or rather, for whatever reason, I get the impression we proved both directions in that, since the substitution can work either way. $$\forall s,t\in A,f(s)=f(t)\Rightarrow s=t$$ Let $g:X\to A,\;\; x\mapsto s$ and $h:X\to A,\;\; x\mapsto t$, then $f(g(x))=f(h(x))\Rightarrow g(x)=h(x)$

@Brody We seemed to assume $f$ is an injection. We can't do that for that direction.

We assumed $f$ is an injection (that's what "$\forall s,t\in A,f(s)=f(t)\Rightarrow s=t$" means), then proved it's a monomorphism.

That's proving injections are monomorphisms, not the other way around.

what I just wrote is definitely injections to monomorphisms, but it's the same as the proof you wrote above?

I guess it is. I see what you mean now sort of

I just parsed it differently.

yeah that confused me for some reason

(aka because I'm dumb)

nah

no, the reason is that I'm dumb :P

working on the other direction

o/

hi

Given $f:A\to B$, suppose $\forall g,h:X\to A, f\circ g=f\circ h\Rightarrow g=h$. Clearly $\forall x\in X, g(x)\in A\wedge h(x)\in A$. Let $g(x)=s$ and $h(x)=t$ then it follows that $$f(s)=f(t)\Rightarrow s=t$$.

I dunno, it feels like I did the exact same thing and it was trivial

There's a flaw in the logic there. Did you prove that for all $s, t \in A$?

None of this is actually complicated, it's just a good exercise in logic.

We assumed $\forall g,h:X\to A$ so declaring $g(x)=s$ and $h(x)=t$ covers all of $A$

@Brody So you have to produce two such functions which cover all of $A$.

But, like, picking any surjective function does the trick so you're through.

I agree this is a harmless detail, but a detail nonetheless :)

Wouldn't that be a restriction (surjectivity of $g$ and $h$) that could violate the quantification $\forall g,h:X\to A$?

No, it's not. You are given that $f \circ g = f \circ h$ implies $g = h$ for all functions $g, h$ from something to $A$. As a special case, you can pick two appropriate functions.

Note that $X$ is not fixed either.

hmm okay. I'm just struggling to make wholesome sense of all the relationships involved

(X blah Y) is satisfied for all X,Y means given any two specific X, Y, (X blah Y) is satisfied. In this case, picking two surjections g, h the monomorphism condition is satisfied for the same reason.

It still seems odd to me; it'll probably make sense to me when I revisit it later and go "oh, duh"

logic is confusing, especially when you're deprived of sleep - don't worry about it.

I hate all-nighters :/

Get some rest :)

Then we can talk about more interesting stuff (say, groups, if you wish).

can't. I sleep like a petrified corpse and I have lecture in 4 hours

yikes

I blame my job (and conveniently take no responsibility for my poor time management and work/sleep schedule)

speaking of groups, I notice a theme since we just began vector spaces. take a set, "equip" it with an operation or two that satisfy a collection of axioms

yeah, I hate that aspect of equipping with operations without explaining why, or brushing past the motivation by saying "because we know things like this exist (eg for vector spaces there is R^n and groups there is Z) so let's be wise and generalize"

Granted, generalizing is sometimes useful to get the big picture but I can't grok generalizations I won't be able to do. I probably wouldn't be able to write down the axioms of a group by just looking at Z.

good point. I can't speak much on the topic of motivation though. Perhaps it's a feature of good textbooks to provide some backstory and then argue why (historically, semantically) something is conceived a certain way

@Brody Here's how I like to think about groups. Denote $X = \{1, 2, \cdots, n\}$ for some $n$. Consider the set of all bijections $f : X \to X$. Denote it by $\text{Aut}(X)$.

Note that for any two elements $f, g \in \text{Aut}(X)$ (aka self-bijections of $X$), $f \circ g$ is also an element of $\text{Aut}(X)$. There's also an inverse (because it's a bijection!). Associativity is easily verified.

So $\text{Aut}(X)$ is a group in the way groups are defined.

Okay, so the set of all permutations together with function composition is a group

Much neater than integers with addition

Yup

But here's the real punchline

ANY group is a subgroup (subset closed under composition) of $\text{Aut}(X)$ for some $n$.

So any group, defined in the abstract binary operation way is actually always realizable as a permutation group.

This is known as Cayley's theorem.

Huh, really? Pretty dope

kewl, eh?

so long as you don't ask me to prove it as an exercise :P

Nah :)

Looking it up. So we're dealing with $\text{Sym}(X)=(\text{Aut}(X),\circ)$

Right, the symmetric group.

The Cayley's theorem is a consequence of a nontrivial idea, called group actions (nontrivial in not the sense of being technical, but much like Newton's three laws - easy to understand, but probably none of us would be able to write it out without knowing it). As an example, $\text{Sym}(X)$ acts on $X$ - the action is a map $\text{Sym}(X) \times X \to X$ given by $(f, x) \mapsto f(x)$. It acts by "permuting" the elements (duh).

Huh, I like the tangibility of the concept, but what's the purpose?

It's a tool for understanding groups. Suppose you have stumbled upon a group, which looks horribly complicated so you have no earthly idea how to understand it better. Group action realizes it as symmetry group of some set/geometric picture/any object that's easier than the group.

Gotcha

@Danu You're here?

Yup.

@Danu Do you know how to explicitly write down a complex line bundle on the torus with the given self-intersection number with the 0 section/Chern class/divisor?

no.

oh well

I'm trying to understand the Albanese map---and specifically why any holomorphic one-form is a pullback of a holomorphic one-form on $Alb(X)$

Albanese $\neq$ Albanian

But okay :)

morning chat

morning everyone

m

​

Does any of you have any comment on the Albanese variety stuff?

noope

not me

right now I'm trying to figure out why my riemann surface looks like it should be genus 3 from the degree-genus formula but comes out as genus 1 when i try to work out the branching

seems like i've got fewer branch points than I think I should have

(8 instead of 12)

I'm trying to figure out how not to get eaten at my meeting today.

@Semiclassical Do some of the branch points have higher ramification?

I don't think so.

That's the only way I can see your claims both being true...

not following you. but here's the specific surface

$p^4 + q^4 + 4 p^2 + 2q^2-1=0$

which obviously has something going on re: $p^2$ and $q^2$

Degree-genus does say it's genus $3$, so something's amiss with your branching calculations.

if i solve for $p$, i get $$p = \pm \sqrt{-2 - q^2 \pm \sqrt{5 + 4 q^2}}$$

so four sheets.

fixed

ack. should've been $2q^2 p^2$ in my statement of my Riemann surface

only thing I can think of is that somehow the branching is wrong. guess i'll check again.

yeah, I'm not seeing it :/

if I've got $G(p,q)=p^2+2q^2 p^2+q^4+4p^2-1$ and solve for all points $(p,q)$ such that $G=0$ and $\partial_p G=0$, I get 8 such points

and unless i'm doing something wrong in my calculations, if I expand $G(p,q)$ in the neighborhood of these points, then in each case it's of the form $Ap^2+B q = 0$

@MikeMiller So if I have some element of $H^0(X,\Omega_X)^*$ given by $\alpha\mapsto\int_\gamma \alpha$, how can I recast it in the form $\alpha\mapsto \int_X\alpha\wedge\beta$ for some fixed $\beta$? I want to say that $\beta$ has to be the Poincare dual of $\gamma$.

I'll need to bug @ted about the above later, I guess :/