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8:15 PM
Heyo
The topology exam went pretty well :3
 
Congrats!
 
I think I just messed up the fundamental group of one of the lens spaces
 
I wrote a short intro to what mathematical gauge theory is. Emphasis on short.
 
@Danu Algtop or pointsettop with short algtop-finish?
(Our pointset topology course has a small algebraic topology part at the end, figured you might have the same.)
 
I'm not sure that deserved to be starred, but thanks
 
8:26 PM
Given some of the other things that gets starred here I'd say its fine.
 
I looked for the thing they were quoting Lee's book on, but wow, it's a one-word citation of gauge theory,.
"...just to get to the point where one can even think about studying specialized applications of manifold theory such as comparison theory, gauge theory, symplectic topology, or Ricci flow"
 
When asked to solve an exercise by a prof and to send her a pdf with the solution, how many corrections can you send before its awkward?
 
@AndrewThompson algtop
(part 1; only fundamental group and some stuff about covering spaces near the end)
(part 2 will be (co)homology)
 
@AndrewThompson I once was working on a project over the summer, sent in a calculation. Realized it was wrong, sent in another. Did this three more times before sending an email saying "That's all garbage. I'm going to stop sending emails. See you on Monday."
 
@Danu Oh, so its half of a course or two different courses?
 
8:29 PM
Two different courses
(intended to be taken in sequence)
One lecturer
 
@MikeMiller Well, for a project I suppose that's fine. This was a really easy identity with Lie derivatives.
 
I'm taking all the guy's courses (next semester will be the 4th :D)
 
I had a sign error, corrected it but in such a way that another sign error popped up, and now I finally corrected that.
 
Meh, sign errors are nbd.
 
As one of my professors always says:
I will take the convention where all my prefactors are correct, and equal to 1.
 
8:32 PM
Hahah. Had my algtop and cohomology exams right before christmas. (Two courses, however cohomology is half the credit.) Did alright in algtop, got my ass kicked in cohomology. Nice sensor gave the same grade for both.
 
Neat-o :)
I'm really excited about (co)homology.
Also math. gauge theory! Btw, @MikeMiller it turns out Kotschick won't be the teacher.
 
I like the algebraic feel of it (co)homology, more so than what one sees in undergrad courses in commutative algebra. (Which might be why I'm studying homological algebra now.)
 
Who, then?
 
I'm trying to find out who it is, but it's apparently some post-doc who is not known to everybody
 
Do you have a department postdoc listing? You could probably just go down it.
 
8:34 PM
I'm excited about it being a post-doc; maybe he'll try hard to make it a nice course.
@MikeMiller Not really a good one, I think, but maybe... Let's see...
Plus, how could I tell who'd be the one teaching it? It's not like they all maintain websites with up-to-date info...
 
I've always wondered: are all junior (i.e. no tenure) professors in the US excellent lecturers, at least in terms of preparation?
 
Ohhhhh, maybe it'll be Swoboda!!! :D
If it's Swoboda I'll be happy.
 
@Danu: I suspect you have a short list of postdocs that do gauge theory.
 
He was the TA for my Riemannian (Chern-Weil) course
 
@AndrewT: No.
 
8:36 PM
and I know he published on Higgs bundles
which is bound to be math. gauge theory
 
@MikeMiller Would've imagined they'd be desperate to keep their jobs.
 
This lists only 3 postdocs, one of which I know for a fact will be TAing top2
So maybe this guy
 
@Semiclassical how is it going?
 
I like a lot
 
oh, alright. quiet
have some grading i should do
 
8:41 PM
18 million dollars lost to Bernie Madoff
 
not difficult to do, but doesn't mean i'm eager to do it
 
I see. :-)
 
it is best to think about a problem for a few days before trying to write a proof
 
@ForeverMozart Lol, depends on how much time you've got on your hands, and how hard the problem is.
 
its a hard problem to me
but that is the question
is it better to try things before you have the proof in your head?
 
8:47 PM
I don't think you can say something in full generality.
 
maybe I will go drive through the country and think about it
 
Perhaps attempt a proof by poetry?
 
you know that feeling when you think like 10 moves ahead in a proof, and you have to stop out of frustration?
can't keep track of all the moves
but then you go down that track again and again
I bet chess players make good mathematicians
 
I'm playing some chess right now ;)
But I'm definitely not a good mathematician! :\
 
9:06 PM
This is the kind of thinking that ruins you a lot in general. I'm not good at that. Really? How do you know that? Why don't you try to work extremely hard for years, 12-16 hours a day and see then who you really are and what you can actually do?
I teach my kids to never use this type of thinking.
 
@I'manartist Ruins me? I feel just fine, knowing that I'm better at physics than mathematics.
 
@Danu It is said in a very general note since I hear many saying these statements.
 
It's funny how you attempt to draw conclusions about my personality from nigh-zero knowledge about me.
But I guess you mean well :)
 
@Danu I only identified a way of thinking, but maybe you just joked and think differently.
Sure, the idea is that after much work we change our opinions about ourselves.
I remember that during some of the interviews I had I was asked to identify negative traits, define, say, 3 (I don't know why they don't ask for 2 or 4, or 5) of them.
But the thing is that one cannot give such an answer. What one may consider a negative trait, it can be a positive one in certain circumstances. Then, I don't like to think I'm the whole life like a statue, but I permanently make improvements.
 
How can I prove if a function is surjective?
 
9:15 PM
They want you to describe yourself as if you're so well defined nothing can be changed, like a statue.
 
@Michael Pick an arbitrary point in the codomain, and prove it is in the image.
 
$\Bbb Z \to \Bbb Z,

f(x)=n^2+1$Prove that this is not surjective
 
What's the idea then? Today I had the interview, right? In one week I'm far more better than I was one week ago. The idea is to see the potential of a human being.
Anyway.
BE EPIC IN EVERYTHING (and to put an end to the small talk)
BBL (some research to do)
 
you cant be epic
maybe rarely epic
cause some problems are difficult
thats why they are unsolved
 
now where is that mindset gonna get you?
 
9:23 PM
I mean, I try some anyway if they interest me.
but you can't just think you're gonna discover something epic every time
 
@Michael Why don't you do it.
 
@Danu I'll try it one more time hahaha
 
Hint: Try some small, positive numbers, and see if they lie in the image.
 
you can't get a negative number
so its not surjective...
 
9:26 PM
Man, why are you spoiling it?
 
lol its obvious
 
That's how you become a Bad Teacherâ„¢
 
sometimes students need a good bashing
 
;)
I knew it couldn't be negative
Wait
facepalm
 
see
facepalms are good
 
9:30 PM
Know its time for me to facepalm at indexed sets
 
Spoiling exercises is never good @ForeverMozart
 
but sometimes I think students need to know it's easy, and that they just have to think in simple logical terms.
 
If $f(x)=x^3$, why is$ f^{-1}(x)=x^{1/3}$
 
@Michael What's the definition of an inverse?
You just need to check that.
 
A one-to-one, onto function f : X → Y has an inverse $f^{-1)$: Y \to X$
−1

if and only if f(x) = y
 
9:35 PM
So, that iff condition you list. Is it non-trivial?
 
What is non-trivial?
 
So let $y\in\mathbb R$...
 
oh ya
 
$y\in Y$ i mean
oh, well your $X$ and $Y$ are both $\mathbb R$
 
yep
I know you replace the x with y and y with x
oh wait
facepalm
 
9:41 PM
you need to evaluate $f^{-1}(y)$. call it $x$. then evaluate $f(x)$...
 
yep yep yep I got it ty
It's just the notation that I kinda need to get used to
What does the big U sign mean? (not union)
 
$\bigcup$
$\cup$ $\bigcup$
the big one?
$A_0\cup A_1$
$\bigcup _{i=0}^1 A_i$
same thing
 
yep
 
they are the same
 
Okay. Can you explain what the i and 1 mean?
 
9:54 PM
$i$ is the index
it goes from $0$ to $1$
so its the union of $A_0$ and $A_1$
 
what if $i $ went to $2$
 
then it would be the union of 3 sets
$A_0\cup A_1 \cup A_2$
 
Oh ok. Kinda like a series
Kinda
 
yeah like $\sum _{i=0} ^1 r_i=r_0+r_1$
you can do infinite number of things with the big symbol
 
God bless highschool notation. /sarcasm
 
9:58 PM
but not with the little one, unless you put "..." or something
 
Man this math is so cool
 
Even though I understand 2% of what this book is saying
 
hello everyone
had a question about harmonic functions
 
Ola
 
10:04 PM
I need to find a function g(x,y) harmonic on function $\{1<x^2+y^2=9\}$
such that $g(x,y)=3$ when $x^2+y^2=1$ and $g(x,y)=8$ when $x^2+y^2=9$
 
should that be harmonic on $\{1<x^2+y^2<9\}$?
 
yes
 
okaly. what i'd point out first is that, while you've stated the boundary conditions in Cartesian coordinates, these aren't necessarily the most useful for this situation
 
ah
honestly, I'm just writing the problem my professor gave me down. I'm a bit lost at the moment
 
sure
well, on a related point, what does your domain look like (geometrically)?
 
10:09 PM
Well, the class is complex analysis (an introduction, anyway)
but as nothing has been specified,
I'm guessing just a standard cartesian plane in $\mathbb{R}^2$
 
sure, but you don't want the entire complex plane.
 
OK
 
all you care about is a certain region in it.
 
OK
 
what region is that (and what does it 'look like')?
 
10:11 PM
As far as I know, the first quadrant
 
then you're not looking at the statement of the problem
 
I am; perhaps I didn't understand your question
Are you referring to the circle with a radius of 8, bounded by the circle with a radius 1 and another with the radius 9?
 
Well, the region $\{1 < x^2 + y^2 < 9\}$ is not the first quadrant
 
well, I agree with the 'bounded by' part. but i don't see why radius 8 matters
 
alright, nevermind that I said that
 
10:14 PM
so it's a circular strip, also known as an annulus
and you're looking for a suitable function on that domain
 
alright
so, what I have from my notes is that this function g(x,y) will have the form $a\cdot ln(x+y)+b = g(x,y)$
 
now, there's two routes to take. one where you make an intelligent guess, and one where you derive it by hand. probably for now i'd suggest the second approach.
well, does that form work?
i.e. can it satisfy the boundary conditions?
if it's not obvious, you should plug in some values and see. so, for example, you can check $g(1,0)$ and $g(-1,0)$---both should equal 3.
 
Yes, it works
 
how so?
 
@daOnlyBG I'm a little dubious. What's $g(-1,-1)$?
 
10:20 PM
before I proceed, how can I get mathjax on this browser?
 
oh, I've been beaten to it
 
(for this chat)
 
See LaTeX in chat on the right
 
@MikeMiller not sure that's the best example, since it doesn't land on the boundary (though perhaps you're aiming somewhere else)
 
OK, one second
A couple things
 
10:21 PM
@Semiclassical: Well, the claimed function was $g(x,y) = a\ln (x+y) + b$...
 
sure?
there's more than one thing wrong with that ansatz, i should say. i'm just focusing on one point for now.
 
Alright, nevermind-
I am incorrect. Sorry, but I'm really frustrated with this problem, and I know I shouldn't be
perhaps if we can step back and see how I can properly derive what g(x,y) should look like, I'll have less trouble going forward
 
yep, and that's the second approach
so, let's recall: what does it mean for a function to be harmonic?
 
for a function to be harmonic, it must satisfy Laplace's equation; that is, the sum of its partial derivatives must equal 0
 
that second statement is true if you're working in Cartesian coordinates
it won't necessarily work if we decide to change coordinates
 
10:26 PM
@Semiclassical Personally, I thought "It's not even defined on the domain" was the biggest one
 
I suppose I am out of luck, then
As far as I know, I am working in Cartesian coordinates
there is nothing to explicitly state otherwise
 
well, that's where you're starting from.
 
Great
 
but what i'm getting at is that that's not necessarily the most useful coordinate system here
 
Okay
 
10:27 PM
one problem with Cartesian coordinates here is that they don't take advantage of the symmetry of the problem
what kind of symmetry does your domain have?
 
My understanding is that my domain has a circular symmetry
 
correct. Cartesian coordinates aren't so helpful there, because both $x$ and $y$ change under rotations about the origin even though the domain doesn't.
Is there a better coordinate system?
 
Polar
(?)
 
yep. then we've got $1<r<3$ as our domain
and then the only thing we need to check re: $\theta$ is that $g(r,\theta)$ is $2\pi$-periodic in $\theta$
this does have one issue, though: I don't remember what Laplace's equation in polar coordinates is off the top of my head!
so unless you do, we'll want to derive it (or look it up)
 
let's derive it
 
10:34 PM
are you trying to find all rotationally symmetric harmonic functions?
 
@daOnlyBG okay. then you'll want to use the Chain rule to express $\partial_{xx}+\partial_{yy}$ in terms of $(r,\theta)$
which is too tedious to type out all here. i suggest going by hand
 
sure. one moment, please
What I understand is
I should find the x partial derivative of $x^2+y^2$ and the y partial derivative
and then let $x=r cos \theta$ and $y=r sin \theta$
correct?
 
well, you can use that to get $\partial_x r,\partial_y r$, which is useful in the chain rule
 
I don't follow
 
well, $r^2=x^2+y^2$
 
10:40 PM
right
 
so if you take $\partial_x$ of that expression, you get $2r \partial_x r = 2x$
 
yes
 
and similarly for $\partial_y r$. so you can obtain those two quantities. but you'll need to plug them into the chain rule
 
sorry- by your partial derivative notation, did you mean
$\frac{dr}{dx}$?
 
no. $\partial_x r= \dfrac{\partial r}{\partial x}$
it's definitely still a partial derivative
 
10:43 PM
okay, proceed
*let's proceed
 
well, then you've got $\partial_x = (\partial_r x )\partial_x+(\partial_\theta x )\partial_\theta$ by the chain rule
hang on, need to sort out nonsense
okay, totally botched that expression
$\partial_x =(\partial_x r)\partial_r +(\partial_x \theta)\partial_\theta$
there
what remains is to compute $\partial_x\theta$, $\partial_y \theta$
need to head out for a bit
 
Thanks anyway
 
11:44 PM
@Hippalectryon slt!
 
i proved an interesting lemma today, hopefully will lead to an interesting result :):)
on a class of homogeneous space
 

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