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6:00 PM
Anyhow, I need to go finish working on the dessert I'm baking for the dinner for 16 people tonight. I'll be back when I'm done.
 
@TedShifrin He didn't even introduce the Chern connection yet!
@TedShifrin Wow, good luck with that---sounds like it has to be a huge dish!
What are you making, actually?
 
good evening
 
6:27 PM
Hi @Alessandro
 
6:43 PM
Hi, does anyone have a reference for the calculation of the cohomology of U(n)?
Using Serre spectral sequence
 
Perhaps Hatcher?
(I have no idea)
 
back
actually nah be back later
 
@Danu, couldn't find it on his book on SS :(
 
@MikeMiller ^?
Or maybe @BalarkaSen?
 
I think it might be in Hatcher's AT using Leray-Hirsch, but I can't say for sure either.
 
6:51 PM
You are right, checked it right now
But my goal is learning SS, so this is only kind of helpful
 
Fair enough. I don't know a thing about spectral sequences though, so I wouldn't be able to give you a reference.
@Danu What's up?
 
Hello.
 
@BalarkaSen Not too much. Hodge decomposition again :P This time for bundle-valued forms.
 
7:06 PM
Ah
 
math.stackexchange.com/questions/1977620/… Please help I'm going mad with this problem.
Sorry for the interruption into the chat I hope you help me.
 
I have to wake up 7 in the morning tomorrow, ugh.
 
@BalarkaSen hii
 
what are you readind now a days?
 
7:14 PM
bit of topology, bit of geometry
 
what??
be more specific
 
vector bundles from the former, geometry of surfaces from the latter
 
btw, what exercises mike gave you?
are there any interesting one related vb?
 
those are exercises about connections (and relevant diffgeom, I suppose), from Taubes' book.
 
you are reading geometry on surface? or Riemannian geometry?
 
7:19 PM
Surfaces, but I do plan to learn some Riemannian geometry.
 
So which books are you following?
 
Ted's differential geometry notes.
 
ohh
I wanted to read the Taubes book, bcz I wanted to know the connection on bundles, which is not very well covered in the book from where I read Rie geo
Any way, chalo good night, I too need to wake up early in the morning
 
me too
night
 
7:45 PM
How's everything going guys?
 
@TedShifrin Serre duality is in the bag
 
8:00 PM
OK, now I'm back.
 
@Rrjrjtlokrthjji Hi
 
my multivariable probability prof yesterday
that's pijamas!
now better
great teacher though.
 
hi chat
 
@Semiclassical Hi
 
8:11 PM
Does anyone want to help me with a challenging problem ? :)
 
as per the room description: "just ask; don't ask to ask."
 
:)
@Mahmoud can you please show some working on your part, this site is not for answering questions for you but for helping you out when you get stuck. — EHH 9 hours ago
What happened ?
@Semiclassical
 
What do you think ?
 
you didn't state an actual question...
what you've quoted wasn't a question, but rather someone else advising you to include your own working.
 
8:15 PM
Prove that for all real x and y if x(x+y−1)+y(y−1)=0 then 0≤x+y≤4/3.
 
okay. that's an actual question.
 
Hi @Randal. You're not here often.
 
My professor gave it as a homework
 
what course?
 
Notions in logic
 
8:16 PM
ah.
one approach is geometric: the equation you gave defines an ellipse (it's quadratic in x and y)
 
I think I should use Proof by contradiction or contraposition but it wasn't working for me.
 
additionally, the equation remains valid if you swap $x$ and $y$.
 
Why ?
 
hence you've got an ellipse which is symmetric along the line $x=y$.
well, you can rewrite your equation as $x(x-1)+y(y-1)+xy=0$
 
Incredible, it took me two days to figure that out
 
8:20 PM
anyways. so you have to have an ellipse with $x=y$ as an axis.
now, you want to know the value of $x+y$ on that ellipse. a
 
You're right but I'm supposed to use logic rules, and I still didn't learn ellipse equations, if you don't mind
 
fair enough. where that reasoning will lead you, anyways, is that $x+y$ will be maximized or minimized when $x=y$
so the max/min values occur when $x(x+y-1)+y(y-1)=x(2x-1)+x(x-1)=3x^2-2x=0\implies x=0,2/3$
and therefore $x+y$ is bounded above by 2*2/3=4/3 and below by 2*0=0.
 
You are amazing
 
eh, this is pretty straightforward if you've got a sense of the geometry
anyways.
even if you don't want to use that geometry directly, you can still get the spirit of it in the following way
first, since you want to maximize/minimize $x+y$, we might as well use that as a variable instead. so let $u=x+y$.
 
Yes
Please continue.
 
8:25 PM
to go further, it's helpful to also define $v=x-y$ to capture the extent to which $x$ and $y$ disagree
(following the logic above, we should end up finding that v=0. but that's getting ahead of ourselves)
with those definitions, we can write $x=(u+v)/2$, $y=(u-v)/2$.
so whatever $x,y$ we have, we can express them in terms of real $u,v$.
if we plug that into the above, we get...
Sorry, have to go now. But try plugging those substitutions in and see what that looks like
 
Is that going to solve it ?
I see that we'll get rid of the term (xy)
But thank you @Semiclassical this was really helpful
I owe you two weeks of thinking.
 
8:43 PM
hello, i need a sequence from $\mathbb{Q}$ but it converge to an element in $\mathbb{R}\setminus \mathbb{Q}$
any idea please ?
 
Can you prove a triangle congruent to itself without stating each individual segment is congruent, just using the reflexive property? Or would it be necessary to state each individual segment is congruent and SSS?
 
@AndrewLi: Have you proved congruence is an equivalence relation? This is part of that proof.
I would, of course, use SSS.
hi @Semiclassic
 
@TedShifrin So I can't directly prove that a triangle is congruent to itself with reflexive property? Each individual side would then have to be proved to be congruent to itself through reflexive right?
 
Hello @TedShifrin
Can I ask you something?
Let $u(x,y), v(x,y)$ be harmonic functions in $\mathbb{R}^2$. I want to find the sign of the difference $u(0,0)-v(0,0)$ given that $u|_{x^2+y^2=1}=\frac{1}{2}-\sin{\phi}$, where $x= \cos{\phi}, y=\sin{\phi}, \phi \in [0,2 \pi)$ and $v|_{(x-2)^2+y^2=3^2} \leq 0$.

Applying the theorem for the solution of the Dirichlet problem defined on a sphere I found that $u(0,0)=\frac{1}{2}$.

In order to find something about $v(0,0)$ I thought to use the following:

If $u \in C^0(\overline{\Omega})$ is harmonic in $\Omega$ and $u \leq 0$ in $\partial{\Omega}$ then $u \leq 0$ in $\overline{\Omega}$.
 
Oh, I see your point
@TedShifrin Thanks!
 
8:52 PM
Sure. :)
 
@TedShifrin hello
can we find a sequence from $\mathbb{Q}$ but it converge to an element in $\mathbb{R}\setminus \mathbb{Q}$
?
 
Why not @evinda?
Of course, @Vrouvrou. You can find a sequence in $\Bbb Q$ that converges to any real number you want.
 
but i don't find an example
 
Ah $v(0,0) \leq 0$ and so $u(0,0)-v(0,0)=\frac{1}{2}-v(0,0) \leq 0$, right? @TedShifrin
 
What is your definition of real numbers, then, @Vrouvrou?
Look again, @evinda.
 
8:55 PM
I meant $\frac{1}{2}-v(0,0) \geq 0$ @TedShifrin
It was a typo
 
In fact, $\ge 1/2$.
 
$\mathbb{R}=\mathbb{Q}\cup \mathbb{R}\setminus\mathbb{Q}$
 
That's not a definition, @Vrouvrou.
What is a real number?
heya @AndrewT
 
@TedShifrin Hello
 
Hello, @Mahmoud.
 
8:56 PM
Hello @TedShifrin. How are things?
 
i don't know a definition of a real number
 
@AndrewT: I'm relaxing a bit. Done preparing for a dinner for 16 people tonight.
 
@TedShifrin Would you mind helping me with my problem ?
 
What is that?
 
Prove that for all real x and y if x(x+y−1)+y(y−1)=0 then 0≤x+y≤4/3.
 
8:58 PM
Oh yes right... Thank you very much!!! @TedShifrin
 
Using mathematical logic.
 
@TedShifrin Wow, that's some work. Family party?
 
Nope, @AndrewT. A friend and I are throwing a party to congratulate a mutual friend on finishing his Ph.D.
@Mahmoud: I doubt just logic will do it. You obviously have to use some algebra, too.
 
I know but it's ... complicated.
@Semiclassical Started it, but he got to go ...
You seem to be busy, just forget about it.
 
Where are you stuck?
 
9:02 PM
@Ted Is it for me?
 
No.
 
I think that I lack the idea that'll solve it.
I'm not mathematician.
a mathematician*
 
What have you been studying?
 
Mathematical logic.
First-order(Basics)
 
This seems much more an exercise in clever algebra than anything to do with logic.
 
9:07 PM
Hi guys I'm back
 
You are right, our teacher likes to push us into thinking, and searching.
 
Aha
$x(x+y-1) + y(y-1) = x(x+y-1) + y(x+y-1 - x)$
Does that give you a small hint?
 
I just did something like that, @Steamy. I don't see where it goes.
I can prove easily that $x+y\ge 0$. But I don't see the upper bound yet.
 
Neither me.
 
Ah, I only have the upper bound actually
heh :P
 
9:09 PM
:)
 
You get $(x+y)(x+y-1) = xy$, then apply AM-GM: $(x+y)(x+y-1) \leq \left(\frac{x+y}{2}\right)^2$
 
AM-GS ? Sorry ...
 
The lower bound comes easily from completing the square.
 
It's the comparison of arithmetic mean (AM) and geometric mean (GM). It's really just $xy \leq \left(\frac{x+y}{2}\right)^2$
 
@Mahmoud: What math exactly do you know? Do you know high school algebra? Has your teacher taught you inequalities like what @Steamy just said?
If I gave you $u^2-u\le 1/2$, could you tell me what you know about $u$?
 
9:11 PM
Yes, I know how to solve single variable inequalities.
Oh wait please
0\le u \le \frac{sqrt(3)+1}{2} ?
 
Well, the upper bound is right, not the lower one. But this turns out not to help.
We don't know what your teacher has taught you, so it's very hard to help you.
I can show $x+y\ge 0$ by using completing the square, and @Steamy can show that $x+y\le 4/3$ by the method he suggested. I don't see that this has anything to do with basic logic. I don't see how to suppose $x+y>4/3$ and arrive at a contradiction, for example, without knowing what Steamy did. Etc.
I suggest you go bother your teacher.
 
It doesn't just continue
WB @Semiclassical
 
@ted the approach i was suggesting to @mahmoud initially was to recognize that said equation is just an ellipse with axes $x=y$ and $x=-y$
well, axes parallel to that. it's not an ellipse centered at the origin
and from that it's easy to see that you'd better have $x=y$ to min-max etc.
 
Right. Again, I don't know what they've been taught.
 
9:25 PM
nor I.
 
I can complete the square, but honestly, I didn't think about rotating the axes, because I thought that was way beyond what they'd be expected to think of.
 
@TedShifrin But actually, who graduated?
 
i mostly thought of that from the fact that it's symmetric w/r/t $x=y$
 
Oh, @MikeM, not a math person. A friend at UCSD finished his Ph.D. in bioengineering.
Sure @Semiclassic.
 
Oh, gotcha.
 
9:26 PM
You're sophisticated :)
 
I should have gone to UCSD for undergrad.
 
@MikeM: Not clear that this guy will appreciate all the effort I've spent cooking for 3 days, but some of the guests will :)
 
Hehe.
 
even without sophistication, i think choosing $u=x+y$ is a fairly smart approach
since that's what you want to maximize.
 
9:27 PM
It simplified the equation
 
taking $v=x-y$ as well is what's harder to justify without some sophistication
 
Sure. But when I complete the square I get $x+y = (x+y/2)^2 + 3y^2/4$, so I can't substitute too early :)
 
@MikeM: At UCSD you too could have been in major classes with 100 students.
 
@mahmoud did it? I should scroll back through the history and see what's been said after me
 
9:28 PM
Maybe so.
But I would have met PVAL.
 
Right.
You would have been pushed much more, but you've managed OK.
 
I'm trying
 
Sorry for annoying you guys
 
you can write $x=(u+v)/2$ and $y=(u-v)/2$, so it should be straightforward if tedious
 
9:29 PM
Actually, I wonder if there are 400 majors how many actually know each other. When I went to MIT as an undergrad, I knew mostly the students I was in classes with, and not all of them.
@Mahmoud: It's not annoying, but sometimes you really just need to bug your teacher. For one thing, your teacher needs to know what you are having difficulty with.
At least, in my 40+ years as a teacher, I wanted to know that.
 
@TedShifrin At any school there's students on different tracks. I suspect the PhD track students mostly knew each other.
 
if you grind through the algebra (or use Mathematica) it looks to become 3/4*u^2+1/4*v^2-u=0
and that's something that's far easier to complete the square of.
 
Actually, @MikeM, one of the guys I ended up rooming with at Berkeley the first year was a classmate when we were undergrads and we were never in classes together and only met because we found out we were both going to Berkeley.
 
Hm.
 
9:31 PM
@Semiclassic: I have yet to comprehend how this has anything to do with learning basic logic in mathematical arguments.
 
My teacher doesn't care about our difficulty, he suppose that we digested all the notions we've been taught before, and blames us if we can't keep on the track
 
actually, if you just write it as v^2=4u-3u^2
eugh.
 
What class is this precisely, @Mahmoud, and at what level?
 
that's not good pedagogy
 
That sounds awful
 
9:32 PM
There's a lot of bad pedagogy going around everywhere, guys. Including MSE, for that matter :P
 
you can then write it as v^2=4u-3u^2=u(4-3u)
 
I don't like that, @Semiclassic. I want to complete the square to get bounds.
Oh, I see.
 
and since the left-hand side is never negative, you need u>0 & 4>3u or u<0 & 4<3u
 
So either $u\ge 0$ and $4-3u\ge 0$ or $u\le 0$ and $4-3u\le 0$.
 
@TedShifrin In your country, is a 16 years old supposed to know how to solve it ?
 
9:33 PM
yeah, should be \ge
 
You're in high school, @Mahmoud?
 
Yes.
 
that's a pretty mean problem :/
at that level and in that context, anyways
 
More likely a high school algebra student who's been taught some of these techniques will solve it than an average college student.
 
true.
 
9:34 PM
Well, @Semiclassic, we don't know what the teacher has taught them. He may have taught them rotation of axes for conics, for all we know.
 
and it's 'obvious' geometrically speaking
true
 
That used to be in the curriculum when I went to high school.
 
could also do it with Lagrangian multipliers :P
 
Plus, knowing that $wz\ge 0$ if and only if both have the same sign is important.
smacks Semiclassic
 
lol, i deserved that
 
9:35 PM
I haven't smacked Balarka in a long time.
8
 
The course is supposed to be about Basics in Mathematical logic, but ...
 
though the geometric notion actually goes well with the multipliers idea, come to think of it.
 
Have you had algebra and trigonometry, @Mahmoud?
 
Yes
 
you imagine sliding the line $x+y=c$ along the ellipse and see that the point of tangency always occurs when $x=y$
okay, time to go catch my bus. later
 
9:37 PM
At c=4/3
 
OK, so this is giving you harder stuff along those lines. But perhaps the key idea you're supposed to be using is what I just said. The square of any real number is non-negative. And if the product of two real numbers is non-negative, either they must both be non-negative or they must both be non-positive.
 
Thanks @Semiclassical See you later
Yes, I understand
 
Anyhow, good luck, @Mahmoud.
 
Thanks for your time I really appreciate it @TedShifrin and @Semiclassical
I need to develop a clever way of thinking, but how ?
 
Doing lots of these exercises means you pick up some "standard" tricks
 
9:42 PM
Hey, if $e^{i\theta}$ plots a point on the unit circle in the complex plane at $\theta$ radians, and $\theta$ were $i$, then we would get $\frac{1}{e}$ on the real line. Kind of interesting, referring to $i$ as a geometric measure. Not that we really get an angle at all, but rather a random-seeming point at 0 radians
@Mahmoud if only there was a clean-cut recipe for "clever thinking"
 
@dsillman2000 That would change everything
 
Tried to sleep for a couple hours. Failed.
 
Try again !
Or do yoga
 
@BalarkaSen There are prescription drugs available for that in California.
They're illegal, though.
 
Not if they're prescription!
Just get them prescribed
I used to have lots of insomnia problems
Turns out they are rooted in stress
Not really easy to stop without a lifestyle change
 
9:49 PM
Some drug laws are more subtle than that.
 
I got to sleep via cold medicine
 
Last thing I'd want to be is a junkie.
 
If someone sees @Semiclassical please tell him that he is genius !
He approaches problem in a very smart and clever approach
 
He sure is a clever person.
@TedShifrin: I'm grateful for that.
 
Is $x+y\le 0$ & $x+y\le 4/3$ A true statement ?
Can I just consider $x+y\le 4/3$ ?
 

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