@Kevin: We certainly teach graduate courses in Lie groups, Lie algebras, and representation theory. Some of the material gets covered, obviously, in graduate differential geometry (especially talking about connections and homogeneous spaces and symmetric spaces).
Grr. Well, neither did any of our grad students taking the algebra qual, but it's a totally standard undergraduate exercise (albeit slightly advanced).
@TedShifrin See, the trouble we were having was with wedging $\mathfrak g_E$-valued 1-forms, since there's some Lie bracket garbage in there, too. We needed to figure out why it was the right action... Nobody really bothers to explain or define.
Oh, that's in some version of my lectures, @Mike, and certainly in Chern's Characteristic classes appendix to Complex Manifolds w/o Potential Theory (yes, you're missing the appendix -- I told you).
I guess I can save one of my two copies of Chern for you. :D
Basically, up to a worrisome sign/constant convention for higher-degree forms, @Mike, you expect $[\omega,\eta](v,w) = [\omega(v),\eta(w)]-[\omega(w),\eta(v)]$.
That's why there are $1/2$'s running around when you do matrix structure equations in terms of the Lie algebra stuff.
Of course, there always questions about 1/2 when you do wedge in the first place. I personally prefer to have $dx\wedge dy$ give signed area and not half of it.
Heya bananas
Well, since I'm overrun with responses, I'll go grade. BBIAB.
sure... the actual context is writing down the curvature of $A+a$, $a \in \Omega^1(\mathfrak g_E)$. The computation goes ahead and ends up giving $(d_A+a)^2 = d_A^2 + d_Aa + a \wedge a$. The first two terms make sense... but to get the last term you need to define how $a$ acts on higher things, which nobody ever does.
@TedShifrin I was writing and thinking.
during lecture Manolescu was having trouble with some of this stuff, incl. $1/2$ conventions, so I don't feel too bad :)
@MikeMiller Does this count as a proof? Suppose you have 2 diagnolizable operators which commute, $A$ and $B$. Let $\lvert \psi \rangle$ be an eigenvector of $B$ with eigenvalue $\psi$, then
$$ B A \lvert \psi \rangle = A B \lvert \psi \rangle = \psi A \lvert \psi \rangle$$
Thus $A \lvert \psi \rangle$ is an eigenvector of $B$ as well.
So, $A$ maps the eigenvector of $B$ to themselves and so $\lvert \psi \rangle$ is also an eigenbasis for $A$. Since $A$ and $B$ have the same eigenbasis, they must be simultaneously diagonalizable.
Given an almost complex manifold (X, J), there is the associated Nijenhuis tensor. In the presence of a hermitian metric h, I'm pretty sure one can obtain a (1, 2)-form \eta. I'm wondering whether there is some (2, 0)-form \phi which can be constructed from $J$ and $h$ such that the (1, 2)-part of d\phi is precisely \eta.
I know that the Nijenhuis tensor is a map TM\otimes TM -> TM, but it is skew symmetric and one can use the isomorphism (TM, J) -> (T^{1,0}M, i) to get a (1, 2)-form (I think).
But you'll sort it out. I have faith in you, and I'll go back to grading students who think that when three vectors are given to be linearly independent, you say that each vector individually is "linearly independent." Sigh.
@MichaelAlbanese Yeah, these javascript links need to be run on the page you want them to modify. As you saw, running it on the installation page just made that page render :-)
I just found that $\displaystyle \sin{\left(\frac{\pi}5\right)}=\sqrt{\frac{5-\sqrt5}8}$. I can safely say that the multiple-angle formula is extremely exhausting to work with on an hour of sleep
I have a question, from some notes I was reading: is $\Bbb{R}\times(\Bbb{R}\times\Bbb{R})=\Bbb{R}^3$? (where $\times$ is the cartesian product). I'd say no, because elements of $\Bbb{R}^3$ are tripes of real $\{x,y,z\}$ while the elements of $\Bbb{R}\times(\Bbb{R}\times\Bbb{R})$ are a pair in which the first element is a real number and the second is another pair of reals: $\{x,\{y,z\}\}$. Does that make sense?
@Alessandro If you define addition and multiplication on your pair of a real plus ordered pair, then I'd guess that the two are isomorphic. but I'm certainly not sure.
I mean if you define addition and multiplication the right way
All I was trying to do was find a "simpler" form for $\Gamma(\frac35)$... It involves using the multiple-angle formula twice. The multiple-angle formula, by the way, is $\displaystyle \sum_{n=0}^k\binom{k}{n}\cos^nx\sin^{k-n}x\sin{\left[\frac{\pi(k-n)}2\right]}$
@DanielFischer Could I ask you something about an algorithm?
@DanielFischer Consider two sets $D=\{ d_1, d_2, \dots, d_n\}$ and $E=\{ e_1, e_2, \dots, e_m \}$ and consider an other variable $K \geq 0$. Show that we can answer in time $O((n+m) \lg (n+m))$ the following question: Is there is a pair of numbers $a,b$ where $a \in D, b \in E$ such that $|a-b| \leq K$?
The algorithm should answer the above question with <<YES>> or <<NO>>. Describe the algorithm, show its correctness and show that its time complexity is $O((n+m) \lg (n+m))$.
@DanielFischer I wrote this algorithm: Algorithm(D,E,K){ found=0; for (i=1; i<=n; i++){ A[i]=D[i]; } for (i=1; i<=m; i++){ A[n+i]=E[i]; } HeapSort(A); j=1; p=2; while (j<=n+m and p<=n+m and found==0){ if (abs(A[j]-A[p]) <= K) found=1; j++; p=j+1; } if (found==0) printf("NO\n"); else printf("YES\n"); }
@evinda When you sort the array A, you lose the information from which set each element comes. Since you want one element of each, you must not lose that information. Keep the sets (arrays) separate. Of course, sorting each is a good idea.
Yes, @teadawg, you should get such formulas using deMoivre's formula (rather than repeated use of angle-sum formulas). You would enjoy learning that. And it's easier to do $\sin(2\pi/5)$ using just the quadratic formula. :)
@Ted Yes I did, and no. All I had to do was wait a few days to have my high school math courses put into the online registration system (dunno why it didn't happen when I first submitted my transcript)
LOL, ok @teadawg. I still don't know how good you are. But I encourage our best students to skip some classes which they really already know. They just have to be able to demonstrate that they do. So be polite but aggressive as you get to the college/university level if you really do know stuff.
The flip side is that a lot of students believe they know stuff and they really do not.
We do allow some "course challenges" at certain levels, but not at all, @Lord. But I'm willing to let students prove to me what they know. When I was associate head of our department for 8 years, I didn't have that many students who could prove it. But students who make it through my multivariable math class (which is highly proof-based) doing A work skip our "intro to proofs" class without a problem.
I understand, @Kevin, but I always thought $\psi$ meant the function itself. I guess that must be chemists' notation :D
So, $\langle x \lvert \psi \rangle = \psi(x)$ is how one would write the wavefunction. Almost never as a bare $\psi$ unless you first say you're suppressing the arguments of the function
Well, the more advanced the student, the more likely they are capable to qualify, @Lord. My biggest problem was with students who were placed into precalculus, who'd already taken high school calculus and who therefore were confident they knew algebra and trig. They didn't.
In low-level courses we may have attendance policies, @Lord. Not typically in upper-level. But I've found that students who miss more than one or two classes a semester in my upper-level courses tend to do very badly.
Anyway @TedShifrin I'll take my baby step. I have no idea what's left to show though. If $A$ maps the eigenbasis of $B$ into itself, why does that set not also form an eigenbasis for $A$?
@TedShifrin This may correlate with the observations of some of my friends who went to the US, noting that the higher-level classes there are much harder than in NL as well.
Basically, I consider it to be way too easy to obtain a Dutch degree.
No, @Kevin, the basis may not be the right one. So you need to argue, first of all, that if you restrict $A$ to an invariant subspace (namely the $B$-eigenspace for $\psi$) that that operator must still be diagonalizable.
I don't know, @Lord. It depends where they went. My impression has been that calculus in Europe is generally proof-based, whereas in the US it is uniformly engineering-style, except for our theoretical, proof-based courses, which fewer than $\epsilon$ of the students take.
@Lord_Farin There may be selection bias. Students who would travel from Netherlands to the US for university probably are going to many of the higher-tier universities, whereas those who say in the Netherlands may go to any level of Dutch school
I sent a brilliant student to Utrecht to do a math-physics program there (about 10 years ago). But he had taken all sorts of graduate courses here as a sophomore and junior (2nd, 3rd year) before he went.
Of course, as far as undergraduate stuff goes, I only experienced Utrecht. That said, Utrecht generally goes as the best Dutch maths university, together with Leiden.
@TedShifrin I think maybe that's my quantum mechanics bias showing. All the operators I can think of span the whole Hilbert space. Or possibly if the Hilbert space is a Cartesian product of complete spaces, then they span one of the factors.
@Lord_Farin And here I thought Leiden was in Germany. Netherlands clearly needs to improve their PR machine.
Anyhow, @Lord, that student did well at Utrecht, ended up with a Ph.D. in probability/statistics at Stanford (although I'd tried to groom him for geometry), but ultimately decided to work for Yahoo. :)
Okay @TedShifrin I think its clear that I've moved too far out of my element. I'm not comprehending even your facially simple statements (an operator can have more than one eigenspace???)
I'm going to community college this semester, but I'm going to a 4-year institution to get a Bachelor's after I get my Associate's degree. My dream is to go to Berkeley for grad school
Yeah, @Lord, college level, at least. I've enjoyed undergraduate teaching far more than graduate. Graduate students for the most part sit there like mute lumps. (I've taught several dozen grad courses.)
@DanielFischer So do we have to look only at the differences $|d_1-e_1|, |d_2-e_1|, |d_3-e_1| $ and so on, since $e_1$ is the smallest element of the set E and so if $|d_i-e_j| \leq K$ for some $1 \leq i \leq n, 1 \leq j \leq m$, it will also be $|d_i-e_1| \leq K$, or am I wrong?
@TedShifrin Interesting. When I was teaching, I enjoyed teaching the upper division courses better than the massive calculus courses. However, I didn't teach any graduate courses, so I don't know about those.
@teadawg1337 I agree with Ted. Also, it's certainly possible to come back from that. My GPA sophomore year of undergrad was like a 2.0 because I had a breakdown due to stress from personal issues. Despite that, I managed to get into a tier 2 PhD program, so it's not the end of the road, for sure.
@Kevin @teadawg: As long as you explain such things in your letter/statement of purpose, sure, but I hope teadawg will stay on an even keel through college :)
@TedShifrin So theres a bike travelling along a path and you're going to give me parametric equation for, say, the center of mass of the back wheel and you want to know the trajectory of the front wheel?
I was bored out of my mind taking Calculus in high school, lol. The only thing that's kept me sane since then has been doing independent math work. I'm still amazed that I've been able to teach myself three semesters of Calc and two semesters of analysis in the past four months
When I switched to French for one lecture, I had a lot of angry students, @robjohn, but the motivation was a silly one ... to get votes for "The Big Screw" at MIT (you can google it — I won it :) )
@teadawg: And, if you're doing well but are bored, I promise to give you something challenging to think about.
@TedShifrin I think I need more information. For example, say the back wheel travels in a line at constant speed, the front wheel could be doing the same thing, or it could be rotating with constant angular speed about the point of contact on the back wheel.
@TedShifrin Ah okay, that makes it easier. I was imagining that your bike was perhaps skidding on ice and so the only constraint was that the bike doesn't fall apart.
@TedShifrin Well you were gonna give me the trajectory! I'm only used to assuming rolling without slipping when you don't know the trajectory because that's the only reasonable assumption which will let you compute it.
I'm not confident at all with my knowledge of vector calculus, it's definitely a subject that's better to learn in a classroom setting than independently
@DanielFischer So, isn't it like that?
Algorithm(D,E,K){
found=0;
HeapSort(D);
HeapSort(E);
i=1;
j=1;
while (i<=n and found==0){
if (abs(D[i]-E[j])<=K) found=1;
else i++;
}
if (found==0) printf("NO\n");
else printf("YES\n");
}
Iff a number $n$ is divisible by $3$, then the sum of its digits is also divisible by $3$.
Proof:
We know $n \mod 3 = 0$. By the basis representation theorem, $n$ can be re-written as $n_k10^k + n_{k-1}10^{k-1} + \cdots + 10n_1 + n_0 \equiv 0 \mod 3$. By the modular arithmetic addition rule, we can take $n_k10^k \mod 3 + n_{k-1}10^{k-1} \mod 3 + \cdots + 10n_1 \mod 3 + n_0 \mod 3$. But then we just get $n_k + n_{k-1} + \cdots + n_1 + n_0$, since $10 \mod 3 \equiv 1$.
Now suppose that the sum of the digits was divisible by 3. How can I prove that the representation in base $10$ is divisib…
@TedShifrin @PedroTamaroff I guess since the back wheel can't turn and the bike is rigid, any force applied ot the back wheel which results in some curvature of the path is due to the front wheel turning. Since the back wheel is rolling without slipping, the torque on the bike about the center of mass must be 0. So then the forces on the front and back wheels are the same. So then you just have to solve newton's 2nd using the derived force as a function of time from the back wheel's path.
Now since Ted is giving us the path parametrically and not necessarily as a function of time, we may have to do some chain rule nonsense to massage the data. But that's not so hard.
@PedroTamaroff We do it for all our introductory courses. You may not realize how many students don't understand the difference between $F = ma$ and $\vec{F} = m \vec{a}$
Once you get to the upper level it's clear which you mean from context. But at the introductory level we force the students to always write the arrows so that their notation is as clear as possible and doesn't lead to mental confusion.
@DanielFischer Ya in practice that's what we often do. But when we don't do it in class we get people trying ot do things like $F_g = G \frac{m_1 m_2}{r^2}$
