@JoshuaA If $b$ is expressible as a power of $a$, then the group is cyclic and may be expressed as $\langle a \rangle$, but otherwise, no, it's not cyclic.
@KajHansen: But that doesn't specify that it can be generated by two elements. The standard term for something that can be generated by at most $n$ elements is, well, $n$-generated.
A fact that I use relatively often @JoshuaA is that $S_p$ can be generated by a $2$-cycle together with a $p$-cycle. So if I wanted to show a polynomial has Galois group $S_5$, I can potentially cut down on my work significantly.
There are all sorts of places where knowing a generating set for a group comes in handy.
IIRC, $A_n$ has a small generating set, a fact you can use when proving $A_n$ is simple for $n \geq 5$. It's been a while since I've thought about that though, so I could very well be wrong.
Well, it depends on the case you're looking at @JoshuaA. $S_n$ itself can be generated by a $2$-cycle together with an $n$-cycle, so long as the $2$-cycle is specifically $(12)$. There are lots of subtleties in the hypotheses that you have to be careful with here, lol
I guess. It just colors all my interactions, and I have a bad habit of wishing that everyone else were just as timid and tactful--bordering on milquetoast--as me. :D
For that question yesterday involving y = |x|, what constitutes a sufficient proof this is not a 1-manifold? I mean, is just the lack of differentiability enough?
No, @Stan. See the first definition of manifold (there are three in section 6.3.)
Well, no, @Fargle, but office hours do help :)
@Stan: This is an answer I've given probably 3 or 4 times on main. A $k$-dimensional submanifold of $\Bbb R^n$ must be a graph of a smooth function over one of the standard coordinate $k$-planes.
In the case of a curve in $\Bbb R^2$, you have only two cases to rule out.
@Fargle: As you can see from reviews on Amazon, my textbook writing is not the handholding, reduce everything to copying from the book, style. Today's students don't like it, but I defend it completely.
I appreciate the usefulness of just doing narrow exercises, computing values on certain integrals or sums and such because frequently in my own work I'm stymied by some computation that has no known solution. But I'd appreciate it a lot more if it were systematically documented and useful, a la, Wolfram
I see. And for a figure in $\Bbb R^2$, $\mathbf{B}$ is always $(0,0,\pm 1)$ (if that point could exist in $\Bbb R^2$), so I can just use $\mathbf{T}$ and $\mathbf{N}$.
@Kevin: Interestingly, the popular books (like Lay) leave dot product to the end of the book and totally de-emphasize geometry. Whereas I emphasize it from the outset.
I think a lot of algebraists and their students hate that, @Kevin.
@Ted I think we go through the same thing. Students get stuck in a particular local minimum of how to solve problems. Often it involves mere memorization and repetition, ends/means reasoning, etc. We try and teach them how to think like a scientist rather than like a student. But they hate it because its not a local maximum of how to get the best grades
@Ted: Well, it's not so much picking here... I need more than the standard one for the de Rham complex. E.g. if I want my favorite moduli spaces to be compact, I need to know that the cohomology of an elliptic complex is finite-dimensional, so I need this version of the theorem.
But I admit I like the ideas here more anyway. More suited towards eg Atiyah-Singer.
@Ted: AFAIK it's only for the Dolbeaut complex. I need/want it for more exotic complexes. That's all. The Hodge theorem's proof seems out of place for most of the other stuff one does in a complex manifolds course, no? so it seems reasonable to skip it.
Well, not necessarily, but it's the same reason I never take the 2-3 lectures to prove Sard's theorem in a diff top course. The proof doesn't give insight into what students should really be learning in the course.
@leo: That sort of question has probably been asked a hundred times.
Anyway, a question about the constant breadth thing: how are we justified in letting $\beta(s)$ be the point opposite $\alpha(s)$? It seems non-trivial to me, at least at first, that $\beta$ would be well-defined.
I heard a few days ago that even for something simple like a scalar $\phi^4$ theory, the number of Feynman graphs at arbitrary order in perturbation theory is not known. This seems like something that mathematicians shouldve figured out 100 years ago.
Probably not, I recall we had quite a mod party for that guy who was in here a while ago..... cant remember the name something to do with 'gay' or 'homosexual'
@MikeMiller It just a simple quantum field theory. There's 1 field which has a single operator-value at each point in spacetime. It's self interaction is proportional to the field amplitude to the 4th, thus $\phi^4$
How should I go about showing that $\lambda = 0$, @Ted, where $\beta(s) = \alpha(s) + \lambda(s)T(s) + \mu N(s)$? I already showed why $\mu$ is the coefficient of $N$. Should I just differentiate?
@MikeMiller That's a tough one because the foundations of QFT aren't rigorous yet. A quantum field theory is defined by a hilbert space where each point represents a certain configuration of the various fields (just a set of operators assocaited with each point on the pseudo-riemannian manifold that is spacetime).
@MikeMiller The state evolves according to the rule that the probability of going from 1 state to any other is given by a functional on that Hilbert space which takes as input ALL the paths connecting the 2 points in the Hilbert space.
Can I assume that $T_{\beta}(s) = -T_{\alpha}(s)$, @Ted? The only other possibility is actual equality, but it seems like convexity eliminates this problem.
@KevinDriscoll Still, the approach you outlined is still ill-defined. The Euler-Lagrange equations for $\phi^4$ are nonlinear and since quantum fields are actually operator valued distributions, you have ill-defined field equations.
@KevinDriscoll The only way of making the theory work is to define it perturbatively via a path integral, although that is marred with problems as well.
@skullpatrol so the difference is 46ms if I divide that by the new time I get 4500 so that means its 450000% as fast? or do I divide by the old time which gives me 99% faster?
@0celo7 If I could explain to Mike a way of understand QFT without that problem, I wouldn't explain it to Mike though. I'd write several papers and collect my Nobel Prize.
@TedShifrin you mention maximal rank on this page in chapter 6 in relation to manifolds? What does maximal rank tell me about a given entity if I am trying to decide if it is a manifold?
@skullpatrol wait that doesn't make sense though, because if something was 100% faster than wouldn't that mean that it would take half the time? the difference between original and new is ~10^4
Symmetry can be surprising. I can give you a function $f(x,y)$ that is symmetric (i.e., $f(x,y)=f(y,x)$, and yet the critical points are not on the diagonal.
Odd because "every" calculus problem that we do that is symmetric has an optimal solution with $x=y$.
first thing is, they're linear combinations of chebyshev polynomials of the second kind. and that falls under the rubric of Bernstein-Szego polynomials---there's a section in Szego's OP book that talks about them
and apparently they're quite important for approximations/universality claims. not sure i actually have a clear sense of that
the latter is basically just "OPUC with only finitely many nonzero verblunsky coefficients," and thus can be seen as truncations of other OPs. so, nice for approximation stuff
@KarlKronenfeld I should add the condition. By the way, the relevance of the Künneth formulas in the post is that the canonical Koszul complex is exact.
suppose I want a sequence of OPs in $x=\cos\theta$ which are of the form $$U_n(x)-t U_{n-k}(x)=(\sin\theta)^{-1}\left(\sin(n\theta+\theta)-t\sin((n-k)\theta+\theta)\right)$$
looking at the Szego reference, i need (up to normalization) some $h(e^{i\theta})$ such that the above combination of sinusoids is the imaginary part of $e^{i(n+1)\theta}\overline{h(e^{i \theta})}$
indeed, i think $h(e^{i\theta})=1-t e^{i k \theta}$ does the job just fine
from that, one gets that the orthogonality weight should have absolutely continuous part $$w(x)\,dx=\frac{\sqrt{1-x^2}}{|h(e^{i \theta}|^2}\,dx=\frac{\sqrt{1-x^2}}{1-2t T_k(x) +t^2}\,dx$$ where i've used $T_n(x)=\cos k\theta$
Sorry @Kevin, I had to drop out. I'll read what you wrote tomorrow. I('m training to) do topology via (mathematical) gauge theory so I'd really like to understand the physics.
@MikeMiller Ah okay, that's really cool. The connection is quite interesting and I think particularly from the physics side not always well-understood. Theres something interesting about what happens when you go from a classical field theory to the corresponding quantum one. The topology of the space that the theory lives in can be quite different such that the symmetries of the classical theory are not preserved in the quantum version. We call this 'an anomaly' and I really don't understand it.
@TedShifrin Alright, I just had the epiphany. You have a bunch of $T_{\alpha}$ terms on both sides, and then the $\kappa(s)\lambda(s)N(s)$ term. Since $N$ is orthogonal to $T$, its coefficient must be identically zero, but $\kappa(s)$ was assumed non-zero, so $\lambda(s) \equiv 0$ as a result.
@TedShifrin That really makes the rest of the problem easy, too, because what falls out of part (a) is exactly the relation I need to manipulate: $\mu \kappa(s) = 1 + u'(s)$, where $u$ is the arclength parameter of $\beta$.
