To prove that the given the discrete topology we have for all $x\in X$ $\pi_1(X,x)=\{1 \}$ I must prove that every loop going from $x$ to $x$ are homotopic right ?
@MikeMiller Take two 3-planes in R^6. Perturb it near the intersection to make it nonsingular (or, work with two RP^3's transversely intersecting in RP^6 if you want). What's the resulting 3-manifold you get?
And if you're going to use degrees rather than radians, you should probably not write $\theta$. You should at least put a degree symbol. Really, get used to working in radians exclusively ... :)
@heather: If you say "the full circle is ..." then it's OK. But remember that angles can be any real number (in radians) ... Radians have no units, but of course degrees is a unit.
You can derive the area of the circle just as a limit, the way the Greeks did. Also, please do not write $\int_0^r f(r)\,dr$. That's one of my biggest pet peeves. Do you see what's wrong?
If you're measuring angles in degrees (as you did before you got to radians), you should write $t^\circ$ or something. (You eventually went back and put $\theta/360^\circ$, but you should have the degree symbol both places.)
@heather: You're using $r$ as a dummy variable (inside the integral) and as a fixed quantity (the upper limit of the integral). PLEASE don't do that.
When you do the fundamental theorem of calculus, differentiating the integral as a function of $r$, it's totally confusing if you have $r$'s everywhere!!!
@MikeMiller You don't get S^2 # S^2 for 2-planes in R^4... I mean link of a singularity for two C's in C^2 looks like a Hopf link and you're gluing in the Siefert surface there.
@heather: I give constructive criticism only to make you better :) Eventually, you might want to use pens and different colors, but I think you're starting off great.
@Adel: Well, there's certainly a "stupid" answer to that question.
@MikeMiller I haven't thought about it all but my basic idea would be to take a nontrivial vector bundle over the sphere of the same dimension as the base; projectivize and take the double of the zero section. That should perturb to a self-intersecting sphere with the singularity replaced by what looks like two n-planes in R^2n perturbed.
Ça va. Les études n'ont jamais arrêter d'aller bien, c'est juste la motivation qui s'est absentée. Le plus dur maintenant ça va être de ne pas succomber au beau temps et à continuer encore 3 mois
@TedShifrin but the polynomial $p_2p_1-p_1p_2$ is not from $F^2->F$, it's from $F->F$ and it is the zero polynomial. I want a non trivial poly $q:F^2->F$ that vanishes on (p_1(t),p_2(t)) for all t.
@BalarkaSen @MikeMiller So, I know that if I have two sets with nonzero distance, and the complement of each set is path-connected, then the complement of their union is path-connected. Is this still true if I replace "path-connected" with "connected"? I've talked about this sort of thing before, but with closed sets specifically, so that path-connected and connected coincide.
(Pinging you because you seem like the people to ping for topology questions)
@AkivaWeinberger In most of the cases, it doesn't matter if you have a path-connected or a connected set. But I am not sure because I only had to deal with it in physics, so maybe this won't help you.
(To prove it for the path-connected version, you overlay a grid onto the plane with mesh size much smaller than the distance between the sets, turning it into a much easier discrete problem.)
@TedShifrin To prove that $\zeta(s)=\sum_{k=0}^{+\infty}\frac{1}{n^s}$ is holomorphic on $U:=\{s; \Re(s)>1\}$ I use Weierstrass convergence theorem; do you know what tools we need to prove that it converges only for $\Re(s)>1$ ?
@Ted Suppose I have a fixed 1-form on a manifold. Do you see a way to use this to "weight" parallel transport? Ideally if it's zero somewhere the holonomy in that direction "isn't counted".
@Lozansky I'm vaguely familiar with this sort of thing, the idea is that p-value is supposed to tell you what the chances are that you get the more extreme results you're getting if you assume that there's no sort of causation, I believe
@MikeM: No, I don't. I can integrate $\omega$ along the closed path, but that has nothing to do with your parallel transport. I can consider $\omega(X_f) - \omega(X_0)$, where $X_f$ is the vector I get parallel transporting $X_0$ along the closed path, but ... meh.
Chat, and one more question: aren't you tired of solving every day the same questions (more or less), and follow the same path in mathematics described in the books you read (well, it's an amazing thing to read books, I do it, but carefully read me to understand my point) without creating your own ways of doing mathematics, or to be more direct, to create your own mathematics?
What can be more noble than discovering mathematics in your own personal style? Following your own ways, creative ways which represent you, define you.
Well say we do some experiment and we're trying to see whether variables $X$ and $Y$ have anything to do with each other. If we run our stuff and find a high p-value, then the variation in results is not enough for us to go and say that there's interesting behavior. If the p-value is low, you get a result which is "statistically significant"
@Adel then you're have to refer to the sylvester resultant. did your problem come from a book? if so, the tools to do the problem come in the section before it...
And @Don'tdisturb I find that with many subjects people see in undergrad (before that you're not even really proving things because school system), there is somewhat less room for creativity.
My school has an IBL Calculus class in first year, where you are given definitions and you figure out all the proofs yourself
@Ted Yeah, me neither. I have an idea about how to do it infinitesimally (if I was really interested in the derivative of holonomy), but not how to actually do it
They even go somewhat off the beaten path, by developing this "continuum" and learning about its topology (spoiler alert: it's $\mathbb{R}$), then building other stuff off it
Say you gamble on a vending machine with the probability $p$ to win. The number of games $X$ till the first win has the probability function $$p_x(k) = p(1-p)^{k-1}$$ for $k\in \mathbb{N^{+}}$. Say that $H_0 : p =0.2$ and $H_1 : p < 0.2$. Can we reject the null hypothesis on a certain significance level (say $0.10$) if we lose the first $10$ games and win the $11$th? The $p$-value would be $$\sum_{k=11}^{\infty}0.2(1-0.2)^{k-1}=0.8^{10} = 0.107$$ so we cannot reject it.
The thing is, though, once you're given those definitions as such, even then it starts leading you somewhat to the proof, and in subjects like analysis it's tricky to just have divine inspiration and come up with the definitions to build up the material. A lot of earlier stages of calculus was inspired by necessity from physics, and later stuff with $\delta-\epsilon$ came to rigorize (heh) it
@JeSuis: Perhaps I'm being stupid, but I don't see how to prove in a global way that $\sum 1/n^s$ diverges when $\Re(s)\le 1$. You can't prove divergence absolutely.
And a lot of early undergrad stuff is analysis. I find that subjects along the lines of graph theory might be more amenable to a more, let's just be creative, treatment
@Lozansky I'm probably not best to answer that, but it seems like the general consensus is that when p-values are low, there's good reason to believe that the results you get in an experiment are indicative of something significant, as opposed to just random
Can anyone please help on this. In power series we use series with center 0 ( we check whether the functions are analytic there or not ). If the function is not analytic at 0 we use Frobenius. Is it ok to not use Frobenius & use power series at another point where the functions are analytic. Or is it necessary to apply Frobenius there at x=0
@Ted I think what I want is a smooth function $S^1 \to [0,1]$, 1 on a small neighborhood of the identity and 0 on a large open set, and a fixed base connection $A_0$, and to compute the holonomy of the time-dependent connection $A+\varphi(t)(A_0-A)$.
I guess the thing to say is: If you pick an ordinary point near the singular point, then you can write down analytic series solutions in the vicinity of that ordinary point.
However, if you analytically continue them around the singular point, they won't necessarily return to themselves.
Hi guys. I have to show that a cube (three-dimensional) has 48 congruences. As of now, I'm counting 20 rotations and 9 reflections. Unless I made a mistake, is there some way to include the compositions of rotations and reflections?
@Semiclassical So can we avoid using Frobenius and use series solution at another point where the functions are analytic. Because I am beginner at series solutions so I am not good at it. And I had asked this question and the comment bay Ian was confusing. Now I don't know whether we can form a series at a point other that 0....
Is it possible to apply the following reasoning; we consider the 6 faces of the cube, and note that each face can be rotated in 4 ways, and rotated+mirrored in 4 ways, so we have 8. If we multiply 8 by 6, we get 48 symmetries.
there's $4$ such pairs, and you can permute them freely, which gives you $S_4$. And then there's inversion through the center, which gives $\mathbb{Z}_2$.
