@PhilipHoskins $X$ is homeomorphic to disjoint union of $X\setminus e$ and $e$ modulo some identifications along the boundary of $e$, not to the disjoint union itself. E.g., a sphere is made up of gluing two disks along their boundary, does that mean it's homeomorphic to disjoint union of two disks?
@MikeMiller OK, so note that since $\pi_1(M) \to \Bbb Z/2$ is given by taking a loop, making it transverse to codim 1 cpt submanifold $N$ and computing mod 2 int. number, it's trivial on the loops not intersecting $N$. This means the cover above is trivial on $M - U$ where $U$ is a small, say, normal bundle nbhd of $N$ (which exists by epsilon-nbhd theorem) So above that the cover is disjoint union of two copies of $M - U$. But note that $\partial U$ is just a two-fold cover of $N$! So I am pretty sure the cover can be obtained by gluing those two copies along $\partial U$ by the nontrivial…
I don't have a rigorous argument, but I am quite sure this is what the cover should look like.
@Huy It's probably 16, but that's likely also true.
yeah, I prefer having harder exams at university level, but anyone should be free to try study what they really want to. maybe they will excel even though on average, they were rather bad in high school
right, i do believe one should be free to choose their stream beforehand (science/arts/whatever) and maybe even particular subjects (math/physics/etc inside science) so high marks on those will matter more than low marks on say, biology for someone who took math.
which, speaking of, today i was told in school to compute the minimum distance between a line and a point using coordinate geom. i used lagrange maximization.
it was good practice. even if i get a bad grade it would be compensated by the knowledge that i didn't forget all the calculus i did through the year :P
@MaryStar If you knew $T_{n,0}$ for $n\ne0$, then you could solve for $k\ge0$, but there is no way to propagate the given information into a unique solution.
Eh, good point, I have never actually constructed that. I just "guessed" that the cover is topologically $(M - U) \times S^0$, modulo pasting the two copies along $\partial U$ by the nontrivial deck transform.
Hm. It should. For any pt $p \in \partial U$, I have an nbhd homeomorphic to a half-ball inside $M - U$. After attaching the other $M - U$ it should glue to it another half of the ball, so that should give an nbhd homeo to ball.
@TedShifrin I just got an invitation to speak as part of a special session at the Joint Mathematics Meeting. Now I am trying to figure out if I am able to go.
@MikeMiller So I am trying to map $(M - U)\times S^0/\partial U \sim \partial U$ to $M$ by mapping $\partial U$ of both copies by two-fold cover to $N \subset M$, and the rest of $M - U$ in two copies by projection. It's unclear if I can do this in a continuous way. :S
so if we have for example $S_{13} $ it can be represented as a disjoint cycles as : (1,2,3)(4,5,6,7)(8,9,10,11,12,13) the order is 12 here ,which means $S_{13}$ has 12 elements?
Is $S_{13}$ symmetric group on 13 elements? You showed that it has at least 12 elements. Order of an element is somewhat different from order of a group. @Sadams
@MikeMiller Sorry, I didn't get to think about that for a while. Suppose I take the manifold I constructed. Then $\Bbb Z/2$ acts naturally on it by taking pts of one copy of $M - U$ to the corresponding one in the other and pts on $\partial U$ to the one the deck transformation of the cover $\tilde{N}$ sends to in $\partial U$. I think the quotient is the covering map you're looking for.
@Semiclassical Hi, I had a question I wanted to ask you. If you're still here I'll ask you in a bit.
@Semiclassical OK, so suppose I have two objects joined by a string, one of mass $m_1$, the other of mass $m_2$. The object of mass $m_1$ is applied the force $F_1$, the object of mass $m_2$ is applied the force $F_2$. Then some calculation tells that the tension on the string is $(m_2F_1 + m_1F_2)/(m_1 + m_2)$ i.e, $m_1m_2/(m_1 + m_2) \cdot (f_1+f_2)$ where $f_1, f_2$ are the accelerations corresponding to $F_1, F_2$.
aka, harmonic mean of $m_1, m_2$ times average of $f_1, f_2$.
Does the last thing have any physical explanation?
This is a pet model I use to work out pulley problems (pulleys don't really matter in pulley problems), and I noticed this weird thing happens and I don't really know why
eh, it isn't entirely formal. after all, the reasoning is 1) if only one mass matters, then only that force matters, and 2) in general, the forces combine linearly
If I write $T = mf$, it's clear that $m$ = harmonic mean and $f$ = arithmetic mean of the $f_i$'s mean it's averaging out between things somehow. i'd like an explanation along that lines.
probably the main thing about tension is to compare it not with a force like gravity but with, say, a contact force from standing somewhere
in both cases, it's not so much an expression of how two bodies interact (as was the case for gravity) but rather than they interact in such a way that some coordinate doesn't change
now, one thing we're not talking about here is work and kinetic energy. that provides a different perspective on this. but if you're not on that yet, don't worry about it for now.
@Semiclassical Here's a more refined question that might explain what I am trying to ask. Suppose I have a system with three masses $m_1, m_2, m_3$, joined by three strings to a single point $p$. I apply force $F_1, F_2, F_3$ (corresponding acceleration $f_1, f_2, f_3$) to each of $m_i$. What does the force given by harmonic mean of $m_1, m_2, m_3$ times arithmetic mean of $f_1, f_2, f_3$ represent in this system?
@GPhys Yes, of course. But why the harmonic mean of $m_1, m_2$, is what I was asking. It's not geometrically apparent why tension works that way.
There should be some visually obvious reason for this.
Right, so we're going to make a surface that maps down to the $z$-plane with two points over most points (the two values of $\sqrt{f(z)}$) and one point over a few (the branch points).
The discussion is helping you understand that you pick up the multi-valued nature when you go around one of the roots of $f$, but that if you go around two of them along a curve, then you come back to the original value.
The "exceptional set" will be $\infty$, plus the poles of the coefficients of $P(T)$, plus the zeros of its discriminant on $\mathbb P^1\setminus\{\text{those other points}\}$
OK, cuz in the general case Forster is considering the coefficients of $p(T)$ to be meromorphic functions on some Riemann surface. OK, very abstract, but OK.
He's talking about an $n$-fold branched covering of a general Riemann surface $X$, @Danu, thinking of it as parametrizing the roots of a meromorphic function of degree $n$ on $X$.
