$(\Omega ,\Sigma )$, what is $ba(\Sigma )$ and $ca(\Sigma )$ Context: I'm supposed to prove that these are Dedikind complete Riesz spaces satisfying a certain critera
@TobiasKildetoft: one definition would be that for every nontrivial element $r \in R$, there exists an ideal $I$ such that $R/I$ is finite and $r \notin I$.
Essentially one is looking at constant vector fields on $\Bbb R^2$, then the induced vector fields in $S^1 \times S^1$ after quotienting by $\Bbb Z^2$, isn't it?
Constant vector fields have integral curves as lines. Lines with rational slopes quotient to circles in the torus, and with irrational slopes quotient to dense circles.
@Huy In your post on MO you have mentioned that you was not able to find Rational Groups by Merzlyakov. In fact, the book can be easily found online. (In Russian, I do not know whether a translation exists.)
Feel free to let me know if you cannot download it for some reason, I can also send it to you by email.
@MartinSleziak: unfortunately, since I don't speak Russian, not really :(
@MartinSleziak: you don't know by any chance if the question is too basic for MO? I only know very little algebra and this comes up in an article about algebraic/geometric topology, so I can't really judge
I can read cyrillic alphabet, but I do not speak Russian either. :-(
@Huy Judging by posts and chat discussions I saw you take, you know certainly more than I do. Personally, I would not worry about that - if some MO users consider it too basic, they will let you know quickly enough by downvoting and/or voting to close.
But my personal opinion is that it is not too basic.
I'm beginning to despair about ever getting up to MO levels. Hopefully my meeting with thesis advisor tomorrow can help me get over the hump in these banach lattices.
I thought that I would be able to find in Merzlyakov's book also the original source where Malcev proved this. But the only entry among references with Malcev's name is Vol. 1 of his collected works (Izbrannye trudy, in Russian - judging by the title).
@BalarkaSen And it should be mod 2pi, since $x$ and $y$ range from $0$ to $2\pi$.
@BalarkaSen Hmm
What multiples
What I did is unfold the torus into a $2\pi\times2\pi$ square, with opposite sides identified
Then the integral curve is a bunch of diagonal lines
In the rational case, the slope of that line will be rational, $a/b$
So you have to solve the simultaneous equations $at\equiv 0\mod 2\pi$, $bt\equiv 0\mod 2\pi$, where we have assumed wlog that the curve starts at the origin.
This has a solution if $a,b\in\Bbb Z$.
In the irrational case, we have $it\equiv0\mod 2\pi, t\equiv 0\mod2\pi$, and this does not have a solution for $i$ irrational
so the orbit never closes
I guess you have to show that the system $it\mod 2\pi, t\mod 2\pi$ gets arbitrarily close to a given $(x_0,y_0)$?
Not sure how to do that
Maybe we can view the torus as a symplectic space and use Poincare recurrence or something.
Right, it's an exercise in Arnold to show this using Poincare recurrence.
@0celo7 I think you might be interested in this question and related ones math.stackexchange.com/questions/450493/… if I understood correctly what you want to prove (but I just quickly read what was going on in chat so that might be unrelated)
@Huy this is a rather deep result in commutative algebra. I don't think a proof just from the definitions makes sense, just because it hides a lot of what is going on
it is best if you take these results for granted if you are not familiar with them
@CRAZYGAYSHERIFF: ok, now let's just say I take it for granted. the answerer now claims that the Jacobson radical/nilradical vanishes due to the Nullstellensatz. I've checked out several versions of the Nullstellensatz but don't really find one which implies this. can you tell me which version I should be looking at?
It is a known fact that if $k$ is a field that is finitely generated as a ring, which is the same as having a surjective ring homomorphism $f:\mathbb{Z}[x_1,\dots,x_n]\to k$ for some $n\in \mathbb{N}$, then $k$ must be finite. Since finite generation as a ring implies finite generation over the p...
@BalarkaSen: yes, I noticed that the nilradical vanishes. do you know an elementary argument that nilradical and Jacobson radical coincide for polynomial rings?
@BalarkaSen: he then claims that taking such a maximal ideal $m$, $A/m$ must be finite again by the Nullstellensatz. I found the version "if $K$ is an algebraically closed field and $m$ a maximal ideal in $A = K[X_1, \dots, X_n]$, then $[A/m : K]$ is finite". I'm guessing he uses this? (just need to think for a second why this implies that $A/m$ is finite)
For groups, I can push a basis at identity around. That's easy. But multiplication by an element using the H-space structure need not give me a self-homeomorphism of the space, need it?
@Semiclassical I guess, but I don't want to define them since nobody will know what it means if I do. I just wanted to know if there was an existing one someone would recognize.
You could also define a derivative with another symbol, but then you have to add in an explanation and it makes it a bit more inconvenient than having something everyone recognizes.
@Owatch True. But even in that regard, it's not atypical for me to see papers which specifically declare that "$\prime$" is used to denote differentiation.
I'm pretty much trying to find some common symbols to represent a couple of the operations my program does. Definite integrals is easy, I've got $\int{f(x)}d(x)$
I came across this definition
Let $U\subset S\subset C$. We say that $U$ is relatively open in $S$ if for every $z_0 \in U$,
there is $r > 0$ such that
$$D(z_0 ;r)\cap S\subset U$$:
Now the author doesnt specify if the subset symbol means that you do not count the trivial subsets ( the em...
there's probably an argument for writing it just as $roots(p)$, since the values of the roots don't depend on whether you label the variable as $x$ or $z$ or w/e
If $p$ is a point in $k^n$ corresponding to a maximal ideal $m \subset A = k[x_1, \cdots, x_n]$, it's ring of functions is $A/m$ (which is, by the way, a field).
That has to be finite-dimensional: there cannot be too many functions on a point when there's not too many functions on the ambient space itself.
let me know if that makes sense
@Huy Re: jacobson an nilradical. I think there's an concrete, equivalent defn of jacobson radical in A-M. that's what you're going to need in that exercise
something like, collection of all $a \in A$ such that $1 - xa$ is a unit for all $x$.
Maybe you should try to work it out, it shouldn't be too hard.
I'm a bit weirded out by the fact that we're working with subring of $\Bbb R$ though, not polynomial rings.
I came across this definition
Let $U\subset S\subset C$. We say that $U$ is relatively open in $S$ if for every $z_0 \in U$,
there is $r > 0$ such that
$$D(z_0 ;r)\cap S\subset U$$:
Now the author doesnt specify if the subset symbol means that you do not count the trivial subsets ( the em...
@archipelago It was asked above why $S^2$ is not an $H$-space. I only know a proof of that for topological groups. Maybe you'd want to tell @iwriteonbananas why it's not one.
@iwriteonbananas @BalarkaSen The first thing that comes to my mind is the following: The rational cohomology of a path-connected H-space of finite type is a finite dimensional Hopf algebra, so it has to be a free graded commutative algebra in odd generators. That's certainly not the case for the rational cohomology ring for S^2.
That proves it for all even spheres at once. Although there should be something more elementary in the case of S^2.
Thanks. I don't really know much about Hopf algebras but I suspected something like that, probably iwriteonbananas would understand it better. Yeah, I also wonder if there is an elementary proof.
before, the authors introduced dihedral groups and I understand them and can list all the elements as products of rotations and reflections and why they are of order 2n for regular polygons, where n is the number of edges of the polygon
You should perhaps study the first few sections of Dummit-Foote again and understand how algebra works in groups. I have got work to do now, so can't really guide you through it.
@BalarkaSen I suppose I didn't need the $A$, since it can be factored out. So now I just want to deal with $\sum\det B^n$. Which your suggestion just turns into the identity matrix with one of the $1$s turned into $1/(1-\det B)$.
@AkivaWeinberger Think about the action of SO(3) on S^2 by rotating it. That gives a homomorphism SO(3) --> S^2, sending g in SO(3) to g times north pole. fiber over an element $x$ then precisely is the collection of all rotations of S^2 which moves the north pole to $x$. That's an isometry of the geodesic circle connecting x to northpole, no? So it's collection of isometries of S^1, which is SO(2).
I think that's the correct description, but I am not sure because I haven't thought about it that way. Probably it's fine.
On the other hand, to see why the unit tangent bundle of S^2 is SO(3), note that the action of SO(3) on S^2 extends fiberwise to the unit tangent bundle. It's transitive, because I can just take a $(x, v)$ to $(y, w)$ in $T_1S^2$ by moving $(x, v)$ to $(y, v)$ along a great circle joining $x$ & $y$, then rotating $(y, v)$ to $(y, w)$ by an axis going through $y$ and it's antipode. It's clearly free everywhere except the axis of rotation, where it just rotates the circle fibers.
Hence, SO(3) is the unit tangent bundle.
@AkivaWeinberger I wondered about that. probably you should think about the Hopf fibration and S^3 covering RP^3.
I don't know off the top of my head how to make that work, but I have heard that the linking # of the fibers in the fiber bundle S^1 --> RP^3 --> S^2 is 2, which makes sort of sense (you're wrapping the circle fibers in the Hopf fibration twice, or something). There should definitely be something like that.
The D^3 description is probably not useful. Use S^3/Z_2.
@AkivaWeinberger Let's look at the Hopf map $S^3 \subset \Bbb C^2 \to \Bbb{CP}^1$, $(z_0, z_1)\mapsto z_0/z_1$.
Antipodal point of (z_0, z_1) in S^3 is (-z_0, -z_1), isn't it? So Hopf map factors through the Z/2-action.
If the last thing I said is true, I can quotient by Z/2 to get a map RP^3 --> S^2, which should definitely still be a fiber bundle, fibers being RP^1's.
Does that sound good to you? I hope I am right about this.
I came across this definition
Let $U\subset S\subset C$. We say that $U$ is relatively open in $S$ if for every $z_0 \in U$,
there is $r > 0$ such that
$$D(z_0 ;r)\cap S\subset U$$:
Now the author doesnt specify if the subset symbol means that you do not count the trivial subsets ( the em...
Can someone please solve this and explain the steps taken to reach this solution ?
