@MikeMiller Am I right in saying the germinal holonomy of the Reeb foliation is identity for every loop? I'd thought it be something more interesting for the longitudinal loops along the leaves but the picture doesn't tell me that
In the direction that contributes to pi_1 of the solid torus (that's what, the longitude?) nothing should change. But pi_1 in the meridian direction should act something like x mapsto x/2.
Right, it should be more interesting for the boundary. If I loop along a longitudinal guy, the diffeomorphism (-1, 1) --> (-1, 1) should be contraction
$n$ is not, but unless $f(x)=n$, $f(x)$ can be in the domain, the point of the $\tau$ is to make sure that $f(x)$ does not lie in the image of $\tau\circ I$
@MikeMiller My loop does contribute to $\pi_1$ of the solid torus; it doesn't bound any disk. I am confused as to why anything should change looping around the opposite direction
I wanted to say: Your loop can't have holonomy, since the holonomy should be the same as the holonomy of the induced path in the universal cover, and up there there's a single chart containing the entire path.
The problem is to make that statement, you have to define holonomy by picking the same induced chart around both of the endpoints around the path, and when you do that, you get a contraction.
The intuition I have is if you take a longitudinal loop and draw a small stick locally transverse to every leaf; slide that leaf around and back to itself stuff gets contracted because the other leaves come closer than it was before
I am a little weirded out tho. Shouldn't that say the Reeb foliation on S^3 has nontrivial holonomy along the longitudinal loops of the torus leaf too?
Ah, homotopy is not actually hard. If I let it run till time very small $\epsilon > 0$, the paths all remain inside the plaque-chain my path was in so nothing happens. So the germ is locally constant wrt time.
So in trying to show that $\tau$ restricted to $\mathbb{N}_{n} - \{f(x)\}$ is a bijection, the range of $\tau$ under that restriction is $\mathbb{N}_{n-1}$. Then, by construction, we see that if $\tau(i) = \tau(k)$, $i,k \in \mathbb{N}_{n}-\{f(x)\}$, then $i = k$.
To show it's surjective...
@s.harp $\forall j \in \mathbb{N}_{n-1}$ is it just obvious that each of these $j$'s has a preimage in $\mathbb{N}_{n}-\{f(x)\}$?
Also @s.harp even if we don't restrict $\tau $ to $\mathbb{N}_{n} - \{ f(x)\}$, it is a bijection, right? As in $\tau: \mathbb{N}_{n} \to \mathbb{N}_{n}$ is a bijection?
Think of the torus as R^2 - 0 mod Z where Z's action is generated by v mapsto 2v say. There's a foliation by the parallel lines (one of them punctured) to the x-axis. If I am not wrong this is exactly the Reeb fella, yes? 'cuz make that R^2 - 0 into a cylinder; the punctured line become two lines and the other ones become parabolic curves asymptotic to the lines
@JessyunBourne Yes, $\tau$ is a bijection and also its own inverse, meaning $\tau\circ\tau$ is the identity on $\Bbb N_n$, the case $n=f(x)$ $\tau$ is actually the identity
@user1952009 after I' ve read one of your last questions, I was asking myself if the abc conjecture was related with probabilistic models of prime numbers in the literature. If you want provide me a response of this comment, any case a good week.
@BalarkaSen not sure what is a lie group, in fact I look at the torus $R^2/Z^2$ and I look at curve in "it". But to look at the length, I need a metric, right
@JeSuis You do, but what does that have to do with "scalar product"? As a quotient R^2/Z^2, the torus admits a (flat) Riemannian metric from R^2; that in turn gives you a usual-metric-space metric on the torus.
A smoothly-varying inner product on the tangent space at each point. This is, intuitively, the least amount of information you need to define length of a curve on a manifold. That actually does give you a (usual) metric.
I don't know how to actually write down a (usual) metric on R^2 which preserves Z^2-action.
@JeSuis: There are lots of different "it"s you can pick up in $\Bbb R^2$, but if you specify a particular point to start at, then "it" will be unique. Regardless, they'll all have the same length because the metric is invariant under $\Bbb Z^2$.
@Kasmir: Depends on the problem. Think about things you've seen with Green's Theorem where the region is a ring ($a\le r\le b$). Then the inside and outside circles are oriented oppositely. The same can happen in 3D with surfaces.
Oh, this is scary. I just got an email from the VP of the US bragging about his tie-breaking vote. I guess the White House has my email from various petitions. They're probably already spying on me.
Ok so when I look at the torus, at a point, looking at the tangent plane, I get an ellipsoïde which give me an dot product, right ? @TedShifrin petitions of what?
@JeSuis: The Riemannian metric that Mike and Balarka were telling you about is not the induced notion of dot product/length coming from the torus sitting in $\Bbb R^3$.
@JeSuis: They're diffeomorphic, @JeSuis. The torus defined as the abstract quotient manifold, the torus in $\Bbb R^3$, and the torus I said in $\Bbb R^4$ are all "the same."
The commutator subgroup of pi_1 of the wedge of circles corresponds to those paths that do not wind around any of the individual circles. Inside the Cayley graph, this is what you get if you superimpose the Cayley graph onto Z^2, and demand that your walk must end up back at the identity,.
Well, I guess I'll go ahead. So the task is to prove that the orthogonal group is a smooth, compact submanifold of $\mathbb{R}^{n^2}$, and to find its dimension.
@JeSuis: Tu n'as pas encore une métrique. Il faut se décider. Si le tore se trouve dans $\Bbb R^3$, ce qu'on discoutait ailleurs ne va pas. Si le tore se trouve dans $\Bbb R^4$, comme je l'ai dit, ça va bien. Il y a beaucoup d'autre choix qu'on pourrait faire, enfin.
If we take the set of symmetric matrices, it's a smooth submanifold of $\mathbb{R}^{n^2}$ with dimension $\frac{n(n+1)}{2}$, by constraint equations on the matrices.
@AliCaglayan The fundamental group of the subspace of R^d consisting of vertices at each integer and edges between vertices that differ by one of the standard unit vectors.
If anyone is interested, I'm hosting a big number contest. Code your number in any language (no experience needed), maximum of 256 characters, not including spaces, and try to reach the largest number you can. First submissions due Saturday. :-) http://chat.stackexchange.com/rooms/51337/this-is-the-realm-of-simply-beautiful-art
@TedShifrin a simple question The equation $ (z - z_0) = f_x(x_0,y_0)(x - x_0) + f_y(x_0,y_0)(y - y_0) $ with $p = (x_0,y_0,z_0) $ gives me a plane tangent to a function right ?
This question is related to one I asked here.
I am trying, overall, to show that $|A - \{x\}| =n-1$ using a bijection given that $A$ is a finite set with $|A|\geq 1$. For the case when $n > 1$, suppose that, since $|A|=n$, $\exists$ a bijection $f: A \to \mathbb{N}_{n}$.
With some help, I came...
to be fair I got interested in differential topology because the physics professor said stuff about differential forms, tangent spaces and jet bundles in a very handwavy way and I wanted to better understand what was going
You should do the exercise I commented on above, @Maks. I said you could write $z=f(x,y)$ as the level set $F(x,y,z)=f(x,y)-z = 0$. Work out the tangent plane equation from the latter and you'll see you get the same thing you had before.
Probably. Soug hadn't actually gone over that definition, but we had to prove that every infinite subset of a compact set had a limit point, and it amounted to establishing that equivalence
@Daminark: Ultimately the right definition is the finite subcover of open covering one. But it's important for concrete applications in Euclidean space to know closed + bounded.
In my course, I used closed + bounded as the definition, but then I had to prove sequential compactness (every sequence has a convergent subsequence), cuz we used that zillions of times to characterize compactness.
Ted sometimes in the in the correction i find the integral of x and y are 0 ( by symmetry ) when the region is circle, can you give me argument so that it stick to my head?
This question is related to one I asked here.
I am trying, overall, to show that $|A - \{x\}| =n-1$ using a bijection given that $A$ is a finite set with $|A|\geq 1$. For the case when $n > 1$, suppose that, since $|A|=n$, $\exists$ a bijection $f: A \to \mathbb{N}_{n}$.
With some help, I came...
