@KevinDriscoll OK, the answer is that there isn't any nontrivial rank > 1 bundle on the circle. But I want to prove all these things
That sums it up for the warm-up for my story. Here comes the actual story
Say you have a rank $k$ vector bundle $E$ on $S^n$. You can write $S^n$ as a union of the upper and lower hemidisks $D_U^n$ and $D_L^n$ intersecting at the equator $S^{n-1}$
Consider the restrictions of $E$ to $D_U^n$ and $D_L^n$ each. Both these restrictions are isomorphic to trivial bundles, because the disk is contractible (there is no nontrivial bundle on $\Bbb R^n$ - it's worthy of a proof, so exercise).
So whatever nontriviality is happening is happening at the equator $S^{n-1}$. You follow me?
If I let G act on its three 2-sylow subgroups by conjugation, then there exists a homomorphism $\phi:G \rightarrow S_3$. We know the kernel of this homomorphism is a proper, non-trivial subgroup because..... why?
At each point $p$ of $S^{n-1}$ there's two things: one, the fiber of $E|_{D_U^n} \cong D_U^n \times \Bbb R^k$ at $p$, other, the fiber of $E|_{D_L^n} \cong D_L^n \times \Bbb R^k$ at $p$ (because both the upper/lower hemidisks contain that point $p$). And, these two fibers of the trivializations of $E$ on the upper/lower hemidisk are glued togather in $E$
Being ejected from your mother is like the least exciting thing you did that day. You started bretahing air for the first time. Some hole flap thing in your heart closed. Lots of excitement
If I'm looking at a group G of order 24 and if I let G act on its three 2-sylow subgroups by conjugation, then there exists a homomorphism ϕ:G→S3. We know the kernel of this homomorphism is a proper, non-trivial subgroup because..... why?
Question (Relations): Can someone tell why in (x,y) \in R iff x \neq y is not transitive? I mean I can think of (1,2) (2,3) (1,3) being transitive and all 3 are different numbers. I feel like I am not understanding something basic here. :*(
@KevinDriscoll In other words, the general recipe to form a vector bundle on $S^n$ is to take two trivial bundles on two individual $n$-disks, glue the disks along their boundary $S^{n-1}$ and glue the fibers according to the map $\varphi : S^{n-1} \to GL_k(\Bbb R)$, where value $\varphi(p)$ of that map at $p$ is precisely the gluing isomorphism of the fibers at $p$
@KevinDriscoll Punchline: Rank $k$ vector bundles on $S^n$ are in 1-1 correspondence with homotopy classes of maps $S^{n-1} \to GL_k(\Bbb R)$. I.e., they are classified by $\pi_{n-1}(GL_k(\Bbb R))$
@hisoka Transitive hear would mean in $(x,y) \in R$ and $(y, z) \in R$ then $(x, z) in R$ but thats clearly wrongsince we could have, say $x=1 \ y=5 \ z=1$
Notice that if $n = 1$, $k = 1$, i.e., we're classifying line bundles on $S^1$, then the group is $\pi_0(GL_1(\Bbb R)) = \pi_0(\Bbb R^\times) \cong \Bbb Z/2$ because $\Bbb R^\times$ has two connected components. That means there are only 2 line bundles on $S^1$
[Random] 3. A natural is finite if it and its predecessors bijects with a standard natural. any natural other than 0 that does not have a predessessor, and its successors are not finite
@BalarkaSen So things with out bundle classification get a lot more complicated if the fibers arent contractible, right? Like, for example, if we have a $U(1)$ bundle
@LeakyNun You do have an isometry of $\Bbb R^3$ that takes the identity embedding to the antipodal embedding (the antipodal map $(x, y, z) \mapsto (-x, -y, -z)$). But this isometry isn't orientation-preserving.
I just ask because in physics we can construct the simple example where you have a continuous symmetry and the system is a so-called integrable system. And then the bundle structure is very simple; itll be $X \times G$ for some $G$ a lie group, just the trivial bundle. But I dont know of any example where the structure is known and its NOT the trivial bundle
@LeakyNun Oh, evert the sphere in some large dimensional Euclidean space? That should be boring and trivial, because $S^1$ can be everted if you embed it in $\Bbb R^3$ :P
You can prove using similar techniques as Whitney embedding theorem that for sufficiently large $N$ you can embed $E$ inside $X \times \Bbb R^N$
That is, you can get a map $E \to X \times \Bbb R^N$ which sends each fiber $\Bbb R^k$ injectively to $\Bbb R^N$
Once you do that, you can get the following map. To each point $p \in X$ associate to it the subspace $\Bbb R^k \subset \Bbb R^N$ that embedding provides you with
This gives a map $X \to \text{Gr}(k, N)$ to the Grassmannian of $k$-subspaces in $\Bbb R^N$ (do you know what that is?)
I barely know what it is. We had 1 homework problem on it and it took me like 6 hours just to really 'git it' and understand what the charts are and such. That was 2 months ago, so Im sure Ive forgotten the details, but refreshing is usually easier than the first time
But yes, the upshot of this is that rank $k$ v.b.'s on $X$ are classified by maps $X \to \text{Gr}(k, \infty)$
And this greatly generalizes, in the sense that if you want to classify $G$-bundles on $X$, the corresponding classifying map is $X \to BG$ where "$BG$" is a mystical object known as classifying space of $G$
If you take a diffeomorphism of $S^1$ it's derivative is a map between the tangent bundles, which takes the tangent field to somewhere. It's not a natural idea to "map a tangent field to another"
because in that sense, mapping the tangent field of S^1 where everything points outward, to the same thing where everything points inward, requires 3 dimensions
Hello, I'm having some difficulty getting a set of solutions for a system of differential equations. I've attempted to use wolframalpha to get a solution so that I can see what I'm looking towards. Unfortunately, the solution it provided only confused me more. Could someone help?
In gibert stang's linear algebra book, as a challenging exercise in the first chapter, it's given to prove if $(a,b)$ is a multiple of $(c,d)$, then prove $(a,c)$ is a multiple of $(b,d)$ . but isn't it trivially obvious unless I'm missing something ?
A while back I saw posted on someone's office door a statement attributed to some famous person, saying that it is an instance of the callousness of youth to think that a theorem is trivial because its proof is trivial.
I don't remember who said that, and the person whose door it was posted on d...
btw, balarka, what is the solution to the paper cube puzzle. You mentioned one of the crease has to be battered enough. Does that mean one of the squares end up rolling up because of the battered crease?
@BalarkaSen I however cannot seemed to fold one of the cubes diagonally about the common line to close it. All the crease become too rigid to do any shear folding...
