So I need to get back to understanding the homogeneous polynomials---my problem is much more basic than this in fact.
I just don't understand how global sections work.
So I have the tautological line bundle $\mathcal O(-1)\subset\Bbb P^n\times\Bbb C^{n+1}$, with trivializations over the standard open covering of $\Bbb P^n$ given by $\psi(\ell,z)=(\ell,z_i)$.
Hmm, had to set the paper width to 50 inches to make room for the tables of cells for type $A_6$. Probably means I have reached the limit of what I should include.
from $D^3$ to the unit tangen bundle of $S^2$ I have this map, $xt$ maps to the matrix which fixes $x$ and rotate the $1$ vector on $Tx(S^2)$ an angle $t\pi$
Well, the unit tangent bundle of $S^n$ is NOT $\Bbb {RP}^n$ for higher $n$. But your idea is good, don't fixate on a particular circle bundle. You just need it to be the total space of some circle bundle (not even with a circle base) (why?).
In general IIRC the unit tangent bundle of $S^n$ is $SO(n)/SO(n-2)$ I think.
" $y$ is an algebraic function of $x$, if it is a function that satisfies an algebraic equation of the form $P_o(x)y^n+...+P_{n-1}(x)y+P_n(x)=0$ where n is a positive integer and $P_o(x), P_1(x)$ etc. are polynomials in $x$" But wikipedia says: en.m.wikipedia.org/wiki/Algebraic_function and I'm not able to correlate and understand the two definitions. Can you please help?
@Danu It's the composition of the $\psi_j$ for $O(-1)$ with the map $U_j \times \Bbb C \to U_j \times \Bbb C$ which dualizes $\Bbb C$ in the latter factor, not?
@mercio Um, no. I build the dual bundle by taking the charts $U \to U \times V$, replacing the fibers $V$ to $\hom(V, \Bbb R)$, i.e. making new charts $U \to U \times \hom(V, \Bbb R)$, same on the first coordinate.
@TobiasKildetoft (Uh, OK, some background info: I've only just graduated high school.) I don't understand all that about polynomials and all given in my textbook. Can you explain in simple words?
@KaumudiHarikumar Hmm, actually I don't think I get what your book is trying to say either. Unless it is missing something, what it says just means that the map is given by some polynomial (the coefficients being polynomials does not change anything)
Did you figure out that you need to actually look at the trivialization of the line bundle and then see why the transition functions tell you that homogeneous linear polynomials transform exactly right?
If you still have to go through an isomorphism (which is provided by the trivialization), then you maybe still have some legwork to do that you really have a section
@Ted I've got some serious misconception going on that's preventing me from understanding the identification of linear homogeneous polynomials with sections of $\mathcal O(1)$ (the higher order case is analogous so once I get this it should be no problem).
What I don't understand is the following: I've got non-trivial transition functions $\psi_{jk}$, but when I restrict a globally defined hom. poly. on $\Bbb C^{n+1}$ to a ray $\ell$, I get one element of $\ell^*$, irrespective of whether I view $\ell$ as lying in the trivialization on $U_j$ or on $U_k$. How do I reconcile this with $s_j=\…
Read all of this keeping in mind that I thought hom poly's map into the trivialization
If you click the permalink you can also read my 2-3 follow-up messages which elaborate
yeah we talked about how $z_2^*$ was indeed a section of the dual bundle, but because it turned out it was so by definition and without chekcing things on the trivializations
I had regular fights with a mathematical friend in real life when I was studying algebraic geometry. I'd ask him something and he'd quickly tell me a proof in $\Bbb C$, but then I'd say I really wanted him to prove it for all algebraically closed fields $k$.
on odd dimensional sphere I can always find a non-vanishing tangent vectro field, normalize it. Also I have a unit normal vector field. Now consider a vector field on the odd diemnsional sphere gen by these two vector. Consider its unit bundle. Then by the same construction like $RP^3$ you can get a homeomorphism from $RP^n$ to here
So for instance that you can get a sphere from quotienting special orthogonal groups, while you get a projective space from quotienting (special?) unitary groups, stuff like that
You should understand that this is the most natural generalization of the proof for $\Bbb{RP}^3$. That is, this is exactly the unit tangent bundle $S^1 \to \Bbb{RP}^3 \to \Bbb{CP}^1$.
Yeah, it's a not so obvious trick. Admittedly I pondered on this a while ago for the case $n = 1$, in the context of trying to prove that the linking number of the circle fibers of the unit tangent bundle is 2.
(Which is obvious once you realize it's a quotient of the Hopf bundle by $\Bbb Z/2$ - the circle fibers in Hopf bundle has linking number 1, and after quotienting your circles "wrap twice around themselves" so you have linking number 2)
The total space of the Hopf fibration is $S^3$. Two generic fibers belong to $S^3$. So you have two embedded circles inside $S^3$. They form a link. You're asking for the linking number of the link.
Nothing so complicated.
In this case, it's $1$. The Hopf fibers form the Hopf link.
You can ask the same thing in the case of the unit tangent bundle, the total space in which case is $\Bbb{RP}^3$ and you have a link inside $\Bbb{RP}^3$.
which is to say, if you actually want to use all that fancy calculus to actually compute something, then coordinates are quite handy.
i mean, it's true that a lot of vector stuff can be done with geometry alone. but the advantage of vectors is that it packages the info in a very succinct and useful way.
I will say that a profusion of dummy indices gets annouying
the only ways I know to get around that are either to find a coordinate-free expression or to use something like diagrammatic notation (which has its own issues, to be sure)
but that just gets to the fact that if you've got a tensor of higher rank, there's a lot of ways to contract it with itself and other tensors.
Remind me; what was the reason behind anomalous expansion of water? More hydrogen bonds than what was supposed to be below 4 degrees (because of electronegativity of O and proximity of the atoms), hence bigger "apparent atoms", so less well-packing?
