So $a$ is quite naturally a map $\Omega^0(E) \to \Omega^1(E)$ - locally, since $a$ is just a $\mathfrak g$-matrix of 1-forms, apply the matrix to the section to get an $E$-valued 1-form. But the obvious way to extend this to a map $\Omega^1(E) \to \Omega^2(E)$ - $1 \otimes a$ (considering these as $T^*M \otimes E \to$ a subthing of $T^*M \otimes T^*M \otimes E$) - is wrong.
@Laters no, the relevant section is 2.3. the one you're thinking of will, hypothetically, consist of a calculation of the maximum # of linearly independent sections on $S^n$ for any $n$, much more general.
@Ramanewbie It doesn't matter. The mathSE webs question site is not a "do my question right now" site. You ask your question, and maybe someone will answer when he wants.
It can take days for some questions to be answered
You can find a proof in Lawson's 'spin geometry' using Clifford algebras that the maximum number is achieved, but not a proof that it is indeed the maximum number.
ie he writes down the right number of independent vector fields, but doesn't prove there aren't more
@TedShifrin: For some reason it hadn't occured to me that I wanted $(a\omega)(x,y) = a(x)\omega(y) - a(y)\omega(x)$... I don't really see why I didn't think of that, but oh well.
@Ramanewbie 1) don't ping like that 2) We'd like it if you could work by yourself a bit before asking questions here. It won't help you at all if we just give you answers.
@Hippalectryon To tell you the truth, I didn't ever imagined there is so much sh*t in the world of mathematics. After I publish my book I'm out of mathematics.
@DanielFischer There is the following proposition in my lecture notes.
Let $X \neq \varnothing$ and $X$ finite set. The following are equivalent: - $X$ is countable. - there is a $f: X \overset{1-1}{\rightarrow} \omega$ - there is a $g: \omega \overset{\text{surjective}}{\rightarrow}X$
Couldn't we pick $f=g$ knowing that a set is countable if it is equinumerous with the set of natural numbers $\omega$, i.e. if there is a $f: A \overset{\text{bijective}}{\rightarrow} \omega$?
@Ted: Here's how I interpreted that comment when you first said it. I kept trying to do this: if $a = \eta \otimes g$, write our $E$-valued 1-form $\omega \otimes \sigma$. Then I wanted $a(\omega \otimes \sigma) = (\eta \wedge \omega) \otimes g\sigma$. This is wrong.
How do you guys deal with reading and reviewing your proofs/answers? I always find myself going mad when I try to do that. I even sometimes hand in tests without going over them too seriously because it just kills me to read what I wrote all over again.
@Hippalectryon Cet exercice je n'arrive toujours pas à trouver la bijection : Montrer qu’un morphisme u : E → F permet de définir une bijection $S \mapsto u (S)$ entre l’ensemble des sev de E contenant ker u et l’ensemble des sev de Im u. Quelle est la bijection réciproque?
@evinda $f$ and $g$ go in different directions, so you can't choose $f = g$ (unless $X = \omega$, then you can take $f = g$ if $f$ [$g$] is bijective).
I tell my students that they should write so that they could hand their paper to an average classmate and that average classmate should learn how to do the problem by reading it.
@Gato Ah, je ne suis pas sur alors. En dimension finie ça a l'air assez direct, mais avec les nouveaux programmes de prépa on fait peu de dimension infinie
@DanielFischer Let $Y$ be a countable set. That means that there is a $h: Y \overset{\text{bijective}}{\rightarrow} \omega$. Which functions do we have to compose?
let's admit in a frame 3 points A, B, C and I such that I is the middle of [BC]. I don't understand why the directive coefficient of the median from A in the triangle ABC is given by $m=\dfrac{yi-ya}{xi-xa}$.
@DanielFischer very little effort: this gives already a good point, for now I only now the definition of holomorphic function. Does I need to 'find' my old course to calmly begin complex analysis?
@Gato You need a bit of real analysis, and if you have a clue about (general) topology [limits, compactness, connectedness, such things], that's good too. But the heavy machinery only enters late in the game.
@DanielFischer In our case $f: X \overset{1-1}{\rightarrow} \omega, g: \omega \overset{\text{surjective}}{\rightarrow} X$ and $h: Y \overset{\text{bijective}}{\rightarrow} \omega$. So: $h^{-1}: \omega \overset{\text{bijective}}{\rightarrow} Y$.
@Ramanewbie "There's a problem with your points" are you saying that my points do not exist... ? or just that your formula does not hold for all points ?
@TedShifrin je m'appele Isabelle is about all I remember from my middle school French :P but I can read a lot of french and spanish since they're very similar to italian
@Pedro: He also said $U$ was open and $U\ne S^1$. Oy. I wrote out such stuff extremely carefully in my diff geo notes (not assuming any knowledge of complex variables or covering spaces or ...)
@DanielFischer A ok... And how do we conclude that a bijection $\beta \colon X \to Y$ exists, although we don't have the same function from X to Y as from Y to X?
I am lost now lol, l’image direct d’un sous-espace de E est un sous-espace de F and here I need that the image direct is included in Im(u) or I want the inverse of $S\mapsto u(S)$?
How do we conclude from $\{ +n: n \in \omega \} \subset \mathbb{Z}$ that $\mathbb{Z}$ is finite? Is it like that? Because $\{ +n: n \in \omega \}$ is isomorhic to $\omega$, it has the same properties, and thus it is finite. A finite set is $\subset \mathbb{Z} \Rightarro \mathbb{Z}$ has to be finite.
@DanielFischer We haven't proven the subtraction between natural numbers. So in order to show that the set $\mathbb{Z}$ is countable would it be better to use an other function?
We also define $\mathbb{Z}$ as follows: $$\mathbb{Z}=\{ +n: n \in \omega\} \cup \{ -n: n \in \omega \}$$ That's what I thought that the equivalence class is a natural number @DanielFischer
