@Jasper it's a neverending job, hundreds of pages, and all the stuff in there is interconnected. I mean I have a clear picture of all is in there together with the network I created.
@Ted: Suppose I look at a ball of radius $r$ at $0$ in $\Bbb C$. There exists germs $g$ at $0$ such that $g$ does not extend to holomorphic functions on all of $\Bbb{B}_r(0)$. I think this is the key observation.
Let's see if I can parse this mathematically. If $p^{-1}(\Bbb{B}_r(0))$ can be written as a disjoint union of neighborhoods above in $|\mathscr{O}(\Bbb C)|$, then there exists a neighborhood $U$ around $g \in \mathscr{O}_0$ such that $p(U) = \mathbb{B}_r(0)$. ($U$ is one of the slices)
@TedShifrin Ah, but by definition, since $U$ is a neighborhood of $g$ it consists of the germs of some holomorphic function $f$ on $p(U)$. Since germ of $f$ at $0$ is $g$, and $\mathbb{B}_r(0)$ is the homeomorphic image of $U$ by $p$, that gives that $f$ is a holomorphic extension of $g$ to $\mathbb{B}_r(0)$, contradicting our assumption about $g$.
(A germ which does not extend to a holomorphic function on the ball of radius $r$ at $0$ can be found by choosing a convergent power series of radius of convergence less than $r$)
@NV-US: I thought about that at the dentist. For induction to work, you need to know that the sum of two projections is again projection, and for that you need that their composition is trivial, I believe.
EiEj = EjEi for all j and i, now we can operate Ei on the equation sum (E(i))=I and add all the equations, and subtract from this each equation one by one to get EiEj =0 for all i and j
@Semiclassical so now I'm trying to find the force which the tree excerts onto the bullet, but this is equal to the force that the bullet exerts onto the tree. So it's just ma?
@Leaky probably a good idea, I remember being told by one of the guys doing algebraic topology talks in the REU that the only things we can really do are linear algebra and kinda combinatorics, so in other parts of math we're trying to turn problems into that
Good exercise for Leaky: If $\theta$ is an irrational multiple of $\pi$, what can we say about the subgroup generated by $e^{i\theta}$ in the unit circle?
@TedShifrin a grad student I talked with today tried to explain me representation theory in 5 minutes, what I got is mostly that linear algebra is easier than group theory so why not try to reduce the latter to the former?
@TedShifrin Hah. No, really, I have understood why linear algebra is crucial as I progressed (major credit goes to you for forcing me to learn multivariable calculus, of course :P).
Good exercise for Leaky: If $\theta$ is an irrational multiple of $\pi$, what can we say about the subgroup generated by $e^{i\theta}$ in the unit circle?
@TedShifrin Hmm, not sure I recall the vectors themselves being called cyclic in that context either. But it has been a long time since I did commutative algebra