BTW, if you've never learned/proved Descartes Law of Signs, I was very proud of myself for figuring out a proof years ago (it's in my algebra book) using basic calculus and induction.
One of my classmates did something quite funny: in German, the word for field (the algebraic structure) is "Körper" which translates to body, so when we had an exercise in our ANT class which involved perfect fields, he googled for "perfekter Körper" (perfect body) and was quite surprised when the search return nothing about algebra, but a bunch of girls in revealing cloths and stuff about fitness
Actually, @Balarka, I think it's a good question for any vector bundle. I don't know why the OP was doing the complexified tangent bundle in particular.
so for exmaple for n=4 it would be 3^2 + 4^2 + 12^2 + 84^2 = 5^2 + 12^2 + 84^2 + 13^2 + 84^2 = 85^2 whiich is easy, since you can apply the fact that a^2 + b^2 = c^2 for some integers
@TedShifrin Hmm, so maybe something like, take the distribution $E$ given by $dz = ydx$ in $\Bbb R^3$, pick the rank 1 subbundle of that bundle consisting of fibers parallel to the y-direction. Then take it's normal bundle inside $E$. Those are both rank 1, so are integrable, right?
@Balarka: So you're generalizing to $E$ rather than complexified tangent bundle? Also, you're sticking just to $\Bbb R$. Be careful. He's complexifying and looking at complex subbundles.
Last night dream does not have any weird maths. Just each countable ordinal up to some $\alpha > \omega$ adjoined with rational intervals $\Bbb{Q} \cap [0,1]$ to form a dense linear order e.g. $0,...,\frac{\omega}{3},...,\frac{\omega}{2},\omega$
for positive integer $n,m$ prove that there is some integer $q$ and sequence of $n$ positive integers $a_1, a_2, ..., a_n$ such that $a_1^m + a_2^m + \dots + a_n^m = q$
prove that for natural $m, n$, there exists a $q$ (sum) such that for every integer $a \leq n$ (length of sequence) there is a sequence of $a$ naturals who's $m$-th powers add up to $q$
@TedShifrin, The definition i have read is that p=>q means if p=>q has a truth value of 0 iff the p has a truth value of 1 and q has a truth value of 1. I am allowed to use properties of 0, properties of 1, absorption law, involution, idempotence law, etc. (the basic laws).
I know nothing about that, but the easiest example, @CaptainBohemian, is to look at the sphere $S^{2n-1}$ sitting inside $\Bbb R^{2n}=\Bbb C^n$. Then there's a smoothly varying $(n-1)$-dimensional complex subspace of the tangent bundle. That's a CR structure.
@TedShifrin With regards to the geometric tangent space question, I don't see how $(a, v) = v^ie_i|_a$ if $e_i|_a = e_i$ where $e_i$ is just the standard basis vectors of $\mathbb{R}^n$
@Perturbative: You're getting lost in notation. $(a,v)$ is just talking about a vector $v$ in the tangent space at $a$. And $v=\sum v^ie_i$ is the expansion of $v$ in terms of the standard basis.
@TedShifrin I have read similar things on a paper about twistor and gravitation. but that paper presumes a lot of prequisite so I still don't have a complete understanding of Cauchy-Riemann structure. It's like I have googled about the structure but didn't find anything litearure introducing CR structure from scretch.
@TedShifrin I think I talked about this before with Akiva, but here's an application of the holonomic approx. theorem. Suppose $A$ is the annulus of inner radius $1$ and outer radius $2$. $f_0(x, y) = x^2 + y^2$ and $f_1(x, y) = -x^2 - y^2$ be these functions on $A$. There is a homotopy $f_t$ joining these such that $\nabla f_t \neq 0$ for all time everywhere.
I mean we do have the Barn but sometimes people reserve it an it's just like, why tho? I guess by 9-10PM everyone's out of here so the whole building is open
You have to use topology or complex analysis, pretty much. The most algebraic one uses Galois theory and still the intermediate value theorem to guarantee an odd degree real polynomial has a real root.
