While working on Heisenberg-Weyl operators (specifically working on their eigenvalues), I have come across the following problem:
Given an $n$, $n$th roots of some complex number $p, |p| = 1$ sum to zero. Let $A$ be the set that contains all these $n$ roots. My intuition was if we form a subset of $A$ whose elements also sum to zero, then this subset is necessarily a complete set of $m$th roots of some $p'$, where $m$ is a factor of $n$. However, this intuition turned out to be wrong
see: https://math.stackexchange.com/questions/490115/summation-of-roots-of-unity-equal-to-zero?rq=1
Let $E$ be a vector bundle of rank $k$ over a smooth manifold $M$ of dimension $m$. Then $E$ can be thought of as a $(m+k)$-dimensional manifold, so the (weak) Whitney immersion theorem states that $E$ can be immersed in $\mathbb{R}^{2(m+k)}$.
Is it possible to immerse $E$ in $\mathbb{R}^{...
I could prove the result for $|\Lambda|$ finite. Here $|\Lambda|$ is arbitrary.
My attempt:-
Let $\langle x_{\alpha}\rangle_{\alpha\in \Lambda}\in B \implies \langle x_{\alpha}\rangle_{\alpha\in \Lambda}\in \pi_{\beta_i}^{-1}(U_{\beta_i}), \forall i=1,2,3,...,n. $ So, $\pi_{\beta_i}(\langle x...
Let $u$ be a integer which has an odd prime $m$ as a divisor. So we let $u=u_1m$ where $u_1=\frac{u}{m}$.
Consider the set of $u$th roots of unity, i.e., $A=\{e^{j\frac{2\pi}{u}n} | 0 \leq n \leq u-1\}=\{e^{j\frac{2\pi}{u_1m}n} | 0 \leq n \leq u_1m-1\}$.
For convenience, let $e^{j\frac{2\pi}...
