Mathematics Asked by Wei Xia on December 5, 2021
Is it possible that a smoothly varying finite dimensional vector subspaces tracing out infinite dimensional locus?
More precisely, Let $E$ be a vertor bundle on a (compact) smooth manifold $M$ and $T_t:Gamma(M,E)to Gamma(M,E)$ is a smooth family of linear operators, where $Gamma(M,E)$ is the space of sections. $T_t$ is smooth on $tin mathbb{R}$ means that if ${sigma_i}$ is a local frame then the coefficients of each $T_tsigma_i$ with respect to the frame ${sigma_i}$ is smooth in $t$.
Now if $Vsubset Gamma(M,E)$ is a linear subspace of finite dimension.
Is it possible that
$$
dim span_{mathbb{R}}left(bigcup_{tinmathbb{R}}{T_t(V)}right)=infty~?
$$
Is it possible that for any sufficiently small $varepsilon>0$,
$$
dim~ span_{mathbb{R}}left(bigcup_{|t|<varepsilon}{T_t(V)}right)=infty~?
$$
A typical example of this question is when $T_t=e^{-f(t)}de^{f(t)}$ ($f(t)$ is a smooth function on $M$ which depends smoothly on $t$), $E=bigwedge^bullet T_M^*$ is the exterior bundle and $V=mathcal{H}^bullet(M)$ is the space of harmonic forms.
Added:If the above possiblities are possible, I want to know whether $dimker(T_tmid_V)$ is upper semicontinuous on $t$ for small $t$? That is,
$$
dimker(T_0mid_V)geq dimker(T_tmid_V),quad |t|<varepsilon~.
$$
It's absolutely possible. Consider the trivial bundle $mathbb{R}timesmathbb{R}$ and the operator defined by $T_tf(x)=cos(tx)f(x)$. Let $V$ be the subspace of constant functions. Since ${cos(tx):tin[0,infty)}$ is linearly independent, the union $bigcup_{tin I}T_t(V)$ is infinite dimensional for any open $Isubsetmathbb{R}$
Answered by Kajelad on December 5, 2021
Get help from others!
Recent Answers
Recent Questions
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP