Let $V$ be a vector space.


Suppose we are given some upper semi-continuous map $\pi:V\rightarrow \bigcup\_{k\le d} Gr(V,d)$, i.e, for any $x\in V$ we specify some subspace $\pi(x) \subseteq V$ of dimension no more than $d$, and these vary in a upper semi-continuous manner.


Must there exist a function $\eta$ from $V$ to the grassmannian $Gr(V,d)$ which is continuous/smooth, so that $\eta(x)$ always contains $\pi(x)$ as a subset?



Lifting Upper Semi-Continuous Functions To Grassmannians
No.


Take $V=\mathbb{R}^2$, $d=1$ and define $\pi$ by $\pi(v)=\text{span}(v)$ if $\|v\|=1$, $\pi(v)=\{0\}$ otherwise. Then $\pi$ is usc and a continuous $\eta$ doesn't exist, as the image of $S^1$ is a nontrivial cocycle in $\text{Gr}(V,1)$.


Note also that $\pi(v)=\text{span}(v)$ for all $v$ is an example of a lsc $\pi$ with no $\eta$.


However, if $\pi:V\to\text{Gr}(V,k)$ is actually continuous for some $k<d$, then such $\eta$ does exist. To see this it is enough to assume $d=k+1$. Then use the contractibility of $V$ to find a no trivial section to the vector bundle $E$ over $V$ given by $E\_v=\pi(v)^\perp$ and add it to $\pi$.


