I'm working on a dynamical systems problem and have arrived at a naive question in differential topology? geometric measure theory? 


I have a smooth path $\gamma\colon \mathbb R\to\mathcal G\_2(\mathbb R^d)$ (that is $\gamma(t)$ is a 2-dimensional subspace of $\mathbb R^d$ for each $t$). I assume that $\mathbb R^d$ is equipped with the standard inner product. Now, let $V(t)$ be the collection of vectors in the hypersphere such that for $v\in V(t)$, $v$ is perpendicular to all vectors in $\gamma(t)$. Of course, $V(t)$ is the intersection of the unit sphere with a subspace of $\mathbb R^d$ of dimension $d-2$, so is a set of dimension $d-3$.


I'd like to show that the union of $V(t)$ is of Hausdorff dimension at most $d-2$. 
I already have a kludgy differential equations proof of this, but would like to find a nicer way to do this.



> 
> Can anyone suggest a suitable framework for a nice proof of this kind of result?
> 



dimension of a union of grassmannians
Denote by $S^{d-1}$ the unit sphere in $\newcommand{\bR}{\mathbb{R}}$ $\bR^d$ and consider the manifold


$$ X= \bigl\lbrace (v,t)\in S^{d-1}\times \bR;\;\; v\perp \gamma(t)\;\bigr\rbrace. $$


The natural map $X\to\bR$ defines a fiber bundle with fiber $S^{d-3}$. The fiber over $t$ is the the $(d-3)$-dimensional sphere consisting of unit vectors orthogonal to $\gamma(t)$. In particular $\dim X=d-2$.


The projection $\pi: X\to S^{d-1}$ is a **smooth** map. According to the Morse-Sard-Federer the image of the map $\pi$ has Hausdorff dimension at most $\dim X=d-2$. (For more details about the Morse-Sard-Federer theorem see Theorem 3.4.3 in Federer's book *Geometric Measure Theory*.)


