Let G be a finite subgroup of U(n), the unitary group acts on $\mathbb{C}^n$. If there is a unit vector $x$ in $\mathbb{C}^n$ such that g(x) is almost orthogonal to x, for all $g\in G$ except the identity, can we perturb x so that g(x) is exactly orthogonal to x, for all $g\in G$ except the identity? More precisely, can we find a very small number $\epsilon>0$, so that if there exist a unit vector $x$ and the inner product $|(g(x),x))|<\epsilon$ for all $g\in G$ \ {1}, then we can find another unit vector $y$, such that $(g(y),y)=0$ for all $g\in G$ \ {1}? Is it possible to further require that $||x-y||$ be small too? 



A Perturbation problem for U(n)
If gx is a bounded distance away from x (which in particular occurs when gx is nearly orthogonal to x), then g is a bounded distance away from the identity. Since U(n) is compact, this and the pigeonhole principle forces the group G to have bounded cardinality; in particular, the set of all such groups is compact (if one chooses closed conditions for properties such as "bounded distance away from origin") in the [Hausdorff distance](http://en.wikipedia.org/wiki/Hausdorff_distance) topology, as the limit of a sequence of finite groups with bounded cardinality in the Hausdorff metric is again a finite group with bounded cardinality. For any single group, the claim is true for some epsilon by continuity (and the compactness of the unit sphere), so the claim is true in general by compactness of the space of groups.


With a bit more effort one can extract an explicit value of epsilon by making the compactness arguments quantitative, though the bounds are likely to be somewhat poor.


(More generally, for studying finite subgroups of compact linear groups, a useful fact to know here is Jordan's lemma, which says that one can always find a bounded index subgroup of such a group which is abelian (the bound can depend on the ambient dimension of the linear group). Here, of course, much more is true, because we are able to exclude group elements from getting too close to the origin, but Jordan's lemma is useful in situations in which we do not have this luxury.)


