A Perturbation problem for U(n) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T02:33:45Z http://mathoverflow.net/feeds/question/59643 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/59643/a-perturbation-problem-for-un A Perturbation problem for U(n) Qingyun 2011-03-26T08:54:59Z 2011-03-27T10:02:52Z <p>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 $|(g(x),x))|&lt;\epsilon$ 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? </p> http://mathoverflow.net/questions/59643/a-perturbation-problem-for-un/59654#59654 Answer by anton for A Perturbation problem for U(n) anton 2011-03-26T13:03:52Z 2011-03-27T10:02:52Z <p>2nd UPDATE: Forget the old solutions. Let's assume that the representation does not contain the trivial representation, then $\sum_{g\in G}gx=0$ for every $x$. Therefore, for a norm one vector $x$, $$1=|(x,x)|=|\sum_{g\ne 1}(gx,x)|\le\sum_{g\ne 1}|(gx,x)|$$ So it will never occur, that $|(gx,x)|&lt;\frac1{|G|-1}$ for every $g$.</p> http://mathoverflow.net/questions/59643/a-perturbation-problem-for-un/59664#59664 Answer by ght for A Perturbation problem for U(n) ght 2011-03-26T15:53:56Z 2011-03-26T22:52:10Z <p>I'm in a hurry but I think that your question can be deduced as a consequence of the fact that every compact groups has the property (T) of <a href="http://en.wikipedia.org/wiki/Kazhdan%27s_property_%28T%29" rel="nofollow">Kazhdan</a>. In particular finite groups have them. Therefore, your question holds in a much more general context. Does it make sense?</p> http://mathoverflow.net/questions/59643/a-perturbation-problem-for-un/59672#59672 Answer by Terry Tao for A Perturbation problem for U(n) Terry Tao 2011-03-26T17:58:27Z 2011-03-26T21:01:05Z <p>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 <a href="http://en.wikipedia.org/wiki/Hausdorff_distance" rel="nofollow">Hausdorff distance</a> 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.</p> <p>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.</p> <p>(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.)</p>