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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?

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3 Answers 3

up vote 9 down vote accepted

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 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.)

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This is a very nice answer. One comment: I was slightly confused by the way you alluded to Jordan's Lemma. For others like me, a precise statement is: for any positive integer $n$, there exists a positive integer $J(n)$ such that any finite subgroup of $\operatorname{GL}_n(\mathbb{C})$ has an abelian normal sub(sub)group of index at most $J(n)$. –  Pete L. Clark Mar 26 '11 at 20:45
    
That's an equivalent formulation (if one has a bounded index abelian subgroup, one also has a bounded index normal abelian subgroup.) I've reworded a little bit to emphasise that the bound does depend on the dimension n. –  Terry Tao Mar 26 '11 at 21:01
    
@Terry: sure, that's true. To put a finer point on what confused me (a little): in your previous version you said "finite subgroups of compact groups", so the word "linear" was missing (and also the dimension, as you say). Moreover you don't need to say "compact linear group", since every finite subgroup of a linear group is contained in a compact linear group. –  Pete L. Clark Mar 27 '11 at 7:57
    
I was trying to restate the question in the following way: For any finite subgroup G of U(n), define $\lambda_G=$$inf_{x\in\mathbb{C}^n}$$\sum_{g\neq1}$|(gx,x)|.So you argument shows that $inf$ {$\lambda_G\neq 0$} is non-zero, but this lower bound depends on n. Am I understanding correctly? –  Qingyun Mar 27 '11 at 19:43
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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 Kazhdan. In particular finite groups have them. Therefore, your question holds in a much more general context. Does it make sense?

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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 $|(gx,x)|<\frac1{|G|-1}$ for every $g$.

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Things may be more complicated than you thought. If the dimension is not 3, then a rotation may not has an axis (or more than one axis?). For example, consider $\mathbb{R}^6=R^2\oplus R^2\oplus R^2$, let $g\in G$ be a rotation of the form $g_1\oplus g_2\ oplus g_3$, where $g_i$ is a rotation of $R^2$ by angle $\theta_i$, the i-th root of unity. If $x=(x_1,\dots,x_6)$, then (gx,x) is a convex combination of $cos(\theta_i)$, which can be arbitrary small without being 0. –  Qingyun Mar 26 '11 at 19:04
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