Show that $G$ ( subgroup of $\mathrm{GL}(E)$) is finite.

Question

I came across with, I think, a difficult problem :

Let E a Hermitian space with a Hermitian norm $||\ ||$. We provide $\mathcal{L}(E)$ with the norm $|||\ \ |||$ subordinated to $||\ ||$.

Let $r \in ]0,2[$ and $G$ a subgroup of $\mathrm{GL}(E)$ such that $G \subset B_o (\mathrm{Id}_ E , r)$ (open ball) . Show that $G$ is finite.

I was able to show the following fact :

• if $\lambda \in \mathbb C$ is an eigenvalue of an element of $G$, then $|\lambda -1| \leq r$

Proof.

For $X$ a eigenvector associated with the eigenvalue $\lambda$ of $f\in G$, then $||f(x)-X||\leq N(f_{Id}) ||X||$.

• The eigenvalues of elements of $G$ are the roots of unity, and there are finitely many eigenvalues.

Proof.

First, The eigenvalues of $f$ are $\vert\ \vert = 1$

Otherwise, $f^n$ element of $G$ (group) and we have $N(f^n)>\vert a\vert ^n$ Thus with $N (f^n) - N(Id) \le N( f^n{_Id}$) we get $\vert \lambda \vert^n \le r + 1$ winch is impossible when $n$ is big enough.

Furthermore $f^-1$ is an element of $G$(group) and its eigenvalues ​​are the inverse of those of $f$.

Therefore we deduce that all the eigenvalues ​​of $f$ are of modulus $1$.

EDIT : I forgot the rest of the proof,(I am not sure it's correct)

Let $\lambda=\exp(i\theta)$ an eigenvalue of $f$, fixed.

If $\theta$ isn't of the form $\frac{2k \pi}n$ then all $\lambda^p$ with $p$ an integer will be dense on the points of the circle of radius 1 and we will find the eigenvalues of $f^p$ near $-1$ wich will contracdict the first inegality.

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How can I continue ?

Thank you in advance

NB: I made this bounty because lie theory it's beyon my reach.

Answer 1

Note that finite dimensionality is necessary for your claim to be true. Indeed, a counterexample can be constructed very simply in infinite dimension, as follows : take $E$ with a countable orthonormal basis $(e_k)_{k\geq 1}$, and take $G$ to be the subgroup of all endomorphisms $f$ satisfying $f(e_k)=\lambda_ke_k (k\geq 1)$, where $(\lambda_k)_{k\geq 1}$ is a sequence of complex numbers such that $\lambda_k \in \lbrace 1,e^{\frac{2\pi i}{3}},e^{\frac{4\pi i}{3}}\rbrace$ for every $k$, and $\lambda_k=1$ for all but finitely many $k$.

In the finite dimensional case, your claim follows from

Lemma. If $G$ is a subgroup of $GL(E)$ such that the set of traces ${\sf Tr}(G)=\lbrace {\sf tr}(g) | g\in G\rbrace$ is finite. Then $G$ is itself finite.

Proof of lemma. Let $g_1,g_2,\ldots ,g_r$ be a maximal linearly independent family in $G$ (thus $r\leq {\sf dim}(E)^2$). Consider the map $\phi : G \to {\sf Tr}(G)^r, g\mapsto ({\sf tr}(g^{*}g_1),{\sf tr}(g^{*}g_2),\ldots,{\sf tr}(g^{*}g_r))$. Then $\phi$ is injective (otherwise the family would not be maximal ; remember that ${\sf tr}(g^*g)=||g||^2$ is zero only when $g=0$), so $G$ must be finite.

In the sequel I show the finiteness property, about which you seemed unsure.

Let $U$ denote the set of the complex numbers with modulus one ; then $U$ is a multiplicative subgroup of ${\mathbb C}^*$. Denote by $\Lambda$ the set of all eigenvalues of elements of $G$, and let $L=\lbrace \theta \in {\mathbb R} \ | \ e^{2\pi i\theta} \in \Lambda \rbrace$. The OP already shows that $\Lambda \subseteq U$, and that

$$ |\lambda-1| \leq r \ (\forall\lambda \in \Lambda)\tag{1} $$

If $\lambda$ is an eigenvalue of some $g\in G$, then for any $j\in{\mathbb Z}$ $\lambda^j$ is an eigenvalue of $g^j$, so that

$$ j\theta \in L \ (\forall\theta \in L,\forall j\in{\mathbb Z})\tag{2} $$

By definition of $L$, we have

$$ k+\theta \in L \ (\forall\theta \in L,\forall k\in{\mathbb Z})\tag{3} $$

Let $\theta\in L$ and $\lambda=e^{2\pi i\theta} \in \Lambda$. Then $\theta$ cannot be too near to $\frac{1}{2}$, else we would contradict (1). So there is an absolute integer constant $N$ such that

$$ \bigg|\theta-\frac{1}{2}\bigg| > \frac{1}{2^N} (\forall\theta \in L) \tag{4} $$

If $L$ contains a $\theta\in ]0,\frac{2^{N-1}-1}{2^{2(N-1)}}]$, then denoting by $p$ the integer part of $\frac{2^{N-1}-1}{2^N\theta}$, we have $p\geq 2^{N-2}$ and also $\frac{2^{N-1}-1}{2^Np} \leq \theta \leq \frac{2^{N-1}-1}{2^N(p+1)}$, whence $\frac{2^{N-1}-1}{2^N} \leq (p+1)\theta \leq \frac{(2^{N-1}-1)p}{2^N(p+1)} \leq \frac{2^{N-1}+1}{2^N}$ so that $|(p+1)\theta-\frac{1}{2}| \leq \frac{1}{2^N}$, contradicting (4). So $L\cap ]0,\frac{2^{N-1}-1}{2^{2(N-1)}}]$ must be empty. As (2) implies $\theta \in L \Leftrightarrow -\theta \in L$, we also see that $L\cap [-\frac{2^{N-1}-1}{2^{2(N-1)}},0[$ must also be empty. To summarize,

$$ |\theta| > \frac{2^{N-1}-1}{2^{2(N-1)}} (\forall\theta \in L\setminus \lbrace 0 \rbrace) \tag{5} $$

Now, for $\theta\in{\mathbb Z}$, let $L_{\theta}={\mathbb Z}+\theta{\mathbb Z}$. Then $L_{\theta} \subseteq L$ by (2) and (3), $L_{\theta}$ is an additive subgroup of $\mathbb R$, and a very famous result (see for example here) states that if $\theta$ is irrational then $L_{\theta}$ is dense in $\mathbb R$. This would clearly contradict (5), so

$$ L \subseteq {\mathbb Q} \tag{6} $$

Let $\theta\in L\cap(0,1)$. We can write $\theta=\frac{a}{b}$ where $a,b$ are positive coprime integers. There are positive integers $u,v$ satisfying the Bezout relation $au-bv=1$. Then $\frac{1}{b}=u\frac{a}{b}-v \in L$ by (2) and (3). By (5) we then have $a \leq b \leq 2^{2(N-1)}$. So there are at most $2^{2(N-1)}$ possible values for $a$ and $b$, and hence

$$ L\cap (0,1) \ \text{is finite} \tag{7} $$

Combining (6) with (7), we have that $\Lambda$ is finite as wished.

Answer 2

There is probably also an elementary proof using Jordan decomposition (I would have to think about it), but I will use Lie theory instead. In particular, this proof will work only in the finite dimensional case.

First of all, since norm is a continuous function, it suffices to consider the case when $G$ is closed (otherwise, replace $G$ with its closure, it will have the same "norm" property as in your question). Note that Closure of a subgroup is again a subgroup. Thus, $G$ is a Lie subgroup of $GL(E)$ (every closed subgroup of a Lie group is a Lie subgroup). Since $G$ is closed and bounded, it is compact. Thus, $G$ has only finite number of connected components. Therefore, it suffices to show that the identity component of $G$ is trivial. (The $G$ would be finite.) I will assume, therefore, that $G$ is closed and connected; I will prove that it is trivial. Arguing by contradiction, assume that $G$ has positive dimension. Then, a maximal torus $T<G$ (a maximal connected abelian subgroup; it is unique up to conjugation) also has positive dimension. Therefore, it suffices to get a contradiction with the case when $G$ is abelian. The maximal torus $T$ splits as the product of circles $S^1\times ... \times S^1$. (In fact, if you like, you can assume that $G\cong U(1)$.) The action of $G$ on $E$ is completely reducible: $E$ splits as a direct sum of $G$-invariant 1-dimensional complex subspaces $L_k$. Since $G$ is nontrivial, its action on at least one of the lines $L=L_k$ is via the full unitary group $U(1)$ of rotations inside of $L$ (unitary with respect to the hermitian norm you are using). By our assumption, the norm of each $g\in U(1)$ (acting on $L$) is $\le r<2$. But this is absurd since $U(1)$ contains $-1$. QED