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Let $\mathfrak{S}$ be a bounded semigroup of complex $d\times d$ matrices. Then there exists an operator norm $\|\cdot \|$ on $\mathbb{C}^d$ such that all elements of $\mathfrak{S}$ have norm $\le 1$.

Notes:

Given a submultiplicative norm on $M_d(\mathbb{C})$ then the unit ball for this norm is a bounded semigroup. We conclude that there exist an operator norm below it.

There is a similar problem about $p$-adic valuations. The method readily transfers to the archimedean case. There we had only one conjugacy class of maximal bounded subgroups, while here we have one for each isometry class of complex $d$-space.

My solution is below. Thank you for your interest!

orangeskid
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1 Answers1

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Let $B$ be a closed ball in $\mathbb{C}^d$ centered at the origin. There exists $\epsilon > 0$ such that $\mathfrak{S} \cdot (\epsilon B) \subset B$. Now consider $$B_1 = \{ v \in \mathbb{C}^d \ | \ \mathfrak{S} \cdot v \subset B\}$$

Clearly $B_1$ is a convex closed body centered at the origin ( non-void interior). Moreover, since $\mathfrak{S}$ is a semigroup, it will invariate $B_1$. We are done.

Note: assume moreover that $\mathfrak{S}$ is a group. For every $s \in \mathfrak{S}$, $s$ and $s^{-1}$ invariate $B_1$, so $s B_1 = B_1$. Consider the interior John ellipsoid $E$ of $B_1$. We have $s E = E$ for all $s \in \mathfrak{S}$. Hence for some Hilbert metric on $\mathbb{C}^d$, $\mathfrak{S}$ will consist of unitary operators ( a standard thing).

orangeskid
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