Let $X$ be a Banach space and let $\mathrm{Iso}(X)$ be its group of isometries, i.e., the set of surjective linear maps $T: X \to X$ with $\|Tx\| = \|x\|$.

**Q: Is $\mathrm{Iso}(X)$ a topological group under the strong topology?**

While it is easy to show that multiplication is continuous, it is not clear to me how to show that inversion is continuous. I did not find a reference in the literature for this statement.

If $X = H$ is a separable Hilbert space, then $\mathrm{Iso(X)} = \mathrm{U}(H)$, the unitary group of $H$, and the statement that this is a topological group is well-established. However, the proof (that I know) uses the fact that the weak and the strong operator topology agree on $\mathrm{U}(H)$ and that the inverse is just given by $u \mapsto u^*$, which is continuous in the weak operator topology.

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