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.

  • 3
    $\begingroup$ I think my answer here (see "question 2" bit) shows how to do this? (That the answer quoted uses Hilbert spaces is irrelevant; you needed bounded, but that's automatic as you ask about isometries...) $\endgroup$ Jan 28 at 10:37
  • 1
    $\begingroup$ Thank you, that argument is indeed easy. Maybe you could just copy it as an answer below so that I can accept it and close the question? I believe it is worthwile to have this thread answered so that others can more easily find it. $\endgroup$ Jan 28 at 10:45
  • 1
    $\begingroup$ I vaguely recall that sometimes people pull the trick of taking the anti-diagonal (with the subspace topology) of $B(X)^2$ with the desired topology (so the strong, in your case). Inversion is then continuous as it is the restriction of the swap map. $\endgroup$ Jan 28 at 10:47
  • $\begingroup$ Aha, cross-posted. Never mind, then! $\endgroup$ Jan 28 at 10:51
  • $\begingroup$ @MatthiasLudewig I don't understand why you say that the (obvious) bounded cases should close the question? Bounded isn't needed, from what you say in the Hilbert case, so isn't the interesting question to which extent it generalizes to more general Banach spaces? $\endgroup$
    – YCor
    Jan 28 at 15:56

1 Answer 1


That the inverse is continuous for the strong topology would actually be true for any bounded subgroup of $GL(X)$, the invertible operators on $X$.

Firstly, as translation is continuous, it suffices to consider continuity at the identity.

Now let $(T_i)$ be a bounded net of invertibles converging strong to $I$, and with $(T_i^{-1})$ also bounded, say a common bound of $K$. Then, for $x\in X$, $$ \|T_i^{-1}(x) - x\| = \|T_i^{-1}(x - T_i(x))\| \leq K \|x-T_i(x)\| \rightarrow 0 $$ by assumption that $T_i\rightarrow I$ strongly. Thus also $T_i^{-1}\rightarrow I$ strongly.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.