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Let $\mathbb H$ be a Hilbert space, and let $\mathcal B(\mathbb H)$ be the space of bounded operators on $\mathbb H$, equipped with the operator-norm topology. Let $\mathbb R\ni t\mapsto A(t)\in \mathcal B(\mathbb H)$ be a continuous mapping such that for all $t$, $A(t)$ is self-adjoint. Let $\mathcal{U}$ be the unique solution of the ODE $$ \dot{\mathcal{U}}(t)+iA(t) \mathcal U(t)=0, \quad \mathcal U(0)=I. \tag{1}$$ Then it is easy to prove that $$\forall t\in \mathbb R, \quad \mathcal U(t)^*\hskip1pt\mathcal U(t)=I, \tag{2}$$ and that for all $t\in \mathbb R$, the operator $\mathcal U(t)\mathcal U(t)^*$ is the orthogonal projection on $K(t)=\operatorname{range}{\mathcal U(t)}$, which is a closed subspace of $\mathbb H$.

Question. Is there a simple sufficient condition ensuring that for all $t\in \mathbb R$, $\mathcal U(t)$ is actually unitary, i.e. such that $K(t)=\mathbb H$ for all $t$?

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This is always true. Consider $P(t)=1-U(t)U(t)^*$. Then $T=\{t\in\mathbb R: P(t)=0\}$ is open and closed and $0\in T$. The set is open because if $t_0\in T$, then $\|P(t)\|<1$ for all $t$ sufficiently close to $t_0$, but a non-zero projection has norm $1$.

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  • $\begingroup$ Thank you very much. Is there an analogous statement with $A(t)$ unbounded self-adjoint. I guess that it is not so easy in that case to formulate the ODE (1). $\endgroup$
    – Bazin
    Commented Jan 16 at 15:57
  • $\begingroup$ @Bazin: I think something along these lines should still be true. This fact is also the key ingredient in the usual proof that operators evolved by the Toda flow are unitarily equivalent, though I didn't remember this initially: the Lax equation is $\dot{T} = [P,T]$, with $P^*=-P$. To see that $T(t), T(0)$ are unitarily equivalent, one solves $\dot{U}=PU$, $U(0)=1$, and verifies that $U(t)T(0)U^*(t)$ also solves the Lax equation, so equals $T(t)$. Of course, this would be pointless unless $U(t)$ is guaranteed to be unitary. (cont'd) $\endgroup$
    – Christian Remling
    Commented Jan 16 at 18:20
  • $\begingroup$ In this situation, $P$ is bounded, but all these statements are also true, with analogous proofs usually, for other hierarchies such as KdV, where now $P,T$ will be unbounded. $\endgroup$
    – Christian Remling
    Commented Jan 16 at 18:20

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