Suppose you have a smooth path of invertible operators $t \rightarrow A(t) $ such that $A(0) = Id$. Is there a way of defining a natural square root $X(t)$ such that the following identity holds ? $$\forall t, X(t)^2 = A(t)$$
Some observations:
-In my problem I can't have any assumption on spectrum of $A(t)$ globaly. Because of continuity, we only know that, for small enough $\epsilon$, the spectrum of $A(t+\epsilon)A(t)^{-1}$ is in the circle $\sigma \in \mathbb{C}, |\sigma-1|<1$
-If $A(t)$ are just numbers or if they are matrices that commutes with each others we have such a result. For exemple the following system define a natural square root associated to the path.
$$ X(0) = 0, \partial_t X(t) = \frac{1}{2}\partial_t A(t)A(t)^{-1}X(t) $$
However I want to deal with the case in which there is, in general, no commutation relation at different times ( $A(t)A(t') \neq A(t')A(t)$). And in the general case, I think that the previous formula fails to define a square root (or at least I'm not able to prove it)
-The operators I deal with are in general self-adjoint/unitary or both. So if we need to assume that kind of hypothesis in order to obtain a result, this would not be a big deal for me.
-For me this relation seems to be linked with the existence of a natural logarithm associated to the same path $e^{Y_t}=A(t)$. Because if it exists, then $X_t= e^{\frac{1}{2}Y_t}$ would be a square root.
Assume that when $t\in (a,b)$,
i) $A(t)\in M_n(\mathbb{C})$ is $C^k,C^{\infty}$ or analytic.
ii) $A(t)$ is invertible and its eigenvalues are simple.
Then the eigenvalues and (unitary) eigenvectors of $A(t)$ can be chosen and numbered as functions with the same regularity; in particular $A(t)=P(t)diag(\lambda_i(t))P^{-1}(t)$ where the functions can be chosen regular.
Finally, $\sqrt{A(t)}=P(t)diag(\sqrt{\lambda_i(t)})P^{-1}(t)$ where the square roots are chosen s.t. the $(\sqrt{\lambda_i(t)})$'s are regular..