Automorphism with no square root

Question

Related to Every normal operator on a separable Hilbert space has a square root that commutes with it

Does it exist an automorphism $f$ in a separable $\mathbb C$ Hilbert space, such that $f$ has no square root?

If so, a concrete example would be useful.

Certainly yes for a real Hilbert space, for example let $H=\Bbb R$ and define $T:H\to H$ by $Tx=-x$. — David C. Ullrich, Nov 08 '19 at 20:49
I should have mentioned over $\mathbb C$. I edited the question. Thanks David. — user384617, Nov 08 '19 at 21:12
The shift operators on $l^2$ do not have square roots: https://math.stackexchange.com/questions/485259/shift-operator-has-no-square-root The proofs there should work for $l^2$ equally well — daw, Nov 08 '19 at 21:33
The shift operator on $\ell_2(\Bbb Z)$ is an automorphism. Of course it has a square root, and in fact it's normal... — David C. Ullrich, Nov 09 '19 at 01:55

Martin Argerami · Accepted Answer · 2019-11-11T16:30:02.937

Yes, such operators exist. This was proven by Halmos, Lumer, and Schäffer, Proc. AMS, 4, 1 (1953), 142-149.

Concretely, given a domain $D\subset\mathbb C$ define $$ D^{1/2}=\{\lambda\in\mathbb C:\ \lambda^2\in D\}. $$ They proved that the multiplication operator $M_z\in B(L^2(D))$ given by $(M_zf)(z)=zf(z)$ has a square root if and only if $D^{1/2}$ is disconnected. This can be seen to be equivalent to $D$ surrounding (but not containing, obviously) the origin.

So, if for instance you take any disk that does not contain the origin, say $D=\{\lambda:\ |\lambda-2|<1\}$, then $M_z\in B(L^2(D))$ is invertible and has no square root.

Automorphism with no square root

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