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In a $*$-algebra $A$, a map $u : A → A$ is an isometry on $A$ if $u^*u = 𝟙$, and a co-isometry if $uu^* = 𝟙$. It is unitary if it is both.

I have to prove that a certain map is unitary, which (in the solutions manual) is done by showing that it is isometric ($\lVert u(x) \rVert^2 = \lVert x \rVert^2 $). I still don't see how that suffices.

The map in question is $u(a + N_\tau) := \alpha(a)+ N_\tau$ for some automorphism $\alpha$ of $A$ and $N_\tau := \{a ∈ A \mid \tau(a^*a) = 0 \}$ for some positive linear functional $\tau$. The objects satisfy $\tau(\alpha(a)) = \tau(a)$ for all $a ∈ A$.
Again, the claim is that since $\lVert u(a+N_\tau) \rVert² = \lVert a+N_\tau\rVert²$, $u$ is unitary, while I can only see that it is an isometry, not also a co-isometry..

(For reference, I am doing Exercise 3.2(c) from Murphy's book $C^*$-Algebras and Operator Theory.)

EDIT
In principle, we don’t know what $u^*$ does, but it will have to, for some $b ∈ A$, map it to $b + N_\tau$. If $u$ is to be unitary (and thus, in particular, a co-isometry), we ought to have $$a+N_\tau = uu^*(a+N_\tau)= u(b+N_\tau) = \alpha(b)+N_\tau,$$ which is true if, but not only if, $b = \alpha^{-1}(a)$. Is this in fact true by necessity? And if so, could we not have applied an analogous argument to $u^*u$? Not having to bother computing the norms?

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    $\begingroup$ Perhaps $u$ is invertible ? $\endgroup$ Jun 20, 2022 at 14:06
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    $\begingroup$ There are too many ingredients missing here. Without the setting in the mentioned book it is hard to start. Is there any avatar of an $\alpha^*$ (for the semi-product introduced by the trace / state $\tau$) and is there a chance that $u^*$ is implemented by $\alpha^*$ (as $u$ is implemented by $\alpha$)? $\endgroup$
    – dan_fulea
    Jun 20, 2022 at 14:27
  • $\begingroup$ I don't think a lot is missing, actually. Also, I don't know what a semi-product is, nor do I think $\tau$ is a 'trace'..? It is not given that it is a state, at least. Moreover, I don't see what you mean by 'implemented', and what I would also like to know is whether we know what $u^*$ is, given what $u$ is. $\endgroup$ Jun 20, 2022 at 14:40
  • $\begingroup$ Ah, I did omit one important datum, perhaps. $u$ is actually a map defined on the Hilbert space completion of $A/N_\tau$. $\endgroup$ Jun 20, 2022 at 14:52
  • $\begingroup$ @dan_fulea sorry, I meant that I didn't know what you meant by 'avatar'. I guess (although I haven't heard it used before) I know what you mean by 'implemented'. $\endgroup$ Jun 20, 2022 at 15:02

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I think the exercise wants you to show that automorphisms $\alpha$ that leave a state $\tau$ invariant induce unitaries on the GNS space $H_\tau$. The unitary $u$ is defined as $u\pi_\tau(a)\Omega_\tau =\pi_\tau(\alpha(a))\Omega_\tau$. This is isometric because of the invariance. From the definition it is clear that $u$ is invertible with the inverse given by the unitary induced by $\alpha^{-1}$ and the inverse $u^{-1}$ is also isometric.

To prove unitarity, you could now simply show that $u^*= u^{-1}$: \begin{align} (u^{-1}\pi_\tau(a)\Omega_\tau,\pi_\tau(b)\Omega_\tau) &=(\pi_\tau(\alpha^{-1}(a))\Omega_\tau,\pi_\tau(b)\Omega_\tau)\\ &=\tau(b^*\alpha^{-1}(a))\\ &=\tau(\alpha(b)^*a)\\ &= (\pi_\tau(a)\Omega_\tau,u\pi_\tau(b)\Omega_\tau) =(u^*\pi_\tau(a)\Omega_\tau,\pi_\tau(b)\Omega_\tau) \end{align} for all $a,b\in A$. This suffices because $\text{span}{\pi_\tau(A)\Omega_\tau}$ is dense in $H_\tau$.

More elegantly, it follows from $u$ and $u^{-1}$ being isometric that you get unitarity (but you do definetly need the isometricity of both $u$ and its inverse!): $$ (ux,y) = (u^{-1}ux,u^{-1}y) = (x,u^{-1}y) \quad \forall{x,y\in H_\tau}. $$ This argument uses the polarization identity which implies that isometries with respect to the norm are also isometries w.r.t. the inner product.

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  • $\begingroup$ Ahhh, that is a nice takeaway! To prove unitarity, instead of checking isometry and co-isometry, one could just prove isometry of the map and its inverse! $\endgroup$ Jun 20, 2022 at 22:21

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