# Unitaries vs isometries and co-isometries in $C^*$-algebras

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?

• Perhaps $u$ is invertible ? Jun 20, 2022 at 14:06
• 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 $\alpha$)? Jun 20, 2022 at 14:27
• 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. Jun 20, 2022 at 14:40
• Ah, I did omit one important datum, perhaps. $u$ is actually a map defined on the Hilbert space completion of $A/N_\tau$. Jun 20, 2022 at 14:52
• @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'. Jun 20, 2022 at 15:02

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.