Pure states and minimal projections

Question

Probably it is trivial or already discussed, but I just wanted to pose a simple question about pure states.

${\bf Lemma \; 1 \colon}$ Let $\mathcal{A}$ be a $C^*$-algebra acting irreducibly on some Hilbert space $\mathcal{H}$ and let $\varphi(x)=(\xi | x \xi)$ be some vector state. If $\xi$ is cyclic, then $\varphi$ is pure.

${\bf Proof \colon}$ It suffices to notice that by $\overline{\mathcal{A} \xi} = \mathcal{H}$ we have that the identity representation is equivalent to the GNS representation $(\pi_\varphi, \mathcal{H}_\varphi, \xi_\varphi)$. But such a representation is irreducible if and only if $\varphi$ is pure and the assertion follows.

A related lemma is the following.

${\bf Lemma \; 2 \colon}$ Let $\mathcal{M}$ be a von Neumann algebra and let $\varphi$ be some faithful normal state in $\mathcal{M}_*^+$. If $\varphi$ is pure then its support $s(\varphi)$ is minimal.

${\bf Proof \colon}$ We can identify $\mathcal{M}$ with its GNS representation associated to $\varphi$, so that $\varphi(x) = (\xi | x \xi)$. By the identity $s(\varphi) = [\mathcal{M}' \xi]$ we have that $s(\varphi)$ is a rank one projection, hence it is minimal.

${\bf Question \colon}$ Is it true that a normal state is pure if and only if its support is minimal?

Clearly it could be useful that a state $\omega$ on a $C^*$-algebra $\mathcal{A}$ is pure if and only if it is the only state vanishing on its left kernel

$$ \mathcal{L}_\omega = \{ x \in \mathcal{A} \colon \omega(x^*x)=0 \} \,.$$

I think that the decomposition of a linear functional $\varphi \in \mathcal{M}^*$ in normal and singular part could be useful too.

Any hint? Thank you in advance.

Answer 1

The result is true, but in a rather trivial fashion. Namely, if a von Neumann algebra has a pure normal state, then the state is supported on a type I central summand. This is simply because minimal projections can only exist in type I central summands.

Let $\varphi:M\to\mathbb C$ be a pure normal state. Because it is pure, its GNS representation $\pi_\varphi$ is irreducible. The normality of $\varphi$ implies that $\pi_\varphi$ is normal; so its image is sot-closed. Thus $\pi_\varphi(M)=B(H_\varphi)$ is type I. As $\pi_\varphi$ is faithful on $pMp$, we get that $pMp$ is isomorphic to $\pi_\varphi(M)$ and so type I. The state $\varphi$ is pure on $pMp$, so it has to be supported on a minimal subprojection of $p$; but $p$ was the support of $\varphi$. It follows that $pMp=\mathbb C p$, so $p$ is minimal.

The takeout is that pure states are largely irrelevant to von Neumann algebra, because a pure state on a von Neumann algebra with no type I summand cannot be normal.

Answer 2

The conjecture about $C^*$-algebras is true, that is, a state $\omega $ on a $C^*$-algebra $A$ is pure iff $\omega$ is the only state vanishing on $\mathcal{L}_\omega $.

Before proving this, let us observe that a state $\varphi $ vanishes on $\mathcal{L}_\omega $ iff $$ \omega (x^*x)=0 \ \Rightarrow \ \varphi (x^*x)=0. \tag 1 $$ The "if" part follows easily from the Cauchy-Schwartz inequality, while the "only if" part is a consequence of the fact that $$ x\in \mathcal{L}_\omega \ \Rightarrow \ x^*x\in \mathcal{L}_\omega . $$

Summarizing, we will prove:

Theorem. Let $\omega $ be a state on the $C^*$-algebra $A$. Then $\omega $ is pure iff, for every state $\varphi $ satisfying (1), one has that $\varphi =\omega $.

Proof. Assume that $\omega $ is not pure. Then there are states $\varphi $ and $\psi $, both distinct from $\omega $, and a real number $a\in (0,1)$, such that $ \omega =a\varphi +(1-a)\psi . $ Given any $x$ in $A$ such that $\omega (x^*x)=0$, we then have that $$ 0\leq a\varphi (x^*x) \leq a\varphi (x^*x)+(1-a)\psi (x^*x) = \omega (x^*x)=0, $$ so also $\varphi (x^*x)=0$, and we see that $\varphi $ satisfies (1).

To prove the converse, assume that $\omega $ is pure and that the state $\varphi $ satisfies (1).

Denoting the GNS representations associated to $\omega $ and $\varphi $ by $(\pi ,H,\xi )$ and $(\rho ,K,\eta )$, respectively, consider the mapping $$ V:\pi (A)\xi ⊆H \to K $$ given by $$ V\big (\pi (a)\xi \big )=\rho (a)\eta ,\quad\forall a\in A. $$ This is well defined because if $\pi (a)\xi =0$, then $$ 0=\langle \pi (a)\xi ,\pi (a)\xi \rangle = \langle \pi (a^*a)\xi ,\xi \rangle = \omega (a^*a), $$ so $\varphi (a^*a)=0$ by hypothesis and hence $\rho (a)\eta =0$ by a similar computation.

Observing that $\pi $ is irreducible, we have by Kadison's transitivity Theorem that in fact $\pi (A)\xi =H$, so $V$ is already globally defined on $H$.

We next claim that $V$ is bounded. To prove this let us first consider the mapping $ Q:A\to H, $ given by $$ Q(a) = \pi (a)\xi ,\quad\forall a\in A. $$ As already observed, $Q$ is onto $H$, hence an open mapping thanks to the Open Mapping Theorem. This says that there exists a constant $c>0$ such that, for every $\zeta $ in $H$, there exists $a$ in $A$ with $\|a\|\leq c\|\zeta \|$, and $Q(a)=\zeta $. We then deduce that $$ \|V(\zeta )\| = \|V\big (\pi (a)\xi \big )\| = \|\rho (a)\eta \| \leq \|a\| \leq c\|\zeta \|, $$ so $V$ is indeed bounded, as claimed.

It is a simple matter to prove that $$ V\pi (a) = \rho (a)V, \quad \forall a\in A, $$ from where it follows that $V^*V$ lies in the commutant of $\pi (A)$. Again because $\pi $ is irreducible, that commutant must coincide with $\mathbb C$, so $V^*V$ is necessarily a scalar, whence $V$ is a scalar multipe of an isometric operator. Observing that $V$ maps the unit vector $\xi $ to the unit vector $\eta $, we see that $V$ must itself be an isometry.

For every $a$ in $A$ we then have that $$ \omega (a) = \langle \pi (a)\xi ,\xi \rangle = \langle V\pi (a)\xi ,V\xi \rangle = \langle \rho (a)\eta ,\eta \rangle = \varphi (a), $$ proving that indeed $\varphi = \omega $. QED