Pure states and minimal projections 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.
 A: 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.
A: 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
