Let $\phi$ be a normal pure state on a w*-algebra $M$ and $\{\pi, \xi, H\}$ its GNS representation associated to $\phi$. Suppose projection $e$ is the support of $\phi$. Show that $\pi(e)$ is a rank one projection onto $\Bbb C \xi$.
My attempt; I suppose $\pi(e)$ is not rank one. There are projections $p,q\in M$ such that $\pi(e) = \pi(p)+\pi(q)$. (Because $\pi(M) = B(H)$, it's possible.) Clearly $\pi(M)\pi(p)\xi$ is a nonzero invariant subspace of $H$, which implies that $\pi(M)\pi(p)\xi = H$. Here I get stuck. I could not find any contradiction. Please regard me.
At first, let us review three points: (1) Since $\phi$ is pure then $s(\phi)$, the support of $\phi$ is a minimal projection in $M$. (2) Let $z_{\pi}$ be the support of $\pi$ (since $\ker\pi$ forms a $w^*$-closed 2-sided ideal then there is central projection $q\in M$ with $\ker\pi=Mq$. Support of $\pi$ is defined by $1-q$. Therefore the restriction of $\pi$ on $Mz_{\pi}$ forms an *-isomeric $w^*-w^*$ continuous isomorphism onto $\pi(M)$). (3) One may check directly that $s(\phi)\leq z_{\pi}$.
We come back now on your problem. Assume that there is projection $w=\pi(p)\in \pi(Mz_{\pi})$ with $\pi(p)\le\pi(s(\phi))$. Therefore $p\le s(\phi)$ which imlies that $p=s(\phi)$.