Diameters of state spaces

Question

The authors mention that if $M$ is not a factor, the diameter of $S_0(M)/Int(M)$ is equal to 2.

When $M$ is not a factor, it is easy to check that the $d(,)\leq 2$, how to check that $d(,)\geq2$?

Answer 1

The idea used by Jonathan Hole can be implemented a bit more concisely.

If $M\cap M'$ is non trivial, then it has a non-trivial projection $p$. Define $v=2p-1$. Then $v$ is a selfadjoint unitary with spectrum $\{-1,1\}$ (this is trivial to see, as $$p=1\times p + (-1)\times (1-p)$$ is the spectral decomposition).

Assuming $M\subset B(H)$, choose unit vectors $\xi,\eta\in H$ with $p\xi=\xi$, $p\eta=0$. Define $$ \omega(x)=\langle x\xi,\xi\rangle,\qquad \phi(x)=\langle x\eta,\eta\rangle. $$ Since $pv=p$, $(1-p)v=p-1$, $$ \omega(v)=\langle v\xi,\xi\rangle=\langle vp\xi,\xi\rangle=\langle \xi,\xi\rangle=1, $$ and $$ \phi(v)=\langle v\eta,\eta\rangle=\langle v(1-p)\eta,\eta\rangle=-\langle (1-p)\eta,\eta\rangle=-1. $$ Then, using that $v$ is central, we have for any unitary $u$ that $$ \|\omega-u\phi u^*\|\geq|\omega(v)-\phi(uvu^*)|=|\omega(v)-\phi(v)|=1-(-1)=2. $$

Answer 2

If $M$ is not a factor then $M\cap M'\neq \mathbb{C}I$. Since the linear span of unitaries is norm-dense in any von Neumann algebra, this means there exists a unitary $u\in M\cap M'$ which is not a scalar multiple of the identity. Consider the self adjoint unitary $v=\frac{u+u^*}{2}\in M\cap M'$. Its spectrum, $\sigma(v)$, is contained in $\{-1,1\}$. We claim that in fact $\sigma(v)=\{-1,1\}$.

Indeed, if $\sigma(v)=\{-1\}$, then $\sigma(v+I)=\{0\}$, hence, since $v+I$ is self adjoint, we get $v+I=0$, i.e. $v=-I$. For all unit vectors $x$ we then obtain $$-1=\langle -x, x\rangle=\langle vx,x\rangle=\frac{1}{2}\langle ux,x \rangle+\frac{1}{2}\langle x,ux \rangle.$$

Since $|\langle ux,x \rangle|, |\langle x,ux \rangle|\leq 1$ this is only possible if $\langle ux,x\rangle=\langle x, ux\rangle=-1$ for all unit vectors $x$ which implies that $u=-I$, which is a contradiction. Hence $\sigma(v)\neq\{-1\}$ and by a similar argument, $\sigma(v)\neq\{1\}$.

Then $\sigma(v)=\{-1,1\}$ and it is a standard fact that this set is contained in $\overline{\{\langle vx,x\rangle: \|x\|=1\}}$. Choose unit vectors $y$ and $z$ such that $\langle vy,y\rangle$ and $\langle vz,z\rangle$ approximate $1$ and $-1$, respectively. Consider the normal states $\omega=\langle \cdot y,y\rangle $ and $\phi=\langle \cdot z,z\rangle$ on $M$.

For any unitaries $u_1, u_2 \in M$ we have, $$\|\omega \circ \text{Ad}(u_1)-\phi\circ \text{Ad}(u_2)\|\geq |\omega(u_1vu_1^*)-\phi(u_2vu_2^*)|=|\omega(v)-\phi(v)|.$$

(The last equality is due to the fact that $v$ commutes with $M$). By choosing $y$ and $z$ appropriately this last quantity can be made as close to $2$ as we wish. We conclude that $\text{diam}(S_0(M)/\text{Int}(M))\geq 2.$