What exactly is "halving" a projection in a von Neumann algebra?

Let $M=\mathbb C\oplus B(\ell^2(\mathbb N))$. Here the identity $1\oplus I$ is infinite, but not properly infinite. The same is true for $1\oplus p$, with $p$ any infinite projection in $B(\ell^2(\mathbb N))$.

Indeed, using that $I\sim I-E_{11}$, we get that $1\oplus I\simeq 1\oplus (I-E_{11})$, so $1\oplus I$ is infinite. Note that the nontrivial central projections in $M$ are $1\oplus0$ and $0\oplus I$.

But $1\oplus I$ is not properly infinite, because $(1\oplus 0)(1\oplus I)=1\oplus 0$ is not infinite.

Note also that the identity in this algebra cannot be halved even in the weaker sense: if $1\oplus I=P+Q$ for projections $P$ and $Q$, then $P$ and $Q$ are not equivalent. Indeed, without loss of generality we may assume that $P=1\oplus p$ and $Q=0\oplus q$ for projections $p$ and $q$. If we had $V^*V=P$, then $V=\lambda\oplus v$ and $$ P=V^*V=|\lambda|^2\oplus v^*v, $$ so $|\lambda|=1$ and $v^*v=p$. But then $$ VV^*=|\lambda|^2\oplus vv^*=1\oplus vv^*\ne Q. $$

Of course the above can be repeated with the von Neumann algebra $N\oplus M$, where $N$ is finite and $M$ is not finite.

Now Q3. Suppose that $M$ is type I$_n$. Then $I=\sum_jE_j$, pairwise orthogonal, each abelian with central carrier $I$. Suppose also that $I=P+Q$, with $P=V^*V$, $Q=VV^*$.

As $P\sim Q$ they have the same central carrier, which has to be necessarily $I$. Then $E_j\preceq P$ for all $j$. In particular there exists $F_1\leq P$, abelian, with central carrier $I$. Let $F_1,\ldots,F_r$ be maximal pairwise orthogonal where each $F_j$ is abelian and has central carrier $I$, such that $F_1+\cdots+F_r\leq P$. Define $G_j=VF_jV^*$, $j=1,\ldots, r$. Then $G_1,\ldots,G_r$ are pairwise orthogonal, abelian, with central carrier $I$, and $G_1+\cdots+G_r\leq Q$. Let $$ R=F_1+\cdots+F_r,\qquad S=G_1+\cdots+G_r. $$ As $R\sim S$ and $P,Q$ are finite, $P-R\sim Q-S\prec F_1$. As $P-R$ and $Q-S$ have the same central carrier and are orthogonal, the projection $I-(R+S)=(P-R)+(Q-S)$ also has the same central carrier. Now $$R+S=F_1+\cdots+F_r+G_1+\cdots+G_r\sim E_1+\cdots+E_{2r}.$$ So $$I-(R+S)\sim I-(E_1+\cdots E_{2r})=E_{2r+1}+\cdots +E_n$$ has central carrier $I$. This means that $P-R$ has central carrier $I$, and so $F_1\preceq P-R$, a contradiction. Thus $R=P$, $S=Q$, and $$ I=F_1+\cdots F_r+G_1+\cdots G_r\sim E_1+\cdots +E_{2r}, $$ and then $I=E_1+\cdots E_{2r}$. Thus $n=2r$.