In Kadison-Ringrose Vol II (Lemma 6.3.3) the authors introduce the concept of "halving" a projection inside a von Neumann algebra. What I understood is that: "halving" a projection $E$ in $\mathscr{R}$ means writing $E$ as a sum of two subprojections both equivalent to $E$ in $\mathscr{R}$, i.e., there are $E_1\le E;~E_2\le E$ such that $E_1\sim E_2 \sim E$ and $E = E_1 + E_2$

In Lemma 6.3.3 they proved that every properly infinite projection $E$ can always be halved. Here, I have one confusion: at the first line of 2nd paragraph of this Lemma 6.3.3 they says "...if $E$ is a properly infinite projection in $\mathscr{R}$, there is a non-zero central projection $P$ in $\mathscr{R}$ such that $PE$ can be "halved"..." But in the paragraph preceding to this, it seems they only used the fact that $E$ is infinite. So isn't that statement true for any infinite projections? I mean,

Question 1. If $E$ is a infinite (not necessarily properly infinite) projection in $\mathscr{R}$, there is a non-zero central projection $P$ in $\mathscr{R}$ such that $PE$ can be "halved"...Doesn't it follow from the first paragraph of the proof of Lemma 6.3.3?

Also I was thinking about the following question. First I tried with $I\in \mathcal{B}(\mathcal{H})$ ...But it is properly infinite...

Question 2. What could be an example of infinite projection which is not halvable?

Then after a few pages later at the bottom of page 425 the authors make the following remark regarding "halving" which I cannot fully understand. They said,

Remark. "...in $\mathcal{B}(\mathcal{H})$, we should not expect to be able to halve a projection whose range has dimension 5. More generally, if $\mathscr{R}$ is a von Neumann algebra of type $I_n$ with $n$ odd, the identity in $\mathscr{R}$ cannot be halved."

I tried to verify the 2nd line (regarding type $I_n$ case) atleast in $M_n(\Bbb{C})$: if $I_n\in M_n(\Bbb{C})$ could be halved then there would exists two projections $E\sim F\sim I$ (equivalently, $\dim \text{ran}E= \dim \text{ran}F=n$) s.t. $I = E + F$ which is impossible since $E, F$ are orhtogonal. There is no need to assume $n$ to be odd to get this contradiction!! Isn't that the case?? or I'm missing something??

Explanations regarding Question 1, 2 and the last statement would be appreciated. Thanks.

EDIT 1. Okay as pointed out in the comments below I realized that in the Remark the authors meant by "halving" just to be able to write the projection equal to two of its equivalent subprojections (the subprojections need not to be equivalent to the initial projection!). With this modification now that Remark is clear to me partially... Because of "rank" equality in finite dimensional case, it is true that you cannot halve a projection of dimension odd in $\mathcal{B}(\mathcal{H})$. So in particular we cannot hope to halve $I\in M_{n}(\Bbb{C})$ with $n$ odd. For if $I = E + F$ with $E\sim F$ and $k = \dim \text{ran}E = \dim \text{ran}F$, we would then $n = 2k$...! Here finite dimensionality and the fact that $E\sim F \text{ in }M_n(\Bbb{C}) \iff \dim \text{ran}E = \dim \text{ran}F$ both helped.

But how to prove this for any general Type $I_n$ algebra with $n$ odd:

Question 3. How exactly to prove that the $I$ cannot be written as sum of two equivalent projections in a type $I_n$ algebra where $n$ is odd?

Because I cannot use the that "finite dimensional" argument here...! Could anybody please explain it . Thanks

EDIT 2. Regarding Question 1; let $E$ be an infinite projection then is it possible that $PE$ is non-zero and finite for some central projection $P$?? It is possible if $E$ is NOT properly infinite (that is exactly the definition of properly infinite). But is there any example? I cannot find any example of such. Precisely,

Question 4. What could be an example of an infinite projection which is not properly infinite?

EDIT 3. Questions 1 and 2 will be solved once an answer of question 4 is found!

  • $\begingroup$ I don't have K-R in front of me so I don't know how they discuss halving. I suspect that the requirement that the "halves" ($E_1$, $E_2$) be equivalent to the whole ("$E$") is somehow an aspect of this definition that is specific to the infinite case. The spirit of the remark in the finite dimensional case is that the identity operator on a space of odd dimension cannot be the sum of two projections having the same rank (e.g. because the non-normalized trace of two projections with the same rank $k$ is $2k$, but the non-normalized trace of the identity is the dimension of the space). $\endgroup$ Dec 31, 2021 at 6:10
  • $\begingroup$ @leslietownes Yes yes that's exactly (in $M_n(\Bbb{C})$ case) what I pointed out at the bottom... But why do they need "odd" to argue this in case of general type $I_n$ algebras?? Thats what I didn't understand!!! In $M_n(\Bbb{C})$ the the rank (or tr) argument works irrespective of the fact that $n$ odd or even!! right?? So why did they mentioned "odd" there in the book (which I mentioned in ittalic at the end of the post)... $\endgroup$
    – sigma
    Dec 31, 2021 at 6:15
  • $\begingroup$ I mean, is it then possible (sometimes) in a type $I_{2n}$ algebras to halve $I$...??? $\endgroup$
    – sigma
    Dec 31, 2021 at 6:17
  • $\begingroup$ I think it is the definition of "halving" that needs adjustment in the finite case. The identity operator on $\mathbb{C}^2$ for example certainly is the sum of two equivalent projections (projection onto the first and second components, respectively). The identity operator on $\mathbb{C}^3$ is not (due to the trace argument). The requirement that "halves" be equivalent to the whole must have been made in K-R in the infinite case, but discarded for the above remark (where I think they might just mean that a projection is 'halved' if it is the sum of two equivalent projections in the algebra) $\endgroup$ Dec 31, 2021 at 6:20
  • 1
    $\begingroup$ Yes, I think so. If you use the K-R definition in connection with halving infinite projections, and apply it to the finite dimensional case, then absolutely nothing halves in the finite dimensional case, irrespective of dimension $\endgroup$ Dec 31, 2021 at 6:31

1 Answer 1


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 and equivalent, 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 $j$ (proof at the end). In particular there exists $F_1\leq P$ (say, a copy of $E_1$), 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 $n=2r$.

Proof that $E_1\preceq P$. By Comparison there exists $R$, central, with $(I-R)E_1\preceq (I-R)P$ and $RP\preceq RE_1$. By equivalence we get $RQ\preceq RE_2$. Then $$ R=R(P+Q)\preceq R(E_1+E_2)\leq R. $$ If $n>2$ this forces $R=0$, as it has to be finite since $M$ is finite. Thus $E_1\preceq P$.

  • $\begingroup$ Thanks for your answer. Sorry but I cannot understand how to answer the Question 3 (the authors remarked it at page 425)...... Is it something that follows immediately?? $\endgroup$
    – sigma
    Jan 2, 2022 at 4:29
  • 1
    $\begingroup$ No, it doesn't. I have edited in an argument. $\endgroup$ Jan 2, 2022 at 8:12

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