1

Let $E$ be a projection in a von Neumann algebra $R$. According to Kadison and Ringrose, $E$ is $\textit{properly infinite}$ if $E$ is infinite and for each central projection $P\in R$ either $PE=0$ or $PE$ is infinite. But I've seen other definitions that require $PE$ to be infinite for each nonzero central projection $P\in R$. It doesn't seem to me that these two definitions are equivalent, and if I am right then I believe there is a proof of a lemma in Kadison and Ringrose which is incorrect as written.

SihOASHoihd
  • 1,886

1 Answers1

1

I don't know where you've seen your second definition, but it is a weird choice. The spirit of "properly infinite" is that it has no finite part. Like, in $B(H)\oplus B(H)$, the projection $E_{11}\oplus I$ is not properly infinite because it has a finite part in the first block; this precludes the possibility of halving it. But saying that $0\oplus I$ is not properly infinite makes little sense, as it satisfies any property a properly-infinite-in-the-second-sense projection would have (other than having central carrier different than the identity).

As for the Halving Lemma in Kadison-Ringrose, the distinction is immaterial as it doesn't change the proof.

Martin Argerami
  • 205,756
  • In the kadison-Ringrose proof of the halving lemma, why is it true that $(C_{E}-\sum Q_{A})E\neq 0$? – SihOASHoihd Aug 06 '21 at 02:47
  • If $(C_E-\sum Q_a)E=0$ then you are done, because this gives you $E=\sum_aQ_aE$, and this is a pairwise orthogonal family of halvable projections. – Martin Argerami Aug 06 '21 at 03:44
  • Ah I see. But do you agree that this part in the proof, where they claim that $(C_{E}-\sum{Q_{a}})E$ is properly infinite, is the only time they use the fact that $E$ Is properly infinite? – SihOASHoihd Aug 06 '21 at 14:59
  • Yes. The idea of the proof is that an infinite projection always has a part that can be halved. If the projection is properly infinite, this can be iterated until you fill everything with halvable parts. – Martin Argerami Aug 06 '21 at 16:50
  • Thank you very much! – SihOASHoihd Aug 07 '21 at 14:43