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I want to know why the following is true: Every semifinite (ie has trivial type III part) von Neumann algebra $R$ is the direct sum of a finite von Neumann algebra and a properly infinite von Neumann algebra. This is clearly true if $R$ is type II. If $R$ is type I and infinite is it necessarily properly infinite (ie has no finite nontrivial central projections)? I cannot find anything that implies this, even for type I$_{\infty}$.

Edit: $R$ is assumed to have a faithful normal state.

9.2.19. LEMMA. If $\omega$ is a faithful normal state of a semi-finite von Neumann algebra $\mathscr{R}$, there is a normal semi-finite faithful tracial weight $\rho$ on $\mathscr{R}$, and a positive element $K$ in the unit ball of $\mathscr{R}$, such that $I-K \in F_{p}$ and $$ \rho((I-K) A)=\omega(K A)=\omega(A K) \quad(A \in \mathscr{R}) . $$ Moreover, both $K$ and $I-K$ are one-to-one mappings. Proof. It suffices to consider separately the two cases in which $R$ (with a faithful normal state $\omega$ ) is either finite or properly infinite but semi-finite; for the general semi-finite $\mathscr{R}$ is a direct sum of two algebras, one of each of these kinds.

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    $\begingroup$ Perhaps you should check the precise statement of the question in the text. If I read it independently of the preamble I feel like answering no: $B(\ell^2)\oplus M_n(\mathbb C)$ is type I and infinite but it does have a finite central projection. $\endgroup$
    – Ruy
    Commented Jun 1, 2022 at 13:52
  • $\begingroup$ @Ruy I added a picture of the statement in question. $\endgroup$ Commented Jun 1, 2022 at 16:21
  • $\begingroup$ @Ruy The existence of a faithful normal state implies that $R$ has no part of type I$_{\infty}$ (similar to how it implies that $R$ is countably decomposable). It must be possible to go further and show that the part of $R$ that is type I is finite. $\endgroup$ Commented Jun 1, 2022 at 17:09
  • $\begingroup$ Actually this doesn't follow... I was assuming that a faithful normal state takes the same value on equivalent projections which I'm pretty sure isn't true. $\endgroup$ Commented Jun 1, 2022 at 18:20
  • $\begingroup$ OK, I guess the point is that you start by splitting your algebra as $R=R_1\oplus R_2$, where $R_1$ is the ideal gerated by the supremum of all finite central projections, so when you deal with $R_2$, all finite central projection are gone! $\endgroup$
    – Ruy
    Commented Jun 1, 2022 at 21:43

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By type decomposition, you know that $$ R=R_0\oplus R_1\oplus R_2\oplus R_\infty, $$ where $R_0$ is a direct sum of algebras of type I$_n$ for $n\in\mathbb N$; $R_1$ is type I$_\infty$; $R_2$ is type II$_1$; and $R_\infty$ is type II$_\infty$. Then $$ R_0\oplus R_2\ \text{ is finite, while }\ R_1\oplus R_\infty\ \text{is properly infinite. } $$


An arbitrary sum of pairwise orthogonal finite central projections is finite. This is easy to see because comparison of projections cannot go outside of a central component. Indeed, you have $\sum_jq_j\leq \sum_jp_j$, with $p_j$ finite and central for all $j$. If $\sum_jq_j\sim\sum_jp_j$, from $q_jp_j=q_j$ and multiplying the partial isometry by $p_j$, we get that $q_j\sim p_j$. As $q_j\leq p_j$ and $p_j$ is finite, then $q_j\sim p_j$ and thus $\sum_jq_j\sim\sum_j p_j$. Thus $\sum_jp_j$ is finite.

The lack of finite central projections in a type I$_\infty$ follows directly from the type decomposition. If $p$ is a finite central projection in $M$ with $M$ type $I$, then $pM$ is type $I$ and finite, so type $I_n$. But then $M$ was not tpe $I_\infty$ to begin with.

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    $\begingroup$ Would $R_{0}$ still be finite if it's an infinite direct sum? And what if $R_{\infty}=0$, why is $R_{1}$ properly infinite? $\endgroup$ Commented Jun 2, 2022 at 5:41
  • $\begingroup$ Yes. And $R_1$ is type $I_\infty$, so it is properly infinite. $\endgroup$ Commented Jun 2, 2022 at 6:46
  • $\begingroup$ Can you explain both of these? In my question I say that I don't know why I$_{\infty}$ would be properly infinite $\endgroup$ Commented Jun 2, 2022 at 16:56
  • $\begingroup$ I wrote a little more. $\endgroup$ Commented Jun 2, 2022 at 17:38

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