# Simple decomposition for semifinite von Neumann algebras

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.

• 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.
– Ruy
Jun 1, 2022 at 13:52
• @Ruy I added a picture of the statement in question. Jun 1, 2022 at 16:21
• @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. Jun 1, 2022 at 17:09
• 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. Jun 1, 2022 at 18:20
• 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!
– Ruy
Jun 1, 2022 at 21:43

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.
• 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? Jun 2, 2022 at 5:41
• Yes. And $R_1$ is type $I_\infty$, so it is properly infinite. Jun 2, 2022 at 6:46
• Can you explain both of these? In my question I say that I don't know why I$_{\infty}$ would be properly infinite Jun 2, 2022 at 16:56