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Let $M$ be a semifinite von Neumann algebra equipped with a semifinite normal faithful trace $\tau$. Let $e \in M$ be a (selfadjoint) projection.

Do we have the equivalence : $e$ is finite $\iff$ $\tau(e)<\infty$ ?

Any explanation or reference is welcome.

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2 Answers 2

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This is not true in general. If $M=L^{\infty}(\mathbb{R},dx)$ is the algebra of essentially bounded functions equipped with the Lebesgue measure then any indicator function gives a finite projection but the trace is finite if and only if the set has finite measure.

Although it fails in general, the result is true for factors. Assume now that $(M,\tau)$ is a semifinite factor endowed with a trace $\tau$. We will show that an orthogonal (from now on I will only deal with orthogonal projections, so I will drop the adjective) projection $p$ is finite if and only if $\tau(p)<\infty$.

We will repeatedly use the notion of a central carrier of a projection, so let me recall the definition. If $p$ is a projection then its central carrier is a projection $C_{p}$ such that $p\leqslant C_{p}$ and $C_{p}$ is a smallest projection belonging to the centre with this property. Note that in a factor central carrier of any non-zero projection is equal to $Id$, as $Id$ is the only non-zero central projection in this case; we will use this a lot.

One implication is true in general: if $\tau(p)<\infty$ then $p$ is finite. Indeed, if $q\leqslant p$ is equivalent to $p$ then they have equal traces, so $\tau(p-q)=0$, which means that $p=q$, by faithfulness of $\tau$.

Assume now that $\tau(p)=\infty$. We will show that there exists a subprojection $q\leqslant p$ such that $q\sim Id$. Note that if $\tau(Id)=\infty$ then by our result the projection $Id$ is infinite because in such a case there is always a non-trivial subprojection $r$ with $\tau(r)=\infty$. Indeed, since there is an increasing sequence $(r_{k})_{k\in\mathbb{N}}$ such that $\tau(r_{k})<\infty$ and $\lim_{k\to\infty} r_{k}=Id$ in the strong operator topology then all projections $Id - r_k$ satisfy $\tau(Id-r_k)=\infty$ and from some point on they are non-trivial, since they tend to zero.

Let us prove our claim. Fix an increasing sequence $(r_k)_{k\in\mathbb{N}}$ of projection such that $\lim_{k\to\infty}r_{k}=Id$ and $\tau(r_{k})<\infty$ and define $q_{1}=r_{1}$, $q_{2}=r_{2}-r_{1}$, $q_{3}=r_{3}-r_{2}$, and so on. Then $(q_{k})_{k\in \mathbb{N}}$ is a sequence of mutually orthogonal projections with finite traces that sums to identity. We will now show that for each $q_k$ there is an equivalent projection $q_k'$ such that $q_{k}'\leqslant p$ and there is enough room in $p$ to ensure that all such projections $q_k'$ are mutually orthogonal; this will allow us to form the infinite sum $\sum_{k\in\mathbb{N}} q_k' \leqslant p$ satisfying $\sum_{k\in\mathbb{N}} q_k' \sim Id$.

Let $e$ be an arbitrary projection with finite trace. We will show that there exists an equivalent projection $e'$ such that $e' \leqslant p$. The proof will be an exhaustion argument. If $e=0$ then there is nothing to do. If $e\neq 0$ then $e$ and $p$ have equal central carriers, so they admit equivalent subprojections $f_{1}$ and $p_{1}$: it means that there exists a partial isometry $v_1$ such that $v_1^{\ast}v_1=f_1 \leqslant e$ and $v_1 v_1^{\ast}=p_1 \leqslant p$. If $f_{1}=e$ then we are done. If not, we apply the same procedure to $e-f_1$ and $p-p_1$; note here that $\tau(p-p_1)=\infty$, so definitely $p_1 \neq p$. We continue in this manner to obtain a partial isometry $v$ such that $v^{\ast}v=e$ and $vv^{\ast}\leqslant p$.

We apply this reasoning to the whole sequence $(q_k)_{k \in \mathbb{N}}$. We first get a partial isometry $w_{1}$ such that $w_{1}^{\ast}w_{1}=q_{1}$ and $w_{1}w_{1}^{\ast} \leqslant p$. Since $\tau(p-w_1 w_1^{\ast})=\infty$, we can always continue this procedure. Therefore we obtain a sequence of partial isometries $(w_k)_{k \in \mathbb{N}}$ with mutually orthogonal ranges such that $w_k^{\ast}w_k=q_k$ and $w_k w_k^{\ast} \leqslant p$. It follows that $w:=\sum_{k \in \mathbb{N}} w_k$ is an isometry such that $ww^{\ast} \leqslant p$; this is exactly what we wanted because now $p$ admits an infinite subprojection, so it is infinite.

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In general, it is not true. When $\tau(P)<\infty$, then $P$ is a so-called $\tau$-finite projection. You may want to read the survey by P.G. Dodds, B. de Pagter, Indag. Math. 2014--- Normed Kothe spaces: A non-commutative viewpoint.

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