Finite von Neumann modules, Hilbert closures, and dimension-zero quotients

A spectral projection of a self-adjoint matrix over $A$ belongs to the same matrix von Neumann algebra: it commutes with every element of that algebra's commutant by [F3], so belongs to its double commutant by [F2]. In the rectangular case apply this observation to $B^{\ast }B$ and $B{B}^{\ast }$; the polar partial isometry is the strong limit of $B(B^{\ast }B+\epsilon I{)}^{-1\mathrm{/}2}$ as $\epsilon \downarrow 0$, and belongs to the rectangular matrix algebra by [F2]–[F4]. Thus kernel, initial and final projections are matrices over $A$. The equality ${\tau }_s(v^{\ast }v)={\tau }_r(v{v}^{\ast })$, obtained by summing the trace identity for the matrix entries of $v$, gives the displayed rank-nullity formula. The algebraic kernel is exactly $P_{\ker B}{A}^s$, and $B$ maps $(I-P_{\ker B})A^s$ bijectively onto its algebraic image.

Every finite projective module has the form $e{A}^r$ for an idempotent $e$. Its Hilbert range is closed because $e^2=e$. Let $p$ be its orthogonal range projection, which lies in $M_r(A)$ by step 1.1. Since $ep=p$ and $pe=e$, one has $e{A}^r=p{A}^r$. Every module map between two such projection modules is given by a rectangular matrix: extend it to the ambient finite free module by composing with the source projection. If the algebraic map is injective, its Hilbert kernel projection is zero, since a nonzero kernel projection has a nonzero column in the algebraic kernel. Polar decomposition then implies that the source projection trace is at most the target projection trace. Applying this to an isomorphism and its inverse proves invariance of projective dimension.

For finitely many vectors of $L$, let $B$ have those vectors as columns. The corresponding range projection $q_B$ lies in $M_r(A)$ by step 1.1. As the finite subsets increase, the ranges increase, and their union has Hilbert closure equal to the closure of $L$: a vector $x\in A^r$ gives its Hilbert representative by applying its entries to ${\delta }_e$, while right translates of those representatives are also representatives of vectors of $L$. Their closed span contains the range of each column operator, because the coordinate vectors span ${\mathrm{\ell }}^2(G)$ densely. The increasing projections converge strongly to their supremum $q\in M_r(A)$ by [F2]. Their ranges lie in the range of $p$, so $q\le p$.

For each such $B$, let $q_{B,\epsilon }={\mathbf{1}}_{[\epsilon ,\mathrm{\infty })}(B{B}^{\ast })$ for $\epsilon >0$. Then $q_{B,\epsilon }\le q_B$ and $q_{B,\epsilon }\uparrow q_B$ as $\epsilon \downarrow 0$. The bounded Borel function $f_{\epsilon }(t)=t^{-1}{\mathbf{1}}_{[\epsilon ,\mathrm{\infty })}(t)$ gives $q_{B,\epsilon }=B{B}^{\ast }{f}_{\epsilon }(B{B}^{\ast })$. Hence every column of $q_{B,\epsilon }$ is a right $A$-linear combination of columns of $B$, and $q_{B,\epsilon }{A}^r\subseteq L$. Normality of the trace first lets these traces approach ${\tau }_r(q_B)$ and then, over finite subsets, ${\tau }_r(q)$. Conversely any finite projective submodule of $L$ embeds in $q{A}^r$, so step 2.1 bounds its dimension by ${\tau }_r(q)$. Taking suprema proves ${\dim }_{A}L={\tau }_r(q)$.

The module $(p-q)A^r$ embeds in $p{A}^r\mathrm{/}L$, giving lower dimension bound ${\tau }_r(p)-{\tau }_r(q)$. For the reverse bound let a finite projective $P$ embed in that quotient. Projectivity lifts this map to a map $f:P\to p{A}^r$. For each projection $q_{B,\epsilon }$ from step 3.1, the map $(p-q_{B,\epsilon })f$ is injective: its kernel maps under $f$ into $q_{B,\epsilon }{A}^r\subseteq L$, and therefore has zero image under the given embedding of $P$ into the quotient. Step 2.1 yields ${\dim }_{A}P\le {\tau }_r(p)-{\tau }_r(q_{B,\epsilon })$. Taking the projection traces arbitrarily close to ${\tau }_r(q)$ proves the quotient formula.

Submodules of a zero-dimensional module are zero-dimensional by the defining supremum. If a finite projective embeds in a quotient of a zero-dimensional module, projectivity lifts it to an injective map into the original module; hence its dimension is zero, proving quotient closure. For an exact sequence $0\to U\to V\to W\to 0$ with $U,W$ zero-dimensional, take any finite projective $P\subseteq V$. Its kernel $K=P\cap U$ and its image $P\mathrm{/}K\subseteq W$ are zero-dimensional. Apply steps 3.1 and 4.1 inside a projection representation of $P$: those formulas give ${\dim }_{A}P={\dim }_{A}K+{\dim }_A(P\mathrm{/}K)=0$. Thus $V$ is zero-dimensional. Finally, each image of a zero-dimensional module in a filtered colimit is zero-dimensional by quotient closure; these images form a directed family. Every finite set of generators of a finite projective submodule of the colimit lies in one member, so that projective has dimension zero. This proves filtered-colimit closure.