Hattori–Stallings pseudo-trace

Let $A$ be a unital associative $\mathbb{C}$-algebra, let $[A,A]\subseteq A$ be the linear span of commutators $ab-ba$, and let $\omega :A\to \mathbb{C}$ be a symmetric linear functional, meaning $\omega (ab)=\omega (ba)$ for all $a,b\in A$. A right $A$-module $P$ is finite projective here if there are an integer $n\ge 0$ and an idempotent $e\in M_n(A)$ with $P\cong e{A}^n$.

A finite dual basis for $P$ is a finite list $p_i\in P$, ${\varphi }_i\in {\mathrm{H}\mathrm{o}\mathrm{m}}_A(P,A)$ satisfying $p={\sum }_i{p}_i{\varphi }_i(p)$ for every $p\in P$; a finite projective module has such a list. For $T\in {\mathrm{E}\mathrm{n}\mathrm{d}}_A(P)$ define the candidate pseudo-trace value set

The basis-independence lemma for this definition proves that this set is a singleton; its unique value is denoted ${\mathrm{T}\mathrm{r}}_{\omega ,A}(T)$ and called the Hattori–Stallings pseudo-trace.

Basis independence of the Hattori–Stallings pseudo-trace

The asserted existence of a finite dual basis follows directly from the chosen definition of finite projective. If $P=e{A}^n$ for an idempotent $e\in M_n(A)$, regard $A^n$ as column vectors, set $p_i=e{e}_i$, and let ${\varphi }_i$ be the $i$th coordinate restricted to $e{A}^n$. For $p\in e{A}^n$, $p=ep={\sum }_{i}e{e}_i{\varphi }_i(p)={\sum }_i{p}_i{\varphi }_i(p)$. An isomorphism $P\cong e{A}^n$ transports this finite dual basis to $P$. The basis-independence lemma then shows the candidate value set in the Definition is a singleton.