I’ve been taking a seminar course on Leavitt path algebras, and our source material is rather laconic as far as explanations are concerned. I am attempting to justify (to myself) the following claim made by the authors:
$L_{K}(\Gamma)$ is a unital $K$-algebra if and only if $E^0$ is finite.
Here, $K$ denotes the base field, $\Gamma$ denotes the base digraph, $E^0$ denotes the set of vertices in $\Gamma$, and $L_{K}\left(\Gamma\right)$ denotes the Leavitt path algebra of $\Gamma$ with coefficients in $K$. The “proof” offered is as follows:
For $E^0$ finite, the element $$1_{L_{K}\left(\Gamma\right)}=\sum_{v\in E^0}{v}$$ acts as the identity by the representation of elements of $L_{K}\left(\Gamma\right)$ as $K$-linear combinations of the set $$\left\{\gamma\lambda^*\ \Big|\ \gamma,\lambda\in\mathrm{Path}\left(\Gamma\right);r\left(\gamma\right)=r\left(\lambda\right)\right\}.$$ $\Big[$Here, $\mathrm{Path}\left(\Gamma\right)$ denotes the set of paths in $\Gamma$ of finite length, and $r\left(\mu\right)$ is the vertex at which the path $\mu$ terminates — the range of $\mu$.$\Big]$ If $E^0$ is infinite, then there is no element of $L_{K}\left(\Gamma\right)$ which acts as the identity on each element of the set $$V=\left\{v\ \Big|\ v\in E^0\right\}.$$
I feel as if I am missing something rather elementary/obvious, but I cannot figure out why that last sentence — the converse of the statement in question — is true. In point of fact, I’m not sure that I understand why the sum of all the vertices (when this is defined) must act as the identity on anything other than the set of vertices. I would appreciate any clarifications or explanations that could be offered.
If $E^0$ were infinite, then $\sum_{v \in E^0} v$ would have finite support. Thus, when $E^0$ is infinite, we only have local units in the path algebra. These local units would be of the form $\sum_{v \in S} v$ where $S \subset E^0$ and $|S| < \infty$.
Hope this answers your question.