Path algebra

Remarks

The product is associative, so $kQ$ is an associative algebra; it has an identity $\sum_{v \in Q_0} e_v$ when $Q_0$ is finite. The $e_v$ are orthogonal idempotents: $e_v e_w = e_v$ if $v = w$ and $0$ otherwise. Example: let $Q$ have vertices $1, 2$ and one arrow $\alpha \colon 1 \to 2$. The paths are $e_1, e_2, \alpha$, and the nonzero products of basis elements are $e_1 e_1 = e_1$, $e_2 e_2 = e_2$ and $e_1 \alpha = \alpha = \alpha e_2$. So $kQ$ has dimension $3$, and it is isomorphic to the algebra of upper triangular $2 \times 2$ matrices over $k$, via $e_1 \mapsto E_{11}$, $e_2 \mapsto E_{22}$, $\alpha \mapsto E_{12}$. Non-example: for the quiver with one vertex and two loops $a, b$, the paths $ab$ and $ba$ are different basis elements. So $kQ$ is the free algebra $k\langle a, b \rangle$ and not the commutative ring $k[a, b]$. $kQ$ is finite-dimensional if and only if $Q$ is finite and has no oriented cycles. Some books compose paths in the opposite order, which replaces $kQ$ by its opposite algebra.