Lie representations as actions before enveloping

From a representation $\rho$, set $xv=\rho (x)v$. Linearity of $\rho$ and of each $\rho (x)$ makes the action bilinear, and bracket preservation expands to $[x,y]v=x(yv)-y(xv)$.

Conversely, a bilinear action defines a linear map $\rho (x)(v)=xv$ into ${\mathrm{End}}_k(V)$. The displayed identity says exactly that $\rho ([x,y])=\rho (x)\rho (y)-\rho (y)\rho (x)$, so $\rho$ is a representation.

The equivalence is therefore exact, but [L2] does not apply directly from the Lie-algebra data in general. The bracket need not be associative; if it is the zero bracket on a nonzero abelian Lie algebra, that multiplication has no unit. There is a genuine exceptional case: if $\mathfrak{g}=0$ and $V=0$, the zero bracket makes $\mathfrak{g}$ the permitted unital zero ring, and its unique action on $V$ is a unital module action. This exception does not give a general ring structure for Lie representations. The canonical associative-module formulation for arbitrary $\mathfrak{g},V$ uses $U(\mathfrak{g})$.