Reductive Lie group

Remarks

Over a field of characteristic $0$, $\mathfrak{g}$ is reductive exactly when its adjoint representation is completely reducible.

Example: $G = GL_n(\R)$ with $n \ge 1$. Here $\mathfrak{g} = \mathfrak{gl}_n(\R)$, whose centre is the scalar matrices $\R I_n$ and whose derived algebra is $\mathfrak{sl}_n(\R)$, the trace-zero matrices. Every matrix $X$ is uniquely $\frac{\tr X}{n} I_n + \left(X - \frac{\tr X}{n} I_n\right)$, so $\mathfrak{g} = \R I_n \oplus \mathfrak{sl}_n(\R)$, and $\mathfrak{sl}_n(\R)$ is $0$ or simple, hence semisimple.

Non-example: the group $B$ of invertible real upper triangular $2 \times 2$ matrices. Its Lie algebra $\mathfrak{b}$ of upper triangular matrices has centre $\R I_2$ and derived algebra $\mathfrak{n}$ of strictly upper triangular matrices. Then $\mathfrak{z}(\mathfrak{b}) \oplus \mathfrak{n}$ has dimension $2$, not $3$, and $\mathfrak{n}$ is abelian and nonzero, so not semisimple; both conditions fail.

Many authors add further conditions, for example that $G$ has finitely many connected components, or that $G$ is the group of real points of a reductive algebraic group. Under the definition above, connected semisimple Lie groups and compact Lie groups are reductive.