- #76

fresh_42

Mentor

- 15,542

- 13,639

Yes, and a bit more detailed:For 3. The kernel of the adjoint map is the centre. Semisimple Lie algebras have trivial centre. For the second part, take a 1d algebra, then the general linear algebra over it is also 1d algebra, hense they are isomorphic. And the abelian algebra is not semisimple.

A semisimple Lie algebra has no Abelian ideals. Its center, however, is an Abelian ideal. Thus we have

\begin{align*}

\mathfrak{Z(g)}&=\{\,Z\in \mathfrak{g}\,|\,[X,Z]=0\;\forall \,X\in \mathfrak{g}\,\}\\&=\bigcap_{X\in \mathfrak{g}}\operatorname{ker}\mathfrak{ad}X = \operatorname{ker}\mathfrak{ad} = \{\,0\,\}

\end{align*}

This means that ##\mathfrak{ad}\, : \,\mathfrak{g}\longrightarrow \mathfrak{gl(g)}## is a monomorphism of Lie algebras and

$$

\mathfrak{g} \cong \mathfrak{ad(g)} \cong \mathfrak{Der(g)} \subseteq \mathfrak{gl(g)}

$$

The adjoint representation cannot be onto, since the center of ##\mathfrak{gl(g)}## are all multiples of the identity matrix

If we consider the non Abelian two dimensional Lie algebra defined by ##[X,Y]=Y##, which is the Borel subalgebra of the simple Lie algebra ##\mathfrak{sl}(2)##, or the Lie algebra of matrices ##\begin{bmatrix}*&*\\0&0\end{bmatrix}##, then we have a solvable and therewith no semisimple Lie algebra which has only a trivial center, too. Hence the condition of semisimplicity is not necessary.

As already mentioned says Ado's theorem, that any finite-dimensional Lie algebra over a field of characteristic zero can be seen as a subalgebra of ##\mathfrak{gl}(n)## for sufficiently large ##n.##