Show that if $L$ is a Lie algebra then $L/Z(L)$ is isomorphic to a subalgebra of $gl(L)$.

Question

In this, $Z(L)$ is the center of $L$. I can't think of any explicit mapping. Any suggestions?

score 1 · Accepted Answer · answered Jul 09 '17 at 10:37

1

Just use the ajoint map. You have the map$$\begin{array}{rccc}\operatorname{ad}\colon&L&\longrightarrow&\mathfrak{gl}(L)\\&X&\mapsto&\left(\begin{array}{ccc}L&\longrightarrow&L\\ Y&\mapsto&[X,Y]\end{array}\right).\end{array}$$It's kernel $Z(L)$ and so $L/Z(L)$ is ismorphic to the image of $\operatorname{ad}$.

answered Jul 09 '17 at 10:37

José Carlos Santos

427,504

Is it correct to argue that $\forall c \in Z(L), \operatorname{ad}_c$ is the identity in $\mathfrak{gl}(L)$ because $\forall c\in Z(L), \forall \lambda \in L$, $[\lambda + c, -]=[\lambda,-]+[c, -]=[\lambda, -]$. Therefore, the adjoint homomorphism maps $Z(L)$ to the identity of $\mathfrak{gl}(L)$? Therefore, $\operatorname{ker}(ad)=Z(L)$. – TheLast Cipher May 06 '19 at 10:25

Show that if $L$ is a Lie algebra then $L/Z(L)$ is isomorphic to a subalgebra of $gl(L)$.

1 Answers1