1

So I am having trouble to understand/see what are the elements in $\mathfrak{gl}(\mathfrak{g})$ for a Lie algebra $\mathfrak{g}$. I know that $\text{ad}(x)$ are elements of $\mathfrak{gl}(\mathfrak{g})$ given $x\in \mathfrak{g}$. But is there a general description of elements in $\mathfrak{gl}(\mathfrak{g})$ given that $\mathfrak{g}$ is any random Lie algebra?

Christina
  • 635
  • 4
  • 8
  • 5
    They're just the endomorphisms of the vector space $\mathfrak{g}$. In general, you can expect lots of linear maps which are not ${\rm ad}(x)$ for any $x \in \mathfrak{g}$ (e.g., for abelian $\mathfrak{g}$). – Ivo Terek Oct 21 '20 at 20:14
  • 5
    In fact, if $\mathfrak g$ has dimension $n$, then $\operatorname{ad}(\mathfrak g)$ has dimension at most $n$, while $\mathfrak{gl}(\mathfrak{g})$ has dimension $n^2$. – WhatsUp Oct 21 '20 at 20:29

1 Answers1

1

Denote by $V$ the underlying $K$-vector space of $\mathfrak{g}$ of dimension $n$. Then $\mathfrak{gl(g)}$ is, by definition, just $\mathfrak{gl}(V)\cong \mathfrak{gl}_n(K)$. It has dimesnion $n^2$. On the other hand, $\operatorname{ad}(x)$ is not only an endomorphism of $V$, but moreover an inner derivation of $\mathfrak{g}$. We write $$ \operatorname{Inn}(\mathfrak g)=\operatorname{ad}(\mathfrak g) \cong \mathfrak{g}/Z(\mathfrak{g}) $$ for this Lie ideal in $\operatorname{Der}(\mathfrak{g})$, which obviously can have dimension at most $n$. So we have the containments $$ \operatorname{Inn}(\mathfrak g)\subseteq \operatorname{Der}(\mathfrak{g}) \subseteq \mathfrak{gl(g)}. $$

Dietrich Burde
  • 130,978