# Lie Groups/Algebras, Notes 2: Representations and Types

Recall a representation of an algebra $A$ on an $R$-module $M$ is an algebra morphism $\rho:A\to End_R(M).$  When we have a Lie algebra $\mathfrak{g},$ a representation of it on an $R$-module $M$ is an algebra morphism

$\rho:\mathfrak{g}\to L(End_R(M)).$

Hence in particular we have $\rho(xy)(m)=\left(\rho(x)\rho(y)-\rho(y)\rho(x)\right)(m).$  We correspondingly call $M$ a $\mathfrak{g}$module.  A representation is faithful if it is injective.  It is simple/irreducible if its $\mathfrak{g}$-module is simple.

Let $\rho_1$ and $\rho_2$ be representations of Lie algebras $\mathfrak{g}_1$ and $\mathfrak{g}_2$ on $M_1$ and $M_2$ respectively.  Then $\rho_1\otimes\rho_2$ is a representation of $\mathfrak{g}_1\times\mathfrak{g}_2$ on $M_1\otimes M_2$ defined by

$\begin{array}{lcl}(\rho_1\otimes\rho_2)(x_1,x_2)(m_1\otimes m_2)&=&(\sigma_1(x_1)\otimes 1+1\otimes\sigma_2(x_2))(m_1\otimes m_2)\\&=&\sigma_1(x_1)(m_1)\otimes m_2+m_1\otimes\sigma_2(x_2)(m_2)\end{array}$

where $\sigma_1,\sigma_2$ are the canonical inclusions of $\mathfrak{g}_1$ in $U(\mathfrak{g}_1)$ and $\mathfrak{g}_2$ in $U(\mathfrak{g}_2).$  Hence $M_1\otimes M_2$ is both a $\mathfrak{g}_1\times\mathfrak{g}_2$-module and a $U(\mathfrak{g}_1)\otimes U(\mathfrak{g}_2)$-module.

Definition 1$\mathfrak{g}$ is solvable if its derived series $D^n\mathfrak{g}=0$ for some $n.$  It is nilpotent if its lower central series $C_n\mathfrak{g}=0$ for some $n.$  The radical of $\mathfrak{g},$ $\sqrt{\mathfrak{g}},$ is the largest solvable ideal in $\mathfrak{g}.$  We say $\mathfrak{g}$ is semisimple if it has no nonzero abelian ideals.

Hence it follows that the center $Z(\mathfrak{g})$ of a nilpotent Lie algebra is nontrivial.  Also note that a nilpotent Lie algebra is solvable.  $\mathfrak{r}=\sqrt{\mathfrak{g}}$ is the smallest ideal such that $\mathfrak{g}/\mathfrak{r}$ has radical $0.$

Theorem 2 (Engel).  Let $V$ be a vector space and $\mathfrak{g}$ be a finite dimensional subalgebra of $\mathfrak{gl}(V).$.  If $V\neq 0,$ then there exists a $u\neq 0$ with $u\in\mathfrak{g}$ such that $xu=0$ for all $x\in\mathfrak{g}.$

It follows that a Lie algebra $\mathfrak{g}$ is nilpotent iff $\mbox{adj}\,x$ is nilpotent for all $x\in\mathfrak{g}.$

Note that $B(x,y)=\mbox{tr}(\mbox{adj}(x)\mbox{adj}(y))$ defines a symmetric bilinear form on $\mathfrak{g},$ called the Killing form of $\mathfrak{g}.$  The Killing form also satisfies $B(xy,z)=B(x,yz).$

Proposition 3.  $\mathfrak{g}$ is solvable iff $B(D\mathfrak{g},\mathfrak{g})=0.$

Proposition 4.  $\mathfrak{g}$ is semisimple iff $\mathfrak{g}=0$ iff $B$ is nondegenerate.

Theorem 5 (Weyl).  Every finite dimensional representation of a semisimple Lie algebra is completely reducible (i.e. the $\mathfrak{g}$-module is semisimple).

A Lie algebra is simple if its only ideals are $\mathfrak{g}$ and $0$ and noncommutative.

Proposition 6.  $\mathfrak{g}$ is semisimple iff it is a product of simple algebras.

A Lie algebra is reductive if its adjoint representation is semisimple.  That is, it is reductive if it can be written $\mathfrak{g}=\mathfrak{s}\oplus\mathfrak{a}$ where $\mathfrak{s}$ is semisimple and $\mathfrak{a}$ is abelian (since here we want $\mathfrak{g}$ as a module to be semisimple–so that as an algebra, it has an an abelian term).  This also turns out to be equivalent to $D\mathfrak{g}$ being semisimple.

Definition 7.  The nilpotent radical of $\mathfrak{g}$ is the intersection of the kernels of all finite dimensional simple representations of $\mathfrak{g}.$

For an element $x$ in the nilpotent radical, it follows that $\mbox{adj}(x)$ is nilpotent.  Now for any nilpotent endomorphism $u$ of an algebra, we have that

$\displaystyle e^u=\sum_{n=0}^\infty\frac{u^n}{n!}$

has a finite number of nonzero terms, and is hence well-defined.  If we can write $\mathfrak{g}=\sqrt{\mathfrak{g}}\ltimes_{e^{u\cdot}}\mathfrak{s}$ with $\mathfrak{s}$ semisimple and $u$ nilpotent, then we will call $\mathfrak{s}$ a Levi subalgebra of $\mathfrak{g}.$

Theorem 8 (Levi-Malcev).  Every Lie algebra has a Levi subalgebra (that is, it can be written $\mathfrak{g}=\sqrt{\mathfrak{g}}\ltimes\mathfrak{s}$ as above).  Moreover, every other Levi subalgebra $\mathfrak{s}'$ is the image of $\mathfrak{s}$ under $\exp(\mbox{adj}(x))$ for some $x$ in the nilpotent radical of $\mathfrak{g}.$

It turns out that a subalgebra of $\mathfrak{g}$ is a Levi subalgebra iff it is a maximal semisimple subalgebra.

Theorem 9 (Ado).  Let $\mathfrak{n}$ be the largest nilpotent ideal in $\mathfrak{g}.$  Then $\mathfrak{g}$ admits a finite dimensional faithful representation $\rho$ such that every element of $\rho(\mathfrak{n})$ is nilpotent.

[1]  Bourbaki, Nicholas.  Lie Groups and Lie Algebras, Chapters 1-3.  Springer-Verlag.  1971.