# Lie Groups/Algebras, Notes 1: Lie Algebras

Definition 1.  A Lie algebra $\mathfrak{g}$ is an $R$-algebra such that $xy+yx=0$ and $x(yz)+z(xy)+y(zx)=0.$

Hence $x^2=-x^2\Rightarrow 2x^2=0\Rightarrow x^2=0$ provided $R$ is not a ring of characteristic $2.$  Also, if $A$ is an $R$-algebra, then we can define a Lie algebra $L(A)$ whose product is defined by

$[x,y]=xy-yx.$

Hence in particular, $L(\mathfrak{g})$ simply doubles the product of $\mathfrak{g}.$  Recall that a derivation $d$ on an algebra satisfies

$d(ax+by)=ad(x)+bd(y)$

and

$d(xy)=d(x)y+xd(y).$

For $x\in\mathfrak{g},$ let us define the adjoint of $x$ as $\mbox{adj}_x\in\mbox{End}\mathfrak{g}$ defined by

$\mbox{adj}_xy=xy.$

Proposition 2.  For $x\in\mathfrak{g},$ $\mbox{adj}_x$ is a derivation on $\mathfrak{g}.$

Proof.  By tensorial properties of algebras we have

$\mbox{adj}_x(ay+bz)=x(ay+bz)=x(ay)+x(bz)=a(xy)+b(xz)=a\mbox{adj}_xy+b\mbox{adj}_xz.$

And

$\mbox{adj}_x(yz)=x(yz)=-z(xy)-y(zx)=(xy)z+y(xz)=(\mbox{adj}_xy)z+y(\mbox{adj}_xz).$

Proposition 3.  Let $\mathfrak{g}$ be a commutative Lie algebra over a ring $R$ with $\mbox{char}(R)\neq 2.$  Then $xy=0.$  If $xy=0,$ then $\mathfrak{g}$ is commutative, regardless of the ring.

Proof.  If $\mathfrak{g}$ is commutative, then $xy=yx=-xy\Rightarrow 2xy=0\Rightarrow xy=0.$  The second claim is trivial.

Let $I$ be a left ideal in $\mathfrak{g},$ $x\in I,$ and $y\in\mathfrak{g}.$  Then $yx=-xy\in I$ and hence $xy\in I.$  So $I$ is also a right ideal.  Hence we simply consider ideals of Lie algebras without reference to leftness or rightness.

Definition 4.  Let $I$ be a submodule of $\mathfrak{g}$ such that $dI\leq I$ for all derivations $d$ on $\mathfrak{g},$ then $I$ is called a characteristic ideal of $\mathfrak{g}.$

It’s easy to see that a characteristic ideal is actually an ideal in $\mathfrak{g}$ since we have

$\mbox{adj}_xI=xI\leq I$

for all $x\in\mathfrak{g}.$

Proposition 5.  Let $\mathfrak{a}$ be an ideal of $\mathfrak{g}$ and $\mathfrak{b}$ be a characteristic ideal of $\mathfrak{a},$ then $\mathfrak{b}$ is an ideal of $\mathfrak{g}.$

Proof.  If $d$ is a derivation on $\mathfrak{a},$ then $d\mathfrak{b}\leq\mathfrak{b}$ since $\mathfrak{b}$ is a characteristic ideal.  But multiplication by $x\in\mathfrak{g}$ is a derivation on $\mathfrak{a}.$  Hence $bx\in\mathfrak{b}$ for $b\in\mathfrak{b}.$

For two ideals $\mathfrak{a},\mathfrak{b}\subseteq R,$ we have that $\mathfrak{a}+\mathfrak{b}$ and $\mathfrak{a}\cap\mathfrak{b}$ are ideals in $R$ and that $\mathfrak{a}\cup\mathfrak{b}\subseteq\mathfrak{a}+\mathfrak{b}$ and $\mathfrak{a}\mathfrak{b}\subseteq\mathfrak{a}\cap\mathfrak{b}.$  In the case of an $R$-algebra $A$, we will let $\mathfrak{a}\mathfrak{b}$ denote the submodule generated by elements $ab.$  This is in fact an ideal in a Lie algebra for if $d$ is a derivation, then

$d(ab)=d(a)b+ad(b)\in\mathfrak{a}\mathfrak{b}.$

Hence it is stable under adjoint action.

Definition 6.  The ideal $D^0\mathfrak{g}=\mathfrak{g}\mathfrak{g}$ is called the derived ideal of $\mathfrak{g}.$  Inductively we can define

$D^{n+1}\mathfrak{g}=D^n\mathfrak{g}D^n\mathfrak{g}.$

The sequence $\{D^n\mathfrak{g}\}$ is called the derived series of $\mathfrak{g}.$  Let us define $C_1\mathfrak{g}=\mathfrak{g}$ and

$C_{n+1}\mathfrak{g}=\mathfrak{g}C_n\mathfrak{g}.$

The sequence $\{C_n\mathfrak{g}\}$ is called the lower central series of $\mathfrak{g}.$

Definition 7.  Let $\mathfrak{g}$ be a Lie algebra, $T(\mathfrak{g})$ be its tensor algebra (with $\mathfrak{g}$ interpreted as a module), and $I$ be the ideal generated by elements of the form $x\otimes y-y\otimes x-xy.$  Then $L^{-1}(\mathfrak{g})=T/I$ is called the enveloping algebra of $\mathfrak{g}.$

Theorem 8.  The enveloping algebra $L^{-1}(\mathfrak{g})$ satisfies the universal property that if $f:\mathfrak{g}\to L(A)$ is an algebra homomorphism for some associative unital algebra $A$ and $\iota:\mathfrak{g}\to L(L^{-1}(\mathfrak{g}))$ is the canonical inclusion, then there exists a unique morphism $g:L^{-1}(\mathfrak{g})\to A$ and $\ell(g):L(L^{-1}(\mathfrak{g}))\to L(A)$ such that $f=\ell(g)\circ\iota.$

Proposition 9.  If $\mathfrak{g}_1$ and $\mathfrak{g}_2$ are Lie algebras, then $L^{-1}(\mathfrak{g}_1\times\mathfrak{g}_2)=L^{-1}(\mathfrak{g}_1)\otimes L^{-1}(\mathfrak{g}_2).$

Let us define $U_n(\mathfrak{g})=T^n(\mathfrak{g})/I$ with $I$ defined as above, $G_n=U_n/U_{n-1},$ and $G=\oplus G_n.$  It follows that $G$ is a filtered algebra, which we will call the associated filtered algebra to $L^{-1}(\mathfrak{g}).$

Theorem 10 (Poincare-Birkhoff-Witt)  Let $G$ be the associated filtered algebra to $L^{-1}(\mathfrak{g})$ and $S(\mathfrak{g})$ be the symmetric algebra of $\mathfrak{g}$ as a module.  Then if $\mathfrak{g}$ is free, $S(\mathfrak{g})=G.$

It follows that if $\mathfrak{g}$ is free, then the canonical homomorphism $f:\mathfrak{g}\to L^{=1}(\mathfrak{g})$ is injective.

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