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


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




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


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

Proof.  By tensorial properties of algebras we have




 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


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


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


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.


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