Enveloping Algebras and The Poincaré-Birkhoff-Witt Theorem

4 Pages·2007·0.095 MB·English

by  Rahbar Virk

Enveloping Algebras and The Poincar´e-Birkhoff-Witt Theorem
Rahbar Virk
Department of Mathematics
University of Wisconsin
Madison, WI 53706, USA
[email protected]
A Lie algebra g is not an algebra in the sense that the Lie bracket is not associative. We
would like to find an associative algebra U(g) such that the modules for g are the same as those
for U(g).
Let g be a Lie algebra and let T be the tensor algebra of the vector space g. Recall that
T = T0⊕T1⊕···⊕Tn⊕··· ,
where Tn = g⊗g⊗···g (n times); in particular T0 = k.1 and T1 = g; the product in T is
simply tensor multiplication. Let J be the two sided ideal of T generated by the tensors
x⊗y−[x,y],
where x,y ∈ g. The associative algebra T/J is called the enveloping algebra (or sometimes the
universal enveloping algebra) of g and is denoted by U(g). The composite mapping σ of the
canonical mappings g → T → U(g) is termed the canonical mapping of g into U(g); observe
that for all x,y ∈ g we have that
σ(x)σ(y)−σ(y)σ(x) = σ([x,y])
We denote the canonical image in U(g) of T0+T1+···+Tq by U (g).
q
Lemma 1.1. Let σ be the canonical mapping of g into U(g), let A be an algebra with unity, and
let τ be a linear mapping of g into A such that
τ(x)τ(y)−τ(y)τ(x) = τ([x,y])
for all x,y ∈ g. There exists one and only one homomorphism τ0 of U(g) into A such that
τ0(1) = 1 and τ0◦σ = τ.
Proof. Note that U(g) is generated by 1 and σ(g) so τ0 must be unique. Now let ϕ be the unique
homomorphism of T into A that extends τ and such that ϕ(1) = 1. For x,y ∈ g, we have
ϕ(x⊗y−y⊗x−[x,y]) = τ(x)τ(y)−τ(y)τ(x)−τ([x,y]) = 0,
hence ϕ(J) = 0 and thus ϕ gives us a homomorphism τ0 of U(g) into A such that τ0(1) = 1 and
τ0◦σ = τ.
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Lemma 1.2. Let a ,...,a ∈ g, σ the canonical mapping of g into U(g), and π be a permutation
1 p
of {1,...,p}. Then
σ(a )···σ(a )−σ(a )···σ(a ) ∈ U (g).
1 p π(1) π(p) p−1
Proof. It suffices to prove the statement when π is the transposition of j and j+1. But observe
that
σ(a )σ(a )−σ(a )σ(a ) = σ([a ,a ]).
j j+1 j+1 j j j+1
The lemma now follows.
We now assume g to a be finite dimensional Lie algebra and fix a basis (x ,...,x ) for g.
1 n
We denote the canonical image of x in U(g) by y . For every finite sequence I = (i ,...,i ) of
i i 1 p
integers between 1 and n, we set y = y y ...y .
I i1 i2 ip
Lemma 1.3. The y , for all increasing sequences I of length ≤ p, generate the vector space
I
U (g).
p
Proof. Clearly the vector space U (g) is generated by y , for all sequences I of length ≤ p. But
p I
now the required statement follows by applying the previous lemma.
Let P be the algebra k[z ,...,z ] of polynomials in n indeterminates z ,...,z . For every
1 n 1 n
i ∈ Z , let P be the set of elements of P of degree ≤ i. If I = (i ,...,i ) is a sequence of
≥0 i 1 p
integers between 1 and n, we set z = z z ···z .
I i1 i2 ip
Lemma 1.4. For every integer p ≥ 0, there exists a unique linear mapping f of the vector
p
space g⊗P into P which satisfies the following conditions:
p
(A ) f (x ⊗z ) = z z for i ≤ I, z ∈ P ;
p p i I i I I p
(B ) f (x ⊗z )−z z ∈ P for z ∈ P , q ≤ p;
p p i I i I q I q
(C ) f (x ⊗f (x ⊗z )) = f (x ⊗f (x ⊗z ))+f ([x ,x ]⊗z ) for z ∈ P . [The terms
p p i p j J p j p i J p i j J J p−1
in (C ) are meaningful by virtue of (B )].
p p
Moreover, the restriction of f to g⊗P is f .
p p−1 p−1
Proof. For p = 0 the mapping given by f (x ⊗1) = z ⊗1, satisfies the conditions (A ),(B )
0 i i 0 0
and (C ), furthemore it is clear that the condition (A ) forces this to be our linear mapping.
0 0
Proceeding by induction, assume the existence and uniqueness of f . If f exists then f
p−1 p p
restricted to g⊗P satisfies (A ),(B ),(C ) and is hence equal to f . Thus, to prove
p−1 p−1 p−1 p−1 p−1
our claim it suffices to show that f has a unique linear extension f to g⊗P which satisfies
p−1 p p
(A ),(B ),(C ). Note that P is generated by z for an increasing sequence I of p elements.
p p p p I
Thus, we must define f (x ⊗z ) for such a sequence I. If i ≤ I, then the choice is dictated by
p i I
(A ). Otherwise we can write I as (j,J) where j < i and j ≤ J. Then we must have that
p
f (x ⊗z ) = f (x ⊗f (z ⊗z )) from (A )
p i I p i p−1 j J p−1
= f (x ⊗f (x ⊗z ))+f ([x ,x ]⊗z ) from (C ).
p j p−1 i J p−1 i j J p
Now f (x ⊗z ) = z z +w, with w ∈ P (from (B )). Hence
p−1 i J i J p−1 p−1
f (x ⊗f (x ⊗z )) = z z z +f (x ⊗w) from (A )
p j p−1 i J j i J p−1 j p
= z z +f (x ⊗w).
j I p−1 j
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The above defines a unique linear extension f of f to g ⊗ P , and by construction this
p p−1 p
extension satisfies (A ) and (B ). All that remains to show is that f when defined this way
p p p
satisfies (C ).
p
Observe that if j ≤ i and j ≤ J then (C ) is satisfied by construction. Since [x ,x ] = −[x ,x ],
p j i i j
it is also satisfied if i < j and i ≤ J. Since (C ) is trivially satisfied if i = j, we see that (C )
p p
is satisfied if i ≤ J or j ≤ J. Otherwise, J = (k,K), where k ≤ K, k < i and k < j. For the
sake of brevity we will write f (x⊗z) = xz for x ∈ g and z ∈ P . Then from the induction
p p
hypothesis we have that
x z = x (x z ) = x (x z )+[x ,x ]z (∗)
i J j k K k j K j k K
Now x z is of the form z z +w, where w ∈ P . As k ≤ K and k < j we can apply (C )
j K j K p−2 p
to x (x (z z )), and to x (x w) from the induction hypothesis and hence also to x (x (x z ));
i k j K i k i k j K
using (∗) this then gives us that
x (x z ) = x (x (x z ))+[x ,x ](x z )+[x ,x ](x z )+[x ,[x ,x ]]z .
i j J k i j K i k j K j k i K i j k k
Interchanging i and j, this gives us that
x (x z )−x (x z ) = x (x (x z )−x (x z )+[x ,[x ,x ]]z −[x ,[x ,x ]]z
i j J j i J k i j K j i K i j k K j i k K
= x ([x ,x ]z )+(x ,[x ,x ]]z +[x ,[x ,x ]]z
k i j K i j k K j k i K
= [x ,x ]x z +[x ,[x ,x ]]z +[x ,[x ,x ]]z +[x ,[x ,x ]]z
i j k K k i j K i j k K j k i K
= [x ,x ]x z
i j k K
= [x ,x ]z ,
i j J
as required.
Lemma 1.5. The y , for every increasing sequence I, form a basis for the vector space U(g).
I
Proof. By the previous lemma (whose notation we will also use), there exists a bilinear mapping
f of g×P into P such that f(x ,z ) = z z for i ≤ I and
i I i I
f(x ,f(x ,z )) = f(x ,f(x ,z ))+f([x ,x ],z ),
i i J j i J i j J
for all i,j,J. Thus, we have a representation % of g in P such that %(x )z = z z for i ≤ I.
i I i I
From lemma 1.1 there exists a homomorphism ϕ of U(g into End((P) such that ϕ(y )z = z z
i I i I
for i ≤ I. We thus obtain that, if i ≤ i ≤ ··· ≤ i , then
1 2 p
ϕ(y y ···y )·1 = z z ···z .
i1 i2 ip i1 i2 ip
Thus, the y are linearly independent, for I increasing. From lemma 1.3 they generate U(g) and
I
hence form a basis.
Proposition 1.6. The canonical mapping of g into U(g) is injective.
Proof. This is immediate from the previous lemma.
We thus have that g is embedded in U(g), so we will identify every element of g with its
canonical image in U(g)
Theorem 1.7 (Poincare´e-Birkhoff-Witt). Let (x ,...,x ) be a basis for the vector space g.
1 n
Then the xλ11xλ22···xλnn, where λ1,...,λn ∈ N, form a basis for U(g).
3
Proof. This is again immediate from the previous lemma.
Taking the Poincar´e-Birkhoff-Witt Theorem into account Lemma 1.1 can be restated as
Lemma 1.8. Let A be an algebra with unity, τ a linear mapping of g into A such that τ(x)τ(y)−
τ(y)τ(x) = τ([x,y]) for all x,y ∈ g. Then τ can be uniquely extended to a homomorphism of
U(g) into A which transforms 1 into 1.
Corollary 1.9. Let V be a vector space, and R and R0 the sets of representations of g and U(g)
in V respectively. For all % ∈ R, there exists one and only one %0 ∈ R0 which extends %, and the
mapping % → %0 is a bijection of R onto R0.
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