Revised Section 9.4. Brian C. Hall. December 29, 2015

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1 Revised Section 9.4 Brian C. Hall December 29, 2015 There were several parts of Section 9.4 (Proof of the PBW Theorem) that were slightly unclear. I have rewritten the section in its entirety. It is notationally convenient to write the elements of the claimed basis for U(g) as i(x j1 )i(x j2 ) i(x jn ), (1) with j 1 j 2 j N, where we interpret the above expression as 1 if N = 0. The easy part of the PBW theorem is to show that these elements span U(g). The proof of this claim is essentially the same as the proof of the reordering lemma (Lemma 6.12). Every element of the tensor algebra T (g), and hence also of the universal enveloping algebra U(g), is a linear combination of products of Lie algebra elements. Expanding each Lie algebra element in our basis shows that every element of U(g) is a linear combination of products of basis elements, but not (so far) necessarily in nondecreasing order. But using the relation XY Y X = [X, Y ], we can reorder any product of basis elements into the desired order, at the expense of introducing several terms that are products of one fewer basis elements. These smaller products can then, inductively, be rewritten as a linear combination of terms that are in the correct order. It may seem obvious that elements of the form (1) are linearly independent. Note, however, that any proof of independence of these elements must make use of the Jacobi identity. After all, if g is a vector space with any skewsymmetric, bilinear bracket operation, we can still construct a universal enveloping algebra by the construction in Section 9.3, and the elements of the form (1) will still span this enveloping algebra. If, however, the bracket does not satisfy the Jacobi identity, the elements in (1) will not be linearly independent. If they were, then, in particular, the map i : g U(g) would be injective. We could then identify g with its image under i, which means that the bracket on g would be given by [X, Y ] = XY Y X, where XY and Y X are computed in the associative algebra U(g). But any bracket of this form does satisfy the Jacobi identity. We now proceed with the proof of the independence of the elements in (1). The reader is encouraged to note the role of the Jacobi identity in our proof. Let D be any vector space having a basis {v (j1,,j N )}. 1

2 indexed by all nondecreasing tuples (j 1,..., j N ). We wish to construct a linear map γ : U(g) D with the property that γ(i(x j1 )i(x j2 ) i(x jn )) = v (j1,,j N ) for each nondecreasing tuple (j 1,..., j N ). Since the elements v (j1,,j N ) are, by construction, linearly independent, if such a map γ exists, the elements X j1 X j2 X jn must be linearly independent as well. (Any linear relation among the X j1 X j2 X jn s would translate under γ into a linear relation among the v (j1,,j N ) s.) Instead of directly constructing γ, we will construct a linear map δ : T (g) D with the properties (1) that δ(x j1 X j2 X jn ) = v (j1,,j N ) (2) for all nondecreasing tuples (j 1,..., j N ), and (2) that δ is zero on the two-sided ideal J. Since δ is zero on J, it gives rise to a map γ of U(g) := T (g)/j into D with the analogous property. To keep our notation compact, we will now omit the tensor product symbol for multiplication in T (g). Since all computations in the remainder of the section are in T (g), there will be no confusion. Suppose we can construct δ in such a way that (2) holds for nondecreasing tuples and that for all tuples (j 1,..., j N ), we have δ(x j1 X jk X jk+1 X jn ) = δ(x j1 X jk+1 X jk X jn ) + δ(x j1 [X jk, X jk+1 ] X jn ). (3) Then δ will indeed by zero on J. After all, J is spanned by elements of the form α(xy Y X [X, Y ])β. After moving all terms in (3) to the other side and taking linear combinations, we can see that δ will be zero on every such element. Define the degree of a monomial X j1 X j2 X jn to be the number N and the index of the monomial to be the number of pairs l < k for which j l > j k. We will construct δ inductively, first on the degree of the monomial, and then on the index of the monomial for a given degree, verifying (3) as we proceed. If N = 0, we set δ(1) = v (0,...,0) and if N = 1, we set δ(x j ) = v (0,...,1,...,0). In both cases, (2) holds by construction and (3) holds vacuously. For a fixed N 2 and p, we now assume that δ has been defined on the span of all monomials of degree less than N, and also on all monomials of degree N and index less than p. If p = 0, this means simply that δ has been defined on the span of all monomials of degree less than N. Our induction hypothesis is that δ, as defined up to this point, satisfies (3) whenever all the terms in (3) have been defined. That is to say, we assume (3) holds whenever both the monomials on the left-hand side of (3) have degree less than N or degree N and index less than p. (Under these assumptions, the argument of δ on the right-hand side of 2

3 (3) will be a linear combination of monomials of degree less than N, so that the left-hand side of (3) has been defined.) We now need to show that we can extend the definition of δ to monomials of degree N and index p in such a way that (3) continues to hold. If p = 0, the new monomials we have to consider are the nondecreasing ones, in which case (2) requires us to set δ(x j1 X jn ) = v (j1,,j N ). Now, the only way both the monomials on the left-hand side of (3) can have degree N and index zero is if j k+1 = j k, in which case, both sides of (3) will be zero. It remains, then, to consider the case p > 0. Let us consider an example that illustrates the most important part of the argument. Suppose N = 3 and p = 3, meaning that we have defined a map δ satisfying (3) on all monomials of degree less than 3 and all monomials of degree 3 and index less than 3. We now attempt to define δ on monomials of degree 3 and index 3 and verify that (3) still holds. A representative such monomial would be X 3 X 2 X 1. Since we want (3) to hold, we may attempt to use (3) as our definition of δ(x 3 X 2 X 1 ). But this strategy gives two possible ways of defining δ(x 3 X 2 X 1 ), either or δ(x 3 X 2 X 1 ) = δ(x 2 X 3 X 1 ) + δ([x 3, X 2 ]X 1 ) (4) δ(x 3 X 2 X 1 ) = δ(x 3 X 1 X 2 ) + δ(x 3 [X 2, X 1 ]). (5) Note that the monomials on the right-hand sides of (4) and (5) have degree 2 or degree 3 and index 2, so that δ has already been defined on these monomials. We now verify that these two expression for δ(x 3 X 2 X 1 ) agree. Since δ has already been defined for the terms on the right-hand side of (4), we may apply our induction hypothesis to these terms. Using induction twice, we may simplify the right-hand side of (4) until we obtain a term in the correct PBW order of X 1 X 2 X 3, plus commutator terms: δ(x 2 X 3 X 1 ) + δ([x 3, X 2 ]X 1 ) = δ(x 2 X 1 X 3 ) + δ(x 2 [X 3, X 1 ]) + δ([x 3, X 2 ]X 1 ) = δ(x 1 X 2 X 3 ) + δ([x 2, X 1 ]X 3 ) + δ(x 2 [X 3, X 1 ]) + δ([x 3, X 2 ]X 1 ). Similarly, the other candidate (5) for δ(x 3 X 2 X 1 ) may be computed by our induction hypothesis as δ(x 3 X 1 X 2 ) + δ(x 3 [X 2, X 1 ]) = δ(x 1 X 3 X 2 ) + δ([x 3, X 1 ]X 2 ) + δ(x 3 [X 2, X 1 ]) = δ(x 1 X 2 X 3 ) + δ(x 1 [X 3, X 2 ]) + δ([x 3, X 1 ]X 2 ) + δ(x 3 [X 2, X 1 ]). 3

4 Subtracting the two expressions gives the quantity δ([x 2, X 1 ]X 3 ) δ(x 3 [X 2, X 1 ]) + δ(x 2 [X 3, X 1 ]) δ([x 3, X 1 ]X 2 ) + δ([x 3, X 2 ]X 1 ) δ(x 1 [X 3, X 2 ]). Since all terms are of degree 2, we can use our induction hypothesis to reduce this quantity to δ([[x 2, X 1 ], X 3 ] + [X 2, [X 3, X 1 ]] + [[X 3, X 2 ], X 1 ]) = δ([x 3, [X 1, X 2 ]] + [X 2, [X 3, X 1 ]] + [X 1, [X 2, X 3 ]]) = 0, by the Jacobi identity. Thus, the two apparently different definitions of δ(x 3 X 2 X 1 ) in (4) and (5) agree. Using this result, it should be apparent that (3) holds when we extend the domain of definition of δ to include the monomial X 3 X 2 X 1 of degree 3 and index 3. We now proceed with the general induction step in the construction of δ, meaning that we assume δ has been constructed on monomials of degree less than N and on monomials of degree N and index less than p, in such a way that (3) holds whenever both monomials on the left-hand side of (3) are in the current domain of definition of δ. Since we have already addressed the p = 0 case, we assume p > 0. We now consider a monomial X j1 X j2 X jn of index p 1. Since the index of the monomial is at least 1, the monomial is not weakly increasing and there must be some j k with j k > j k+1. Pick such a k and define δ on the monomial by δ(x j1 X jk X jk+1 X jn ) = δ(x j1 X jk+1 X jk X jn ) + δ(x j1 [X jk, X jk+1 ] X jn ). (6) Note that the first term on the right-hand side of (6) has index p 1 and the second term on the right-hand side has degree N 1, which means that both of these terms have been previously defined. The crux of the matter is to show that the value of δ on a monomial of index p is independent of the choice of k in (6). Suppose, then, that there is some l < k such that j l > j l+1 and j k > j k+1. We now proceed to check that the value of the right-hand side of (6) is unchanged if we replace k by l. Case 1: l k 2. In this case, the numbers l, l + 1, k, k + 1 are all distinct. Let us consider the two apparently different ways of calculating δ. If we use l, then we have δ( X l X l+1 X k X k+1 ) = δ( X l+1 X l X k X k+1 ) + δ( [X l, X l+1 ] X k X k+1 ). (7) Now, the second term on the right-hand side of (7) has degree N 1. The first term has index p 1, and if we reverse X k and X k+1 we obtain a term of index 4

5 p 2. Thus, we can apply our induction hypothesis to reverse the order of X k and X k+1 in both terms on the right-hand side of (7), giving δ( X l X l+1 X k X k+1 ) = δ( X l+1 X l X k+1 X k ) + δ( X l+1 X l [X k, X k+1 ] ) + δ( [X l, X l+1 ] X k+1 X k ) + δ( [X l, X l+1 ] [X k, X k+1 ] ). (8) (Note that on the right-hand side of (8), all terms have both X l and X l+1 and X k and X k+1 back in their correct PBW order, with X l+1 to the left of X l and X k+1 to the left of X k.) Since the right-hand side of (8) is symmetric in k and l, we would get the same result if we started with k instead of l. Case 2: l = k 1. In this case, the indices j l, j l+1 = j k, and j l+2 = j k+1 are in the completely wrong order, j l > j l+1 > j l+2. Let us use the notation X = X l, Y = X l+1, and Z = X l+2. We wish to show that the value of δ( XY Z ) is the same whether we use l (that is, interchanging X and Y ) or we use k (that is, interchanging Y and Z). But the argument is then precisely the same as in the special case of δ(x 3 X 2 X 1 ) considered at the beginning of our proof, with some extra factors, indicated by, tagging along for the ride. Once we have verified that the value of δ is independent of the choice of k in (6), it should be clear that (3) holds, since we have used (3) with respect to a given pair of indices as our definition of δ. We have, therefore, completed the construction of δ and the proof of the PBW theorem. 5