AP07_2-3.vp 1 Introduction Let g be a finite dimensional complex semisimple Lie al- gebra and U g U( ) � its corresponding enveloping algebra. If we want to study all two-sided ideals of the associative algebra U (see [1] chapters 6–10), i.e. to find all the vector spaces I U� which satisfy for all a b U aIb I, :� � we can easily see that such a two-sided ideal is invariant with respect to adjoint action of U: if we take adjoint action � : ( )g L U� , i.e. �( ) [ , ]x a x a� for x g� and a U� and extend this definition to all U by � �� �� � �( ) ( ) ( ) , , [ , ]u x x x x an n� �1 1� � � for all u x x x gn i� �1 � , we see that �( )U I I� . It is well known that such adjoint action � is completely re- ducible for g semisimple: it is possible to cut the enveloping algebra U into pieces U k� such that U U k k � � � 0 � and each piece U k� is an invariant subspace (corresponding to an irreducible representation (component) of �) generated by its highest weight vector �k: �( )U Uk k � �� Now if we take any ideal I U� and find any highest weight vector �k I� , we automatically know that �( )U U Ik k � �� � , i.e. we know a large set of vectors which must be contained also in the ideal I. This can help considerably in the classification of all ideals of U. The main aim of this paper is to give an explicit decompo- sition � � k U k0 � , i.e. to give a list of highest weight vectors �k, k � 0 1, ,� for the simplest cases of Lie algebras and to show that even in more complicated examples the decomposition does not look too weird and is relatively easyli obtained by a generally described procedure. 2 The simplest example – algebra sl(2) Let us have algebra sl(2) which is a complex linear span of three vectors E E H E E12 21 1 11 22, , � � , where by Eij we denote the matrix having 1 in position (i, j) and 0s elsewhere. If we compute the commutation relations of these elements we obtain [ , ] , [ , ] , [ , ]E E H H E E H E E12 21 1 1 12 12 1 21 212 2� � � � . The enveloping algebra U U sl� ( ( ))2 fulfils the Poincare- -Birkhoff-Witt theorem and consists of all complex linear combinations of monomials E E H12 21 1 0 � � � � � �, , , � . It has a natural filtration Un, n � 0 given by the degree n of the elements in Un. It is easily seen that the adjoint action has Un as its invariant subspace (by applying a commutator we cannot obtain an element of higher degree). Therefore it is completely reducible on each Un, i.e. we can see Un as a di- rect sum of invariant subspaces generated by certain highest weight vectors. The highest weight vector v of weight m satisfies the relations [ , ] , [ , ]E v H v mv12 10 2� � . From this we can directly find that all highest weight vec- tors of small degree: degree 0 1 degree 1 E12 degree 2 E C12 2 1, degree 3 E C E12 3 1 12, … … Here C H E E H1 1 2 12 21 14 2� � is a Casimir element from the center of U. By commuting these vectors we see that © Czech Technical University Publishing House http://ctn.cvut.cz/ap/ 25 Acta Polytechnica Vol. 47 No. 2–3/2007 Structure of the Enveloping Algebras Č. Burdík, O. Navrátil, S. Pošta The adjoint representations of several small dimensional Lie algebras on their universal enveloping algebras are explicitly decomposed. It is shown that commutants of raising operators are generated as polynomials in several basic elements. The explicit form of these elements is given and the general method for obtaining these elements is described. Keywords: adjoint representation, enveloping algebra, Lie algebra. U U U U E U U U E U C U 0 1 12 0 2 12 2 1 1 1� � � � � � � � � � ( ) , ( ) , ( ) ( ) , � From this we can claim that for any n � 0 U U C En k m k m n k m� � � , ( ) 0 2 1 12� (1) The proof of this general formula is based on a dimen- sional check: First we see that the sum � �( ) ( )U v U v1 2 , where v1, v2 are highest weight vectors, is direct if and only if v1, v2 are linear independent. Thus in our example we must check that the vectors C Ek m1 12 are linear independent for different k, m. But this is easily seen from the definition, since C E E Hk m m k1 12 12 1 2� lower terms where “lower terms“ contain monomials with H k1 2 1� and lower. Next it is well known from representation theory that the dimension of the irreducible representation with highest weight m is 2 1m . On the other hand the dimension of Un is also easy to de- termine, because it is well known that it is isomorphic (as a vector space) to the direct sum � �k n kS g 0 ( ) of the k-th symmetric powers of Lie algebra g. We can construct the symmetric alge- bra S(g) from the tensor algebra T(g) by taking the quotient algebra of T(g) by the ideal generated by all differences of products v w w v� � � for v w g, � ; then there is a direct sum decomposition of S(g) as a graded algebra into the summands Sk(g) which consist of the linear span of the monomials in vec- tors of g of degree k. The symmetric algebra S(g) is in effect the same as the polynomial ring in the indeterminates that are basis vectors for g. Therefore we see that dim ( ) dim S g g k k k � �� � � � � � 1 and from this we have dim dim dim dim U g k k n g gn k n � �� � � � � � � � � � � � � � � 1 0 , thus in our example dim U n n � � � � � � � 3 3 . Now it is sufficent to prove that the dimensions of two spaces on both sides (1) are equal, i.e. that n m k m k m n � � � � � � � � � 3 3 2 1 0 2 ( ) , . We have ( ) ( ) ( ) , 2 1 2 1 2 1 0 2 0 2 0 0 m m m k m k m n m n k k m n � � � � � � � � �� � � � � �� � � � � � � � � � � 2 0 2 2 2 1 1 6 1 2 3 3 3 k k k n n k n n n n ( ) ( )( (( ) � � � � 0 2 n . 3 Moving to the commutative case There is an alternative approach to obtain this decomposi- tion – move to the commutative case: It is a well-known fact (see [2]) that the mapping e c x x i ij k k jj k n � � � � � , 1 is a representation of an n-dimensional Lie algebra with basis ei, i n� 1, ,� on a space of polynomials of the variables xi, i n� 1, ,� ; if we view S(g) as the polynomial ring, this repre- sentation is equivalent as a representation of the enveloping algebra to the adjoint representation via canonical isomorph- ism U g S g( ) ( )� . Therefore if we return to our example and represent the basis vectors of the Lie algebra sl(2) by operators E e e h E e e h H e e 12 21 12 1 21 12 21 1 1 12 12 2 2 2 2 � � � � � � � � � � � � � � � � e e21 21 � � acting on the space of polynomials in three variables e12, e21, h1, we can find all highest weight vectors by solving this system of differential equations: h f e e f h e f e e f e mf 1 21 12 1 12 12 21 21 2 0 2 2 2 � � � � � � � � � � � � , . This system can be easily solved, and the general solution is f e e h e F cm( , , ) ( )12 21 1 12 1� , where F is any differentiable function of the variable c h e e1 1 2 12 214� . We see that when F x x k( ) � , the solution is polynomial and we can transfer back to the enveloping algebra using the canonical isomorphism S g U g( ) ( )� : if a1, …, a gk � , this isomorphism sends a a k A Ak k1 1 1 � �� �! ( ) ( )� � � 26 © Czech Technical University Publishing House http://ctn.cvut.cz/ap/ Acta Polytechnica Vol. 47 No. 2–3/2007 (we sum over all permutations of the indices 1, …, k). In our example c H E E E E H E E H C 1 1 2 12 21 21 12 1 2 12 21 1 1 4 1 2 1 2 4 2 � � � � � � � � � � . Doing the dimension check as before the we see that de- composition is complete. 4 The algebras sl(3) and gl(3) The case of algebra sl(3) is more complicated. The decom- position of this case is done in [3]. Let us show where the com- plications in the decomposition arise, using the similar case of Lie algebra gl(3) with the basis Eij, i j, , ,� 1 2 3 and commuta- tion relations [ , ]E E E Eij kl ij jk kj il� � . We turn directly into the commutative case. First we set up, as before, the system of differential equations for the highest weight vectors of weight (N1, N2, N3): E x f x x f x f x x x f 12 13 23 12 11 22 11 22 � � � � � �� � � �� � � � � � � � � ( ) � � � � � � � � � x x f x E x f x x f x f x 21 32 31 23 13 12 23 22 33 0� � � � � � � � �� � � �� � � � � � ( ) ,x x f x x f x E x f x 22 33 32 21 31 13 13 11 0 � � � � � � �f x x f x x f x x x f x � � � � � � � 33 23 21 12 32 11 33 31 � � �� � � �� � � �( ) 0 11 12 12 13 13 21 21 31 31 1 , ,E x f x x f x x f x x f x N f� � � � � � � � � � � � E x f x x f x x f x x f x N f E 22 12 12 23 23 21 21 32 32 2� � � � � � � � � � � � , 33 13 13 23 23 32 32 31 31 3� � � �x f x x f x x f x x f x N f � � � � � � � � . From this immediately follows N N N1 2 3 0 � . If we define new variables x x1 13� , x x x x x x x x x3 12 2 23 12 13 33 22 13 2 32� � �( ) then any solution f can be written as f x x x Fij N N N( ) � � �1 3 2 3 2 , where the new unknown function F satisfies the following sys- tem of differential equations: x F x x F x F x x x F x13 23 12 11 22 11 22 21 � � � � � � � � � � � � �� � � �� �( ) � � � � � �� � � �� � x F x x F x x F x F x x 32 31 13 12 23 22 33 0 � � � � � � � � , ( 22 33 32 21 31 13 11 33 0� � � � � � �� � � �� x F x x F x x F x F x ) , � � � � � � � � � � � �x F x x F x x x F x23 21 12 32 11 33 31 0 � � � � � � ( ) , x F x x F x x F x x F x x F x 12 12 13 13 21 21 31 31 13 1 0 � � � � � � � � � � � � � , 3 23 23 32 32 31 31 0 � � �x F x x F x x F x � � � � � � . Solving this system, we obtain that the general solution of the system has the form: f x x x F x x c c cij N N N( ) , , ,� � � �� � � �� � � 1 3 2 1 1 2 3 2 3 2 , where x x xk k2 1 3� , c xkk1 � , c x xik ki2 � , c x x xik kl li3 � . (we omit the summation of repeated indices going over 1, 2, 3). We now encounter the first problem which arises in the process of decomposition. Although we know the general so- lution of this system of differential equations, an important question is how to obtain enough polynomial solutions in vari- ables xij from this general solution. This can be generally difficult. For example, the expression x x x x c x x c c x x c c 4 1 3 2 3 1 3 1 2 2 1 2 1 2 2 2 1 1 3 32 1 2 3 1 3 � � � � � � ( ) ( ) � � � � � � � � � � x x x x x x x x x 3 12 23 2 13 23 11 22 13 2 21( ) although rational in variables xj, cj is polynomial in variables xij! The second question which arises is, in some sense, the opposite kind of problem. We can generate too many polyno- mial solutions and their polynomial combinations will no lon- ger be linear independent (or, equivalently, the polynomial solutions will be functionally dependent). For example, in our case of algebra gl(3) we have x x x c x x c c x x c c x3 4 2 3 1 2 2 1 1 2 2 2 1 2 1 3 3 1 32 1 2 3 1 3 � � � � �( ) ( ) . This dependency signals that it is forbidden to have prod- uct x3x4 in the decomposition. Luckily, both above-mentioned problems are easily found out by a dimension check. If there are not enough solutions we will not have enough linear independent vectors to con- struct the decomposition, so that the number of linear inde- pendent vectors we generate is less than the dimension of the corresponding filtration part Un. In the opposite case, if the number is greater, it clearly indicates that there is some linear dependency between the considered vectors. © Czech Technical University Publishing House http://ctn.cvut.cz/ap/ 27 Acta Polytechnica Vol. 47 No. 2–3/2007 Summing up these facts, if we now anti-symmetrize xj, cj into elements Xj, Cj of the enveloping algebra U, i.e. x X E x X E E E E E E 1 1 13 2 2 12 23 13 11 33 13 1 2 � � � � �( ) etc. we can construct vectors C C C X X X Xk k k n n n n1 2 3 1 2 3 4 1 2 3 1 2 3 4 , k nj j, � 0, n n3 4 0� and compute their weights. The weight of the product of any elements is obtained by the sum of the weights of the corre- sponding elements; we have element weight X1 (1,0,�1) X2 (1,0,�1) X3 (2,�1,�1) X4 (1,1,�2) C1,2,3 (0,0,0) therefore weight ( ) ( C C C X X X X n n n n k k k n n n n 1 2 3 1 2 3 4 1 2 3 4 1 2 3 1 2 3 4 2� , , ).n n n n n n4 3 1 2 3 42� � � � � The degree of the general element is n k k k n n n n� 1 2 3 1 2 3 42 3 2 3 3 . Now we can perform a dimension check to see if our hy- pothesis is correct: from representation theory, the dimension of the representation with highest weight (N1, N2, N3) is d N N N N N N N N N( , , ) ( )( )( )1 2 3 1 2 1 3 2 3 1 2 1 2 1� � � � , hence the dimension of the representation generated by our general element is ~ ( , , , ) ( )( ) ( d n n n n n n n n n n n 1 2 3 4 1 2 3 1 2 4 1 1 2 3 1 3 1 2 2 � n n n2 3 43 3 2 ). It follows that we must prove that ~ ( , , , ) , , , , , , d n n n n k k k n n n n n n k k 1 2 3 4 0 0 2 1 2 3 1 2 3 4 3 3 1 2 � � 3 2 3 33 1 2 3 4 9 k n n n n n n n � � � � � � � � . The first look leads us of course to check if it is correct for small n: we make a table n sum on the left side n � � � � � � 9 9 1 10 10 2 55 55 3 220 220 4 715 715 … … from which we see that it should work. A general proof of the formula can be performed for example using generating functions: because n n n � � � � � � � � � � � � � � � � � � � � � 9 9 1 9 9 8 8 ( ) it is sufficient to show ~ ( , , , ) , , , , , , d n n n n k k k n n n n n n k k 1 2 3 4 0 0 2 1 2 3 1 2 3 4 3 4 1 2 � � 3 2 3 33 1 2 3 4 8 8 k n n n n n n � � � � � � � � � or, equivalently, ~ ( , , , ) , , , , , , d n n n n xn k k k n n n n n n k k 1 2 3 4 0 0 2 1 2 3 1 2 3 4 3 4 1 � � 2 3 1 2 3 43 2 3 3 0 0 8 8 � � � �� �� � � � � � � k n n n n n n n n n x . This is easily done with the help of the sum of the geomet- ric series, namely 1 1 0 � � � �x x n n which can be differentiated several times to get 1 1 2 0( )� � � �x n x n n , x x x n xn n ( ) ( ) � � � �11 3 2 0 , … etc. This way, after expanding the right hand side we get n x x n n � � � � � � � � � � 8 8 1 10 9( ) . 28 © Czech Technical University Publishing House http://ctn.cvut.cz/ap/ Acta Polytechnica Vol. 47 No. 2–3/2007 To remove the unwanted condition n n3 4 0� we can rewrite the inner sum as ~ ( , , , ) , , , , , , d n n n n k k k n n n n n n k k 1 2 3 4 0 0 2 1 2 3 1 2 3 4 3 4 1 2 � � 3 2 3 3 1 2 3 3 1 2 3 4 1 2 3 1 2 0 k n n n n n k k k n n d n n n � � � ~ ( , , , ) , , , , , , ~ ( , , , n n k k k n n n n n d n n n 3 4 1 2 3 1 2 3 4 0 2 3 2 3 3 1 2 40 � � � ) , , , , , ,k k k n n n n k k k n n n n n 1 2 3 1 2 3 4 1 2 3 1 2 3 4 0 2 3 2 3 3 � � � � � ~ ( , , , ) , , , , , , d n n k k k n n n n k k k n 1 2 0 2 3 0 0 1 2 3 1 2 3 4 1 2 3 1 � � 2 3 32 3 4n n n n © Czech Technical University Publishing House http://ctn.cvut.cz/ap/ 29 Acta Polytechnica Vol. 47 No. 2–3/2007 and sum each of the summands separately. For the first one we have ~ ( , , , ) , , , , , d n n n xn k k k n n n k k k n 1 2 3 0 2 3 2 0 1 2 3 1 2 3 1 2 3 1 � n n n n n n n n n n n 2 33 0 1 2 1 2 3 1 2 1 2 1 3 1 2 2 3 � � �� � ( )( )( n xk k k n n n k k k n n n 3 2 3 2 3 0 2 1 2 3 1 2 3 1 2 3 1 2 3 � � � ) , , , , , 10 5 1 1 1 2 9 5 x x x x � ( ) ( ) The second term is obviously equal to the first term. For the third sum we have and sum each of the summands separately. For the first one we have ~ ( , , , ) , , , , d n n xn k k k n n k k k n n n 1 2 0 2 3 2 0 0 1 2 3 1 2 1 2 3 1 2 � � �� � � n k k k n n k n n n n x 0 1 2 2 1 2 2 3 21 2 1 2 2 2 1 2 3 1 2( ) ( ) 1 2 3 1 2 3 0 4 3 2 8 5 6 16 6 1 1 1, , , , , ( ) ( )k k n n n x x x x x x � � � � . Now 2 10 5 1 1 1 6 16 6 1 1 1 2 9 5 4 3 2 8 x x x x x x x x x x � � � ( ) ( ) ( ) ( )5 9 1 1 � �( )x and the dimension check succeeds. 5 Lie algebra gl(4) Finally let us demonstrate how the procedure goes in the case of Lie algebra gl(4). For the vectors with highest weights (N1, N2, N3, N4) we obtain the following equations E x f x x f x x f x x f x x f 11 12 12 13 13 14 14 21 21 31� � � � � � � � � � � � �x x f x N f E x f x x f x x f x 31 41 41 1 22 21 21 23 23 24 2 � � � � � � � � � � � , 4 12 12 32 32 42 42 2 33 31 31 � � � � � x f x x f x x f x N f E x f x � � � � � � � � , x f x x f x x f x x f x x f x32 32 34 34 13 13 23 23 43 43 � � � � � � � � � � � � � � N f E x f x x f x x f x x f x x 3 44 41 41 42 42 43 43 14 14 2 , � � � � � � � � � � � 4 24 34 34 4 12 11 22 21 12 2 � � � � � � � � f x x f x N f E x x f x x f x � � � � , ( ) 2 13 23 14 24 12 11 32 31 42 4 � � �x f x x f x x f x x f x x f x � � � � � � � � � � 1 23 21 31 22 33 32 23 33 24 0� � � , ( )E x f x x x f x x f x x f� � � � � � � �x x f x x f x x f x E x f x 34 13 12 23 22 43 42 34 31 41 0� � � � � � � � � � � � � , x f x x x f x x f x x f x x32 42 33 44 43 34 44 14 13 24 � � � � � � � � � � � �( ) f x x f x E x x f x x f x x � � � � � � � 23 34 33 13 11 33 31 12 32 1 0� � � � , ( ) 3 33 14 34 13 11 23 21 43 41 0 � � � � � � � � � � f x x f x x f x x f x x f x � � � � , E x f x x x f x x f x x f x24 21 41 22 44 42 23 43 24 44 � � � � � � � � � � � ( ) x f x x f x x f x E x x f x 14 12 24 22 34 32 14 11 44 4 0 � � � � � � � � � � � � � , ( ) 1 12 42 13 43 14 44 14 11 24 2 � �x f x x f x x f x x f x x f x � � � � � � � � � � 1 34 31 0� �x f x � � . Now let us denote e xik ik ( )1 � , e x xik in nk ( )2 � , e x x xik in nm mk ( )3 � . Using these elements we define the following solutions of the system (summation of repeated indices going over 1… 4): c ekk1 1� ( ), c ekk2 2� ( ), c ekk3 3� ( ), c e Ekl lk4 3 1� ( ) ( ), The highest weights of the corresponding elements of the enveloping algebra are element weight Ck (0,0,0,0) Xk (1,0,0,�1) Yk (1,1,�1,�1) Zk (2,0,�1,�1) Zk � (1,1,0,�2) W (3,�1,�1,�1) W� (1,1,1,�3) T (2,1,�1,�2) The dimension of the representation with highest weight (N1, N2, N3, N4) is d N N N N N N N N N N � � � � � � � 1 12 1 2 3 1 2 1 2 1 3 1 4 2 3 2 4 ( )( )( ) ( )( )( ).N N3 4 1� Now we start as in the case of gl(3) to count the dimensions of the filtration subspaces to see the conditions which arise as the degree of the elements grows. Elements of degree 1 are element dimension C1 1 X1 15 total � 16 Elements of degree 2 are element dimension C1 2 1 C1X1 15 X1 2 84 C2 1 X2 15 Y1 20 total � 136 … etc. Analyzing degree 6, we encounter the first linear de- pendency which signals some polynomial relations (syzygies) between the generators. Namely we have Z Z X Y X Y C X X Y1 1 1 2 2 2 2 1 1 1 2 1 ( ) ( ) � � � , X Z X Z X Z1 3 2 2 3 1 0 ( ) ( ) ( ) � � X Z X Z X Z1 3 2 2 3 1 0 ( ) ( ) ( )� � �� � 8 4 2 2 03 2 1 2 2 1 4 1 2 2 4 2 2 1Y C C Y C C C C C Y � � � �( ) ( ) . Higher degrees give us many other relations of this type which lead to the system of conditions that must be fulfilled to keep the set of highest weight vectors linear independent. Going up to degree 12 finishes the analysis, and we no longer encounter new relations. Thus we are ready to formulate the hypothesis and after a successful dimension check we can state the final result, which is contained in the following theorem: Theorem The set of elements C C C X X X Y Y T Z Z k k k n n n m m p r 2 3 4 1 2 3 1 2 1 2 2 3 4 1 2 3 1 2 1 � ( ) (( ) ( )) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )r r s s s t tZ Z Z Z W W2 3 1 2 3 1 23 1 2 3 � � � � with conditions r n r s r s r s r p r t s n s r s r s p1 3 1 1 1 2 1 3 1 1 2 1 3 1 2 1 3 1 0� � � � � � � � � � r s r s r t r p s r s t r s s p s t2 2 2 3 2 2 2 2 3 2 1 3 3 2 1 1 0� � � � � � � � � , r t r p s t s p t p t p t t3 2 3 3 1 3 1 2 1 2 0� � � � � � � , r2 0 1� , , s2 0 1� , , r3 0 1� , , s 3 0 1� , , p � 0 1, form a desired decomposition of the enveloping algebra U( gl(4)). 6 Conclusion We have studied the structure of enveloping algebras U(g) where g sl gl gl� ( ), ( ), ( )2 3 4 and we have decomposed the adjoint representation into its irreducible components. We have found explicitly the highest weight vector in any such component. Our result can be useful for a further study of the tensor products of representations and ideals of enveloping 30 © Czech Technical University Publishing House http://ctn.cvut.cz/ap/ Acta Polytechnica Vol. 47 No. 2–3/2007 y e e e e 1 13 1 14 1 23 1 24 1� � � � � � � � � det ( ) ( ) ( ) ( ) , y e e e e 2 13 2 14 2 23 2 24 2� � � � � � � � � det ( ) ( ) ( ) ( ) , y e e e e 3 13 3 14 3 23 3 24 3� � � � � � � � � det ( ) ( ) ( ) ( ) , z e e e e 1 13 1 14 1 13 2 14 2 � � � � � � � � � det ( ) ( ) ( ) ( ) , z e e e e 2 13 1 14 1 13 3 14 3 � � � � � � � � � det ( ) ( ) ( ) ( ) , z e e e e 3 13 2 14 2 13 3 14 3 � � � � � � � � � det ( ) ( ) ( ) ( ) , z e e e e 1 14 2 24 2 14 1 24 1 � � � � � � � � � � det ( ) ( ) ( ) ( ) , z e e e e 2 14 3 24 3 14 1 24 1 � � � � � � � � � � det ( ) ( ) ( ) ( ) , z e e e e 3 14 3 24 3 14 2 24 2 � � � � � � � � � � det ( ) ( ) ( ) ( ) , w e e e e e e e e � det ( ) ( ) ( ) ( ) ( ) ( ) ( ) 12 1 13 1 14 1 12 2 13 2 14 2 12 3 13 3 14 3( ) ( )e � � � � �� � � � � �� , w e e e e e e e e � � det ( ) ( ) ( ) ( ) ( ) ( ) ( ) 14 3 24 3 34 3 14 2 24 2 34 2 14 1 24 1 34 1( ) ( )e � � � � �� � � � � �� , t e x e x e e x e x� � � 2 24 2 13 14 3 14 13 3 14 2 23 14 3 1 ( ) ( ) ( ) ( ) ( )( ) ( 3 24 3 14 23 3 24 13 3e x e x e( ) ( ) ( ))� � . algebras. The method can also be used for other simple Lie algebras (see [4]). Acknowledgment The authors are grateful to M. Havlíček for valuable and useful discussions. This work was partially supported by GAČR 201/05/0857 and by research plan MSM6840770039. References [1] Dixmier, J.: Algebres Enveloppantes. Gauthier-Villars Editeur, Paris 1974. [2] Kirillov, A.: Elementy Teorii Predstavlenij. Nauka, Moscow 1972. [3] Flath, D.: Decomposition of the Enveloping Algebra of sl3. J. Math. Phys. Vol. 31 (1990), No. 5, p. 1076–1077. [4] Burdík, Č., Navrátil, O.: Decomposition of the Envelop- ing Algebra so(5), submitted to J. Gener. Lie Theory and App. Prof. RNDr. Čestmír Burdík, DrSc. phone:+420 224 358 549 e-mail: burdik@kmlinux.fjfi.cvut.cz Ing. Severin Pošta, Ph.D. phone: +420 224 358 562 e-mail: severin@km1.fjfi.cvut.cz Department of Mathematics Czech Technical University in Prague Faculty of Nuclear Sciences and Physical Engineering Trojanova 13 120 00 Prague 2, Czech Republic Doc. RNDr. Ondřej Navrátil, Ph.D. phone: +420 224 890 714 e-mail: navratil@fd.cvut.cz Department of Mathematics Czech Technical University in Prague Faculty of Transportation Sciences Na Florenci 25 110 00 Prague 1, Czech Republic © Czech Technical University Publishing House http://ctn.cvut.cz/ap/ 31 Acta Polytechnica Vol. 47 No. 2–3/2007