XIII. Weyl's Dimension Formula. In this Section we derive the celebrated formula of Weyl for the dimensionality

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1 10 III. Weyl's Dimension Formula III. Weyl's Dimension Formula In this Section we derive the celebrated formula of Weyl for the dimensionality of an irreducible representation of a simple Lie algebra in terms of its highest weight. Our derivation is essentially that of Jacobson, which is based on the technique of Freudenthal. We shall be considering functions dened on H0. Instead of parameterizing elements of H 0 in terms of the simple roots, it is convenient toover-parameterize by writing H 0 as = : (III:1) We can dene the action of an element, S, oftheweyl group on a function of by (SF)() =F (S ;1 ) : (III:)

2 III. Weyl's Dimension Formula 103 As an example, consider the function known as the character of the representation () = M n M exphm i (III:3) where the sum is over all the weights of a representation and n M is the dimensionality oftheweight space for M. Nowwe calculate the action of S W on : (S)() = M n M exphm S ;1 i = M n M exphsm i = M n M exphm i : (III:4) Here we have used the orthogonality property of the Weyl group and the relation n M = n SM.Thus we see that S =, that is is invariant under the Weyl group. Consider next the function Y Q() = >0[exp( 1 h i) ; exp(;1 h i)] : (III:5) We want to determine the behavior of this function when acted upon by elements of the Weyl group. It suces to determine the eect of the S i = S i, the reections associated with simple roots. (S i Q)() = Y >0 = Y >0 exp exp 1 h S;1 i i ; exp ; 1 1 hs i i h S;1 i i ; exp ; 1 hs i i : (III:6)

3 104 III. Weyl's Dimension Formula We have already seen that S i interchanges all the positive roots except i whose sign it changes. Thus we see directly that (S i Q)() =;Q() : (III:7) Now S i reverses the sign of i, but leaves every vector orthogonal to i unchanged. Thus dets i = ;1 and we can write Indeed, every S W is a product of S i 's, so (S i Q)=(detS i )Q: (III:8) SQ = detsq : (III:9) Functions with this property are called alternating. We can make alternating functions by applying the operator = SW(detS)S (III:10) for we have S 0 = SW S 0 (dets)s = dets 0 det(s 0 S)S 0 S SW =dets 0 : (III:11) It is convenient to nd a representation of the alternating function Q() in the form F(). From the denition of Q() it is clear that there must be an expansion of the form

4 III. Weyl's Dimension Formula 105 Q() = c exph ; i (III:1) where is half the sum of all the positive roots and where is a sum of distinct positiveroots. Now in such an expansion it is redundant to include both exphm i and exphsm i since they are equal up to a factor dets. Wehave already seen that among all the SM, S W, the largest one has only non-negative Dynkin coecients. Thus we need only consider terms where ; has only non-negative Dynkin coecients. In fact, we can restrict this further because if M has a Dynkin coecient which is zero, then exphm i = 0. This is easy to establish since if M(h i ) = 0, then S i M = M and S i exphm i = exphs i M i = exphm i = dets i exphm i = ; exphm i =0. However, we have seen that has Dynkin coecients which are all unity. Now since is a sum of positive roots it cannot have only negative Dynkin coecients. Thus we see that the sum, Eq. (III.1) need include only = 0. Comparing the coecients of exph i we see that c =0 =1so Y Q() = >0[exp 1 h i;exp ;1 h i] = SW(detS) exphs i : (III:13) We shall now use these results to analyze the character. P We begin with 1 the Freudenthal recursion relation, Eq. (I.45), together with k=;1 n M+khM + k i =0: [h+ +i;h i;hm Mi]n M = 6=0 1 k=0 n M+k hm + i: (III:14) Mulitplying by exphm i and summing over M, wehave

5 106 III. Weyl's Dimension Formula [h+ +i;h i] ; M n m hm Mi exphm i = M 1 6=0 k=0 n M+ hm + i exphm i : (III:15) Remembering that (see Eq. (IV.3)) hm Ni = (h M )(h N )= hm ih Ni (III:16) we derive the relations exphm i = h Mi exphm i exphm i = hm Mi exphm i : (III:17a) (III:17b) Inserting this into Eq. (III.15) gives [h+ +i;h i; ] = M 1 6=0 k=0 n M+ hm + i exphm i : (III:18) To analyze the right hand side of Eq. (III.18), let us rst x 6= 0 and consider the SU() generated by E E ; and H. The irreducible representation of the full algebra is, in general, a reducible representation of this SU(). The dimensionality n M is just the number of SU()-irreducible representations present attheweight M. Thus we can proceed by calculating the contribution of each SU()-irreducible representation to the sum for xed M and. The string of weights containing M which correspond to an SU() irreducible representation are distributed symmetrically about a center point, M 0 which can be expressed in terms of the highest weight in the sequence, M,as

6 III. Weyl's Dimension Formula 107 M 0 = M ; hm i h i : (III:19) Note that hm 0 i =0. Thus each weight in the sequence is of the form M 0 + m where m is an integer or half-integer. The range of m is from ;j to j, where again j is an integer or half-integer. Now we can write the contribution of a single SU() irreducible representation to the sum as M 1 k=0 hm + i exphm i = m = m j;m k=0 j;m k=0 hm 0 + m + k i exphm 0 + m i h i(m + k) exphm 0 i exp(mh i) = h i exphm 0 i m j;m k=0 (m + k) exp(mh i) : (III:0)

7 108 III. Weyl's Dimension Formula The double sum is most easily evaluated by multiplying rst by exph i;1: j j;m m=;j k=0 = (k + m)exphm i(exph i;1) j j;m m=;j k=0 ; (m + k) exph(m +1) i j;1 m=;j;1 = j exph(j +1) i + = j m=;j + j k=0 j;m;1 k=0 j;1 m=;j (k ; j)exph;j i m exph(m +1) i : (m + k + 1) exph(m +1) i [j ; (j ; m)] exph(m +1) i (III:1) Thus the contribution of one SU() irreducible representation to the original sum is h i exphm 0 i = j m=;j j m=;j m exph(m +1) i[exph i;1] ;1 hm i exphm + i[exph i;1] ;1 : (III:) Summing over all SU() irreducible representations, and over all the roots, we have [h+ +i;h i; ] = 6=0 M n M h Mi exphm + i[exph i;1] ;1 : (III:3)

8 III. Weyl's Dimension Formula 109 From the denition of Q, we see that Y 6=0 [exph i;1] = Q () (III:4) where is +1 if the number of positive roots is even and ;1 ifitisodd.thus and so M 6=0 exph i log Q () = h i exph i;1 (III:5) 6=0 exphm i = h Mi exphm i n M h Mi exphm + i[exph i;1] ;1 (III:6) = log Q () =Q ;1 = Q ;1 4 Q (Q) ; Q ; 3 Q5 : (III:7) Combining these results, we have the dierential equation h+ +iq = 4 h i; 3 5 Q + Q : (III:8) From the relation Q() = SW(detS)exphS i (III:9) it follows that Q() =h iq() where we have used the orthogonality of the S's. Altogether then we have Q = h+ +iq : (III:30) (III:31)

9 110 III. Weyl's Dimension Formula Now the function Q()() is alternating since it is the product of an alternating function and an invariant one. Since () = M n M exphm i (III:3) and Q() = SW(detS)exphS i (III:33) the product must be of the form Q()() = N c N exphn i (III:34) where N is of the form M + S where M is a weight and where S is in the Weyl group. Substituting into the dierential equation, we see that M contributes only if hs ;1 M + S ;1 M + i = h+ +i : (III:35)

10 III. Weyl's Dimension Formula 111 In fact, we can show that Eq. (III.35) is satised only when S ;1 M =. We rst note that hsm + SM + i is maximized for xed M when SM has only non-negative Dynkin coecients. Indeed if h i =h i=h i i i and if M(h i ) < 0, then hs i M + S i M + i ;hm + M + i = h;m(h i ) i M + ; M(h i ) i i = ;h i i im(h i ) > 0. Now consider M < withm(h i ) 0. Then, by similar arguments, it is easy to show that h+ +i > hm + M + i. It follows that the sum in Eq. (III.34) need contain only the single term for N =+. By comparison with the denitions of Q() and (), it is easy to see that the overall coecient isunity, so Q()() = SW(detS)exph+ Si : (III:36) This then yields Weyl's character formula () = P (dets) exph+ Si PSW (dets) exph Si : (III:37) More useful for our purposes is the less general formula which gives the dimension of an irreducible representation. It is clear that this dimension is the value of ( = 0). This cannot be obtained simply by setting =0,butmust be obtained as a limit. We choose = t and let t! 0. This gives (t) = P SW P SW (dets) exphs( + ) ti (dets) exphs ti = Q(t( + )) Q(t) =exph; ti Y >0 exph t( + )i;1 exph ti;1 : (III:38) In this expression we can let t! 0 and nd the dimensionality,dimr = (0): dim R = Y >0 h +i h i : (III:39)

11 11 III. Weyl's Dimension Formula Toevaluate this expression, we write each positive root,, in terms of the simple roots i : = k i i : (III:40) i Suppose the Dynkin coecients of are (h i )= i, where h i =h i=h i i i. Then we have Y P i dim R = ki ( i +1)h i i i P : (III:41) >0 i ki h i i i The algebras A n D n E 6 E 7 and E 8 have simple roots all of one size, so for them we can drop the factors of h i i i in Eq. (III.41). Let us illustrate this marvelous formulawith anumber of examples. Consider rst SU(3), that is, A. The simple roots are all the same size so we ignore the factor h i i i. The positive roots are 1 and 1 +, which we shall abbreviate here by (1), (), and (1). Suppose the irreducible representation at hand has a highest weight with Dynkin coecients (m 1 m ), then we compute m1 +1 m +1 m1 + m + dim R = : (III:4) 1 1 From this example and the fundamental formula, Eq. (III.41), we see that the rule for nding the dimensionality of an irreducible representation may be phrased as follows: The dimension is a product of factors, one for each positive root of the algebra. Each factor has a denominator which is the number of simple roots which compose the positive root. The numerator is a sum over the simple roots in the positive root, with each simple root contributing unity plus the value of the Dynkin coecient corresponding to the simple root. If the simple roots are not all the same size, each contribution to the numerator and to the denominator must be weighted by h i i i.

12 III. Weyl's Dimension Formula 113 Let us consider a more complicated application of Weyl's formula. The algebra G has, as we have seen, fourteen roots, of which six are positive. If the simple roots are denoted 1 and with the latter being the smaller, then the square of 1 is three times larger than that of. The positive roots are and 1 +3 which we denote here (1) () (1) (1 ) (1 3 ), and (1 3 ). We compute below the dimensions of the representations with the highest weights (0,1) and (1,0), where the rst entry pertains to 1 and the second to. (0 1) (1 0) (1) () (1) (1 ) (1 3 ) () () () () (1 3 ) ()+3() dim = 7 dim = 14 As yet another example, we consider SO(10), that is, D 5. The simple roots are numbered so that 4 and 5 are the ones which form the fork at the end of the Dynkin diagram. There are 45 roots of which 0 are positive. Below we calculate the dimensionality of the representations with highest weights (1,0,0,0,0), (0,0,0,0,1), and (0,0,0,0,). ( ) ( ) ( ) (1) () (3) (4) (5) 3 (1) (3) (34) (35) 3 (13) (34) (35) (345) 4 3 (134) (345) (135) (1345) (3 45) 6 5 (13 45) (1 3 45) dim R =10 dim R =16 dim R =16 9 7

13 114 III. Weyl's Dimension Formula With little eort, we can derive a general formula for the dimensionality of an irreducible representation of SU(n), that is A n;1. In the same notation as above, the roots are (1) () :::(1) (3) :::(13) (34) ::: (13 :::n). We compute the dimensionality for the representation whose highest weight is(m 1 m :::m n ): m (1) () :::(n) 1+1 m +1 ::: mn (1) (3) :::(n ; 1 n) ::: m 1+m+ (1 :::n) m 1+m+:::m n+n n m +m3+ mn;1+mn + It is a simple matter to multiply all these factors to nd the dimension of the representation. We can recognize the correspondence between the Dynkin notation and the more familiar Young tableaux if we start with the fundamental representation, (1 0 :::0), and take the k-times anti-symmetric product of this representation with itself to obtain (0 0 ::: m k =1 0 :::0). This corresponds to the tableau with one column of k boxes. More generally, (m 1 m :::m n ) corresponds to the tableau with m k columns of k boxes.

14 III. Weyl's Dimension Formula 115 References We follow closely Jacobson's version of the Freudenthal derivation of the Weyl formula, except that we have adopted a less formal language. See JACOBSON,pp. 49{59. Exercises 1. Determine the dimensionality of the SO(10) representations ( ) and ( ).. Determine the dimensionality ofthee 6 representations ( ) and ( ). 3. Show that the E 6 representation ( ) has a dimensionality divisible by 137.

116 XIV. Reducing Product Representations

116 XIV. Reducing Product Representations XIV. Reducing Product Representations In Chapter X we began the consideration of nding the irreducible components of a product representation. The procedure for