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Home > Vol 2, No 1 (1993)

Paul Terwilliger

This is a continuation of an article from the previous issue. In this section, we determine the structure of a thin, irreducible module for the subconstituent algebra of a P- and Q- polynomial association scheme. Such a module is naturally associated with a Leonard system. The isomorphism class of the module is determined by this Leonard system, which in turn is determined by four parameters: the endpoint, the dual endpoint, the diameter, and an additional parameter f. If the module has sufficiently large dimension, the parameter f takes one of a certain set of values indexed by a bounded integer parameter e.

for part II appear in part I. TERWILLIGER in (51), (56), and Definition 3.3. Let W denote a thin, irreducible T-module, with endpoint u, dual-endpoint v, and diameter d, as defined in (79), (72), and Definition 3.5. Then (ii)-(viii) hold: (ii) Pick a nonzero u e EuW, and a nonzero v EUR E*W. Then are bases for W. We call S (resp. S*) a standard basis (resp. dual basis) for W. is a Leonard system over R, where A = A1, A* = A*, and where [ a ] B denotes the matrix representing a with respect to the basis B. LS(W) is uniquely determined by W (once the orderings of the associate matrices and the primitive idempotents are fixed). Consequently, LS(W), LS(Y) are related as follows if d > 1: Case I Case IA Case II Case IIA Case IIB Case IIC Case III (v) If d = 0 set f = I and if d > 1 let f be as in part (iv) above, where we interpret f = (f1 ,f2) (unordered pair) in Cases I, II, and Case III(d odd), and f = (f1, f2) (ordered pair) in Case III(d even). Then f is uniquely determined by LS(W). We refer to the 4-tuple (u, v, d, f) as the data sequence of W (with respect to the given orderings of the associate matrices and primitive idempotents). (vi) The statements are all equivalent, where M is the Bose-Mesner algebra of Y. If p is some object associated with LS(W), we will occasionally write p(W) to distinguish it from the corresponding object associated with LS(Y). TERWILLIGER Note 4.2. p10 = 1, q10 = 1, p10 = 0, g10 = 0 by (31), (39), and these equations give relationships among the constants q, h, h*, r1, r 2 , . . . that appear in part (iv) above. However, we make no use of these relationships until Corollary 4.12. Proof of Theorem 4.1. It is convenient to prove the parts in the order (ii), (iii), (vi), (i), (iv), (v), (vii), (viii). Proof of (ii). This is immediate from parts (ii) and (v) of Lemmas 3.9 and 3.12. Proof of (iii). First, we show the 4-tuple LS(W) is a Leonard system over C. Certainly the matrices B := [A]s, B* := [A*]s are tridiagonal, and have nonzero entries directly above and below the main diagonal, by parts (i)-(iii) of Lemmas 3.9, 3.12. The matrices H := [A]s, H* := [A*]s. are diagonal, for indeed H = d i a g ( T u , Tu+1,. . . , Tu+d ) and H* = d i a g ( T * , T * , . . . , T*) by construction. Also, H, H* each has distinct entries on the main diagonal by part (iii) of Lemmas 3.8, 3.11. So far we have (4)-(7). Now let Q denote the transition matrix from the basis S to the basis S* that is, the matrix whose columns represent the elements of S with respect to S*. Then by linear algebra Note by (53) and part (ii) of the present theorem that the sum of the elements of S* is a scalar multiple of the first element in S. It follows that the entries in the leftmost column of Q are all equal. Replacing (Q, S, S*) by ( Q - 1 , S*, S) in the above argument, we find the entries in the leftmost column of Q-1 are all equal. Now conditions (8)-(11) of Theorem 2.1 are satisfied, so LS(W) is a Leonard system over C. In fact LS(W) is over R. Certainly H e Matd+1(R) by part (iii) of Lemma 3.8, so consider the entries a;(W), bi(W), and ci(W) of B. We have since this is an eigenvalue of the real symmetric matrix Ei+v AEi+v. From (12), we find where c0(W) = bd(W) =
0. In particular b0(W) e R, and But also for this is an eigenvalue of the real symmetric matrix Ei+v A E i + v + 1 AEi+v. Now since the product in (92) is never
0. Combining the above implications we find b i - 1 ( W ) , Ci(W) EUR R (1 < i < d), so B e Matd+1(R) in view of (91). A similar argument shows H*, B' E Matd+1(R), so LS(W) is over R. Proof of (vi). It is immediate from (72), (79), and Definition 3.5 that Combining this with Lemma 3.6, we find the first four statements of (vi) are equivalent. Certainly the last statement implies the first and, hence, the first four, so now suppose W = MS. Observe by (68), (69), and part (ii) of the present theorem, that is a standard basis for W, and that is a dual basis for W. Now [A]s> = (p1j)0<i,j<D by (30), (97), (A]s = diag (T0, T1, . . . , TD) by (46), (94), [A*]s = (q1j)0<i,j<D by (61), (95), and [A*]S = diag(T*, T*,. . ., T*) by (60), (96), so the 4-tuples LS(W), LS(Y) are indentical. Proof of (i). The two 4-tuples LS(Y), LS(MX) are identical by part (vi) of the present theorem, and the 4-tuple LS(MX) is a Leonard system over R by part (iii) of the present theorem. Proof of (iv). The first statement is immediate from part (iii) of the present theorem. Now let LS' denote the Leonard system on the right side of (82)-(88). Then one may readily verify using the data in Theorem 2.1 that LS' has eigenvalue sequence Tu, Tu+1, . . . , Tu+d and dual eigenvalue sequence T*, TV+d, ..., TV+d. It follows from Lemma 2.4 that LS(W) = LS' for a suitable choice of the / parameters. Proof of (v). This is immediate from Lemma 2.4. Proof of (vii). We may assume v =
0. Then since A* is real symmetric, and since the basis S := (Euv, Eu+1v, ..., Eu+d v) of W is orthogonal, it follows from linear algebra that TERWILLIGER [A*]s is real by part (iii) of the present theorem, so we may eliminate the complex conjugate. Now computing the entries just above the main diagonal in the above products we find The result is immediate from this and induction. Proof of (viii). Similar to the proof of (vii). LEMMA 4.3. Let Y be as in Theorem 4.1, pick any x e X, and let W, W' denote any thin irreducible T(x)-modules. Then the following are equivalent. (i) W, W' are isomorphic as T(x)-modules. (ii) LS(W) = LS(W'). (iii) W, W have the same data sequence. Proof. Write E* = E*(x) (0 < i < D), A* = A*(x), T = T(x). (i) -> (ii). Let a : W -> W' denote an isomorphism of T-modules, and let S (resp. S*) denote a standard basis (resp. dual basis) for W. Since SEi = EiS, aE* = E*S (0 < i < D) by (3), we find aS (resp. aS*) is a standard basis (resp. dual basis) for W. But now (ii) -> (i). Let S, S' denote standard bases for W, W', respectively, and define the linear transformation a : W -{\tt>\hskip-.5e>} W' so that SS = S'. Then for B e {A, A*}, so a A - AS, a A* - A*a vanish on W. But A, A* generate T by part (ii) of Lemmas 3.8, 3.11, so aa - aa vanishes on W for every a EUR T. Now a is an isomorphism of T-modules by (3). (ii) -> (iii). The diameter of W is determined by the sizes of the matrices in LS(W). The endpoint of W is determined by the eigenvalue sequence of LS(W), and the dual endpoint of W is determined by the dual eigenvalue sequence of LS(W). The parameter f in the data sequence of W is determined by LS(W) according to part (v) of Theorem 4.1. (iii) -> (ii). This is immediate from part (iv) of Theorem 4.1. In Theorem 4.10 we will show that if the parameters u, v, d in a data sequence satisfy certain general inequalities, then the parameter f in the data sequence takes the following special form. Definition 4.4. Let the scheme Y = (X, {Ri}0<i<D) be as in Theorem 4.1. Pick any x e X, and let W denote a thin irreducible T(x)-module, with data sequence (u, v, d, f). Then W is said to be strong whenever d > 1, and there exists an integer e satisfying such that, [referring to part (iv) of Theorem 4.1], Case I f1, f2 is a permutation of Case IA Case II f1, f2 is a permutation of Case IIA, IIB Case IIC Case III f1, f2 is a permutation of (d odd) Case III (d even) TERWILLIGER The parameter e may not be unique. If W is strong, the auxiliary parameter of W is the integer e with subject to (98)-(106). (The auxiliary parameter is unique by the first condition of (98).) On our way toward Theorem 4.10, our next task is to consider how the data sequences of the various modules are related. Theorems 4.6, 4.9 are our main results on this subject. They are proceeded by the technical lemmas 4.5, 4.8. Recall nonempty subsets W, W' of the standard module V are said to be orthogonal whenever (w, w') = 0 for all w e W and all w' e W'. LEMMA 4.5. Let the scheme Y = (X, {Ri }0<i<D ) be as in Theorem 4.1. Pick any x, y EUR X, any thin irreducible T(x)-module W, and any thin irreducible T(y)-module W', such that W, W are not orthogonal. Let v, v' denote the dual endpoints of W, W', respectively, and pick nonzero v e E*(x)W, v' e E*(y)W'. Then there exist nonzero polynomials S, S' e C[A] such that and where (x, y) e Rp. (The supports Ws, W' are pom Definition 3.5.) Proof. This will consist of two claims. Claim
1. There exist nonzero polynomials S, S' e C[A] such that and where A = A1 is the first associate matrix of Y and projBa denotes the orthogonal projection of a onto
0. Proof of Claim
1. By symmetry, it suffices to show there exists a nonzero polynomial S e C[A] satisfying (112) and (113). To do this, it suffices to show pro]wv' is nonzero and contained in Span{v, Av, ..., Ap-V+V'v}. Now by assumption, there exist w EUR W, w' \sterling W' with (w, w')= 0, and by (74) we may write w' = aV' for some element a of the Bose-Mesner algebra M. Since a is symmetric we obtain and since aw E W, we observe projwv' =
0. Now write where d denotes the diameter of W, and where Vi e E*(x)W (v < i < v + d). We will now show To see this, note since (V = standard module), and E*(x)V, E*(y)V are orthogonal whenever piv' = 0 (0 < i < D). Now by (74), (114), and (115) we have as desired. This proves Claim
1. Claim
2. TERWILLIGER In particular, the polynomial Proof of Claim
2. By symmetry it suffices to prove (116) and (118). But since v' - projwv' is orthogonal to W, we have, for each integer i (0 < i < D), which gives (116). Now pick any f EUR Ws\W' , so that Eev = 0, Eev' =
0. Then from (116) and (117) we find so A - T\sterling divides tp. Thus (118) holds, and we have proved Claim
2. Now set Observe S, S' are nonzero by Claim 1, and contained in C[A] by (118) and (119). They satisfy (108) and (109) by (116) and (117), and (110) and (111) by Claim
1. This proves Lemma 4.5. THEOREM 4.6. Let the scheme Y = (X, {Ri}0<i<D ) be as in Theorem 4.1. Pick any x, y e X, any thin irreducible T(x)-module W, and any thin irreducible T(y)- module W', such that W, W' are not orthogonal. Let (u, v, d, f), (u', v', d', f') denote the data sequences of W, W, respectively, and suppose (x, y) e Rp for some P (0 < P < D). Then the following statements (i)-(v) hold. (iii) Assume |WsUW's|>2p +
2. Then d, d' >
1. Furthermore, there exists an integer e satisfying such that [referring to part (iv) of Theorem 4.1] Case I f1 f2 is a permutation of Case IA Case II f1 f2 is a permutation of Case IIA, IIB Case IIC Case III (iv) Suppose Then (v) Suppose W is strong, and that TERWILLIGER Then W is strong. Proof of (i). Ws n W' = Q, for otherwise at least one of EiW, EiW' is zero for each integer i (0 < i < D), contradicting the assumption that W, W' are not orthogonal. To simplify the notation for the rest of the proof, set and note Proof of (ii). Observe m, n are nonnegative by (110) and (111), so using (136) and (137) we find Proof of (iii). Combining (134), (135), and the assumption |Ws U W's| > 2p + 2, we have so d, d' >
1. Let v, v',s,s' be as in Lemma 4.5. Then, comparing the right sides of (108) and (109), we find where We observe si = 0 for all i e Ws n W'. Claim
1. Assume Case I (ss* = 0). Then and where Proof of Claim
1. From (82) we find so Evaluating this using (142), we obtain (140). To obtain (141), it suffices to show By (139) and (89) and the definition of Ws, W's, we find To evaluate this, we use the following notation. Set Then TERWILLIGER as long as ((a/B)), ((B/G)) are defined. Evaluating the data in Case I of Theorem 2.1 using (82), we find that for all (i - 1, i e Ws n W'): where the (( )) expressions in (146)-(149) are all defined. From (15) we also have where the (( )) expressions in (150) and (151) are defined. Now using (145), we find the product of the (( )) expressions in (146)-(151) is
1. Multiplying together the remaining factors in (146)-(151), we find But (152) equals (144) upon applying (136) and (142). To complete the proof of Theorem 4.6, we will need the following identity. See the given reference for a proof. LEMMA 4.7. (Terwilliger [68]). Let m, n denote any nonnegative integers with m < n, and pick any nonzero scalars a, b, c, d, e, q e C such that Then where We now return to the proof of Theorem 4.6. Claim
2. Suppose m < n. Then TERWILLIGER where Ak (0 < k < m) is given as follows: Case I (ss* = 0) Case I (s*=f1,=f1= 0) Case I (s* =f1,=f1 = 0) Case IA Case II Case IIA Case IIB Case IIC Case III Proof of Claim
2. First assume Case I (ss* =0), and consider the matrix in Lemma 4.7, where (a, b, c, d, e, q) are from (142). Note the determinant formula in that lemma remains valid if we replace vi by avi + B, (0 < i < m + n + 1), where a, B are any complex numbers. Choose and observe by (14), (142) that Thus the determinant formula in Lemma 4.7 remains valid if we replace vi by Ti+T, (0 < i < m + n+1). But after this replacement, the matrix (155) is obtained from the matrix (154) by dividing column i by pi(0 < i < m + n + 1). Now the determinant (155) can be readily determined from Lemma 4.7. In Case I (ss* = 0), IA, II, IIA, IIB, IIC, III, the determinant (155) is obtained by taking limits as indicated in Note 2.6. Claim
3. There exists an integer rj such that and such that Case I f1, f2 is a permutation of Case LA Case II f1, f2 is a permutation of TERWILLIGER Case IIA, IIB Case IIC Case III Proof of Claim
3. Interchanging the roles of W, W', if necessary, we may assume m < n. Also observe by (133) and (138) that Now the rows of the matrix (155) are linearily dependent by (139), (110), and (111), so the determinant (156) of that matrix is
0. We have observed the constants sT, sT+1, ..., sr+m are nonzero, and the Vandermonde determinants in (156) are nonzero, so Ak = 0 for some integer k (0 < k < m). Now assume Case I (ss* = 0), and consider the factors in the numerator of (157). The first two factors a, q are assumed to be nonzero. The next factor is where the product is over all integers \sterling such that Recall by (17) that sq =1 (2 < \sterling < 2D). Therefore, the product (166) is nonzero if we can show and The bound (168) is immediate, since r, m, k are nonnegative by (133) and (110), and the definition of k. To see (169), observe by (111) and (138), that The next factor in the numerator of (157) is where the product is over all integers ( such that Just as above, by (17) we have s*q^ =1 (2 < C < 2D). Therefore, the product (170) is nonzero if we can show and Line (171) holds, since n < d - 1 by (138) and v, k are nonnegative. To see (172), observe The remaining factor in the numerator of (157) is where the product is over all integers n such that The product (173) must be 0 since Ak is 0, so for some integer n that satisfies (174). Now (158) holds since k is nonnegative, and line (159) follows from (175) and (143). We have now proved Claim 3 for Case I (ss* = 0). The remaining cases are very similar. Now set TERWILLIGER where 77 is from Claim
3. Let us check that s satisfies (123). Certainly e + d + d' is even. Also, by (136), (158) and so by (137), (177) and (178). Thus (123) holds. Solving (176) for n, we find Now (124)-(130) are obtained upon evaluating the data in Claim 3 using (180). This proves (iii). Proof of (iv). Assuming p < v - v' in (122), we find so equality holds in (181)-(183). From (181) we find p = v - v'. Comparing (181) and (183), we see the three terms in parenthesis in (183) are equal to their absolute value, and are, hence, nonnegative. This implies (131). Proof of (v). Note Ws C W' by (131), so and part (iii) of the present theorem applies. Let the integer e be from that part, let e' denote the auxiliary parameter of W', and set To show W is strong, it suffices to show e satisfies (98)-(106). First consider (98). Certainly e + d + D is even by (184), since e' + d' + D is even by Definition 4.4 and e + d + d' is even by (123). By Definition 4.4, (123), (131), (132), and (184) we also have so (98) holds. Now for the moment assume Case I. Then by (124), f1, f2 is a permutation of where we may assume by Definition 4.4. Combining this information, we obtain (99). Lines (100)-(106) are obtained in a similar manner. This proves (v), and the theorem. We now give "dual" versions of Lemma 4.5 and Theorem 4.6. LEMMA 4.8. Let the scheme Y = (X, {Ri}0<i<D) be as in Theorem 4.1. Pick any x, y e X, any thin irreducible T(x)-modules W, W', and suppose W, A*p(y)W' are not orthogonal for some integer p (0 < p < D). Let u, u' denote the endpoints of W, W, respectively, and pick nonzero vectors u e EuW, u' e Eu'W'. Write E* = E*(x) (0 < i < D). Then there exist nonzero polynomials V*, V*' EUR C[l] such that and Proof. Write A* = A*(x), and observe A*, E* (0 < i < D) commute with A*(y). TERWILLIGER Claim
1. There exist nonzero polynomials <p*, p*' EUR C[A] such that and Proof of Claim
1. By symmetry, it suffices to show (189) and (190). To do this, it suffices to show projwA*(y)u' is a nonzero and contained in Span{u, A*u, ..., (A*)p-u+u'u}. NOW by assumption, there exists w e W, w' e W' such that (w, A*p(y)w') = 0, and by (81) we may write w' = au' for some element a of the dual Bose-Mesner algebra M*(x). Since a is symmetric and commutes with A*p(y), we obtain and since aw e W, we observe projwA*(y)u' =
0. Now write where d denotes the diameter of W, and where ui EUR EiW (u < i < p + d). Then since by (65). Now by (81), (191), and (192), we have as desired. This proves Claim
1. Claim
2. In particular Proof of Claim
2. Since A'p(y)u' - projwA*p(y)u' is orthogonal to W, we have, for each integer i (0 < i < D), which gives (193). The remaining assertions of the claim are obtained as in Claim 2 of Theorem 4.6. Now set Observe s*, s*' are nonzero by Claim 1, and contained in C[A] by (195) and (196). They satisfy (185) and (186) by (193) and (194), and satisfy (187) and (188) by Claim
1. This proves Lemma 4.8. THEOREM 4.9. Let the scheme Y = (X, {Ri}0<i><D) be as in Theorem 4.1. Pick any x, y e X, any thin irreducible T(x)-modules W, W', and suppose A*p(y)W', W are not orthogonal for some integer p (0 < p < D). Let (u, v, d, f), (u', v', d', f') denote the data sequences of W, W', respectively. Then the following statements (i)-(v) hold. such that Case I f1, f2 is a permutation of TERWILLIGER Case LA Case II f1, f2 is a permutation of Case IIA, IIB Case IIC Case III (iv) Suppose Then (v) Suppose W' is strong, and that Then W is strong. Proof. Similar to the proof of Theorem 4.6. THEOREM 4.10. Let the scheme Y = (X, {Ri}0<i<D) be as in Theorem 4.1. Pick any x EUR X, and let W denote a thin irreducible T(x)-module, with some endpoint u, dual endpoint v, and diameter d (0 <u, v < D - d < D). Then (i) W is strong whenever (iii) Suppose Y is thin. Then W is strong whenever Proof of (i). Assume (203). Then one of the following (a)-(c) holds. First suppose (a), so that W = MS by part (vi) of Theorem 4.1. Then, using Definition 4.4 one may readily check that M\sterling is strong (with auxiliary parameter e = 0), so we are done in this case. Next suppose (b), and set W = M$, where y is any element in X with (x, y) e Rv, such that \sterling is not orthogonal to W (y exists by the definition of v). Then W' is a thin irreducible T(y)-module by Lemma 3.6, W is strong by case (a) above, and W, W' are not orthogonal by construction. Now W, W' satisfy the conditions of part (v) of Theorem 4.6 (with p = v, v' = 0, d' = D), so W is strong. Next assume (c), and set W = MX. Then W is a thin, irreducible T(x)-module, and strong by part (a) above. Also, there exists y e X such that W, A*(y)W' are not orthogonal, since the all 1s vector d e W, A*(y)d = |X|Euy by (69), and Span {Euy| y e X} = EuV is not orthogonal to W by the definition of u. Now W, W', and y satisfy the conditions of part (v) of Theorem 4.9 (with u' = 0, d' = D, and p = u), so W is strong. Thus W is strong in general, and we are done. Proof of (ii). Suppose n < (D-d)/2 or v < (D-d)/2. Then (203) holds, so W is strong. But then 2u - D + d, 2v - D + d are nonnegative by (98), a contradiction. This proves (ii). Proof of (iii). In view of part (i) above, it suffices to prove W is strong under the assumption d >
3. The proof is by induction on /j + v. First assume v < p. Pick any y' EUR X such that (x, y') EUR Rv, and such that y is not orthogonal to W. Now pick any y e X such that (x, y) e R1 and TERWILLIGER (y, y') EUR Rv-1. Now Ev-1(y)V contains ff, and is therefore not orthogonal to W. But now there exists a (thin) irreducible T(y)-module W', with endpoint v' < v - 1, that is not orthogonal to W. Applying part (iv) of Theorem 4.6 to W, W' with p = 1, we find by (131) that u' < u, d' > d, and v' = v - 1, where u', d' are, respectively, the endpoint and diameter of W. In particular d' > 3 and u' + v' < u + v, so W is strong by induction. But now W, W' satisfy the conditions of part (v) of Theorem 4.6, so W is strong. Next assume u < v. Since EuV is contained in the column space of E1 o Eu-1 by (38), there exists y e X such that A*(y)Eu-1$ is not orthogonal to W. But then there exists a (thin) irreducible T(z)-module W, with endpoint u' < u - 1, such that A*(y)W' is not orthogonal to W. Applying part (iv) of Theorem 4.9 to W, W' with p = 1, we find v' < v, d' > d, and u' = u - 1, where v' and d' are, respectively, the dual endpoint and diameter of W. In particular d' > 3 and u' + v' < u + v, so W is strong by induction. But now W, W satisfy the conditions of part (v) of Theorem 4.9, so W is strong. COROLLARY 4.11. Let the scheme Y = (X, {Ri}0<i<D ) be as in Theorem 4.1, and pick any x e X. Pick any integers u, v, d (0 < u, v < D - d < D), and assume (u, < D/2 or v < D/2. Then the number of pairwise nonisomorphic, thin, irreducible T(x)-modules with (i) endpoint u, dual endpoint v, diameter d (ii) endpoint u, dual endpoint v (iii) endpoint n (iv) dual endpoint v is at most Proof of (i). Immediate from Lemma 4.3, Definition 4.4, and (203). Proof of (ii). To obtain (208), sum (206) over all integers d (D - 2u < d < D - v). To obtain (209), sum (207) over all integers d (D - 2v < d < D - u). Proof of (Hi). Line (210) is the sum of (208) over all integers v ( u / 2 < v < p), plus the sum of (209) over all integers v (n < v < 2u). Proof of (iv). Similar to (iii). COROLLARY 4.12. Let the scheme Y = (X, {Ri}0<i<D) be as in Theorem 4.1. Pick any x e X, pick any integer i (0 < i < D), and write E* = E* (x), T = T(x). (Observe E*AE* : E*V -> E*V is the adjacency map for the undirected graph with vertex set X n E*V, and edge set {(y, z) | (y, z) e R1 y, z e X n E*V}.) Let W denote an irreducible T-module with E*W=
0. Then the following statements (l)-(5) hold. (1) E*W is an E*AE*-invariant subspace of E*V. (2) Suppose W is thin. Then E*W is a (one-dimensional) eigenspace of E*AE*. The eigenvalue is A := a i - v ( W ) , where v is the dual endpoint of W. An E* AE*-eigenvalue of this form will be said to be of thin type. (3) E*V is an orthogonal direct sum E*W0 + E*W1 +. . .+ E* Wn, where W0, W1, . . ., Wn are irreducible T-modules that intersect E*V nontrivially. In particular, if Y is thin with respect to x then every eigenvalue of E*AE* : E*V -> E*V is of thin type. (4) Suppose i < D/2, and that W is thin. Then there are at most a* ways to choose W up to isomorphism of T(x)-modules, where In particular, E*AE* has at most si distinct eigenvalues of thin type. We note T0 = 1, s1 = 5, s2 = 16, a3 = 39,. . . (5) Suppose i = 1, and that W is thin, with some endpoint u, diameter d, and auxiliary parameter e. Then (u, v, d, e), A is given in one of (i)-(v) below (here we use the notation of part (iv) of Theorem 4.1). TERWILLIGER (i) (u,v, d, e) = (0, 0, D, 0), A = p11. (ii) (u, v,d, e) = (1, 1,D-1, 1), A equals Case I(S* = 0) Case I(S* = r1 = 0) - 1 Case IA -1 Case II Case II4 -1 Case IIB Case IIC does not occur Case III(D even) does not occur Case III(D odd) (in) (u, v, d, e) = (1, 1, D - 1, -1), A equals Case I(S* = 0) Case I(S* = r1 = 0) Case IA Case II Case IIA Case IIB Case IIC -1 Case III(D even) Case III(D odd) (iv) (u, v, d, e) = (1, 1, D - 2, 0), A equals Case I(S* = 0) Case I(a* = r1 = 0) Case IA Case II TERWILLIGER Case IIA Case IIB Case IIC Case III(D even) Case III(D odd) does not occur (v) (u, v, d, e) - (2, 1, D - 2, 0), A equals Case I(s* = 0) Case I(s* = r1 = 0) Case IA Case // Case IIA is -2 Case IIB Case IIC is -2 Case III Proof of (1). Immediate. Proof of (2). Immediate from Theorem 2.1. Proof of (3). Immediate from Lemma 3.4. Proof of (4). Si is the sum of (211) over v = 0, 1, . . . , i. Proof of (5). The given values for (u, v, d, e) represent all the integer solutions to (93) and (98) that satisfy v <
1. In case (i), we have A = p11, by parts (i) and (vi) of Theorem 4.1. In case (ii)-(v), A = a0(W) is computed using Theorem 2.1 and Note 4.2. Acknowledgments This research was partially supported by NSF grant DMS-880-0764.