ITEP/TH-38/17 IITP/TH-23/17
On the block structure of the quantum R-matrix in the three-strand braids L. Bishlera,b∗, An. Morozovb,c†, A. Sleptsovb,c,d‡, Sh. Shakirove§
a
arXiv:1712.07034v1 [hep-th] 19 Dec 2017
d
Moscow State University, Moscow 119991, Russia b ITEP, Moscow 117218, Russia c Institute for Information Transmission Problems, Moscow 127994, Russia Laboratory of Quantum Topology, Chelyabinsk State University, Chelyabinsk 454001, Russia, e Society of Fellows, Harvard University, Cambridge, MA 20138, USA
Abstract Quantum R-matrices are the building blocks for the colored HOMFLY polynomials. In the case of threestrand braids with an identical finite-dimensional irreducible representation T of SUq (N ) associated with each strand one needs two matrices: R1 and R2 . They are related by the Racah matrices R2 = UR1 U † . Since we can always choose the basis so that R1 is diagonal, the problem is reduced to evaluation of R2 matrices. This paper is one more step on the road to simplification of such calculations. We found out and proved for some cases that R2 -matrices could be transformed into a block-diagonal ones. The essential condition is that there is a pair of accidentally coinciding eigenvalues among eigenvalues of R1 -matrix. The angle of the rotation in the sectors corresponding to accidentally coinciding eigenvalues from the basis defined by the Racah matrix to the basis in which R2 is block-diagonal is ± π4 .
1
Introduction
HOMFLY polynomials (Wilson loop averages in Chern-Simons theory [1]), especially colored HOMFLY polynomials [2, 3], receive now a lot of attention. In part it is due to the connections to other theories such as [4, 5], also provided by such alternatives to HOMFLY polynomials as hyper- [6, 7, 8] and super- [9, 10, 20] polynomials. There are different approaches to evaluation of HOMFLY polynomials. Among them the Reshetikhin-Turaev (RT) group-theoretical method [11], the approach connected with conformal theories [12], useful in calculations of two-bridge and arborescent knots [13]-[19], and the evolution method [20, 21, 22], which can be very effective for series of knots. For very promising alternative approaches, see [8, 23, 24]. In this paper the RT method is used. The polynomial in this method is given by the formula: Y HTK1 ⊗T2 ... = TrT1 ⊗T2 ... Rα , (1) α
where α enumerates all crossings in the braid. To use the RT approach one should redraw a projection of a knot as a braid. Each line in the braid is associated with a finite-dimensional irreducible representation Ti of the quantum group SUq (N ). In our paper we worked with three-strand braids. Since we want to calculate polynomials of knots (not links), we chose the same representation T for each strand. Matrices R1 and R2 represent crossings of the first and the second strands and the second and the third strands correspondingly. They are related to each other through a Racah mixing matrix (also known as 6j-symbol and the Racah-Wigner coefficients) U: R2 = UR1 U † . ∗
[email protected] †
[email protected] ‡
[email protected] §
[email protected]
1
By definition Racah matrices [25] relate the intertwiners: UTT14,T2 ,T3 : ((T1 ⊗ T2 ) ⊗ T3 → T4 ) −→ (T1 ⊗ (T2 ⊗ T3 ) → T4 ).
(2)
i.e. describe deviations from the associativity in the product of representations. There are different ways to calculate Racah matrices, such as via the pentagon relation [25, 26, 14] or the highest wight vector method [27, 28]. A number of Racah matrices were calculated with the highest weight vector method for different representations T of SUq (N ) [29]: for T = [2, 1] in [31], for T = [3, 1] in [32] and for T = [3, 3] in [33]. These examples [29] enabled us to study certain properties of Racah matrices and therefore R2 -matrices. R-matrices by definition satisfy the Yang-Baxter (YB) equation R1 R2 R1 = R2 R1 R2 (which is a mathematical form of the third Reidemeister move). It also relates R1 and U matrices: R1 UR1 U † R1 = UR1 U † R1 UR1 U † . Eigenvalues λi of the R-matrix [34] are defined by irreducible representations Qi from T ⊗2 = λi = Qi q κQi ,
(3) L
i
ai Qi . (4)
P where κQi = (m,n)∈Qi (m − n) is the image of the quadratic Casimir operator in the representation Qi , Qi = ±1 depending whether Qi belongs to the symmetric or antisymmetric square of T ⊗2 . There are five possible sets of eigenvalues in R-matrices: 1. all eigenvalues are different, 2. some eigenvalues coincide because of the multiplicity (when ai > 1 or/and bi > 1 in recompositions L L T ⊗2 = i ai Qi and T ⊗3 = i bi Qi ) −→ repetitive eigenvalues 3. some eigenvalues accidentally coincide: Qi q κQi = Qj q κQj , but Qi 6= Qj −→ accidentally coinciding eigenvalues, 4. there are both repetitive (λ) and accidentally coinciding eigenvalues (µ), which don’t coincide between each other (@ λi , µj : λi = µj ), 5. there are both repetitive (λ) and accidentally coinciding eigenvalues (µ), which also coincide (∃ λi , µj : λi = µj ). In the case of the first set of eigenvalues, the equation (3) can be solved for matrices up to 6 × 6 size, as it was done in [35]. As a result the eigenvalue hypothesis and its generalisation [28] were formulated. It claims that eigenvalues of R-matrix fully define elements of the corresponding Racah matrix. It’s also very important to consider other sets. In this paper we considered sets 2, 3 and 4. The equation (1) has the following symmetry: one can arbitrary change the basis of R1 - and R2 -matrices: R0 1 = QR1 Q† , R0 2 = QR2 Q† (one should apply the same transformation to the corresponding Racah matrix so that it can be used to calculate HOMFLY polynomials). When R-matrix contains coinciding eigenvalues, we can change just vectors corresponding to them. Such rotation doesn’t change R1 -matrix: R0 1 = OR1 O† = R1 , R0 2 = OR2 O† . We used such rotations to obtain the results of this paper. In experimental part of the present paper we rotated Racah matrices from [29] in the sectors of coinciding eigenvalues (repetitive and accidental) of R-matrices. In case of repetitive eigenvalues such rotations represent just the change of the vectors in the corresponding representations. The resulting matrix still satisfies the definition (2). Rotations in the sectors corresponding to accidentally coinciding eigenvalues mix vectors from different irreducible representations. Thus the resulting matrix is no longer a Racah matrix. But it can still be used to calculate HOMFLY polynomials, because equation (1) is invariant under such rotations. This paper is devoted to studying the properties of Racah matrices and R2 -matrices which come from these transformations. The main results of the paper are two hypotheses.
2
1. R2 -matrix can be transformed into a block diagonal form, if its eigenvalues satisfy two conditions: a number of pairs of accidentally coinciding eigenvalues ≥ Na , a number of pairs of multiple eigenvalues ≥ Nr , where • Na = 1, Nr = 0 or Na = 0, Nr = 1 for matrices up to size 5 × 5, • Na = 1, Nr = 1 for 6 × 6 matrices, • Na = 1, Nr = 2 for 8 × 8. 2. The angle of the rotation in the sectors corresponding to accidentally coinciding eigenvalues from the basis defined by the Racah matrix to the basis in which R2 is block-diagonal is ± π4 .
The first hypothesis was proven for R2 -matrices up to size 3×3. It can also be formulated for a Racah matrix which has a block-diagonal structure under certain transformations. In fact all calculations in experimental part were made for Racah matrices. The results can be found in the table 1. The resulting blocks can be found from the eigenvalues hypothesis since no eigenvalues in each block coincide. This is a very important for calculations fact and it follows from the first hypothesis. The second hypothesis is the purely experimental observation. It is still an open question why the angle of the rotation in the “accidental” sector is precisely ± π4 . The paper is organized as follows. In the section 2 there is the detailed information about R-matrices and how they act on irreducible representations. In the section 3 we considered general case Racah matrices with one pair of coinciding eigenvalues. We straightforwardly solved the YB equation and showed that such R2 -matrices have a block-diagonal structure. In the section 4 we described all rotations that were used in the experimental part. In sections 5 - 9 we listed all particular examples that were used to formulate the results of our work.
2
R-matrices in three-strand braids
R-matrix is an operator which acts on a tensor product of two representations Ti ⊗ Tj (Tk is a representation of SUq (N )) which are associated with lines of a braid. As it was mentioned earlier in the case of three strand braids and identical representations there are two R-matrices which are used in the RT formalism: R1 = R ⊗ I ∈ M at(T ⊗ T ⊗ T ),
(5)
R2 = I ⊗ R ∈ M at(T ⊗ T ⊗ T ),
(6)
where I is the unit operator. L In general case vectors of irreducible representations Qk (Ti ⊗ Tj = k ak Qk ) are eigenvectors of R-matrix. It follows from the fact that R-matrix commutes with the coproduct. To show this fact, let’s consider how R-matrix acts on vectors vi and uj of two irreducible representations Q1 and Q2 (dim(Q1 ) < dim(Q2 )). v1 and u1 are the highest weight vectors of representations Q1 and Q2 , T + and T − are the raising and the lowering operators. Pdim(Q ) Pdim(Q ) Rv1 = i=1 1 ai vi + j=1 2 bj uj , P Pdim(Q ) dim(Q ) T + Rv1 = i=2 1 ai vi−1 + j=2 2 bj uj−1
(7)
T + Rv1 = RT + v1 = 0,
(8)
From the other point of view it means that ai = bj = 0 for i = 2, . . . , dim(Q1 ), j = 1, . . . , dim(Q2 ) and now Rv1 = a1 v1 + b1 u1 . (T − )dim(Q1 ) Rv1 = b1 udim(Q1 )+1 = R(T − )dim(Q1 ) v1 = 0 ⇒ b1 = 0 ⇒ Rv1 = a1 v1
(9)
Acting on this expression with the raising operator one can show that Rvi = a1 vi . One can use the same procedure to show that it’s true for each irreducible representation. 3
(10)
size
repr
eigenvalues repetitive accidental
res
blocks
block 1
block 2
[2,1] 6
[5,3,1]
λ1 = λ2
λ3 = λ4
+
(3,3)
λ1 ,λ∗3 , λ6
λ2 , λ∗4 , λ5
8
[4,3,2]
λ4 = λ5 , λ6 = λ7
λ2 = λ3
+
(3,5)
λ∗3 , λ5 , λ6
λ1 , λ∗2 , λ4 λ7 , λ8
9
[4,3,1,1]
λ4 = λ5 , λ6 = λ7
λ2 = λ3
–
[3,1] 4
[4,3,2,2,1]
λ2 = λ3
+
(1,3)
λ∗3
λ1 , λ2 , λ∗4
5
[4,4,2,2]
λ1 = λ2
+
(2,3)
λ∗1 , λ4
λ∗2 , λ3 , λ5
6
[4,3,3,1,1]
λ1 = λ2
λ3 = λ4
+
(3,3)
λ1 , λ∗4 , λ6
λ2 , λ∗3 , λ5
9
[5,3,2,2]
λ4 = λ5
λ2 = λ3
–
[4,4,3,1]
λ5 = λ6 , λ7 = λ8
λ2 = λ3
+
(3,6)
λ∗3 , λ5 , λ8
λ1 , λ∗2 , λ4 λ6 , λ7 , λ9
[3,2] 6
λ1 , λ∗4 , λ5
λ2 , λ∗3 , λ6
[4,3,3,2,2,1]
λ5 = λ6
λ3 = λ4
+
(3,3)
[6,4,2,1,1,1]
λ1 = λ2
λ3 = λ4
+
(3,3)
λ1 , λ∗4 , λ6
λ2 , λ∗3 , λ5
[4,1] 4
[5,4,3,2,1]
-
λ2 = λ3
+
(1,3)
λ∗3
λ1 , λ∗2 , λ4
6
[5,5,3,2]
λ3 = λ4
λ5 = λ6
+
(3,3)
λ1 , λ3 , λ∗5
λ2 , λ4 , λ∗6
[5,4,4,1,1]
λ2 = λ3 , λ4 = λ5
λ6 = λ7
+
(3,3)
λ1 , λ∗3 , λ6
λ2 , λ∗4 , λ5
[6,4,2,2,1]
-
λ4 = λ5
-
[4,2] 4
[5,4,3,3,2,1]
-
λ2 = λ3
+
(1,3)
λ∗3
λ1 , λ∗2 , λ4
5
[5,5,3,3,1,1]
-
λ4 = λ5
+
(2,3)
λ1 , λ∗4
λ2 , λ3 , λ∗5
6
[5,4,4,2,2,1]
λ5 = λ6
λ3 = λ4
+
(3,3)
λ1 , λ∗3 , λ6
λ2 , λ∗4 , λ5
[11,6,1]
λ1 = λ2
λ3 = λ4
+
(3,3)
λ1 , λ∗3 , λ5
λ2 , λ∗4 , λ6
Table 1: Matrices with accidentally coinciding eigenvalues. In the last two columns accidentally coinciding eigenvalues are marked with stars, repetitive eigenvalues are underlined (and twice underlined),“symmetric – antisymmetric” eigenvalues are boxed.
4
RQi = λi Qi = ±q κi Qi
(11)
That’s why there is a basis where R1 is diagonal. R1 (T1 ⊗ T2 ⊗ T3 ) = R(T1 ⊗ T2 ) ⊗ I(T3 ) = R
M
Qi ⊗ T3 =
M
i
T1 ⊗ T2 ⊗ T3 =
M
λi Qi ⊗ T3 .
(12)
i
ak Q0k .
(13)
k
R1 and R2 have block-diagonal forms with blocks R1;Q0k and R2;Q0k : R1;Q0k = diag(λj1 , . . . λjaj ), R2;Q0k = (UQ0k )† R1;Q0k UQ0k ,
(14)
where λjl are eigenvalues of representations Qjl (from the sum T1 ⊗ T2 = this sum there is Q0ml = Q0k ). The equation (1) is also modified [27]: (m1 ,n1 |m2 ,n2 |... )
HT
i
Qi : Qjl ⊗ T3 =
† † n1 m2 n2 ∗ 1 SQ Tr Q (Rm 1;Q UQ R1;Q UQ R1;Q UQ R1;Q UQ . . . ),
X
=
L
L
m
Q0m and in
(15)
Q∈T ⊗3 ∗ where mi and ni enumerate the crossings in the braid, SQ are the Schur polynomials1 .
3
The proof of the first hypothesis
In the experimental part we have considered Racah matrices up to six dimensions. Among all the examples we encountered just cases with one pair of accidentally coinciding eigenvalues. That’s why, we’ll consider now a general case of R-matrices with a pair of coinciding eigenvalues. 1. n = 2 In this case R=
λ
λ
=λ
1
1
,
(17)
it means that R2 is proportional to the identity matrix too, because R2 = U RU † . 2. n = 3 Here λ1 R=
λ1
(18)
γ1 γ2 γ3
(19)
λ3 R2 is symmetric because R is diagonal and U is real. α1 R2 = β1 γ1
β1 β2 γ2
Since one can always use a rotation to make β1 = 0: 1 there
also exists a constructive hook formula, which allows one to calculate them easily: k i Y Aq i−j − A−1 q j−i ∗ SQ (A, q) = , x j q hi,j − q −hi,j (i,j)∈Q
[n]q denotes the quantum number n, i.e., [n]q ≡
q n −q −n . q−q −1
(16) hi,j = k + l + 1.
5
O(β1 =0)
c −s = s c
, 1
where
(20)
√
c=
1 v 2 u √ u 2 +α −β (α1 −β2 )2 +4β1 1 2 t 2 4β1
,
s=
+1
2β1
(α1 −β2 )2 +4β12 +α1 −β2 v 2 u √ u 2 +α −β (α1 −β2 )2 +4β1 1 2 t 2 4β1
,
(21)
+1
˜ 2 = O(β =0) R2 O† it’s reasonable to consider R 1 (β1 =0) . Such rotation doesn’t change R1 -matrix because it’s in the sector of coinciding eigenvalues. Renaming elements we get
a1 ˜2 = 0 R c1
0 a2 c2
c1 c2 c3
(22)
˜ 2 obey the Yang-Baxter equation (YB). The element YB12 of this equation is R and R λ3 c1 c2 = 0.
(23)
˜ 2 block-diagonal. There are two possible solutions (c1 = 0 and c2 = 0) and both of them make R
4
Calculations
We studied a number of R-matrices in different representations of SUq (N ) and picked up matrices with coinciding eigenvalues. Then we rotated corresponding Racah matrices [29] and found out that almost all matrices rotated in a sector of accidentally coinciding eigenvalues became block diagonal, therefore corresponding R2 matrices should be block-diagonal too. One can see the results in the table 1. If R-matrix has coinciding eigenvalues (λi = λj ) .. . λi . .. (24) R= λj .. . we can rotate corresponding Racah matrix U in the sector i − j with the rotation matrix O: .. . c −s . .. O= s c .. .
(25)
U 0 = OUOT ,
c 0 0 s
0 1 0 0
··· ··· .. .
0 0
··· ···
1 0
−s ui,i ui,i+1 ui+1,i ui+1,i+1 0 0 uj−1,i uj−1,i+1 uj,i uj,i+1 c 0 ui,i u0i,i+1 0 ui+1,i ui+1,i+1 u0j−1,i uj−1,i+1 u0j,i u0j,i+1
··· ··· .. .
ui,j−1 ui+1,j−1
··· ···
uj−1,j−1 uj,j−1
··· ··· .. . ··· ··· 6
u0i,j−1 ui+1,j−1 uj−1,j−1 u0j,j−1
(26) ui,j c 0 ui+1,j uj−1,j 0 uj,j −s 0 ui,j u0i+1,j . 0 uj−1,j u0j,j
0 1 0 0
... ... .. .
0 0
... ...
1 0
s 0 = 0 c
(27)
(28)
To make U 0 block-diagonal one can determine the angle of rotation from the requirement that diagonal elements inside the sector of final matrix equal zero: u0i,j = u0j,i = 0. (29) This method works only if ui,j = uj,i in the initial matrix, otherwise equations are unsolvable. If ui,j 6= uj,i we can use two different rotation matrices: .. .. . . c2 −s2 c1 −s1 .. .. (30) O1 = and O2 = . . s c s c 2 2 1 1 .. .. . . to get a new mixing matrix: U 0 = O1 UO2T ,
c1 0 0 s1
0 ··· 1 ··· .. .
0 0
0 ··· 0 ···
1 0
(31)
c2 ui,i ui,i+1 ··· ui,j−1 ui,j −s1 ui+1,i ui+1,i+1 · · · ui+1,j−1 ui+1,j 0 0 .. . uj−1,i uj−1,i+1 · · · uj−1,j−1 uj−1,j 0 0 −s2 uj,i uj,i+1 ··· uj,j−1 uj,j c1 0 ui,i ui,i+1 ··· u0i,j−1 u0i,j 0 ui+1,i ui+1,i+1 · · · ui+1,j−1 u0i+1,j .. . . u0j−1,i uj−1,i+1 · · · uj−1,j−1 u0j−1,j u0j,i u0j,i+1 ··· u0j,j−1 u0j,j
0 ... 1 ... .. .
0 0
0 ... 0 ...
1 0
s2 0 = 0 c2
(32)
(33)
The angle of the rotation can be found from the same requirement (29) as in the previous case.
5
Representation [2,1]
We know that [2, 1] ⊗ [2, 1] = [4, 2] ⊕ [4, 1, 1] ⊕ [3, 3] ⊕ [3, 2, 1] ⊕ [3, 2, 1] ⊕ [3, 1, 1, 1] ⊕ [2, 2, 2] ⊕ [2, 2, 1, 1],
(34)
where there are two types of irreducible representations: symmetric and antisymmetric (underlined). Corresponding R-matrices have eigenvalues: λ[4,2] = q15 , λ[3,1,1,1] = q 3 ,
λ[4,1,1] = − q13 , λ[3,3] = − q13 , 3 λ[2,2,2] = q , λ[2,2,1,1] = −q 5 ,
λ[3,2,1]± = ±1,
(35)
and one can see that there are two pairs of accidentally coinciding eigenvalues:
5.1
λ[4,1,1] = λ[3,3] = − q13 , λ[3,1,1,1] = λ[2,2,2] = q 3 .
(36)
R[5,3,1] = diag(q κ[2,2,1,1] , q κ[2,2,1,1] , −q κ[3,1,1,1] , −q κ[2,2,2] , −q κ[3,2,1] , q κ[3,2,1] ) = diag(−q −5 , −q −5 , q −3 , q −3 , −1, 1).
(37)
[5,3,1]
We see that λ1 = λ2 and λ3 = λ4 , that’s why we rotate U[5,3,1] in sectors 1 − 2 and 3 − 4. And we have q [5,3,1] q(1+q 2 )(1+q 4 )2 = √12 1 + (1+q2 (1+(−1+q)q)(1+q+q c12 2 )(2+q 4 +q 6 ) , (38) q [5,3,1]
s12
=
√1 2
(1+(−1+q)q)(1+q+q 2 +q 3 +q 4 )(1+(−1+q)q(1+(−1+q)q)(1+q+q 2 )) , (1+q 2 (1+(−1+q)q)(1+q+q 2 )(2+q 4 +q 6 )
7
Table of matrices with accidentally coinciding eigenvalues [2, 1] size
repr
eigenvalues repetitive accidental
result
blocks
6
[5,3,1]
λ1 = λ2
λ3 = λ4
+
(3,3)
8
[4,3,2]
λ4 = λ5 , λ6 = λ7
λ2 = λ3
+
(3,5)
9
[4,3,1,1]
λ4 = λ5 , λ6 = λ7
λ2 = λ3
-
and [5,3,1]
c34
=
[5,3,1]
√1 , 2
s34
= − √12 .
(39)
After the rotations U[5,3,1] becomes block-diagonal:
U 0 [5,3,1]
q2 (1+q+q2 ) − (1+q2 )y1 0 x13 = 0 0
0
x13
0
0
x16
q 2 (1−q+q 2 ) (1+q 2 )y2
0
x24
x25
0
−1+q−q 2 +q 3 −q 4 (1+q 2 )(1−q+q 2 )
0
0
y1 y1+q 2
0 x45
x36 0
x52
0
x45
− 1+q2 −qq3 +q4 +q6
0
0
x36
0
0
q − y1+q 6
,
0 x24
x16 where x13 =
x16 =
q 1+q 2
1+q+q 2 +q 3 +q 4 r
x52 =
1+q+q 2 +q 3 +q 4 +q 5 +q 6 , 1+q 2 +q 3 +q 4 +q 6
r 2 +q 3 +q 4 +q 5 +q 6 (1+q 4 ) 1+q+q 1−q+q 2 +q 4 −q 5 +q 6
q
x36 =
q
1+q 4 1+q+q 2 +q 3 +q 4 1−q+q 2
q x24 = − 1+q 2 r q
,
x45 = √
,
x25 =
r 2 −q 3 +q 4 −q 5 +q 6 (1+q 4 ) 1−q+q 1+q+q 2 +q 4 +q 5 +q 6 1−q+q 2 −q 3 +q 4
q
3
3
(40)
1−q+q 2 −q 3 +q 4 −q 5 +q 6 , 1+q 2 −q 3 +q 4 +q 6
1+q 4 1−q+q 2 −q 3 +q 4 1+q+q 2
,
(1+q 4 )(1+q 2 −q 3 +q 4 +q 6 )(1−q+q 2 −q 3 +q 4 −q 5 +q 6 ) , (1+q+q 2 )(1−q+q 2 −q 3 +q 4 )3/2
(41)
,
y1 = 1 + q + q 2 + q 3 + q 4 ,
y2 = 1 − q + q 2 − q 3 + q 4 .
Changing lines of the matrix (40) we can see that it has a block-diagonal structure with blocks 3 × 3.
5.2
[4,3,2] R[4,3,2] = diag (q κ[2,2,1,1] , −q κ[3,1,1,1] , −q κ[2,2,2] , −q κ[3,2,1] , −q κ[3,2,1] , q κ[3,2,1] , q κ[3,2,1] , q κ[3,3] ) = diag (−q 5 , q 3 , q 3 , −1, −1, 1, 1, −q −3 )
(42)
Despite the fact that here we have tree pairs of coinciding eigenvalues: λ2 = λ3 , λ4 = λ5 and λ6 = λ7 , we need just one rotation in sector 2 − 3 with [4,3,2]
c23
=
√1 , 2
[4,3,2]
s23
8
= − √12 .
(43)
to make U[5,3,1] block-diagonal: x11 x12 0 2q 2 x12 0 (1+q 2 )2 0 0 0 x14 x24 0 U 0 [4,3,2] = 0 0 − √12 0 0 − √12 x x27 0 17 x18 x28 0
x14 x24 0
0 0
0 0
− √12
− √12
0
0
1 2
0 0
1 2 − 21
− 12
0 0
1 2
0
0
0
0
1+q−3q 2 +q 3 +q 4 2(1−q+q 2 −q 3 +q 4 )
−
2 q√ (1+q 2 ) 1+q 4
x17 x27 0
1 2
√
x18 x28 0
(−1+q+3q 2 +q 3 −q 4 )2 2(1+q+q 2 +q 3 +q 4 )
−
2 q√ (1+q 2 ) 1+q 4
2 q√ − (1+q 2 ) 1+q 4 , 0 0 2 q√ − (1+q 2 ) 1+q 4 4 q (1+q 2 )2 (1+q 4 )
(44) where q 4 (1+q 2 +3q 4 +q 6 +q 8 ) (1+q 2 )2 (1+q 2 +2q 4 +2q 6 +2q 8 +q 10 +q 12 ) ,
x11 =
x12 =
q
2q 6 (1+q 4 +q 8 ) (1+q 2 )4 (1+q 2 +2q 4 +2q 6 +2q 8 +q 10 +q 12 ) ,
x14 =
q
q 2 (1−q+q 2 )3 (1+q+q 2 )(1−q 2 +q 4 ) (1+q 2 )2 (1+q 4 )(1−q+q 2 −q 3 +q 4 )2 ,
x17 = −
q
q 2 (1−q+q 2 )(1+q+q 2 )3 (1−q 2 +q 4 ) (1+q 2 )2 (1+q 4 )(1+q+q 2 +q 3 +q 4 )2 ,
x24 =
q
(q−1)4 (1+q+q 2 +q 3 +q 4 ) 2(1+q 2 )2 (1−q+q 2 −q 3 +q 4 ) ,
x27 = −
q
(1+q)4 (1−q+q 2 −q 3 +q 4 ) 2(1+q 2 )2 (1+q+q 2 +q 3 +q 4 ) ,
x18 =
q
(1−q 2 +q 4 )(1+q 2 +q 4 )3
√ (1+q 2 )4 (1+q 4 )2
r 2q
x28 = −
,
1+q 2 +q 4 +q 6 +q 8 1+q 4 (1+q 2 )2
(45)
.
Changing lines of the matrix (45) we can see that it has a block-diagonal structure with blocks 5 × 5 and 3 × 3.
6
Representation [3,1]
[3, 1]⊗2 = [6, 2] ⊕ [6, 1, 1] ⊕ [5, 3] ⊕ 2[5, 2, 1] ⊕ [5, 1, 1, 1] ⊕ [4, 4] ⊕ 2[4, 3, 1] ⊕ [4, 2, 2] ⊕ [4, 2, 1, 1] ⊕ [3, 3, 2] ⊕ [3, 3, 1, 1]. (46) Eigenvalues are: λ[6,2] = q −14 , λ[6,1,1] = −q −12 , λ[5,3] = −q −10 , λ[5,2,1]± = ±q −7 , −4 −4 −8 λ[5,1,1,1] = q , λ[4,4] = q , λ[4,3,1]± = ±q , λ[4,2,2] = q −2 , λ[4,2,1,1] = −1, λ[3,3,2] = −1, λ[3,3,1,1] = q 2 .
(47)
And here we have two pairs of eigenvalues which coincide:
6.1
λ[5,1,1,1] = λ[4,3,1]+ = q −4 λ[4,2,1,1] = λ[3,3,2] = −1
(48)
R[4,3,2,2,1] = diag(q κ[3,3,1,1] , −q κ[4,2,1,1] , −q κ[3,3,2] , q κ[4,2,2] ) = diag(q 2 , 1, 1, q −2 ).
(49)
[4,3,2,2,1]
After the rotation in the sector 2 − 3 with [4,3,2,2,1]
c23
=
√1 , 2
[4,3,2,2,1]
s23
= − √12 .
(50)
R[4,3,2,2,1] becomes block-diagonal: √ √ ( 2q 1+q 2 +q 4 (1+q 2 )2 (−1−q 4 ) (1+q 2 )2
[4,3,2,2,1]
U 0 [4,3,2,2,1] = O23
[4,3,2,2,1] †
U[4,3,2,2,1] (O23
) =
q2 √ (1+q √ 2 )2 2q 1+q 2 +q 4 − (1+q2 )2
0
1+q 2 +q 4 (1+q 2 )2
√0
√ ( 2q 1+q 2 +q 4 − (1+q 2 )2
0 0 1 0
1+q 2 +q 4 (1+q 2 )2 √ √ 2q 1+q 2 +q 4 (1+q 2 )2 .
0
q2 (1+q 2 )2
(51) Changing lines of the matrix U 0 [4,3,2,2,1] we get blocks 3 × 3 and 1 × 1. 9
Table of matrices with accidentally coinciding eigenvalues [3, 1]
6.2
size
repr
eigenvalues repetitive accidental
result
blocks
4
[4,3,2,2,1]
λ2 = λ3
+
(3, 1)
5
[4,4,2,2]
λ1 = λ2
+
(3, 2)
6
[4,3,3,1,1]
λ1 = λ2
λ3 = λ4
+
(3, 3)
9
[5,3,2,2]
λ4 = λ5
λ2 = λ3
-
-
[4,4,3,1]
λ5 = λ6 , λ7 = λ8
λ2 = λ3
+
(3, 6)
[4,4,2,2]
R[4,4,2,2] = diag(−q κ[4,2,1,1] , −q κ[3,3,2] , q κ[4,2,2] , q κ[4,3,1] , −q κ[4,3,1] ) = diag(−1, −1, q −2 , q −4 , −q −4 )
(52)
R[4,4,2,2] has a pair of accidentally coinciding eigenvalues. We rotate R[4,4,2,2] in the sector 1 − 2 with [4,4,2,2]
c12
=
[4,4,2,2]
√1 , 2
s12
= − √12 .
(53)
and get
U 0 [4,4,2,2]
x11 0 = 0 −x14 0
0 x22 x23 0 x25
0 x23 x33 0 x23
x14 0 0 x11 0
0 −x25 x23 , 0 −x22
where x11 =
q2 1+q 4 ,
x25 =
1+q 2 +q 4 (1+q 2 )2 ,
√
2
q x22 = − (1+q 2 )2 , √ √ √ 1−q+q 2 1+q+q 2 1−q 2 +q 4 x14 = , 1+q 4
x23 = −
(54)
√ 2q
√
1−q+q 2 1+q+q 2 , (1+q 2 )2
4
1+q x33 = − (1+q 2 )2 .
(55)
Changing lines in the matrix we get blocks 3 × 3 and 2 × 2.
6.3
[4,3,3,1,1]
R[4,3,3,1,1] = diag(q κ[3,3,1,1] , q κ[3,3,1,1] , −q κ[4,2,1,1] , −q κ[3,3,2] , q κ[4,3,1] , −q κ[4,3,1] ) = diag(q 2 , q 2 , −1, −1, q −4 , −q −4 )
(56)
R[4,3,3,1,1] has two pairs of coinciding eigenvalues. We use rotation matrices in sectors 1 − 2 and 3 − 4 with: [4,3,3,1,1]
=
[4,3,3,1,1]
=
c12
s12
q
1+q 2 +q 4 √1 (1 + q 4 ) 1+2q 4 +2q 8 +q 1 2 , 2q 1−q 2 +q 4 +q 8 −q 1 0+q 1 2 √1 1+2q 4 +q 6 +2q 8 +q 1 2 , 2
(57)
and [4,3,3,1,1]
c34
=
√1 , 2
[4,3,3,1,1]
s34
and get
10
=
√1 . 2
(58)
U 0 [4,3,3,1,1] = 2 +q 6 − 1+qq2 +q 6 +q 8 0 0 A41 0 A61
0
0
q2 1+q 2 +q 4
−√
√ q2 4 1+q +q
q 1+q 2 +q 4
√
q
1+q 2 +q 4 +q 6 +q 8 1+q 2 +q 4
1+q 2 +q 4 (1+q 2 )2
0
0
q4 1+q 2 +2q 4 +q 6 +q 8
1+q 2 +q 4 +q 6 +q 8 1+q 2 +q 4
0
0
where
A14 = −
(1+q 2 )
q+q 9 1−q 2 +q 4 +q 8 −q 10 +q 12
√ 2
(1+q 2 +q 4 +q 6 +q 8 ) 1+q 2 +q 4 1+q 4
0,
q
(1+q 2 )
1+q 2 +q 4 +q 6 +q 8 1+q 2 +q 4 r
q(1+q 8 ) 1−q 2 +q 4 +q 8 −q 10 +q 12
−q
1+q 2 +q 4 +q 6 +q 8 1−q 2 +q 4 (1+q 2 )2
A16
0
1+q 4
√ 2
√
0
r
A41 =
0
−1+q 2 −q 4 1+q 4
0 r
A14
0
0 √ 1+q 2 +q 4 +q 6 +q 8 (1−q 2 +q 4 ) q (1+q 2 )2 0 4 q − 1+q2 +q 6 +q 8 (59) 0
√ (1+q 8 )(1+q 2 +q 4 +q 6 +q 8 ) A61 = − 1+q 2 +q 6 +q 8 √ (1+q 8 )(1+q 2 +q 4 +q 6 +q 8 ) A16 = 1+q 2 +q 6 +q 8
(60)
Changing lines of the matrix U 0 [4,3,3,1,1] we can see that it has a block-diagonal structure with two blocks 3 × 3
6.4
[4,4,3,1]
R[4,4,3,1] = diag(q κ[3,3,1,1] , −q κ[4,2,1,1] , −q κ[3,3,2] , q κ[4,2,2] , q κ[4,3,1] , q κ[4,3,1] , −q κ[4,3,1] , −q κ[4,3,1] , q κ[4,4] ) = diag(q 2 , −1, −1, q −2 , q −4 , q −4 , −q −4 , −q −4 , q −8 )
(61)
R[4,4,3,1] has three pairs of coinciding eigenvalues: λ2 = λ3 (accidentally), λ5 = λ6 , λ7 = λ8 . We rotate [4,4,3,1]
[4,4,3,1]
corresponding Racah matrix U [4,4,3,1] just in sectors 2-3 and 7-8, because U56 = U65 = 0. To get just one rotation matrix in the sector 2 − 3 we multiply the third line of U [4,4,3,∞] with −1. And we have [4,4,3,1] [4,4,3,1] (62) c23 = √12 , s23 = − √12 . Second rotation is in the sector 7 − 8. Rotation matrices are with q q 2 +q 4 [4,4,3,1] [4,4,3,1] 1−q 2 +q 4 , s = − c1,78 = 1+q 4 1,78 2(1+q ) 2(1+q 4 )
(63)
and [4,4,3,1]
c2,78
[4,4,3,1]
= 1 , s2,78
(64)
= 0.
Finally we get:
x11 0 0 0 0 −x13 x14 0 0 − √12 −x16 0 −x17 0 0 − √12 x19 0
x13 0
x14 0
2
0 √1 2
− (q22q+1)2
−x34
0
x34
q 8 +q 6 +3q 4 +q 2 +1 (q 2 +1)2 (q 4 +q 2 +1)
0
0
0
− 12
−r
1 4q 2 +2 q 4 −q 2 +1
x27 0 −x39
√
q q 4 +q 2 +1 5 q√ +q q 4 +q 2 +1
(q 2 +1)2
0 x49
−r
x16 0
x17 0
− √12
1
x37
0
4q 2 +2 q 4 −q 2 +1
−√
q q 4 +q 2 +1
q (q 4 +1) (q 2 +1)2
0 4
2
0
−q +q −1 2(q 4 +q 2 +1)
0
1 2
1 2
0 x69
0
−
−q
8
√
q 4 +q 2 +1
0
0
0
− 12
1 2
0
+3q 6 −4q 4 +3q 2 +1 2(q 8 +q 6 +q 2 +1)
0 −x79
0 1 2
0
x19 0 x39 x49 0 , −x69 x79 0 x99 (65)
11
where
√
x11 =
q 4 ((q 2 +1)(q 8 +2q 4 +q 2 +1)q 2 +1) (q 2r +1)2 ((q 4 +1)(q 10 +q 8 +q 6 +q 4 +2q 2 +1)q 2 +1) q
q 12 +q 10 +q 8 +q 6 +q 4 +q 2 +1 (q2 +1)(q8 +q6 +2q4 +q2 +2)q2 +1
2
x14 = x17 =
(q √
x34 = x39 =
r
+q )
x16 =
q 12 +q 10 +q 8 +q 6 +q 4 +q 2 +1
(q4 +1)(q10 +q8 +q6 +q4 +2q2 +1)q2 +1 (q 2 +1)2
q 12 +q 10 +q 8 +q 6 +q 4 +q 2 +1 q 8 +q 6 +q 4 +q 2 +1 q 4 +q 2 +1
r
q 12 +q 10 +q 8 +q 6 +q 4 +q 2 +1 q 8 +q 6 +q 4 +q 2 +1 q 8 +q 6 +q 2 +1
x19 =
2q q 4 −q 2 +1 2 (q 2 +1) r
(q4 +1)
x37 = −
q 12 +q 10 +q 8 +q 6 +q 4 +q 2 +1 q 4 +q 2 +1
x49 =
q 8 +q 6 +q 4 +q 2 +1 q3 (q 2 +1)(q 8 +q 6 +2q 4 +q 2 +2)q 2 +1 q6 (q 2 +1)(q 8 +q 6 +2q 4 +q 2 +2)q 2 +1
x69 = √ x99 =
r
√
(q4 +1)
r
x13 = q
(q 2 +1)2 5
2q 3
x79 =
(q
2
q 12 +q 10 +q 8 +q 6 +q 4 +q 2 +1 q 4 +q 2 +1
q 8r +q 6 +q 4 +q 2 +1 2
2
2q −1) +1 q 4 −q 2 +1 √ 2(q 2 +1)2
(66)
6
+q 2 q 8 +q 6 +q 4 +q 2 +1 q3 √ 4 2 (q +q +1)(q 8 +q 6 +q 4 +q 2 +1) (q 4 +q 2 +1)
√q
Changing lines of the matrix (65) we can see that it has a block-diagonal structure with blocks 3 × 3 and 6 × 6.
7
Representation [3,2] [3, 2]⊗2 = [6, 4] ⊕ [6, 3, 1] ⊕ [6, 2, 2] ⊕ [5, 5] ⊕ [5, 4, 1] ⊕ [5, 4, 1] ⊕ [5, 3, 2] ⊕ [5, 3, 2] ⊕ [5, 3, 1, 1]⊕ (67) [5, 2, 2, 1]⊕, [4, 4, 2] ⊕ [4, 4, 1, 1] ⊕ [4, 3, 3] ⊕ [4, 3, 2, 1] ⊕ [4, 3, 2, 1] ⊕ [4, 2, 2, 2] ⊕ [3, 3, 3, 1] ⊕ [3, 3, 2, 2],
where antisymmetric irreducible representations are underlined. Corresponding R−matrices have eigenvalues: λ[6,4] = q −17 , λ[6,3,1] = −q −13 , λ[6,2,2] = q −11 , λ[5,5] = −q −15 , λ[5,4,1]± = ±q −10 , λ[5,3,2]± = ±q −7 , λ[5,3,1,1] = q −5 , λ[5,2,2,1] = −q −3 , λ[4,4,2] = q −5 , λ[4,4,1,1] = −q −3 , λ[4,3,3] = −q −3 , λ[4,3,2,1]± = ±1, λ[4,2,2,2] = q 3 , λ[3,3,3,1] = q 3 , λ[3,3,2,2] = −q 5 .
(68)
One can see that several eigenvalues accidentally coincide: λ[5,3,1,1] = λ[4,4,2] = q −5 λ[5,2,2,1] = λ[4,4,1,1] = λ[4,3,3] = −q −3 λ[4,2,2,2] = λ[3,3,3,1] = q 3 .
(69)
Table of matrices with accidentally coinciding eigenvalues [3, 2]
7.1
size
repr
eigenvalues repetitive accidental
6
[4,3,3,2,2,1]
λ5 = λ6
[6,4,2,1,1,1]
λ1 = λ2
result
blocks
λ3 = λ4
+
(3,3)
λ3 = λ4
+
(3,3)
[4,3,3,2,2,1]
R[4,3,3,2,2,1] = diag(−q κ[4,3,2,1] , q κ[4,3,2,1] , q κ[4,2,2,2] , q κ[3,3,3,1] , −q κ[3,3,2,2] , −q κ[3,3,2,2] ) = diag(−1, 1, q 3 , q 3 , −q 5 , −q 5 ). (70) R[4,3,3,2,2,1] has two pairs of coinciding eigenvalues: λ3 = λ4 (accidentally) and λ5 = λ6 . To get just one rotation matrix in the sector 3 − 4 we change the sign of the third line of the Racah matrix U[4,3,3,2,2,1] . For the [4,3,3,2,2,1]
O34
we use 12
[4,3,3,2,2,1]
c34
=
[4,3,3,2,2,1]
√1 , 2
s34
= − √12 .
(71)
For the second rotation in the sector 5 − 6 we use two rotation matrices with r [4,3,3,2,2,1]
c1,56
=
√
r [4,3,3,2,2,1] c2,56
=
q (q 2 +1)(q 8 +1) +1 q 12 +2q 10 +2q 8 +q 6 +2q 4 +2q 2 +1
2 q (q 2 +1)(q 8 +1) 1− q12 +2q10 +2q8 +q6 +2q4 +2q2 +1 √ 2
r [4,3,3,2,2,1]
,
s1,56
=
[4,3,3,2,2,1] s2,56
=
q (q 2 +1)(q 8 +1) 1− q12 +2q10 +2q8 +q6 +2q4 +2q2 +1 √ 2 r q (q 2 +1)(q 8 +1) +1 q 12 +2q 10 +2q 8 +q 6 +2q 4 +2q 2 +1 √ 2
(72)
Finally, we get a matrix: U 0 [4,3,3,2,2,1] = 3 q
0
x14
−x15
0
q − q6 +q4 +q 3 +q 2 +1 −x23 0 0
x23 −x33 0 0
0 0 x44 −x45
0 0 x45 q 2 ((q−1)q+1) (q 2 +1)((q−1)q(q 2 +1)+1)
−x26
−x36
0
0
x26 −x36 0 0 q 2 (q 2 +q+1)
q 6 +q 4 −q 3 +q 2 +1
0
0 0 x14 x15 0
where
x14 = x23 = x33 = x44 =
3
√
, (73)
(q 2 +1)(q 4 +q 3 +q 2 +q+1)
√
(q 4 +1)((q−1)q(q 2 +1)+1) , 3 +q 2 +1 √ 4q6 +q4 −q q (q +1)(q 4 +q 3 +q 2 +q+1) , q 6 +q 4 +q 3 +q 2 +1 2 q ((q−1)q+1)(q 2 +1) − 1, q2 q 4 +q 3 +2q 2 +q+1 − 1, q
x15 = x26 = x36 = x45 =
(q 4 +1)((q 2 +1)(q 4 −q+1)q 2 +1) , q 6 +q 4 −q 3 +q 2 +1 (q 4 +1)((q 2 +1)(q 4 +q+1)q 2 +1) , q 6 +q 4 +q 3 +q 2 +1 √ q (q 4 +q 3 +q 2 +q+1)((q 2 +1)(q 4 +q+1)q 2 +1) , (q 6 +2q 4 +q 3 +2q 2 +q+2)q 2 +1 q (−(q−1)q((q−1)q+1)(q 2 +q+1)−1)
√
√
(q 2 +1)
((q−1)q(q 2 +1)+1)((q 2 +1)(q 4 −q+1)q 2 +1)
(74) .
One can see that there are blocks 3 × 3 and 3 × 3 in U 0 [4,3,3,2,2,1] .
7.2
[6,4,2,1,1,1]
R[6,4,2,1,1,1] = diag(q κ[5,3,1,1] , q κ[5,3,1,1] , −q κ[5,2,2,1] , −q κ[4,4,1,1] , −q κ[4,3,2,1] , q κ[4,3,2,1] ) = diag(q −5 , q −5 , −q −3 , −q −3 , −1, 1). (75) R[6,4,2,1,1,1] has two pairs of coinciding eigenvalues: λ1 = λ2 and λ3 = λ4 (accidentally). The rotation in the sector 3 − 4 should be done with [6,4,2,1,1,1]
c34
=
√1 , 2
[6,4,2,1,1,1]
s34
= − √12 .
(76)
and rotation in the sector 1 − 2 requires two rotation matrices with q [6,4,2,1,1,1] c1,12
[6,4,2,1,1,1] s1,12
=
= q
(q 14 +q 10 +q 8 +q 7 +q 6 +q 4 +1)2 ((q−1)q+1)(q 6 +q 5 +q 4 +q 3 +q 2 +q+1)((q−1)q(q 2 +1)+1) (q 4 +1)(q 34 +2q 32 +3q 30 +5q 28 +8q 26 +11q 24 +14q 22 +17q 20 +19q 18 +20q 16 +18q 14 +17q 12 +15q 10 +11q 8 +7q 6 +5q 4 +4q 2 +2)q 2 +1
√
2
q(q 14 +q 10 +q 8 −q 7 +q 6 +q 4 +1)2 (q 2 +1)(q 9 +q 8 +q 7 +q 5 +2q+1)+1 (q 4 +1)(q 34 +2q 32 +3q 30 +5q 28 +8q 26 +11q 24 +14q 22 +17q 20 +19q 18 +20q 16 +18q 14 +17q 12 +15q 10 +11q 8 +7q 6 +5q 4 +4q 2 +2)q 2 +1
√
q [6,4,2,1,1,1] c2,12
=
q(q 2 +1)(q 8 +1) q 12 +2q 10 +2q 8 +q 6 +2q 4 +2q 2 +1
√
q [6,4,2,1,1,1] s2,12
=−
2
1−
2
+1 ,
q(q 2 +1)(q 8 +1) q 12 +2q 10 +2q 8 +q 6 +2q 4 +2q 2 +1
√
2
. (77)
13
,
,
U 0 [6,4,2,1,1,1] = q 2 (−q 2 +q−1) (q2 +1)((q−1)q(q2 +1)+1) 0 0 x 14 0
0
0
q 2 (q 2 +q+1) − (q2 +1)(q4 +q3 +q2 +q+1)
x16
x14
−x16
0 −x25 x35 0
x23 0 x25
x23 x33 0 −x35
0 0 x44 0
q3 q 6 +q 4 +q 3 +q 2 +1
0 0 x46 0
0
0
x46
0
q − q6 +q4 −q 3 +q 2 +1
3
, (78)
where x14 =
(q 2 +1)
x23 = − x33 =
q −(q−1)q((q−1)q+1)(q 2 +q+1)−1) √( 2 2 4
((q−1)q(q +1)+1)((q +1)(q −q+1)q 2 +1) q (q 6 +q 5 +q 4 +q 3 +q 2 +q+1)
√
(q 2 +1)
,
(q 4 +q 3 +q 2 +q+1)((q 2 +1)(q 4 +q+1)q 2 +1)
q2 ((q−1)q+1)(q 2 +1)
x44 = 1 −
√
− 1,
x16 = ,
x35 =
q2 q 4 +q 3 +2q 2 +q+1 ,
√
x25 =
x46 =
(q 4 +1)((q 2 +1)(q 4 −q+1)q 2 +1) , q 6 +q 4 −q 3 +q 2 +1 (q 4 +1)((q 2 +1)(q 4 +q+1)q 2 +1) , q 6 +q 4 +q 3 +q 2 +1
√
(79)
(q 4 +1)(q 4 +q 3 +q 2 +q+1) , √q6 4+q4 +q3 +q2 +12 q (q +1)((q−1)q(q +1)+1) − . q 6 +q 4 −q 3 +q 2 +1 q
One can see that U 0 [6,4,2,1,1,1] has a block-diagonal form with blocks 3 × 3 and 3 × 3.
8
Representation [4,1] [4, 1]⊗2 = [8, 2] ⊕ [8, 1, 1] ⊕ [7, 3] ⊕ [7, 2, 1] ⊕ [7, 2, 1] ⊕ [7, 1, 1, 1] ⊕ [6, 4] ⊕ [6, 3, 1] ⊕ [6, 3, 1] ⊕ [6, 2, 2]⊕ (80) [6, 2, 1, 1] ⊕ [5, 5] ⊕ [5, 4, 1] ⊕ [5, 4, 1] ⊕ [5, 3, 2] ⊕ [5, 3, 1, 1] ⊕ [4, 4, 2] ⊕ [4, 4, 1, 1].
Eigenvalues are: λ[8,2] = q −27 , λ[8,1,1] = −q −25 , λ[7,3] = −q −21 , −18 −15 , λ[7,1,1,1] = q , λ[6,4] = q −17 , λ[7,2,1]± = ±q −13 −11 , λ[6,2,2] = q , λ[6,2,1,1] = −q −9 , λ[6,3,1]± = ±q −10 −15 , λ[5,3,2] = −q −7 , λ[5,5] = −q , λ[5,4,1]± = ±q −5 −5 λ[5,3,1,1] = q , λ[4,4,2] = q , λ[4,4,1,1] = −q −3 ,
(81)
where there is just one pair of accidentally coinciding eigenvalues: λ[5,3,1,1] = λ[4,4,2] = q −5 .
(82)
Table of matrices with accidentally coinciding eigenvalues [4, 1] size
repr
eigenvalues repetitive accidental
result
blocks
4
[5,4,3,2,1]
-
λ2 = λ3
+
(3,1)
6
[5,5,3,2]
λ3 = λ4
λ5 = λ6
+
(3,3)
[5,4,4,1,1]
λ2 = λ3 , λ4 = λ5
λ6 = λ7
+
(3,3)
[6,4,2,2,1]
-
λ4 = λ5
-
14
8.1
[5,4,3,2,1]
R[5,4,3,2,1] = diag(−q κ[5,3,2] , q κ[5,3,1,1] , q κ[4,4,2] , −q κ[4,4,1,1] ) = diag(−q −7 , q −5 , q −5 , −q −3 ). (83) There are just one pair of coinciding eigenvalues: λ2 = λ3 . The rotation matrix is with [5,4,3,2,1]
c23
=
[5,4,3,2,1]
√1 , 2
s23
= − √12 .
(84)
After the rotation U[5,4,3,2,1] goes to U 0 [5,4,3,2,1]
√
q2 √
=
(q 2 +1)2
√
√
2q q 4 +q 2 +1 (q 2 +1)2
2q q 4 +q 2 +1 (q 2 +1)2
0
q 4 +1 (q 2 +1)2
0
√
q 4 +q 2 +1 (q 2 +1)2
−
0 −1
√0 4
2q q +q 2 +1 (q 2 +1)2
q 4 +q 2 +1 (q 2 +1)2 √ √ 2q q 4 +q 2 +1 − (q2 +1)2
0 q2 (q 2 +1)2
0
,
(85)
where one can see 3 × 3 and 1 × 1 blocks.
8.2
[5,5,3,2]
R[5,5,3,2] = diag(−q κ[5,4,1] , q κ[5,4,1] , −q κ[5,3,2] , −q κ[5,3,2] , q κ[5,3,1,1] , q κ[4,4,2] ) = diag(−q −10 , q −10 , −q −7 , −q −7 , q −5 , q −5 ). (86) There are two pairs of coinciding eigenvalues: λ3 = λ4 and λ5 = λ6 (accidentally). cos and sin for rotation in the sector 5 − 6 are [5,5,3,2]
c56
=
[5,5,3,2]
√1 , 2
= − √12 .
s56
(87)
For the second rotation one needs two matrices with r [5,5,3,2] c1,34
=
√
r [5,5,3,2]
c2,34
q (q 4 +1)(q 2 +1)3 +1 ((q−1)q+1)(q 2 +1)(q 2 +q+1)(q 4 +2)q 2 +1
=
2
q (q 2 +1)(q 4 +1)2 +1 ((q−1)q+1)(q 2 +q+1)(q 6 +q 4 +2)q 2 +1
√
2
r
,
[5,5,3,2] s1,34
1−
=−
√
r [5,5,3,2]
,
s2,34
=−
q (q 2 +1)3 (q 4 +1) ((q−1)q+1)(q 2 +1)(q 2 +q+1)(q 4 +2)q 2 +1
2
,
((q−1)q+1)(q 4 +q 3 +q 2 +q+1)((q−1)q((q−1)q+1)(q 2 +q+1)+1) ((q−1)q+1)(q 2 +q+1)(q 6 +q 4 +2)q 2 +1
√
. (88)
2
After the rotations we get U 0 [5,5,3,2] =
q3 q 6 +q 4 +q 3 +q 2 +1
0
0 x13 0 x15
x13
0
x15
0
q − q6 +q4 −q 3 +q 2 +1 0 x24
0 x33 0
x24 0 x44
0 x35 0
−x26 0 −x46
0
x35
0
q 2 (q 2 +q+1) (q 2 +1)(q 4 +q 3 +q 2 +q+1)
0
0
q 2 (−q 2 +q−1) (q 2 +1)((q−1)q(q 2 +1)+1)
3
0
x26
0
x46
, (89)
where x13 = x24 =
√
(q 4 +1)(q 4 +q 3 +q 2 +q+1) , √q6 4+q4 +q3 +q2 +12 q (q +1)((q−1)q(q +1)+1) − , q 6 +q 4 −q 3 +q 2 +1 q
x33 = q
1 q 2 +1
x44 = 1 −
+
1 −q 2 +q−1 q2
q 4 +q 3 +2q 2 +q+1
,
+ 1,
√
x15 = x26 = x35 =
((q−1)q+1)(q 4 +1)(q 6 +q 5 +q 4 +q 3 +q 2 +q+1) , q 6 +q 4 +q 3 +q 2 +1 (q 2 +q+1)(q 4 +1)((q−1)q((q−1)q+1)(q 2 +q+1)+1) , q 6 +q 4 −q 3 +q 2 +1 √ q ((q−1)q+1)(q 4 +q 3 +q 2 +q+1)(q 6 +q 5 +q 4 +q 3 +q 2 +q+1) − , (q 6 +2q 4 +q 3 +2q 2 +q+2)q 2 +1
√
x46 = q
q
(q−1)q((q−1)q+1)(q 2 +q+1)+1 . (q 2 +1)(q 6 +q 4 −q 3 +q 2 +1)
U 0 [5,5,3,2] has blocks 3 × 3 and 3 × 3. 15
(90)
8.3
[5,4,4,1,1]
R[5,4,4,1,1] = diag(−q κ[5,4,1] , q κ[5,4,1] , q κ[5,3,1,1] , q κ[4,4,2] , −q κ[4,4,1,1] , −q κ[4,4,1,1] ) = diag(−q −10 , q −10 , q −5 , q −5 , −q −3 , −q −3 ). (91) There are two pairs of coinciding eigenvalues: λ3 = λ4 (accidentally) and λ5 = λ6 . For the first rotation we use [5,4,4,1,1]
c34
=
√1 , 2
[5,4,4,1,1]
s34
= − √12 .
(92)
For the rotation in the sector 5 − 6 we need two matrices with q 16 14 13 12 10 9 8 7 6 4 3 2 [5,4,4,1,1] −q +q +2q −q +q −q +2q +q −q +q +1 , c1,56 = q +q 2(q 16 +q 14 +q 12 +2q 10 +q 8 +2q 6 +q 4 +q 2 +1) q 6 3 [5,4,4,1,1] +1)(q 10 +q 8 +q 6 −q 5 +q 4 +q 2 +1) s1,56 = − 2(−q26 −2q24 −3q22 +q21 −5q20 +q19 −6q18 +q(q17+q 16 15 −8q 14 +q 13 −8q 12 +2q 11 −8q 10 +q 9 −6q 8 +q 7 −5q 6 +q 5 −3q 4 −2q 2 −1) , −8q q 16 14 13 12 11 10 9 8 7 6 5 +2q [5,4,4,1,1] q +q −q +q −2q +2q −q +3q −q +2q −2q +q 4 −q 3 +q 2 +1 , c2,56 = 16 14 12 10 +3q 8 +2q 6 +q 4 +q 2 +1) q 16 14 132(q 12+q 11+q +2q [5,4,4,1,1] q +q +q +q +2q +2q 10 +q 9 +3q 8 +q 7 +2q 6 +2q 5 +q 4 +q 3 +q 2 +1 s2,56 = , 2(q 16 +q 14 +q 12 +2q 10 +3q 8 +2q 6 +q 4 +q 2 +1) (93) x11 0 x13 0 0 x16 0 x 0 x x 0 22 24 25 x 0 x 0 0 x 13 33 36 0 (94) U [5,4,4,1,1] = 0 −x 0 x x 0 24 44 45 0 x25 0 −x45 x55 0 x16 0 x36 0 0 x66 Elements of U 0 [5,4,4,1,1] are pretty complicated. That’s why we don’t list them here. U 0 [5,4,4,1,1] has blockdiagonal form with two blocks 3 × 3.
9
Representation [4,2] [4, 2]⊗2 = [8, 4] ⊕ [8, 3, 1] ⊕ [8, 2, 2] ⊕ [7, 5] ⊕ [7, 4, 1] ⊕ [7, 4, 1] ⊕ [7, 3, 2] ⊕ [7, 3, 2] ⊕ [7, 3, 1, 1]⊕ [7, 2, 2, 1] ⊕ [6, 6] ⊕ [6, 5, 1] ⊕ [6, 5, 1] ⊕ [6, 4, 2] ⊕ 2[6, 4, 2] ⊕ 6, 4, 1, 1 ⊕ [6, 3, 3] ⊕ [6, 3, 2, 1]⊕ (95) [6, 3, 2, 1] ⊕ [6, 2, 2, 2] ⊕ [5, 5, 2] ⊕ [5, 5, 1, 1] ⊕ [5, 4, 3] ⊕ [5, 4, 3] ⊕ [5, 4, 2, 1] ⊕ [5, 4, 2, 1] ⊕ [5, 3, 3, 1]⊕ [5, 3, 2, 2] ⊕ [4, 4, 4] ⊕ [4, 4, 3, 1] ⊕ [4, 4, 2, 2].
Eigenvalues are: λ[8,4] = q −30 , λ[8,3,1] = −q −26 , λ[8,2,2] = q −24 , λ[7,5] = −q −26 , λ[7,4,1]± = ±q −21 , −18 −16 −14 −24 , λ[7,3,1,1] = q , λ[7,2,2,1] = −q , λ[6,6] = q , λ[6,5,1]± = ±q −18 , λ[7,3,2]± = ±q −14 −12 −12 −9 , λ[6,4,1,1] = −q , λ[6,3,3] = −q , λ[6,3,2,1]± = ±q , λ[6,2,2,2] = q −6 , λ[6,4,2]± = ±q −12 −10 −9 , λ[5,5,1,1] = q , λ[5,4,3]± = ±q , λ[5,4,2,1]± = ±q −6 , λ[5,3,3,1] = q −4 , λ[5,5,2] = −q −2 −6 λ[5,3,3,2] = −q , λ[4,4,4] = q , λ[4,4,3,1] = −q −2 , λ[4,4,2,2] = 1.
(96)
Accidentally coinciding eigenvalues are: λ[8,2,2] = λ[6,6] = q −24 , λ[7,3,2]± = λ[6,5,1]± = ±q −18 , λ[8,3,1] = λ[7,5] = −q −26 , −14 −12 λ[7,2,2,1] = λ[6,4,2]− = −q , λ[6,4,1,1] = λ[6,3,3] = λ[5,5,2] = −q , λ[6,3,2,1]± = λ[5,4,3]± = ±q −9 , −6 −2 λ[6,2,2,2] = λ[5,4,2,1]+ = λ[4,4,4] = q , λ[5,3,2,2] = λ[4,4,3,1] = −q . (97)
16
Table of matrices with accidentally coinciding eigenvalues [4, 2]
9.1
size
repr
eigenvalues repetitive accidental
result
blocks
4
[5,4,3,3,2,1]
-
λ2 = λ3
+
(3,1)
5
[5,5,3,3,1,1]
-
λ4 = λ5
+
(3,2)
6
[5,4,4,2,2,1]
λ5 = λ6
λ3 = λ4
+
(3,3)
[11,6,1]
λ1 = λ2
λ3 = λ4
+
(3,3)
[5,4,3,3,2,1]
R[5,4,3,3,2,1] = diag(q κ[5,3,3,1] , −q κ[5,3,2,2] , −q κ[5,5,3,1] , q κ[4,4,2,2] ) = diag(q −4 , −q −2 , −q −2 , 1). (98) There is one pair of coinciding eigenvalues: λ2 = λ3 . We use a rotation matrix with [5,4,3,3,2,1]
c23
=
[5,4,3,3,2,1]
√1 , 2
s23
= − √12
(99)
and get a block-diagonal Racah matrix: U
0
[5,4,3,3,2,1]
=
q2 2 +1)2 (q√ √ 2q q 4 +q 2 +1 − (q2 +1)2
0 4
√
+1 − q(q+q 2 +1)2
4
0
4
√
2
√
2q q 4 +q 2 +1 (q 2 +1)2 +1 − (qq2 +1) 2 √0
0 1
2q q 4 +q 2 +1 (q 2 +1)2
−
0
2
+1 − q(q+q 2 +1)2 √ √ 4 2
2q q +q +1 (q 2 +1)2
0 q2 (q 2 +1)2
.
(100)
Changing the order of lines in U 0 [5,4,3,3,2,1] one can get blocks 3 × 3 and 1 × 1.
9.2
[5,5,3,3,1,1]
R[5,5,3,3,1,1] = diag(−q κ[5,4,2,1] , q κ[5,4,2,1] , q κ[5,3,3,1] , −q κ[5,3,2,2] , −q κ[4,4,3,1] ) = diag(−q −6 , q −6 , q −4 , −q −2 , −q −2 ) (101) There is just one pair of coinciding eigenvalues (λ4 = λ5 accidentally). After a rotation with [5,5,3,3,1,1]
c45
=
[5,5,3,3,1,1]
√1 , 2
s45
= − √12
(102)
we get
U 0 [5,5,3,3,1,1]
√
2
− q4q+1
0 = 0 √ √ q 4 −q 2 +1 q 4 +q 2 +1 − q 4 +1 0
0
√
q2 √
−
−
(q 2 +1)2
√
2q q 4 +q 2 +1 (q 2 +1)2
2q q +q 2 +1 (q 2 +1)2 q 4 +1 (q 2 +1)2
0
√
4 2 +1 − q(q+q 2 +1)2
0 √4
−
0 √4
2q q +q 2 +1 (q 2 +1)2
√
q 4 −q 2 +1 q 4 +q 2 +1 q 4 +1 4
0 −
0 2
0 2
+1 − q(q+q 2 +1)2 √ √ 4 2
2q q +q +1 (q 2 +1)2
− q4q+1
0
0
q2 (q 2 +1)2
(103) One can see 3 × 3 and 2 × 2 blocks in this matrix.
17
.
9.3
[5,4,4,2,2,1]
R[5,4,4,2,2,1] = diag(−q κ[5,4,2,1] , q κ[5,4,2,1] , −q κ[5,3,2,2] , −q κ[4,4,3,1] , q κ[4,4,2,2] , q κ[4,4,2,2] ) = diag(−q −6 , q −6 , −q −2 , −q −2 , 1, 1). (104) There are two pairs of coinciding eigenvalues: λ3 = λ4 (accidentally), λ5 = λ6 . [5,4,4,2,2,1]
c34 [5,4,4,2,2,1]
=
[5,4,4,2,2,1]
=
c1,56 c2,56
q
(q 4 +q 2 +1)(q 8 +1) 2(q 12 +2q 8 −q 6 +2q 4 +1) ,
q 4 +1 √ q 8 6 2 q +q +2q 4 + q4 −q12 +1
U 0 [5,4,4,2,2,1] = q4 − q8 +q6 +2q 4 +q 2 +1 0 x 13 0 0 −x16 where x13 = x24 =
=
s34
[5,4,4,2,2,1]
= −√
[5,4,4,2,2,1]
=
s1,56 ,
s2,56
√
2
= − √12 ,
(105)
q 4 +1 q 8 6 2 q +q +2q 4 + q4 −q12 +1
√
, (106)
(q4 +q2 +1)(q8 +1) (q 12 +2q 8 −q 6 +2q 4 +1)(q 12 +q 10 +q 8 +q 4 +q 2 +1)
0
x13
0
0
x16
q4 q 8 +q 6 +q 2 +1
0
x24
x25
0
0
0
q2 q 4 +1
0 −x24
−1
q 4 +q 2 +1 2 2 (q r +1)
0
q
−
x25
0
0
−√ 4 q 2 q +q +1
q
√
q 8 +1 q 4 −q 2 +1 2 2
√ x25 = −
2
q q 4 +q 2 +1
0
q +q − q8 +q 6 +q 2 +1
0
0
q2 q 4 +q 2 +1
0
x16 =
√
(q 4 −q 2 +1)(q 8 +1) q 8 +q 6 +q 2 +1 6
(q +1)
√
(q 2 +1)(q 8 +q 6 +2q 4 +q 2 +2)q 2 +1 , √ 12 q88 +q66 +2q4 4 +q2 +1 q q +q +q +q +1 , q 8 +q 6 +q 2 +1
q
[5,4,4,2,2,1]
√1 , 2
q 8 +q 6 +q 4 +q 2 +1
√ q44+q22+1
,
(q −q +1)(q 8 +1)(q 12 +q 8 +q 6 +q 4 +1) . (q 6 +1)2
.
, (107)
(108)
One can see that U 0 [5,4,4,2,2,1] has two blocks 3 × 3.
9.4
[11,6,1]
R[11,6,1] = diag(q κ[8,4] , q κ[8,4] , −q κ[8,3,1] , −q κ[7,5] , −q κ[7,4,1] , q κ[7,4,1] ) = diag(q −30 , q −30 , −q −26 , −q −26 , −q −21 , q −21 ). (109) There are two pairs of coinciding eigenvalues: λ1 = λ2 and λ3 = λ4 (accidentally). q 2 4 3 2 +q+1)(q 6 +q 3 +1)((q−1)q((q−1)q+1)(q 2 +1)(q 2 +q+1)(q 4 −q 2 +1)+1) [11,6,1] c1,12 = (q +q+1)(q +q +q , 2 +1)(q 4 +1)(q 8 −q 4 +1)(q 8 +q 6 +2q 4 +q 2 +2)q 2 +1) 2((q q [11,6,1] ((q−1)q+1)(q 6 −q 3 +1)((q−1)q(q 2 +1)+1)(q(q+1)((q−1)q+1)(q 2 +1)(q 2 +q+1)(q 4 −q 2 +1)+1) s1,12 = , 2((q 2 +1)(q 4 +1)(q 8 −q 4 +1)(q 8 +q 6 +2q 4 +q 2 +2)q 2 +1) q [11,6,1] ((q−1)q+1)(q 6 −q 3 +1)((q−1)q(q 2 +1)+1)(q(q+1)((q−1)q+1)(q 2 +1)(q 2 +q+1)(q 4 −q 2 +1)+1) c2,12 = , 8 4 8 +q 6 +2q 4 +q 2 +2)q 2 +1) 2((q 2 +1)(q 4 +1)(q √ 2 −q +1)(q 2 4 8 3 2 6 3 (q +q+1)(q +q +1)((q +1)(q +1)(q −q +1)q +1) [11,6,1] s2,12 = √ √ 4 3 2 2 2 4 2 2 4 8 4 2
(q +q +q +q+1)((q−1)q((q−1)q+1)(q +1)(q +q+1)(q −q +1)+1)((q +1)(q +1)(q −q +1)(q 8 +q 6 +2q 4 +q 2 +2)q 2 +1)
(110) and [11,6,1]
c34
=
√1 , 2
[11,6,1]
s34
= − √12 .
(111)
After the rotations we’ll get a matrix U 0 [11,6,1]
=
x11 0 x13 0 x15 0
0 x22 0 −x24 0 x26
x13 0 x33 0 x35 0 18
0 x24 0 x44 0 x46
x15 0 x35 0 x55 0
0 x26 0 −x46 0 x66
,
(112)
.
where 4
x11 = x15 = x24 = x35 =
4
3
q (q +q +q +q+1) (q 2 +q+1)(q 4 +1)(q 6 +q 3 +1) , √ (q 16 +q 14 +2q 12 +2q 10 +2q 8 +2q 6 +2q 4 +q 2 +1)((q 2 +1)(q 4 +1)(q 8 +q 3 +1)q 2 +1) , (q 2 +q+1)(q 4 +1)(q 6 +q 3 +1)((q−1)q(q 2 +1)+1) r (q2 +1)(q4 +1)(q8 −q3 +1)q2 +1 2 q ((q−1)q+1)(q 6 −q 3 +1) , (q 4 +1)(q 4 +q 3 +q 2 +q+1) 2 q r , (q2 +q+1)(q4 +1)(q6 +q3 +1) ((q−1)q(q 2 +1)+1) q 12 +q 10 +q 8 +q 6 +q 4 +q 2 +1 q2
x46 =
r (q 4 +q 3 +q 2 +q+1) 7
((q−1)q+1)(q 4 +1)(q 6 −q 3 +1) q 12 +q 10 +q 8 +q 6 +q 4 +q 2 +1 5
,
r
(q2 +1)(q4 +1)(q8 +q3 +1)q2 +1 (q2 +q+1)(q6 +q3 +1) (q 4 +1)((q−1)q(q 2 +1)+1) q 4 ((q−1)q (q 2 +1)+1) ((q−1)q+1)(q 4 +1)(q 6 −q 3 +1) , q2
2
x13 = − x22 =
x44 =
((q−1)q+1)(q 6 −q 3 +1) (q 4 +1)((q−1)q(q 2 +1)+1) , (q2 +q+1)(q6 +q3 +1) − (q4 +1)(q4 +q3 +q2 +q+1) ,
x55 =
q 7 +q 5 (q 2 +1)(q 8 +q 4 +q 3 +1)q 2 +1 ,
x33 =
,
+q x66 = − q12 +q10 +q8 −qq7 +q 6 −q 5 +q 4 +q 2 +1 .
(113) U 0 [11,6,1] contains two blocks 3 × 3.
10
Conclusion
We investigated all Racah matrices up to size 6 × 6 and some matrices of size 8 × 8 and 9 × 9 from cubes of representations [2, 1], [3, 1], [3, 2], [4, 1] and [4, 2]. We found out that some of them could be transformed into block-diagonal ones. The essential condition was that the corresponding R-matrix contained a pair of accidentally coinciding eigenvalues. The additional condition emerged for matrices of size 6 × 6 and larger. In this case corresponding R-matrices should also contain repetitive eigenvalues: one pair for 6 × 6 matrices and two pairs for 8 × 8. We couldn’t make such conclusion for 9 × 9 matrices because we had both successful and unsuccessful examples in the case of two pairs of repetitive eigenvalues and one pair of accidentally coinciding eigenvalues. The transformations that made the Racah matrices (and therefore R2 -matrices) block-diagonal were rotations in the sectors of coinciding eigenvalues (accidental and repetitive). Particular examples showed that the angle of rotation in ”accidental” sector was always ± π4 . Since such rotation mixed vectors from different irreducible representations, the resulting matrix didn’t satisfy the definition (2) and no longer was a Racah matrix, but the R2 -matrix it produced was suitable for calculations. The practical impact of this work is that it gives us hope to use the eigenvalue hypothesis [35] to calculate Racah matrices even in the case of coinciding eigenvalues. It’s possible because the resulting blocks in the corresponding R-matrix don’t contain coinciding eigenvalues. It’s still a question how to determine which eigenvalues go to each particular block. In case of large matrices (of size 9 × 9 and larger) more complex cases emerged: when more than a pair of eigenvalues coincided. Such groups of coinciding eigenvalues could consist of both accidental and repeating eigenvalues. The problem was that there was more than one angle of rotation in this case and it’s a lot more difficult to determine the right rotation matrix.
11
Acknowledgements
We would like to thank A.Morozov and A.Mironov for useful discussions. This work was funded by the Russian Science Foundation (Grant No. 16-12-10344).
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