Irreducible characters of the icosahedral group Текст научной статьи по специальности «Математика»

NANOSYSTEMS: Kanemitsu S., et al. Nanosystems:

PHYSICS, CHEMISTRY, MATHEMATICS Phys. Chem. Math., 2023,14 (4), 405-412.

http://nanojournal.ifmo.ru

Original article DOI 10.17586/2220-8054-2023-14-4-405-412

Irreducible characters of the icosahedral group

S. Kanemitsu1", Jay Mehta2'6, Y. Sun3 c

1Sanmenxia SUDA New Energy Research Institute, Sanmenxia, Henan, P. R. China 2Department of Mathematics, Sardar Patel University, Vallabh Vidyanagar, Gujarat, India 3Graduate School of Engrg., Kyushu Inst. Tech., 1-1Sensuicho Tobata, Kitakyushu, Japan

aomnikanemitsu@yahoo.com, bjay_mehta@spuvvn.edu, csun@ele.kyutech.ac.jp

Corresponding author: Jay Mehta, jay_mehta@spuvvn.edu

PACS 02.20.Bb

Dedicated to Professor Dr. Vladimir N. Chubarikov

Abstract To study point groups, their irreducible characters are essential. The table of irreducible characters of the icosahedral group A5 is usually obtained by using its duality to the dodecahedral group. It seems that there is no literature which gives a routine computational way to complete it. In the works of Harter and Allen, a computational method is given and the character table up to the tetrahedral group A4 using the group algebra table and linear algebra. In this paper, we employ their method with the aid of computer programming to complete the table. The method is applicable to any other more complicated groups.

Keywords icosahedral group, irreducible representation, simple characters, regular representation, eigenvalues.

For citation S. Kanemitsu, Jay Mehta, Y. Sun Irreducible characters of the icosahedral group. Nanosystems: Phys. Chem. Math., 2023, 14 (4), 405-412.

1. Introduction and irreducible characters of S3

The icosahedral group A5 also denoted C60 is important in the light of recent developments of fullerene structures, cf. [1-4]. The character table of the icosahedral group A5 is usually obtained by using its duality to the dodecahedral group [5](pp. 216-219). Alternatively, it is simply stated without any indication of proof, cf. e.g. [6,7]. It seems that there is no literature which gives a routine computational way to find it with the aid of computers. We describe the method of Harter [8,9] and Allen [10] and determine the character table of the icosahedral group. The method is rather a light-hearted one and without much knowledge, one can construct character tables. The procedure is described in the following subsections. For some algebraic preliminaries, see, e.g. [11].

1.1. Algebra table and regular representation matrices

[10](p. 27) gives one the table of irreducible characters up to the tetrahedral group Td and we shall give one for the icosahedral group. We need to form the table of conjugate classes and their algebra table. We illustrate the procedure by the 3rd symmetric group (Table 1).

Table 1. Conjugate classes of S3

label representative type cardinality

Ci (1)(2)(3) (1, 0, 0) 1

C2 (1, 2)(3) (0,1, 0) 3

C3 (1, 2,3) (0, 0,1) 2

Here C- 1 is the conjugate class consisting of all the inverses of elements of Cj (Table 2).

The (right) regular representation matrix R(Ca) has the (i, j)-entry cja, which are the structure constants defined by

n

C-Ca = £ cjaCj, (1)

j = 1

where n is the number of conjugate classes of G.

© S. Kanemitsu, Jay Mehta, Y. Sun, 2023

Table 2. Class algebra table for S3

classes Ci C2 Cf

Ci Ci C2 Cf

C2 i = C2 C2 3Ci + 3Cf 2 2

C- = Cf C3 2C2 2Ci + C3

Looking at each column in Table 2, we see immediately that

R(C2)

1

3 0 3

y0 2 0y

R(C3)

^0 0 020 201

Note that we always have R(Ci) = E, where E is the identity matrix. 1.2. Eigenvalues and eigenspaces of regular representations

R(C2) has eigenvalues 0, ±3 with the following eigenspaces

Er(C2 )(0) = R

1

\-1/

Efl(02)(±3)= R

1

±3 2

and R(C3) has eigenvalues -1,2,2 with the following eigenspaces

er(C3)(-1)= R

1

\-1/

Efl(C3)(2)= R

1

2

© R

0

0

R

1

2

©R

1

-3 2

Remark 1. To find eigenvalues of R(Cj), j = 2, 3 Allen uses the method of raising-to-powers.

c0 = Ci, c2 = C2, c2 = 3C + 3C3, c3 = 3Ci + 3C3 = 3C2 + 3C3C2 = 3C2 + 6C2

whence the Cayley-Hamilton equation resp. the characteristic equation

9C2,

Cf - 9C2 =0, A3 - 9A = 0

and the eigenvalues are 0, ±3.

(2)

(3)

(4)

(5)

(6)

C0

Ci, C3 — C3, C3 — 2Ci + C3, C3 — 2Ci + 3C3,

(7)

whence the Cayley-Hamilton equation Cf — 4Cf + 4C2 characteristic equation

C| - C3 - 2Ci =0, A2 - A - 2 = 0 and the eigenvalues are 2, —1. But this is practical only for lower degree matrices. This process may be automated.

1.3. Matching the eigenvalues

0, But we already have a lower order equation resp. the

(8)

This process may remain manual and depends on inspection.

A character table (CT) is in effect a collection of traces of IR's (Irreducible representation) of the group. As such, all of the entries in a given row of a CT belong to the same IR. Up to now the eigenvalues are arranged in sets according to classes Q. For a specific IR, P, say, the character x(a) assigned to class Cj is associated with a specific member of the set {Aj}. It is therefore required that for a given P(a), a single eigenvalue be picked from each of the n sets Aj and that these eigenvalues be arranged in a new set

{A(a)} = A'

(a)

,A(a)

all of which are associated with the given . This procedure is called matching the eigenvalues.

The collection of eigenvalues A(a) has a single column vector v(a) associated with it, which has the property

R[Cj]v(a)

A(a)v(a),

(9)

(10)

0

0

0

1

3

i

The vector v(a) is, simultaneously, an eigenvector of every R(Cj). When this property is used in conjunction with (3) and (4), we see that A = -1 from R(C2), and A = 0 from R(C3) belong to the same set {A(1)}, where the common eigenvectors are

V-1/

,(2)

,(3)

\2/

1

-3 2

(11)

Here v(2) is a common eigenvector of R(C2) and R(C3) belonging to the eigenvalue a22) = 3 resp. a3.2) = 2. Similarly, v(3) is a common eigenvector belonging to the eigenvalue a22) = -3 resp. a3.3) = 2. Every set {A(a)} contains the n-fold multiple eigenvalues A =1 from R(C1), so that the complete set found is represented in Table 3.

Table 3. Eigenvalues arranged

0

3

eigenvalue set Ci C2 C3 eigenvector

{ A(1)} a11) = 1 a21) = 0 a21) = -1 V1)

{ A(2)} a12) = 1 a22) = 3 a32) = 2 v(2)

{A(3)} a13) = 1 a23) = -3 A33) = 2 v(3)

1.4. Finding values of irreducible characters

To find CT we accommodate the values of A(a) and arrange the characters in the order of increasing dimension of IR. The following formula appears as the coefficients of (49') in [9](p. 747, l. 2):

j = Aja), (12)

J card(Cj) J

where x(a) is the character value x(a)(ord(Cj)) of the jth class in the ath irreducible representation (IR), l(a) the dimension of the ath IR and ord(Cj) is the order of the jth class. We appeal to the formula ( [9](p. 747))

ivjl = (13)

|G| j card(Cj) (,(»))2 ■ ( '

It follows that l(2) = 2 and other two are 1. We rearrange Table 3 in the order of dimensions and label them as follows (Table 4) (so as to compare with [10](p. 23)).

TABLE 4. Eigenvalues arranged

IR C1 C2 C3 dim

p (0) a10) = 1 a20) = 3 A30) = 2 l(0) = 1

p (1) a11) = 1 a21) = -3 A21) = 2 l(1) = 1

p (2) a12) = 1 a22) = 0 a32) = -1 l(2) = 2

We stretch the interpretation of (12) to mean

A(a)

Then

Similarly,

j=caj <14)

J = 2 (. . ) = f 1. 3. -2) = (2.0. -1)- (1-5)

J \card(6i) card(C2) card(C3)J \1 3 2J

J = (carder, carler, carder) =1'a J =(11)- (16)

Table 5. Values of irreducible characters

IR Ci C2 C3

p (0) x10) = 1 x20) = i x30) = 1

p (1) x11) = 1 x21) = -i x21) = 1

p (2) x12) = 2 x22) = 0 x32) = -1

Table 6. Conjugate classes of S5

label representative type cardinality

Ci (1)(2)(3)(4)(5) (5,0, 0, 0, 0) 1

C2 (1, 2)(3)(4)(5) (3,1, 0, 0, 0) 10

C2 (1, 2)(3,4)(5) (1, 2, 0, 0, 0) 15

C4 (1, 2, 3)(4)(5) (2,0,1, 0, 0) 20

C5 (1, 2, 3)(4, 5) (0,1,1, 0, 0) 20

Ce (1, 2, 3,4)(5) (1,0, 0,1, 0) 30

C7 (1, 2, 3,4, 5) (0,0, 0, 0,1) 24

Table 7. Conjugacy classes of A5

class type representative kj

Ci (5,0,0,0, 0) (1) 1

C2 (2,0,1.0,0) (1, 2, 3) 20

C3 (1, 2,0,0, 0) (1, 2)(3, 4) 15

C4 (0,0,0,0,1) (1, 2, 3,4, 5) 12

C5 (0,0,0,0,1) (2,1, 3,4, 5) 12

2. Irreducible characters of A5

This will be much harder. We need to prepare the class algebra table (Table 6). The goal is to establish the following theorem.

Theorem 1. All simple characters of A5 are given by Table 8:

Table 8. All simple characters of A5 (t indicates 1 - the golden ratio)

Ci C2 C3 C4 C5

Xi 1 1 1 1 1

X2 3 0 -1 T -T-1

X3 3 0 -1 -T-1 T

X4 4 1 0 -1 -1

X5 5 -1 1 0 0

2.1. Class algebra table and regular representation matrices

For this, we need the class algebra table (Table 9)

Table 9. Class algebra table for A5

classes Ci C2 c3 C4 C5

Ci C1 C2 C3 C4 C5

C*2 1=C2 C2 20C1+7C2+8C3+5C4+5C5 6C2+4C3+5C4+5C5 3C2+4C3+5C4+5C5 3C2+4C3+5C4+5C5

C3-1=C3 C3 6C2+4C3+5C4+5C5 15C1+3C2+2C3+5C4+5C5 3C2+4C3+5C5 3C2+4C3+5C4

C4 1=C4 C4 3C2+4C3+5C4+5C5 3C2+4C3+5C5 12C1+3C2+5C4+C5 3C2+4C3+C4+C5

C5-1=C5 C5 3C2+4C3+5C4+5C5 3C2+4C3+5C4 3C2+4C3+C4+C5 12C1+3C2+C4+5C5

R(C2)=

0 10 0 0

20 7 8 5 5 0 6 4 5 5

R(C3)=

R(C4)=

R(C5) =

0

3 4 5 5 3 4 5 5

0 0 10 0

0 6 4 5 5

15 3 2 5 5

0 3 4 0 5

0 3 4 5 0

0 0 0 1 0

0 3 4 5 5

0 3 4 0 5

12 3 0 5 1

0 3 4 1 1

0 0 0 0 1

0 3 4 5 5

0 3 4 5 0

3 4 11

12 3 0 1 5

2.2. Finding eigenvalues and eigenvectors by a computer

(1) The eigenvalues of the matrix R(C2) are obtained by a python program:

20, 5, -4, 0, 0 . Their corresponding eigenvectors are as follows

and the polynomial is

v1 = (0.03, -0.15, -0.196, -0.012, -0.016) v2 = (0.661, -0.753, -0.784,0,0) v3 = (0.496, 0, 0.588, 0.063, 0.849) v4 = (0.396, 0.452, 0, -0.73, -0.437) v5 = (0396,0.452,0,0.679, -0.241)

A5 - 21 A4 +400A2.

(17)

(18)

(19)

(20)

(2) The eigenvalues of the matrix R(C3) obtained by a python program are:

15, -5, 3,0, -5

and the their corresponding eigenvectors are

v1 = (-0.033, -0.171,0.196,0.15,0.016) v2 = (-0.661,0, -0.784,0.753,0) v3 = (-0.496,0.857,0.588,0, -0.08) v4 = (-0.396, -0.342,0, -0.452, -0.671) v5 = (-0.396, -0.342,0, -0.452, 0.736)

and the polynomial is

A5 - 8A4 - 110A3 + 1125A.

(3) The eigenvalues of the matrix R(C4) obtained by a python program are:

12, 6.47, 0, -3, -2.47 and the their corresponding eigenvectors are

v1 = (0.033,0.116, -0.196, -0.15,0.116) v2 = (0.661,0, 0.784, -0.753, 0) v3 = (0.496, -0.581, -0.588,0, -0.581) v4 = (0.396,0.752,0, 0.452, -0.287) v5 = (0.396, -0.287, 0, 0.452, 0.752)

and the polynomial is

A5 - 13A4 - 16A3 + 288A2 + 576A.

(4) The eigenvalues of the matrix R(C5) obtained by a python program are:

12, 6.47, 0, -3, -2.47 and the their corresponding eigenvectors are

v1 = (0.033, 0.116, -0.196,0.15,0.116) v2 = (0.661, 0,0.784,0.753,0) v3 = (0.496, -0.581, -0.588,0, -0.581) v4 = (0.396, -0.287,0, -0.452, 0.752) v5 = (0.396, 0.752, 0, -0.452, -0.287)

and the polynomial is

A5 - 13A4 - 16A3 + 288A2 + 576A = A(A - 12)(A + 3)(A2 - 4A - 16) The eigenvalues of R(C5) are:

12, 4t, 0, -3, -4t-1.

2.3. Matched eigenvalues

This section combines §1.2 and §1.3 to give the table corresponding to Table 4.

Table 10. Eigenvalues arranged

Ci C2 C3 C4 C5 eigenvectors dim

{A(1)} 1 20 15 12 12 vC1) l(1) = 1

{A(2)} 1 0 -5 4T -4T-1 v(2) l(2) = 3

{A(3)} 1 0 -5 —4T-1 4t v(3) l(3) = 3

|A(4)| 1 5 0 -1 -3 v(4) l(4) = 4

{A(2)} 1 -4 3 0 0 v(5) l(5) = 5

Here

v(1) =(0.03, -0.15, -0.196, -0.012, -0.016) v(2) =(0396,0.452,0,0.679, -0.241) v(3) =(0.661, -0.753, -0.784,0,0) v(4) =(0.396, 0.452, 0, -0.73, -0.437) v(5) =(0.496, 0, 0.588, 0.063, 0.849).

2.4. Proof of Theorem 1

Using the method in §1.4, we find the values of all ICs.

ivcard(Ci) > card(C2) > card(C3) > card(C4) > card(C5) ^ ( 1

1>20>15>12>12 \ = a> D.

1 > 20 > 15 > 12 > 12 / ( > > > > )

1 20 15 12 12

ivcard(C1) > card(C2) > card(C3) > card(C4) > card(C5) ^ ( 1

= (1> 0> -1>t> -T-1).

-5 4t —4t

1

(3) _ 3 i 1 0 -5 — 4t-1 4t (23)

( ycard(Ci)' card(C2)' card(C3)' card(C4)' card(C5)

= (1> 0> —1> —t-1>t).

( = 4 ( 1 > 5 > 0 > —1 > —3 ) (24)

( ycard(Ci)' card(C2)' card(C3)' card(C4)' card(C5)

= (4> 1> 0> —1> —1).

( = 5 1 > —4 > 3 > 0 > 0 ) (25)

( \card(C1^ card(C2)' card(C3)' card(C4)' card(C5)

= 5 | 1 > —> — > 0> 0 ) = (5> —1> 1> 0> 0).

20 ' 15' ' J v ' ' ' ' 7

as in Table 7. This proves Theorem 1. □

3. Conclusion

The method described here of Harter and Allen may be applied to any other interesting finite groups which will be conducted elsewhere.

References

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Submitted 11 August 2023; revised 15 August 2023; accepted 16 August 2023

Information about the authors:

S. Kanemitsu - Sanmenxia SUDA New Energy Research Institute, No. 1, Taiyang Road, Sanmenxia Economic Development Zone, Sanmenxia, Henan, 472000, P. R. China; ORCID 0000-0002-8489-7665; omnikanemitsu@yahoo.com

Jay Mehta - Department of Mathematics, Sardar Patel University, Vallabh Vidyanagar, Gujarat 388 120, India; ORCID 0000-0003-1739-9639; jay_mehta@spuvvn.edu

Y Sun - Graduate School of Engrg., Kyushu Inst. Tech., 1-1Sensuicho Tobata, Kitakyushu 804-8555, Japan; ORCID 0000-0003-0433-8917; sun@ele.kyutech.ac.jp

Conflict of interest: the authors declare no conflict of interest.