Remark The material in this section is not made use of in this book until Section16.5, in the chapter on space groups. Consequently, readers may choose to postpone their study of Section4.8until they reach Section16.5.
Let G¼ {gj} be a group of order g with a subgroup H¼ {hl} of order h. The left coset expansion of G on H is
G¼Pt
r¼1
gr H, t¼ g=h, g1 ¼ E, (1)
where the coset representatives grfor r¼ 2, . . . , t, are 2 G but 62 H. By closure in G, gjgs(gs2 {gr}) is2 G (gksay) and thus a member of one of the cosets, say grH. Therefore, for some hl2 H,
gj gs¼ gk ¼ gr hl: (2)
(2) gj gsH¼ gr hlH¼ gr H; (3)
(3) gjhgsHj ¼ hgr Hj ¼ hgs Hj gðgjÞ: (4)
In eq. (4) the cosets themselves are used as a basis for G, and from eq. (3) gs H is transformed into grH by gj. Since the operator gjsimply re-orders the basis, each matrix representation in the ground representation gis a permutation matrix (AppendixA1.2).
Thus the sth column of ghas only one non-zero element, (4), (2) ½gðgjÞus¼ 1, when u ¼ r, gj gs¼ grhl
¼ 0, when u 6¼ r: (5)
Because binary composition is unique (rearrangement theorem) the same restriction of only one non-zero element applies to the rows of g.
Exercise 4.8-1 What is the dimension of the ground representation?
Example 4.8-1 The multiplication table of the permutation group S(3), which has the cyclic subgroup H¼ Cð3Þ, is given in Table 1.3. Using the coset representatives {gs}¼ {P0P3}, write the left coset expansion of S(3) on C(3). Using eq. (2) find grhlfor 8 gj2 G. [Hint: gr2 {gs} and hlare determined uniquely by gi, gs.] Hence write down the matrices of the ground representation.
The left coset expansion of S(3) on C(3) is G¼Pt
s¼1
gs H¼ P0 H P3 H¼ fP0 P1 P2g fP3 P4 P5g, (6) with grand hl, determined from gj gs¼ gk¼ gr hl, given in Table4.9. With the cosets as a basis,
gjhP0 H, P3 Hj ¼ hP00
H, P30 Hj ¼ hP0H, P3Hj gðgjÞ: (7)
The matrices of the ground representation are in Table4.10. Each choice of gjand gsin eq. (2) leads to a particular hlso that eq. (2) describes a mapping of G on to its subgroup H in which hlis the image of gj.
The purpose of this section is to show how the representations of G may be constructed from those of its subgroup H. Let {eq}, q¼ 1, . . . , li, be a subset of {eq}, q¼ 1, . . . , h, that is an irreducible basis for H. Then
Table 4.9. The values of gkand hldetermined from eq. (4.8.2) for G¼ S(3) and H ¼ C(3).
Table 4.10. The ground representation Ggdetermined from the cosets P0H, P3H by using the cosets as a basis,
hl eq¼Pli
p¼1
ep ~iðhlÞpq, (11)
where ~iis the ith IR of the subgroup H. Define the set of vectors {erq} by
erq¼ gr eq, r¼ 1, . . . , t; q ¼ 1, . . . , li: (12) Thenherq| is a basis for a representation of G:
(2), (5) gjesq¼ gj gseq¼ gr hl eq ¼P
u
gu ½gðgjÞus hl eq: (13)
In eq. (13) grhas been replaced by P
u
gu ½gðgjÞus¼ gr (14)
since the sth column of gconsists of zeros except u¼ r.
(13), (11) gj esq ¼P
u
gu ½gðgjÞusP
p
ep ~iðhslÞpq (15)
¼P
u
P
p
eup ððgjÞ½u sÞpq: (16)
In the supermatrix in eq. (16) each element [u s] is itself a matrix, in this case ~iðhslÞ multiplied by g(gj)us.
(16), (15) ðgjÞup, sq¼ gðgjÞus ~iðhslÞpq, (17) in which u, p label the rows and s, q label the columns; (gj) is the matrix representation of gj in the induced representation ¼ ~i" G. Because gis a permutation matrix, with
g(gj)us¼ 0 unless u ¼ r, an alternative way of describing the structure of is as follows:
(15), (16), (5), (10) ðgjÞup, sq¼ ~iðg1u gjgsÞpq ur: (18)
~iðgu1gj gsÞ is the matrix that lies at the junction of the uth row and the sth column of [u s], and the Kronecker in eq. (18) ensures that ~iis replaced by the null matrix except for [r s]. Table 4.11. This table confirms that for gs¼ P1, grand hl
are determined by the choice of gj, where gjgs¼ gk¼ grhl.
gj gs gk gr hl
P0 P1 P1 P1 P0
P1 P1 P2 P2 P0
P2 P1 P0 P0 P0
P3 P1 P4 P2 P3
P4 P1 P5 P1 P3
P5 P1 P3 P0 P3
Example 4.8-3 Construct the induced representations of S(3) from those of its subgroup C(3).
The cyclic subgroup C(3) has three 1-D IRs so that ~iðhslÞ has just one element (p ¼ 1, q¼ 1). The character table of C(3) is given in Table4.13, along with that of S(3), which will be needed to check our results. The subelements hsland coset representatives grdepend on gjand gs, and our first task is to extract them from Table4.8. They are listed in Table4.13.
Multiplying the ½~iðhslÞ11¼ ~iðhslÞ by the elements of g(gj) in Table 4.12 gives the representations of S(3). An example should help clarify the procedure. In Table4.13, when gj¼ P4, gs¼ P3, and gr¼ P0, the subelement hsl(gj)¼ P2. (In rows 3 and 4 of Table4.13the subelements are located in positions that correspond to the non-zero elements of g(gj).) From Table4.10, [g(P4)]12¼ 1, and in Table4.12~2ðP2Þ ¼ ", so that½~2" G12¼ ", as entered in the sixth row of Table4.13.
From the character systems in Table 4.13 we see that for the IRs of S(3),
½~1" G ¼ A1 A2and½~2" G ¼ E. We could continue the table by finding ~3" G, but since we already have all the representations of S(3), this could only yield an equivalent Table 4.13. Subelements hsl(gj) and MRs (gj) of two representations of S(3), ~" G, obtained by the method of induced representations.
The third and fourth rows contain the subelements hsl(gj) as determined by the values of gs
(in row 2), gj, and gr(in the first column). The g(gj) matrices were taken from Table4.9.
"¼ exp (i2p / 3). Using Table4.11, we see that the two representations of S(3) are
~1" G ¼ A1 A2and ~2" G ¼ E.
Table 4.12. Character table of the cyclic group C(3) and of the permutation group S(3).
representation. Note that while this procedure ~" G does not necessarily yield IRs, it does give all the IRs of G, after reduction. A proof of this statement may be found in Altmann (1977).
4.8.1 Character system of an induced representation
We begin withExercise 4.8-2 Verify closure in Hr. Is this sufficient reason to say that Hris a group?
The character of the matrix representation of gjin the representation induced from ~iis
(20) ðgjÞ ¼P
and so the second term in eq. (25) must be zero if is irreducible. The irreducibility criterion eq. (25) thus becomes
(25), (23) P
fgkg
rðgkÞ sðgkÞ ¼ 0, 8 r 6¼ s, fgkg ¼ Hr\ Hs: (27)
Equation (27) is known as Johnston’s irreducibility criterion (Johnston (1960)).
The number of times cithat the IR ioccurs in the reducible representation ¼P
i
ci i of a group G¼ {gk}, or frequency of iin , is
(4.4.20) ci¼ g1P
k
iðgkÞ ðgkÞ, (28)
where (gk) is the character of the matrix representation of gkin the reducible represen-tation . If iis a reducible representation, we may still calculate the RS of eq. (28), in which case it is called the intertwining number I of iand ,
Iði, Þ ¼ g1P
The frequency cmof an IR mof G in the induced representation ~i" G with characters
m(gj) is equal to the frequency ~ci of ~i in the subduced m # H. The tilde is used to emphasize that the ~i are representations of H. It will not generally be necessary in practical applications when the Mulliken symbols are usually sufficient identification.
For example the IRs of S(3) are A1, A2, and E, but those of its subgroup C(3) are A,1E, and
2E. Subduction means the restriction of the elements of G to those of H (as occurs, for example, in a lowering of symmetry). Normally this will mean that an IR mof G becomes a direct sum of IRs in H,
although if this sum contains a single term, only re-labeling to the IR of the subgroup is necessary. For example, in the subduction of the IRs of the point group T to D2, the IR T becomes the direct sum of three 1-D IRs B1 B2 B3in D2, while A1is re-labeled
(31) ¼ h1P
l
P
p
c~p~pðhlÞ ~iðhlÞ (33)
¼ ~ci: (34)
The tildes are not standard notation and are not generally needed in applications, but are used in this proof to identify IRs of the subgroup. In writing eq. (32), the sum over j is restricted to a sum over l (subduction) because the elements gr hl gr1belong to the class of hl. In substituting eq. (31) in eq. (32) we use the fact that {cp} is a set of real numbers.
Equation (34) follows from eq. (33) because of the OT for the characters. When H is an invariant subgroup of G, Hr¼ Hs¼ H, 8 r, s. Then
(27), (22) P
l
rðhlÞ sðhlÞ ¼ 0, 8 r, s, r 6¼ s , (35) where r, sare representations of H but are not necessarily IRs.
(29) P
l
rðhlÞ sðhlÞ ¼ h Iðr, sÞ: (36)
Therefore, when H is an invariant subgroup of G,
(35), (36) Iðr, sÞ ¼ 0; (37)
that is, the representations r, sof H have, when reduced, no IRs in common.
Exercise 4.8-3 Test eq. (27) using the representations ~1" G and ~2" G of S(3), induced from C(3).