Invariant tensors

Ephra[iuml ]m, M.; Janssen, T.; Janner, A.; Thiers, A.

doi:10.1107/97809553602060000629

International
Tables for
Crystallography
Volume D
Physical properties of crystals
Edited by A. Authier

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International Tables for Crystallography (2006). Vol. D. ch. 1.2, pp. 68-70

Section 1.2.7.4.6. Invariant tensors

M. Ephraïm,^b T. Janssen,^a A. Janner^c and A. Thiers^d

1.2.7.4.6. Invariant tensors

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Once one has the character of the properly symmetrized tensor, the number of invariants is just $[m_{1}]$ , the number of times the trivial representation occurs in the decomposition.

Example (1). Dimension 3, rank 3, symmetry type (123), group 3. Basis: [xxx] , [xxy] , [xxz] , [xyy] , [xyz] , [xzz] , [yyy] , [yyz] , [yzz] , [zzz] . Under $[\pmatrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 }]$ the basis elements go to [yyy,] [yyz,] [yyx,] [yzz,] [yzx,] [yxx,] [zzz,] [zzx,] [zxx,] [xxx] , respectively, and these are equivalent to [yyy,] [yyz,] [xyy,] [yzz,] [xyz,] [xxy,] [zzz,] [xzz,] [xxz,] [xxx] , respectively. This gives the ten-dimensional matrix $[M = \pmatrix{ 0&0&0&0&0&0&0&0&0&1\cr 0&0&0&0&0&1&0&0&0&0\cr 0&0&0&0&0&0&0&0&1&0\cr 0&0&1&0&0&0&0&0&0&0\cr 0&0&0&0&1&0&0&0&0&0\cr 0&0&0&0&0&0&0&1&0&0\cr 1&0&0&0&0&0&0&0&0&0\cr 0&1&0&0&0&0&0&0&0&0\cr 0&0&0&1&0&0&0&0&0&0\cr 0&0&0&0&0&0&1&0&0&0 }.]$ Then [P(M-E)Q=D] , with D diagonal. There are four diagonal elements of D which are zero, and the invariant tensors correspond to the corresponding four columns of the matrix Q. The invariant polynomials are $[xxx+yyy+zzz,\quad xxy+xzz+yyz,\quad xxz+yzz+xyy,\quad xyz. ]$

Example (2). Dimension 2, rank 2, symmetry type (12). Group generated by $[ \pmatrix{ 0 & -1 \cr 1 & -1 }.]$ Basis [xx] , [xy] , [yy] goes to [yy] , [-xy+yy] , [xx-2xy+yy] . This gives $[ M = \pmatrix{ 0 & 0 & 1 \cr 0 & -1 & -2 \cr 1 & 1 & 1 }. ]$ Because $[ \pmatrix{ 1&0&0 \cr 2&1&2 \cr 1&0&1 } (M-E) \pmatrix{1&-1&1 \cr 0&1&0 \cr 0&-1&1 } = \pmatrix{ -1&0&0 \cr 0&0&0 \cr 0&0&2 } ,]$ the invariant tensor corresponds to the second column of Q, which as a polynomial reads [-xx+xy-yy] . This can be written with the tensor $[{\rm T}_{ij}]$ as $[-xx+xy-yy = - \textstyle\sum\limits_{i,j}T_{ij}x_{i}x_{j},\quad T_{ij} = \pmatrix{1&-{{1}\over{2}} \cr -{{1}\over{2}} & 1 } ]$ This tensor T is invariant under the group.

Example (3). Dimension 3, rank 2, tensor type (12). Group generated by matrix([[0 −1 0][1 0 0][0 0 1]]). The basis [xx] , [xy] , [xz] , [yy] , [yz] , [zz] goes under the generator to [yy] , [-xy] , [-yz] , [xx] , [xz] , [zz] . The solution of [(M-E)v=0] is $[ \alpha_1 (xx+yy) + \alpha_2 zz. ]$ The matrix D has two zeros on the diagonal.

Example (4). Dimension 3, rank 3, type (123). Same group as in Example (3). Basis [xxx] , [xxy] , [xxz] , [xyy] , [xyz] , [xzz] , [yyy] , [yyz] , [yzz] , [zzz] . The solution $[ \alpha_1 (xxz+yyz) + \alpha_2 zzz ]$ corresponds to a tensor with relations $[{\rm T}_{113}={\rm T}_{223}]$ , $[{\rm T}_{111}={\rm T}_{112}={\rm T}_{122}={\rm T}_{123} ={\rm T}_{133}={\rm T}_{222}={\rm T}_{233}=0]$ .

Example (5). Dimension 3, rank 4, type ((12)(34)). Not only $[i_{1}\leq i_{2}]$ and $[i_{3}\leq i_{4}]$ , but also ( $[i_{1}i_{2})]$ , should come lexicographically before ( $[i_{3}i_{4}]$ ). Basis [xxxx,] [xxxy,] [xxxz,] [xxyy,] [xxyz,] [xxzz,] [xyxy,] [xyxz,] [xyyy,] [xyyz,] [xyzz,] [xzxz,] [xzyy,] [xzyz,] [xzzz,] [yyyy,] [yyyz,] [yyzz,] [yzyz,] [yzzz,] [zzzz] . Under the same group as in example (3), there are seven invariants. Invariant polynomial: $[\eqalign{& \alpha_1 (xxxx+yyyy)+\alpha_2 (xxxy-xyyy)+\alpha_3 xxyy +\alpha_4 xyxy \cr &\quad +\alpha_5 zzzz+\alpha_6 (xxzz+yyzz)+ \alpha_7 (xzxz+yzyz). }]$ This corresponds to the tensor relations $[\displaylines{\matrix{ {\rm T}_{xxxx}=-{\rm T}_{yyyy}\hfill &{\rm T}_{xxxy}={\rm T}_{xyyy}\hfill & {\rm T}_{xxxz}=0\hfill \cr {\rm T}_{xxyz}=0 \hfill & {\rm T}_{xxzz}={\rm T}_{yyzz} \hfill & {\rm T}_{xyxz}=0\hfill \cr {\rm T}_{xyyz}=0 \hfill & {\rm T}_{xyzz}=0 \hfill & {\rm T}_{xzxz}={\rm T}_{yzyz} \hfill \cr {\rm T}_{xzyy}=0 \hfill & {\rm T}_{xzyz}=0 \hfill & {\rm T}_{xzzz}=0 \hfill\cr {\rm T}_{yyyz}=0 \hfill & {\rm T}_{yzzz}=0 \hfill& \cr}\cr \rightarrow\pmatrix{ \alpha_{1} & \alpha_{3} & \alpha_6 & 0 & 0 & \alpha_2 \cr \alpha_3 & \alpha_1 & \alpha_6 & 0 & 0 & -\alpha_2 \cr \alpha_6 & \alpha_6 & \alpha_5 & 0 & 0 & 0 \cr 0 & 0 & 0 & \alpha_7 & 0 & 0 \cr 0 & 0 & 0 & 0 & \alpha_7 & 0 \cr \alpha_2 & -\alpha_2 & 0 & 0 & 0 & \alpha_4 }. }]$ The latter form is that of an elastic tensor with the usual convention [1=xx] , [2=yy] , [3=zz] , [4=yz] , [5=xz] , [6=xy] .

Example (6). Dimension 3, rank 2, type [12]. The same group as in example (3). Basis $[xy,xz,yz \rightarrow -yx,-yz,xz]$ , which are equivalent to [xy,-yz,xz] . The transformation in the tensor space is $[\eqalign{ M &= \pmatrix{ 1&0&0 \cr 0&0&1 \cr 0&-1 &0 } \rightarrow \pmatrix{ 0&0&0 \cr 0&-1&1 \cr 0&-1&-1 } v=0 \!:\cr v&=\pmatrix{ 1\cr 0\cr 0 } \sim xy. }]$ There is just one invariant antisymmetric polynomial [xy=-yx] corresponding to the tensor $[T = \pmatrix{ 0&1 \cr -1 & 0 }. ]$

Example (7). Dimension 3, rank 3, type [123]. Basis [xyz] invariant under the group: $[xyz\rightarrow -yxz \sim xyz]$ .The corresponding tensor is the fully antisymmetric rank 3 tensor: $[{\rm T}_{ijk} = 1]$ if [ijk] is an even permutation of 123, [= -1] if [ijk] is an odd permutation, and [=0] if two or three indices are equal (permutation tensor, see Section 1.1.3.7.2 ).

Example (8). Calculation with characters. See Table 1.2.7.2.

Table 1.2.7.2 | top | pdf |
Calculation with characters

Generator	Composite character	Characters
$[\pmatrix{ 0&1&0\cr 0&0&1\cr 1&0&0 }]$	R	E	A	AA
	$[\chi (R)]$	3	0	0
	$[\chi (R)^3]$	27	0	0
	$[\chi (R^2)]$	3	0	0
	$[\chi (R^2)\chi (R)]$	9	0	0
	$[\chi (R^3)]$	3	3	3
Example (1)	$[{{1}\over{6}}(\chi (R)^3 +3\chi (R^2)\chi (R)+2\chi (R^3))]$	10	1	1
$[\pmatrix{ 0&-1 \cr 1& -1 } ]$	R	E	A	AA
	$[\chi (R)]$	2	−1	−1
	$[\chi (R)^2]$	4	1	1
	$[\chi (R^2)]$	2	−1	−1
Example (2)	$[{{1}\over{2}} (\chi (R)^2 +\chi (R^2)]$ )	3	0	0
$[\pmatrix{ 0&-1&0\cr 1&0&0 \cr 0&0&1 }]$	R	E	A	AA	AAA
	$[\chi (R)]$	3	1	−1	1
	$[\chi (R)^2]$	9	1	1	1
	$[\chi (R^2)]$	3	−1	3	−1
Example (3)	$[{{1}\over{2}} (\chi (R)^2 +\chi (R^2)]$ )	6	0	2	0
As above	$[\chi (R)]$	3	1	−1	1
Example (4)	$[\chi (R)^3]$	27	1	−1	1
	$[\chi (R^2)]$	3	−1	3	−1
	$[\chi (R^2)\chi (R)]$	9	−1	−3	−1
	$[\chi (R^3)]$	3	1	−1	1
	$[{{1}\over{6}}(\chi (R)^3 +3\chi (R^2)\chi (R)+2\chi (R^3))]$	10	0	−2	0
As above	$[\chi (R)]$	3	1	−1	1
Example (5)	$[{{1}\over{2}} (\chi (R)^2 +\chi (R^2)]$ )= $[\chi_s (R)]$	6	0	2	0
	$[\chi_{s}(R)^2]$	36	0	4	0
	$[\chi_{s}(R^2)]$	6	2	6	2
	((12)(34))	21	1	5	1
As above, example (6)	$[{{1}\over{2}} (\chi (R)^2 -\chi (R^2)]$ )	3	1	−1	1
As above, example (7)	$[{{1}\over{6}}(\chi (R)^3 -3\chi (R^2)\chi (R)+2\chi (R^3))]$	1	1	1	1

Example (9). The action matrix for a pseudotensor.

Take the group [4/m] with generators $[\pmatrix{0&-1&0\cr 1&0&0 \cr 0&0&1 }, \quad \pmatrix{ 1&0&0\cr 0&1&0\cr 0&0&-1 }. ]$ Consider the rank 3 pseudotensor (123). The action matrix is determined from the action of the generators A and B on the basis: $[\matrix{&A &B\cr \phantom{-}xxx &-yyy & -xxx\cr \phantom{-}xxy & \phantom{-}xyy & -xxy\cr \phantom{-}xxz & \phantom{-}yyz & \phantom{-}xxz\cr \phantom{-}xyy & -xxy & -xyy\cr \phantom{-}xyz & -xyz & \phantom{-}xyz\cr \phantom{-}xzz & -yzz & -xzz \cr \phantom{-}yyy & \phantom{-}xxx & -yyy\cr \phantom{-}yyz & \phantom{-}xxz & \phantom{-}yyz \cr \phantom{-}yzz & \phantom{-}xzz & -yzz \cr \phantom{-}zzz & \phantom{-}zzz & \phantom{-}zzz }]$ Therefore, the action matrix becomes $[ \pmatrix{ 0&0&0&0&0&0&1&0&0&0\cr 0&0&0&-1&0&0&0&0&0&0\cr 0&0&0&0&0&0&0&1&0&0\cr 0&1&0&0&0&0&0&0&0&0\cr 0&0&0&0&-1&0&0&0&0&0\cr 0&0&0&0&0&0&0&0&1&0\cr -1&0&0&0&0&0&0&0&0&0\cr 0&0&1&0&0&0&0&0&0&0\cr 0&0&0&0&0&-1&0&0&0&0\cr 0&0&0&0&0&0&0&0&0&1\cr -1&0&0&0&0&0&0&0&0&0\cr 0&-1&0&0&0&0&0&0&0&0\cr 0&0&1&0&0&0&0&0&0&0\cr 0&0&0&-1&0&0&0&0&0&0\cr 0&0&0&0&1&0&0&0&0&0\cr 0&0&0&0&0&-1&0&0&0&0\cr 0&0&0&0&0&0&-1&0&0&0\cr 0&0&0&0&0&0&0&1&0&0\cr 0&0&0&0&0&0&0&0&-1&0\cr 0&0&0&0&0&0&0&0&0&1 }.]$ After diagonalization, one finds two nonzero elements on the diagonal: $[\displaylines{zzz=a;\quad xxz=yyz=b;\cr xxx=xxy=xyy=xyz=xzz=yyy=yzz=0. }]$

References

International Tables for Crystallography (2006). Vol. D. ch. 1.2, pp. 68-70