Determination of tensor products and their decomposition

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, p. 68

Section 1.2.7.4.5. Determination of tensor products and their decomposition

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

1.2.7.4.5. Determination of tensor products and their decomposition

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Given a character (for an irreducible representation from the character table, or for the vector representation, for example), the character of the standard rank n tensor is the nth power of the character and can be decomposed with the multiplicity formula for $[m_{\alpha}]$ given above.

Fully symmetrized or antisymmetrized tensor products have characters given by $[\eqalign{ n=2: \chi^\pm(R) &={1 \over 2!}\big(\chi (R)^2\pm\chi (R^2)\big)\cr n=3: \chi^\pm(R) &={1 \over 3!}\big(\chi (R)^3\pm 3\chi (R^2)\chi (R) +2\chi (R^3)\big)\cr n=4: \chi^\pm(R) &= {1 \over 4!}\big(\chi (R)^4\pm 6\chi (R^2)\chi (R)^2+3\chi (R^2)^2 \cr &\quad +8\chi (R^3)\chi (R)\pm 6\chi (R^4)\big)\cr n=5: \chi^\pm(R) &= {1 \over 5!}\big(\chi (R)^5\pm 10\chi (R^2)\chi (R)^3+15\chi (R^2)^2\chi (R)\cr&\quad +20\chi (R^3)\chi (R)^2 \pm 20\chi (R^3)\chi (R^2)\cr& \quad \pm 30\chi (R^4)\chi (R) +24\chi(R^5)\big)\cr n=6: \chi^\pm(R) &= {1 \over 6!}\big(\chi (R)^6\pm 15\chi(R^2)\chi (R)^4+45\chi (R^2)^2\chi (R)^2 \cr &\quad +40\chi (R^3)^2 \pm 15\chi(R^2)^3+ 40\chi (R^3)\chi (R)^3\cr &\quad \pm 120 \chi (R^3)\chi (R^2)\chi (R) \pm 90 \chi (R^4)\chi (R)^2\cr &\quad +90\chi (R^4)\chi (R^2) +144\chi (R^5)\chi (R)\cr&\quad\pm120 \chi (R^6)\big). }]$ From this follows immediately the dimension of the subspaces of symmetric and antisymmetric tensors : $[\eqalign{n&=2: {1 \over 2}(d^2\pm d)\cr n&=3: {1 \over 6}(d^3\pm 3d^2+2d)\cr n&=4:{1 \over 24}(d^4\pm 6d^3+11 d^2\pm 6d)\cr n&=5: {1 \over 120}(d^5\pm 10d^4+35d^3\pm 50 d^2+24d)\cr n&=6: {1 \over 720}(d^6\pm 15d^5+85d^4\pm 225 d^3+274 d^2\pm 120 d). }]$

The general expression for arbitrary rank can be determined as follows. (See also Section 1.2.2.7)

(1) If n is the rank, the first step is to determine all possible decompositions $[n = \textstyle\sum\limits_{i=1}^{n} f_{i}]$ with non-negative integers satisfying $[f_i \leq f_{i-1}]$ .
(2) For each such decomposition $[m=1,\ldots n_{\rm tot}]$ there is a term $[P_m = \prod\limits_{i=1}^{p} \pmatrix{ N_i \cr f_i }(f_i -1)!, ]$ where , $[N_i=N_{i-1}-f_{i-1}]$ $[(i\,\gt\,1)]$ and p is the number of nonzero integers .
(3) If there are equal values of in the mth decomposition, should be divided by for each t-tuple of equal values ( $[f_{k+1}=\ldots =f_{k+t}]$ ).
(4) The sign of the term is for a symmetrized power and $[\textstyle\prod\limits_{i=1}^{p} (-1)^{(f_i -1)}]$ for an antisymmetrized power.
(5) The expression for the character of the (anti)symmetrized power then is $[\chi^{\pm} (R) = (1/M!)\textstyle\sum\limits_{m=1}^{n_{\rm tot}}{\rm sign}_m P_m \textstyle\prod\limits_{i=1}^{p} \chi (R^{f_i}). ]$