Isotropic Tensors Under Non-compact Rotation Groups

In the recent past, linearly independent isotropic tensors of rank up to 6, under the compact rotation groups SO(2), SO(3) and SO(4) have been studied in some detail. The present paper extends these studies to the case of linearly independent isotropic tensors under the non-compact rotation groups SO(1, 1), SO(1, 2), SO(1, 3) and SO(2, 2). This is done by using the direct method of explicitly constructing these tensors, proving their linear independence and counting their numbers. Interestingly, it is found that these numbers are identical with the corresponding numbers for the case of the compact groups SO(2), SO(3) and SO(4).


Introduction
Linear invariants and isotropic tensors of the rotation groups SO(2), SO (3) and SO (4), have recently attracted the attention of many researchers. Thus Faiz and Riaz [1] have studied eigenvectors of isotropic tensors while Hirth and Lothe [2], Juret Schke [3], Hearrmon [4] and Norris [5] have considered the linear invariants of the stiffness tensor C ijkl , appearing in elasticity, under SO (3). Similarly Zhou et. al. [6], Lazar [7], Itin et. al. [8] and Gusev and Lurie [9] have used them in isotropic strain gradient elasticity theory. Next, Ahmad and Rashid [10] first showed that the number of linearly independent linear invariants of tensors of any rank under any rotation group SO(p), is equal to the dimension of the space of isotropic tensors of SO(p), of that rank, and then obtained all the linearly independent linear invariants under SO(2) of tensors up to rank 6 and those under SO(3) also up to rank 6. They also found the number of linearly independent linear invariants of a tensor of general rank, under the group of rotations about a fixed axis, say the 3axis. Recently, Naila Amir [11] has found the number of linearly independent isotropic tensors of ranks r = 2, 4, 6,8,under SO(4) and has also obtained explicit expressions for complete sets of basis elements in the space of isotropic tensors of these ranks.
After having considered the isotropic tensors under these compact rotation groups, the question that immediately arises is: what happens when we move on to the non-compact rotation groups SO(1, 1), SO(2, 1), SO (3,1) and SO (2,2). This is the motivation for the present work in which, using the direct method, we explicitly construct a set of isotropic tensors, of rank up to 6, under each of these non-compact groups, select a linearly independent subset of each of these sets, and then show that each of these selected subsets is complete in the sense that every isotropic tensor of the relevant type, is a linear combination of elements of this subset. This means that the current work is an immediate generalization of the work of Ahmad and Rashid [10] for SO(2) and SO(3) and of Naila Amir [11] for SO(4). An interesting fact which appears from this work, is that the number of elements of a complete set of linearly independent isotropic tensor of any particular type, under a non-compact group, is identical with the corresponding number, under the corresponding compact group. The basic reason for this appears to be the fact that the metric tensor δ ij of the Euclidean spaces and η ij of the Minkowski spaces, both satisfy the identical relations δ ij = 0 i ≠ j, and η ij = 0, i ≠ j.
As in references [10,11], we denote by d the dimensions of the space in which rotations take place, and by r the rank of the tensor. To proceed systematically, we start with the case d = 2, and then move on to the higher values of d = 3, 4, one by one.

d = 2; Isotropic Tensors Under SO(1, 1)
We take SO(1, 1) to be the set of hyperbolic rotations (with determinant +1) in the Minkowski space M(1, 1), with metric x 2 = x 0 2 -x 1 2 i.e. the metric tensor η ij , i, j = 0, 1, is the matrix As no tensors of rank 3 can be formed out of η i j and ɛ i j and ɛ i j k is identically zero for d = 2, it follows that there do not exist any isotropic tensors of rank 3 under SO (1, 1); obviously, the same will be true for isotropic tensors of any ODD rank.

THE CASE r = 4:
Here the possible isotropic tensors, are η η , η , , , However, not all of these are linearly independent as one can find a number of linear relations between them. To obtain these relations, one has to use the following result of Appendix B: Taking n = 2, one gets One can similarly obtain expressions for , in terms of product of 2 s, so that we conclude that each of the 3 tensors (2.2) can be expressed as a linear combination of the 3 tensors (2.1). Next, taking n= 3 in (3), one gets As the LHS is zero for d = 2 and ≠ 0, we get Choose now i = j, m = k and then replace n by l, to get - This means that the first tensor of (2.4) is expressible as a linear combination of the 3 tensors (2.3); as the same will be true for the other 2 tensors of (2.4), it follows that we have proved, using an obvious notation, that "All tensors of the form •  = 0 then taking i = j = 0, k = l = 1, one gets a 1 = 0, and similarly, one can show that a 2 = a 3 = 0. Next, if one takes i=0, j = k = l = 1, one gets a 6 = 0, and a 4 = a 5 = 0 are similarly proved. Hence the 6 tensors (2.1, 2.3) are indeed linearly independent. Thus in the case d = 2, r = 4, there are 6 linearly independent isotropic tensors forming a complete set, which may be chosen to be (2.1, 2.3).

THE CASE r = 6:-
Here we choose the indices as i, j, k, l, m, n; the possible candidates for linearly independent isotropic tensors, are now , , , , in the obvious short hand notation introduced earlier.
However, as identities (4) and others obtained by choosing other sets of 4 indices out of the 6 indices i, j, k, l, m, n, remain valid here, one sees that can always be expressed as a linear combination of tensors of the form , while can be expressed as a linear combination of tensors of the form . It follows that in order to get all the linearly independent isotropic tensors of rank 6, we need to consider only those which are of the form

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; we consider these one by one. The total number of isotropic tensors of the form we write them explicitly as follows.
On the other hand, the total number of tensors of the form explicitly, these are 9 : ; With the help of a technique used by Rashid et. al. in [12] to obtain the number of linearly independent isotropic tensors of rank 6 under SO (2), it is possible to show that there exist just 20 linearly independent tensors among these 60, and that 10 of them can be taken from the 15 tensors of the form while the other 10 can be taken from the 45 tensors of the form . Let us write a linear equation we then obtain 15 equations between a 1 to a 1 5 as follows. Agreeing that " i j "means i = j = 0, k = l = m = n = 1, one gets ij ⇒ a . + a 0 + a 1 = 0, ik ⇒ a 2 + a 3 + a 4 = 0, il ⇒ a M + a N + a O = 0, im ⇒ a .F + a .. + a .0 = 0, in ⇒ a .1 + a .2 + a .3 = 0, jk ⇒ a M + a .F + a .1 = 0, jl ⇒ a 2 + a .. + a .2 = 0, jm ⇒ a 3 + a N + a .3 = 0, jn ⇒ a 4 + a O + a .0 = 0, kl ⇒ a . + a .0 + a .3 = 0, km ⇒ a 0 + a O + a .2 = 0, kn ⇒ a 1 + a N + a .. = 0, lm ⇒ a 1 + a 4 + a .1 = 0 ln ⇒ a 0 + a 3 + a .F = 0, mn ⇒ a . + a 2 + a M = 0.
Hence if one considers the linear equation then it follows from equations (6) that the 10 a's a 1 , a 2 , a 3 ,... a 8 , a 10 , a 15 , vanish when a 9 = a 1 1 = a 1 2 = a 1 3 = a 1 4 = 0; this implies that the 10 isotropic tensors of which the a's of (8) are the coefficient, are linearly independent. Further, if any one of the above 5 a's is nonzero, then some of the a's of (8) will also be nonzero, so that equation (7) will show that each of the isotropic tensor corresponding to one of the above 5 a's is a linear combination of those corresponding to a's of (8). It follows that among the 15 isotropic tensors of the form , precisely 10 are linearly independent and the rest are linear combination of these 10.
Next, consider the tensors of the form . Using the statement given on lines 8-10 of first column of page 3, with l replaced by n and i, j, k, replaced by i, j, k, l, m, we see that all the tensors of the form can be expressed in terms of the 15 tensors of the form •) . It follows that among the 45 tensors of the form , all the linearly independent ones may be chosen from those of the type •) . To find these linearly independent tensors, consider the equation It is interesting to note that these are identical with equations (6) obtained earlier. Applying now the same arguments as were used after equation (6), we conclude that out of the 15 isotropic tensors of the form • ) , precisely 10 are linearly independent and these can be chosen to be the ones whose coefficients are . − N , . F , . 3 in the above linear equation.
We have thus completed the proof of our assertion that out of the 60 isotropic tensors of rank 6 of the form , , precisely 20 are linearly independent (and form a complete set) and that 10 of these may be chosen to be of the form while the rest of 10 can be chosen to be of the form • ) . , .

d = 3; Isotropic Tensors
These are easily checked, as in the case d = 2, to be linearly independent; thus there are just 3 linearly independent isotropic tensors of rank 4. THE CASE r = 5. Here, the possible isotropic tensors are This shows that ( can be expressed as a linear combination of tensors in which the subscripts of , do not contain m. As the same will obviously be true for the 3 tensors The linear independence of these 6 tensors is easily checked by writing an equation one gets a 1 = 0, and one can similarly show that the rest of them are also 0. This proves the linear independence of the set so that we conclude that there exist precisely 15 linearly independent isotropic tensors of rank 6, forming a complete set, under SO(2, 1).

d=4: Isotropic Tensors Under SO(3, 1) and SO(2, 2)
Here, the matrix is given by for SO(2, 2); both these cases can be considered together as the arguments are independent of the difference between the two ′k. THE CASE r = 2: As in the case of SO(2, 1), the only isotropic tensor of rank 2, in the present case, is . THE CASE r = 3: As already mentioned, there are no isotropic tensors of odd rank when d = 4.
THE CASE r = 4: Here, the possible candidates for the linearly independent isotropic tensors are , , , .
It is easy to check that these are all independent; for, if Choosing i = j = 0, k = l = 1, m = n = 2, the above equation gives a 1 = 0; the rest of the a's can be similarly proved to be equal to 0, so that one may conclude that all the tensors of the form are indeed linearly independent.
Next, to consider the linear independence of tensors of the form , we see that from Appendix B, we get gives a 1 = 0; all other a's are proved to be equal to zero in exactly the same way. It follows that the ten tensors are indeed linearly independent. Thus we conclude that there are 15 + 10 = 25 linearly independent isotropic tensors of rank 6, forming a complete set, under SO(3, 1) and SO(2, 2).

Conclusion
We have obtained all the linearly independent isotropic tensors of ranks from 2 to 6, under the non-compact rotation groups SO(1, 1), SO(2, 1), SO(3, 1) and SO (2,2). We find that their number is exactly the same as that of corresponding linearly independent isotropic tensors under the compact rotation groups SO(2), SO(3) and SO(4) as given by Naila [11] and Faiz and Rashid [10,12]. showing that is indeed isotropic underSL(3, ℝ ). In particular, it will be isotropic under both SO(3) and SO(2, 1) as these are just subgroups of SL (3, ℝ). In fact, the result holds for arbitrary n≥ 2 rather than for just n = 3; thus we have # ...# is an isotropic tensor under SL(k, ℝ), k ≥ 2. The fact that is isotropic under both SO(2) and SO(1, 1) follows directly from this result.