ISSN 0025-6544, Mechanics of Solids, 2009, Vol. 44, No. 1, pp. 14–30. c  Allerton Press, Inc., 2009. Original Russian Text c  M.U. Nikabadze, 2009, published in Izvestiya Akademii Nauk. Mekhanika Tverdogo Tela, 2009, No. 1, pp. 17–36. On the Construction of Linearly Independent Tensors M. U. Nikabadze * Lomonosov Moscow State University, GSP-2, Leninskie Gory, Moscow, 119992 Russia Received June 25, 2007 Abstract—We consider various methods for constructing linearly independent isotropic, gyrotropic, orthotropic, and transversally isotropic tensors. We state assertions and theorem that permit one to construct these tensors. We find linearly independent above-mentioned tensors up to and including rank six. The components of the tensor may have no symmetry or have symmetries of various types. DOI: 10.3103/S0025654409010026 1. ON ISOTROPIC TENSORS IN R 3 It is known [1–5] that E ˜ E = r i r i = g ij r i r j is the only isotropic tensor of rank 2 that can be used to represent any other isotropic tensor a ˜ a of rank 2 in the form a ˜ a = aE ˜ E, where a is a scalar; i.e., an arbitrary isotropic tensor of rank 2 is a spherical tensor. The tensors C ˜ C ˜ C (1) = E ˜ EE ˜ E = r i r i r j r j , C ˜ C ˜ C (2) = r i r j r i r j , C ˜ C ˜ C (3) = r i E ˜ Er i = r i r j r j r i (1.1) are three linearly independent (irreducible to each other) tensors of rank 4. The general expression for an arbitrary isotropic tensor of rank 4 is their linear combination C ˜ C ˜ C = 3  k=1 a k C ˜ C ˜ C (k) . If we pay attention to the structure of isotropic tensors of rank 2 and rank 4 in (1.1), then we easily see that they can be obtained from the corresponding multiplicative bases by pairwise convolution (contraction) of indices of the basis vectors and by exhausting all possible cases of such contraction. By way of example, let us also construct all linearly independent isotropic tensors of rank 6. The multiplicative basis of a tensor of rank 6 is r i r j r k r l r m r n . By contracting the indices pairwise arbitrarily, we obtain some isotropic tensor of rank 6. For example, r i r i r k r k r m r m = E ˜ EE ˜ EE ˜ E. (1.2) All other isotropic tensors of rank 6 can be obtained from (1.2) by permutations of basis vectors. Obviously, by rearranging the basis vectors in (1.2), we obtain 6! = 720 permutations (isotropic tensors of rank 6) in the general case. Of these tensors, only fifteen are linearly independent (irreducible to each other) [3, 5]. To obtain these linearly independent tensors of rank 6, it suffices, for example, to consider the following tensors: r i r i r k r k r m r m , r i r k r i r k r m r m , r i r k r k r i r m r m , r i r k r k r m r i r m , r i r k r k r m r m r i . (1.3) It is clear that the basis vector r i occupies all possible positions in (1.3). Now by keeping the vectors r i and r i at their positions in (1.3) and by permuting the other vectors, we obtain two additional tensors * E-mail: munikabadze@mail.ru 14