Counting the number of independent components in tensors with symmetry constrains is easy to conceptualize, but it took me days to learn a systematic algorithm for solving the problem. Here I present one method for enumerating the components of tensors with arbitrary rank, in arbitrarily high dimensional spaces, with arbitrary number of symmetry and antisymmetry constraints.
The Schur-Weyl duality is a connection between the general linear group and the symmetry group on the same tensor space. The duality connects the irreducible representations of these two groups in the sense that it says that the symmetries determine each other on a similar tensor space. Let's unpack.
First, lets understand the general linear group. This is the group of all invertible transforms on a given tensor space. The general linear GL group is often specified in a certain degree n, which corresponds to the dimension of the underlying tensor space. So the general linear group GL(n) is filled with invertible tensors of size n by n. Invertible matrices clearly form a group under tensor multiplication: the product of two invertible matrices is again invertible, the inverse of an invertible matrix is itself invertible, and the identity element is the identity matrix.
Second, the symmetry group on a finite tensor space consists of all permutations over the indices of a tensor. Again, this operation forms a group.
Finally, what is an irreducible representation? This is one of the more difficult aspects of this duality for me to explain (even to myself). But I will try and break it down simply. A representation in general is a mapping between elements of a group (say, the symmetric group) and a specific matrix that performs the action. The group may include abstract ideas; for instance, interchange of the first and second indices. For tensors in a given space, this action can be encoded into a matrix operation.
If matrix operations we call representations are block diagonalizable, then they can be decomposed into subrepresentations. In other words, if one can find a basis transformation for the tensor space where the representation can be block-diagonalized, then the representation is said to be reducible. Irreducible representations are a space that cannot be further decomposed into block diagonal form. To put this another way, they have no proper (trivial) invariant subspaces. The irreducible representations are the atomic building blocks of our group actions.
An irreducible representation will always exist for the general linear group and the symmetric group. The irreducible representation is not necessarily unique; however, the multiplicities (or size) of the irreducible representation is unique. This is a result due to Maschke.
Schur-Weyl duality says that the irreducible representation of the general linear group and the symmetry group commute with each other on their respective tensor space. This can be seen if one considers how each group acts on the tensor space. The general linear group acts locally within each slot of the tensor space. The symmetry group acts globally, swapping the slots themselves. Both groups act on the same tensor space, but they never step on each other's toes. So the order in which they are performed does not matter — that's what commutativity means. And because the irreducible representations are commutative, they can be block-diagonalized simultaneously. Because these irreducible representations share blocks, we will find that they can be naturally described by Young diagrams.
Schur-Weyl duality establishes more than commutativity. The duality shows that the two groups determine each other. They determine each other in the sense that knowing the general linear groups determines the symmetry groups, and visa versa. This can be shown with the double centralizer theorem, but I think it can be explained more simply.
First grant that a linear transform (an element of the group GL(n)) and a tensor-rank permutation operator commute for all potential coefficients of the linear transform. Actually, let's say any potential coefficients of the linear transform such that the transform remains invertible. If these two operations commute, then the linear transform must act identically on all slots of tensors. Otherwise, we could permute the tensor such that the linear transform would not commute. So permutations between ranks restricts linear transforms to and within ranks. The duality establishes that this relationship works both ways: linear transforms determine the symmetries available, and symmetries determine the linear transforms.
This last point, that known symmetries on a tensor space determine the group of linear transforms, is critical to our application of the Schur-Weyl duality. We would like to use this result to count the number of independent components of a tensor with arbitrary symmetries. By encoding our symmetries as irreducible representations, we can determine which linear transforms are necessary. The sum of the products of the dimensions of the irreducible representations of the symmetry group and the irreducible general linear group equals the dimension of the underlying tensor space.
In order to determine the dimension of the tensor space (or the number of independent components) we need to compute the dimensions of the irreducible symmetry groups and the irreducible general linear group elements. This can be accomplished in various ways — I will utilize Young's diagram, which are also referred to as Young's tableaux.
Clearly I am not a fan of typesetting equations and graphics. To be frank, I just find it painful in html. So I will go relatively quickly over Young's diagram, because they are a highly visual counting tool. Thankfully, they are well documented elsewhere, so hopefully the reader who wants to see the formula and tables can find them. If not, send me an email and I will find a couple links for you.
Young's diagram is a way to determine the dimension of the irreducible representations of the general linear and symmetry groups. To construct the Young's diagram, first list all the partitions of the rank of the tensor. For instance, a rank four tensor will have five partitions: (4), (3,1), (2,2), (2,1,1), (1,1,1,1). Each partition of four corresponds to an irreducible representation. (Recall that the irreducible representations of the symmetry group and the general linear group will overlap. Therefore, when I refer to groups I refer the general linear group and the symmetry group simultaneously.
The partitions each have a physical interpretation in terms of tensor symmetries. (4) corresponds to a fully-symmetric fourth rank tensor. (3,1) corresponds to a tensor symmetric in three components and antisymmetric in one pair of indices. (2,2) corresponds to a tensor with two pairs of symmetric indices and simultaneous antisymmetry between these two groups. (2,1,1) is a symmetric in one one pair of indices and antisymmetric in the other three indices. (1,1,1,1) corresponds to a fully antisymmetric tensor (i.e., a tensor of Levi-Cevita type). So in terms of partition notation, commas split antisymmetric groups of indices. Symmetric groups of tensors are indicated by numbers greater than one in the partition.
The number of standard Young's tableaux for each partition corresponds to the dimension of the symmetry group corresponding to that partition. Formula to compute this number can be found online. There is only one point of note: the number of standard Young's diagrams can be computed without knowledge of the dimension of the space that the tensor lives in. This should make sense — the symmetric group only cares about how many slots there are for a particular tensor (the rank), not necessarily how many values the slots individually can take (the dimension).
The formula for the dimension of the irreducible representation of the general linear group associated with a particular partition is a little more involves. One has to compute the hook lengths and offsets of every entry in the Young diagram, as well as have knowledge of the dimension of the tensor space. Again, I will not list the explicit formula. These are just counting details. Both formulas for the dimensions of the irreducible representations require nothing more than algebra. From a certain perspective, that is the beauty of Young's diagram — they turn complex counting tasks into a sequence of algebraic operations.
For an arbitrary tensor of rank r living in a space with n dimensions, the number of independent components is nr. For example, a fourth rank tensor in three dimensions will have 81 independent components (assuming no restrictions on the components). If one has correctly computed the dimension of the irreducible representations, the sum of the products of the dimensions for the symmetric and general linear groups should yield the same number as the above formula. This is an easy way to confirm that the formula being used to compute the dimensions of the irreducible representations associated with each partition are correct, and the algebra has been done correctly.
Up to this point, no symmetry constraints have been leveraged to decrease the number of independent components of a tensor. We will leverage the Young's diagram one last time to decrease the dimension of the symmetry parts associated with the Schur-Weyl duality.
Symmetries associated with a tensor will reduce the number of independent components. In the language of the Schur-Weyl duality, the dimension of the symmetry group associated with the irreducible representation of the tensor will be decreased. We decrease the dimension of the irreducible representation associated with each partition by showing that certain symmetries between indices imply that our tensor can be represented without certain irreducible representation.
From the perspective of a geophysicist, this process feels like orthogonal projection. We are going to identify irreducible representations of the symmetry group that our tensor symmetries cannot exist in. Analogously, these are directions in a model space orthogonal that we cannot reach with our linear transform (the null space). We are going to construct our tensor only our of the necessary irreducible representations. This is just like constructing our model out of a basis that spans our column space and setting the part of the basis that spans the null space to zero. Because we are incapable of representing models in the null space, we can just forget about it without loosing any information. Hopefully the analogy makes more sense as we move forward.
The dimension of the symmetry group (or multiplicity of the symmetry group) indicates how many non trivial combinations of indices each Young tableaux can hold. For instance, consider the partition (4). This corresponds to a fully symmetric tensor. Four indices can only be arranged in one non trivial way in a fully symmetric tensor, because the order of indices is completely symmetric. Therefore the multiplicity of the irreducible representation of (4) is one. Note this result is true only for fourth rank tensor.
Another example. The (2,2) partition corresponds a fourth rank tensor with two pairs of symmetric indices, with the pairs being antisymmetric. There are only two non trivial ways to arrange these groups: into the pairs (i,j) and (k,l) or (i,k) and (j,l). Any other combination is rendered trivial by symmetry within groups. So the multiplicity of the partition (2,2) is two. Again, this result is true only for fourth rank tensors. But since the size of the symmetry group associated with a given partition is insensitive to the dimension the tensor is in, it is true for fourth rank tensors in any dimension.
Now the job becomes mechanical. For every non trivial arrangement of indices, we need to see whether or not the symmetry constraints we impose on our tensor require us to include this symmetry group. Remember that for the general, unrestricted tensor all of the irreducible representations of the symmetry group are necessary. After imposing symmetry constraints, we will invalidate some of the non trivial representations — in other words, we can prove that our symmetry constraints do not allow our tensor to be represented in some of the irreducible representations. Such proofs can be constructed by assuming that the constraints and the symmetry corresponding to the particular partition are both simultaneously true. If we can but the tensor into an inconsistent state, then we can say the irreducible irreducible representation is unnecessary for our tensor and reduce the associated multiplicity.
Each non trivial symmetry in each group must be separately tested. For instance, we return to our example of for the partition (2,2). Both cases of (2,2) must be independently confirmed. If both cases are consistent with our constraints, then the multiplicity stays at two. It only one of the cases will hold and the other cases will not, we reduce the associated multiplicity to one. If both cases lead to contradictions, the multiplicity is reduced to zero. In practice, any one of these situations can happen: for the elastic tensor (rank four) in three dimensions, with symmetries following from Hooke's law and adiabatic arguments, only one of the two cases of the (2,2) irreducible representation is necessary. All cases of partitions (3,1), (2,1,1), and (1,1,1,1) can be shown to be unnecessary. And the partition (4) is entirely necessary. Tallying up the resulting dimensions with the Schur-Weyl duality, the total number of independent components the elastic tensor with symmetry demands in three dimensions is 21.
Now that we can see the entire workflow, maybe the analogy between the irreducible representations and orthogonal projection makes more sense. The Schur-Weyl duality acts like a linear combination of irreducible representations of the general linear group (basis vectors) weighted by irreducible representations of the symmetry group sizes (weights). Whenever we are orthogonal to a symmetry group, we can reduce the necessary size of the space required to express our tensor in — these unnecessary symmetry groups are in this sense orthogonal to our tensor.