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Section 5.1 Tensor representations of \(SO(3)\)

In this section we study intuitively what representations of \(SO(3)\) look like. This will give us an intuitive understanding of what the representation theory of \(SO(3)\) should be, which we will then construct explicitly using representations of its Lie algebra in the following sections.

Subsection 5.1.1 Representations of \(SO(2)\)

Let us start by giving a brief and naive construction of the irreducible representations of \(SO(2)\text{.}\)

We start with the cyclic group \(\mathbb{Z}_N\text{,}\) which we can think of a discrete set of rotations in two dimensions by angle \(2 \pi / N\text{.}\) We know its irreducible representations: there are \(N\) of them, all one-dimensional, given by sending the generator of \(\mathbb{Z}_N\) to \(e^{2 \pi i k / N}\) for some \(k \in \{0,1,\ldots,N-1 \}\text{.}\)

Well, as \(N\) goes to infinity, we can think of the group \(\mathbb{Z}_N\) as turning into the continuous group \(SO(2)\) of rotations in two dimensions. So we could expect the irreducible representations of \(SO(2)\) to be all one-dimensional (which is true since \(SO(2)\) is abelian), and to be given by the complex numbers \(e^{i k \theta}\) for \(\theta \in [0,2 \pi)\text{,}\) and with \(k\) a non-negative integer. So there would be an infinite number of them, indexed by \(k\text{.}\)

While this construction is of course very naive (since it is not clear what precisely is meant by the limit \(N \to \infty\) here), the intuition is correct. Those are the irreducible representations of \(SO(2)\text{,}\) and they are indexed by a non-negative integer \(k\text{.}\)

Subsection 5.1.2 Representations of \(SO(3)\)

Let us now move on to the more interesting case of rotations in three dimensions, that is, \(SO(3)\text{.}\) What are the irreducible representations? After looking at \(SO(2)\text{,}\) we expect an infinite number of them. But here we do not expect them to be all one-dimensional, since \(SO(3)\) is not abelian.

The simplest representation is of course the trivial one, where we sent every element \(R \in SO(3)\) to \(T(R) = 1\text{.}\) This is as usual a rather boring one. It is sometimes called the scalar representation in physics, since, if you think of it as acting on a one-dimensional vector space \(V\text{,}\) then \(T(R) v = v\) for all \(v \in V\text{,}\) that is, it leaves everything invariant. So a scalar object (which is an object that is invariant under rotations) transforms according to the trivial representation of \(SO(3)\text{.}\)

We have already studied another representation of \(SO(3)\text{:}\) its defining representation, which associates to every \(R \in SO(3)\) a three-dimensional rotation matrix \(T(R)\text{.}\) In physics, we sometimes call this representation the vector representation. This is because it defines how vectors in \(\mathbb{R}^3\) transform under rotations. Indeed, if \(v \in \mathbb{R}^3\text{,}\) then under a rotation we get \(v' = R v\text{,}\) where \(R\) is a \(3 \times 3\) rotation matrix. In index notation, we would write

\begin{equation*} v'_a = \sum_{j=1}^3 R_{aj} v_j. \end{equation*}

Turing this around, one could think of vectors as being objects that transform according to the three-dimensional defining representation of \(SO(3)\text{.}\)

Remark 5.1.1.

This is a good time to introduce a notation that is standard in physics, which is to denote the representation of a Lie group by its dimension in bold face. For instance, we would write \(\mathbf{3}\) for the defining representation of \(SO(3)\text{.}\) The notation is slightly ambiguous though, because there may be more than one representation of the same dimension. For instance, if a representation is complex, then its complex conjugate has the same dimension (we would then write, say, \(\mathbf{5}\) and \(\overline{\mathbf{5}}\) for the complex conjugate representations). Nevertheless, this is common notation, and it is usually clear from context what representation is being discussed.

Now you probably see a pattern. How can we get more representations of \(SO(3)\text{?}\) A natural way to think about representations is to think about how they act. Namely, we think of objects that transform in certain ways under rotations. We have already studied how scalars and vectors transform, which gave us the one-dimensional and three-dimensional representation of \(SO(3)\text{.}\) What next?

The next objects to look at are matrices. How do matrices tranform under rotations? Think of a matrix \(M\) as being a linear operator on \(\mathbb{R}^3\text{.}\) Then under a rotation \(R\) (which is a change of basis for \(\mathbb{R}^3\)) the matrix \(M\) will change by a similarity transformation \(M' = R M R^T\text{,}\) since \(R\) is orthogonal. In index notation, this becomes

\begin{equation*} M'_{ab} = \sum_{i,j=1}^3 R_{a i} M_{ij} R^T_{j b} = \sum_{i,j=1}^3 R_{a i} R_{b j} M_{ij}. \end{equation*}

This defines a new representation of \(SO(3)\text{.}\) Indeed, we could place the nine components of the matrix \(M\) in a vector, and then the transformation rule above would give us a nine-dimensional representation of \(SO(3)\text{,}\) where we associate to every group element of \(SO(3)\) the \(9 \times 9\) matrix with components given by \(R_{ai} R_{bj}\text{,}\) where \(R\) is the corresponding \(3 \times 3\) rotation matrix. In other words, if you recall the construction of tensor product representations, what we are constructing here is the nine-dimensional representation that is the tensor product of the defining representation with itself:

\begin{equation*} \mathbf{9} = \mathbf{3} \otimes \mathbf{3}. \end{equation*}

We know that matrices transform according to this representation of \(SO(3).\)

We could keep going and define higher-dimensional representations by looking at how objects with more than two indices transform under rotations. Those objects are known as tensors, and the corresponding representations are known as tensor representations. The rank of a tensor is the number of indices. For instance, a rank \(3\) tensor \(T_{ijk}\) transforms as

\begin{equation*} T'_{abc} = \sum_{i,j,k=1}^3 R_{a i} R_{b j} R_{c k} T_{ijk}, \end{equation*}

which defines a \(27\)-dimensional representation of \(SO(3)\text{,}\) namely \(\mathbf{27} = \mathbf{3} \otimes \mathbf{3} \otimes \mathbf{3}\text{.}\) In fact, we can think of tensors as being objects that transform according to representations of \(SO(3)\text{.}\)

Subsection 5.1.3 Irreducible representations of \(SO(3)\)

In the previous section we constructed an infinite class of representations of \(SO(3)\) by looking at how tensors transform under rotations. But are these representations irreducible? We will not answer this question rigorously here. We will instead look at the rank \(2\) case to get an intuition about what to expect, and then state the general result without proof.

Let us look at the rank \(2\) tensor representation (the matrix one) as a starting point. This \(9\)-dimensional representation maps elements of \(SO(3)\) to operators that act on matrices as

\begin{equation*} M'_{ab} = \sum_{i,j=1}^3 R_{a i} R_{b j} M_{ij}. \end{equation*}

This representation will be reducible if there exists a proper subspace of the space of matrices that is invariant under the action of the representation.

The key here is to notice that the transformation preserves the symmetry property of tensors under permutation of indices. In the case of a matrix, we can always write a matrix as a sum of its symmetric and antisymmetric parts:

\begin{equation*} M_{ij} = \frac{1}{2} (M_{ij} - M_{ji} ) + \frac{1}{2} (M_{ij} + M_{ji} ) := A_{ij} + S_{ij}, \end{equation*}

where \(A_{ij} = - A_{ji}\) and \(S_{ij} = S_{ji}\text{.}\) The subspaces given by antisymmetric and symmetric matrices respectively are invariant under the action of the representation. Indeed, we get that

\begin{align*} A'_{ab} = \amp \frac{1}{2} (M'_{ab} - M'_{ba} ) \\ = \amp \frac{1}{2} \sum_{i,j=1}^3\left(R_{a i} R_{b j} M_{ij} - R_{b i} R_{a j} M_{ij} \right)\\ = \amp \frac{1}{2} \sum_{i,j=1}^3 R_{a i} R_{b j}\left( M_{ij} - M_{ji} \right)\\ = \amp \sum_{i,j=1}^3 R_{a i} R_{b j} A_{i j}. \end{align*}

A similar argument holds for \(S_{ij}\text{.}\) Thus the subspaces of symmetric and anti-symmetric rank \(2\) tensors are invariant subspaces, and thus provide subrepresentations. The number of independent components in a rank \(2\) anti-symmetric tensor in three dimensions is \(3\text{,}\) while the number of independent components in a rank \(2\) symmetric tensor in three dimensions is \(6\text{.}\) We thus obtain the decomposition \(\mathbf{9} = \mathbf{6} \oplus \mathbf{3}\) for the nine-dimensional tensor representation.

Now, are the \(\mathbf{6}\) and \(\mathbf{3}\) representations irreducible? It turns out that the three-dimensional one is, and is in fact dual to the vector representation (since in three dimensions rank \(2\) anti-symmetric tensors are dual to vectors, see Zee's book, section IV.1, for more on this). However the six-dimensional representation is still reducible.

Indeed, consider the subspace consisting of the trace \(S = \sum_{i=1}^3 S_{i i}.\) Then one sees that

\begin{align*} S' = \amp \sum_{a=1}^3 S'_{aa}\\ = \amp \sum_{a=1}^d \sum_{i,j=1}^3 R_{a i} R_{a j} S_{i j}\\ = \amp \sum_{i,j=1}^3 \left(\sum_{a=1}^d R^T_{i a} R_{a j} \right) S_{ij}\\ = \amp \sum_{i,j=1}^3 \delta_{ij} S_{ij}\\ = S, \end{align*}

where we used the fact that \(R\) is an orthogonal matrix, and hence \(R^T R = I\text{.}\) Therefore the trace is invariant! Thus the subspace spanned by the trace is a one-dimensional invariant subspace, which transforms according to the trivial one-dimensional representation. We thus get the further decomposition \(\mathbf{6} = \mathbf{5} \oplus \mathbf{1}\text{.}\) Overall, the nine-dimensional tensor representation decomposes as \(\mathbf{9} = \mathbf{5} \oplus \mathbf{3} \oplus \mathbf{1}.\) It turns out that these summands are now all irreducible. The objects that transform according to the five-dimensional representation are the symmetric traceless tensors, which can be written as

\begin{equation*} S_{ij} - \delta_{ij} \frac{S}{3}, \end{equation*}

and are indeed traceless.

A similar decomposition, albeit slightly more complicated, is possible for higher rank tensors. For \(SO(3)\text{,}\) it turns out that there is a special duality that can be used to show that for higher rank tensors, only the symmetric traceless tensors give rise to new irreducible representations (see Zee's book, section IV.1). The end result, which we will not prove here, is that all irreducible representations of \(SO(3)\) can be constructed by looking at how symmetric traceless tensors transform.

Our last question for this section is: what is the dimension of these irreducible representations? To find the dimension we need to find the number of independent components of symmetric traceless tensors. This will give the dimensions of the irreducible representations of \(SO(3)\text{.}\)

We need to count the number of independent components of \(S\text{.}\) First, let us count the number of components of a symmetric rank \(j\) tensor. Suppose first that each index can only take values \(1\) and \(2\text{.}\) There the possibilities are \(2 \cdots 2\text{,}\) \(2 \cdots 2 1\text{,}\) \(2 \cdots 2 1 1\text{,}\) and so on. This gives \(j+1\) possibilities. Then we add a \(3\text{.}\) If we have a \(3\) for the first index, then we have \(j\) possibilities for \(2\)s and \(1\)s in the remaining \(j-1\) indices. If we have two \(3\)s, then we have \(j-1\) possibilities for the remaining indices. And so on. Overall we get: \(\sum_{k=0}^j (k+1) = \left(\sum_{k=0}^j k\right) + (j+1) = \frac{1}{2}j (j+1) + (j+1) = \frac{1}{2} (j+1)(j+2) \) possibilities. So a symmetric rank \(j\) tensor has \(\frac{1}{2}(j+1)(j+2)\) independent components.

The tracelessness condition \(\sum_{k_1, k_2=1}^3 \delta_{k_1 k_2} S_{k_1 k_2 \ldots k_j} = 0\) consists in a number of independent conditions. The number of conditions here is the number of values that the indices \(k_3 \cdots k_n\) can take. From the previous paragraph, this is the number of independent components of a rank \(j-2\) symmetric tensor, which is \(\frac{1}{2} j (j-1)\text{.}\) Therefore, the total number of independent components of a symmetric traceless rank \(j\) tensor is

\begin{equation*} \frac{1}{2}(j+1)(j+2) -\frac{1}{2} j (j-1) = 2 j + 1. \end{equation*}

The end result is that there is an infinite number of irreducible representations for \(SO(3)\text{,}\) indexed by a non-negative integer \(j\text{,}\) with dimensions \(2j+1\text{.}\) The objects that transform according to these representations are symmetric traceless rank \(j\) tensors.

This is the key result of this section. And, if you have done some physics, you may have encountered the formula \(2j+1\) before. It is rather famous in the history of quantum mechanics and atomic physics. It corresponds to the degeneracy of states of the hydrogen atom (or spherical harmonics). It also corresponds to the multiplicity, or quantum states, of particles with integer spin. This is certainly not a coincidence, as we will see!

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