In Zee's book, in section 4.1 under the heading "The tensors of $SO(2)$," he states: "Hence the dimensions of the irreducible representations are $(j + 1) − (j − 2 + 1) = 2$. All of them are 2-dimensional!" He then goes on to answer Confusio's question that previously we said that all irreducible representations of $SO(2)$ are one dimensional. He answers this by stating that the two-dimensional representation is actually reducible to two one-dimensional representations.

I feel that this is an inadequate answer to the question because he had just stated that "all irreducible representations have dimension 2". His answer to Confusio's question contradicts this statement because he claims now that the representation is reducible.

What am I missing here?


1 Answer 1


The issue lies with what precisely is meant by "reducible." A reducible representation is one for which the group action on the vector space leaves a subspace of the vector space invariant. However, this definition does not specify what kind of vector space is involved—in particular, for the case of interest here, whether we are talking about a vector space over ${\mathbb R}$ or over ${\mathbb C}$. (You can, mathematically have representations over other fields as well, but those are typically the two possibilities of interest in physics.) The resolution to your question is that the faithful irreducible representations of $SO(2)\approx U(1)$ are two-dimensional over ${\mathbb R}$, but only one-dimensional over ${\mathbb C}$.

The Lie group $SO(2)$ is the group of orthogonal matrices $$g(\theta)=\left[ \begin{array}{cc} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array} \right].$$ The irreducible representations over ${\mathbb R}$ are indexed by $n$ and formed by taking $n$th powers of $g(\theta)$, corresponding to rotation by $n\theta$, $$\rho_{n}(g)=\left[ \begin{array}{cc} \cos n\theta & -\sin n\theta \\ \sin n\theta & \cos n\theta \end{array} \right].$$ None of these leaves a proper linear subspace of ${\mathbb R}^{2}$ invariant, except for $n=0$, which is actually two copies of the trivial representation (and the trivial representation is always one dimensional).

The presentation of the group in terms of $2\times 2$ matrices is one way of describing it. However, the same isomorphism class of group can also be expressed as $U(1)$: the group of $1\times 1$ complex matrices $h$ with $h^{\dagger}h$ equal to the identity. Of course, a $1\times 1$ matrix can simply be identified with its single element, so this is the group of complex number $z$ of unit magnitude $z^{*}z=|z|^{2}=1$. This presentation of the group provides a natural form for the representation theory over ${\mathbb C}$. The representations, again indexed by $n$, act on the vector space ${\mathbb C}^{1}$ by multiplication by $z^{n}$, $$\bar{\rho}_{n}(z)=z^{n}.$$ There representations are manifestly one dimensional.

The equivalence between the two descriptions of the group comes simply from writing $z=\cos\theta+i\sin\theta$. This provides a one-to-one mapping between $z$ and $g(\theta)$, which is, in fact, an isometry. The representation function $\rho_{n}$ can then be written as a complex combination of $\bar{\rho}_{n}$ and $\bar{\rho}_{-n}$, but this cannot be done using only real numbers. In other words, $\rho_{n}$ (for $n\neq 0$) preserves no invariant subspace of ${\mathbb R}^{2}$, but it does split ${\mathbb C}^{2}$ into two invariant subspaces. The two invariant subspaces are spanned by the eigenvectors of $g(\theta)$.

We can see explicitly which subspaces these are. Look at $\rho_{n}(g)$ acting on a particular eigenvector, $$\rho_{n}(g)\left[ \begin{array}{c} 1 \\ i \end{array} \right]= \left[ \begin{array}{cc} \cos n\theta & -\sin n\theta \\ \sin n\theta & \cos n\theta \end{array} \right]\left[ \begin{array}{c} 1 \\ i \end{array} \right]=\left[ \begin{array}{c} \cos n\theta-i\sin n\theta \\ \sin n\theta+i\cos n\theta \end{array} \right]=\left[ \begin{array}{c} e^{-in\theta} \\ ie^{-in\theta} \end{array} \right]=e^{-in\theta}\left[ \begin{array}{c} 1 \\ i \end{array} \right],$$ we see that the subspace spanned by the eigenvector $[1,\,i]^{T}$ is left invariant by $\rho_{n}$. So is the subspace spanned by the other eigenvector, $[1,\,-i]^{T}$. This splits the two-dimensional faithful representation into two one-dimensional faithful representations, but the splitting is only possible over the field ${\mathbb C}$.

  • 2
    $\begingroup$ "None of these leaves a proper linear subspace of $\mathbb R^2$ invariant, except for $n=1$...". Did you mean $n=0$? $\endgroup$
    – Christophe
    Aug 9, 2021 at 7:53
  • $\begingroup$ @Christophe Thanks. Yes, of course I meant $n=0$. I guess I’m more used to the representation theory of $SU(2)$, where $n=1$ is the trivial representation. $\endgroup$
    – Buzz
    Aug 9, 2021 at 17:56

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