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As per the title I wish to describe all of the irreducible representations of the group $SO(2)$. As a first guess I would say that the following rep is irreducible. $$\text{Let }~~~\mathcal{D}:SO(2)\rightarrow GL(2,\mathbb{R}), ~~ \mathcal{D}(R):=R.$$ This representation acts on the vector space $\mathbb{R}^2$, and as one can imagine, there are no vectors which lie in the plane and that are invariant under all rotations in the plane (except for the trivial zero vector) thus there is no non-trivial invariant subspace and this representation is irreducible. But then since $SO(2)$ is an Abelian group, by Schurs lemma its irreps are one-dimensional, which is in clear contradiction to the above. I suspect this is because Schur's lemma applies only to complex vector spaces. So is it correct to say that with regards to $\textit{real}$ representations $\mathcal{D}$ is irreducible, but not with regards to complex representations.

So is the question as stated in the title ill-defined, and the word real or complex representations must be inserted? This question appeared in an exam, when phrased in this way does one usually assume real representations is meant?

If so then would i be correct in using $\mathcal{D}$ or the isomorphism $SO(2)\cong U(1)$ and thus irreps are labelled by an integer?

EDIT

Let $$\Psi:U(1)\rightarrow SO(2);~~\Psi(e^{i\phi})=\begin{pmatrix}\cos(\phi)&&\sin(\phi)\\-\sin(\phi)&&\cos(\phi)\end{pmatrix}$$ be the map that establishes the isomorphism $U(1)\cong SO(2)$.

Denote the (complex) irreps of $U(1)$ by $\mathcal{T}^{(n)}$ where $$\mathcal{T}^{(n)}:U(1)\rightarrow GL(1,\mathbb{C})\cong\mathbb{C}^*;~~~\mathcal{T}^{(n)}(e^{i\phi})=e^{in\phi},~~n\in\mathbb{Z}.$$ Denote the (real) irreps of $SO(2)$ by $\mathcal{D}^{(n)}$ where $$\mathcal{D}^{(n)}:SO(2)\rightarrow GL(2,\mathbb{R})\\D^{(n)}\bigg(\begin{pmatrix}\cos(\phi)&&\sin(\phi)\\-\sin(\phi)&&\cos(\phi)\end{pmatrix}\bigg)=\begin{pmatrix}\cos(n\phi)&&\sin(n\phi)\\-\sin(n\phi)&&\cos(n\phi)\end{pmatrix}.$$ Now $\mathcal{T}^{(n)}$ acts on the vector space $\mathbb{C}$ whilst $\mathcal{D}^{(n)}$ acts on the vector space $\mathbb{R}^2$ and since $\mathbb{C}\cong\mathbb{R}^2$ we can use $\Psi$ to show that the irreps are the same: $$\mathcal{D}^{(n)}\cong \Psi \circ\mathcal{T}^{(n)}.$$ Does this show that the complex and real irreps of $SO(2)$ are the same, or have I shown something else? Does this result generalize?

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    $\begingroup$ Yes, you need to distinguish the real and complex cases, in the latter the irreps are 1-dimensional but not necessarily the former. $\endgroup$ – Qiaochu Yuan Nov 17 '17 at 18:06
  • $\begingroup$ @QiaochuYuan May I please have your opinion on the recent edit to my question. $\endgroup$ – NormalsNotFar Nov 18 '17 at 4:18

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