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Vectors $\vec A=(A_1,A_2)$ and $\vec B=(B_1,B_2)$ are 2-dimensional representations of $SO(2)$. I want to understand the decomposition $$2\otimes 2=1\oplus 1\oplus 2.$$ I can easy identify that the two 1D irreducible representations are $A_iB_i=A_1B_1+A_2B_2$ and $\epsilon_{ij}A_iB_j=A_1B_2-A_2B_1$. What is the 2 on the RHS of the decomposition?

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  • $\begingroup$ SO(2) is Abelian. All irreps are 1-dimensional. Do you mean O(2)? $\endgroup$ Jun 5 at 21:34
  • $\begingroup$ or else are you working over the reals? $\endgroup$ Jun 5 at 22:44
  • $\begingroup$ @ZeroTheHero That's a nice point. All irreps of SO(2) are 1-dimensional which would mean $2\times 2=1+1+1+1$, I guess. But Michael Seifert's and others' answers seem to prove my assertion that $2\times 2=1+1+2$. I would appreciate it if you take the pain the clarify this point. $\endgroup$ Jun 6 at 14:30
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    $\begingroup$ well if you are working over the complex field then you can use $e^{i\theta}=\cos\theta+i\sin\theta$ so the matrix $\left( \begin{array}{cc} \cos (\theta ) & -\sin (\theta ) \\ \sin (\theta ) & \cos (\theta ) \\ \end{array} \right)$ can be reduced to its diagonal form using a similarity transformation with complex entries. If you are limited to the real the diagonalization cannot be done, so the field matters. Note that O(2) has two-dimensional irreps (with basis vectors $\vert m\rangle$ and $\vert -m\rangle$ which go one into the other under the reflection in O(2)... $\endgroup$ Jun 6 at 14:51
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    $\begingroup$ but this 2-dim irrep "collapses" to a 1-d irrep when $m=0$ so O(2) has both 2-d irreps (based on parity) and a 1-dim irrep. $\endgroup$ Jun 6 at 14:52

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You can parameterize a general $2\times 2$ matrix of real numbers in terms of 4 real numbers $t$, $a$, $s_1$, and $s_2$ as \begin{equation} A = \left( \begin{array} _ \frac{t}{2} + s_1 & s_2 + a \\ s_2 - a & \frac{t}{2} - s_1 \end{array} \right) \end{equation} Note that

  • The trace of $A$ is given by $$ {\rm tr} A = t $$
  • The antisymmetric part of $A$ is given by $$ \frac{1}{2}\left(A-A^T\right) = \left( \begin{array} _ 0 & a \\ -a & 0 \end{array} \right) $$
  • The traceless symmetric part of $A$ is given by $$ \frac{1}{2}\left(A+A^T\right) - \frac{1}{2}{\rm tr}A \mathbb{1} = \left( \begin{array} _ s_1 & s_2 \\ s_2 & -s_1 \end{array} \right) $$ where $\mathbb{1}$ is the ($2\times 2$) identity matrix.

You can check (or have WolframAlpha check) that under an arbitrary rotation \begin{equation} A \rightarrow A' = R(\theta) A R(\theta)^T \end{equation} where \begin{equation} R(\theta) = \left( \begin{array} _ \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array} \right) \end{equation} that \begin{eqnarray} t &\rightarrow& t' = t \\ a &\rightarrow& a' = a \\ s_1 &\rightarrow& s_1' = s_1 \cos 2\theta - s_2 \sin 2\theta \\ s_2 &\rightarrow& s_2' = s_1 \sin 2\theta + s_2 \cos 2\theta \end{eqnarray} Since $t$ and $a$ do not change under a rotation they are scalars (or "1" representations). Since $s_1$ and $s_2$ get "mixed up" under rotations, they combine to form a different representation (in this case, the "2" representation).

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The "2" is the space of all trace-free symmetric rank-2 tensors. So an arbitrary tensor product of two vectors can be decomposed as $$ A_i B_j = \frac{1}{2} \delta_{ij} A_k B_k + \frac{1}{2} \left( A_i B_j - B_i A_j \right) + \frac{1}{2} \left( A_i B_j + B_i A_j - \delta_{ij} A_k B_k \right) $$ and the three "pieces" above do not mix with each other when an $SO(2)$ rotation is applied to the vectors $A_i$ and $B_i$.

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  • $\begingroup$ The third piece should have 2 components. Am I right? If so, what are they explicitly? $\endgroup$ Jun 5 at 22:33
  • $\begingroup$ @Solidification: It has two independent components (and hence the irreducible subspace is 2-dimensional); but it has four components overall because it's a rank-2 tensor. You can shown that the components of the tensor will always be of the form$$\begin{bmatrix} p & q \\ q & -p\end{bmatrix}$$ for some $p$ and $q$. $\endgroup$ Jun 6 at 2:28
  • $\begingroup$ @Solidification: Further: if $\vec{A}$ and $\vec{B}$ have components $A_1$, $A_2$, $B_1$, and $B_2$ respectively, then the values of $p$ and $q$ above work out to be $p = A_1 B_1 - A_2 B_2$ and $q = A_1 B_2 + A_2 B_1$. $\endgroup$ Jun 6 at 13:13
  • $\begingroup$ Thanks! So under a general SO(2) rotation, $p=A_1B_1-A_2B_2$ and $1=A_1B_2+A_2B_1$ will mix with each other. Is that right? $\endgroup$ Jun 6 at 14:23
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    $\begingroup$ @Solidification: Yup, that's right. The exact mixing law is given in @Andrew's excellent answer. $\endgroup$ Jun 6 at 15:33
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Any $2\times2$ matrix (4 dof) can be broken up into its trace part (1 dof), an antisymmetric part (1 dof) and a symmetric traceless part (2 dof). Hence, $2\times 2=4=1+1+2$.

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