Help needed in understanding $2\otimes 2=1\oplus 1\oplus 2$ of $SO(2)$ 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?
 A: 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).
A: 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 $B_i$.
A: 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, $2\times 2=4=1+1+2$.
