# 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?

• SO(2) is Abelian. All irreps are 1-dimensional. Do you mean O(2)? Jun 5 at 21:34
• or else are you working over the reals? Jun 5 at 22:44
• @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. Jun 6 at 14:30
• 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)... Jun 6 at 14:51
• 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. Jun 6 at 14:52

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 $$$$A = \left( \begin{array} _ \frac{t}{2} + s_1 & s_2 + a \\ s_2 - a & \frac{t}{2} - s_1 \end{array} \right)$$$$ 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 $$$$A \rightarrow A' = R(\theta) A R(\theta)^T$$$$ where $$$$R(\theta) = \left( \begin{array} _ \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array} \right)$$$$ that $$\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).

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 $$B_i$$.

• The third piece should have 2 components. Am I right? If so, what are they explicitly? Jun 5 at 22:33
• @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 $q$. Jun 6 at 2:28
• @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 $q = A_1 B_2 + A_2 B_1$. Jun 6 at 13:13
• 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? Jun 6 at 14:23
• @Solidification: Yup, that's right. The exact mixing law is given in @Andrew's excellent answer. Jun 6 at 15:33

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, $$2\times 2=4=1+1+2$$.