Why am I getting this tensor rotation wrong?

Question

Let $$\rho_\theta \equiv \rho(R_\theta) = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\tag{1}$$ be a representation of $SU(2)$, and consider the tensor product representation $\rho_\theta \otimes \rho_\theta$. We want to write it in the basis $$E_1 = e_{11} + e_{22}, \quad E_2 = e_{11} - e_{22}, \quad E_3 = e_{12} + e_{21}, \quad E_4 = e_{12} - e_{21},\tag{2}$$ where $e_{ij} \equiv \mathbf e_i \otimes \mathbf e_j$, and $\{\mathbf e_1, \mathbf e_2\}$ is a standard cartesian basis. The book I'm reading (Blennow - Mathematical Methods for Physics and Engineering, Example 4.35) states that $$\rho_\theta \otimes \rho_\theta E_1 = E_1,\tag{3}$$ $$\rho_\theta \otimes \rho_\theta E_2 = \cos(2\theta)E_2 - \sin(2\theta)E_3,\tag{4}$$ $$\rho_\theta \otimes \rho_\theta E_3 = \sin(2\theta)E_2 + \cos(2\theta)E_3,\tag{5}$$ $$\rho_\theta \otimes \rho_\theta E_4 = E_4.\tag{6}$$ However, I get a different result for (4) and (5), namely $$\rho_\theta \otimes \rho_\theta E_2 = \cos(2\theta)E_2 + \sin(2\theta)E_3,\tag{7}$$ $$\rho_\theta \otimes \rho_\theta E_3 = -\sin(2\theta)E_2 + \cos(2\theta)E_3.\tag{8}$$ It looks like a mixup with active and passive rotations, but I don't think that should be applicable here, seeing as we are given the matrix (1) explicitly. Let me show you how I computed (7), and maybe you can spot a mistake? Here goes: $$\begin{split} \rho_\theta \otimes \rho_\theta E_2 &= \rho_\theta \otimes \rho_\theta(\mathbf e_1 \otimes \mathbf e_1 - \mathbf e_2 \otimes \mathbf e_2)\\ &= (\rho_\theta \mathbf e_1) \otimes (\rho_\theta \mathbf e_1) - (\rho_\theta \mathbf e_2) \otimes (\rho_\theta \mathbf e_2)\\ &= (\cos(\theta) \mathbf e_1 + \sin(\theta) \mathbf e_2) \otimes (\cos(\theta) \mathbf e_1 + \sin(\theta) \mathbf e_2)\\ &\quad - (-\sin(\theta) \mathbf e_1 + \cos(\theta) \mathbf e_2) \otimes (-\sin(\theta) \mathbf e_1 + \cos(\theta) \mathbf e_2)\\ &= \cos^2(\theta) e_{11} + \cos(\theta)\sin(\theta)(e_{12} + e_{21}) + \sin^2(\theta)e_{22}\\ &\quad - \sin^2(\theta) e_{11} + \sin(\theta)\cos(\theta)(e_{12} + e_{21}) - \cos^2(\theta)e_{22}\\ &= (\cos^2(\theta) - \sin^2(\theta)) e_{11} + 2\cos(\theta)\sin(\theta)(e_{12} + e_{21}) - (\cos^2(\theta) - \sin^2(\theta)) e_{22}\\ &= \cos(2\theta) e_{11} + \sin(2\theta)(e_{12} + e_{21}) - \cos(2\theta) e_{22}\\ &= \cos(2\theta) E_2 + \sin(2\theta) E_3. \end{split}\tag{9}$$ Could it be that the matrix representation in (1) is meant to indicate how contravariant components transform, and the basis vectors, being covariant, transform inversely? If so, what is a practical way of computing the matrix elements of $\rho_\theta \otimes \rho_\theta$ in the given basis, seeing as the inverse might not always be as obvious as it is here?

Answer 1

I just heard from the author of the textbook who confirms that it is indeed a typo. So equations (7) and (8) in my question are the correct ones.