Rank-2 isotropic tensors: one or two? Isotropic tensor: components are unchanged after rotation.
MathWorld says there is only one rank-2 isotropic tensor, $\delta_{ij}$.
But with
$$\epsilon_{ij}=\left(\begin{matrix}0&1\\-1&0\end{matrix}\right)$$
a rotation
$$R(a)=\left(\begin{matrix}\cos a&-\sin a\\ \sin a&\cos a\end{matrix}\right)$$
gives
$$\epsilon_{ij}\rightarrow\epsilon_{ij}'=r_{ia}r_{jb}\epsilon_{ab}=\epsilon_{ij}$$
So it seems to me that $\epsilon_{ij}$ is also a rank-2 isotropic tensor, in addition to $\delta_{ij}$.
What am I getting wrong?
 A: You are right: in fact, any 2D rotation, or really any linear combination of the (linearly) independent (indeed, orthogonal wrt the Frobenius inner product) matrices $(\begin{smallmatrix}1&0\\0&1\end{smallmatrix})$ and $(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix})$ is invariant under orthogonal coordinates changes which preserve orientation - equivalently, are invariant under conjugation by rotation matrices. This basically illustrates the fact the complex numbers $\mathbb{C}$ are commutative, since this is a matrix representation of it.
What I found when looking at multiple sources, including the Classical Mechanics book MathWorld links, but which is missing from the MathWorld entry and even not explicitly stated in other physics sources is that we're considering the situation in three dimensions, not two. (As illustrated by the fact they mention the permutation symbol for rank 3, which only makes sense in 3D.)
So it appears they are talking about the dimension of the $\mathrm{SO}(3)$-invariant subspace of $(\mathbb{R}^3)^{\otimes n}$. (Although the OEIS page that MathWorld mentions talks about the $\mathfrak{sl}_2\mathbb{C}$-invariant subspace of $(\mathbb{C}^2)^{\otimes n}$, which I guess must be equivalent because of $\mathrm{SU}(2)\approx\mathrm{SO}(3)$ and spinors somehow?)
