Isn't $\epsilon_{ij}$ an isotropic, rank-2 tensor? Definition of isotropic tensor: components are unchanged after rotation: $T_{ij}\rightarrow T_{ij}' \equiv R_{ia}R_{jb}T_{ij} = T_{ij}$
MathWorld says there is only one rank-2 isotropic tensor, $\delta_{ij}$.
But with
$$\epsilon_{ij}=\left(\begin{matrix}0&1\\-1&0\end{matrix}\right)$$
there is no change either:
$$\epsilon_{ij}\rightarrow\epsilon_{ij}'=R_{ia}R_{jb}\epsilon_{ab}=\epsilon_{ij}$$
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?
Notes:

*

*$R_{ij}=\left(\begin{matrix}\cos a&-\sin a\\ \sin a&\cos a\end{matrix}\right)$

*I asked at math.stackexchange, but got no answer. Maybe this is more Phys Math Met as in Boas, which I was reading when this question came up.

 A: I think @mikestone's comment is correct. The MathWorld site is talking about tensors in 3 dimensions, not 2D. In 3 dimensions (actually, in all dimensions $\geq3$), it can be shown that any isotropic rank-2 tensor is proportional to the identity ($\delta_{ij}$), see Richard Fitzpatrick's notes here, for example.
In two dimensions, there are two rank-2 isotropic tensors, $\delta_{ij}$, and what you have called $\epsilon_{ij}$. Here's a quick way to list all the isotropic tensors in 2D that I have drawn heavily from the fantastic analyses here and here. There are two equivalent ways to define an isotropic tensor: one is the way you have defined it, saying that $A$ is an isotropic tensor iff $$A = R\cdot A\cdot R^T,$$
but an equivalent way is to say that $A$ is isotropic iff it commutes with the generators of rotations, $L_i$, for example $[A,L_z] = 0$. Now, let's look at this condition in 2D. It turns out -- if you do the calculations -- that $$L_z = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix},$$ precisely (the negative of) what you called $\epsilon_{ij}$!
Now consider an arbitrary 2D tensor $A$, and demand that it satisfy the commutation relation given above:
$$\Bigg[ \begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{pmatrix}, \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\Bigg] = 0,$$
and you should be able to see that this just means such an arbitrary isotropic tensor in 2D should look like: $$A = \begin{pmatrix}a & -b \\ b & a\end{pmatrix},$$ and you should be able to see two things:

*

*An arbitrary isotropic tensor in 2D is itself a rotation, since $$A = \frac{1}{\sqrt{a^2+b^2}}\begin{pmatrix}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{pmatrix}, \quad \quad \alpha=\arctan{\left(\frac{b}{a}\right)}$$


*An arbitrary isotropic tensor in 2D can be written as: $$A_{ij} = a \delta_{ij} -b \epsilon_{ij}.$$
Thus, there are precisely two independent isotropic rank-2 tensors in 2D, $\delta_{ij}$ and $\epsilon_{ij}$.

Note: I've not spoken too much about generators and how to derive $L_z$ in 2D since the answer would get too long, but I'd be happy to explain it if necessary.
A: This is only true for 2D space not in general
3D space:
$$S=\left[ \begin {array}{ccc} \cos \left( a \right) &-\sin \left( a
 \right) &0\\ \sin \left( a \right) &\cos \left( a
 \right) &0\\ 0&0&1\end {array} \right] 
$$
and
$$\epsilon=\left[ \begin {array}{ccc} 0&-1&1\\ 1&0&-1
\\ -1&1&0\end {array} \right] 
$$
$\Rightarrow$
$$\epsilon'=S\,\epsilon\,S^T=\left[ \begin {array}{ccc} 0&-1&\cos \left( a \right) +\sin \left( a
 \right) \\ 1&0&\sin \left( a \right) -\cos \left( a
 \right) \\ -\cos \left( a \right) -\sin \left( a
 \right) &-\sin \left( a \right) +\cos \left( a \right) &0\end {array}
 \right] 
$$
thus only if $~a=\pi/2$ you get $\epsilon'=\epsilon$
