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?


  • $R_{ij}=\left(\begin{matrix}\cos a&-\sin a\\ \sin a&\cos a\end{matrix}\right)$
  • 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.
  • 3
    $\begingroup$ I think they mean in 3 dimensions. $\epsilon_{ij}$ is not defined in three dimnsions. $\endgroup$
    – mike stone
    Nov 25 '20 at 14:32
  • $\begingroup$ They might be additionally imposing invariance under reflections? Under a general orthongonal transform $T$, $\epsilon_{ij}\mapsto \text{det}(T) \epsilon_{ij}$. $\endgroup$
    – jacob1729
    Nov 25 '20 at 14:57
  • $\begingroup$ @jacob1729 That was my first thought too, but it seems to be mentioned in many places that $\delta_{ij}$ is the only rank-2 isotropic tensor, and the definition of isotropic tensors seem only to mention rotations, for example, see here or here. Besides, the same site later mentions $\epsilon_{ijk}$ to be a rank-3 isotropic tensor, so they can't be using invariance under reflection to define it (could they?). $\endgroup$
    – Philip
    Nov 25 '20 at 15:20
  • 1
    $\begingroup$ Crossposted from math.stackexchange.com/q/3919304/11127 $\endgroup$
    – Qmechanic
    Nov 25 '20 at 16:20
  • $\begingroup$ @Qmechanic, should I remove the original at math.stackexchange, where it got no attention? Not sure on the etiquette... $\endgroup$
    – Travis Lee
    Nov 25 '20 at 16:29

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:

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

  2. 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.

  • 1
    $\begingroup$ Very nice insights. Thanks for the answer. $\endgroup$
    – Travis Lee
    Nov 26 '20 at 8:01
  • $\begingroup$ Thank you! Though a lot of them are from the notes that I linked in the answer. :) $\endgroup$
    – Philip
    Nov 26 '20 at 8:02

This is only true for 2D space not in general

3D space:

$$S=\left[ \begin {array}{ccc} \cos \left( a \right) &-\sin \left( a \right) &0\\ \sin \left( a \right) &\cos \left( a \right) &0\\ 0&0&1\end {array} \right] $$


$$\epsilon=\left[ \begin {array}{ccc} 0&-1&1\\ 1&0&-1 \\ -1&1&0\end {array} \right] $$


$$\epsilon'=S\,\epsilon\,S^T=\left[ \begin {array}{ccc} 0&-1&\cos \left( a \right) +\sin \left( a \right) \\ 1&0&\sin \left( a \right) -\cos \left( a \right) \\ -\cos \left( a \right) -\sin \left( a \right) &-\sin \left( a \right) +\cos \left( a \right) &0\end {array} \right] $$

thus only if $~a=\pi/2$ you get $\epsilon'=\epsilon$


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