# Is every linear operator which is $SO(n)$-invariant necessarily isotropic?

Let $M_n$ be the vector space of $n \times n$ real matrices.

We say a linear operator $\alpha:M_n \to M_n$ is hemitropic* if:

$(*) \, \, \alpha(S^TXS)=S^T\alpha(X)S \, , \, \forall S \in SO(n)$

and isotropic, if the above formula holds $\forall S \in O(n)$

My question: Is every hemitropic operator necessarily isotropic? Does the answer change if we assume $\alpha$ is injective?

In odd dimensions, the answer is yes, since for any $Q \in O(n) \setminus SO(n)$, $\det(-Q)=1$ so $-Q \in SO(n)$ and by (*):

$\alpha(Q^TXQ)=\alpha\big((-Q)^TX(-Q)\big)=(-Q)^T\alpha(X)(-Q)=Q^T\alpha(X)Q$

I suspect the answer in even dimensions is negative, but so far I didn't find a way to construct a hemitropic non-isotropic operator. (Though I guess this can be done even in dimension $2$).

Update:

For dimension $2$ I have constructed (see answer below) a hemitropic non-isotropic operator. However, my construction relied on the fact that all $2$-dimensional rotation commute. This is not the case for higher dimensions. (See here).

This leaves open the question for even dimensions above two.

The terminology "hemitropic" comes from Elasticity theory.


Well, it turns out the answer is indeed negative.

Even for dimension $2$, there are hemiropic, non-isotropic operators. (In particular injective ones).

Here is an example:

$\alpha:M_2 \to M_2, \al(A)=R_{\theta} \cdot A$ where $R_{\theta}=\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix}$ is a rotation matrix.

$\al$ is linear and invertible.

We show $\al$ is hemitropic ($SO(n)$-invariant):

This is essentially the fact that any two $2$-dimensonal rotations commute.

$SO(2)=\{R_{\be}| \be \in [0,2\pi) \} \, , \, R_{\be}^T=R_{\be}^{-1}=R_{-\be}$

$\al(R_{\be}^TAR_{\be})=\al(R_{-\be}AR_{\be})=R_{\the}R_{-\be}AR_{\be}\stackrel{(*)}{=} R_{-\be}R_{\the}AR_{\be}=R_{\be}^T\al(A)R_{\be}$

as required. (Where (*) follows from the fact that any two rotations commute).

We show $\al$ is not isotropic ($O(n)$-invariant):

This is essentially because two $2$-dimensonal rotations do not commute with reflections.

Denote $R_{\theta}=\begin{pmatrix} q & -s \\ s & q \end{pmatrix} \, , q=\cos(\the),s=\sin(\the)$

Choose $A=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, O=O^T=\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \in O(n) \setminus SO(n)$

$\al(A)=\begin{pmatrix} q & -s \\ s & q \end{pmatrix} \cdot \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} -s & 0 \\ q & 0 \end{pmatrix}$

In general: $O \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot O = \begin{pmatrix} a & -b \\ -c & d \end{pmatrix}$, so:

$(1): \, \, O \al(A) O = O \begin{pmatrix} -s & 0 \\ q & 0 \end{pmatrix} O=\begin{pmatrix} -s & 0 \\ -q & 0 \end{pmatrix}$, but

$(2): \, \, \al(OAO)=\al(\begin{pmatrix} 0 & 0 \\ -1 & 0 \end{pmatrix})=\begin{pmatrix} q & -s \\ s & q \end{pmatrix} \cdot \begin{pmatrix} 0 & 0 \\ -1 & 0 \end{pmatrix} =\begin{pmatrix} s & 0 \\ -q & 0 \end{pmatrix}$

So, $(1),(2) \Rightarrow O \al(A) O \neq \al(OAO)$ so $\al$ is not isotropic, as required.