I have recently learned about the representation theorem for isotropic, linear operators, which says the following:


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 isotropic if:

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


Let $\alpha$ be a linear isotropic operator. Then there exists 3 real scalars $a,b,c$ such that:

$\alpha(X)=a \text{tr}(X)I + bX + cX^T$

A proof is given for example here.

My question:

Is there an a priori way, to see in advance that the number of degrees of freedom left are exactly 3? Or at least that this number is independent of $n$?

Trivial Note: $\dim (\operatorname{Hom}(M_n,M_n))=n^4, \dim O(n) = \frac{n(n-1)}{2}$

What I would like to see is some simple heuristic, which does not involve repeating the proof. (which is not very enlightening I am afraid)


This might be more challenging than I originally thought; I hoped for some heuristc concerning only the dimensions of $\operatorname{Hom}(M_n,M_n),O(n)$. However, any argument that will consider only these numbers will not be delicate enough to distinguish between $O(n)$-invariance and $SO(n)$-invariance.

$\dim(SO(n))=\dim(O(n))$, but if we replace the requirement for $O(n)$-invariance in $(*)$ with $SO(n)$-invariance, the theorem no longer holds (in even dimensions).

There are $SO(n)$-invariant operators which are not in the above form. For example ($n=2$), take $\alpha(A)=R_{\theta} \cdot A$ where $R_{\theta}$ is a rotation matrix. (See full details here). This does not work for even $n>2$ since the $SO(2)$ invariance is based on the fact that all $2$ dimensional rotations commute which is false for higher dimensions.

So for the case $\dim=2$ there are at least $5$ degrees of freedom if we consider $SO(n)$-invariance:

$\alpha(X)=a_1 \text{tr}(X)I + a_2X + a_3X^T+a_4(R_{a_5}\cdot X)$

Is there a representation theorem for $SO(n)$-invariant operators? How many degrees of freedom do we have in this case?


Let $V$ be a finite-dimensional real inner product space. You want to know the dimension of $\text{End}_G(V \otimes V^{\ast})$ where $G = O(V)$. Using the inner product we have an isomorphism $V \cong V^{\ast}$, so

$$V \otimes V^{\ast} \cong V \otimes V \cong S^2(V) + \wedge^2(V)$$

(where $+$ denotes direct sum). The inner product is itself a $G$-invariant element of $S^2(V)$, so as a $G$-representation $S^2(V)$ further splits up as the direct sum of the trivial rep and another rep. If this other rep is always irreducible and nonisomorphic to $\wedge^2(V)$ then the dimension of $\text{End}_G$ is always $3$ as desired, and this is probably not hard to prove directly.

Heuristics based on dimension counts probably won't say much; already $SO(3)$ has irreducible representations of arbitrarily high dimension.


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