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


**Defintion:**


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:


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


**Theorem:**


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](http://link.springer.com/article/10.1007%2FBF00041995#page-1).


**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)


**Update:**


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](https://math.stackexchange.com/a/1603096/104576)). 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?** 



Heuristics for counting degrees of freedom
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. 


