Confusion about group action? I want to show that $$(T\cdot L)\epsilon=T(L(T^{-1}\epsilon T))T^{-1}$$(Here $T\cdot L$ means group action.)
where $L$ is a symmetric operator mapping symmetric $3\times 3$ matrices into symmetric $3\times 3$ matrices, and $\epsilon \in Sym(\Bbb R^3)$.
I tried to use the fact that $T\cdot A=TAT^{-1}$, but I ended up getting $TLT^{-1}\epsilon$, which is not right. 
Where am I wrong? Could someone help look at this? Thanks! 
 A: This seems like an unnecessarily complicated group action to construct, but okay...
Previously we observed that if a group $G$ acts on the domain and codomain of a function space then it acts on functions via conjugation, i.e. $(g\cdot f)(\sigma)=gf(g^{-1}\sigma)$.
Here I assume the orthogonal group ${\rm O}(3)$ is acting on the space $V$ of symmetric matrices. Since we know that matrices are functions, and ${\rm O}(3)$ acts on the domain/codomain of these functions, if we are using our function action we should get $g\cdot A:=gAg^{-1}$ where $g\in{\rm O}(3)$ and $A$ is a $3\times 3$ symmetric matrix in $V$. (Now concatenation implies matrix multiplication.)
Say $L:V\to V$ is a map, and we currently have ${\rm O}(3)$ acting on $V$, so the induced action is
$$(g\bullet L)(A):=g\cdot L(g^{-1}\cdot A)=gL(g^{-1}Ag)g^{-1}.$$
We're using concatenation for matrix multiplication, $\cdot$ for ${\rm O}(3)$ acting on $V$, and $\bullet$ for ${\rm O}(3)$ acting on the new function space ${\rm End}(V)$, whilst $V$ is itself a space of functions! Be sure to distinguish the different actions with different notations so you don't get lost as much in the future.
Note: I'm having ${\rm O}(3)$ act on $V$ (the space of symmetric $3\times 3$ matrices) since it's the largest group of invertible $3\times3$ matrices that preserves the property of being symmetric under conjugation (or at least I think it's the largest group of matrices that does that).
