UPD Equivallent formulation --- how do you find all the independent isometric invariants of a tensor?
In what follows $V$ is a real inner product space.
I want to understand how does one find all (scalar) invariants of tensors under isometries. Wikipedia has a relevant article, but it confines itself only with tensors of type $V^* \otimes V$ or equivalent.
What I mean by an invariant of a tensor under the isometry. Invariant of a $V \otimes V^*$ tensor is a function $I : V \otimes V^* \to \mathbb R$ if
$$I\left( Q(A) \right) = I(A)$$
for any isometry $Q$ (transformation) properly lifted to act on tensors.
I've asked a question about the invariants of tensor of type $V$, but couldn't get the intuition to extend the reasoning to higher rank tensors
So far as I understand vectors have invariants of the form
$$f(\langle v, v \rangle)$$
whereas tensors in $V^* \otimes V$ have
$$f(\operatorname{Tr} A, \operatorname{Tr} A^2, \operatorname{Tr} A^3)$$
$f$ is meant to be an arbitrary function to $\mathbb R$ of appropriate type.
I don't even see why there are three independent invariants for rank two tensors.
One might be be puzzled of the importance of the question and why I'm asking it. The above mentioned wikipedia article provides a good example. Potential energy (scalar) in an elastic material is a function of the strain tensor. The coupling between them should be natural in the sense that it should not involve arbitrary tensors or vectors (because there are simply none), one have to construct a scalar just from the strain tensor. Theories in continuum mechanics abounds in dozens of various tensor fields of different ranks and one need to seek the forms of natural couplings between them.