Let's call an algebraic invariant (under the action of $\text{GL}_n(\mathbb{C})$) a tensor of order $N$ to an algebraic combination of the components of that tensor. In other words, a homogeneous polynomial in the components of the tensor.
- For $N = 2$ algebraic invariants are well known: given a second-order tensor, defined over vector spaces of dimension $n$, it is very simple to see that there are $n$ independent invariants that are coefficients of the Cayley–Hamilton polynomial (linear or trace invariant, quadratic invariant, and n-degree invariant or determinant). Any other algebraic invariant can be computed as a polynomial in the above invariants.
- It is also simple to see that for a tensor of order $2N$ there are at least $n^N$ algebraic invariants, since the equation: $$\sum_{j_1,\dots,j_n = 1}^n A_{i_1 \dots i_n j_1 \dots j_n}\ v_{ j_1 \dots j_n } = \lambda\ v_{ i_1 \dots i_n}$$ with $ v_{ j_1 \dots j_n }\neq 0$. It is still a problem of eigenvalues over a space of dimension $n^N$ and therefore that gives rise to a Cayley-Hamilton polynomial of degree $n^N$, whose coefficients must be algebraic invariants. However, the above is a very non-general procedure. In particular, it does not seem to be applicable to third-order tensors, or tensors of order $2N+1$. What about the algebraic invariants of a third-order tensor, for example?