Is there an invariant similar to the characteristic polynomial for (0,2) and (2,0) tensors?

Question

The characteristic polynomial of a matrix - a (1,1) tensor - is its invariant (independent on basis transformation). Is there a similar invariant for (0,2) and (2,0) tensors? The characteristic polynomial on its own doesn't work, for example its last coefficient is the determinant of a matrix, and determinant isn't preserved under basis transformations for (0,2) and (2,0) tensors.

Answer 1

No. In general, let $G$ be a group (say a Lie group) and let $V$ be a representation of it. One thing we might mean by "invariants" of $v \in V$ are polynomial invariants: that is, elements of the algebra $S(V^{\ast})^G$ of $G$-invariant polynomial functions on $V$.

This algebra sometimes consists of only the constant functions. In particular, this is true for for $G = GL(W)$ and $V = W \otimes W$ or $W^{\ast} \otimes W^{\ast}$ (so tensors of type $(2, 0)$ and $(0, 2)$), for the simple reason that already the subgroup of $G$ consisting of diagonal matrices with respect to some basis never acts trivially on any subspace of any tensor power $V^{\otimes n}$ of $V$, hence never acts trivially on any subspace of any symmetric power $S^n(V)$ of $V$.

By contrast, if $G = GL(W)$ and $V = W \otimes W^{\ast}$ (so tensors of type $(1, 1)$) then the algebra of $G$-invariant polynomials is the free polynomial algebra on the coefficients of the characteristic polynomial.