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Let $1<k<n$ be a fixed integer. I am trying to understand "what can be said" about the $k$-degree minors of a linear map $T:V \to V$, when $\dim V=n$.

Unlike the determinant, the $k$-minors are not invariant under conjugation. So, we cannot associate an ordered sequence of minors to $T$, independently of the basis we choose for representing it. This raises the following question:

Consider the action of $\text{GL}(n)$ on $\text{End}(\bigwedge^k \mathbb{R}^n)$:

$$ (M , A) \to \bigwedge^k M \circ A \circ \bigwedge^k M^{-1}.$$

After choosing a basis for $\mathbb{R}^n$, we can identify $\text{End}(\bigwedge^k \mathbb{R}^n) $ with $\mathbb{R}^{\binom{n}{k}^2}$.

Can we classify all the polynomials $P:\text{End}(\bigwedge^k \mathbb{R}^n) \to \mathbb{R}$ which are invariant under the "conjugation-action" by $\text{GL}(n)$ described above?

Of course, every polynomial on $\text{End}(\bigwedge^k \mathbb{R}^n)$ which is invariant under the conjugation action of $\text{GL}(\bigwedge^k \mathbb{R}^n)$ would be invariant under conjugation by the smaller subgroup which "comes from the copy of $\text{GL}(n)$ below".

These polynomials are classified.

A starting point would be to know whether there are any other invariant polynomials, besides the $\text{GL}(\bigwedge^k \mathbb{R}^n)$-invariant.

Asaf Shachar
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    Cross-posted at https://mathoverflow.net/questions/311151/which-polynomials-in-the-minors-of-a-matrix-are-invariant-under-conjugation. When you cross-post, you should make sure that each post contains a link to the other, to minimize the chances of someone spending a lot of time trying to solve your questions when it's already answered on the other site. – Nate Eldredge Sep 22 '18 at 14:39
  • Thanks, you are right. I shall try to remember that for next time. – Asaf Shachar Sep 22 '18 at 14:40

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