This is a cross-post.

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. Thus 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 $\text{End}(\bigwedge^k \mathbb{R}^n)$ given by

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

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 $\text{GL}(\bigwedge^k \mathbb{R}^n)$ -invariant polynomials are classified.

For even $n$ and $k=\frac{n}{2}$, the determinant of an element $A \in \text{GL}(n)$ can be expressed as a quadratic polynomial in its $\frac{n}{2}$-minors. This polynomial is clearly $\text{GL}(n)$-invariant, but probably not $\text{GL}(\bigwedge^k \mathbb{R}^n)$-invariant. (Is there a nice way to see that it's not $\text{GL}(\bigwedge^k \mathbb{R}^n)$-invariant?).

I think this can be generalized: If $d | n$, the determinant of $A \in \text{GL}(n)$ can be expressed as a polynomial in its $\frac{n}{d}$-minors. This polynomial is clearly $\text{GL}(n)$-invariant, but probably not $\text{GL}(\bigwedge^{\frac{n}{d}} \mathbb{R}^n)$-invariant.

Sub-question: Except for the determinant exception mentioned above, are there any other $\text{GL}(n)$-invariant polynomials which are not $\text{GL}(\bigwedge^k \mathbb{R}^n)$-invariant?

$\newcommand{\Cof}{\operatorname{cof}}$

This is a cross-post.

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. Thus 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 $\text{End}(\bigwedge^k \mathbb{R}^n)$ given by

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

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 $\text{GL}(\bigwedge^k \mathbb{R}^n)$ -invariant polynomials are classified.

For even $n$ and $k=\frac{n}{2}$, the determinant of an element $A \in \text{GL}(n)$ can be expressed as a quadratic polynomial in its $\frac{n}{2}$-minors. This polynomial is clearly $\text{GL}(n)$-invariant, but probably not $\text{GL}(\bigwedge^k \mathbb{R}^n)$-invariant. (Is there a nice way to see that it's not $\text{GL}(\bigwedge^k \mathbb{R}^n)$-invariant?).

I think this can be generalized: If $d | n$, the determinant of $A \in \text{GL}(n)$ can be expressed as a polynomial in its $\frac{n}{d}$-minors. This polynomial is clearly $\text{GL}(n)$-invariant, but probably not $\text{GL}(\bigwedge^{\frac{n}{d}} \mathbb{R}^n)$-invariant.

Edit:

Actually, there are more $\text{GL}(n)$-invariant polynomials; these are related to the reconstruction of a matrix from its minors. Here is an explicit special case:

Let $k=n-1$. Write $B=\Cof A$. Then $\Cof B=(\det A)^{n-2}A$ is a $\text{GL}(n)$-invariant polynomial of degree $n-1$. (It also holds that $\det(B)=(\det(A))^{n-1}$ and this allows (when $n$ is even) to recover $A$ in terms of $B$).

This can be generalized to whenever $k$ is relatively prime to $n$. (See, this comment, and this answer for details).

This raises the following

Sub-question: Except for the two examples mentioned above, are there any other $\text{GL}(n)$-invariant polynomials which are not $\text{GL}(\bigwedge^k \mathbb{R}^n)$-invariant?

Interestingly, both examples were related to the determinant. Perhaps this is not a coincidence-maybe all $\text{GL}(n)$-invariant polynomials of the $k$-minors are "factored" through the determinant in some way.

This is a cross-post.

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. Thus 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 $\text{End}(\bigwedge^k \mathbb{R}^n)$ given by

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

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 $\text{GL}(\bigwedge^k \mathbb{R}^n)$ -invariant polynomials are classified.

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

This is a cross-post.

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. Thus 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 $\text{End}(\bigwedge^k \mathbb{R}^n)$ given by

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

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 $\text{GL}(\bigwedge^k \mathbb{R}^n)$ -invariant polynomials are classified.

For even $n$ and $k=\frac{n}{2}$, the determinant of an element $A \in \text{GL}(n)$ can be expressed as a quadratic polynomial in its $\frac{n}{2}$-minors. This polynomial is clearly $\text{GL}(n)$-invariant, but probably not $\text{GL}(\bigwedge^k \mathbb{R}^n)$-invariant. (Is there a nice way to see that it's not $\text{GL}(\bigwedge^k \mathbb{R}^n)$-invariant?).

I think this can be generalized: If $d | n$, the determinant of $A \in \text{GL}(n)$ can be expressed as a polynomial in its $\frac{n}{d}$-minors. This polynomial is clearly $\text{GL}(n)$-invariant, but probably not $\text{GL}(\bigwedge^{\frac{n}{d}} \mathbb{R}^n)$-invariant.

Sub-question: Except for the determinant exception mentioned above, are there any other $\text{GL}(n)$-invariant polynomials which are not $\text{GL}(\bigwedge^k \mathbb{R}^n)$-invariant?