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

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.

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    $\begingroup$ For the "starting point": it is possible to express the determinant of an element of $GL_4$ quadratically in terms of its $2$-minors. This isn't $GL(\wedge^2 \mathbb{R}^4)$-invariant, but obviously is $GL_4$-invariant. $\endgroup$
    – user44191
    Sep 22, 2018 at 6:15
  • $\begingroup$ I shouldn't say that it is $GL_4$-invariant; instead, it represents a class of $GL_4$ invariants. Of that class, at least one is necessarily $GL_4$-invariant, as $GL_4$ is reductive. $\endgroup$
    – user44191
    Sep 22, 2018 at 22:33
  • $\begingroup$ Thanks. I have some more questions now about what you wrote, I will appreciate your help: (1) I think that I see now what is wrong with the following explanation: We have a quadratic polynomial in $\binom{4}{2}^2$ variables, which have the following property: When we apply this polynomial to the $2$-minors of $4 \times 4$ matrix, it gives the determinant of that matrix. So, in some sense it is $GL_4$-invariant, when you only look at orbits in $\text{End}(\bigwedge^2 \mathbb{R}^4)$ which come from endomorphisms below. (only on orbits of $2$-minors...). $\endgroup$ Sep 23, 2018 at 7:47
  • $\begingroup$ So, it is not necessarily true that this polynomial is $GL_4$-invariant "on the whole space" (when we do not constraint the variables to be minors of anything). Is this the problem? (Do you see an argument showing that it isn't strictly $GL_4$-invariant?). (2) I am not sure what do you mean by a "class of $GL_4$-invariants. Can you elaborate on that? (My knowledge in representation theory is lacking, unfortunately). (3) How does $GL_4$ being reductive implies there is an element in the class which is really $GL_4$-invariant? $\endgroup$ Sep 23, 2018 at 7:47
  • $\begingroup$ (4) Finally, I think this discussion made me realize I was asking the wrong question. I do not really care about the $GL_n$-invariance on the whole space of variables "$\text{End}(\bigwedge^k \mathbb{R}^n)$"- I am only interested in polynomials which are conjugation-invariant when restricted to the $k$-minors of a matrix-so from this perspective the quadratic polynomial which gives the determinant is legitimate. (So, I guess we now have two different questions, each might be interesting on its own). I will edit the question to emphasize this. Thanks for your help. $\endgroup$ Sep 23, 2018 at 7:56

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