Timeline for Which polynomials in the minors of a matrix are invariant under conjugation?
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| Sep 24, 2018 at 20:29 | comment | added | François Brunault | The other point I wanted to make is about your subquestion. I'm not sure what is the quotient $W/\mathrm{GL}(n)$ but we know the dimension is at least $\dim W - n^2 = \begin{pmatrix} n \\ k \end{pmatrix}^2 - n^2$. On the other hand we know $W/\mathrm{GL}(\wedge^k \mathbb{R}^n)$ has dimension $\begin{pmatrix} n \\ k\end{pmatrix}$ as the link you mention shows. This means that there will be lot of polynomials which are $\mathrm{GL}(n)$-invariant but not $\mathrm{GL}$-invariant. | |
| Sep 24, 2018 at 18:32 | comment | added | François Brunault | @user44191 you're right, $\phi $ is not linear but polynomial and homogeneous of degree $k$, sorry. So we have a linear map $\phi^* : W^* \to \mathrm{Sym}^k V^*$ which you can extend to the symmetric algebra of $W^*$. | |
| Sep 24, 2018 at 16:18 | comment | added | user44191 | @FrançoisBrunault Why is $\phi$ linear? | |
| Sep 24, 2018 at 12:14 | comment | added | François Brunault | @AsafShachar It seems you are looking at the following. If $V$ is a vector space then the ring of polynomials on $V$ is the symmetric algebra $\mathrm{Sym} V^*$ where $V^*$ is the dual of $V$. There is a natural linear map $\phi : V \to W$ with $V=\mathrm{End}(\mathbf{R}^n)$ and $W=\mathrm{End}(\bigwedge^k \mathbf{R}^n)$. You are looking at the preimage of $(\mathrm{Sym} V^*)^{\mathrm{GL}_n}$ (which is known) under the canonical morphism $\mathrm{Sym} W^* \to \mathrm{Sym} V^*$. (...) | |
| Sep 23, 2018 at 13:56 | comment | added | user44191 | (1) is more an algebraic geometry thing than anything else. The ring of polynomials on a subvariety is a quotient of the whole space. So the ring of polynomials on the space of minors (i.e. numbers that follow Plucker relations) is a quotient of the ring of polynomials on $\wedge V$. And so you get a class, not just a single polynomial. (2) was bad phrasing on my part; I should have said "class of polynomials on $\wedge V$ that give the determinant on the subvariety". (3) is that reductive groups allow "averaging" - so take any polynomial that gives the det, and "average" it over an orbit. | |
| Sep 23, 2018 at 7:56 | comment | added | Asaf Shachar | (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. | |
| Sep 23, 2018 at 7:47 | comment | added | Asaf Shachar | 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? | |
| Sep 23, 2018 at 7:47 | comment | added | Asaf Shachar | 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...). | |
| Sep 22, 2018 at 22:33 | comment | added | user44191 | 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. | |
| Sep 22, 2018 at 14:33 | history | edited | Asaf Shachar | CC BY-SA 4.0 |
I have added another class of invariant polynomials.
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| Sep 22, 2018 at 8:16 | history | edited | Asaf Shachar | CC BY-SA 4.0 |
added 756 characters in body
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| Sep 22, 2018 at 6:15 | comment | added | user44191 | 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. | |
| Sep 22, 2018 at 5:43 | history | asked | Asaf Shachar | CC BY-SA 4.0 |