8
$\begingroup$

Let $K$ be a field, $1 \leq d \leq n$ integers and $V$ an $n$-dimensional vector space. The Grassmann–Plücker relations are quadratic forms on $\bigwedge^d V$ whose zero set is exactly the set of decomposable vectors in $\bigwedge^d V$ (i.e. which are of the form $v_1 \wedge \dotsb \wedge v_d$), thus describing the ideal corresponding to the Plücker embedding $\operatorname{Gr}_d(V) \to \mathbb{P}(\bigwedge^d V)$.

In other words, given an $d$ by $n$ matrix these are the relations between the determinants of $d$ by $d$ minors of the matrix.

What is known about the variety of decomposable elements for the symmetric power instead of the exterior power? (The case where $K$ is the field of complex numbers is sufficiently interesting.) In other words, given an $d$ by $n$ (say, complex) matrix what can be said about the relations between the ${{n+d-1} \choose {d}}$ permanents of $d$ by $d$ (multi) submatrices of the matrix (we allow repeated columns but not repeated rows)?

I made a quick Google search and I found one possibly-related paper entitled "Permanental ideals" by Reinhard C. Laubenbacher and Irena Swanson. They point out that there there is a strong dependence on the characteristics. (I am not sure this is indeed related but this was the closeset I found.)

$\endgroup$
6
  • 2
    $\begingroup$ Do permanents really have a relation to symmetric powers as close as determinants have to exterior powers? I'm not saying they don't, just wondering. $\endgroup$ Apr 28, 2014 at 17:01
  • 2
    $\begingroup$ @Darij: yes they do. First consider the "split tensor" $v_1\otimes\cdots\otimes v_d$ in the tensor product. If you write things in coordinates you will see that if you antisymmetrize you get the determinants (for the matrix of v's) and if you symmetrize then you get permanents. $\endgroup$ Apr 28, 2014 at 18:03
  • $\begingroup$ Could you clarify the question? I was not sure whether you were asking the question Abdesselam addressed, or for the syzygies of the ideal generated by subpermanents, or something else. $\endgroup$ Apr 28, 2014 at 18:19
  • $\begingroup$ @Abdelmalek, do you ultimately just mean the result: $\mbox{per}(\langle x_i, y_j\rangle) = \langle x_1 \vee \cdots x_k, y_1 \vee \cdots \vee y_k\rangle$ $\endgroup$
    – Suvrit
    Apr 28, 2014 at 21:16
  • $\begingroup$ @Survit: yes essentially. $\endgroup$ Apr 28, 2014 at 21:32

1 Answer 1

8
+200
$\begingroup$

The corresponding variety of homogeneous polynomials that split as a product of linear forms goes under several names: Brill-Gordan loci, Chow variety for zero-cycles, the split variety... Set theoretic equations (of degree $d+1$) were discovered by Brill and Gordan. The defining ideal is still unknown. See, e.g., this MO question: which homogeneous polynomials split into linear factors?

$\endgroup$
3
  • $\begingroup$ Dear Abdelmalek, many thanks for the answer. You wrote "Set theoretic equations (of degree d+1) were discovered by Brill and Gordan." Can you elaborate on these equations? $\endgroup$
    – Gil Kalai
    Apr 30, 2014 at 19:34
  • 1
    $\begingroup$ The equations are complicated. They can be found, e.g., on page 2 of emmanuel.jean.briand.free.fr/publications/eaca2004/eaca2004.pdf also please follow the MO link I indicated $\endgroup$ Apr 30, 2014 at 20:13
  • $\begingroup$ @GilKalai: I just saw the bounty awarded to my answer. Thank you! This is very kind of you. I hope the references I pointed out were at least somehow helpful to your research endeavors. If so, please keep me posted about your related work. $\endgroup$ Dec 29, 2021 at 22:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.