Grassmann-Plücker relations for permanents

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 $\wedge^d V$ whose zero set is exactly the set of decomposable vectors in $\wedge^d V$ (i.e. which are of the form $v_1 \wedge ... \wedge v_d$), thus describing the ideal corresponding to the Plücker embedding $\text{Gr}_d(V) \to \mathbb{P}(\wedge^d V)$.

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, comlplex) 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 here there is a strong dependence on the characteristics. (I am not sure this is indeed related but this was the closeset I found.)

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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. –  darij grinberg Apr 28 '14 at 17:01
@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. –  Abdelmalek Abdesselam Apr 28 '14 at 18:03
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. –  JM Landsberg Apr 28 '14 at 18:19
@Abdelmalek, do you ultimately just mean the result: $\mbox{per}(\langle x_i, y_j\rangle) = \langle x_1 \vee \cdots x_k, y_1 \vee \cdots \vee y_k\rangle$ –  Suvrit Apr 28 '14 at 21:16
@Survit: yes essentially. –  Abdelmalek Abdesselam Apr 28 '14 at 21:32

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?