I am unfamiliar with algebraic geometry, yet I am faced with calculating three special cases of the following (the full text can be found at http://arxiv.org/abs/1107.4659)
The $n^{\times d}$-hyperdeterminant of a symmetric tensor of degree $d \ge 2$ on $n \ge 2$ variables splits as the product $\prod_\lambda \Xi_{\lambda,n}^{M_\lambda}$ where $\Xi_{\lambda,n}$ is the equation of the dual variety of the Chow variety $Chow_\lambda \mathbb{P}^{n-1}$ when it is a hypersurface in $\mathbb{P}^{\binom{n-1+d}{d}-1}$, $\lambda = (\lambda_1, \ldots , \lambda_s)$ is a partition of $d$, and the multiplicity $M_\lambda = \binom{d}{\lambda_1, \ldots , \lambda_s}$ is the multinomial coefficient.
This is theorem 1.1 of the paper by Oeding, L., (2012) titled “Hyperdeterminants of Polynomials” in Advances in Mathematics, 231, 1308-1326. The theorem is followed by this explanation:
“Geometrically, this theorem is essentially a statement about the symmetrization of the dual variety of the Segre variety. It says that the symmetrization of this dual variety becomes the union of several other varieties (with multiplicities).” I am told that for a given $n$ and $d$ this formula produces a single polynomial.
I need to know what these specific polynomials look like for the cases where $n = 2$ and $d = 2, 3, 4$. Any help would be greatly appreciated.