M Theory Lesson 350

A general hyperdeterminant begins with an $n$ dimensional matrix of shape

$(k_1 + 1) \times (k_2 + 1) \times \cdots \times (k_n + 1)$

and the special set $A$ is now the full set of entries $a_{i_{1} i_{2} \cdots i_{n}}$. The first main theorem (page 446) says that the hyperdeterminant of this matrix is non trivial if and only if

$k_{j} \leq \sum_{i \neq j} k_{i}$

for all possible $j$. For a hypercubic matrix, where all the $k_{i}$ are the same, this is certainly true. That is, the hypercube for $n$ qudits always gives us something. The next proposition says that this hyperdeterminant is invariant (in a suitable sense) under what could be called a SLOCC group: $SL(V_1) \times \cdots \times SL(V_{n})$. Another cool result (page 454) tells us that the degree of our hyperdeterminant is given by a sum over partitions $\lambda$

$N(k_1, \cdots, k_{n}) = \sum_{\lambda} (d_{\lambda k}) (1 + m_2 + \cdots + m_p)! \prod_{i = 2}^{p} \frac{(i - 1)^{m_i}}{(m_i)!}$

where the $m_i$ are the indices of the partition $\lambda$ (so that $m_1 = 0$) and $d_{\lambda k}$ counts binary matrices with row sums in the ordered partition $k$ (of the $k_i$) and column sums in $\lambda$. The cubic case $(k + 1)^{3}$ is given by

$N(k,k,k) = \sum 2^{k - 2j} \frac{(j + k + 1)!}{(j!)^{3} (k - 2j)!}$

for a sum over $0 \leq j \leq k/2$. We see that for three qubits, where $k = 1$, there is only one term and it gives the degree $4$ of the classical polynomial.