What is the general variety corresponding to the Segre embedding for $n$-qubit systems?

Question

It is well known that entanglement is precisely the difference between the Cartesian product and the tensor product. The space where every point corresponds to a state is the projective Hilbert space, and the Segre embedding from the projective space of the Cartesian product to the projective space of the tensor product. The image of this embedding is the set of non-entangled states.

In the case of two-qubit systems, this is:

$$\sigma: \mathbb{C}P^1 \times \mathbb{C}P^1 \to \mathcal{P}(\mathcal{H}_1 \otimes \mathcal{H}_2) \cong \mathbb{C}P^3$$ $$([x_1 : y_1],[x_2 : y_2]) \mapsto [x_1x_2:x_1y_2:y_1x_2:y_1y_2].$$

The image of this embedding is the variety $$Z_0Z_3 - Z_1Z_2 = 0$$ where $$[Z_0:Z_1:Z_2:Z_3] \in \mathbb{C}P^3.$$

For two qubits systems, the solution set is the above quadric.

Question:

Is it possible to write down the general variety for $n$-qubit systems? If yes, what will it be?

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^{Note: Folks may be interested in this thread on Physics SE: Should it be obvious that independent quantum states are composed by taking the tensor product? It took me quite some time to appreciate why exactly we choose the tensor product rather than the direct sum for composite quantum systems; it makes a lot more sense now. The tensor product space allows for the notion of entanglement whereas the direct product space does not!}

Answer 1

It is possible, but a bit annoying.

For each subset S of $[1\cdots n]$ containing 1 that isn't all of $[1 \cdots n]$. We get a Segre embedding $\mathbb{CP}^{2^{\mid S \mid} - 1} \times \mathbb{CP}^{2^{\mid S^c \mid} - 1} \to \mathbb{CP}^{2^n-1}$. $S^c$ stands for the complementary subset of $n$

The reason we restrict so that $S$ contains 1 is so that we don't repeat between $(S,S^c)$ and another term which is the same as $(S^c,S)$. That is we don't want to repeat when $S=[1,2]$ giving $(S,S^c)=([1,2],[3])$ and $S=[3]$ giving $([3],[1,2])$

The equations that cut out a Segre embedding are similar to the 2 qubit example you state. They are still quadratic equations.

$$ Z_{i,j} Z_{k,l} - Z_{i,l} Z_{k,j} $$

where $Z_{i,j}$ means you write the $2^n$ homogenous coordinates of the $CP^{2^n-1}$ in a rectangular matrix which is $2^{\mid S \mid}$ by $2^{\mid S^c \mid}$. $i$ runs 1 through $2^{\mid S \mid}$ and $j$ runs 1 through $2^{\mid S^c \mid}$. You have to be careful keeping track of these indices because we have to do this many times for each $S$.

Once you do that, that takes care of each possible bipartitioning. If you want the states on n qubits that have no way to break up into two or more factors. That is no way to write $|\psi \rangle = R_\sigma (|\phi \rangle \otimes | \alpha \rangle)$ with $R_\sigma$ the action of a permutation of $n$. I did that because the way the state broke up could be like a bell state with 1 and 3 tensored with 0 for qubit 2. Just needed to straighten everything out.

So it can't be in the image of any of those Segre embeddings above. So take the union of all the Segre embeddings and avoid that subspace. Call the space of these genuinely entangled states $X$, the complement of all these Segre embeddings.

We have one of the embeddings cut out by some polynomial $f$ and another embedding cut out by $g$. So if you look at the zeros of $fg$, both of those will be included as well. Naturally, this generalizes to the many embeddings you are trying to avoid with this example. Don't want to use too much of the algebraic geometry vocabulary here, because I don't know how much others are familiar with it, but that is what is happening here so keeping it a little vague for future readers.

I can send something else I wrote up that used this to calculate certain combinations of cohomology groups of $X$. It's basically the procedure I describe here, but followed by only remembering the spaces as elements of a certain ring instead of as varieties/complements thereof. It means you don't have to do as much tedious work of writing those quadratic equations for each embedding, but you still get some information out. But that is a different question.