We analysed Cayley’s 2x2x2 hyperdeterminant earlierand now it’s time to look at Schläfli’s 2x2x2x2 hyperdeterminant. I’ll sketch how this can be done using suitable elimination argumentsto reduce the number of non-zero elements in the hypermatrix. I am not going to go though all cases because it would not be instructive. You can complete it for yourself.

For a 2x2x2x2 matrix with components t_{ijkl}to be singular we require the existence of four vectors **w**_{1},**x**_{1},**y**_{1},**z**_{1}** **such that simultaneously

Σ t_{ijkl}x_{1}^{j}y_{1}^{k}z_{1}^{l} = 0

Σ t_{ijkl}w_{1}^{i}y_{1}^{k}z_{1}^{l} = 0

Σ t_{ijkl}w_{1}^{i}x_{1}^{j}z_{1}^{l} = 0

Σ t_{ijkl}w_{1}^{i}x_{1}^{j}y_{1}^{k} = 0

If we concentrate on the 2x2x2 hypermatrix given by s_{ijk}= Σ t_{ijkl}z_{1}^{l }we then need from the first three conditions that

det(s(**z**_{1})) = 0

and from the last condition

Σ s_{ijk}w_{1}^{i}x_{1}^{j}y_{1}^{k} = 0

The determinant of s(**z**) is Cayley’s hyperdeterminant of degree 4 so it forms a homogeneous quartic in the two components of **z**. It must have a root at **z** = **z**_{1 }Let’s assume that it has at least one other distinct root **z**_{2}(We should consider separately the case where it has four equivalent roots but I’ll skip that.)

becuase det(s(**z**_{2})) = 0 there must also be vectors **w**_{2},**x**_{2},**y**_{2} such that

Σ t_{ijkl}x_{2}^{j}y_{2}^{k}z_{2}^{l} = 0

Σ t_{ijkl}w_{2}^{i}y_{2}^{k}z_{2}^{l} = 0

Σ t_{ijkl}w_{2}^{i}x_{2}^{j}z_{2}^{l} = 0

Transform the hypermatrix to a basis using these vectors suitably normalised, so that **x**_{1}= (1,0), **x**_{2 }= (0,1), **y**_{1}= (1,0), **y**_{2 }= (0,1) etc. In this basis the equations now simply tell us that certain components are zero

t_{1111}= t_{2111}= t_{1211}= t_{1121}= t_{1112}= t_{1222}= t_{2222}= t_{2122}= t_{2212}= 0

With this simplification the quartic det(s(**z**)) can be wroked out in terms of the remaining components. We find that some of the terms are zero and with z = (u,v) it takes the form

det(s(**z**)) = cu^{2}v^{2}+ duv^{3}

This tells us that v = 0 is in fact a double root of the quartic. By consideration of all other possible cases it is possible to show that in fact the 2x2x2x2 hypermatrix is singular iff the quartic has a double root. This is then equivalent to the requirement that the discriminant of the quartic is zero. So finally the hyperdteriminant must be given by this discriminant.

The discriminant of a quartic is of degree 6 in the coefficients, and the coefficient are degree four in the components of the hypermatrix. Therefore the degree of the 2x2x2x2 hyperdeterminant is 24.

It may not be obvious at first but if you think about it you will realise that Schläfli’s hyperdeterminant has quite a large number of terms. For most purposes we do not need to compute the whole thing but it can be done and the total number of terms is 2894276 (see arXiv:math/0602149)

Another observation is that the hyperdteriminant is not a fundamental generator in the ring of invariants on the 2x2x2x2 hypermatrix. We know that because we saw that the discriminant of the quartic can be written in terms of two simpler Sl(2) invariants Δ_{4 }= g_{2}^{3} – 27g_{3}^{2} . These invariants also provide full invariants S and T of the hypermatrix A such that det(t) = S^{3} – 27T^{2}, but that is not the end of the story. We have already seen that there is a much simpler invariant of degree 2 with just eight terms. In fact it is not difficult to construct other invariants of degree 4 and 6. then with some work all invariants including the hyperdeterminant can be expressed in terms of those (see arXiv:quant-ph/0212069)

This demonstrates that the analysis of hyperdeterminants becomes very complex even for quite small formats. Nevertheless we are going to become interested in some significantly larger examples quite soon.

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