A cross product is a bilinear operation which, given two input vectors $x, y$, produces a vector $x \times y$ orthogonal to both, whose length equals the area of the parallelogram spanned by $x$ and $y$.
The components of a cross product can be expressed using Einstein notation as $(x\times y)_i = X_{ijk} x_j y_k$, where $X_{ijk}$ is a rank-3 totally antisymmetric tensor satisfying the following area condition:
$$X_{ijm} X_{mkl} + X_{jkm} X_{mli} = 2 \delta_{ik}\delta_{jl}-\delta_{ij}\delta_{kl}-\delta_{il}\delta_{kj},$$
where $\delta_{ij}$ denotes the Kronecker delta.
It is a known fact that cross products only exist in dimensions $0, 1, 3$ and $7$. My question is essentially how close can a given totally antisymmetric tensor get to a cross product in dimensions other than these. One way to measure for some $X_{ijk}$ the failure to satisfy the area condition above is to set
$$\Delta_{ijkl} = (X_{ijm} X_{mkl} + X_{jkm} X_{mli}) - (2 \delta_{ik}\delta_{jl}-\delta_{ij}\delta_{kl}-\delta_{il}\delta_{kj}),$$
and define the scalar $\Delta = \Delta_{ijkl} \Delta_{ijkl}$. In practical terms, $\Delta$ can be thought of as the average value of the squared deviation $(|x\times y|^2-\operatorname{area}(x,y)^2)^2$ over all unit vectors $x, y \in S^{n-1}$, where $\times$ is the bilinear product associated to $X_{ijk}$.
Note that if $X_{ijk}$ defines a true cross product, we clearly have $\Delta(X_{ijk}) = 0$; otherwise, $\Delta(X_{ijk})$ will take some positive value (since it is a sum of squares). We now define the minimum average deviation $\Delta_{\mathrm{min}}$ by minimizing $\Delta$ over all possible rank-3 totally antisymmetric tensors, i.e.,
$$\Delta_{\mathrm{min}} = \min_{X_{ijk} \in \Lambda^3\mathbb{R}^n} \Delta(X_{ijk}).$$
This quantity only depends on the dimension $n$. Here is a table with some values I've computed in Mathematica (unfortunately the computations take more than half an hour to complete after $n=12$. I can provide my code if needed):
| $n$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ | $13$ | $14$ |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| $\Delta_{\mathrm{min}}(n)$ | $0$ | $0$ | $6$ | $0$ | $36$ | $\frac{672}{11}$ | $72$ | $0$ | $84$ | $\frac{351}{2}$ | $269.207$ | $326.737$ | $448.972$ | $528$ | $588$ |
Surprisingly, the values up to $n=9$ turn out to be rational numbers (to a very good approximation at least). However, for $n\ge 10$ this doesn't seem to necessarily hold anymore, so there probably isn't a nice formula for the minimum average deviation as a function of $n$. Instead of that, I would like to ask about its growth properties.
My questions are:
- Is there any (nontrivial) asymptotic lower bound for $\Delta_{\mathrm{min}}(n)$?
Note that there is an obvious upper bound $\Delta_{\mathrm{min}}(n) \le 6n(n-1)$ obtained by setting $X_{ijk}$ to zero. Apparently the actual values tend to be close to half of that, that is, $\Delta_{\mathrm{min}}(n) \simeq 3n(n-1)$ seems to hold.
- Is $\Delta_{\mathrm{min}}(n)$ nondecreasing for all $n>7$? If not, what are the dimensions where it decreases?
EDIT: Here is the Mathematica code I'm using:
n = 5; (* the dimension is specified here *)
xarray =
SymmetrizedArray[
pos_ :> Subscript[x, StringJoin[ToString /@ pos]], {n, n, n},
Antisymmetric[All]];
x2 = (Activate@
TensorContract[
Inactive[TensorProduct][xarray, xarray], {{1, 4}}])/2;
id2 = 2 Transpose[
Symmetrize[TensorProduct[IdentityMatrix[n], IdentityMatrix[n]],
Antisymmetric[{1, 3}]], {1, 3, 2, 4}];
delta = Flatten[
Flatten /@ (x2 - id2 + Transpose[x2 - id2, {2, 3, 4, 1}])];
deltan = N[DeleteCases[delta, 0]];
AbsoluteTiming[
NMinimize[Total[Map[#^2 &, deltan]], Variables@Normal[xarray],
WorkingPrecision -> 4]]
(* output should be: { Computation time, { minimum Delta, X_ijk that achieves it } } *)
(* increasing WorkingPrecision gives more digits, but takes more time *)
EDIT 2: I managed to make it work for $n=13$ and $n=14$ by pre-simplifying the minimization function and leaving it running all night; I updated the table with these two new values. Interestingly, they are rational numbers again.
I don't have the $n=15$ case yet, but following the discussion in the comments I have computed $\Delta(X^{\mathbb{S}}_{ijk})$ for a single $15$-dimensional tensor $X^{\mathbb{S}}_{ijk}$ defined in the following way: if $a$ and $b$ are two purely imaginary sedenions and $\Im ab$ denotes the imaginary part of their multiplication, we have $(\Im ab)_i = X^{\mathbb{S}}_{ijk} a_j b_k$.
Since all previous Cayley-Dickson algebras (real numbers $\mathbb{R}$, complex numbers $\mathbb{C}$, quaternions $\mathbb{H}$ and octonions $\mathbb{O}$) satisfy $\Delta(X^{\mathbb{R}}_{ijk})=\Delta(X^{\mathbb{C}}_{ijk})=\Delta(X^{\mathbb{H}}_{ijk})=\Delta(X^{\mathbb{O}}_{ijk})=0$, it is in principle reasonable to think that $\Delta(X^{\mathbb{S}}_{ijk})$ will also be relatively small, perhaps a local minimum. However, it turns out that
$$\Delta(X^{\mathbb{S}}_{ijk})=1152,$$
which almost saturates the bound $6\cdot15(15-1)=1260$. So contrary to my expectations, sedenions spectacularly fail to define anything close to a cross product. Even a simple rescaling $\Delta(0.5 X^{\mathbb{S}}_{ijk}) = 891$ does it better.
There is a notion of 'circulation', which applies to every dimension subset to another dimension. This equates to if one imagines some kind of circulation in a closed loop, an equivalent circulation exists in the orthogonal.
For example, if one supposes that the faces (ie N-1 d) of a polytope has an out-vector that is normal to the surface, then the removal of a number of faces of that polytope will leave a 'ring' (ie surface N-2 without interior N-1), that has a net vector equal to the sum of vectors spanning the hole, and any alternate cover of this hole will have the same out-vector sum. This is a generalisation of the magnetic dipole = current × vector-area.
Not all of these constitute an algebraic space, but it is evident that any open loop (of any dimension m), has a circulation that matches the space orthogonal to it (ie n-m), in such a way that the circulation or direction of space is transferred.
I believe this is something that Clifford may have looked at. I am also looking at this identical matter from first principles.