# Fluid Mechanics in Seven Dimensions?

Cayley-Dickson Algebras:
The Cayley-Dickson algebras $\mathbb{A}_n$ are special real algebras satisfying

$\dim_\mathbb{R}\mathbb{A}_n=2^n$

for $n\geq 0.$  If we let $\{e_0,...,e_{2^n-1}\}$ be the basis of $\mathbb{A}_n,$ then the algebra is generated by the relations

$e_i^2=e_1\cdots e_{2^n-1}=-1$

where $i\geq 1.$  Hence $\mathbb{A}_0=\mathbb{R}$ and $\mathbb{A}_1=\mathbb{C}.$  $\mathbb{A}_2$ is called the quaternions and is often denoted $\mathbb{H}.$  $\mathbb{A}_3$ and $\mathbb{A}_4$ are called the octonions and sedenions respectively and are respectively also denoted $\mathbb{O}$ and $\mathbb{S}.$

The Cayley-Dickson algebras can also be constructed inductively using a recursive unary operation (called conjugation).  It proceeds as follows, let $\mathbb{A}_0=\mathbb{R}.$  We define $\mathbb{A}_1$ constructively.  Let $\mathbb{A}_1=\mathbb{R}\oplus\mathbb{R}$ and define the following operations:

$\begin{array}{rcl}(a,b)+(c,d)&=&(a+b,c+d)\\(a,b)(c,d)&=&(ac-bd,ad+bc)\\(a,b)^*&=&(a,-b).\end{array}$

One can verify this structure is isomorphic to $\mathbb{C}.$  Next we let $\mathbb{A}_2=\mathbb{A}_1\oplus\mathbb{A}_1=\mathbb{C}\oplus\mathbb{C}$ and define its structure

$\begin{array}{rcl}(a,b)(c,d)&=&(ac-d^*b,da+bc^*)\\(a,b)^*&=&(a^*,-b)\end{array}$

and addition remaining the same and the inner conjugation being complex conjugation.  How does is this retain isomorphism to our original definition of $\mathbb{A}_2?$  Note the dimension is retained by the fact that

$\mathbb{A}_2=\mathbb{C}\oplus\mathbb{C}=\mathbb{R}\oplus\mathbb{R}\oplus\mathbb{R}\oplus\mathbb{R}.$

Hence using this as our inclusion, we show the original requirements that

$i^2=j^2=k^2=ijk=-1.$

We have

$\begin{array}{rcl}i^2&=&(0,1,0,0)^2\\&=&\left((0,1),(0,0)\right)^2\\&=&\left((0,1),(0,0)\right)\left((0,1),(0,0)\right)\\&=&\left((0,1)(0,1)-(0,0)^*(0,0),(0,0)(0,1)+(0,0)(0,1)^*\right)\\&=&\left((-1,0)-(0,0),(0,0)+(0,0)(0,-1)\right)\\&=&\left((-1,0),(0,0)\right)\\&=&\left(-1,0,0,0\right).\end{array}$

Similar computations can be done for $j$ and $k$ and for showing that

$(0,1,0,0)(0,0,1,0)(0,0,0,1)=(-1,0,0,0).$

It turns out $\mathbb{A}_3$ and beyond are not associative, so in our initial requirement that

$e_1\cdots e_{2^n-1}=-1,$

we will clarify that we mean

$\left(\cdots((e_1e_2)e_3)\cdots e_{2^n-2}\right)e_{2^n-1}=-1.$

We continue inductively be setting $\mathbb{A}_{n+1}=\mathbb{A}_n\oplus\mathbb{A}_n$ and defining product and conjugation as before (and componentwise addition).  One can establish the following table of properties

$\begin{array}{|c|c|c|c|c|}\hline\mathbb{R}&\mathbb{C}&\mathbb{H}&\mathbb{O}&\mathbb{S}\\\hline\mbox{division algebra}&\mbox{division algebra}&\mbox{division algebra}&\mbox{division algebra}&\phantom{normed}\\\mbox{associative}&\mbox{associative}&\mbox{associative}&&\\\mbox{commutative}&\mbox{commutative}&&&\\\mbox{trivial conjugation}&&&&\\\end{array}$

where associativity and commutativity refer to the multiplication (the addition is always both).  All of the above properties (save the division algebra structure) can at this point (albeit with some tediousness) be shown.  If $a=(a_0,...,a_{2^n-1})\in\mathbb{A}_n,$ we can define its real part as

$\mbox{Re}\,a=a_0.$

Note for $\mathbb{R}$ we have

$a^*a=aa=a^2.$

For $a\in\mathbb{R}$ we can define $\|a\|=\sqrt{\mbox{Re}\,a^*a}=\sqrt{a^2}=|a|.$  For $\mathbb{C}$ we have

$(a,b)^*(a,b)=(a,-b)(a,b)=(a^2+b^2,ab-ba)=(a^2+b^2,0).$

Similarly we can define $\|(a,b)\|=\sqrt{\mbox{Re}\,(a,b)^*(a,b)}=\sqrt{a^2+b^2}.$  For $\mathbb{H}$ we have

$(a,b)^*(a,b)=(a^*,-b)(a,b)=(a^*a+b^*b,a^*b-a^*b)=(\|a\|^2+\|b\|^2,0).$

Hence $\|(a,b,c,d)\|=\sqrt{a^2+b^2+c^2+d^2}.$  And similarly for $\mathbb{O}$ we have

$(a,b)^*(a,b)=(a^*,-b)(a,b)=(\|a\|^2+\|b\|^2,0),$

so $\|(a_0,...,a_7)\|=\sqrt{a_0^2+\cdots+a_7^2}.$  Inductively, one can see that the norm on $\mathbb{A}_n$ coincides with the euclidean norm in $\mathbb{R}^{2^n}.$  We can also note that the multiplicative inverse of an element $x\in\mathbb{A}_n$ is

$x^{-1}=\frac{x^*}{\|x\|^2}.$

Although it turns out that in the sedenions and above ($\mathbb{A}_i$ for $i\geq 4$), there are zero divisors (for example, $(e_3+e_{10})(e_6-e_{15})=0$), so the algebra fails to be a division algebra.

Theorem 1.  (Hurwitz’ Theorem)  If $A$ is a real normed division algebra with identity, then $A\in\{\mathbb{R},\mathbb{C},\mathbb{H},\mathbb{O}\}.$

Corollary 2.  (Frobenius’ Theorem)  If $A$ is an associative real division algebra, then $A\in\{\mathbb{R},\mathbb{C},\mathbb{H}\}.$

Generalized Cross Products:
The cross product of two elements $x,y\in\mathbb{R}^3$ can be written

$x\times y=\begin{vmatrix}e_1&e_2&e_3\\x_1&x_2&x_3\\y_1&y_2&y_3\end{vmatrix}=(x_2y_3-x_3y_2,x_3y_1-x_1y_3,x_1y_2-x_2y_1).$

There is a natural way to define a cross product of $n-1$ elements in $\mathbb{R}^n$ by

$\mathsf{X}(v_1,...,v_{n-1})=\begin{vmatrix}e_1&\cdots&e_n\\v_{1_1}&\cdots&v_{1_n}\\\vdots&\ddots&\vdots\\v_{{n-1}_1}&\cdots&v_{{n-1}_n}\end{vmatrix}.$

However, our ultimate objective is to define vorticity in other dimensions.  So we will want a binary cross product that we can apply to our differential operator and a vector field: $\nabla\times f.$  We want $\times$ to satisfy some conditions

$\begin{array}{rcl}(w+x)\times(y+z)&=&(w+x)\times y+(w+x)\times z=w\times y+x\times y+w\times z+x\times z\\ 0&=&x\cdot(x\times y)=(x\times y)\cdot y\\\|x\times y\|^2&=&\|x\|^2\|y\|^2-(x\cdot y)^2.\end{array}$

These conditions are respectively called bilinearity, orthogonality, and magnitude.  Hence we want $\times$ to be the product in a real normed algebra.  Moreover magnitude tells us that we want $x\times x=0.$  We also have $e_i^2=-1$ in our Cayley-Dickson algebras.  It turns out that the cross product in $\mathbb{R}^3$ can be modeled in $\mathbb{H}$ as follows.  Define $C:\mathbb{R}^3\to\mathbb{H}$ by

$C(x_1,x_2,x_3)=x_1i+x_2j+x_3k$

and $C^{-1}:\mathbb{H}\to\mathbb{R}^3$ by

$C^{-1}(a+bi+cj+dk)=(b,c,d).$

Then one can verify that for $x,y\in\mathbb{R}^3$

$x\times y=C^{-1}\left(C(x)C(y)\right).$

Note that in $\mathbb{R},$ the cross product is trivial $r\times s=0$ for all $r,s\in\mathbb{R}.$  This is obvious if we require it to satisfy orthogonality—where the dot product is just multiplication.  So the zero-product property implies the cross product must be $0.$  It is compatible with our map as well (in this case $C:\mathbb{R}\to\mathbb{C}$):

$r\times s=C^{-1}(C(r)C(s))=\mbox{Im}\,(ri)(si)=\mbox{Im}\,(-rs)=0.$

Hence we are left with one remaining cross product: the cross product in $\mathbb{R}^7$ where, hereafter, $C:\mathbb{R}^7\to\mathbb{O}$ is defined by

$C(x_1,...,x_7)=x_1e_1+\cdots+x_7e_7$

and $C^{-1}:\mathbb{O}\to\mathbb{R}^7$ is the imaginary map

$C^{-1}(a_0e_0+\cdots+a_7e_7)=(a_1,...,a_7).$

The cross product of $x,y\in\mathbb{R}^7$ is defined by

$x\times y=C^{-1}(C(x)C(y)).$

We will now define

$C(\nabla)=e_1\frac{\partial}{\partial x_1}+\cdots+e_7\frac{\partial}{\partial x_7}.$

If $f:\mathbb{R}^7\to\mathbb{R}^7$ is differentiable, then we define its curl by

$\mbox{curl}\,f=\nabla\times f=C^{-1}(C(\nabla)C(f)).$