Cross Exponentiation

Today, in this blursed mathematics post, we are going to investigate what happens if you do a repeated cross product on a vector with another vector. This process is what I define as cross exponentiation. However, first, we have to describe a way to find a way to cross exponentiate vectors when using the multiplication format (Hereafter, I may refer to cross product as cross multiplication, because it fits with the analogy described below:

• repeated multiplication is exponentiation
• repeated cross multiplication is cross exponentiation

To put it in short, the cross product exhibits a non-associative property, namely the fact that ${\displaystyle ({\vec {a}}\times {\vec {a}})\times {\vec {b}}\neq {\vec {a}}\times ({\vec {a}}\times {\vec {b}})}$ . Because of the anti-associativity, the LHS is the zero vector (which is trivially boring), but the RHS provides something that we can exploit for later. Because of this, for cross exponentiation, if we expand it, we must perform it from right to left.

To maximize the blursedness this post has, we will use the notation ${\displaystyle {\vec {a}}^{c}{\vec {b}}}$ to represent such an operation.

It can be shown that the recurrence relation from this function is the follows:

• ${\displaystyle {\vec {a}}^{1}{\vec {b}}={\vec {a}}\times {\vec {b}}}$
• ${\displaystyle {\vec {a}}^{c}{\vec {b}}={\vec {a}}\times {\vec {a}}^{c-1}{\vec {b}}}$

With this set, let us work some of the examples:

Note: The cross exponentiation, like the cross product, only is valid for vectors in ${\displaystyle \mathbb {A} ^{3}}$, where ${\displaystyle \mathbb{A}}$ is a field whose multiplication is commutative and outputs ${\displaystyle \mathbb{A}}$. To simplify the examples that follow, we will assume that ${\displaystyle \mathbb{A}}$ is ${\displaystyle \R}$, though it can be generalized as shown above. The bureaucrat is currently finding a way to generalize the cross product to ${\displaystyle \mathbb{A}^n}$ where ${\displaystyle n\ge 2}$ so he can use it in later blog posts.

Example 1:

${\displaystyle {\vec {a}}=\left({\frac {1}{4}},-{\frac {1}{2}},1\right)}$

${\displaystyle {\vec {b}}=\left({\frac {1}{3}},1,-{\frac {2}{3}}\right)}$

${\displaystyle {\vec {a}}^{1}{\vec {b}}=\left(-{\frac {2}{3}},{\frac {1}{2}},{\frac {5}{12}}\right)}$

${\displaystyle {\vec {a}}^{2}{\vec {b}}=\left(-{\frac {17}{24}},-{\frac {37}{48}},-{\frac {5}{24}}\right)}$

${\displaystyle {\vec {a}}^{3}{\vec {b}}=\left({\frac {7}{8}},-{\frac {21}{32}},-{\frac {35}{64}}\right)}$

Other Exponents?

For this to work on all other exponents, we need to translate our operation into something that tolerates having other exponents. One way is to do it is via matrix exponentiation. To convert our operation to a matrix, we do the operation on the columns of the identity matrix ${\displaystyle I_3}$, namely ${\displaystyle \left(1,0,0\right)}$,${\displaystyle \left(0,1,0\right)}$, and ${\displaystyle \left(0,0,1\right)}$. We log those into our transformation matrix, which we will call ${\displaystyle A}$, and we will have ourselves a way to cross product that tolerates exponentiation! We will use ${\displaystyle {\vec {a}}=\left({\frac {1}{4}},-{\frac {1}{2}},1\right)}$ as an example before moving to a general case.

${\displaystyle A_{1}=\left({\frac {1}{4}},-{\frac {1}{2}},1\right)\times \left(1,0,0\right)=\left(0,1,{\frac {1}{2}}\right)}$

${\displaystyle A_{2}=\left({\frac {1}{4}},-{\frac {1}{2}},1\right)\times \left(0,1,0\right)=\left(-1,0,{\frac {1}{4}}\right)}$

${\displaystyle A_{3}=\left({\frac {1}{4}},-{\frac {1}{2}},1\right)\times \left(0,0,1\right)=\left(-{\frac {1}{2}},-{\frac {1}{4}},0\right)}$

${\displaystyle A={\begin{bmatrix}0&-1&-{\frac {1}{2}}\\1&0&-{\frac {1}{4}}\\{\frac {1}{2}}&{\frac {1}{4}}&0\end{bmatrix}}}$

({(1 4^(-1), -1 2^(-1), 1)x(1, 0, 0), (1 4^(-1), -1 2^(-1), 1)x(0, 1, 0), (1 4^(-1), -1 2^(-1), 1)x(0, 0, 1)})^T

Command: Transpose[Cross[{1/4, -1/2, 1},{1,0,0}],Cross[{1/4, -1/2, 1},{0,1,0}],Cross[{1/4, -1/2, 1},{0,0,1}]]

The properties of the matrix are as follows:

The characteristic equation is ${\displaystyle -\lambda ^{3}-{\frac {21\lambda }{16}}}$.

The eigenvalues are ${\displaystyle {\frac {i{\sqrt {21}}}{4}},-{\frac {i{\sqrt {21}}}{4}},0}$.

The eigenvectors are ${\displaystyle \left({\frac {1}{5}}\left(-4+2i{\sqrt {21}}\right),{\frac {1}{5}}\left(8+i{\sqrt {21}}\right),1\right),\left({\frac {1}{5}}\left(-4-2i{\sqrt {21}}\right),{\frac {1}{5}}\left(8-i{\sqrt {21}}\right),1\right),\left({\frac {1}{4}},-{\frac {1}{2}},1\right)}$.

Surprisingly, the matrix can be diagonalized, but not the way one thinks it can be done: We will not list the resulting diagonalization here due to length concerns; that is left to the reader as an exercise.

With the diagonalized matrix, one can easily find the fractional powers. If ${\displaystyle M=PDP^{-1}}$, then ${\displaystyle M^{n}=PD^{n}P^{-1}}$.

And yes, if one wonders why all of this is possible, consider this fact:

${\displaystyle {\vec {a}}\times \sum _{n=1}^{k}c_{n}{\vec {b_{n}}}=\sum _{n=1}^{k}c_{n}\left({\vec {a}}\times {\vec {b_{n}}}\right)}$.

Challenging Problems involving Cross Exponentiation:

There are some limitations to cross exponentiation, namely the following:

What is ${\displaystyle {\vec {a}}^{0}{\vec {b}}}$?

Per our two definitions ${\displaystyle {\vec {a}}^{1}{\vec {b}}={\vec {a}}\times {\vec {b}}}$ and ${\displaystyle {\vec {a}}^{1}{\vec {b}}={\vec {a}}\times {\vec {a}}^{1-1}{\vec {b}}={\vec {a}}\times {\vec {a}}^{0}{\vec {b}}}$. However, due to the fact that we cannot cancel a vector in cross multiplication, we would not be able to conclude that ${\displaystyle {\vec {a}}^{0}{\vec {b}}={\vec {b}}}$. Our matrix definition will not help us because we would have a singular matrix. However, mathematically, by convention, any matrix raised to the zeroth power is the identity matrix. Therefore ${\displaystyle {\vec {a}}^{0}{\vec {b}}={\vec {b}}}$ by convention.

What is ${\displaystyle {\vec {a}}^{-1}{\vec {b}}}$?

If we allow this operation to occur, this will mean we need to find a vector such that this is true: ${\displaystyle {\vec {b}}={\vec {a}}\times {\vec {b}}}$.

However, by definition, the right vector in the RHS is parallel to the LHS, therefore, it could not be a cross product by definition. Other reasons include the fact that the diagonal matrix in the diagonalization matrices is singular.

What is ${\displaystyle {\vec {a}}^{-n}{\vec {b}}}$?

Does not exist by inductive proof. See the above for the base case.

Future Work:

Since ${\displaystyle {\vec {a}}^{n}{\vec {b}}}$ is valid for ${\displaystyle n\geq 0}$, perhaps the next step is to see if we take this further by defining a vector exponentiation series, such as ${\displaystyle f\left({\vec {a}}\right)=\sum _{n=0}^{\infty }{\vec {a}}^{n}{\vec {b}}}$. Maybe a multivector generalization would be ${\displaystyle f\left({\vec {a}},{\vec {b}}\right)=\sum _{m=0}^{\infty }\sum _{n=0}^{\infty }{\vec {a}}^{n}{\vec {b}}^{m}}$. Maybe we would be able to see whether identities like ${\displaystyle \left({\vec {a}}^{n}{\vec {b}}+{\vec {a}}^{m}{\vec {b}}\right)\times \left({\vec {a}}^{q}{\vec {b}}+{\vec {a}}^{p}{\vec {b}}\right)={\vec {a}}^{m+p}{\vec {b}}^{2}+{\vec {a}}^{n+p}{\vec {b}}^{2}+{\vec {a}}^{m+q}{\vec {b}}^{2}+{\vec {a}}^{n+q}{\vec {b}}^{2}}$ are true. This might open to a new world of vector algebra, so everything is speculative right now.