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According to the Wikipedia page on bivectors:

...if $B$ is a bivector, then the rotor $R$ is $e^{B/2}$ and rotations are generated [by] $v'=RvR^{-1}$.

But how do you take an exponent between a scalar and a bivector? I've seen expressions like $B^2$ before, meaning to take the geometric product of $B$ with itself, but the blade was always in the mantissa. Is it just an element-wise operation, like $\lambda B$ for scalar multiplication?

Draconis
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1 Answers1

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First, if $B$ is a bivector, then so is $B/2$. So the problem is to define the exponential of a bivector.

Every bivector $B$ can be factored $B = (B/|B|)\,|B| \equiv \mathbf{i}\theta$, where $\mathbf{i}$ is a bivector with $|\mathbf{i}| = 1$, and $\theta$ is a scalar.

Then the definition is $e^{\mathbf{i}\theta} = \cos\theta + \mathbf{i}\sin\theta$.

An equivalent definition is to use a power series -- the same series from scalar calculus.

Alan Macdonald
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  • This makes sense, but why do we define the exponential like that? Isn't the imaginary unit more analogous to the unit pseudoscalar, i.e. the n-vector in n dimensions, than to a unit bivector? – Draconis Oct 18 '15 at 06:37
  • This only applies in $\mathbb{R}^3$, where $|B|$ is defined – Eric Jan 13 '20 at 12:41