What is an imaginary number?

In the following analysis, we trace the development of geometric algebra which allows a unified approach for tensors, quaternions, differential forms, spinors and lie algebras.

Aidan Rocke https://github.com/AidanRocke


In the following analysis, we trace the development of geometric algebra which allows a unified approach for tensors, quaternions, differential forms, spinors and lie algebras. The fundamental mathematical notion that allows such a powerful generalisation is the multivector, or what others more aptly call an oriented surface.

Curiously, the oriented surface, a subtle topological notion from mathematical physics elucidates the geometric meaning of the imaginary number as the square root of geometry.

The Geometric Product:

The fundamental operation that lies at the heart of Geometric Algebra is the Geometric Product, which is defined in terms of the interior product and exterior product for any pair of vectors \(\vec{u},\vec{v} \in \mathbb{R}^n\) as follows:

\[\begin{equation} \vec{u}\vec{v} = \vec{u} \cdot \vec{v} + \vec{u} \land \vec{v} \tag{1} \end{equation}\]

\[\begin{equation} \vec{v}\vec{u} = \vec{u} \cdot \vec{v} - \vec{u} \land \vec{v} \tag{2} \end{equation}\]

This allows us to recover formulas for the inner and exterior product in terms of the geometric product:

\[\begin{equation} \vec{u} \cdot \vec{v} = \frac{\vec{u}\vec{v} + \vec{v}\vec{u}}{2} \tag{3} \end{equation}\]

\[\begin{equation} \vec{u} \land \vec{v} = \frac{\vec{u}\vec{v} - \vec{v}\vec{u}}{2} \tag{4} \end{equation}\]

The reader is no doubt familiar with the inner product on Euclidean spaces whereas the exterior product merits a special introduction.

The Exterior product:

The exterior product of two vectors encodes the properties of an oriented surface: its area as well as its orientation.

If \(\theta\) denotes the angle between \(\vec{u}\) and \(\vec{v}\), the area associated with the exterior product of \(\vec{u}\) and \(\vec{v}\) is given by:

\[\begin{equation} \lVert \vec{u} \land \vec{v} \rVert = \lVert \vec{u} \rVert \lVert \vec{v} \rVert \sin \theta \tag{5} \end{equation}\]

so the area is maximal when the vectors are orthogonal.

On the other hand, in order to understand how the exterior product encodes orientation it is necessary and sufficient to consider positive and negative orientations. Given two distinct vectors \(e_i\) and \(e_j\) that are part of an orthonormal basis, we may succinctly represent both orientations using the algebraic formula:

\[\begin{equation} e_i \land e_j = -e_j \land e_i = +e_ie_j \tag{6} \end{equation}\]

where \(e_i^2=e_j^2 = 1\) implies:

\[\begin{equation} (e_i \land e_j)^2 = e_ie_je_ie_j = -(e_je_i)e_ie_j = -1 \tag{7} \end{equation}\]

Complex Numbers, or Geometric Algebra in two dimensions:

The general form of an oriented surface in two dimensions is given by:

\[\begin{equation} V = a + b\hat{x} + c\hat{y} + d\hat{x}\hat{y} \tag{8} \end{equation}\]

which has \(2^2=4\) dimensions, with the basis \((1,\hat{x},\hat{y},\hat{x}\hat{y})\) where \(\lVert \hat{x} \rVert^2 = \lVert \hat{y} \rVert^2 = 1\).

Using formula (7), we will note that:

\[\begin{equation} (\hat{x}\hat{y})^2 = \hat{x}\hat{y}(-\hat{y}\hat{x})=-1 \implies \hat{x}\hat{y}=i \tag{9} \end{equation}\]

which brings us to an intuition that we shall revisit; the imaginary number as an oriented surface or an operator that rotates vectors.

In fact, we may represent the geometric product of two vectors:

\[\begin{equation} \vec{u} = a \hat{x} + b\hat{y} \tag{10} \end{equation}\]

\[\begin{equation} \vec{v} = c\hat{x} + d\hat{y} \tag{11} \end{equation}\]

using Euler’s formula:

\[\begin{equation} \vec{u}\vec{v} = \vec{u} \cdot \vec{v} + \vec{u} \land \vec{v} = \lVert \vec{u} \rVert \lVert \vec{v} \rVert \cos \theta + \lVert \vec{u} \rVert \lVert \vec{v} \rVert i \sin \theta = \lVert \vec{u} \rVert \lVert \vec{v} \rVert e^{i \theta} \tag{12} \end{equation}\]

The Geometric Product as a generalisation of the Cross Product:

Given that \(e_i^2 = \lVert e_i \rVert^2 = 1\), we may derive the following formula for the Geometric Product of \(\vec{u},\vec{v} \in \mathbb{R}^3\):

\[\begin{equation} \vec{u} = a_1 \hat{x} + b_1 \hat{y} + c_1 \hat{z} \tag{13} \end{equation}\]

\[\begin{equation} \vec{v} = a_2 \hat{x} + b_2 \hat{y} + c_2 \hat{z} \tag{14} \end{equation}\]

where \(\{\hat{x},\hat{y},\hat{z}\}\) defines an orthonormal basis so we have:

\[\begin{equation} \vec{u}\vec{v} = a_1a_2 + b_1b_2 + c_1c_2 + a_1b_2\hat{x}\hat{y}-b_1a_2\hat{x}\hat{y} + a_1c_2\hat{x}\hat{z} -c_1a_2\hat{x}\hat{z} + b_1c_2\hat{y}\hat{z}-c_1b_2\hat{y}\hat{z} \tag{15} \end{equation}\]

which may be decomposed into:

\[\begin{equation} \vec{u} \cdot \vec{v} = a_1a_2 + b_1b_2 + c_1c_2 \tag{16} \end{equation}\]

\[\begin{equation} \vec{u} \land \vec{v} = (a_1b_2 - b_1a_2)\hat{x}\hat{y} + (b_1c_2 - c_1b_2)\hat{y}\hat{z} +(a_1c_2 -c_1a_2)\hat{x}\hat{z} \tag{17} \end{equation}\]

so the geometric product may be understood as a powerful generalization of the usual cross product in three dimensions.

Geometric Algebra as a model for Rotation Algebra:

Rotation algebra without rotation matrices:

From the Pythagorean theorem, given any unit vector \(\alpha\) we may resolve an arbitrary vector into parts parallel and perpendicular to \(\alpha\):

\[\begin{equation} \alpha: x = x_{\perp} + x_{\parallel} \tag{18} \end{equation}\]

Using the properties of the geometric product, we find:

\[\begin{equation} \alpha x_{\parallel} = x_{\parallel}\alpha \tag{19} \end{equation}\]

\[\begin{equation} \alpha x_{\perp} = -x_{\perp}\alpha \tag{20} \end{equation}\]

which allows us to express a reflection in a plane perpendicular to \(\alpha\):

\[\begin{equation} x_{\perp}-x_{\parallel} = \alpha(\alpha x_{\perp} -\alpha x_{\parallel}) = -\alpha x \alpha \tag{21} \end{equation}\]

\[\begin{equation} x \rightarrow -\alpha x \alpha \tag{22} \end{equation}\]

To produce a rotation, we need two reflections \(x \rightarrow \alpha \beta x \beta \alpha = R x \bar{R}\) where \(R = \alpha \beta\) is an oriented surface of even grade known as a rotor. Since \(\alpha^2 = \beta^2 = 1\), we find that:

\[\begin{equation} R\bar{R} = \alpha \beta \beta \alpha = 1 \tag{23} \end{equation}\]

For concreteness, let’s rotate the unit vector \(\alpha\) into the unit vector \(\beta\). A simple diagram reveals that this may be accomplished via a reflection perpendicular to the unit vector that is halfway between \(\alpha\) and \(\beta\):

\[\begin{equation} \gamma = \frac{\alpha + \beta}{\lvert \alpha + \beta \rvert} \tag{24} \end{equation}\]

\[\begin{equation} \gamma \alpha \gamma = -\beta \tag{25} \end{equation}\]

To rotate \(\alpha\) into \(\beta\), we need a second reflection perpendicular to \(\beta\):

\[\begin{equation} x \rightarrow \beta \frac{\alpha + \beta}{\lvert \alpha + \beta \rvert} x \frac{\alpha + \beta}{\lvert \alpha + \beta \rvert} \beta \tag{26} \end{equation}\]

where \(\lvert \alpha + \beta \rvert = \sqrt{(\alpha + \beta)(\alpha + \beta)}\), so we find that:

\[\begin{equation} R\bar{R} = \frac{(\beta \alpha + 1)(\alpha \beta + 1)}{\lvert \alpha + \beta \rvert^2} = 1 \tag{27} \end{equation}\]


\[\begin{equation} \beta = R \alpha \bar{R} \tag{27} \end{equation}\]

\[\begin{equation} \alpha = \bar{R} \beta R \tag{28} \end{equation}\]

Rotations as a Complex Exponential:

Using the fact that \(R\) and \(\bar{R}\) both induce half a rotation:

\[\begin{equation} \alpha \cdot \beta = \lVert \alpha \rVert \lVert \beta \rVert \cos \frac{\theta}{2} = \cos \frac{\theta}{2} \tag{29} \end{equation}\]

\[\begin{equation} \lVert \alpha \land \beta \rVert = \lVert \alpha \rVert \lVert \beta \rVert \sin \frac{\theta}{2} = \sin \frac{\theta}{2} \tag{30} \end{equation}\]

and we may resolve \(\alpha\) and \(\beta\) into orthogonal components using the Pythagorean theorem:

\[\begin{equation} \alpha = a\hat{x} + b\hat{y} \tag{31} \end{equation}\]

\[\begin{equation} \beta = c\hat{x} + d\hat{y} \tag{32} \end{equation}\]

so we may deduce that:

\[\begin{equation} \alpha \cdot \beta = ac + bd = \cos \frac{\theta}{2} \tag{33} \end{equation}\]

\[\begin{equation} \alpha \land \beta = (ad-bc)\hat{x}\hat{y} = \sin \frac{\theta}{2} \hat{x}\hat{y} \tag{34} \end{equation}\]

Setting \(i = \hat{x}\hat{y}\), we find that the rotor has a natural formulation as a complex exponential:

\[\begin{equation} R = \alpha \beta = \alpha \cdot \beta + \alpha \land \beta = e^{i \frac{\theta}{2}} \tag{35} \end{equation}\]

Moreover, since \(\hat{x}e^{i\frac{\theta}{2}}= e^{-i\frac{\theta}{2}}\hat{x}\), we may deduce the formula:

\[\begin{equation} R\hat{x}\bar{R} = e^{i\frac{\theta}{2}}\hat{x}e^{-i\frac{\theta}{2}}=e^{i\theta}\hat{x} \tag{36} \end{equation}\]


The power of Geometric Algebra is that it clarifies the geometric and topological origins of the imaginary number from an algebraic perspective. By encoding an oriented surface, the rotation algebra of the imaginary number transcends the unit circle into higher dimensions.

Finally, this subtle topological property allows us to reformulate complex analysis from a geometric point of view and gives deeper insight into the topological origins of Euler’s formula:

\[\begin{equation} e^{i\pi} = i^2 = -1 \tag{37} \end{equation}\]


  1. S. Gull, A. Lasenby and C. Doran. Imaginary Numbers are not Real - the Geometric Algebra of Spacetime. Foundations of Physics. 1993.

  2. David Hestenes. Spacetime Physics with Geometric Algebra. American Journal of Physics. 2003.

  3. C. Doran & A. Lasenby, Geometric Algebra for Physicists. Cambridge U Press: Cambridge. 2003.


For attribution, please cite this work as

Rocke (2022, Nov. 23). Kepler Lounge: What is an imaginary number?. Retrieved from keplerlounge.com

BibTeX citation

  author = {Rocke, Aidan},
  title = {Kepler Lounge: What is an imaginary number?},
  url = {keplerlounge.com},
  year = {2022}