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When reading introductory texts on geometric algebra, the author usually introduces two kinds of products and provides a geometric interpretation for understanding them:

  • the dot product or inner product $\vec{a} \cdot \vec{b}$ represents projecting $\vec{a}$ on to $\vec{b}$ and scaling by the magnitude of $\vec{b}$, or vice versa.
  • the wedge product or outer product $\vec{a} \wedge \vec{b}$ represents the bivector or “plane segment” produced by displacing $\vec{a}$ along $\vec{b}$

Both products can be visualized, and have properties that follow intuitively from their geometric interpretations like commutativity or anti-commutativity and relationships with $\sin$ and $\cos$.

However, when the geometric product $\vec{a}\vec{b} = \vec{a} \cdot \vec{b} + \vec{a} \wedge \vec{b}$ is introduced, there is no explanation of what it even means to add a scalar and a bivector or how you would visualize that compound object, and therefore, it isn't as easy to reason intuitively about the properties of the geometric product.

Is there a good way of visualizing such a multivector? And what is the purpose of combining the two products like this when they seem to be represent separate ideas?

Edit In response to a comment, I'm also curious about Clifford algebra: what is it, what is its relationship to geometric algebra, and what insights it does it provide?

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    $\begingroup$ The keyword to look up is "Clifford algebra." I think this is not a good way to think about this operation (which I would call Clifford multiplication). $\endgroup$
    – Qiaochu Yuan
    May 23, 2016 at 3:37
  • $\begingroup$ Geometric algebra de-emphasizes the fact that the elements can be represented as linear combinations of basis elements and focuses on the geometry. When thinking in terms of unit basis elements, addition of the scalar and outer products makes sense. When thinking about them geometrically, one could just as well write the product as an ordered pair rather than a sum. $\endgroup$
    – John Wayland Bales
    May 23, 2016 at 3:51
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    $\begingroup$ @jjw5432 The term "geometric algebra" is mostly a relatively recent rebranding of a part of "Clifford algebra" by some math-physics people. Clifford algebra is an abstract study of algebras generated by vector spaces with bilinear forms, and geometric algebra focuses on using the forms real and complex vector spaces which lend themselves to geometric interpretations. $\endgroup$
    – rschwieb
    May 27, 2016 at 13:45

1 Answer 1

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A product $\mathbf{uv}$ of vectors can be written $r(\cos\theta + \mathbf{i}\sin\theta) = r\exp{(\mathbf{i}\theta)}$, where $\mathbf{i}$ is the unit pseudoscalar of the plane containing $\mathbf{u}$ and $\mathbf{v}$ and $r = |\textbf{u}||\textbf{v}|$ and $\theta$ is the angle between the vectors.

This is interpreted geometrically as an arc of a circle of radius $r$ subtending the angle $\theta$. The arc can be slid around the circumference without changing the interpretation, just as a vector can be moved parallel to itself without changing it.

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  • $\begingroup$ How does that include both the bivector and scalar components of $\mathbf{u}\mathbf{v}$? It seems like an arc segment is only the bivector component. $\endgroup$
    – JJW5432
    May 26, 2016 at 14:12
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    $\begingroup$ Do a Google search "geometric algebra addition of arcs on a sphere". Click on the "Geometric Algebra for Computer Science". Scroll down to the diagrams at the bottom of the displayed page. $\endgroup$
    – Alan Macdonald
    May 29, 2016 at 3:41
  • $\begingroup$ I think that you do not ask a good question. Instead I will answer: how is the arc representation \emph{useful}? Perhaps you know that $\mathbf{uv}$ is called a \emph{rotor}, and is used to represent rotations. And the rotor for a composition of 2 rotations is the product of the rotors for the 2 rotations. Now do a Google search "geometric algebra addition of arcs on a sphere". Click on the "Geometric Algebra for Computer Science" entry. Scroll down to the diagrams at the bottom of the displayed page. You will see the composition above is found by "adding" their arcs on a sphere. $\endgroup$
    – Alan Macdonald
    May 29, 2016 at 3:52
  • $\begingroup$ Thank you so much for your answer. I'm still struggling to understand specifically the relationship between the interpretation of $\mathbf{u}\mathbf{v}$ and the individual parts, the bivector and scalar or the wedge and dot product respectively. And if this is not a good question, can you explain why not? $\endgroup$
    – JJW5432
    May 30, 2016 at 21:00
  • $\begingroup$ Can you explain what you mean by "It seems like an arc segment is only the bivector component."? $\endgroup$
    – Alan Macdonald
    Jun 10, 2016 at 4:37

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