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Geometric Algebra

Vectors can be combined to build parallelograms, parallelograms can be combined with vectors to build parallelepipeds, and so on. And then one can scale all of these to obtain new objects of the same kind. Geometric Algebra gives a precise meaning to this composition operation, called the geometric product, which can serve as the basis for performing most geometric operations on geometric objects, in any dimension.

Most of the following is adapted from the online resource by Alan MacDonald1.

  1. Geometric Product
  2. Basis
  3. Outer Product
  4. Norm
  5. Blades
  6. References

Geometric Product

As mentioned in the introduction, the geometric product is essentially a composition operation on geometric objects. As such, it should enjoy a few basic properties to be well-behaved:

\[1 x = x 1 = x\]

Furthermore, the space of \(n\)-dimensional geometric objects \(\GG^n\) (whatever they are, at this point) should be closed under the geometric product, making \(\GG^n\) a monoid.

Let us populate \(\GG^n\) with vectors from \(\RR^n\), and examine how the geometric product plays with the existing composition on vectors: vectors can be added and scaled by real numbers to obtain new vectors. So we can define the sum of two parallelograms by summing their constituent vectors pairwise:

\[(u_1 v_1) + (u_2 v_2) \eqdef (u_1 + u_2) (v_1 + v_2)\]

This generalises to arbitrary geometric objects, and the geometric product is distributive over the (commutative) sum as just defined. Similarly, we can extend scalar multiplication on vectors as:

\[(\lambda u) v = \lambda (u v)\]

One can easily check that the sum and scalar multiplication make \(\GG^n\) a vector space. Also, we see that scalars end up being geometric objects in their own right, spanned by the unit element \(1\) (called \(0\)-vectors). So far, we’ve seen how the geometric product behaves internally and with respect to the vector space structure, but we still don’t know how to compute it in practice. However, we do know that it is bilinear so it has a symmetric part and an antisymmetric part:

\[uv = \half\block{uv + vu} + \half\block{uv - vu}\]

Let us now add that tiny bit of Euclidean geometry giving the geometric product all its power (and name). Consider an inner product \(\cdot\) over \(\RR^n\): it is a symmetric positive bilinear form. All that we’re asking of the geometric product of two vectors is that its symmetric part be the inner product:

\[\half\block{uv + vu} = u\cdot v\]

From this, we immediately obtain \(u^2 = u\cdot u = \norm{u}^2\), and every non-zero vector \(u \neq 0\) has an inverse:

\[\quad u^{-1} = \frac{u}{\ \norm{u}^2}\]

Likewise, when \(u\cdot v = 0\) only the anti-symmetric part remains, hence:

\[uv = -vu\]

Summarizing, we obtained that:

Basis

Given an orthonormal orthogonal basis \(e_1, e_2, e_3\) for \(\RR^3\), we can quickly check than the following family forms a basis for \(\GG^3\):

\[\begin{matrix} 1 & \text{scalars} \\ e_1,\ e_2,\ e_3 & \text{vectors} \\ e_1 e_2,\ e_1 e_3,\ e_2 e_3 & \text{bi-vectors} \\ e_1 e_2 e_3 & \text{tri-vectors} \\ \end{matrix}\]

Scalars are called \(0\)-vectors, vectors of \(\RR^n\) are called \(1\)-vectors and so on. General elements of \(\GG^n\) are called multi-vectors. This basis is called the canonical basis due to the arrangement of the indices in strict increasing order. Similarly, for \(\GG^4\) one obtains:

\[\begin{matrix} 1 & \text{scalars} \\ e_1,\ e_2,\ e_3,\ e_4 & \text{vectors} \\ e_1 e_2,\ e_1 e_3,\ e_1 e_4,\ e_2 e_3,\ e_2 e_4,\ e_3 e_4 & \text{bi-vectors} \\ e_1 e_2 e_3,\ e_1 e_2 e_4,\ e_1 e_3 e_4,\ e_2 e_3 e_4 & \text{tri-vectors} \\ e_1 e_2 e_3 e_4 & \text{quadri-vectors} \\ \end{matrix}\]

A quick induction shows that \(\GG^n\) has dimension \(2^n\). Since the geometric product is associative, finding inverses for basis \(k\)-vectors is easy:

\[\block{e_{i_k} \ldots e_{i_1}}\block{e_{i_1} \ldots e_{i_k}} = 1\]

The inverse is another \(k\)-vector which can be reordered to the original canonical basis \(k\)-vector by a number \(m\) of swaps, giving:

\[\block{e_{i_1} \ldots e_{i_k}}^{-1} = (-1)^{m} \block{e_{i_1} \ldots e_{i_k}}\]

In particular, basis bi-vectors satisfy:

\[\block{e_j e_k}^2 = -1\]

which provides the connection with complex numbers and quaternions.

Outer Product

Since the symmetric part of the geometric product for two vectors is the inner product, the anti-symmetric part is logically called the outer product (or exterior product), denoted by \(\wedge\)2:

\[uv = u\cdot v + u \wedge v\]

For two vectors \(u, v\) in the plane \(e_1e_2\), their geometric product is:

\[\underbrace{(a e_1 + b e_2)}_u\underbrace{(c e_1 + d e_2)}_v = \underbrace{(ac + bd)}_{u\cdot v} + \underbrace{(ac - bd) e_1 e_2}_{u \wedge v}\]

hence the outer product \(u\wedge v\) has coordinate \(ac - bd=\norm{u}\norm{v}\sin(\alpha)\) along \(e_1e_2\), which is the signed area of the parallelogram delimited by \(u\) and \(v\) (where \(\alpha\) is the angle between them). The following formula hints at the connection with complex numbers and quaternions:

\[uv = \norm{u}\norm{v}\block{\cos(\alpha) + \sin(\alpha) e_1e_2}\]

The outer product can be extended from vectors to the whole geometric algebra by defining:

\[x \wedge y = \inner{x y}_{m + n}\]

for an \(m\)-vector \(x\) and an \(n\)-vector \(y\), and where \(\inner{z}_k\) is the grade projection operator selecting the \(k\)-vector part of a multivector \(z\). The outer product as defined remains associative, antisymmetric and bilinear. In particular, the outer product of linearly dependent vectors is zero.

Norm

The Euclidean norm on vectors extends naturally to multi-vectors: let \(x = \sum x_I e_I\), where \(e_I=\Pi_{i \in I} e_i\) and \(I\) ranges over the set of canonical basis index sets3, the norm of \(x\) is defined by:

\[\norm{x}^2 = \sum x_I^2\]

One can verify that it is indeed a norm, and that it does not depend on the choice of a canonical basis for \(\GG^n\).

Blades

Blades represent linear subspaces of \(\RR^n\) as the geometric product of orthogonal \(1\)-vectors. Let \(B \subseteq \RR^n\) be a subspace of dimension \(k\) and \(\block{b_i}_{i\leq k}\) an orthogonal basis of \(B\), then the product:

\[b = b_1 \ldots b_k\]

is called a blade of grade \(k\), or \(k\)-blade. In particular, the canonical basis vectors are all blades. Of particular interest is the canonical \(n\)-vector, called the pseudo-scalar:

\[1^* \eqdef e_1 \ldots e_n\]

An important property is that the pseudo-scalar does not depend on the orthonormal basis (except for sign). If \(e'_1 \ldots e'_n\) is another orthonormal basis for \(\RR^n\), then \(e'_1 \ldots e'_n\) is a unit \(n\)-vector, therefore:

\[e'_1 \ldots e'_n = \pm 1^*\]

The multiplication of a basis \(k\)-vector \(e\) by \(1^*\) cancels-out every occurence of basis \(1\)-vector in \(e\) and adds-in every missing basis \(1\)-vector, up to a sign. For instance, in \(\GG^3\):

\[e_1 e_3 1^* = e_2\]

This operation defines the dual of a multi-vector \(x\in\GG^n\):

\[x^* = x 1^*\]

One can easily show that the dual of a \(k\)-blade is an \((n-k)\)-blade representing the orthogonal complement of the original subspace: our \(k\)-blade can be completed into an orthonormal basis of the whole space, and multiplication by the pseudo-scalar for this basis (\(\pm 1^*\)) gives the result. Finally, we note that the sum of \(k\)-blades is generally not a blade, except when \(k=1\) and \(k=n-1\)4.

References

  1. Alan MacDonald, A Survey of Geometric Algebra and Geometric Calculus, Adv. Appl. Cliff. Alg. 27, 853–891 (2017). ↩

  2. It is quite unfortunate that the \(\wedge\) symbol has been used for something acting like a join in lattice theory, which has symbol \(\vee\) (like a union) ↩

  3. e.g. for \(\GG^3\): \(I \in \set{\set{0}, \set{1}, \set{2}, \set{3}, \set{1, 2}, \set{1, 3}, \set{2, 3}, \set{1, 2, 3}}\) ↩

  4. Use the dual twice. ↩