Geometric Algebra: getting the geometric interpretation of $k$-blades

Question

I'm approaching "Geometric Algebra for Computer Science" coming from "Abastract algebra" by Dummit and Foote. I learned exterior algebra, from an abstract point of view, as quotient of the tensor algebra. Now I'm facing some difficoulties getting a full understanding of the proposed practical and succint treatment.

How exactly the k-blades $a_1\wedge \dots \wedge a_k$ gets associated with its attitude, orientation and weight, as an element of $\bigwedge \mathbb{R}^n$ and how these attributes depends on $n$?

The book says that the attitude is the subspace of $\mathbb{R}^n$ spanned by $a_1, ..., a_k$, and that seems enough to me. Then it only defines orientation and weight for the case $n=k=2$ or $n=k=3$ saying also that orientation has not a complete generalization for all the cases, namely, for every $k,n$. The weight is the determinant of the matrix with $a1,..,a_k$ as columns.

One particular thing that I have been asking is: how can I calculate the weight of $a_1 \wedge a_2$ in the case $k=2, n=3$. Here the matrix with column $a_1, a_2$ is not squared thus it has not a determinant but the concept of area for a 2-D surface still makes sense. I guess I need some projection into the subspace, do I?

Answer 1

According to the Amazon preview of the book “Geometric Algebra for Computer Science (Revised Edition): An Object-Oriented Approach to Geometry” by Leo Dorst, Daniel Fontijne, and Stephen Mann (GACS), it seems like an answer to your questions (as well as a potential error that led to the OP's confusion) can be found in sections 2.8 and 3.1.

Another reference for an introduction to geometric algebra is “Linear and Geometric Algebra” (LaGA) by MSE's own Alan Macdonald.

Attitude

In the table referenced in 2.8.2 of GACS, they write that the attitude of a blade $\mathbf{A}$ is “The equivalence class $\lambda\mathbf{A}$, for any $\lambda\in\mathbb{R}$”. That should probably not allow $\lambda=0$, but it is good to think of the attitude as defined in the OP: “the subspace of $\mathbb{R}^{n}$ spanned by $a_{1},\ldots,a_{k}$”. Section 6.2 of LaGA says “A blade and a nonzero scalar multiple of it (another blade) represent the same subspace.”

Weight (and Norm)

In the table referenced in 2.8.2 of GACS, they write that the (relative) weight is “The value of $\lambda$ in $\mathbf{A}=\lambda\mathbf{I}$ (where $\mathbf{I}$ is a selected standard subspace with the same attitude)”. So “relative weight” can be positive or negative, and depends crucially on the choice of another blade $\mathbf{I}$ with the same attitude. This definition seems to throw out any absolute sense of scale - if $\mathbf{I}_{1}$ is one choice giving a certain relative weight $\lambda_{1}$ for $\mathbf{A}$, then it seems that $\mathbf{I}_{2}=2\mathbf{I}_{1}$ would give a new weight $\lambda_{2}=\lambda_{1}/2$. But I am not sure whether that was intended.

In 3.1.3 of GACS, they write “[the norm of a blade] indeed gives us a sensible geometric measure of the weight of the subspace represented by the blade, as an area or (hyper)volume”. So it seems, at the very least, that GACS supports a definition of “(absolute) weight” being the norm $\left\Vert \mathbf{A}\right\Vert $. It is defined in 3.1.3 by $$ \left\Vert \mathbf{a}_{1}\wedge\cdots\wedge\mathbf{a}_{k}\right\Vert ^{2} $$

$$ =\left(\mathbf{a}_{1}\wedge\cdots\wedge\mathbf{a}_{k}\right)*\widetilde{\left(\mathbf{a}_{1}\wedge\cdots\wedge\mathbf{a}_{k}\right)} $$

$$ =\left(\mathbf{a}_{1}\wedge\cdots\wedge\mathbf{a}_{k}\right)*\left(\mathbf{a}_{k}\wedge\cdots\wedge\mathbf{a}_{1}\right) $$ $$ =\det\begin{bmatrix}\mathbf{a}_{1}\cdot\mathbf{a}_{1} & \mathbf{a}_{1}\cdot\mathbf{a}_{2} & \cdots & \mathbf{a}_{1}\cdot\mathbf{a}_{k}\\ \mathbf{a}_{2}\cdot\mathbf{a}_{1} & \mathbf{a}_{2}\cdot\mathbf{a}_{2} & \cdots & \mathbf{a}_{2}\cdot\mathbf{a}_{k}\\ \vdots & & \ddots & \vdots\\ \mathbf{a}_{k}\cdot\mathbf{a}_{1} & \mathbf{a}_{k}\cdot\mathbf{a}_{2} & \cdots & \mathbf{a}_{k}\cdot\mathbf{a}_{k} \end{bmatrix}\text{ by eq. (3.2)} $$ This definition is essentially the same as the definition in a comment by mr_e_man.

That said, I think there is some language towards the end of 3.1.3 of GACS which is misleading, if not outright incorrect; and that may have led to the OP's confusion. It states “We know...that the $k$-volume associated with a $k$-blade $\mathbf{A}=\mathbf{a_{1}}\wedge\cdots\wedge\mathbf{a}_{k}$ is proportional to the determinant of the matrix $[[\mathbf{A}]] =[[ \mathbf{a}_{1}\cdots\mathbf{a}_{k}]] $. ...$\left\Vert \mathbf{A}\right\Vert ^{2}=\det\left([[ \mathbf{A}]] ^{T}[[ \mathbf{A}]] \right)$, you can...simplify this to $\det\left([[ \mathbf{A}]] \right)^{2}$. So indeed, for $k$-blades, we do compute the squared $k$-dimensional hypervolume.” But as the OP points out, if $k<n$, then $[[\mathbf{A}]]$ so defined is not a square matrix, so that $\det[[ \mathbf{A}]] $ is not even defined, so this discussion does not apply. As an aside, “is proportional to” is a bit vague - assuming the determinant is nonzero, every number (of the same sign?) is proportional to it. I think the intent might have been for a factor of $\pm1$ only.

Here is another equivalent definition of norm that is more explicitly coordinate-based: The standard basis (or your favorite basis)$\mathbf{e}_{1},\ldots,\mathbf{e}_{n}$ of $\mathbb{R}^{n}$ gives a basis for $\bigwedge\mathbb{R}^{n}$ along the lines of $\left(1,\mathbf{e}_{1},\ldots,\mathbf{e}_{n},\mathbf{e}_{1}\wedge\mathbf{e}_{2},\mathbf{e}_{1}\wedge\mathbf{e}_{3},\ldots,\mathbf{e}_{\left(n-1\right)}\wedge\mathbf{e}_{n},\mathbf{e}_{1}\wedge\mathbf{e}_{2}\wedge\mathbf{e}_{3},\ldots\right)$. Then the norm of a $k$-blade (or any multivector that is a sum of various blades) can be expanded in terms of this basis of $\bigwedge\mathbb{R}^{n}$ with a finite collection of coefficients. Analogous to the norm in $\mathbb{R}^{n}$, we can then define the norm to be the square root of the sum of the squares of these coefficients. This expression for norm is Definition 6.10 in LaGA.

Finally, the first definition of norm of a blade in LaGA is Definition 6.8, which requires writing the blade in a special form, using Gram-Schmidt or similar. If we have nonzero orthogonal vectors $\mathbf{b}_{1},\mathbf{b}_{2},\cdots,\mathbf{b}_{k}$, then “the norm of the blade $\mathbf{B}=\mathbf{b}_{1}\mathbf{b}_{2}\cdots\mathbf{b}_{k}$ is $\left|\mathbf{B}\right|=\left|\mathbf{b}_{1}\right|\left|\mathbf{b}_{2}\right|\cdots\left|\mathbf{b}_{k}\right|$. This is the volume of the rectangular parallelogram with edges $\mathbf{b}_{1},\mathbf{b}_{2},\ldots,\mathbf{b}_{k}$.”

Orientation

In the table referenced in 2.8.2 of GACS, they write that the (relative) orientation is “the sign of the weight relative to $\mathbf{I}$” “where $\mathbf{I}$ is a selected standard subspace with the same attitude”. When $n=k$, we have a conventional standard choice for $\mathbf{I}$: e.g. $\mathbf{e}_{1}\wedge\mathbf{e}_{2}$ or $\mathbf{e}_{1}\wedge\mathbf{e}_{2}\wedge\mathbf{e}_{3}$, where $\mathbf{e}_{i}$ are standard basis vectors. But if $n>k>0$, there's not really a canonical choice; so for a given $k$-dimensional subspace of $\mathbb{R}^{n}$ you'd have to make a somewhat arbitrary choice for $\mathbf{I}$. But, no matter what, you could still determine whether two blades with the same attitude/subspace had the same orientation or opposite orientations. Section 6.2 of LaGA says “A positive (respectively negative) scalar multiple [of a blade] retains (respectively reverses) the orientation of the subspace.”