Why use geometric algebra and not differential forms?

Question

This is somewhat similar to Are Clifford algebras and differential forms equivalent frameworks for differential geometry?, but I want to restrict discussion to $\mathbb{R}^n$, not arbitrary manifolds.

Moreover, I am interested specifically in whether

$$(\text{differential forms on }\mathbb{R}^n\text{ + a notion of inner product defined on them}) \simeq \text{geometric algebra over }\mathbb{R}^n$$

where the isomorphism is as Clifford algebras. (I.e., is geometric algebra just the description of the algebraic properties of differential forms when endowed with a suitable notion of inner product?)

1. Is any geometric algebra over $\mathbb{R}^n$ isomorphic to the exterior algebra over $\mathbb{R}^n$ in the following senses:

• as a vector space? (Should be yes.)
• as an exterior algebra?

(Obviously they are not isomorphic as Clifford algebras unless our quadratic form is the zero quadratic form.)

Since the basis of the geometric algebra (as a vector space) is the same (or at least isomorphic to) the basis of the exterior algebra over $\mathbb{R}^n$, the answer seems to be yes. Also because the standard embedding of any geometric algebra over $\mathbb{R}^n$ into the tensor algebra over $\mathbb{R}^n$ always "piggybacks" on the embedding of the exterior algebra over $\mathbb{R}^n$, see this MathOverflow question.

2. Are differential forms the standard construction of an object satisfying the algebraic properties of the exterior algebra over $\mathbb{R}^n$?

3. Does the answers to 1. and 2. being yes imply that the part in yellow is true?

EDIT: It seems like the only problem might be that differential forms are covariant tensors, whereas I imagine that multivectors are generally assumed to be contravariant. However, distinguishing between co- and contravariant tensors is a standard issue in tensor analysis, so this doesn't really seem like an important issue to me.

Assuming that I am reading this correctly, it seems like the elementary construction of the geometric algebra with respect to the standard inner product over $\mathbb{R}^n$ given by Alan MacDonald here is exactly just the exterior algebra over $\mathbb{R}^n$ with inner product.

David Hestenes seems to try and explain some of this somewhat here and here, although I don't quite understand what he is getting at.

(Also his claim in the first document that matrix algebra is subsumed by geometric algebra seems completely false, since he only addresses those aspects which relate to alternating tensors.)

Answer 1

I just want to point out that GA can be used to make covariant multivectors (or differential forms) on $\mathbb R^n$ without forcing a metric onto it. In other words, the distinction between vectors and covectors (or between $\mathbb R^n$ and its dual) can be maintained.

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This is done with a pseudo-Euclidean space $\mathbb R^{n,n}$.

Take an orthonormal set of spacelike vectors $\{\sigma_i\}$ (which square to ${^+}1$) and timelike vectors $\{\tau_i\}$ (which square to ${^-}1$). Define null vectors

$$\Big\{\nu_i=\frac{\sigma_i+\tau_i}{\sqrt2}\Big\}$$

$$\Big\{\mu_i=\frac{\sigma_i-\tau_i}{\sqrt2}\Big\};$$

they're null because

$${\nu_i}^2=\frac{{\sigma_i}^2+2\sigma_i\cdot\tau_i+{\tau_i}^2}{2}=\frac{(1)+2(0)+({^-}1)}{2}=0$$

$${\mu_i}^2=\frac{{\sigma_i}^2-2\sigma_i\cdot\tau_i+{\tau_i}^2}{2}=\frac{(1)-2(0)+({^-}1)}{2}=0.$$

More generally,

$$\nu_i\cdot\nu_j=\frac{\sigma_i\cdot\sigma_j+\sigma_i\cdot\tau_j+\tau_i\cdot\sigma_j+\tau_i\cdot\tau_j}{2}=\frac{(\delta_{i,j})+0+0+({^-}\delta_{i,j})}{2}=0$$

and

$$\mu_i\cdot\mu_j=0.$$

So the spaces spanned by $\{\nu_i\}$ or $\{\mu_i\}$ each have degenerate quadratic forms. But the dot product between them is non-degenerate:

$$\nu_i\cdot\mu_i=\frac{\sigma_i\cdot\sigma_i-\sigma_i\cdot\tau_i+\tau_i\cdot\sigma_i-\tau_i\cdot\tau_i}{2}=\frac{(1)-0+0-({^-}1)}{2}=1$$

$$\nu_i\cdot\mu_j=\frac{\sigma_i\cdot\sigma_j-\sigma_i\cdot\tau_j+\tau_i\cdot\sigma_j-\tau_i\cdot\tau_j}{2}=\frac{(\delta_{i,j})-0+0-({^-}\delta_{i,j})}{2}=\delta_{i,j}$$

Of course, we could have just started with the definition that $\mu_i\cdot\nu_j=\delta_{i,j}=\nu_i\cdot\mu_j$, and $\nu_i\cdot\nu_j=0=\mu_i\cdot\mu_j$, instead of going through "spacetime".

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The space $V$ will be generated by $\{\nu_i\}$, and its dual $V^*$ by $\{\mu_i=\nu^i\}$. You can take the dot product of something in $V^*$ with something in $V$, which will be a differential 1-form. You can make contravariant multivectors from wedge products of things in $V$, and covariant multivectors from wedge products of things in $V^*$.

You can also take the wedge product of something in $V^*$ with something in $V$.

$$\mu_i\wedge\nu_i=\frac{\sigma_i\wedge\sigma_i+\sigma_i\wedge\tau_i-\tau_i\wedge\sigma_i-\tau_i\wedge\tau_i}{2}=\frac{0+\sigma_i\tau_i-\tau_i\sigma_i-0}{2}=\sigma_i\wedge\tau_i$$

$$\mu_i\wedge\nu_j=\frac{\sigma_i\sigma_j+\sigma_i\tau_j-\tau_i\sigma_j-\tau_i\tau_j}{2},\quad i\neq j$$

What does this mean? ...I suppose it could be a matrix (a mixed variance tensor)!

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A matrix can be defined as a bivector:

$$M = \sum_{i,j} M^i\!_j\;\nu_i\wedge\mu_j = \sum_{i,j} M^i\!_j\;\nu_i\wedge\nu^j$$

where each $M^i_j$ is a scalar. Note that $(\nu_i\wedge\mu_j)\neq{^-}(\nu_j\wedge\mu_i)$, so $M$ is not necessarily antisymmetric. The corresponding linear function $f:V\to V$ is (with $\cdot$ the "fat dot product")

$$f(x) = M\cdot x = \frac{Mx-xM}{2}$$

$$= \sum_{i,j} M^i_j(\nu_i\wedge\mu_j)\cdot\sum_k x^k\nu_k$$

$$= \sum_{i,j,k} M^i_jx^k\frac{\nu_i\mu_j-\mu_j\nu_i}{2}\cdot\nu_k$$

$$= \sum_{i,j,k} M^i_jx^k\frac{(\nu_i\mu_j)\nu_k-\nu_k(\nu_i\mu_j)-(\mu_j\nu_i)\nu_k+\nu_k(\mu_j\nu_i)}{4}$$

(the $\nu$'s anticommute because their dot product is zero:)

$$= \sum_{i,j,k} M^i_jx^k\frac{\nu_i\mu_j\nu_k+\nu_i\nu_k\mu_j+\mu_j\nu_k\nu_i+\nu_k\mu_j\nu_i}{4}$$

$$= \sum_{i,j,k} M^i_jx^k\frac{\nu_i(\mu_j\nu_k+\nu_k\mu_j)+(\mu_j\nu_k+\nu_k\mu_j)\nu_i}{4}$$

$$= \sum_{i,j,k} M^i_jx^k\frac{\nu_i(\mu_j\cdot\nu_k)+(\mu_j\cdot\nu_k)\nu_i}{2}$$

$$= \sum_{i,j,k} M^i_jx^k\frac{\nu_i(\delta_{j,k})+(\delta_{j,k})\nu_i}{2}$$

$$= \sum_{i,j,k} M^i_jx^k\big(\delta_{j,k}\nu_i\big)$$

$$= \sum_{i,j} M^i_jx^j\nu_i$$

This agrees with the conventional definition of matrix multiplication.

In fact, it even works for non-square matrices; the above calculations work the same if the $\nu_i$'s on the left in $M$ are basis vectors for a different space. A bonus is that it also works for a non-degenerate quadratic form; the calculations don't rely on ${\mu_i}^2=0$, nor ${\nu_i}^2=0$, but only on $\nu_i$ being orthogonal to $\nu_k$, and $\mu_j$ being reciprocal to $\nu_k$. So you could instead have $\mu_j$ (the right factors in $M$) be in the same space as $\nu_k$ (the generators of $x$), and $\nu_i$ (the left factors in $M$) in a different space. A downside is that it won't map a non-degenerate space to itself.

I admit that this is worse than the standard matrix algebra; the dot product is not invertible, nor associative. Still, it's good to have this connection between the different algebras. And it's interesting to think of a matrix as a bivector that "rotates" a vector through the dual space and back to a different point in the original space (or a new space).

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Speaking of matrix transformations, I should discuss the underlying principle for "contra/co variance": that the basis vectors may vary.

We want to be able to take any (invertible) linear transformation of the null space $V$, and expect that the opposite transformation applies to $V^*$. Arbitrary linear transformations of the external $\mathbb R^{n,n}$ will not preserve $V$; the transformed $\nu_i$ may not be null. It suffices to consider transformations that preserve the dot product on $\mathbb R^{n,n}$. One obvious type is the hyperbolic rotation

$$\sigma_1\mapsto\sigma_1\cosh\phi+\tau_1\sinh\phi={\sigma_1}'$$

$$\tau_1\mapsto\sigma_1\sinh\phi+\tau_1\cosh\phi={\tau_1}'$$

$$\sigma_2={\sigma_2}',\quad\sigma_3={\sigma_3}',\quad\cdots$$

$$\tau_2={\tau_2}',\quad\tau_3={\tau_3}',\quad\cdots$$

(or, more compactly, $x\mapsto\exp(-\sigma_1\tau_1\phi/2)x\exp(\sigma_1\tau_1\phi/2)$ ).

The induced transformation of the null vectors is

$${\nu_1}'=\frac{{\sigma_1}'+{\tau_1}'}{\sqrt2}=\exp(\phi)\nu_1$$

$${\mu_1}'=\frac{{\sigma_1}'-{\tau_1}'}{\sqrt2}=\exp(-\phi)\mu_1$$

$${\nu_2}'=\nu_2,\quad{\nu_3}'=\nu_3,\quad\cdots$$

$${\mu_2}'=\mu_2,\quad{\mu_3}'=\mu_3,\quad\cdots$$

The vector $\nu_1$ is multiplied by some positive number $e^\phi$, and the covector $\mu_1$ is divided by the same number. The dot product is still ${\mu_1}'\cdot{\nu_1}'=1$.

You can get a negative multiplier for $\nu_1$ simply by the inversion $\sigma_1\mapsto{^-}\sigma_1,\quad\tau_1\mapsto{^-}\tau_1$; this will also negate $\mu_1$. The result is that you can multiply $\nu_1$ by any non-zero Real number, and $\mu_1$ will be divided by the same number.

Of course, this only varies one basis vector in one direction. You could try to rotate the vectors, but a simple rotation in a $\sigma_i\sigma_j$ plane will mix $V$ and $V^*$ together. This problem is solved by an isoclinic rotation in $\sigma_i\sigma_j$ and $\tau_i\tau_j$, which causes the same rotation in $\nu_i\nu_j$ and $\mu_i\mu_j$ (while keeping them separate).

Combine these stretches, reflections, and rotations, and you can generate any invertible linear transformation on $V$, all while maintaining the degeneracy ${\nu_i}^2=0$ and the duality $\mu_i\cdot\nu_j=\delta_{i,j}$. This shows that $V$ and $V^*$ do have the correct "variance".

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See also Hestenes' Tutorial, page 5 ("Quadratic forms vs contractions").

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Answer 2

This seems to be best answered by Lounesto's paper "Marcel Riesz's Work on Clifford Algebras" (see here or here). In what follows:

$\bigwedge V=$ the exterior algebra over $V$
$C\ell(Q)=$ the Clifford (geometric) algebra over $V$ w.r.t. the quadratic form $Q$

Note in particular that we always have $C\ell(0)=\bigwedge V$, $0$ being the degenerate quadratic form.

On p. 221, Professor Lounesto discusses, given a non-degenerate quadratic form $Q$, how to define an "inner product" (contraction $\rfloor$) on the exterior algebra $\bigwedge V$.

On p. 223, Professor Lounesto discusses how to extend the inner product (by combining it with the wedge product of the exterior algebra) to produce a Clifford/geometric product on $\bigwedge V$, which makes $\bigwedge V$ isomorphic to $C\ell(Q)$ (the Clifford algebra w.r.t. the quadratic form $Q$).

One can also go the other way around, as M. Riesz originally did in 1958 (see section 1.3, beginning on p. 230, "Riesz's Introduction of an Exterior Product in $C\ell(Q)$ "), and use the Clifford product to define a notion of exterior product which makes $C\ell(Q)$ isomorphic to $\bigwedge V$.

In other words, we do indeed have:

(exterior algebra over $\mathbb{R}^n +$ inner product) $\simeq$ geometric algebra over $\mathbb{R}^n$

One should note that $\bigwedge \mathbb{R}^n$, the exterior algebra over $\mathbb{R}^n$, consists of alternating contravariant tensors of rank $k$ over $\mathbb{R}^n$. However, differential forms are alternating covariant tensors of rank $k$ over $\mathbb{R}^n$. So in general they behave differently.

Nevertheless, an inner product on $V$ gives a linear isomorphism between any vector space $V$ and its dual $V^*$ to argue that covariant tensors of rank $k$ and contravariant tensors of rank $k$ are "similar". (Mixed variance tensors complicate things somewhat further, but are not relevant to this question.)

Thus, differential forms are "similar" to $\bigwedge \mathbb{R}^n$ (since they are essentially $\bigwedge (\mathbb{R}^n)^*$). Also, we can just as easily construct a Clifford algebra from $\bigwedge V$ as from $\bigwedge V^*$, so we can extend differential forms to "covariant geometric algebras" by introducing an inner product based on a quadratic form $Q$.

So, perhaps less convincingly, we also do have (at least in an algebraic sense):

(differential forms + inner product) $\simeq$ "covariant geometric algebra" over $\mathbb{R}^n$

It is also worth noting that, according to Professor Lounesto on p. 218, Elie Cartan also studied Clifford algebras, in addition to introducing the modern notions of differential form and exterior derivative. So it is not all too surprising that they should actually be related to one another.

In fact, thinking about (covariant) geometric algebra in terms of "differential forms + inner product", while using the geometric intuition afforded by geometric algebra, actually makes the ideas behind differential forms much more clear. See for example here. I'm only beginning to process all of the implications, but as an example, a $k-$blade represents a $k-$dimensional subspace, and its Hodge dual is the differential form of rank $n-k$ which represents its orthogonal complement. The reason why orthogonal complements are represented in the dual space is because the inner product between two vectors can also be defined as the product of a vector and a covector (w.r.t. our choice of non-degenerate quadratic form $Q$).

All of this should be generalizable from $\mathbb{R}^n$ to the tangent and cotangent spaces of arbitrary smooth manifolds, unless I am missing something. This is especially the case for Riemannian manifolds, where we also get a non-degenerate quadratic form for each (co)tangent space for free.

(Which raises the question of why David Hestenes wants us to throw out smooth manifolds in favor of vector manifolds, a topic for future research.)

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As to the answer to "why use geometric algebra and not differential forms", for now my answer is:

Use the tensor algebras over $\mathbb{R}^n$ and $(\mathbb{R}^n)^*$, while appreciating the special properties of their exterior sub-algebras and remembering that, given our favorite quadratic form $Q$, we can always introduce an additional notion of "contraction" or "inner product" to make them into Clifford (geometric) algebras.

Hyping geometric algebra alone misses the importance of linear duals and arbitrary tensors. Likewise, focusing on differential forms alone seems like a good way to do differential geometry without geometric intuition (i.e. with a mathematical lobotomy). Sensible minds may disagree.

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Note: There are a lot of differences in the theory in the case that the base field is $\mathbb{F}_2$. To some extent we should expect this, since in that case we don't even have "alternating = anti-symmetric".

In particular, we don't have bivectors for fields of characteristic two, and defining/identifying a grading of the Clifford algebra via an isomorphism with the exterior algebra is impossible (at least if I am interpreting Professor Lounesto's paper correctly).

In any case, when I say "geometric algebra", I essentially mean "Clifford algebras of vector spaces with base field the real numbers", so the exceptions thrown up in the case that the characteristic equals 2 don't really matter for the purposes of this question; we are dealing exclusively with characteristic 0, although generalizations are possible.

Answer 3

$ \newcommand\lcontr{\mathbin\rfloor} \newcommand\lintr{\mathbin\lrcorner} $I suppose this is like your own answer you wrote, but perhaps more explicit.

1. Yes, and "yes". There is a canonical linear isomorphism, and the second "yes" is because the wedge product of geometric algebra is exactly the product you get by exploiting this isomorphism and using the exterior product.

2. No. "Differential forms" (over flat space $\mathbb R^n$) to me means ${\bigwedge}(\mathbb R^n)^*$ however you construct it, so the answer is no since the space of multivectors ${\bigwedge}\mathbb R^n$ (constructed by, say, constructing the tensor algebra and taking a quotient) is not the same thing. Canonically ${\bigwedge}\mathbb R^n\cong({\bigwedge}(\mathbb R^n)^*)^*$, i.e. the exterior algebra of $\mathbb R^n$ is the dual space of the space of differential forms. Another note: if you conceptualize ${\bigwedge}(\mathbb R^n)^*$ as the space of alternating covariant tensors, this is not quite right since this description fails if we replace $\mathbb R$ with a field with nonzero characteristic.

For (3), there is nothing in "yellow" and your links are dead, but I assume your referring to your block-quoted "rough isomorphism". The answer is essentially yes with caveats.

By "inner product" I assume you might really mean "symmetric bilinear form", since you acknowledge the possibility of multiple different quadratic forms when constructing the Clifford algebra; I will avoid the term "inner product" and just say "quadratic form". A nondegenerate quadratic form gives a canonical isomorphism $\mathbb R^n\cong(\mathbb R^n)^*$ and indeed ${\bigwedge}\mathbb R^n\cong{\bigwedge}(\mathbb R^n)^*$, so in this case we can just replace "differential forms" with multivectors which are more proper here anyway and what I will talk about from here on.

The main caveat is the degenerate case, where it really does matter whether or not you're talking about $\mathbb R^n$ or its dual. This is why plane-based geometric algebra uses hyperplanes as it's vectors—what that really means is that we're talking about a Clifford algebra over $(\mathbb R^n)^*$.

Finally, let me make a precise connection between a Clifford algebra and the exterior algebra by making (1) more precise and more general by connecting all Clifford algebras. Let $Q$ be a quadratic form, $\beta$ be any (not necessarily symmetric) bilinear form, and define $Q+\beta$ to be the quadratic form $x\mapsto Q(x)+\beta(x,x)$. Then there is a canonical linear isomorphism $$ \psi_{Q\beta}:\mathrm{Cl}(Q) \cong \mathrm{Cl}(Q+\beta) $$ called the Chevalley map, or the deformation of $\mathrm{Cl}(Q)$ by $\beta$. Furthermore, when the underlying field has characteristic not two we may choose $\beta$ to be symmetric; it is then uniquely determined by $Q$ and $Q+\beta$, so together with the case $Q=0$ we see that all (real) Clifford algebras are canonically isomorphic as vector spaces and are deformations of ${\bigwedge}\mathbb R^n$.

What do these deformations look like? You're probably familiar with the geometric algebra identity $$ vX = v\lcontr X + v\wedge X $$ where $v$ is a vector, $X$ arbitrary, and $\lcontr$ is the left contraction (which perhaps you know as the "dot product" or "inner product"). This is (essentially) the Chevalley map $\psi_{Q(-\beta)} = \psi_{0\beta}^{-1}$ with $\beta(x,y) = \frac12(Q(x+y)-Q(x)-Q(y))$. This can be made precise in the reverse direction as follows.

Over the exterior algebra there is an interior product $\lintr : (\mathbb R^n)^*\times{\bigwedge}\mathbb R^n\to{\bigwedge}\mathbb R^n$ "dual" to the one you are probably more familiar with; I feel no need to comment on it further. If $v$ is a vector then let $v^\beta$ be the covector $x\mapsto \beta(v,x)$. For any multivector $X\in{\bigwedge}\mathbb R^n$ $$ \psi_{0\beta}(v\wedge X) = v\psi_{0\beta}(X) - \psi_{\beta}(v^\beta\lintr X) $$ which is well-defined by induction on grade and the fact that when $X = v\wedge X'$ we have $$ \psi_{0\beta}(v\wedge(v\wedge X')) = v(v\psi_{0\beta}(X') - \psi_{0\beta}(v^\beta\lintr X')) - \beta(v,v)\psi_{0\beta}(X) + \psi_{0\beta}(v\wedge(v^\beta\lintr X')) = \psi_{0\beta}(v^\beta\lintr(v^\beta\lintr X')) = 0. $$

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I feel like I should also comment on mr_e_man's answer. His split Clifford $\mathrm{Cl}(\mathfrak W)$ algebra is the canonical Clifford algebra for the quadratic form $$ Q(\omega+v) = 2\omega(v) $$ over $\mathfrak W = (\mathbb R^n)^*\oplus\mathbb R^n$. You can indeed represent $\mathrm{GL}(n)$ in this algebra using bivectors and interior products. However, there is a much more Clifford algebraic way: for any $T\in\mathrm{GL}(n)$ note that $\omega+v\mapsto T^{-*}(\omega) + T(v)$ is an isometry of $Q$, where $T^{-*}(\omega) = \omega\circ T^{-1}$. Even better, $\det(T^{-*} + T)=1$ so these are "rotations".

Thus there is a double cover of $\mathrm{GL}(n)$ which is a subgroup of the Spin group of $\mathrm{Cl}(\frak W)$. (I seem to remember we can do even better and get $\mathrm{GL}(n)$ as a subgroup of $\mathrm{Spin}(\mathfrak W)$, but I am not recalling how and need to look into it.)

The relationship between this and the bivector representation is as far as I know an open question.

Answer 4

The most basic reason is that the "Geometric Algebra (GA)" developed and appropriated by David Hestenes and others is not just another name for Clifford Algebra because it integrates differential forms into a unified mathematical formalism that allows one to work with vectors, tensors, spinors and differential forms without having to deal with multiple, different notation systems and fragmentation of knowledge silo issues.

"The literature relating Clifford algebra to fiber bundles and differential forms is rapidly growing into a monstrous, muddled mountain. I hold that the muddle arises mainly from the convergence of mathematical traditions in domains where they are uncritically mixed by individuals who are not fully cognizant of their conceptual and historical roots. As I have noted before [1], the result is a highly redundant literature, with the same results appearing over and over again in different notational guises. The only way out of this muddle, I think, is to establish a consensus on the issues."

Source: https://davidhestenes.net/geocalc/pdf/DIF_FORM.pdf

Reference: 1. Hestenes, D.: 1986, ‘A Unified Language for Mathematics and Physics,’ Clifford Algebras and their Applications in Mathematical Physics, J.S.R. Chisholm/A.K. Common(eds.), Reidel, Dordrecht/Boston, pp. 1–23. https://davidhestenes.net/geocalc/pdf/UnifiedLang.pdf https://davidhestenes.net/geocalc/html/GeoCalc.html

"Integration with differential forms: The entire calculus of differential forms in GC [Geometric Calculus] is built around the vector derivative, so no distinct concept of “exterior derivative” is needed, and the integral theorems are manifestly complementary to the vector derivative." - The Genesis of Geometric Algebra: A Personal Retrospective, David Hestenes https://d-nb.info/1096286491/34

The problem David Hestenes was trying to address was that in Physics, the notation systems for vectors, matrices, tensors, quaternions, spinors and differential forms existed in disconnected knowledge silos and sometimes two or more different notations would appear in one Physics equation with overlapping geometric content.Geometric Algebra does subsume matrix notation because the GA notation system allows one to algebraically represent the same geometric content or system of equations as a matrix.

Over the last three decades, it has become necessary to note the existence of a disambiguation issue regarding use of the term "Geometric Algebra". It's true that "Geometric Algebra" has been used as another name for "Clifford Algebra" as can be seen in the name of the 1976 MIT course of Laszlo Tisza. However, in order to reduce confusion, some consideration should be given to allowing David Hestenes to appropriate the name "Geometric Algebra" exclusively for the unified mathematical system he designed for use in physics and mathematics that has Clifford Algebra at its core and no longer using the term "Geometric Algebra" as another name for Clifford Algebra. Then there is a third usage of the term "Geometric Algebra" by E. Artin and others that should be discouraged because of the confusion it causes that is similar to two businesses having the same name.Such confusion also occurs with use of the term "Modeling Instruction" by teachers.

"László Tisza was Professor of Physics Emeritus at MIT, where he began teaching in 1941. This online publication is a reproduction the original lecture notes for the course “Applied Geometric Algebra” taught by Professor Tisza in the Spring of 1976." https://ocw.mit.edu/courses/res-8-001-applied-geometric-algebra-spring-2009/

"6) What are the main advantages that geometric calculus has over differential forms? Should one learn differential forms if geometric calculus is superior?" "With only minor adjustments in notation, GC generalizes the standard theory of differential forms to include spinor and tensor fields. It reduces differential and integral calculus on manifolds to properties of a single differential operator D, a natural generalization of the Dirac operator already recognized as fundamental in Maxwell’s equation."

Reference: https://www.physicsforums.com/insights/interview-mathematician-david-hestenes/#6-What-are-the-main-advantages-that-geometric-calculus-has-over-differential-forms-Should-one-learn-differential-forms-if-geometric-calculus-is-superior