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.)
This seems to be best answered by Lounesto's paper "Marcel Riesz's Work on Clifford Algebras" (see here or here). In what follows:
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:
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):
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.)
As to the answer to "why use geometric algebra and not differential forms", for now my answer is:
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.
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.