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.
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".
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)!
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).
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".
See also Hestenes' Tutorial, page 5 ("Quadratic forms vs contractions").
$\newcommand{\Cl}{\mathscr{Cl}(V)}$The problem with your definition of the wedge product in $\Cl$ is that it is an $n$-ary product of vectors. And so it's not clear to me how one would even discuss associativity of it (i.e. it doesn't allow one to distinguish between the product of a bivector and a vector vs the product of a vector and a bivector). So while $\Cl$ with the wedge product is an algebra -- it isn't an associative algebra like the exterior algebra is. Let me instead recommend a different definition based on grade projection (basically the same definition given in the works of Alan Macdonald):
To define the wedge product, we first define it in the case of single-grade elements. Let $A,B$ be an $r$-vector and $s$-vector, respectively. Then
$$A\wedge B := \langle AB\rangle_{r+s}$$
Now using this special case, we define the wedge product for all multivectors. Let $C,D$ be (potentially mixed grade) multivectors in $\Cl$. Then we define
$$C \wedge D := \sum_{j,k} \langle C\rangle_j \wedge \langle D\rangle_k$$
Proof that for all $A,B,C\in \Cl$, we have $A\wedge (B\wedge C) = (A\wedge B)\wedge C$:
Consider a $j$-vector $A$, $k$-vector $B$, and $l$-vector $C$. Then
$$(A\wedge B)\wedge C = \langle AB\rangle_{j+k}\wedge C = \langle \langle AB\rangle_{j+k}C\rangle_{j+k+l} = \langle (AB)C\rangle_{j+k+l}$$
where that last step holds because only the \langle AB\rangle_{j+k} part of $AB$ can contribute to $\langle (AB)C\rangle_{j+k+l}$.$^\dagger$.
Then $$(A\wedge B)\wedge C = \langle (AB)C\rangle_{j+k+l} = \langle A(BC)\rangle_{j+k+l} = A\wedge\langle BC\rangle_{k+l} = A\wedge(B\wedge C)$$
To finish the proof by allowing $A,B,C$ to be any multivectors in $\Cl$, just note that our definition of the wedge product is bilinear.$^\ddagger$$\ \ \ \ \square$
Now we'll show that with this definition, your proposition holds. Take $\wedge_C$ to be the wedge product in $\Cl$ and $\wedge_T$ to be the wedge product in $\Lambda V$ (i.e. the coproduct of the antisymmetric tensor powers of $V$).
Let $\{e_1, \dots, e_n\}$ be an orthogonal basis for $V$ wrt the symmetric bilinear form and construct bases for $\Cl$ and $\Lambda V$ in the obvious way from it. Let $\Gamma: \Cl \to \Lambda V$ be given by $$\Gamma(A_0 + A_1e_1 + A_2e_2 + \cdots + A_{12\cdots n}e_1\wedge_C e_2\wedge_C \cdots \wedge_C e_n) = A_0 + A_1e_1 + A_2e_2 + \cdots + A_{12\cdots n}e_1\wedge_T e_2\wedge_T \cdots \wedge_T e_n$$
By definition this is linear. It's also clearly invertible. To see that it's multiplicative, it suffices (due to linearity of $\Gamma$ and bilinearity of $\wedge_C$) to show that $\Gamma$ is multiplicative on blades. Let $A=A_{i_1\cdots i_p}e_{i_1}\wedge_C\cdots\wedge_C e_{i_p}$ be a $p$-blade and $B=B_{j_1\cdots j_r}e_{j_1}\wedge_C\cdots\wedge_C e_{j_r}$ an $r$-vector. Then $$\begin{align}\Gamma(A\wedge_C B) &= \Gamma(A_{i_1\cdots i_p}e_{i_1}\wedge_C\cdots\wedge_C e_{i_p}\wedge_CB_{j_1\cdots j_r}e_{j_1}\wedge_C\cdots\wedge_C e_{j_r}) \\ &= A_{i_1\cdots i_p}e_{i_1}\wedge_T\cdots\wedge_T e_{i_p}\wedge_T(B_{j_1\cdots j_r}e_{j_1})\wedge_T\cdots\wedge_T e_{j_r} \\ &= (A_{i_1\cdots i_p}e_{i_1}\wedge_T\cdots\wedge_T e_{i_p})\wedge_T(B_{j_1\cdots j_r}e_{j_1}\wedge_T\cdots\wedge_T e_{j_r}) \\ &=\Gamma(A_{i_1\cdots i_p}e_{i_1}\wedge_C\cdots\wedge_C e_{i_p})\wedge_T\Gamma(B_{j_1\cdots j_r}e_{j_1}\wedge_C\cdots\wedge_C e_{j_r}) \\ &= \Gamma(A)\wedge_T\Gamma(B)\end{align}$$
A linear, invertible, multiplicative map is by definition an isomorphism of algebras. Hence $\Cl$ is isomorphic to $\Lambda V$, as desired.
Another Interesting Tidbit:
$\Lambda V$ is actually (or at least can be) defined by a set of axioms like other types of vector spaces.
From nlab:
Suppose $V$ is a vector space over a field $K$. Then the exterior algebra $\Lambda V$ is generated by the elements of $V$ using these operations:
- addition and scalar multiplication
- an associative binary operation $\wedge$ called the exterior product or wedge product,
subject to these identities:
- the identities necessary for $\Lambda V$ to be an associative algebra
- the identity $v\wedge v = 0$ for all $v\in V$.
It is easily confirmed that $\Cl$ satisfies these properties under the definition of $\wedge$ given above (and thus why associativity is so important for $\wedge$). Thus, not only can we construct an isomorphism, but we can even consider $\Cl$ to just be $\Lambda V$ with extra structure (the Clifford product).
$\dagger:$ To prove this, note that it is known that if $\Bbb F$ is not characteristic $2$, then $V$ has an orthogonal basis wrt any symmetric bilinear form. Then you can construct a "standard" basis for $\Cl$ and decompose your multivectors wrt to it.
$\ddagger:$ By the linearity of the grade projection operator.
Best Answer
Part 1
The differential forms approach is indeed very powerful. What Hestenes points out in his From Clifford Algebra to Geometric Calculus is that to give a complete treatment of differential geometry of manifolds you need various structures. In the book, you will find an alternative. The starting point (as was pointed out above) is the notion of a vector manifold.
A vector manifold is an abstract set contained within an infinite dimensional abstract algebra. This can be given different interpretations. There are special elements of this algebra called vectors and others called pseudoscalars. The set of elements that define a vector manifold is a set of vectors. These vectors generate tangent space that are not part of the set. This is customary: one often invokes tangent spaces on manifolds. Notice that the vectors which define the vector manifold and the vectors which form the tangent space are very different. For visualization purposes, the elements of a vector manifold are called "points." The space tangent to a point generates a tangent algebra; this algebra contains ONE element called its pseudoscalar.
The function $I_m(x)$, which is pseudoscalar-valued and takes as arguments "points" of the vector manifold, characterizes the vector manifold. If this function is differentiable then one says that the vector manifold is differentiable. If so all the differential geometry of a vector manifold can be carried out using only $I_m(x)$. Here, $m$ is the dimension of the tangent spaces, this also defines the dimension of the vector manifold.
So the idea behind the definition is the following. If you have a manifold, its points lack algebraic structure, and so one imposes other structures as needed. If, on the other hand, you start with a vector manifold which is algebraically rich, then no further structures are needed later on. A manifold can be defined as a space which is isomorphic to a vector manifold. This isomorphism can be thought of as a mere stripping of the algebraic structure. A manifold can be treated as a completely abstract object, but for the purposes for which it was constructed, a geometrical interpretation is rewarding. This is also true for vector manifolds.
Some may think that vector manifolds are embedded in a Euclidean space or that one needs a metric, but this is not true. The main reason could be (this is my opinion) the nomenclature of GA. For example a vector manifold sounds like a special kind of manifold. Another case is the inner product. This product is defined algebraically and does not need a metric. It can be interpreted metrically and thus lead to very important developments, but it is only an interpretation. I use "only" in the sense that a metric does not define the abstract algebra or anything about it.
Part 2
Differential forms can live in the GA framework as follows. There is a unique $k$-vector for every $k$-form. A one-to-one correspondence.
Take the following example. Suppose you want to manipulate an expression before integrating. You do so and right before you integrate, you "multiply" by a differential. Then you integrate. You can "multiply" by the differential from the very beginning, but you do not need to do so and would probably not do so. The difference between a $k$-vector and a $k$-form is somewhat similar. If you already know what a $k$-form is, you could think of a $k$-vector as a $k$-form without the differential. If you don't know what a $k$-form is, you can study $k$-vectors and some of their mathematics and right before you need to integrate a $k$-vector, you turn it into a $k$-form. (I point this out because it is more or less how Hestenes proceeds in the book I mentioned above: forms are only strictly needed when integration is to be carried out.)
What I am pictorially trying to answer with the above example is "are these frameworks strictly equivalent?" When scalar-valued integrals are considered, the answer is YES. GA can also handle vector-valued integrals or in general multivector integrals. And integrals are Riemannian! Outside of integration theory, GA is more general. For example, the exterior derivative of differential forms has a counterpart in GA. Both are grade raising. GA also has a grade lowering derivative, but in differential forms, this can only be achieved with a metric. When comparing the two, one might get confused and think GA also needs a metric for this. This is no true. Furthermore, the sum of these grade raising and lowering operators is in actuality the fundamental derivative of GA this derivative is invertible while the exterior derivative is in general not.
The overall answer to your question is geometric algebra (technically geometric calculus) is more general.
Comment. There are also other alternatives to geometric calculus which do not use the concept of a vector manifold. They are also more general by treating any manifold and use Clifford algebra structures and do not use coordinates. Some other approaches exist which use coordinates here and there but also use this algebra structure and are more general.