1. It is not clear to me that linear duals, and not just Hodge duals, can be represented in geometric algebra at all. See, for example, here.
Can linear duals (i.e. linear functionals) be represented using the geometric algebra formalism?
2. It also seems like most tensors cannot be represented, see for example here. This makes intuitive sense since any geometric algebra is a quotient of the corresponding tensor algebra. Also it seems like only some contravariant tensors (and no covariant tensors whatsoever) can be represented unless the answer to 1. is no.
Which types of tensors admit a representation using geometric algebra?
3. The exterior algebra under which differential forms operate can clearly be represented by geometric algebra and its outer product.
However, do objects "sufficiently isomorphic" to differential forms admit a representation in geometric algebra?
This would be preferred given how geometric algebra is more geometrically intuitive than differential forms. See also here.
4. This question seemingly depends on the answers to 1. and 3., since derivations=vector fields are the linear dual of differential forms.
Can vector fields=derivations be represented using geometric algebra?
This paper seems to suggest that the answer is yes, although it was unclear to me. It also listed as references Snygg's and Hestene's books for representing derivations=vector fields via geometric algebra. However, I quickly searched Snygg's book and could not even find the use of the word "bundle" once, which seems to cast doubt on the claim.
Moreover, derivations are just the Lie algebra of smooth functions between manifolds, correct? Since Lie algebras are non-associative, it seems doubtful to me that derivations could be represented effectively by the associative geometric algebra. On the other hand, quaternions are somehow also Lie algebras, and they can be represented in geometric algebra, so I am not sure.
5. This probably a duplicate of 4. but I am asking it anyway.
Do tangent/cotangent spaces/bundles admit a representation using geometric algebra?
This one is especially unclear to me, since using "ctrl-f" the word "bundle" is not used even once in Snygg's book "Differential Geometry via Geometric Algebra", which appears to be the most thorough treatment of the subject.
(Incidentally, the word "dual" also only appears once, in reference to Pyotr Kapitza's dual British and Russian citizenship.)
Basically I am wondering if differential geometry can be "translated" completely using the language of geometric algebra. I think the answer is no because Hestene's conjecture regarding smooth and vector manifolds has yet to be proved (see the comments here), but it seems like we would run up with barriers even sooner than that. Although I probably am misunderstanding the comment.
I have found differential geometry difficult to understand at times, and would like to learn it by translating it as much into geometric algebra and then back. The extent to which the two is "equivalent" obviously presents a barrier to how much this is possible. Still, I already feel like I understand the concepts and motivations of multilinear algebra and related fields much better after having just learned a little geometric algebra, and would like to apply this as much as possible to the rest of differential geometry.
These questions are also related: symmetric products are the inner product from geometric algebra, and wedge products are the outer product from geometric algebra; geometric algebra is a special type of Clifford algebra which contains the exterior algebra over the reals; and this question discusses derivations in algebras in detail.
On p.4 of this document we can see that multivectors over a given Euclidean space $\mathbb{R}^n$ do not have arbitrarily high grade/order; instead the highest order possible is $n$ (the determinant/volume element). This is because of the two products available in geometric algebra, the inner (symmetric) algebra and the outer (exterior) algebra, the inner product is grade-reducing, while the outer product is always degenerate for $k>n$.
However, general tensors can have arbitrarily high grade, so it therefore follows that the tensor algebra on $\mathbb{R}^n$ cannot be embedded into the geometric algebra on the same space. And algebraically we would have expected this result anyway, since as a Clifford algebra on $\mathbb{R}^n$, the geometric algebra is isomorphic to a quotient of the tensor algebra of Euclidean space. From this is doesn't necessarily follow that the cardinality is strictly lesser and thus that no injection exists, but from this fact alone it would seem implausible that any multilinear embedding could exist, and indeed, as the first paragraph of this answer shows, it is impossible.
For grades $1-n$, it seems clear to me that the tensors which are elements of the symmetric and exterior algebras on $\mathbb{R}^n$ should all be representable in the geometric algebra, and those tensors which can be represented as linear combinations or geometric products of those elements, but otherwise I don't believe (although I am not 100% certain) that any other elements of the tensor algebra could exist inside of the geometric algebra.
Here is a similar question on Math.SE: what's the relationship of tensor and multivector. Basically all multivectors are tensors, but not all tensors are multivectors. These documents (1) (2) outline how to represent multivectors of the geometric algebra for $\mathbb{R}^3$ and the regular inner product as tensors.
Differential forms are tensors on $\mathbb{R}^n$, but not just any tensors, they are alternating tensors, i.e. the exterior algebra on $\mathbb{R}^n$. Since the geometric algebra contains the exterior algebra as a sub-algebra, it follows that geometric algebra completely subsumes differential forms (perhaps not their vector space duals, but most certainly differential forms themselves). A detailed explanation of how that can be done is to be found here.
The dual operation for geometric algebras is equivalent to what is called the Hodge dual in the language of differential forms. See page 13 of this document. In particular, like the simpler exterior algebra on $\mathbb{R}^n$, the geometric algebra is not an involutive algebra under its dual operation, in contrast to the vector space/tensor dual on the corresponding tensor algebra of $\mathbb{R}^n$. See here.
However, the tensor algebra on $\mathbb{R}^n$ is involutive with respect to its dual operation (transposition, the vector space dual operation). Thus, the Hodge/geometric algebra dual and the vector space dual are clearly not equivalent.
Since the geometric algebra is isomorphic to a subset of the tensor algebra, it should be possible to define dual (transpose/adjoint) objects for all of its elements in the vector space sense. However, I believe that we should only expect these dual/transpose/adjoint objects to exist inside the geometric algebra for even-grade bivectors, or at least for all dimensions greater than or equal to two. This is for at least two reasons.
1: Column vectors (tensors/multivectors of rank 1) are clearly not closed under transposition, whereas square matrices (tensors/multivectors of rank 2) clearly are. I am not familiar enough with higher order tensors to make any definitive statements about them.
2: The spinor algebra on $\mathbb{R}^n$, which is a sub-algebra of the geometric algebra consisting of all multivectors of even grade, appears to be closed under transposition (i.e. vector space duals), at least as far as I can tell. See here.