Related: math.stackexchange.com/q/1184671/169085

Do you realize that your metric is a symmetric form and your symplectic structure is a differential form? What are you planning to integrate if you refuse to use differential forms? What your Stokes' theorem will be like without forms? What your curvature form for vector bundles will be?....

It’s not wrong to be skeptical. My suggestion as you learn everything carefully, is to try to reformulate everything without using forms. Sometimes you’ll like your version better. But if not, this is a good way to really appreciate the power of using the dual vector space and differential forms

There is no exterior derivative in the land of vector fields.

@TedShifrin For scalar-valued functions the gradient and exterior derivative are isomorphic, no? Why can't you just define the wedge product and exterior derivative on vectors? This has a clear visualization for me in terms of constructing n+1-dimensional parallelotopes out of lower dimensional ones and in some sense generalizing the gradient.

I’m a big fan of multivectors, but even a bigger fan of Stokes’s Theorem and its power with connections and curvature. The gradient, by the way, is only a vector field once you introduce a Riemannian metric. You fall way short.

@TedShifrin You're saying that the proof for the G. Stokes Theorem can't be written in terms of multivectors and some abstract binary operation on them? If I go through Lee or Tu's book (maybe even Rudin) I won't be able to write or prove it without invoking forms? There is no binary operation that can be used to derive prove the same result?

Correct. You can’t even get off the ground. You can’t define the exterior derivative, and integration over submanifolds will need measure theory, but still you’ll get stuck with the analogues of flux.

@TedShifrin Is there already some reference you know of that shows this not working? Else I'll go through Lee/Tu/Rudin and see if I can't do it myself.

@TedShifrin I read through Lee's book for several hours and saw nothing that prevents the construction for exterior derivative. On the other hand (re-reading page 274): "Because of Proposition 11.8, the real number $\omega(v)$ obtained by applying a covector $\omega$ to a vector $v$ is sometimes denoted by either of the more symmetric-looking notations and $\langle \omega, v \rangle$ both expressions can be thought of either as the action of the covector $\omega \in V^*$ on the vector $v\in V^*$ or the action of the linear functional $\xi(v)\in V^{**}$" (continued)...

@TedShifrin Yeah, the choice of isomorphism to the dual space is arbitrary. But it's still strange that I can't take a family of isomorphisms $\{\iota\}_i:V \rightarrow V^*$ and induce a family of bilinear pairings $\langle v, w\rangle_i = \langle \iota_i(v) , w \rangle$ on multivectors. I never called this the inner product (I never specified that requirement). Does the lack of a natural isomorphism throw off any expectation of symmetry (conjugate or otherwise)? I still don't see the problem after reading.

@TedShifrin Either way I'm out of time to do this. If you or someone can explicitly screw it up by showing that every vector in the goes into the domain of a single form (so it's hard to define a good isomorphism, that you can't get an exterior derivative on multivectors, or something like flux falls apart, then I'll mark it as an answer.

I'm going disagree with @TedShifrin. It is possible to do multilinear algebra and differential geometry without explicitly using the concept of a dual vector space or differential forms or the exterior derivative. Many physicists still do. It's not so bad in dimensions 2 and 3, because you can use the cross product and curl. But the calculations are messy and hard to understand This is particularly true if there is no Riemannian metric. Differential geometers didn't adopt this fancy stuff for nothing. It both simplifies calculations and clarifies the geometry.

What turns out to be particularly important is to distinguish which aspects of differential geometry do not require a Riemannian metric and which do, all while keeping track of how things depend on coordinates. Physicists in fact understand this. They know that the cross product of two vector fields is not an honest vector field. So they call it a pseudovector field (en.wikipedia.org/wiki/Pseudovector).

Here's another way to express this: The concepts and theorems of geometry and physics should not depend on the coordinates or their infinitesimal analogues (namely a basis of the tangent space). If you choose a pointwise isomorphism between the tangent and cotangent space, then it does induce isomorphisms between exterior multivectors and differential forms. But in the end, whatever you do, you'll have to prove that your final result is independent of this isomorphism.

@Deane I admit that my patience with this entire concept is wearing thin. I talk to several geometers (beyond this chat) and I keep getting conflicting/different statements from different people. What does the covector do in particular that makes it coordinate or metric independent? What is meant rigorously by metric and coordinate independence? I found no precise definition in Lee's book. Is there some hidden canonical property or equivalence class?. Can you give a very concrete and simple example/answer, with no abuse of notation or weasel words? I'll mark your precise example as the answer.

It seems to me a lot of your arguments above, while reasonable, imply a dislike for duality on the whole. Restricting ourselves to the simplest case of vector spaces and their duals, I feel like a lot of intuition for why one studies cotangent bundles can come just from doing some explicit calculations in linear algebra concerning dual spaces. In my experience, the isomorphisms tend to get rather messy in coordinates, and that motivates separating the concepts for me. Additionally, I'd point out a cotangent space is sometimes more algebraic, being $I/I^2$ (here the tangent space is dual! )

@BrevanEllefsen Could you provide me links or reference to such exercises in linear algebra concerning dual spaces? Reference to problem sets, practice exams, old exams, or textbook problems? I'd like to see a kind of concrete example that shows me concretely some of the utility of dual spaces using relatively elementary linear algebra. I couldn't find such problems within the books I was reading and practice exams I used to grind the material.

The first thing that comes to mind is my answer here, specifically the calculation near the end. Beyond that, simple calculations would be something like considering a varying inner product (e.g. take the dot product plus a varying term on $\mathbb R^n$) and calculating the reciprocal basis in this changing coordinate system. One can then do the same with various non-positive definite metrics. I am constantly changing metrics when doing geometry (I am a low dim topologist), so I find comfort in knowing duality lets me avoid such computations.

I'm sure you know this, but when I teach linear algebra I often stress to my students the importance of duality by speaking of the transpose map. It's very arcane at first, until one realizes it's just the dual map in a specific coordinate system. To do that, I lead my students through the commutative diagram formed by taking $f: V \to W$ along the top and $f^*: W^* \to V^*$ along the bottom, with vertical maps being isomorphisms. One shows how different inner products yield different adjoints, which allow identification of dual map with adjoint map (or tranpose in standard coords).