Is 4-momentum a vector or a 1-form? This is a follow-on to https://physics.stackexchange.com/a/576885/117014.  If we should not consider a vector and its "canonically" dual 1-form to represent the same object, then it seems that we should be able to say whether 4-momentum, for example, is a vector or a 1-form.  I could ask the same thing about the electromagnetic field tensor, but momentum seems to be a good place to start.  I tend to think of force as a 1-form because in pure classical mechanics it is the negative of the gradient of a potential (Ask Susskind.)  Since force is also the time derivative of momentum, it seems reasonable to consider momentum to be a 1-form.  But, we can also think of momentum as the product of mass with 4-velocity, which is a 4-vector.
So if a vector and its dual 1-form are not to be viewed as different representations of the same object, momentum must be one or the other.  Either that, or we have two distinct geometric objects representing momentum.
Which is it?
 A: To me, the most natural definition of momentum is via the Lagrangian formalism, which yields the one-form $p_\mu = \frac{\partial L}{\partial \dot x^\mu}$.  Taking the standard Lagrangian
$$L(x, \dot x) = m\sqrt{g_{\mu\nu} \dot x^\mu \dot x^\nu}$$
(where differentiation is taken with respect to the proper time), it then follows that $p_\mu = g_{\mu\nu} m\dot x^\nu$.  That being said, this is clearly the brother of the 4-vector $\tilde{p}^\mu = m \dot x^\mu$, with the index raised/lowered via the metric.
From the Lagrangian standpoint, if we add a potential energy term then the Lagrangian equations of motion take the form
$$\frac{d}{d\tau} p_\mu = -\frac{\partial U}{\partial x^\mu} \equiv f_\mu$$
so as you say, from this perspective force is naturally a one-form.  But again, the metric provides us with an isomorphism, so solving
$$\frac{d}{dt}p_\mu = f_\mu$$
and
$$\frac{d}{d\tau} \tilde{p}^\mu =g^{\mu\nu} f_\nu \equiv \tilde{f}^\mu$$
are ultimately equivalent.


If we consider the mythical classical point mass, it has a 4-momentum determined by its mass and its world-line. I call such a thing a "priority" object. It exists prior to any manifold parameterization, or metric (or observation).

Okay, that's fine.  You're talking about $p^\mu = m \dot x^\mu$.  This expression is perfectly well-defined with no additional structure needed.

Whether we express it covariantly or contravariantly, the expression refers to the same physical entity.

Without a metric (or some other structure which provides a similar isomorphism), you can't "express it covariantly."  The momentum you referred to before is well-defined on its own, but you can't map it to a covector without implicitly making a choice of metric (or other index-lowering map).

I typically write momentum covariantly. But I don't have an ontological argument to consider that to be an inherent property of momentum.

For that, you'll need to be more specific about what you mean by momentum.  If you're talking about the mass times the 4-velocity, that is a 4-vector.  If you're talking about the canonical momentum which is conjugate to position in the Lagrangian or Hamiltonian pictures, and whose spatial components (i) act as the infinitesimal generators of spatial translations, and (ii) are conserved in the presence of spatial translation symmetry, then that object is a covector.
As a concrete example, consider the flat space FLRW spacetime in which
$$ds^2= c^2dt^2 - a^2(t)\big(dx^2+dy^2+dz^2\big)$$
This metric is homogeneous and isotropic, which implies 3-momentum conservation.  However, it is not $p^k = m \dot x^k,\ k=1,2,3$ which is conserved, but rather $p_k = -a^2(t)m\dot x^k$.
