Are dual vectors not intrinsic to the manifold?

Question

I'm saying this based on their transformation. Say, we change the co-ordinate chart of a manifold according to $x'=f(x,y)$, $y'=g(x,y)$. Let $A$ be the Jacobian matrix of this transformation.

Vectors transform as:

$$v'=A^{-1}v$$

This looks like a passive transformation. As if we're trying to describe some abstract entity, attached to the manifold, after a change of basis

Dual-vectors transform as:

$$v'^{*}=Av^{*}$$

This looks like an active transformation. As if dual vectors are entities attached to the co-ordinate system instead of the manifold. For example, if $A$ is a rotation, the dual vectors rotate exactly the same as the co-ordinates.

I'm visualising an abstract manifold with vectors attached on it (looking like squishy vomit). Brushing against it is a co-ordinate chart ($R^n$), with dual vectors attached on it. Each abstract point on the manifold is touching a point on $R^n$ according to the chart. When we change the chart, the $R^n$ space transforms, dragging the dual-vectoes along with it. The manifold stays resting with its vectors. (This visualisation requires a second chart: The chart mapping all the points to the background of the visualisation (say, a computer screen))

Am I wrong?

Answer 1

I think you will be less confused if you consider a map $\phi:M\rightarrow N$ between two distinct manifolds, understand what's going on in that more general context, and then specialize to the case $M=N$.

Pick a point $m\in M$ and put $n=\phi(m)$.

Let $df$ be a cotangent vector at $n$. Then we get an associated cotangent vector $\phi^*(df) = d(f\circ\phi)$ at $m$.

Let $v$ be a tangent vector at $m$ (so that $v$ acts on cotangent vectors). Then we get an associated tangent vector $\phi_*(v)$ at $n$, defined by $$\phi_*(v)(df)=v(\phi^*(df))$$

So the cotangent vectors get pulled back from $N$ to $M$ and the tangent vectors get pushed forward from $M$ to $N$.

Again, this can be confusing to think about if you start with the case $M=N$ as you've done. But once you understand the general case, the special case should be clearer.