Angle between one-form and vector

Question

How do you compute the angle between a vector and one-form? Does it make sense to even talk about it since inner product between the arguments (vector and one-form) doesn’t require a metric? If so, how do you show that such angle is not computable in mathematical sense?

Edit: Can I assume one-form works exactly like a surface (a plane), which then I could compute the angle between a line and a surface by: $$\theta=\arcsin\left(\frac{\langle \textbf{N},\widetilde{dx}\rangle}{|\textbf{N}||\widetilde{dx}|}\right)$$ where $N$ is a vector and $\widetilde{dx}$ is a one-form.

(Note that I used arcsin rather than arccos considering this case: E.g: Say I have a vector $V$ in the $y$-direction and a one-form X which represents surfaces of constant x . Obviously, the vector V is tangent to the one-form X , therefore by applying inner product I get 0 since there’s no intersection. However that doesn’t make sense because you would obtain θ=90 if you used arccos, which contradicts the fact that the one-form and vector is tangent to each other. And this is just an analogy to the equation for the angle between a line and a plane.)

But since the inner product between vector and one form doesn't require a metric tensor, does it make sense to define angle as how I did above? A more realistic situation is: given there's a flow of river water across a control surface (in the context of fluid mechanics) in $x$-direction, is it necessary if the flow is in x-direction?

Answer 1

Note that while the inner product of vector and one-form indeed doesn't need a metric, your formula actually does, in order to calculate the norms in the denominator.

Note that as soon as you do have a metric, you can use it to associate vectors and one-forms, associating the vector $v$ with the one-form mapping other vectors to the inner product of those vectors with $v$.

And then you can e.g. ask what is the angle between the vector associated to the one-form through the metric, and another vector. Which turns to be given by your formula with arcsin replaced by arccos.

What remains is the question what this angle means. Remember, the one-form is nothing else but a linear function on the vector space. The vector associated to it through the metric points in the direction of steepest ascent. So the formula just tells you the vector's angle relative to the direction of steepest ascend of the one-form.

Now your formula as is also makes sense, and indeed tells the vector's angle to the one-form's equipotential surface, which indeed is a (hyper-)plane. But the arccos variant is more in line with the angle definitions between two vectors or two one-forms.

Note that I'm throughout assuming a real vector space; for complex vector spaces, I don't think the angle definition would make sense.