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The interpretation of vectors as partial derivatives of the position vector and dual vectors as stacks is visually very appealing. I understand that in the book Gravitation by Misner the passage from one leave of the stack of a covector is made even more intuitive by pretending to be associated to some sort of sound effect ("bongs of bell").

Incredibly, though, I couldn't find any good diagrams of how this would work out to visualize the coordinates of a tangent vector $\vec v$ (in red below) to a point $P$ (central point below) on a manifold $M$ (in turquoise), where $P$ is located through the positional vector $\vec R = x^1(y_1,y_2,y_3)\vec e_1+x^2(y_1,y_2,y_3)\vec e_2+x^3(y_1,y_2,y_3)\vec e_3$ on a generalized curvilinear coordinate system.

The vector $\vec v$ is NOT intended to belong to a vector field. It is a velocity vector tangent to the parameterized curve $f.$

Just dweling on one of the curvilinear coordinates in which only the $y_2$ component is allowed to change (in magenta) - $\vec R(c_1, y_2, c_3)$ - and with the stack (covector) represented as the orthogonal layers (also in magenta), I imagine that at point $P$ the tangent vector to a curve $f$ (in red) would be projected on one of the basis vectors, $\frac{\partial \vec R}{\partial y_2}\in T_pM$ (in black) as on the image to the left below.

In the event that there is vector field (or scalar field?) defined on the manifold, the number of layers of the covector (1-form) pierced by the projected vector $\vec v$ on $\frac{\partial \vec R}{\partial y_2},$ if read-out directly on the stack (covector $\mathrm dy_2 \in T^*_pM$) as portrayed in the diagram (right side picture) it would amount to $2$ - i.e. the projection onto the black curvilinear coordinate basis vector pierces through exactly $2$ layers. This 1-forms underpin the integration along the curve.

enter image description here

Is it correct to say that to introduce the 1-form (covector) a vector field on the manifold (or possibly a scalar field) needs to be defined (not drawn on the diagram)?

Is it true that there is no role for 1-forms if all that was needed is to calculate the arc length of the curve $f$ on these generalized curvilinear coordinates, which would instead call for concepts like Lamé coefficients?


The question remains unanswered, likely due to the imprecisions in its formulation. However I tend to think that the correct answer is likely (I look forward to confirmation or corrections):

In the case of integration to get the arc length, and since the differential distances, i.e. $dS= \sqrt{1 +\left(\frac{\partial \vec R}{\partial y_i}\right)^2}\mathrm dy,$ are 1-forms, the graphic correct (with recent modifications): the velocity vector $\vec v$ along the integration curve $f$ at point $P$ is part of the tangent space. The component along the curvilinear coordinate $\vec R(c_1,y_2,c_3)$ would give rise to the black vector - i.e. the projection of the vector $\vec v$ with its magnitude given by the number of magenta layers pierced orthogonal to the coordinate $\vec R(c_1,y_2,c_3)$.

If there was a scalar or vector field defined on the manifold there would be a gradient or a field vector at point $P.$ The integration along the curve $f$ would implicitly involve for the tangent vector $\vec v$ to the curve $f$ at point $P$ to be dotted with that field vector as in any other point along the curve:

$$\int_C \vec F \cdot \mathrm d\vec R =\int_a^b \vec F(\vec R(y_1(t),y_2(t),y_3(t))\cdot \vec R'(y_1(t),y_2(t),y_3(t))\mathrm dt.$$

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  • $\begingroup$ I don't understand what precisely your question is? I agree that you can visualize a one-form as the magenta "stack" you've drawn on the manifold (though it seems this visualization only really makes sense if your one-form is exact.) $\endgroup$ – user7530 May 5 '18 at 5:15
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I'm not sure I understand your question 100%, but: no, a one-form does not presuppose you also possess any particular vector field. You can visualize a one-form $\omega$ as a set of infinitesimal magenta "stacks," as you've drawn, at every point on $M$. However you cannot glue together these infinitesimal "stacks" together into a global "front" crossing $M$ unless $\omega$ is exact.

$\omega$ exists independently of any vector field, but of course $\omega$ is intimately related to vector fields in the sense that if $v$ is a vector field on $M$, you can compute $\omega(v)\in\mathbb{R}$, and the answer is bilinear in $\omega$ and $v$ (you can think of the resulting number as the integral over the surface of the number of stacks of $\omega(p)$ the vector $v(p)$ pierces at each point $p$).

If you have a metric on the surface, you can also convert vector fields and one-forms to each other with the musical isomorphisms, i.e. you can define a vector field $\omega^\sharp$ by the relation $$\langle \omega^\sharp, v\rangle = \omega(v)$$ for all vector fields $v$. Visually: for every magenta "stack" you can find a magenta vector so that the inner product of the magenta vector and red vector measures the number of stacks pierced by the red vector.

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