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Suppose a tangent vector field is given on a planar curve and one asks the following question: What is the condition on this tangent vector field that it comes from the ordinary Euclidean projection of a constant vector field defined on a planar 2 dimensional neighbourhood of this curve? A simple necessary and sufficient condition is if one parallelly translates, in the usual Euclidean sense, each tangential vector of the given field to a common point on the plane, the tips of the translated vectors describe a circular arc. This can be easily proven by elementary Euclidean geometry.

My question is if it can be seen in a more matured way, I mean, in terms of parallelism, connection, covariant derivatives, in these languages. Specifically, what I am looking for is some specially defined connection with respect to which the covariant derivative of the given tangential field being zero implies, and is implied by, the field be a projection of a constant field in the embedding space.

Sorry for this late amendments which I should have pointed out earlier. There are the following general versions of this problem:

Generalization to Surfaces in $\mathbb{R}^3$

Suppose we have an oriented smooth surface $S$ embedded in $\mathbb{R}^3$ and a sufficiently smooth tangent vector field $\mathbf{w}$ defined over $S$. Then it can be shown, using elementary analysis, that the vector field $\mathbf{w}$ comes from the Euclidean projection of a constant vector field $\mathbf{C}$ defined over a tubular neighbourhood of $S$ in $\mathbb{R}^3$, if, and only if, the surface described by the vectors in the given field $\mathbf{w}$, when translated parallely to the origin (let's take the common point as the origin), is part of a 2-sphere in $\mathbb{R}^3$ that passes through the origin. In fact, $\mathbf{w}$ satisfies the following equation:

$$(\mathbf{w}-\frac{\mathbf{C}}{2})\cdot(\mathbf{w}-\frac{\mathbf{C}}{2})=\frac{\mathbf{C}\cdot\mathbf{C}}{4}.$$

The most general version

Let $S$ be a $p$-dimensional sufficiently smooth submanifold embedded in $\mathbb{R}^n$ and $\mathbf{w}$ be a tangential vector field defined on $S$. Then, $\mathbf{w}$ comes from the Euclidean projection of a constant vector field $\mathbf{C}$ defined on a neighbourhood of $S$ in $\mathbb{R}^n$ if and only if the submanifold described by the vectors in the given field $\mathbf{w}$ when translated to a common origin is part of an $(n-1)$-dimensional sphere passing through the origin.

I am looking for an interpretation of these general results in terms of connection on some appropriately defined bundle over the embedded submanifold $S$.

Note: We only are given with the tangential component of the field and know nothing about the transverse component. The above results, also, involve only the tangential component.

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This is an elementary exercise in the differential geometry of curves and really belongs in MathStackExchange.

Assume that the curve is parametrized by arclength, say $X(s)$, and that your given vector field is $f(s) T(s)$, where $T(s) = X'(s)$ is the unit tangent vector field and $N(s)$ is the oriented unit normal along the curve. Then one has the equations $T'(s) = \kappa(s) N(s)$ and $N'(s) = -\kappa(s) T(s)$ where $\kappa$ is the curvature of the curve.

If $f(s)\ T(s) + g(s)\ N(s)$ is to be a constant vector field along the curve, then $f$ and $g$ must satisfy the differential equations $$ f'(s) = \kappa(s) g(s)\qquad\text{and}\qquad g'(s) = -\kappa(s) f(s) $$ and this is sufficient. (You can regard this as defining a connection on an $\mathbb{R}^2$-plane bundle over the curve if you want.) Eliminating $g$ from these equations yields that $f$ must satisfy the linear second-order equation $$ f''(s) - \bigl(\kappa'(s)/\kappa(s)\bigr)\ f'(s) + \kappa(s)^2\ f(s) = 0. $$

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  • $\begingroup$ Thank you very much Sir for the reply. This really seems to be an elementary exercise and should belong to MathStackExchange. I have added two generalizations to my original question and in these general cases, the interpretation that I am looking for is, I think, would be slightly involved and not so straightforward. $\endgroup$ – Ayan Apr 26 '14 at 12:56
  • $\begingroup$ Further, your answer involves an interplay between the tangential and normal components of the field. But what is given is the field $f(s)\mathbf{T}(s)$, a purely tangential field. The objective is to determine the condition, using only $f(s)$ and properties of the curve, if there exists a constant $\mathbf{C}$ such that $\mathbf{C}\cdot\mathbf{T}(s)=f(s)$. The higher order ode seems to be such a condition, but what would be the appropriate connection then? $\endgroup$ – Ayan Apr 26 '14 at 14:31
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    $\begingroup$ Just define a connection on the tangent bundle by $\nabla\bigl(f(s) \mathbf{T}(s)\bigr) = f'(s)\ \mathbf{T}(s)$. Then the equation above for the projection $\mathbf{F}$ of a constant vector field just becomes $\nabla^2\mathbf{F} = \kappa'(s)/\kappa(s)\ \nabla \mathbf{F} - \kappa(s)^2\ \mathbf{F}$. (You obviously cannot get a first order equation for this condition involving any covariant derivative on the tangent bundle because the space of parallel sections has dimension $2$.) $\endgroup$ – Robert Bryant Apr 27 '14 at 1:41
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Edited: I misunderstood the question. My answer (which is now starts after ``old version of the answer'' below ) answers the following question: suppose there exists a vector field along a curve $c(t)$ given as linear combination of the velocity and the normal vectors to a curve (i.e., what you we know is two functions, $a(t)$ and $b(t)$, and your vector field is actually $a(t)*T(t) + b(t)*\nu(t)$, where $T(t)$ is the velocity vector and $\mu(t)$ is the normal vector). How to understand, without knowing the tangent vector and the normal but possibly knowing the curvature of the curve, that this vector field is constant in the sence that the usual coordinates of this vector field in $R^2$ do not depend on $t$?

One can of course relatively straightforward obtain the answer on your initial question which was done by Robert Bryant (clearly independet since he posted his answer few minutes before I did it)

Old version of the answer: If I understand your question correctly, such connection exists and is closed related to Frenet formulas.

More precisely, the bundle where the connection lives is a bundle over 1-dimensional manifold (the curve) with 2-dimensional fiber.

The geometric picture behind the bundle is as follows: the fiber at each point is identified with the plane where the curve lies, where we have chosen the following basis: the first vector is the velocity vector of at this point and the second vector is the normal vector at this point.

Since your manifold (the curve) is one-dimensional, the connection can be thought to be the field of $2\times 2$ matrices that depend on the parameter $t$ of the curve.

If you curve is which is arc length parameterised, the connection is given by the formula
$$ \begin{pmatrix} 0 & \kappa \\ - \kappa & 0 \end{pmatrix}, $$ where $\kappa(t)$ is the curvature of the curve.

The required condition that a parallel section in this connection corresponds to a constant vector field follows from Frenet formulas.

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  • $\begingroup$ Dear Sir, thank you so much for this illuminating answer which goes hand in hand with the answer by Prof. Bryant. I have added two generalized versions of my original problem, and I think, similar interpretations in terms of bundles is much involved for these general cases. I would really appreciate if you kindly throw some light on these general versions. $\endgroup$ – Ayan Apr 26 '14 at 13:10
  • $\begingroup$ Dear Sir, I have a little doubt with your answer. A connection on a fibre bundle is a horizontal distribution which has the same dimension as the base manifold, and in this case it is one. So, is the appropriate horizontal distribution is the 1-dimensional axial vector corresponding to the above skew-symmetric matrix? $\endgroup$ – Ayan Apr 26 '14 at 13:56

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