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Edit: I have reworded the question.

This may be a basic question but I am having trouble figuring out the correct answer. I want to find a local coordinate chart that fits a d-dimensional submanifold in $\mathbb{R}^N$. I am given two points $p_1, p_2 \in \mathbb{R}^N$ and corresponding orthonormal bases $(\phi_1, \phi_2, \ldots, \phi_d)$, $(\tau_1, \tau_2, \ldots, \tau_d) \subseteq \mathbb{R}^N$ for their tangent spaces. I would like to find an algorithmic method for finding functions $(f_1(x_1,x_2,\ldots,x_d), f_2(x_1,x_2,\ldots,x_d),\ldots f_N(x_1,x_2,\ldots,x_d))$ that satisfy these conditions.

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The question is meaningless as stated. My guess is that you are trying to find a d-dimensional submanifold in $R^n$ passing through two points and whose tangent space at the given points is prescribed. It seems also that you want the submanifold to be diffemorphic to a disk and to admit a polynomial parameterization. Existence of such submanifold follows from Weierstrass approximation theorem. Lastly, you want to have the submanifold to be optimal, but it is unclear what is your notion of optimality. –  Misha Oct 14 '12 at 20:35
If I'm understanding correctly, then you know the mapping $f$ that takes coordinates in one chart to coordinates in the overlapping chart. You want to approximate $f$ with a polynomial. Without any further information about your criteria, it's irrelevant that $f$ is being used as a change of coordinates on a manifold, that you know of certain orthonormal bases, or that there may be curvature. It's simply the generic problem of making a polynomial fit to a smooth function from $\mathbb{R}^n$ to $\mathbb{R}^n$. This is a problem that has many different answers, e.g., truncated Taylor series, ... –  Ben Crowell Oct 14 '12 at 22:52
Sorry I have tried to edit the equation. I would like to just find the functions in one chart. The problem I find is what is an appropriate polynomial form to fit given this data. There is ambiguity in how I choose my coordinates and then I need to fit the right polynomial given my constraints. I just wasn't sure if there was a canonical way of doing this type of thing (something like a generalization of a spline). –  Chirag Lakhani Oct 14 '12 at 23:18
I mean edit the question. –  Chirag Lakhani Oct 14 '12 at 23:19

2 Answers 2

up vote 1 down vote accepted

This isn't really a formula so much as a conceptual construction.

First suppose that $d=n-1.$

Let $T_1=span(\phi_1,..., \phi_{n-1})$ and $T_2=span(\tau_1,...,\tau_{n-1})$ be the tangent subspaces of $T{\mathbb R}^N$ at $p_1$ and $p_2,$ respectively. We can think of these as hyperplanes in real N-space so that there is an intersection of half-spaces created by these hyperplanes containing the line segment $l=p_1 p_2.$ Choose $r>0$ small enough that the spheres tangent to $T_1$ and $T_2$ at $p_1$ and $p_2$ with radii equal to $r$ and centers located on $l$ do not intersect. Call these spheres $S_1$ and $S_2$. Define a path of spheres by $S_{1+\lambda}=(1-\lambda)S_1 + \lambda S_2$ for $0\leq \lambda \leq 1,$ where addition is given by Minkowski addition of subsets in $N$-space.

The envelope of this path of spheres should be the boundary of a tubular neighborhood of a line segment, which is tangent to $T_1$ and $T_2.$

Second suppose that $d$ is strictly less than $n-1.$ Fill in the given orthonormal bases so that we have the above case and make the above construction. Then we let $T_1$ and $T_2$ be the subspaces at $p_1$ and $p_2$ spanned by the original given orthonormal bases and set $T_{1+\lambda}=(1-\lambda)T_1 + \lambda T_2$. The intersection of $T_{1+\lambda}$ with the corresponding $(1+\lambda)$-element from the envelope as constructed above should determine a $d$-dimensional submanifold satisfying the desired conditions.

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Thanks! This looks like an interesting construction. I will have to play around with this and see if I can get nice explicite formulas for my case. –  Chirag Lakhani Oct 20 '12 at 14:05

There isn't a formula for what you're looking for. At least, the formula can't make sense for all initial data and depend continuously on that initial data.

Your initial data is two points in the Stiefel manifold $V_{n,d}$, together with two points in $\mathbb R^n$. If you were to find an embedding $B^d \to \mathbb R^n$ whose derivatives agreed at two points with these points of your Stiefel space, you'd have produced a section of the fibration:

$$ Emb(B^d, \mathbb R^n) \to V_{n,d}^2 \times C_2 \mathbb R^n$$

where this map is given by restriction to the endpoints. $C_2 \mathbb R^n$ is the configuration space of two points in $\mathbb R^n$, it has the homotopy-type of a sphere.

$Emb(B^d, \mathbb R^n)$ has the homotopy-type of $V_{n,d}$. So if your section existed, the homotopy and homology groups of $V_{n,d}^2 \times C_2 \mathbb R^n$ would embed in the homotopy and homology groups of $V_{n,d}$. Since the first non-trivial homotopy group of $V_{n,d}$ is cyclic, this is impossible.

Admittedly, I'm making some extra assumptions that you did not specify, but I think this shows you that whatever your solution is, it has to be somewhat nuanced.

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A great example of technology solving a problem that most likely just did not see it coming :-) –  Mariano Suárez-Alvarez Oct 15 '12 at 3:46
Thanks! The use of algebraic topology is really interesting, even if it provides a nonexistent solution! –  Chirag Lakhani Oct 20 '12 at 14:07

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