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The following question might be connected to Embed curves in the plane

Given a curve in $\mathbb{R}^n$ (smooth and so on), what is the smallest basis in which every point of the curve can be written as a linear combination of the basis vectors?

For example $g(t) = (t, t^2, t)$ is defined in $\mathbb{R}^3$, but lies completely in the subspace spanned by vectors $(1,0,1)$ and $(0,1,0)$.

I'm not experienced in topology, so I can't tell whether what I want is equivalent to what "embedding" normally means. My question though is, how to explicitly construct the smallest linear (sub)space that contains a given curve.

Or is it only possible for trivial cases, like the one I chose above?

EDIT: Maybe to put the question in more symbolic Notation

For a given differentiable function $g:\mathbb{R}^+ \rightarrow \mathbb{R}^n$, with $g(\mathbb{R}^+) \subset [0,1]^n$ we define $\mathcal{B}_g = \{v_i \in \mathbb{R}^n \,, \forall \, i = 1,\ldots,k\}$ to be a Basis iff

$(\forall \, t \in \mathbb{R}^+)(\exists \, \alpha_1,\ldots,\alpha_k \in \mathbb{R}) \, \, g(t) = \sum\limits_{i=1}^k \alpha_i v_i $

Question: How to construct the smallest Basis for a given curve $g$?

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Kenji
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  • curves are 1d objects, so probably the smallest one will be 1? – Kiryl Pesotski Feb 28 '17 at 09:12
  • thanks for the Suggestion, but can you Elaborate it a bit? My example curve contains the Points $(1,1,1)$ and $(2,4,2)$. These vectors are lineary Independent, so I need at least a two dimensional basis to write them both as linear combination of that Basis' vectors. – Kenji Feb 28 '17 at 09:53
  • You can define a single basis vector $\hat{t}=\frac{\nabla{t}}{||\nabla{t}||}=\frac{2t(\hat{i}+\hat{k})+\hat{j}}{\sqrt{8t^{2}+1}}$. So, for example you can write the gradient operator on that 1d space as $\nabla_{t}=\hat{t}\frac{\sqrt{8t^{2}+1}}{6t}\frac{\partial}{\partial{t}}$. – Kiryl Pesotski Feb 28 '17 at 10:17
  • I have some trouble with the Notation. $\Delta t$ = $\Delta g(t)$, i.e. $\hat{t} = \frac{(1, 2t, 1)}{\sqrt{2 + 4t^2}}$? And what does $i, k$ and $j$ stand for? I'm rather used to write the differential operator before the term that is supposed to be differentiated, so I would write $\frac{\partial}{\partial t}\hat{t}\frac{\sqrt{8t^2+1}}{6t}$. Is this what you meant? As far as i can tell, your vector depends on $t$?! I need a constant basis that doesn't depend on $t$ though. – Kenji Feb 28 '17 at 11:56
  • First $||\nabla{t}||\hat{t}=\nabla{t(x, y, z)}=\partial_{x}t(x, y, z)\hat{i}+\partial_{y}t(x, y, z)\hat{j}+\partial_{z}t(x, y, z)\hat{k}$ So, nabla is the gradient in cartesian coordiantes. The vector $||\nabla{t}||\hat{t}$ is along the curve, vector $\hat{t}=\frac{\nabla{t}}{||\nabla{t}||}$ is the unit vector along the curve, and clearly it is the only basis vector you need. – Kiryl Pesotski Feb 28 '17 at 12:25
  • Second, $\nabla_{t}$ is the gradient operator in the basis of $\hat{t}$. It differentiates the functions of $t$ with exist on on the 1d space which is the curve itself. So the above written actually means $\nabla_{t}f(t)=\frac{\sqrt{8t^{2}+1}}{6t}\frac{df(t)}{dt}\hat{t}$, or if you are used to the differential operators $\nabla_{t}=\hat{t}\frac{\sqrt{8t^{2}+1}}{6t}\frac{\partial}{\partial{t}}$ – Kiryl Pesotski Feb 28 '17 at 12:27
  • Thanks for the detailed answer. Could it be that you have mistaken my function $g(t) = (t^2, t, t^2)$?. Then the term $\sqrt{8t^2+1}$ would make sense to me. Could you please give an example on how to express, say $(4,2,4)$ by a multiple of your Basis vector $\hat{t}$? – Kenji Feb 28 '17 at 12:49
  • @KirylPesotski, I might be mistaken, but to me OP asks the following: given a particular curve in $\mathbb{R}^n$ (i.e., a mapping $g(t): \mathbb{R} \rightarrow \mathbb{R}^n$), how to tell if this particular curve is lying in some smaller linear subspace of $\mathbb{R}^n$? – Evgeny Feb 28 '17 at 16:06
  • @Kenji By the way, the answer for curves in $\mathbb{R}^3$ is given here. Frenet-Serra formulas for higher dimensions exist too, so maybe the approach in this answer could be generalized. – Evgeny Feb 28 '17 at 16:21

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