# Isometric embedding of 1-manifold

Suppose the unit circle $\gamma$ in $R^2$ is not endowed with the canonical inner product of $R^2$. Let the riemannian metric defined be $g:=[2, 1;1, 1]$. So the length is measured with this metric on the circle.
Q.1 Is there a way to construct the isometric embedding of this $(\gamma,g)$ in some $(R^n,can)$?
Q.2 In general given any such $g$ can we always construct an isometric embedding?

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Your $g$ is not a Riemannian metric as it is not symmetric. – Sergei Ivanov Dec 24 '10 at 10:37
Your metric has constant coefficients, hence there is a linear map that transforms it into the standard Euclidean metric. Restrict this map to the circle and you are done. For the specific example $g=[2,1;1,1]$, the linear map $(x,y)\mapsto(x+y,y)$ does the job. – Sergei Ivanov Dec 24 '10 at 11:59
As for Q2, every Riemannian metric on the circle is isometric to a Euclidean circle of a suitable radius and hence admits an isometric embedding to $\mathbb R^2$. The said isometry can be written explicitly in terms of an arc-length parameterization. – Sergei Ivanov Dec 24 '10 at 12:07
I suggest you provide some background or motivation. Right now the question looks like it is from someone stuck at the first paragraph of a textbook. (And if this indeed is the case, there is nothing wrong with it, but math.stackexchange.com is a better place for the question.) – Sergei Ivanov Dec 24 '10 at 12:26
Motivation: I work on curve reconstruction in $SE(3)$. And the focus is on the simple closed curves in $SE(3)$. The exp map for $SE(3)$ is known and the riemannian metric defined on $se(3)$ is $g:=[\alpha I_3 0,0,\beta I_3]$. Given a curve in it, I want to find an isometric embeddig of this curve in some $R^n$. Furthermore, I want to construct an isometric embedding of $SE(3)$ to some $R^n$?. – Pratik 0 secs ago – Pratik Dec 24 '10 at 13:08

$g$ is symmetric, now is there a tractable construction possible for such an embedding? If yes how to proceed about constructing one in the case stated here. – Pratik Dec 24 '10 at 11:54