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I am a beginner to Riemannian geometry. Following is my question.

In the Euclidean space, say $\mathbb{R}^3$, let us consider a plane, for simplicity, say one passing through the origin, $\mathbb{P}=\{\textbf{x}=(x_1,x_2,x_3):ax_1+bx_2+cx_3=0\}$ and consider the origin $\textbf{0}=(0,0,0)\in \mathbb{P}$.

Now, let us consider the set of all points ($\mathbb{L}$) in $\mathbb{R}^3$ whose best approximation with respect to $\ell_2$-distance on $\mathbb{P}$ is $\textbf{0}$, i.e., $$\{\textbf{y}:\arg \min_{\textbf{x}\in \mathbb{P}}\|\textbf{x}-\textbf{y}\|_2=\textbf{0}\}$$ which is precisely the line $$\mathbb{L}:~~~~~~~\frac{x_1}{a}=\frac{x_2}{b}=\frac{x_3}{c}$$

This line $\mathbb{L}$ can also be characterised by $\mathbb{L}=\{\textbf{y}:\langle \textbf{x},\textbf{y}\rangle=0 \text{ for all } \textbf{x}\in \mathbb{P}\}$.

My question is, how to define a Riemannian metric $g$ with components $g_{i,j}$ and a connection $\nabla$ with connection coefficients $\Gamma_{i,j}^k$ so that $\mathbb{L}$ is a $\nabla$-geodesic and prove that it intersects the plane $\mathbb{P}$ at the single point $\textbf{0}$?

Thank you.

Kumara
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  • The Riemannian metric of what? – Mariano Suárez-Álvarez Feb 16 '13 at 06:31
  • Do you wish to equip $ \mathbb{L} $ with a Riemannian metric? – Haskell Curry Feb 16 '13 at 06:33
  • For $\mathbb{R}^3$. – Kumara Feb 16 '13 at 06:33
  • @MarianoSuárez-Alvarez and Haskell Curry: I haven't understood Riemannian geometry yet. This is question is towards understanding it better. I just wanted to define a Riemannian metric on $\mathbb{R}^3$ and prove that result in that framework. Am I making sense? – Kumara Feb 16 '13 at 06:38
  • Let $e_1, e_2, e_3$ be constant, orthonormal basis vectors for $\mathbb R^3$. The metric $g$ is just $g(e_i, e_j) = \delta_{ij} = \langle e_i, e_j \rangle$. – Muphrid Feb 16 '13 at 07:10
  • @HaskellCurry: I wish to see $\mathbb{L}$ as a $\nabla$-geodesic for some connection $\nabla$ with connection coefficients $\Gamma_{i,j}^k$. May I edit the question accordingly? – Kumara Feb 16 '13 at 07:17
  • @kumara $\mathbb{L}$ is a geodesic for the Levi-Civita connection associated with the standard euclidean metric $g(\frac{\partial}{\partial x_i}, \frac{\partial}{\partial x_j}) = \delta_{ij}$. The coefficients for this connection simply vanishes (i.e. $\Gamma_{ij}^k = 0$ ) – achille hui Feb 16 '13 at 08:36
  • @achillehui Thank you. But, how to prove the result I mentioned in my question, i.e., $\mathbb{P}\cap \mathbb{L}={\textbf{0}}$ with respect to the metric and connection you have mentioned? – Kumara Feb 16 '13 at 12:38
  • When $\Gamma_{ij}^k = 0$, the equation of geodesic becomes $\frac{d^2 x^\mu}{ds^2} = 0$. The general solution of it are straight lines in $\mathbb{R}^3$ and $\mathbb{L}$ is a straight line! – achille hui Feb 16 '13 at 12:44
  • @achillehui Can you tell me how to prove the pythagorean relation from this framework? $|\textbf{x}-\textbf{y}|^2=|\textbf{x}-\textbf{0}|^2+|\textbf{0}-\textbf{y}|^2$ for any $\textbf{x}\in \mathbb{P}$ and $\textbf{y}\in \mathbb{L}$. Can you also suggest a good reference for all these? – Kumara Feb 16 '13 at 13:19
  • What is your $|\cdot|$? If it is the geodesic distance (corresponding to $g$) between two points, then follow the definition by construct a geodesic joining them, compute the path length, show it is equal to the ordinary euclidean distance and apply the regular Pythagorean theorem. – achille hui Feb 16 '13 at 13:44
  • @achillehui By $|\cdot |$, I mean the ordinary $\ell_2$-distance. I wanted to get the ordinary Pythagorean theorem from the Riemaanian geometric framework. I am a beginner in Riemannian gemetry. If you can provide the details as an answer, it would be of great help. – Kumara Feb 18 '13 at 04:37

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