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Consider the metric space $(M, d \,)$ where set $M$ contains sufficiently many (at least five) distinct elements,
and consider the assignment $c_f$ of coordinates to (the elements of) set $M$,

$c_f \, : \, M \leftrightarrow {\mathbb{R}}^3; \, c_f[ P ] := \{ x_P, y_P, z_P \}$

such that distinct coordinates values are assigned to distinct elements of set $M$, and
such that for the function

$f \, : \, ({\mathbb{R}}^3 \times {\mathbb{R}}^3) \rightarrow {\mathbb{R}};$
$f[ \{ x_P, y_P, z_P \}, \{ x_Q, y_Q, z_Q \} ] := $ ${\sqrt{ (x_Q - x_P)^2 + (y_Q - y_P)^2 + (z_Q - z_P)^2 }} \equiv {\sqrt{ \sum_{ k \in \{ x \, y \, z \} } (k_Q - k_P)^2 }}$

and for any three distinct elements $A$, $B$, and $J$ $\in M$ holds

$f[ c_f[ A ], c_f[ J ] ] \, d[ B, J ] = f[ c_f[ B ], c_f[ J ] ] \, d[ A, J ]$.

Is the metric space $(M, d \,)$ therefore flat?

(i.e. in the sense of vanishing Cayley-Menger determinants of distance ratios between any five elements of set $M$.)

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This probably belongs on math.se or even on mathoverflow. –  Emilio Pisanty Feb 21 '13 at 17:35
    
Aren't coordinates being used in physics; thus requiring a physics-based definition? –  user12262 Feb 21 '13 at 20:47
    
They are. But addition is also used in physics and that doesn't make it more physics than math. Your question is purely mathematical (geometrical, at that) in nature, I think. –  Emilio Pisanty Feb 21 '13 at 21:48
    
I admit that the physics import of my question is subtle, namely to point out that determining and considering distance ratios is paramount and indispensible for characterizing a metric space (such as determining possible flatness, or evaluating curvature) while any assignment of coordinates is at best secondary. Surely there are various ways of putting this point (more or less) in the required form of a question. I hope to still do better myself; for instance generalizing from $\kappa = 0$ (flatness) to any value $\kappa$; eventually asking ... –  user12262 Feb 23 '13 at 14:30
    
... under which conditions the assignment of coordinate-tuples is sufficient for establishing that a given metric space constitutes a manifold. p.s. "addition is also used in physics and that doesn't make it more physics than math" Right, per se. However, it remains to define what to add; or whether to use any other operation instead. p.p.s. Gotta go, look up again whether/how subtraction is defined for Dirichlet cuts ... –  user12262 Feb 23 '13 at 14:30

1 Answer 1

Yes -- if the coordinates (real number triples) $c_f$ can be assigend to the elements of set $M$ as required in the statement of the question, given the distances (ratios) $d$ and function $f$ as described above then the metric space $(M, d \,)$ is flat.

Because: for any fifteen (real) numbers, $\{ x_\alpha, y_\alpha, z_\alpha \}$, $\{ x_\beta, y_\beta, z_\beta \}$, $\{ x_\gamma, y_\gamma, z_\gamma \}$, $\{ x_\phi, y_\phi, z_\phi \}$ and $\{ x_\lambda, y_\lambda, z_\lambda \}$ the following determinant vanishes

0 = $ \begin{array}{|cccccc|} 0 & {\small{\sum\limits_{ k \in \{ x \, y \, z \} } (k_\alpha - k_\beta)^2}} & {\small{\sum\limits_{ k \in \{ x \, y \, z \} } (k_\alpha - k_\gamma)^2}} & {\small{\sum\limits_{ k \in \{ x \, y \, z \} } (k_\alpha - k_\phi)^2}} & {\small{\sum\limits_{ k \in \{ x \, y \, z \} } (k_\alpha - k_\lambda)^2}} & 1 & \\ {\small{\sum\limits_{ k \in \{ x \, y \, z \} } (k_\beta - k_\alpha)^2}} & 0 & {\small{\sum\limits_{ k \in \{ x \, y \, z \} } (k_\beta - k_\gamma)^2}} & {\small{\sum\limits_{ k \in \{ x \, y \, z \} } (k_\beta - k_\phi)^2}} & {\small{\sum\limits_{ k \in \{ x \, y \, z \} } (k_\beta - k_\lambda)^2}} & 1 & \\ {\small{\sum\limits_{ k \in \{ x \, y \, z \} } (k_\gamma - k_\alpha)^2}} & {\small{\sum\limits_{ k \in \{ x \, y \, z \} } (k_\gamma - k_\beta)^2}} & 0 & {\small{\sum\limits_{ k \in \{ x \, y \, z \} } (k_\gamma - k_\phi)^2}} & {\small{\sum\limits_{ k \in \{ x \, y \, z \} } (k_\gamma - k_\lambda)^2}} & 1 & \\ {\small{\sum\limits_{ k \in \{ x \, y \, z \} } (k_\phi - k_\alpha)^2}} & {\small{\sum\limits_{ k \in \{ x \, y \, z \} } (k_\phi - k_\beta)^2}} & {\small{\sum\limits_{ k \in \{ x \, y \, z \} } (k_\phi - k_\gamma)^2}} & 0 & {\small{\sum\limits_{ k \in \{ x \, y \, z \} } (k_\phi - k_\lambda)^2}} & 1 & \\ {\small{\sum\limits_{ k \in \{ x \, y \, z \} } (k_\lambda - k_\alpha)^2}} & {\small{\sum\limits_{ k \in \{ x \, y \, z \} } (k_\lambda - k_\beta)^2}} & {\small{\sum\limits_{ k \in \{ x \, y \, z \} } (k_\lambda - k_\gamma)^2}} & {\small{\sum\limits_{ k \in \{ x \, y \, z \} } (k_\lambda - k_\phi)^2}} & 0 & 1 & \\ 1 & 1 & 1 & 1 & 1 & 0 & \end{array}$.

Consequently, for any five distinct elements $A$, $B$, $J$, $K$ and $Q$ $\in M$ holds

0 = $ \begin{array}{|cccccc|} 0 & \left(\frac{f[ c_f[ A ], c_f[ B ] ]}{f[ c_f[ A ], c_f[ B ] ]}\right)^2 & \left(\frac{f[ c_f[ A ], c_f[ J ] ]}{f[ c_f[ A ], c_f[ B ] ]}\right)^2 & \left(\frac{f[ c_f[ A ], c_f[ K ] ]}{f[ c_f[ A ], c_f[ B ] ]}\right)^2 & \left(\frac{f[ c_f[ A ], c_f[ Q ] ]}{f[ c_f[ A ], c_f[ B ] ]}\right)^2 & 1 & \\ \left(\frac{f[ c_f[ B ], c_f[ A ] ]}{f[ c_f[ A ], c_f[ B ] ]}\right)^2 & 0 & \left(\frac{f[ c_f[ B ], c_f[ J ] ]}{f[ c_f[ A ], c_f[ B ] ]}\right)^2 & \left(\frac{f[ c_f[ B ], c_f[ K ] ]}{f[ c_f[ A ], c_f[ B ] ]}\right)^2 & \left(\frac{f[ c_f[ B ], c_f[ Q ] ]}{f[ c_f[ A ], c_f[ B ] ]}\right)^2 & 1 & \\ \left(\frac{f[ c_f[ J ], c_f[ A ] ]}{f[ c_f[ A ], c_f[ B ] ]}\right)^2 & \left(\frac{f[ c_f[ J ], c_f[ B ] ]}{f[ c_f[ A ], c_f[ B ] ]}\right)^2 & 0 & \left(\frac{f[ c_f[ J ], c_f[ K ] ]}{f[ c_f[ A ], c_f[ B ] ]}\right)^2 & \left(\frac{f[ c_f[ J ], c_f[ Q ] ]}{f[ c_f[ A ], c_f[ B ] ]}\right)^2 & 1 & \\ \left(\frac{f[ c_f[ K ], c_f[ A ] ]}{f[ c_f[ A ], c_f[ B ] ]}\right)^2 & \left(\frac{f[ c_f[ K ], c_f[ B ] ]}{f[ c_f[ A ], c_f[ B ] ]}\right)^2 & \left(\frac{f[ c_f[ K ], c_f[ J ] ]}{f[ c_f[ A ], c_f[ B ] ]}\right)^2 & 0 & \left(\frac{f[ c_f[ K ], c_f[ Q ] ]}{f[ c_f[ A ], c_f[ B ] ]}\right)^2 & 1 & \\ \left(\frac{f[ c_f[ Q ], c_f[ A ] ]}{f[ c_f[ A ], c_f[ B ] ]}\right)^2 & \left(\frac{f[ c_f[ Q ], c_f[ B ] ]}{f[ c_f[ A ], c_f[ B ] ]}\right)^2 & \left(\frac{f[ c_f[ Q ], c_f[ J ] ]}{f[ c_f[ A ], c_f[ B ] ]}\right)^2 & \left(\frac{f[ c_f[ Q ], c_f[ K ] ]}{f[ c_f[ A ], c_f[ B ] ]}\right)^2 & 0 & 1 & \\ 1 & 1 & 1 & 1 & 1 & 0 & \end{array}$;

and therefore also

0 = $ \begin{array}{|cccccc|} 0 & \left(\frac{d[ A, B ]}{d[ A, B ]}\right)^2 & \left(\frac{d[ A, J ]}{d[ A, B ]}\right)^2 & \left(\frac{d[ A, K ]}{d[ A, B ]}\right)^2 & \left(\frac{d[ A, Q ]}{d[ A, B ]}\right)^2 & 1 & \\ \left(\frac{d[ B, A ]}{d[ A, B ]}\right)^2 & 0 & \left(\frac{d[ B, J ]}{d[ A, B ]}\right)^2 & \left(\frac{d[ B, K ]}{d[ A, B ]}\right)^2 & \left(\frac{d[ B, Q ]}{d[ A, B ]}\right)^2 & 1 & \\ \left(\frac{d[ J, A ]}{d[ A, B ]}\right)^2 & \left(\frac{d[ J, B ]}{d[ A, B ]}\right)^2 & 0 & \left(\frac{d[ J, K ]}{d[ A, B ]}\right)^2 & \left(\frac{d[ J, Q ]}{d[ A, B ]}\right)^2 & 1 & \\ \left(\frac{d[ K, A ]}{d[ A, B ]}\right)^2 & \left(\frac{d[ K, B ]}{d[ A, B ]}\right)^2 & \left(\frac{d[ K, J ]}{d[ A, B ]}\right)^2 & 0 & \left(\frac{d[ K, Q ]}{d[ A, B ]}\right)^2 & 1 & \\ \left(\frac{d[ Q, A ]}{d[ A, B ]}\right)^2 & \left(\frac{d[ Q, B ]}{d[ A, B ]}\right)^2 & \left(\frac{d[ Q, J ]}{d[ A, B ]}\right)^2 & \left(\frac{d[ Q, K ]}{d[ A, B ]}\right)^2 & 0 & 1 & \\ 1 & 1 & 1 & 1 & 1 & 0 & \end{array}$.

Thus, the (normalized) Cayley-Menger determinants of distance ratios between any five elements of set $M$ vanishes; the metric space $(M, d \,)$ is flat. (However, the metric space $(M, d \,)$ is then still not necessarily plane, or even straight.)

The suitable assignment of real number triples $c_f$ to elements of any flat metric space, together with the described function $f$ therefore provides a good (scaled-isometric) representation of the given flat metric space.

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