I was trying to compare the metric tensor at the wikipedia pages of the

Beltrami Klein model https://en.wikipedia.org/wiki/Klein_disk_model

and the metric tensor of the Poincare disk model at https://en.wikipedia.org/wiki/Poincar%C3%A9_disk_model

A the moment (22 may 2015)

The formulas for the metric tensor of the Beltrami Klein model is given as:

$$ g (x, dx) = \frac{4 (x \cdot dx)^2}{(1 - \left\Vert x \right\Vert^2)^2} + \frac{4 \left\Vert dx \right\Vert^2}{(1 - \left\Vert x \right\Vert^2)}. $$

While for Poincare disk model the metric tensor is given as:

$$ ds^2 = 4 \frac{\sum_i dx_i^2}{(1-\sum_i x_i^2)^2}$$

(where the $x_i$ are the Cartesian coordinates of the ambient Euclidean space)

While I understand the two metric tensers should be different, here there also seems to be a difference in notation, and that makes it for me impossible even to compare the two tensors.

How can I convert one tensor into the notation of the other?

OR (maybe more practical) :

What are the metric tensors of the Beltrami Klein model in the notation $ds^2 = ... $

What are the metric tensors of the Poincare disk model in the notation $ g (x, dx) = ... $


Are are these imposssible questions and are the tensors incommensurable.? (the tensors are just not the same kind of beast)

PS I now hardly anything about tensors, (let alone metric tensors) so any basic info about them should be welcome. (what is the easiest formula?)

  • $\begingroup$ Those notations with $dx$ have to be interpreted formally, treating differential forms as if they were numbers. The symbol $dx_i^2$ reads in more rigorous notation as $dx_i\otimes dx_i$, of this I am entirely sure. The symbol $x\cdot dx$ might be a synonim of $\sum_i x_idx_i$. And similarly $\lvert dx\rvert^2$ might be a synonim of $\sum_i dx_i^2$. $\endgroup$ – Giuseppe Negro May 21 '15 at 13:39

The two notations are only slightly different. The metric tensor in the Klein model is $$ ds^2 \;=\; \frac{{dx_1}^2 + \cdots + {dx_n}^2 }{1 - {x_1}^2-\cdots - {x_n}^2}+\frac{(x_1\,dx_1 + \cdots + x_n\, dx_n)^2}{\bigl(1-{x_1}^2-\cdots-{x_n}^2\bigr)^2} $$ and the metric tensor in the Poincaré model is $$ ds^2 \;=\; 4\frac{{dx_1}^2+ \cdots + {dx_n}^2}{\bigl(1-{x_1}^2-\cdots -{x_n}^2\bigr)^2} $$ If we let $\textbf{x} = (x_1,\ldots,x_n)$ and $\textbf{dx} = (dx_1,\ldots,dx_n)$, the metric tensor for the Klein model can be written $$ ds^2 \;=\; \frac{\|\mathbf{dx}\|^2}{1-\|\mathbf{x}\|^2} + \frac{(\textbf{x}\cdot\textbf{dx})^2}{\bigl(1-\|\mathbf{x}\|^2\bigr)^2} $$ and the metric tensor for the Poincaré model can be written $$ ds^2 \;=\; \frac{4\|\mathbf{dx}\|^2}{\bigl(1-\|\mathbf{x}\|^2\bigr)^2} $$ If you prefer to think in terms of matrices/components, the $ij$'th component of the metric tensor for the Klein model is $$ g_{ij} \;=\; \frac{\delta_{ij}}{1-{x_1}^2-\cdots-{x_n}^2} + \frac{x_ix_j}{\bigl(1-{x_1}^2-\cdots-{x_n}^2\bigr)^2} $$ where $\delta_{ij}$ is the Kronecker delta, and the $ij$'th component of the metric tensor for the Poincaré model is $$ g_{ij} \;=\; \frac{4\,\delta_{ij}}{\bigl(1-{x_1}^2-\cdots-{x_n}^2\bigr)^2} $$

  • $\begingroup$ thanks but could you give references (i want to incorporate them on wikipedia) also are they in fact directly comparable (does the x mean the same in each formula) $\endgroup$ – Willemien May 24 '15 at 5:43
  • $\begingroup$ The x means the same in each formula. My reference is section 7 of Cannon, James W., William J. Floyd, Richard Kenyon, and Walter R. Parry. "Hyperbolic geometry." Flavors of geometry 31 (1997): 59-115. $\endgroup$ – Jim Belk May 24 '15 at 6:36
  • $\begingroup$ thanks , but that paper only mentions the formulas do you know somewhere where they are deducted? (I am just wondering about the klein disk formula, it should be siilar to the Jemisphere formula (just replacing $(d)x_{n+1}$ by some formula but i cannot find the connection. $\endgroup$ – Willemien May 26 '15 at 9:14
  • $\begingroup$ The paper proves the equivalence of (what it calls) the $H$, $J$, and $K$ models on pg. 71 - 72, where $K$ is the Klein model. To prove that the Poincare model $I$ is isometric to these, it is only necessary to show that $\beta^*(ds_I^2) = ds_J^2$, using methods similar to the computations on pg. 72, where $\beta\colon J\to I$ is the map defined on pg. 71. $\endgroup$ – Jim Belk May 26 '15 at 15:32
  • $\begingroup$ if I compare the metric tensors of the hemisphere(J) and the Klein model (K) as given on page 71 (they have the same coordinates except the last but $x_{n+1} = \sqrt{( 1- x_1 - x_2 .... -x_n)} $ then $ dx^2_{n+1}$ equals $ \frac{(x_1\,dx_1 + \cdots + x_n\, dx_n)^2}{1-{x_1}^2-\cdots-{x_n}^2 } $ and this bit I don't understand. $\endgroup$ – Willemien May 27 '15 at 17:33

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.