Timeline for Geometry of the space of circles in the Euclidean plane
Current License: CC BY-SA 3.0
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| Dec 12, 2014 at 17:05 | comment | added | James Griffin | That's a nice proof, you can also see it immediately from Robert's answer above, the projective space of $\mathbb{R}^{3,1}$ is decomposed via its inner product into three pieces, depending on whether vectors have strictly positive, strictly negative or zero norm. The negative vectors form an open ball (and a model for hyperbolic space), the "0" vectors are its boundary, so the positive vectors are $\mathbb{R}P^3$ take a closed ball. | |
| Dec 11, 2014 at 20:23 | history | answered | Benoît Kloeckner | CC BY-SA 3.0 |