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I'm aware that the complex projective line is a sphere, the Riemann sphere, and the dual projective line is a cylinder; but I can't find anything mentioning what shape the split-complex projective line would be. I probably could figure it out for myself, of course, but I'm surprised that it doesn't come up in searching.

Seth Schmidt-O'Hainle
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  • It's probably not that straightforward, since it's not a field. What definition of projective line would you propose? – Berci Apr 16 '20 at 16:20
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    Duals aren't a field either, but they have a projective line. The definition is rather standard, isn't it? Ordered pairs (x,y) together with an equivalence relation ~ where for all invertible elements n you have (x,y) ~ (xn,yn). – Seth Schmidt-O'Hainle Apr 17 '20 at 01:51
  • It's a hyperboloid of one sheet, see for example the third paragraph here. More generally, if I'm not mistaken the geometry of the projective line over the ring $\mathbb{R}[j]/(j^2-t)$ is the surface of revolution of the curve $|x+yj|^2=x^2-ty^2=1$ around the $y$-axis. – pregunton Jun 14 '20 at 14:21
  • Thanks! I somehow never found that. If you could write that up as a full answer (and if possible include references to why that thing about surfaces of revolution is true) that would be great! – Seth Schmidt-O'Hainle Jun 14 '20 at 14:38

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Duals aren't a field either, but they have a projective line. The definition is rather standard, isn't it? Ordered pairs (x,y) together with an equivalence relation ~ where for all invertible elements n you have (x,y) ~ (xn,yn). – Seth Schmidt-O'Hainle Commented Apr 17, 2020 at 1:51

Let $\mathbb{D}$ be the split–complex numbers $$ \mathbb{D} = \{x + y j \mid x,y \in \mathbb{R},\ j^2 = +1\}. $$

For a commutative ring $R$, the projective line is $$ \mathbf{P}^1(R) = \{ (x,y) \in R^2 \setminus \{(0,0)\} \}/\sim, $$ where $$ (x,y) \sim (x n, y n) \quad \text{for every invertible } n \in R^\times. $$ Each equivalence class is denoted $[x:y]$.

Identifying $\mathbb{D}$ with $\mathbb{R}\times\mathbb{R}$

Using the idempotents $e_\pm = \tfrac{1}{2}(1\pm j)$, any element $$ x + y j = (x+y)e_+ + (x-y)e_-, $$ so $$ \mathbb{D} \;\cong\; \mathbb{R}\times\mathbb{R}, \qquad x + y j \longmapsto (x+y,\;x-y), $$ and $\mathbb{D}^\times \cong \mathbb{R}^\times \times \mathbb{R}^\times$.

Computing $\mathbf{P}^1(\mathbb{D})$

Under this isomorphism, a pair $(x,y)\in\mathbb{D}^2$ corresponds to $((x_1,x_2),(y_1,y_2))\in(\mathbb{R}\times\mathbb{R})^2$, and the equivalence relation becomes $$ (x_1,x_2;\;y_1,y_2) \sim (x_1 n_1,\;x_2 n_2;\;y_1 n_1,\;y_2 n_2), \qquad (n_1,n_2)\in\mathbb{R}^\times\times\mathbb{R}^\times. $$ Hence the quotient splits componentwise: $$ \mathbf{P}^1(\mathbb{D}) \;\cong\; \mathbf{P}^1(\mathbb{R}) \times \mathbf{P}^1(\mathbb{R}). $$

Topology and conclusion

Since $\mathbf{P}^1(\mathbb{R}) \cong \mathbb{RP}^1 \cong S^1$, we get $$ \mathbf{P}^1(\mathbb{D}) \;\cong\; \mathbb{RP}^1 \times \mathbb{RP}^1 \;\cong\; S^1 \times S^1. $$

Therefore the projective line over the split–complex numbers is a 2-torus.

user1693987
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