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different specializations of $\mathbb{RP}^2$

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hbghlyj posted 2022-5-2 06:06 |Read mode
Let's consider what's happening to conics in passing from $\Bbb R^2$ to $\Bbb{RP}^2$. A conic is defined by a second order equation. In $\Bbb R^2$ this is$$ax^{2}+by^{2}+dxy+ex+fy+c=0.$$The analogue of which in $\Bbb{RP}^2$ is$$ax^{2}+by^{2}+cz^{2}+dxy+exz+fyz=0$$that arise when we replace $x$ and $y$ with $x/z$ and $y/z$. In $\Bbb R^2$, there are distinct classes of conics, say, ellipses, parabolas, hyperbolas, each with its own characteristic properties - some similar but many distinct. (E.g., ellipses are bounded, hyperbolas have asymptotes, all parabolas are similar.) That is not the case in $\Bbb{RP}^2$.
As an example, consider the equation $x^{2}-yz=0$ and different specializations of $\mathbb{RP}^2.$ If we choose $x=0$ as the ideal line at infinity, the equation reduces to $yz=1$ (with the intermediate step $\displaystyle 1-\frac{y}{x}\frac{z}{x}=0.)$ This equation describes a hyperbola in $\mathbb{R}^2.$ If $y=0$ is taken to be the ideal element, the equation reduces to first $\displaystyle\bigg(\frac{x}{y}\bigg)^{2}-\frac{z}{y}=0$ and then to $x^{2}=z$ which is a parabola in $\mathbb{R}^2.$ Finally, let's choose $y+z=0$ as the line at infinity. To see what happens, let's make a substitution $y=y'+z',$ $z=y'-z':$ $x^{2}-(y')^{2}+(z')^{2}=0.$ Drop now the apostrophe and choose $y=0$ as the ideal line: $\displaystyle\bigg(\frac{x}{y}\bigg)^{2}-1+\bigg(\frac{z}{y}\bigg)^{2}=0,$ or, in $\mathbb{R}^2,$ $x^{2}+z^{2}=1,$ a circle.

Note that collinearity, concurrence, tangency are preserved in embedding of $\mathbb{R}^2$ into $\mathbb{RP}^2$ and the converse specialization (although, this begs for a more detailed treatment). This is why, say, the validity of the three tangents theorem for circles implies its validity for all non-degenerate conics.

Copied from cut-the-knot.org/m/Geometry/ThreeTangentsTheorem.shtml

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