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The zero locus of $x^2+y^2-z^2$ is a cone in $\mathbb{R}^3$.

What is the projective version of this cone? That is, what is the homogeneous polynomial whose zero locus is a cone in $\mathbb{RP}^3$?


Edit:

It seems that the same equation describes a cone in $\mathbb{RP}^3$ as well. What I am confused by, then, is that I thought $x^2+y^2-z^2$ describes a circle (being the homogenized form of $x^2+y^2-1$). So the equation for a circle in $\mathbb{RP}^2$ becomes the equation for a cone when viewed in $\mathbb{RP}^3$. Could someone please elaborate on this?

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The same equation works. –  Zhen Lin Mar 19 '12 at 10:28
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What's confusing? Equations interpreted in different environments mean different things. For example, $x^2 + y^2 - 1 = 0$ defines a circle in $\mathbb{R}^2$ but a cylinder in $\mathbb{R}^3$. I guess the point is that a cylinder and a cone are the same thing in $\mathbb{R P}^3$. –  Zhen Lin Mar 19 '12 at 18:51
    
@ZhenLin Ahhh I see. Yes, a cylinder in $\mathbb{RP}^3$ is a cone whose apex is on the plane at infinity. Thanks! –  Will Mar 19 '12 at 20:24
    
More precisely, the projective cone is $\mathrm{Proj}{\mathbb R}[x,y,z,t]/(x^2+y^2-z^2)$ with the vertex $[0,0,0,1]$. –  user18119 Mar 19 '12 at 22:23

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