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I need to comprehend projective plane as a prerequisite for some other topic. But I can't understand what it really looks like. How can I make this plane natural to me?

Moreover I want to understand why homogenous equations like $X^2+Y^2=Z^2$ boil down to like $x^2+y^2=1$ in this plane.

I know the definition of the plane as the set of all lines passing through the origin and also have some experience working with projective transformations in Olympiad geometry problems.

Grobber
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  • Similar question here, http://math.stackexchange.com/questions/344597/learning-projective-geometry – GFR Feb 07 '14 at 19:37

1 Answers1

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From the definition as lines you have that a point $(X,Y,Z)\neq(0,0,0)$ determines a line through the origin.

If the point happens to have $Z\neq0$ then the point $(X/Z,Y/Z,1)$ determines the same line through the origin.

This means that $(X,Y,Z)$ and $(x,y,1)$, with $x=X/Z, y=Y/Z$, are representing the same point of the projective plane.

If we take $X^2+Y^2=Z^2$ and look only in the part of the projective plane where $Z\neq0$ then we can divide both side by $Z$ and get $(X/Z)^2+(Y/Z)^2=1$, which is the equation $x^2+y^2=1$ in the coordinates $(x,y)$.

Of course, the points such that $x^2+y^2=1$ are only those points for which $Z\neq0$ (the complements of a line in the projective plane).

To have the complete picture we also need to study it at points of the projective plane where $Z=0$. So the solutions of $x^2+y^2=1$ are not the whole picture.

If $Z=0$,then either $X\neq0$ or $Y\neq$. Since the equation is symmetric with respect to $X,Y$, it is enough to look at the case $Y\neq0$. In this case we can proceed similarly. We divide by $Y^2$ and get $(X/Y)^2+1=(Z/Y)^2$. Calling $\tilde{x}:=X/Y$ and $\tilde{z}:=Z/Y$ we get $\tilde{x}^2+1=\tilde{z}^2$.

  • How does $X^2=Y^2+Z^2$ look in Projective plane? – Grobber Feb 07 '14 at 19:40
  • Use the same idea. Look at the overlapping pieces of the projective plane where $X\neq0$, $Y\neq0$ and $Z\neq0$. Divide both sides by the coordinate that is not vanishing in each case, raised to the power $2$ (the degree of the homogeneous polynomial). Use the quotients as new coordinates in that portion of the projective plane. –  Feb 07 '14 at 19:46
  • Yeah, but why are they different? $X^2+Y^2=Z^2$ and $X^2=Y^2+Z^2$ look the same in $\mathbb{R}^3$ but $x^2+y^2=1$ and $x^2=y^2+1$ are entirely different in $\mathbb{R}^2$. – Grobber Feb 07 '14 at 19:53
  • Isn't $Z/Y=0$ if $Z=0$? Sorry, if I'm talking nonsense, but I'm not familiar with projective geometry. – Stefan Hamcke Feb 07 '14 at 19:55
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    The equations $x^2+y^2=1$ and $x^2=y^2+1$ are not the entire set of solutions of $X^2+Y^2=Z^2$. Moreover, since you are beginning to learn this you should call different coordinates with different names. You obtain $x^2+y^2=1$ for $x:=X/Z$, $y:=Y/Z$, and you get $\tilde{x}^2+1=\tilde{z}^2$ for $\tilde{x}:=X/Y$ and $\tilde{z}:=Z/Y$. On purpose I gave them different names. They are (part of) the original equation written in two different sets of coordinates. The equations look different for the same reason your shadow looks different if you are lighten from above or form the side. –  Feb 07 '14 at 20:15
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    @Grobber: Dear Grobber, If you intersect $X^2 +Y^2 = Z^2$ and $X^2 = Y^2 + Z^2$ with the plane $Z = 0$, then they look different! When you choose an affine patch and dehomogenize the equation, you are intersecting the curve in $\mathbb RP^2$ with the $\mathbb RP^1$ which is at infinity (from the point of view of the given affine patch), and throwing away those intersection points. In the standard affine patch, the circle $x^2 +y^2 = 1$ has no points at infinity, so no points are missing. When you change coords. so that the line $y = 0$ gets put at infinity, then in the new coordinates ... – Matt E Feb 07 '14 at 20:29
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    ... (namely $\bar{x} := x/y$ and $1/y$) the equation becomes $$\bar{x}^2 + 1 = \bar{y}^2$, and there is a point at infinity, namely the point labelled $(1,0)$ in the original $(x,y)$-coords. Regards, – Matt E Feb 07 '14 at 20:32
  • @Lily, in $\mathbb R\mathbb P^2$ you can not have solutions of $X^2+Y^2=Z^2$ with $Z=0$, since $X^2+Y^2=0$ implies $X=Y=0$ and $(0,0,0)$ does not represent a point in $\mathbb R\mathbb P^2$. If you look at the equation in $\mathbb C\mathbb P^2$ you get $(X/Z)^2+(Y/Z)^2=1$ in the chart $Z\neq 0$ and $X^2+Y^2=0$ in the line at $Z=0$, where we now have complex solutions to this equation. – Christoph Feb 08 '14 at 09:22
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    Have a look at http://pages.bangor.ac.uk/~mas010/outofline/motion.html#motion for a sequence of stills showing how the projective plane represents rotations in 3-space. – Ronnie Brown Feb 08 '14 at 16:34