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Given a plane normal $n=(n_x,n_y,n_z)$ and a point on that plane $p_0\in\mathbb{R}^3$, testing whether another point $p\in\mathbb{R}^3$ is "above" the plane is easy:

$n\cdot(p-p_0) > 0$

Is there an equivalent in projective geometry?

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2 Answers 2

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I am not sure if I understood your question correctly but if I did, here is my answer.

Projective space is constructed by identifying points on any one line passing through the origin. Which means $ a \equiv -a $. So, if a "point" is above a plane then it is also below it. Similarly, if it is below a plane then it is also above it. The point is not above if it lies on the plane.

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An hyperplane in projective geometry doesn't split the space in two parts, but two hyperplanes do. When you split the regular space by one plane or the regular plane by one line, it's like splitting the projective space by that plane together with the plane at infinity, or the projective space by that line and the line at infinity. So the natural problem is to describe how two hyperplanes split projective space in two.

If you represent points as vectors with $n+1$ coordinates modulo scaling, there is a notion of "orthogonalness" given by the dot product : $x \dot {} y = 0$. The sign (nor the magnitude) of $x\dot {} y$ has no meaning because you identify $x$ with $-x$ and with any $r.x$ with $r \in \mathbb R^*$. The only meaning to this number is whether it is $0$ or not.

But when you have $3$ points, you can talk about the sign of the quantity $(x \dot {} y)(y \dot {} z)(z \dot {} x)$, because whenever you change a representant, it multiplies this by a nonzero square, which doesn't change the sign.

Now, there is a correspondance between hyperplanes and point in projective space : if you have a point $y$, you have a corresponding hyperplane containing the points $x$ satisfying the equation $x \dot {} y = 0$. So if you have two hyperplanes corresponding to $y_1,y_2$, you can talk about the points $x$ such that $(x \dot {} y_1)(x \dot {} y_2)(y_1 \dot {} y_2) > 0$. This quantity is $0$ at least when $x$ belongs to one of the hyperplanes, so it does say on "which side" the point $x$ is, relatively to the two hyperplanes. In fact it is positive if $x$ belong to the half-space whose hyperplane borders make an obtuse angle, and it is negative if the borders make an acute angle.

One special case arise when your two hyperplanes are "orthogonal". Well, you can remove the $(y_1 \dot {} y_2)$ factor, but that costs a bit because now the sign of the result depends on what equation you pick for your hyperplanes. This problem is dependant on our choice of dot product, and it happens because there is no locally constant global function (hyperplane, hyperplane, point) -> {-1,+1} that does what we want for all possible triplets. If you make one hyperplane do a rotation of $\pi$ around itself, and keep track of which section of the projective space goes where, you will find that they get switched around after completing the rotation. So the space of triplets (hyperplane, hyperplane, point not in the first two hyperplanes) is connected : we can't have a nice formula working everywhere nicely.

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