I have problems finding the equations of a line l in a projective space $\mathbb{RP}^3$ given two points in homogenous coordinates as there exists no cross product. The points: $(1:0:0:0)$ and $(0:1:1:0)$.

Also given these equations of line $l$ and the ones of a line $k$: $x+y+z-t=0$ and $x-y=0$ i shall determine a line $m$ which passes through $(2:1:1:1)$ and intersects with both lines $l$ and $k$.

For the first problem i thougt i could just calculate the cross product in the first three dimensions and obtain the following equations: $y+z=0$ and $t=0$ and then calculate the intersection of the lines $l$ and $k$ and use the obtained point to calculate $m$. I thougt the intersecting point would be calculated solving the system of equations but unfortunately this only leads to $(0:0:0:0)$ which i am unsure if it is correct (at least it makes no sense to me so far...)?

Appreciate any help!

  • $\begingroup$ What do you think is a generic equation of a line in projective 4-space? $\endgroup$ – Somos May 4 '18 at 16:26
  • $\begingroup$ @Somos what is ‘generic‘? As for equation, i‘d say you need two and these are of the form ax+by+cz+dt=0 because the projective line would be a real plane (which is not a hyperplane) in R4, but am unsure how this helps me $\endgroup$ – user526159 May 4 '18 at 16:49
  • $\begingroup$ By “projective 4 space” do you mean $\mathbb{RP}^3$, as your use of homogeneous coordinates with four components suggests? $\endgroup$ – amd May 4 '18 at 20:12
  • $\begingroup$ @amd oh, correct, my mistake! $\endgroup$ – user526159 May 5 '18 at 10:29
  • $\begingroup$ In the first problem, the second point doesn’t satisfy $y+z=0$. A correct equation is $y-z=0$. $\endgroup$ – amd May 6 '18 at 3:02

Just as there are two basic ways to specify a vector subspace in linear algebra, there are two basic ways to specify a flat in projective geometry: as the join of a set of lower-dimensional flats (a spanning set) or as the meet of a set of higher-dimensional flats (intersection of null spaces). Converting from one to the other is a straightforward matter of computing the null space of a matrix.

A line through distinct points $\mathbf p$ and $\mathbf q$ is their join: $\lambda\mathbf p+\mu\mathbf q$, in matrix form $$\begin{bmatrix}\mathbf p&\mathbf q\end{bmatrix} \begin{bmatrix}\lambda\\\mu\end{bmatrix}.$$ When expressed in homogeneous coordinates, the line is effectively the column space of the matrix $M=\small{\begin{bmatrix}\mathbf p&\mathbf q\end{bmatrix}}$. In $\mathbb{RP}^3$, every line has an associated pencil of planes that contain the line. By linearity, these planes are the solutions of $\mathbf\pi^T M = 0$, i.e., the set of all planes that contain the line is the null space of $M^T$. Any two of these planes suffice to uniquely identify the line, and the nullity of $M^T$ is two, so you can find your two equations by computing a basis for the null space of $M^T$. This method generalizes to higher-dimensional spaces, but the nullity of $M^T$ will be greater and thus you will need correspondingly more hyperplanes to identify the line uniquely.

In your first problem we have two points at infinity. $$M^T=\begin{bmatrix}1&0&0&0\\0&1&1&0\end{bmatrix}$$ is already in row-reduced echelon form, so we can read a null space basis from it without further ado: $\mathbf\pi_1=[0:-1:1:0]$ and $\mathbf\pi_2=[0:0:0:1]$. The line is therefore the intersection of the planes $y=z$ and $t=1$ (the plane at infinity), which is a line at infinity as expected. This doesn’t quite match the equation $y+z=0$ that you came up with. I suspect that you’ve made a sign error somewhere.

For the second problem, you can make use of the fact that two lines that intersect must be coplanar. The line $\mathbf m$ must lie in the plane defined by $\mathbf l$ and the point $\mathbf p = [2:1:1:1]$ as well as the plane defined by $\mathbf k$ and $\mathbf p$, i.e., it can be computed as $meet[join[\mathbf l,\mathbf p],join[\mathbf k,\mathbf p]]$. For $join[\mathbf l,\mathbf p]$ you can either go back to the points originally used to define $\mathbf l$ and compute the unique plane defined by the three points (another null space computation), or use Plücker’s mu with the two planes computed in the first part for a direct calculation: $$join[\mathbf l,\mathbf p] = (\mathbf\pi_2^T\mathbf p)\mathbf\pi_1-(\mathbf\pi_1^T\mathbf p)\mathbf\pi_2 = [0:-1:1:0].$$ Slightly rearranged, this expression can also be read as the product of the line’s dual Plücker matrix $\mathbf\pi_1\mathbf\pi_2^T-\mathbf\pi_2\mathbf\pi_1^T$ and the point $\mathbf p$. For $join[\mathbf k,\mathbf p]$, we’re given a pair of planes that intersect in $\mathbf k$, so another application of Plücker’s mu seems convenient, yielding $[2:-4:-1:1]$. These two planes, with respective equations $y=z$ and $2x-4y-z+t=0$ uniquely identify the line, but we can also convert to the spanning set representation with another null space computation that yields $\lambda[-1:0:0:2]+\mu[5:2:2:0]$. When converted to inhomogeneous coordinates in $\mathbb R^3$, we get the parameterization $(-1/2,0,0)+\mu (5,2,2)$ for $\mathbf m$.

  • $\begingroup$ Thank you very much. This solution seems convenient to use, will study it thoroughly! $\endgroup$ – user526159 May 6 '18 at 8:31

Given two points $\;p_1,p_2\;$ in space $P$. Pick two other points $\;p_3,p_4\;$ in $P$ such that the four points are not linearly dependent. The determinant of the matrices having four rows $\;p_1,p_2,p_3,p\;$ and $\;p_1,p_2,p_4,p\;$ are each zero if $\;p=p_1\;$ or $\;p=p_2\;$ and thus they are two linear equations of planes and their intersection is line $\;k.\;$ This easily generalizes to higher dimensions.

In your case, the two points $\;p_1:=(1:0:0:0)\;$ and $\;p_2:=(0:1:1:0)\;$ with two more points $\;p_3:=(0:0:1:0)\;$ and $\;p_4:=(0:0:0:1)\;$ lead to the equations $\;l: y-z=0,\;t=0\;$ for the line $\;l=p_1p_2.\;$ This solves part of your problem. For the line $k$: $x+y+z-t=0$ and $x-y=0$ take two points on it, for example, $\;p_3:=(0:0:1:1)\;$ and $\;p_4:=(1:1:-1:1).\;$

Given a fifth point $\;p_5:=(2:1:1:1),\;$ to find a line $\;m\;$ incident with $\;k,l,p_5\;$ we use a similar method. The determinant of the matrices having four rows $\;p_1,p_2,p_5,p\;$ and $\;p_3,p_4,p_5,p\;$ being zero determine two planes each containing $\;p_5\;$ and in the first case containing line $\;k\;$ and the second case containing line $\;l.\;$ They determine the line $\;m\!: y-z=0\;$ and $\;2x-4y-z+t=0.\;$

  • $\begingroup$ Thank you very much! So $y+z=0$ and $t=0$ are indeed the equationa of line $l$ which apparently lies on the plane at infinity (?). I just realized since the space is $\mathbb{RP}^3$ two lines do not necessarily intersect. So i am stuck with the second problem again. Do you also happen to know how to calculate a line that goes through a point and intersects two other lines? $\endgroup$ – user526159 May 5 '18 at 12:20
  • $\begingroup$ @user526159 Ok. I now have the complete solution to your problem. $\endgroup$ – Somos May 5 '18 at 13:54

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