I need to show that there is a unique line (in $\mathbb{P}^4$ I assume, or could they also mean in $\mathbb{R}^5$?) that intersects three lines $L,M,N$ which are pairwise non-intersecting and not in the same hyperplane.

In a previous exercise I worked out that $\dim(\langle L, M\rangle \cap N)=0$, so a single point in $\mathbb{P}^4$.

I think I have to use this dimension, but I'm not sure what it means. It says $\langle L, M\rangle$ is the union of all lines through $L$ and $M$. So is $\langle L, M\rangle \cap N$ then the set of all lines through all three lines?

Other things I tried:

I know that $\langle L, M, N \rangle = 4$, since these lines are not in the same hyperplane.

Let $P$ be a point on $L$. Let $\Pi$ be the plane containing $M$ and $P$. In the same way we can construct a plane $\Pi'$ through $N$ and $P$. But I don't know if I'm going in the good direction, I simply don't know how to visualize this stuff in higher dimensions..


The key here is that $\langle L,M\rangle$ here is a hyperplane, three-dimensional in $\mathbb{P}^4$. There is, as you have noted, a single point $x$ where $N$ meets this hyperplane.

Now, the union of all lines through $x$ that intersect $L$ is a 2-d plane in $\langle L,M\rangle$. Similarly, the union of lines through $x$ intersecting $M$ is another 2-d plane in $\langle L,M\rangle$. The intersection of these planes will be a line; it can't be a plane because then $L$ and $M$ would intersect, and a dimension argument says it's at least a line. That line passes through $x$, so it's one of the lines that defines each of the planes, and thus it's the line we seek.

Now, seeing how this fits together, an explicit construction:

Define the lines $L$, $M$, $N$ as follows: $L=\{ax_1+bx_2\mid a,b\in \mathbb{R}\}$, $M=\{cy_1+dy_2\mid c,d\in \mathbb{R}\}$, $N=\{ez_1+fz_2\mid e,f\in \mathbb{R}\}$ where $x_1,x_2,y_1,y_2,z_1,z_2$ are pairs of points on each line, with coordinates in $\mathbb{R}^5\setminus \{0\}$. All of these use the description of $\mathbb{P}^4$ as the quotient of $\mathbb{R}^5$ minus the origin by $\mathbb{R}^*$.

Now, the six 5-tuples $x_1,x_2,y_1,y_2,z_1,z_2$ in $\mathbb{R}^5$ must have a nontrivial linear dependence relation; there are constants $A,B,C,D,E,F$ not all zero so that $Ax_1+Bx_2+Cy_1+Dy_2+Ez_1+Fz_2 = 0$. These constants can be found by Gaussian elimination on the matrix with rows $x_1,x_2,y_1,y_2,z_1,z_2$ augmented with a $6\times 6$ identity.

Since the lines don't lie in a hyperplane, our six points must span $\mathbb{R}^5$, and the constants $A,B,C,D,E,F$ are unique up to constant multiples. Also, since $L$ and $M$ don't intersect, there's no nontrivial linear combination of $x_1,x_2,y_1,y_2$ that's zero - which means that at least one of $E$ and $F$ is nonzero. Similarly, at least one of $A$ and $B$ is nonzero, and at least one of $C$ and $D$ is nonzero.

The line we seek is the set of linear combinations $$\{\alpha(Ax_1+Bx_2)+\beta(Cy_1+Dy_2)+\gamma(Ez_1+Fz_2)\mid (\alpha,\beta,\gamma)\in \mathbb{R}^3\}$$ Since the three vectors $Ax_1+Bx_2$, $Cy_1+Dy_2$, $Ez_1+Fz_2$ are linearly dependent, that's a $2$-dimensional subset of $\mathbb{R}^5$, which projects to a line. It intersects $L$ at $Ax_1+Bx_2$ for $\alpha=1,\beta=0,\gamma=0$ and similarly for the other two lines. We could also eliminate the redundancy by restricting to $\alpha+\beta+\gamma=0$.

  • $\begingroup$ How do you see all lines from $L$ through $x$ lie in $\langle L, M \rangle$? Wouldn't that mean that all these lines also go through $M$, since $\langle L, M \rangle$ is the set of all lines going through $L$ and $M$? Wouldn't that then mean that there are infinitely many lines through $L, M$ and $N$? $\endgroup$ – The Coding Wombat Mar 28 at 22:03
  • $\begingroup$ $\langle L, M\rangle$ isn't a set of lines. It's a set of points - each point in it lies in some line intersecting both $L$ and $M$. That set of points is a 3-d hyperplane. If we take some random line composed of points in it, that probably won't be one of the lines we used to define it, so there's no reason to expect it will intersect either $L$ or $M$. $\endgroup$ – jmerry Mar 28 at 22:13
  • $\begingroup$ How do we know the line is unique? Couldn't there be a line through L,M and N, not going through $x$? $\endgroup$ – The Coding Wombat Mar 28 at 22:19
  • 1
    $\begingroup$ No. $x$ is the only point on $N$ that lies on any line intersecting $L$ and $M$. That's what your previous exercise result is telling you. We have to work a bit constructing the line and showing that it's the only line that works, but that any such line goes through $x$? We know that from the start. (In my explicit construction, $x$ is the point $Ez_1+Fz_2$.) $\endgroup$ – jmerry Mar 28 at 22:48

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