Geometric representation of a rank 4 matroid I am working through Oxley's notes on Matroid Theory (https://www.math.lsu.edu/~oxley/survey4.pdf). Exercise 5.8 asks for a geometric representation of the matroid associated to the graph $K_5$, the complete graph on 5 vertices. In this situation, a geometric representation is a set of points representing the elements of the matroids and a set of "lines" and "planes" such that the following holds:

*

*Independent sets are sets of 2 or fewer vertices, sets of 3 non-collinear vertices and sets of 4 non-coplanar vertices


*2 lines intersect in at most 1 point


*2 planes intersect in at most 1 line (i.e any points in the intersection of 2 planes are colinear)
(It may be possible to do this with real lines and planes so that conditions 2 and 3 are trivial; I am unclear on this from the problem statement.)
I am having trouble doing this problem specifically (i.e for the matorid associated to $K_5$) but would also appreciate some pointers for general techniques for drawing out geometric representations of matroids because I find it difficult even in 2 dimensions.
 A: The construction is essentially: start with a circuit corresponding to a triangle in $K_5$. This circuit is represented by a line with 3 points on it. Now choose a second triangle circuit that shares an edge with the first circuit. In the representation then the corresponding line intersects with the line representing the first circuit. But now you have 4 additional points which themselves sit in some other circuits. Find these circuits, draw their lines and add points of intersection.
I suggest you follow the construction with drawing software such as Cinderella.2 or GeoGebra.
The triangles in $K_5$ are circuits (but not all circuits are triangles), so the corresponding points in the representation must be on the same lines. There are 10 triangles in $K_5$, so there are $10$ lines in the representation.
Let the vertices of $K_5$ be labelled by $\{1,2,3,4,5\}$, counter-clockwise.
We can start with one triangle circuit, e.g. $\{12,23,13\}$. These correspond to three points $p_{12}$, $p_{23}$, $p_{13}$ on the same line, call it $l_1$.
The point $p_{12}$ lies on three more lines, corresponding to the circuits $\{12,25,15\}$ and $\{12,24,14\}$. Let $l_2$ be the line with points $p_{12}$, $p_{25}$, $p_{15}$.
Now we have to check the dependencies between points on $l_1$ and $l_2$. The edges $23$ and $25$ belong to the circuit $\{23,25,35\}$, and $13$ and $15$ belong to the circuit $\{13,15,35\}$. Therefore we have to introduce a new point $p_{35}$ as the intersection of the two lines $l_3=\{13,15,35\}$ and $l_4=\{23,25,35\}$.
So far we haven't left the plane as no basis can be found in $\{12,13,15,23,25,35\}$.
We still haven't exhausted all lines through $p_{12}$, so let $l_5=\{12,24,14\}$. This line does not go through any of the present vertices, except obviously $p_{12}$. Furthremore, the tetrahedron $\{12,13,15,14\}$ is non-flat because it is a basis, i.e. it corresponds to a spanning tree of $K_5$, so with this line we are moving into $\mathbb R^3$.
So far we have 8 points. We're missing the points $p_{34}$ and $p_{45}$. The point $p_{45}$ is on the intersection of the lines defined by the circuits $\{14,15,45\}$ and $\{24,25,45\}$, so we add the lines $l_6=\{14,15\}$ and $l_7=\{24,25\}$ with $p_{45}$ as their intersection.
Similarly, the point $p_{34}$ is on the intersection of the lines defined by the circuits $\{13,14,34\}$ and $\{23,24,34\}$, so we add the lines $l_8=\{13,14\}$ and $l_9=\{23,24\}$ with $p_{35}$ as their intersection.
So we have 10 points in the representation but only 9 lines. The missing line is $l_{10}$ corresponding to the circuit $\{34,35,45\}$. That this is actually a line in $\mathbb R^3$ is guaranteed by Desargues' theorem.
