I am trying to understand the following picture of a matroid and its dual but I found myself not understanding exactly what we are doing to find the dual:
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Roughly speaking, according to some document online , Matroid duality in a nutshell is:
1-contraction $\longrightsquigarrow$ deletion
2-base $\longrightsquigarrow$ complement of a base
3-independent set $\longrightsquigarrow$ complement of a spanning set
4-circuit (minimal dependent set) $\longrightsquigarrow$ cut (minimal, intersects every base)
5-loop (no independent set contains it) $\longrightsquigarrow$ bridge(coloop) (every base contains it)
(The terminology comes from graphic matroids.)
I still can not see how the sets in 3 & 4 are complements of one another. Also, I can not see how is this applied in the picture below.
If someone clarify the exact steps that I should follow to dualize a matroid I would appreciate it very much.
I take it that we’re talking about the graphic matroid corresponding to the depicted graph $G$.
The dual of a graphic matroid is a graphic matroid if and only if the graph is planar (see Wikipedia here and here). $G$ is obviously planar. In this case, the dual of the graphic matroid of $G$ is the graphic matroid of the dual of $G$. Actually, of a dual of $G$, since $G$ may have various duals depending on how we embed it in the plane, but the graphic matroids of all these dual graphs are isomorphic. So let’s stick with the depicted embedding.
The dual graph $G^*$ has a vertex for each face of $G$. In this case, $G$ has three faces (including the unbounded exterior face). And for each edge $e$ of $G$, the dual graph $G^*$ has an edge $e^*$ that connects the faces that $e$ separates. In your image of $G^*$, the vertex on the left corresponds to the exterior face, the top right vertex corresponds to the triangle and the bottom right vertex corresponds to the square. Each of the six edges in $G^*$ corresponds to the edge with the same label in $G$ and connects the faces that that edge separates.