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I want to find a generalization of the idea that, in a graphic matroid, every base can be decomposed on the stars (edges adjacent to a vertex).

For example one could say that a matroid $M$ of rank $k$ has "generating independent sets" $I_1, \dots, I_k$ if every base $B$ of $M$ can be written $B = \{b_1, ..., b_k\}$ with $b_i \in I_i$. I want to know if these matroids have been studied and if they have a name.

I found the notion of "combinatorial decomposable" (http://www.math.univ-montp2.fr/~ramirez/Oxford.pdf) which is close but not exactly the same. There is also the Matroid covering theorem (http://math.mit.edu/~goemans/18438F09/lec13.pdf) but again it is not exactly the same.

I would also be interested in the stronger property that there is a unique way to write the base $B$ as a system of distinct representative in $I_1, \cdots , I_k$.

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  • $\begingroup$ But a set of edges incident to a vertex is not a base. $\endgroup$ Commented Jun 9, 2016 at 7:22
  • $\begingroup$ If the matroid graph is not complete, certainly not indeed. Let's say that I don't need B_1, ... B_k to be bases but only independent. $\endgroup$ Commented Jun 9, 2016 at 7:30

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Let $G = (V,E)$ be a connected graph on $n$ vertices. The star at vertex $v$ described in the question is the cut-set associated the partition $v \uplus V\setminus v$ of the vertices. We remark that these cut-sets will be independent sets when $G$ is a simple graph, but will not always be independent sets for general (multi)graphs. Take any $n-1$ vertices, the cut-sets associated the these vertices given a basis of the cut-space of $G$. The cut-space is the orthgonal complement of the cycle-space of $G$ in the edge-space which is $\mathbb{F}_2^E$ where addition is given by symmetric difference. We now give a generalization of the OP's observation that every spanning tree can be decomposed over the cut-sets of vertices.

If $T = \{e_1, \dots, e_{n-1}\}$ is a spanning tree and $C^*_1, \dots, C^*_{n-1}$ be a basis of the cut-space, then there exists $\sigma \in S_{n-1}$ such that $e_i \in C^*_{\sigma(i)}$.

The spanning tree $T$ gives us another basis $C^*_{e_1}, \dots, C^*_{e_{n-1}}$ of the cut-space consisting of fundamental cuts. That is, $C^*_{e_i}$ is the set of edges of $G$ crossing the partition of vertices given by the connected components of $T \setminus e$. We observe that $e_i \in C^*_{e_i}$ for all $i$ and $e_i \not\in C^*_{e_j}$ for all $i \ne j$.

Since the fundamental cuts form a basis we have $$ \begin{bmatrix} & \vdots &\\ \cdots & a_{ij} & \cdots\\ & \vdots & \end{bmatrix} \begin{bmatrix} C^*_{e_1}\\ \vdots \\ C^*_{e_{n-1}} \end{bmatrix} = \begin{bmatrix} C^*_1\\ \vdots \\ C^*_{n-1} \end{bmatrix} $$

for $A = [a_{ij}] \in \mathbb{F}_2^{(n-1) \times (n-1)}$. Notice $a_{ij} = 1$ if and only if $e_j \in C^*_i$. So, we must find a permutation $\sigma \in S_{n-1}$ such that $a_{i,\sigma(i)} = 1$ for all $i$. The matrix $A$ is invertible since $C^*_1, \dots, C^*_{n-1}$ is also a basis. So, $$\det A = \sum_{\sigma \in S_{n-1}} \prod_i a_{i,\sigma(i)} \ne 0$$ and it follows such a $\sigma$ exists.

Now this generalizes the OP's observation within the graphic matroid setting. For general matroids cut-sets become cocircuits. However, the cocircutis of general matroids do not form a vector space with symmetric difference. So, the argument above does not extend, but perhaps something can be said looking that cocircuits or fundamental cocircuits

For an example consider the non-graphic matroid $U_{2,4}$. The symmetric difference of the cociruits $123$ and $234$ is $14$ which is not a cocircuit but a base. However, the fundamental cociruits of $12$ are $134$ and $234$, and every base of $U_{2,4}$ can be decomposed over these cocircuits.

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  • $\begingroup$ Thank you. Indeed the fundamentals cocircuits associated to a base of a general matroid have the property that every base is a transversal of these cocircuits (ON FUNDAMENTAL TRANSVERSAL MATROID, R. A. BRU ALDI, Corollary 1). I am not sure in which extent these cocircuits have "good" properties. $\endgroup$ Commented Jun 13, 2016 at 13:40
  • $\begingroup$ Ok so my intuition was right about the cocircuits, thanks for the reference! Yes, I was not sure how helpful it is to you, but I though I might offer some explanation of the phenomenon you observed for graphs. $\endgroup$ Commented Jun 13, 2016 at 20:28
  • $\begingroup$ It was useful, and your proof also prove that every spanning tree has an odd number of representations (i.e there are an odd number of possible sigmas in your statement). $\endgroup$ Commented Jun 14, 2016 at 7:52

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