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Let $M$ be a matroid on the set $\{1,\dots,m\}$. The Bergman fan $\tilde{B}(M)$ is defined in literature to be the set in $\mathbb{R}^{m}$ consisting of the vectors $v=(w_1,\dots,w_m)$ such that for every circuit $c\in C(M)$, the minimum $\min_{i\in c}w_i$ is attained at least twice.

It is then mostly left implicit what the fan structure on this set should be. Sometimes people mention equivalence classes formed by $v_1\sim v_1$ iff $M_{v_1}=M_{v_2}$, where $M_v$ is the matroid with as bases those bases $B$ of $M$ attaining the minimum $\min_{B\in B(M)} \sum_{b\in B}v_b$. For what I gather it is implied that these should define a fan structure on $\tilde{B}(M)$, but simple examples show that it doesn't:

Let $M=U_{3,3}$. Then there are no circuits, so the support of $\tilde{B}(M)$ is the whole space. Since there is only a single basis, there is only a single equivalence class. So the equivalence classes do not define a fan structure on $\tilde{B}(M)$.

What am I not getting here?

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From Matroid polytopes, nested sets and Bergman fans, Feichtner and Sturmfels, just above Proposition 2.5 on the bottom of page 4:

Two vectors $w,w' \in \mathbb{R}^n$ are considered equivalent for the matroid $M$ if $M_w = M_{w'}$. The equivalence classes are relatively open convex polyhedral cones. These cones form a complete fan in $\mathbb{R}^n$. This fan is the normal fan of $P_M$. If $\Gamma$ is a cone in the normal fan of $P_M$ and $w \in \Gamma$ then we write $M_\Gamma = M_w$. The following proposition shows that the Bergman fan $\tilde{\mathcal{B}}(M)$ is a subfan of the normal fan of the matroid polytope $P_M$.

For $M = U(3,3)$ the one equivalence class is $\mathbb{R}^3$, which is indeed open, convex and a cone (it is the positive hull of $\{\pm e_1, \pm e_2, \pm e_3\}$). The set $\{\mathbb{R}^3\}$ is a fan:

  • every face of a cone in the fan is in the fan (check: $\mathbb{R}^3$ is the only face)
  • the intersection of any two cones in the fan is a face of each (check: $\mathbb{R}^3$ is the only intersection)

I tried working out another example but I'm still new to this so bear with me. For another example, let $M = U(1,3)$. Then the equivalence classes can be parameterized in terms of subsets of the bases of $M$:

  • $M_v$ has only $1$ as a basis iff $v = (a,b,b)$ with $a > b$.

  • $M_v$ has $1,2$ as bases iff $v = (a,a,b)$ with $a > b$

  • $M_v$ has $1,2,3$ as bases iff $v = (a,a,a)$

and we have similar conditions for permutations of the variables.

Therefore the equivalence classes are $$[1,0,0], [0,1,0], [0,0,1], [1,1,0], [1,0,1], [0,1,1], [1,1,1]$$ where $[1,0,0] = -[0,1,1], [0,1,0] = -[1,0,1], [0,0,1] = -[1,1,0]$. The line $[1,1,1]$ is a face of every cone. The intersecton of any two cones $C_1, C_2$ is either the line $[1,1,1]$ or if $C_1 = C_2$ then $C_1 \cap C_2 = C_1$.

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    $\begingroup$ Thanks for the answer. The reason why I said that $\mathbb{R}^3$ was not a fan, is that I always believed that cones cannot contain any line through the origin. I see now that those are called "strongly convex" instead of "convex" cones. That clears it up. $\endgroup$ Commented Aug 8, 2017 at 15:52

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