[proofplan]
We prove that the Bergman circuit condition is equivalent to the condition that all strict superlevel sets of the weight vector are flats. This converts an arbitrary Bergman vector into a chain of flats by grouping the ground-set elements according to their weights. Conversely, any vector in a cone attached to a flag of flats has strict superlevel sets among those flats, so every circuit has its minimum weight at least twice. The final dependence statement follows because the construction uses only containment among flats.
[/proofplan]
custom_env
admin
[step:Define strict superlevel sets and record their chain structure]Let $w \in \mathbb{R}^E$ be a weight vector. For each real number $t \in \mathbb{R}$, define the strict superlevel set
\begin{align*}
F_t(w):=\{e \in E : w_e > t\}.
\end{align*}
Since $E$ is finite, the distinct values taken by $w$ may be written as
\begin{align*}
\alpha_1 > \alpha_2 > \cdots > \alpha_m.
\end{align*}
For each index $j \in \{1,\dots,m-1\}$, define
\begin{align*}
F_j:=\{e \in E : w_e > \alpha_{j+1}\}.
\end{align*}
Then
\begin{align*}
\varnothing \subsetneq F_1 \subsetneq F_2 \subsetneq \cdots \subsetneq F_{m-1} \subsetneq E.
\end{align*}
Moreover, every proper nonempty strict superlevel set $F_t(w)$ is one of the sets $F_j$. Indeed, if $j \in \{1,\dots,m-1\}$ and $\alpha_{j+1}\leq t < \alpha_j$, then $F_t(w)=F_j$; if $t\geq \alpha_1$, then $F_t(w)=\varnothing$; and if $t<\alpha_m$, then $F_t(w)=E$.[/step]
custom_env
admin
[guided]The useful way to read a vector $w \in \mathbb{R}^E$ is not coordinate by coordinate, but by its level sets. Because $E$ is finite, the set of weights $\{w_e : e \in E\}$ is finite, so we can list the distinct weights in strictly decreasing order:
\begin{align*}
\alpha_1 > \alpha_2 > \cdots > \alpha_m.
\end{align*}
For a real number $t$, we define
\begin{align*}
F_t(w):=\{e \in E : w_e > t\}.
\end{align*}
This is the subset of elements whose weights lie strictly above the threshold $t$.
The only thresholds that matter are those between two consecutive weight levels. Thus, for $j \in \{1,\dots,m-1\}$, define
\begin{align*}
F_j:=\{e \in E : w_e > \alpha_{j+1}\}.
\end{align*}
This set consists exactly of the elements with weights $\alpha_1,\dots,\alpha_j$. Hence the sets form a strictly increasing chain:
\begin{align*}
\varnothing \subsetneq F_1 \subsetneq F_2 \subsetneq \cdots \subsetneq F_{m-1} \subsetneq E.
\end{align*}
The inclusions are strict because each weight level $\alpha_j$ occurs for at least one element of $E$. Every nonempty proper strict superlevel set is one of these $F_j$, since changing $t$ without crossing a weight level does not change the set $\{e : w_e>t\}$.[/guided]
custom_env
admin
[step:Show that the circuit condition forces every strict superlevel set to be a flat]
Assume that $w \in \widetilde{\mathcal B}(M)$. Let
\begin{align*}
\operatorname{cl}_M:2^E \to 2^E
\end{align*}
denote the matroid closure operator of $M$. We prove that every proper nonempty strict superlevel set $F_t(w)$ is a flat of $M$.
Fix $t \in \mathbb{R}$, and write $F:=F_t(w)$. Suppose, for contradiction, that $F$ is not a flat. Then there exists an element $e \in \operatorname{cl}_M(F)\setminus F$. Since $e \in \operatorname{cl}_M(F)$ and $e \notin F$, the matroid closure criterion gives a circuit $C$ of $M$ such that
\begin{align*}
e \in C \subset F \cup \{e\}.
\end{align*}
For every $x \in C\setminus\{e\}$, we have $x \in F$, hence $w_x>t$. Since $e \notin F$, we have $w_e \leq t$. Therefore
\begin{align*}
w_e < w_x
\end{align*}
for every $x \in C\setminus\{e\}$. Thus $e$ is the unique element of $C$ on which $w$ attains its minimum, contradicting $w \in \widetilde{\mathcal B}(M)$. Hence $F$ is a flat.
[/step]
custom_env
admin
[step:Show that flat strict superlevel sets imply the circuit condition]Assume that every proper nonempty strict superlevel set $F_t(w)$ is a flat of $M$. Let $C$ be a circuit of $M$. Suppose, for contradiction, that the minimum of $w$ on $C$ is attained at a unique element $e \in C$. Define the real number
\begin{align*}
\beta:=w_e
\end{align*}
and define the strict superlevel set
\begin{align*}
F:=F_\beta(w)=\{x \in E : w_x>\beta\}.
\end{align*}
Since $e$ is the unique minimum-weight element of $C$, every element of $C\setminus\{e\}$ belongs to $F$, while $e \notin F$. Thus
\begin{align*}
C\setminus\{e\}\subset F
\end{align*}
and
\begin{align*}
e\notin F.
\end{align*}
The set $F$ is proper because $e \notin F$. If $F=\varnothing$, then $C\setminus\{e\}=\varnothing$, so $C=\{e\}$ would be a loop, contradicting that $M$ is loopless. Hence $F$ is nonempty, and by hypothesis $F$ is a flat.
Because $C$ is a circuit and $e \in C$, the circuit closure property gives
\begin{align*}
e \in \operatorname{cl}_M(C\setminus\{e\}).
\end{align*}
Since $C\setminus\{e\}\subset F$ and closure is monotone,
\begin{align*}
e \in \operatorname{cl}_M(F).
\end{align*}
But $F$ is a flat, so $\operatorname{cl}_M(F)=F$, contradicting $e \notin F$. Therefore no circuit has a unique minimum-weight element, and $w \in \widetilde{\mathcal B}(M)$.[/step]
custom_env
admin
[guided]We now prove the converse direction. Assume every proper nonempty strict superlevel set of $w$ is a flat. To show that $w$ lies in the Bergman fan, we must check the defining circuit condition: for every circuit $C$, the minimum of $w$ on $C$ is attained at least twice.
Fix a circuit $C$. Suppose the desired conclusion fails. Then there is a unique element $e \in C$ where $w$ is minimal on $C$. Define
\begin{align*}
\beta:=w_e.
\end{align*}
Now form the strict superlevel set at the level $\beta$:
\begin{align*}
F:=F_\beta(w)=\{x \in E : w_x>\beta\}.
\end{align*}
Because $e$ is the unique minimum of $w$ on $C$, every other element of $C$ has weight strictly larger than $\beta$. Hence
\begin{align*}
C\setminus\{e\}\subset F.
\end{align*}
At the same time, $e\notin F$ because $w_e=\beta$, not $w_e>\beta$.
We must check that the hypothesis about superlevel sets applies to $F$. The set $F$ is proper since $e\notin F$. It is also nonempty: if $F=\varnothing$, then $C\setminus\{e\}=\varnothing$, so $C=\{e\}$ would be a one-element circuit, meaning that $e$ is a loop. This contradicts the assumption that $M$ is loopless. Therefore $F$ is a proper nonempty strict superlevel set, and by hypothesis $F$ is a flat.
Now we use the defining closure property of circuits. For any circuit $C$ and any element $e\in C$, the element $e$ lies in the matroid closure of $C\setminus\{e\}$:
\begin{align*}
e \in \operatorname{cl}_M(C\setminus\{e\}).
\end{align*}
Since $C\setminus\{e\}\subset F$, monotonicity of matroid closure gives
\begin{align*}
e \in \operatorname{cl}_M(F).
\end{align*}
But $F$ is a flat, so $\operatorname{cl}_M(F)=F$. Hence $e\in F$, contradicting the already established fact that $e\notin F$. The contradiction shows that the minimum on $C$ cannot be unique. Since $C$ was arbitrary, $w$ satisfies the Bergman circuit condition.[/guided]
custom_env
admin
[step:Express a Bergman vector as a nonnegative combination of indicator vectors of flats]
Let $w \in \widetilde{\mathcal B}(M)$, and let
\begin{align*}
\alpha_1 > \alpha_2 > \cdots > \alpha_m
\end{align*}
be the distinct values of $w$. For $j \in \{1,\dots,m-1\}$, let
\begin{align*}
F_j:=\{e \in E : w_e>\alpha_{j+1}\}.
\end{align*}
By the previous steps, each $F_j$ is a proper nonempty flat. Define nonnegative [real numbers](/page/Real%20Numbers)
\begin{align*}
\lambda_j:=\alpha_j-\alpha_{j+1}
\end{align*}
for $j \in \{1,\dots,m-1\}$. Then, for every $e \in E$,
\begin{align*}
w_e=\alpha_m+\sum_{j=1}^{m-1}\lambda_j(\mathbb{1}_{F_j})_e.
\end{align*}
Equivalently,
\begin{align*}
w=\alpha_m\mathbb{1}_E+\sum_{j=1}^{m-1}\lambda_j\mathbb{1}_{F_j}.
\end{align*}
Thus $w$ belongs to the cone
\begin{align*}
\mathbb{R}\mathbb{1}_E+\mathbb{R}_{\geq 0}\{\mathbb{1}_{F_1},\dots,\mathbb{1}_{F_{m-1}}\}
\end{align*}
associated to the chain of flats $\varnothing \subsetneq F_1 \subsetneq \cdots \subsetneq F_{m-1}\subsetneq E$.
[/step]
custom_env
admin
[step:Verify that every flat flag cone satisfies the circuit condition]
Let
\begin{align*}
\varnothing \subsetneq F_1 \subsetneq \cdots \subsetneq F_k \subsetneq E
\end{align*}
be a chain of proper nonempty flats of $M$. Let $a \in \mathbb{R}$ and let $\lambda_1,\dots,\lambda_k \in \mathbb{R}_{\geq 0}$. Define
\begin{align*}
w:=a\mathbb{1}_E+\sum_{i=1}^k \lambda_i\mathbb{1}_{F_i}\in \mathbb{R}^E.
\end{align*}
For each $t \in \mathbb{R}$, the strict superlevel set $F_t(w)$ is one of $\varnothing$, $E$, or one of the flats $F_i$ with zero-coefficient repetitions omitted. Indeed, the value of $w_e$ is constant on each difference $F_i\setminus F_{i-1}$, where $F_0:=\varnothing$, and on $E\setminus F_k$; moreover these values weakly decrease as one moves from $F_1$ outward to $E\setminus F_k$ because each $\lambda_i\geq 0$. Therefore every proper nonempty strict superlevel set of $w$ is a flat. By the equivalence proved above, $w \in \widetilde{\mathcal B}(M)$.
[/step]
custom_env
admin
[step:Conclude the fan decomposition and dependence on the lattice of flats]
The previous two steps prove both inclusions:
\begin{align*}
\widetilde{\mathcal B}(M)\subset \bigcup_{\varnothing \subsetneq F_1 \subsetneq \cdots \subsetneq F_k \subsetneq E} \left(\mathbb{R}\mathbb{1}_E+\mathbb{R}_{\geq 0}\{\mathbb{1}_{F_1},\dots,\mathbb{1}_{F_k}\}\right)
\end{align*}
and
\begin{align*}
\bigcup_{\varnothing \subsetneq F_1 \subsetneq \cdots \subsetneq F_k \subsetneq E} \left(\mathbb{R}\mathbb{1}_E+\mathbb{R}_{\geq 0}\{\mathbb{1}_{F_1},\dots,\mathbb{1}_{F_k}\}\right)\subset \widetilde{\mathcal B}(M).
\end{align*}
Hence the asserted equality holds. The line $\mathbb{R}\mathbb{1}_E$ is common to every cone, so quotienting by it gives the corresponding flag cones modulo diagonal lineality. Since the construction uses only the collection of flats and their inclusion chains, the Bergman fan depends only on the lattice of flats of $M$.
[/step]