State/The domain topograph has no circuits

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Proposition
Proof

Proposition: (The domain topograph has no circuits) There do not exist circuits in the domain topograph.

Logical Connections

This statement logically relies on the following definitions and statements: Def/Circuit, Def/Domain topograph, State/Binary quadratic forms are uniquely determined by their values at a superbasis, State/Arithmetic progression rule for binary quadratic forms, State/Climbing in topographs

The following statements and definitions rely on the material of this page:

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State/The domain topograph has no circuits

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	Suppose that $p$ is a point in the domain topograph, and $(b_1, \ldots, b_\ell)$ is a circuit of length $\ell \geq 1$ consisting of edges in the topograph, based at $p$. The point $p$ corresponds to a lax superbasis $\{ \pm \vec u, \pm \vec v, \pm \vec w \}$. There exists a unique quadratic form $Q$, such that: $Q(\vec u) = 1$, $Q(\vec v) = 1$, and $Q(\vec w) = 1$. In the range topograph of the quadratic form $Q$, the edges adjacent to $p$ are directed away from $p$, as seen from the arithmetic progression rule, as illustrated. The range topograph near $p$ Since the path $b_1, \ldots, b_\ell$ is a path that begins at $p$, and $b_1$ climbs away from $p$, the climbing property of range topographs implies that every oriented edge $b_i$ climbs from its source to its target. However, the final edge $b_\ell$ has target $p$, and all edges climb away from $p$.
	This contradiction implies that there are no circuits in the domain topograph.