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Twisting gains real ground. The resulting bound is strictly stronger than the untwisted tropical bounds obtained from the usual characteristic varieties: for a one-relator group whose $\Sigma^1$ was computed by Brown, the untwisted bound excludes only two directions in $H^1(G;\mathbb{R})$, whereas the twisted bound determines $\Sigma^1(G)$ exactly. For a compact orientable $3$-manifold $M$ with toroidal or empty boundary, the twisted bound is sharp: the union of the twisted tropical varieties over all finite-image integral representations of $\pi_1(M)$ computes $\Sigma^1(\pi_1(M))$, and hence recovers the fibered faces of the Thurston norm ball. Sharpness genuinely requires twisting: a non-fibered class enters the tropical variety through the vanishing of a twisted Alexander polynomial along it, and the untwisted polynomial need not vanish. For a compact Kähler manifold $X$, we prove that the first twisted Alexander polynomial $\Delta^{\sigma}(X)$ is either $0$ or $1$, for every representation $\sigma$ over every field, and that $\Sigma^1(\pi_1(X))$ is controlled by the hyperbolic orbifold fibrations of $X$ for every $\sigma$. The obstruction to Kählerianity that comes out of this is strictly finer than its untwisted counterpart: we exhibit groups $G$ with $\Delta(G)=1$ but $\Delta^{\sigma}(G)\ne 1$, which the twisted test excludes from being Kähler and the classical one does not.
From: Yongqiang Liu [view email]
[v1]
Wed, 27 May 2026 15:14:40 UTC (36 KB)
[v2]
Tue, 11 Aug 2026 03:14:47 UTC (38 KB)
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