ScalingStacks

Theorem 1.11 [014B]

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Theorem 1.11

Let f:X→Bf:X\rightarrow B be a well-behaved T3T^{3}-fibration, b0∈Δdb_{0}\in\Delta_{d}, with Xb0X_{b_{0}} a semi-stable fibre, U,T1,…,TnU,T_{1},\ldots,T_{n} as above. Then

(1) If Xb0X_{b_{0}} is of type (2,2)(2,2), then n=2n=2 and in a suitable basis, T1=(101010001)T_{1}=\pmatrix{1&0&1\cr 0&1&0\cr 0&0&1\cr} and T2=T1−1T_{2}=T_{1}^{-1}.

(2) If Xb0X_{b_{0}} is of type (1,2)(1,2) or (2,1)(2,1), then n=3n=3 and in a suitable basis, T1=(101010001)T_{1}=\pmatrix{1&0&1\cr 0&1&0\cr 0&0&1\cr}, T2=(100011001)T_{2}=\pmatrix{1&0&0\cr 0&1&1\cr 0&0&1\cr} and T3=(10−101−1001)T_{3}=\pmatrix{1&0&-1\cr 0&1&-1\cr 0&0&1\cr} in the (1,2)(1,2) case, and the transpose of these matrices in the (2,1)(2,1) case.

(3) If Xb0X_{b_{0}} is a fibre of type (1,1)(1,1), then n=4n=4. Furthermore, in a suitable basis, possibly after relabelling the pip_{i}’s, T1=(110010001)T_{1}=\pmatrix{1&1&0\cr 0&1&0\cr 0&0&1\cr}, T2=(100011001)T_{2}=\pmatrix{1&0&0\cr 0&1&1\cr 0&0&1\cr}, T3=(1−1a010001)T_{3}=\pmatrix{1&-1&a\cr 0&1&0\cr 0&0&1\cr}, and T4=(10−a−101−1001)T_{4}=\pmatrix{1&0&-a-1\cr 0&1&-1\cr 0&0&1\cr}.

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