Tensor refinement of tame atoms

Proof of Theorem 5.1

Apply Mackey decomposition [23]. The double cosets

are the orbits of simultaneous translation on \({\mathbb Z}/n{\mathbb Z}\times{\mathbb Z}/m{\mathbb Z}\). There are \(d\) orbits, each of length \(L\). The stabilizer of a point in each orbit is \(W_{K_L}\). The second character is shifted by the chosen Frobenius representative. Summing the induced terms over the double cosets gives the decomposition.

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Proof of Corollary 5.2

The double coset calculation in Theorem 5.1 gives exactly \(d\) summands, and every stabilizer is \(W_{K_L}\), so every induced packet has dimension \(L\). The identity \(dL=nm\) follows from the definitions of greatest common divisor and least common multiple.

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Proof of Proposition 5.3

Norm lifting a character exponent from \({\mathbb F}_{q^n}^{\times}\) to \({\mathbb F}_{q^L}^{\times}\) multiplies the exponent by \((q^L-1)/(q^n-1)\). The Frobenius shift multiplies the second exponent by \(q^i\). Multiplication of characters adds exponents. Triviality is equivalent to the resulting exponent being zero modulo \(q^L-1\).

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Proof of Proposition 5.4

For nontrivial inertia, the invariant space is zero. In the trivial case, Frobenius on the invariant packet is the companion matrix of \(X^r-a\). Substituting \(X=u^{-1}\) and multiplying by \(u^r\) gives \(1-a u^r\).

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Proof of Corollary 5.5

The invariant space is the direct sum of the invariant refined packets, and determinant is multiplicative on direct sums.

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Proof of Proposition 5.6

A nonzero fixed vector must lie in an inertia trivial packet, so \(c_i=0\). On that packet, Frobenius is a companion cycle with characteristic polynomial \(X^{r_i}-a_i\). Its fixed space is the eigenspace for eigenvalue one. This eigenspace is one dimensional exactly when \(a_i=1\), and it is zero otherwise. Summing over the refined packets proves the formula.

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Proof of Corollary 5.7

The Hom space is the fixed space of \(\tau_\xi\otimes\tau_\eta\). Schur's lemma gives dimension zero or one, and Proposition 5.6 gives the packet criterion.

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Proof of Proposition 5.8

Wild inertia acts trivially, so the Swan conductor is zero. An inertia fixed vector would generate a nonzero Frobenius stable sum of trivial character packets. Irreducibility would then make inertia act trivially on all of \(\tau\). An irreducible representation of the quotient generated by Frobenius would be one dimensional, contrary to \(n>1\). The Artin conductor is therefore

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