Lemma 15.112.7. With assumptions and notation as in Lemma 15.112.5. The inertia character $\theta : I \to \mu _ e(\kappa (\mathfrak m))$ satisfies the following property

for $\tau \in D$ and $\sigma \in I$.

Lemma 15.112.7. With assumptions and notation as in Lemma 15.112.5. The inertia character $\theta : I \to \mu _ e(\kappa (\mathfrak m))$ satisfies the following property

\[ \theta (\tau \sigma \tau ^{-1}) = \tau (\theta (\sigma )) \]

for $\tau \in D$ and $\sigma \in I$.

**Proof.**
The formula makes sense as $I$ is a normal subgroup of $D$ and as $\tau $ acts on $\kappa (\mathfrak m)$ via the map $D \to \text{Aut}(\kappa (\mathfrak m))$ discussed in Lemma 15.112.4 for example. Recall the construction of $\theta $. Choose a uniformizer $\pi $ of $B_\mathfrak m$ and for $\sigma \in I$ write $\sigma (\pi ) = \theta _\sigma \pi $. Then $\theta (\sigma )$ is the image $\overline{\theta }_\sigma $ of $\theta _\sigma $ in the residue field. For any $\tau \in D$ we can write $\tau (\pi ) = \theta _\tau \pi $ for some unit $\theta _\tau $. Then $\theta _{\tau ^{-1}} = \tau ^{-1}(\theta _\tau ^{-1})$. We compute

\begin{align*} \theta _{\tau \sigma \tau ^{-1}} & = \tau (\sigma (\tau ^{-1}(\pi )))/\pi \\ & = \tau (\sigma (\tau ^{-1}(\theta _\tau ^{-1}) \pi ))/\pi \\ & = \tau (\sigma (\tau ^{-1}(\theta _\tau ^{-1})) \theta _\sigma \pi )/\pi \\ & = \tau (\sigma (\tau ^{-1}(\theta _\tau ^{-1}))) \tau (\theta _\sigma ) \theta _\tau \end{align*}

However, since $\sigma $ acts trivially modulo $\pi $ we see that the product $\tau (\sigma (\tau ^{-1}(\theta _\tau ^{-1}))) \theta _\tau $ maps to $1$ in the residue field. This proves the lemma. $\square$

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