Inflation and Restriction

Suppose $ H$ is a subgroup of a finite group $ G$ and $ A$ is a $ G$-module. For each $ n\geq 0$, there is a natural map

$\displaystyle \res _H : \H^n(G,A) \to \H^n(H,A)

called restriction. Elements of $ \H^n(G,A)$ can be viewed as classes of $ n$-cocycles, which are certain maps $ G \times
\cdots \times G \to A$, and the restriction maps restricts these cocycles to $ H
\times \cdots \times H$.

If $ H$ is a normal subgroup of $ G$, there is also an inflation map

$\displaystyle {\mathrm{inf}}_H: \H^n(G/H, A^H) \to \H^n(G,A),

given by taking a cocycle $ f : G/H \times \cdots \times G/H \to A^H$ and precomposing with the quotient map $ G\to G/H$ to obtain a cocycle for $ G$.

Proposition 11.3.1   Suppose $ H$ is a normal subgroup of $ G$. Then there is an exact sequence

$\displaystyle 0 \to \H^1(G/H, A^H)\xrightarrow{{\mathrm{inf}}_H} \H^1(G,A)\xrightarrow{\res _H} \H^1(H,A).

Proof. Our proof follows [Ser79, pg. 117] closely.

We see that $ \res \circ {\mathrm{inf}}= 0$ by looking at cochains. It remains to prove that $ {\mathrm{inf}}_H$ is injective and that the image of $ {\mathrm{inf}}_H$ is the kernel of $ \res _H$.

  1. That $ {\mathrm{inf}}_H$ is injective: Suppose $ f:G/H\to A^H$ is a cocycle whose image in $ \H^1(G,A)$ is equivalent to 0 modulo coboundaries. Then there is an $ a\in A$ such that $ f(\sigma) =
\sigma a - a$, where we identify $ f$ with the map $ G\to A$ that is constant on the costs of $ H$. But $ f$ depends only on the costs of $ \sigma$ modulo $ H$, so $ \sigma a - a = \sigma \tau a - a$ for all $ \tau \in H$, i.e., $ \tau a = a$ (as we see by adding $ a$ to both sides and multiplying by $ \sigma^{-1}$).Thus $ a\in A^H$, so $ f$ is equivalent to 0 in $ \H^1(H,A^H)$.

  2. The image of $ {\mathrm{inf}}_H$ contains the kernel of $ \res _H$: Suppose $ f:G\to A$ is a cocycle whose restriction to $ H$ is a coboundary, i.e., there is $ a\in A$ such that $ f(\tau) = \tau a - a$ for all $ \tau \in H$. Subtracting the coboundary $ g(\sigma) = \sigma a - a$ for $ \sigma\in G$ from $ f$, we may assume $ f(\tau) = 0$ for all $ \tau \in H$. Examing the equation $ f(\sigma\tau) = f(\sigma) + \sigma f(\tau)$ with $ \tau \in H$ shows that $ f$ is constant on the cosets of $ H$. Again using this formula, but with $ \sigma\in H$ and $ \tau\in G$, we see that

    $\displaystyle f(\tau) = f(\sigma \tau) = f(\sigma) + \sigma f(\tau) = \sigma f(\tau),

    so the image of $ f$ is contained in $ A^H$. Thus $ f$ defines a cocycle $ G/H \to A^H$, i.e., is in the image of $ {\mathrm{inf}}_H$.

$ \qedsymbol$

This proposition will be useful when proving the weak Mordell-Weil theorem.

Example 11.3.2   The sequence of Proposition 11.3.1 need not be surjective on the right. For example, suppose $ H=A_3 \subset S_3$, and let $ S_3$ act trivially on the cyclic group $ C = \mathbf{Z}/3\mathbf{Z}$. Using the $ \Hom $ interpretation of $ \H^1$, we see that $ \H^1(S_3/A_3, C) = \H^1(S_3, C) = 0$, but $ \H^1(A_3, C)$ has order $ 3$.

Remark 11.3.3   On generalization of Proposition 11.3.1 is to a more complicated exact sequence involving the ``transgression map'' tr:

$\displaystyle 0 \to \H^1(G/H, A^H)\xrightarrow{{\mathrm{inf}}_H} \H^1(G,A)\xrig...
...w{\res _H} \H^1(H,A)^{G/H}
\xrightarrow{{\rm tr}} \H^2(G/H,A^H) \to \H^2(G,A).

Another generalization of Proposition 11.3.1 is that if $ \H^m(H,A) = 0$ for $ 1\leq m < n$, then there is an exact sequence

$\displaystyle 0 \to \H^n(G/H, A^H)\xrightarrow{{\mathrm{inf}}_H} \H^n(G,A)\xrightarrow{\res _H} \H^n(H,A).

Remark 11.3.4   If $ H$ is a not-necessarily-normal subgroup of $ G$, there are also maps

$\displaystyle \cores _H: \H^n(H,A) \to \H^n(G,A)

for each $ n$. For $ n=0$ this is the trace map $ a\mapsto \sum_{\sigma
\in G/H} \sigma a$, but the definition for $ n\geq 1$ is more involved. One has $ \cores _H \circ \res _H = [\char93 (G/H)]$. Taking $ H=1$ we see that for each $ n\geq 1$ the group $ \H^n(G,A)$ is annihilated by $ \char93 G$.

William Stein 2012-09-24