2
$\begingroup$

Let $G$ be a connected,reductive group over a $p$-adic field $k$. Let $M_0$ be a minimal Levi subgroup of $G$, and define $M_0^{\operatorname{der}}$ to be $M_0 \cap G_{\operatorname{der}}$. Lemma 2.1 of this paper by Takuya Konno on the Langlands classification for $p$-adic groups comes down to showing that

$$H^1(k, M_0^{\operatorname{der}}) \rightarrow H^1(k, G_{\operatorname{der}})$$

is injective. He claims that this is equivalent to Proposition 4.7 and Theorem 4.13 of the article Groupes Reductifs by Borel and Tits. These results are about conjugacy of minimal parabolics and Levis for a reductive group $G$ over an arbitrary field $k$:

Propostion 4.7: Let $P,P'$ be two parabolic $k$-subgroups of $G$. Then $P \cap P'$ contains a maximal torus defined over $k$. Every connected $k$-closed subgroup of $P \cap P'$ is defined over $k$. In particular, $P \cap P', \mathscr R(P \cap P'), \mathscr R_u(P \cap P')$ are defined over $k$. The group $P \cap P'$ has Levi $k$-subgroups, and any two of them are conjugate by a unique element of $\mathscr R_u(P \cap P')(k)$.

Theorem 4.13: Let $P, P'$ be two parabolic $k$-subgroups of $G$.

a) The fibration of $G$ by $P$ has a local section defined over $k$; the projection $G(k) \rightarrow G/P(k)$ is surjective.

b) If $P$ and $P'$ are minimal (among parabolic $k$-subgroups), they are conjugate over $k$.

c) If $P$ and $P'$ are conjugate over an extension of $k$, then they are conjugate over $k$.

I don't understand the connection between these results and the claim in the lemma. How can one interpret these results about the conjugacy of minimal parabolics as as a statement of injectivity of the $H^1$ sets?

$\endgroup$
3
$\begingroup$

For simplicity we write $G$ for $G^{\rm der}$ and $M_0$ for $M_0^{\rm der}$, then $G$ is a connected reductive group, and $M_0$ is a Levi subgroup of a minimal parabolic $P_0$ of $G$. Let $$\xi\in{\rm ker}[H^1(k,M_0)\to H^1(k, G)]$$ be a cohomology class, represented by some 1-cocycle $c\in Z^1(k,M_0)$. Since $\xi=[c]$ is contained in the kernel, the exists $g_1\in G(\bar k)$ such that $c_\sigma=g_1^{-1}\cdot \,^\sigma g_1$ for all $\sigma\in \Gamma:={\rm Gal}(\bar k/k)$. We set $$M_1=g_1\cdot M_0\cdot g_1^{-1},\quad P_1=g_1\cdot P_0\cdot g_1^{-1},$$ then one can easily check that $M_1$ and $P_1$ are defined over $k$. Since $(P_1,M_1)$ is conjugate to $(P_0,M_0)$ over $\bar k$, we see that $P_1$ is a parabolic of $G$ and that $M_1$ is a Levi subgroup of $P_1$. By Theorem 4.13(c) $P_0$ is conjugate to $P_1$ over $k$: $$P_0=g\cdot P_1\cdot g^{-1}\quad\text{for some }g\in G(k).$$ Set $g_2=g g_1\in G(\bar k)$, then $$g_2 \cdot P_0 \cdot g_2^{-1}=P_0\,,$$ hence, $g_2\in N_G(P_0)=P_0$. Now since $^\sigma g=g$, we notice that $$c_\sigma=g_2^{-1}\cdot \,^\sigma g_2\,,$$ so we see that $$\xi\in{\rm ker}[H^1(k,M_0)\to H^1(k,P_0)].$$ Since $H^1(k,P_0)= H^1(k,M_0)$, see e.g. Lemma 1.13 in Sansuc's paper, we conclude that $\xi=1$, hence $${\rm ker}[H^1(k,M_0)\to H^1(k, G)]=1.$$ A twisting argument shows that every fiber has only one element.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.