4
$\begingroup$

Let $T$ be a torus over a $p$-adic field $F$. Let $q = f(F/\mathbb{Q}_p)$, and normalize the absolute value $| \cdot |$ on $F$ so that a uniformizer has value $\frac{1}{q}$.

Let $X(T)_F$ be the group of rational characters of $T$ which are defined over $F$. Define a homomorphism

$$H_T: T(F) \rightarrow \textrm{Hom}_{\mathbb{Z}}(X(T)_F,\mathbb{Z})$$

by $$H_T(t)(\chi) = \log_q |\chi(t)|$$

Giving the free abelian group $\textrm{Hom}_{\mathbb{Z}}(X(T)_F,\mathbb{Z})$ the discrete topology, it is easy to see that $H_T$ is continuous: the kernel is open as a finite intersection of open subgroups $\chi^{-1}\mathcal O_F^{\ast}$, as $\chi$ runs through a basis of $X(T)_F$.

It follows that $\textrm{Ker}(H_T)$ contains every compact subgroup of $T(F)$, because if $E$ is any compact subgroup of $T(F)$, then $H_T(E)$ is isomorphic to a compact, hence finite, subgroup of $\mathbb{Z}^n$, and every nontrivial subgroup of $\mathbb{Z}^n$ is infinite.

On the other hand, I have read that $\textrm{Ker}(H_T)$ itself is compact. How can one see this?

$\endgroup$
2
  • $\begingroup$ I would have guessed that a locally compact abelian group has a unique maximal compact subgroup. Is that not true? $\endgroup$
    – user94041
    Commented May 10, 2017 at 17:52
  • 2
    $\begingroup$ @user94041 No, consider for example $\mathbf{Q}_p$. $\endgroup$ Commented May 10, 2017 at 22:05

1 Answer 1

3
$\begingroup$

There might be an easier way to see this but here is one way.

Any $F$-torus $T$ can be written as $T = T_s T_a$ with finite intersection, where $T_s$ (resp. $T_a$) is the unique maximal split (resp. anisotropic) subtorus of $T$ (Springer, LAG, Proposition 13.2.4). Further, the co-characters all come only from the split part; indeed (Springer, LAG, Proposition 13.3.3(iii)), $T$ is anisotropic iff any $F$-morphism $\mathbb G_m \to T$ is constant. Since $F$ is locally compact, $T_a(F)$ is compact (Springer, somewhere). Thus it suffices to restrict $H_T$ to $T_s$ where the kernel is the $\mathcal O_F$-points of $T_s$, a compact group. Up to finite index, the kernel is a product of $T_a(F)$ with $T_s(\mathcal O_F)$ and is hence compact.

$\endgroup$
2
  • $\begingroup$ Of course, a little work is necessary to speak of "the $\mathcal O_F$-points of $T_{\mathrm s}$" (since it is not a priori defined over $\mathcal O_F$). It is perhaps easier to replace this little work by the observation that, for any basis $B$ of $X^*(T)_F$, the map $\ker(H_T, T_{\mathrm s}) \to (\mathcal O_F^\times)^B$ given by $t \mapsto (\chi(t))_\chi$ is a homeomorphism. $\endgroup$
    – LSpice
    Commented Nov 27, 2017 at 1:44
  • $\begingroup$ (I meant $\ker(H_T, T_{\mathrm s}(F))$, not just $\ker(H_T, T_{\mathrm s})$.) $\endgroup$
    – LSpice
    Commented Nov 27, 2017 at 22:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .