This is moved from MSE, where I asked and didn't receive an answer (see https://math.stackexchange.com/questions/1145151/lattices-in-mathbbq-pn-with-the-same-stabilizer)

Let $T$ be the diagonal torus in $G = GL_n(\mathbb{Q}_p)$, and consider the action of $G$ on the set of (full-rank) lattices $\Lambda\subset \mathbb{Q}_p^n$. Let $\Lambda$ and $\Lambda'$ be full-rank sublattices ($\mathbb{Z}_p$-submodules of rank $n$) such $Stab_T(\Lambda) = Stab_T(\Lambda')$. Is it true that $\Lambda' = t\cdot \Lambda$ for some $t\in T$?

This is true when $n = 2$ and $p > 2$; to see this, if $\{e_1,\, e_2\}$ is the standard basis and $\Lambda$ any lattice, we can always take a basis of the form $$\{p^re_1,\, \alpha e_1 + p^s e_2\}$$ where $r,\, s\in \mathbb{Z}$ and $\alpha$ is chosen modulo $p^s$; at this point you can check directly that $\Lambda$ and $\Lambda'$ have the same stabilizer if and only if $v_p(\alpha) - r = v_p(\alpha') - r'$, and any two such lattices are in the same $T$-orbit.

If this is known, I'd love a reference.

EDITED: When $p = 2$, this is false, even when $n = 2$, in view of JWitte's example below. I'm still interested in the case where $p > 2$.


First a counter example:

Look at the case $\mathbb{Q}_2$ and $n=2$:

Take $\Lambda = \langle e_1,e_2\rangle$ and $\Lambda'= \langle pe_1, e_1+e_2\rangle$. Let $t_{\alpha,\beta}$ be the diagonal matrix with $\alpha$ and $\beta$ on the diagonal: $t_{\alpha,\beta}e_1 = \alpha e_1$ and $t_{\alpha,\beta}e_2 = \beta e_2$. The stabilizer of $\Lambda$ in $T$ is $\{ t_{\alpha,\beta} : v(\alpha)=v(\beta)=0\}$.

Now we show that this is also the stabilizer of $\Lambda'$:

Assume that $v(\alpha)=v(\beta)=0$, then $\alpha\beta^{-1} = 1 + \gamma p$ with $v(\gamma)\geq 0$ (p=2!). Now $t_{\alpha,\beta}\Lambda' = \langle p\alpha e_1, \alpha e_1+\beta e_2\rangle = \langle pe_1, \alpha\beta^{-1} e_1+e_2\rangle = \langle pe_1, \gamma p e_1+e_1+e_2\rangle=\langle pe_1,e_1+e_2\rangle=\Lambda'$.

Let $\pi_i$ be the projection to $e_i$ for $i\in\{1,2\}$. Then $\pi_i(\Lambda') = \mathbb{Z}_p e_i$. Now $\pi_1(t_{\alpha,\beta} \Lambda') = \alpha\mathbb{Z}_p$ and $\pi_2(t_{\alpha,\beta}\Lambda') = \beta\mathbb{Z}_p$. Therefore if $t_{\alpha,\beta}$ fixes $\Lambda'$, then $v(\alpha)=v(\beta)=0$.

Moreover if $t_{\alpha,\beta}\Lambda' =\Lambda$, then $\pi_i(t_{\alpha,\beta}\Lambda') = \pi_i(\Lambda) = \mathbb{Z}_p e_i$. Thus $v(\alpha)=v(\beta)=0$ by the preceding calculations. But then $t_{\alpha,\beta}$ is in the stabilizer of $\Lambda$. Thus $\Lambda$ and $\Lambda'$ are not in the same $T$ orbit.

A general comment:

If you know something about buildings, you could see that this example is based on 3.6.1 of Reductive groups over local fields, J. Tits. The stabilizer of a point in the apartment of $T$ is the set of diagonal matrices with units. And 3.6.1 states that if the residual characteristic is 2 this stabilizer fixes also points in the building that are not in the apartment of $T$. Because $T$ acts on its own apartment the points outside the apartment of $T$ are not in the same orbit as a point in the apartment. So the answer to your question is negative for all $n$ in the case $p=2$. However for odd $p$ I don't know the answer.

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  • $\begingroup$ Great example. One small point; when $\alpha\beta^{-1} = 1 + \gamma p$, I believe all we know is $\nu(\gamma) \geq 0$. This doesn't affect the proof though. $\endgroup$ – John Binder Nov 18 '15 at 17:55

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