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This is moved from MSE, where I asked and didn't receive an answer (see http://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 $p > 2$.

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 $p > 2$.

This is moved from MSE, where I asked and didn't receive an answer (see http://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 KWitte's example below. I'm still interested in the case where $p > 2$.

This is moved from MSE, where I asked and didn't receive an answer (see http://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 $p > 2$.

This is moved from MSE, where I asked and didn't receive an answer (see http://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 $n = 2$; 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.

This is moved from MSE, where I asked and didn't receive an answer (see http://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 KWitte's example below. I'm still interested in the case where $p > 2$.