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Following Garrett's book, we have the usual description of affine buildings of SL(n) in terms of homothety classes of lattices. So let $F$ be a local field, $\mathcal{O}$ be its ring of integers and $\mathfrak{m} = <\pi>$ be its unique maximal ideal. Let $V = F^{4}$ and $G = SL(V)$. Let $e_{1}, \ldots, e_{4}$ be the standard basis vectors. Then one of the chambers is given by the following lattices: $$ L_{0} = \mathcal{O}e_{1} + \mathcal{O}e_{2} + \mathcal{O}e_{3} + \mathcal{O}e_{4},$$ $$ L_{1} = \mathcal{O}\pi^{-1}e_{1} + \mathcal{O}e_{2} + \mathcal{O}e_{3} + \mathcal{O}e_{4},$$ $$ L_{2} = \mathcal{O}\pi^{-1}e_{1} + \mathcal{O}\pi^{-1}e_{2} + \mathcal{O}e_{3} + \mathcal{O}e_{4},$$ $$ L_{3} = \mathcal{O}\pi^{-1}e_{1} + \mathcal{O}\pi^{-1}e_{2} + \mathcal{O}\pi^{-1}e_{3} + \mathcal{O}e_{4}. $$

The type of lattice $L_{i}$ is $i$ according to typing given in Garrett on page 329. Hence they are all of different type (which they have to be since they form a chamber).

Now $G$ acts in a type preserving manner (claimed in Garrett on Page $329-330$) on the building. However, I do not see it. Also, I seem to run into a problem with the following example.

Let $g \in G$ be given by $$ g = \pmatrix{1 & 0 & \pi^{-1} & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\} $$

Then $gL_{0} = L_{1}$ which have different types. Where am I making the mistake?


EDIT: To see the last statement, observe that $gL_{0}$ is the lattice

$$ gL_{0} = \mathcal{O}(1+\pi^{-1})e_{1} + \mathcal{O}e_{2} + \mathcal{O}e_{3} + \mathcal{O}e_{4}.$$

Clearly, $gL_{0} \subseteq L_{1}$. To see the other inequality, we observe that $(1 + \pi)$ is a unit in the ring (otherwise $1$ belongs to the maximal ideal)and $(1+\pi)^{-1}(1+\pi^{-1}) = \pi^{-1}$.

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    $\begingroup$ I see that $gL_0$ is contained in $L_1$, but i don't see equality. $\endgroup$ – user1688 Aug 11 '16 at 10:03
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    $\begingroup$ Since $\det g=1$, you can't have $g L_0 = L_1$. $\endgroup$ – David E Speyer Aug 11 '16 at 12:15
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    $\begingroup$ It is also not equal up to homothety. Further, any base-change matrix from $L_0$ to $L_1$ must have determinant of absolute value $|\pi^{-1}|$. $\endgroup$ – user1688 Aug 12 '16 at 8:01
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    $\begingroup$ We have $gL_0 = \mathcal{O}e_1 + \mathcal{O}e_2+\mathcal{O}(e_3+\pi^{-1}e_1)+\mathcal{O}e_4$. The expression for $gL_0$ that you wrote is incorrect. $\endgroup$ – Aurel Aug 18 '16 at 10:22
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The groups $G={\rm SL}(V)$ and $G'={\rm GL}(V) \supset G$ have the same affine building. If the action of $G$ is type preserving, the action of $G'$ is not! Your particular element $g$ lies in $G'$ but not in $G$.

If an element of $G'$ stabilizes a chamber globally then it stabilizes it pointwise. This is not true for $G$: given a chamber $C$, on can find an element $\Pi$ of $G'$ stablizing $C$ and inducing a cyclic permutation of the vertices of $C$.

More generally the action of a simply connected reductive group (like ${\rm SL}(N)$, ${\rm Sp}(2N)$, ...) on its building is type preserving. But this is not true in general.

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