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The answer is negative even for ${\rm{PGL}}_2$: the stabilizer of an edge in the building is a counterexample (with Iwahori preimage in ${\rm{SL}}_2(F)$). It likewise fails for ${\rm{PGL}}_n$ for every $n > 2$ by using stabilizers of chambers in the building.

Clearly $\pi(f_k(K'_0)) = k K_0 k^{-1} \subset K$, so $f_k(K'_0) \subset \pi^{-1}(K)$. But $\pi^{-1}(K)$ is a compact open subgroup of ${\rm{SL}}_n(F)$ which contains the maximal compact open subgroup $K'_0$ and hence is equal to $K'_0$, so $f_k(K'_0) \subset K'_0$. Hence, a lift $\widetilde{k} \in {\rm{GL}}_n(F)$ of $k$ normalizes $K'_0 = {\rm{SL}}_n(O_F)$; more precisely, $\widetilde{k} K'_0 \widetilde{k}^{-1} = K'_0$ by volume considerations (using that ${\rm{SL}}_n$ inis "unimodular" as an algebraic group over $F$). In terms of the standard $O_F$-lattice $\Lambda = O_F^n$ inside $F^n$, this says that ${\rm{SL}}(\widetilde{k}(\Lambda)) = {\rm{SL}}(\Lambda)$ ${\rm{SL}}(\widetilde{k}.\Lambda) = {\rm{SL}}(\Lambda)$. But rather generally, a pair of $O_F$-lattices $\Lambda_1$ and $\Lambda_2$ in $F^n$ satisfies ${\rm{SL}}(\Lambda_1) = {\rm{SL}}(\Lambda_2)$ inside ${\rm{GL}}(F^n) = {\rm{GL}}_n(F)$ if and only if $\Lambda_1$ and $\Lambda_2$ are $F^{\times}$-multiples of each other, so $\widetilde{k} \in F^{\times} {\rm{GL}}(\Lambda) = F^{\times} {\rm{GL}}_n(O_F)$ and hence $k \in {\rm{PGL}}_n(O_F) = K_0$ as desired.

The conclusion is that a maximal compact open subgroup $K$ of ${\rm{PGL}}_n(F)$ that is not conjugate to ${\rm{PGL}}_n(O_F)$ has preimage in ${\rm{SL}}_n(F)$ that is not maximal as a compact open subgroup. Any maximal compact open subgroup of the unimodular ${\rm{PGL}}_n(F)$ having volume (relative to a choice of Haar measure) distinct from that of ${\rm{PGL}}_n(O_F)$ is an example of such a $K$. Such $K$ exist for any $n \ge 2$, as stabilizers of a chamber in the building. But since the building can be most readily visualized and defined for $n = 2$, we now focus on describing an explicit counterexample for $n=2$.

Finally, we return toconsider ${\rm{PGL}}_2$. Let $K$ be the stabilizer of an edge in the building. Explicitly, this is described as follows. For the lattices $\Lambda = O_F^2$ and $\Lambda' = O_F \oplus \varpi O_F$ (with $\varpi$ a uniformizer of $F$) we choose some $g \in {\rm{GL}}_2(F)$ such that $g(\Lambda') = \Lambda$, so $g(\Lambda)$ corresponds to a line in the plane $\overline{\Lambda} := \varpi^{-1}\Lambda/\Lambda$ over the residue field. By surjectivity of ${\rm{PGL}}_2(\Lambda) \rightarrow {\rm{PGL}}_2(\overline{\Lambda})$, we can adjust the choice of $g$ so that also $g(\Lambda) = \varpi^{-1}O_F \oplus O_F = \varpi^{-1}\Lambda'$. Then $g$ swaps the homothety classes of the lattices $\Lambda$ and $\Lambda'$, so $g^2$ preserves these homothety classes; i.e., $g^2$ lies in the compact open subgroup $\mathbf{K} := {\rm{PGL}}(\Lambda) \cap {\rm{PGL}}(\Lambda')$$$\mathbf{K} := {\rm{PGL}}(\Lambda) \cap {\rm{PGL}}(\Lambda')$$ that is the intersection of the stabilizers of the endpoints of an edge of the building whereas $g \not\in \mathbf{K}$. Clearly $$K = \mathbf{K} \coprod g \mathbf{K} = N_{{\rm{PGL}}_2(F)}(\mathbf{K}).$$ This is exactly the stabilizer of the midpoint of an edge in the building, and that midpoint is its unique fixed point. Since any compact subgroup of ${\rm{PGL}}_2(F)$ must fix some point of the building, it follows that $K$ must be maximal as a compact open subgroup of ${\rm{PGL}}_2(F)$.

The answer is negative even for ${\rm{PGL}}_2$: the stabilizer of an edge in the building is a counterexample (with Iwahori preimage in ${\rm{SL}}_2(F)$).

Clearly $\pi(f_k(K'_0)) = k K_0 k^{-1} \subset K$, so $f_k(K'_0) \subset \pi^{-1}(K)$. But $\pi^{-1}(K)$ is a compact open subgroup of ${\rm{SL}}_n(F)$ which contains the maximal compact open subgroup $K'_0$ and hence is equal to $K'_0$, so $f_k(K'_0) \subset K'_0$. Hence, a lift $\widetilde{k} \in {\rm{GL}}_n(F)$ of $k$ normalizes $K'_0 = {\rm{SL}}_n(O_F)$; more precisely, $\widetilde{k} K'_0 \widetilde{k}^{-1} = K'_0$ by volume considerations (using that ${\rm{SL}}_n$ in "unimodular" as an algebraic group over $F$). In terms of the standard $O_F$-lattice $\Lambda = O_F^n$ inside $F^n$, this says that ${\rm{SL}}(\widetilde{k}(\Lambda)) = {\rm{SL}}(\Lambda)$. But rather generally, a pair of $O_F$-lattices $\Lambda_1$ and $\Lambda_2$ in $F^n$ satisfies ${\rm{SL}}(\Lambda_1) = {\rm{SL}}(\Lambda_2)$ inside ${\rm{GL}}(F^n) = {\rm{GL}}_n(F)$ if and only if $\Lambda_1$ and $\Lambda_2$ are $F^{\times}$-multiples of each other, so $\widetilde{k} \in F^{\times} {\rm{GL}}(\Lambda) = F^{\times} {\rm{GL}}_n(O_F)$ and hence $k \in {\rm{PGL}}_n(O_F) = K_0$ as desired.

The conclusion is that a maximal compact open subgroup $K$ of ${\rm{PGL}}_n(F)$ that is not conjugate to ${\rm{PGL}}_n(O_F)$ has preimage in ${\rm{SL}}_n(F)$ that is not maximal as a compact open subgroup. Any maximal compact open subgroup of the unimodular ${\rm{PGL}}_n(F)$ having volume (relative to a choice of Haar measure) distinct from that of ${\rm{PGL}}_n(O_F)$ is an example of such a $K$.

Finally, we return to ${\rm{PGL}}_2$. Let $K$ be the stabilizer of an edge in the building. Explicitly, this is described as follows. For the lattices $\Lambda = O_F^2$ and $\Lambda' = O_F \oplus \varpi O_F$ (with $\varpi$ a uniformizer of $F$) we choose some $g \in {\rm{GL}}_2(F)$ such that $g(\Lambda') = \Lambda$, so $g(\Lambda)$ corresponds to a line in the plane $\overline{\Lambda} := \varpi^{-1}\Lambda/\Lambda$ over the residue field. By surjectivity of ${\rm{PGL}}_2(\Lambda) \rightarrow {\rm{PGL}}_2(\overline{\Lambda})$, we can adjust the choice of $g$ so that also $g(\Lambda) = \varpi^{-1}O_F \oplus O_F = \varpi^{-1}\Lambda'$. Then $g$ swaps the homothety classes of the lattices $\Lambda$ and $\Lambda'$, so $g^2$ preserves these homothety classes; i.e., $g^2$ lies in the compact open subgroup $\mathbf{K} := {\rm{PGL}}(\Lambda) \cap {\rm{PGL}}(\Lambda')$ that is the intersection of the stabilizers of the endpoints of an edge of the building whereas $g \not\in \mathbf{K}$. Clearly $$K = \mathbf{K} \coprod g \mathbf{K} = N_{{\rm{PGL}}_2(F)}(\mathbf{K}).$$ This is exactly the stabilizer of the midpoint of an edge in the building, and that midpoint is its unique fixed point. Since any compact subgroup of ${\rm{PGL}}_2(F)$ must fix some point of the building, it follows that $K$ must be maximal as a compact open subgroup of ${\rm{PGL}}_2(F)$.

The answer is negative even for ${\rm{PGL}}_2$: the stabilizer of an edge in the building is a counterexample (with Iwahori preimage in ${\rm{SL}}_2(F)$). It likewise fails for ${\rm{PGL}}_n$ for every $n > 2$ by using stabilizers of chambers in the building.

Clearly $\pi(f_k(K'_0)) = k K_0 k^{-1} \subset K$, so $f_k(K'_0) \subset \pi^{-1}(K)$. But $\pi^{-1}(K)$ is a compact open subgroup of ${\rm{SL}}_n(F)$ which contains the maximal compact open subgroup $K'_0$ and hence is equal to $K'_0$, so $f_k(K'_0) \subset K'_0$. Hence, a lift $\widetilde{k} \in {\rm{GL}}_n(F)$ of $k$ normalizes $K'_0 = {\rm{SL}}_n(O_F)$; more precisely, $\widetilde{k} K'_0 \widetilde{k}^{-1} = K'_0$ by volume considerations (using that ${\rm{SL}}_n$ is "unimodular" as an algebraic group over $F$). In terms of the standard $O_F$-lattice $\Lambda = O_F^n$ inside $F^n$, this says that ${\rm{SL}}(\widetilde{k}.\Lambda) = {\rm{SL}}(\Lambda)$. But rather generally, a pair of $O_F$-lattices $\Lambda_1$ and $\Lambda_2$ in $F^n$ satisfies ${\rm{SL}}(\Lambda_1) = {\rm{SL}}(\Lambda_2)$ inside ${\rm{GL}}(F^n) = {\rm{GL}}_n(F)$ if and only if $\Lambda_1$ and $\Lambda_2$ are $F^{\times}$-multiples of each other, so $\widetilde{k} \in F^{\times} {\rm{GL}}(\Lambda) = F^{\times} {\rm{GL}}_n(O_F)$ and hence $k \in {\rm{PGL}}_n(O_F) = K_0$ as desired.

The conclusion is that a maximal compact open subgroup $K$ of ${\rm{PGL}}_n(F)$ that is not conjugate to ${\rm{PGL}}_n(O_F)$ has preimage in ${\rm{SL}}_n(F)$ that is not maximal as a compact open subgroup. Any maximal compact open subgroup of the unimodular ${\rm{PGL}}_n(F)$ having volume (relative to a choice of Haar measure) distinct from that of ${\rm{PGL}}_n(O_F)$ is an example of such a $K$. Such $K$ exist for any $n \ge 2$, as stabilizers of a chamber in the building. But since the building can be most readily visualized and defined for $n = 2$, we now focus on describing an explicit counterexample for $n=2$.

Finally, consider ${\rm{PGL}}_2$. Let $K$ be the stabilizer of an edge in the building. Explicitly, this is described as follows. For the lattices $\Lambda = O_F^2$ and $\Lambda' = O_F \oplus \varpi O_F$ (with $\varpi$ a uniformizer of $F$) we choose some $g \in {\rm{GL}}_2(F)$ such that $g(\Lambda') = \Lambda$, so $g(\Lambda)$ corresponds to a line in the plane $\overline{\Lambda} := \varpi^{-1}\Lambda/\Lambda$ over the residue field. By surjectivity of ${\rm{PGL}}_2(\Lambda) \rightarrow {\rm{PGL}}_2(\overline{\Lambda})$, we can adjust the choice of $g$ so that also $g(\Lambda) = \varpi^{-1}O_F \oplus O_F = \varpi^{-1}\Lambda'$. Then $g$ swaps the homothety classes of the lattices $\Lambda$ and $\Lambda'$, so $g^2$ preserves these homothety classes; i.e., $g^2$ lies in the compact open subgroup $$\mathbf{K} := {\rm{PGL}}(\Lambda) \cap {\rm{PGL}}(\Lambda')$$ that is the intersection of the stabilizers of the endpoints of an edge of the building whereas $g \not\in \mathbf{K}$. Clearly $$K = \mathbf{K} \coprod g \mathbf{K} = N_{{\rm{PGL}}_2(F)}(\mathbf{K}).$$ This is exactly the stabilizer of the midpoint of an edge in the building, and that midpoint is its unique fixed point. Since any compact subgroup of ${\rm{PGL}}_2(F)$ must fix some point of the building, it follows that $K$ must be maximal as a compact open subgroup of ${\rm{PGL}}_2(F)$.

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user76758
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The conclusion is that a maximal compact open subgroup $K$ of ${\rm{PGL}}_n(F)$ that is not conjugate to ${\rm{PGL}}_n(O_F)$ has preimage in ${\rm{SL}}_n(F)$ that is not maximal as a compact open subgroup. Any maximal compact open subgroup of the unimodular ${\rm{PGL}}_n(F)$ having volume (relative to a choice of Haar measure) is distinct from that of ${\rm{PGL}}_n(O_F)$ is an example of such a $K$.

The conclusion is that a maximal compact open subgroup $K$ of ${\rm{PGL}}_n(F)$ that is not conjugate to ${\rm{PGL}}_n(O_F)$ has preimage in ${\rm{SL}}_n(F)$ that is not maximal as a compact open subgroup. Any maximal compact open subgroup of the unimodular ${\rm{PGL}}_n(F)$ having volume (relative to a choice of Haar measure) is distinct from that of ${\rm{PGL}}_n(O_F)$ is an example of such a $K$.

The conclusion is that a maximal compact open subgroup $K$ of ${\rm{PGL}}_n(F)$ that is not conjugate to ${\rm{PGL}}_n(O_F)$ has preimage in ${\rm{SL}}_n(F)$ that is not maximal as a compact open subgroup. Any maximal compact open subgroup of the unimodular ${\rm{PGL}}_n(F)$ having volume (relative to a choice of Haar measure) distinct from that of ${\rm{PGL}}_n(O_F)$ is an example of such a $K$.

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user76758
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In the work of Bruhat-Tits, analyzing maximal compact open subgroups of $G(F)$ (for connected semisimple $G$ over a non-archimedean local field $F$, say) is "easiest" when $G$ is split and simply connected. To understand more general split cases, especially without the simply connected property, involves additional work precisely to deal with the failure of surjectivity of $\widetilde{G}(F) \rightarrow G(F)$. To To understand the origin of counterexamples with $G = {\rm{PGL}}_n$ (for $n > 1$), it is convenient to first carry out some general considerations (or jump ahead to the 2nd to last paragraph below if you are impatient).

In the work of Bruhat-Tits, analyzing maximal compact open subgroups of $G(F)$ (for connected semisimple $G$ over a non-archimedean local field $F$, say) is "easiest" when $G$ is split and simply connected. To understand more general cases, especially without the simply connected property, involves additional work precisely to deal with the failure of surjectivity of $\widetilde{G}(F) \rightarrow G(F)$. To understand the origin of counterexamples with $G = {\rm{PGL}}_n$ (for $n > 1$), it is convenient to first carry out some general considerations (or jump ahead to the 2nd to last paragraph below if you are impatient).

In the work of Bruhat-Tits, analyzing maximal compact open subgroups of $G(F)$ (for connected semisimple $G$ over a non-archimedean local field $F$, say) is "easiest" when $G$ is split and simply connected. To understand more general split cases without the simply connected property involves additional work precisely to deal with the failure of surjectivity of $\widetilde{G}(F) \rightarrow G(F)$. To understand the origin of counterexamples with $G = {\rm{PGL}}_n$ (for $n > 1$), it is convenient to first carry out some general considerations (or jump ahead to the 2nd to last paragraph below if you are impatient).

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