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It is indeed well-known that the category of smooth admissible representations of $G$ (and other reductive $p$-adic groups) is not semi-simple. The principal series, that is the representations induced from a character of the Borel of $G=GL_2(\mathbb Q_p)$, are always indecomposable, but they may nor be irreducible -- think of the case of the trivial character.

Another nice way to construct an example is with trees. You may know that $G$ acts transitively on the regular infinite tree $T$ of arity $p+1$ (i.e. every vertex has exactly $p+1$ neighbors -- this tree is the Bruhat-Tits tree of $G$). The stabilizer of a vertex is $ZK$ where $Z$ is the center and $K$ a compact maximal subgroup (= a conjugate of $GL_2(\mathbb Z_p)$). If you take for $W$ the set of functions with finite support on the set of vertices of the tree $T$, then $W$ is a representation of $G$ (since $G$ acts on $T$) which is smooth (stabilizers are intersection of finitely many $K_iZ$, with $K_i$ compact open), admits a non-trivial $G$-morphism to the trivial representation (the linear form which to a function $f$ on $T$ attaches the sum of its value), but certainly does not contain the trivial representation as sub-representation, because $W$ has no function invariant by $G$ except 0 -- they would be of infinite support.

Small problem, $W$ is not admissible. But now take let $D: W \rightarrow W$ the operator which to a function $f$ attaches the function $g(x)=\sum f(y)$ or $y$ running mont the $p+1$ neighbors of $x$. And define $V = W/ (D-(p+1))W$. You can check easily that the trivial representation is still a quotient of $V$, and with some little geometric reasoning on the tree that I leave as an exercise, that the trivial representation is not a sub-representation of $V$, and that $V$ is admissible. Hence a second example of smooth-admissible non-semi-simple reducible representation which shows that the answer to your question is yes. (Actually this example is the dual of the first one).

It is indeed well-known that the category of smooth admissible representations of $G$ (and other reductive $p$-adic groups) is not semi-simple. The principal series, that is the representations induced from a character of the Borel of $G=GL_2(\mathbb Q_p)$, are always indecomposable, but they may nor be irreducible -- think of the case of the trivial character.

Another nice way to construct an example is with trees. You may know that $G$ acts transitively on the regular infinite tree $T$ of arity $p+1$ (i.e. every vertex has exactly $p+1$ neighbors -- this tree is the Bruhat-Tits tree of $G$). The stabilizer of a vertex is $ZK$ where $Z$ is the center and $K$ a compact maximal subgroup (= a conjugate of $GL_2(\mathbb Z_p)$). If you take for $W$ the set of functions with finite support on the set of vertices of the tree $T$, then $W$ is a representation of $G$ (since $G$ acts on $T$) which is smooth (stabilizers are intersection of finitely many $K_iZ$, with $K_i$ compact open), admits a non-trivial $G$-morphism to the trivial representation (the linear form which to a function $f$ on $T$ attaches the sum of its value), but certainly does not contain the trivial representation as sub-representation, because $W$ has no function invariant by $G$ except 0 -- they would be of infinite support.

Small problem, $W$ is not admissible. But now let $D: W \rightarrow W$ be the operator which to a function $f$ attaches the function $g(x)=\sum f(y)$ with $y$ running over the $p+1$ neighbors of $x$, and define $V = W/ (D-(p+1))W$. You can easily check that the trivial representation is still a quotient of $V$, and with some little geometric reasoning on the tree that I leave as an exercise, that the trivial representation is not a sub-representation of $V$, and that $V$ is admissible. Hence a second example of smooth-admissible non-semi-simple reducible representation which shows that the answer to your question is yes. (Actually this example is the dual of the first one).

It is indeed well-known that the category of smooth admissible representations of $G$ (and other reductive $p$-adic groups) is not semi-simple. The principal series, that is the representations induced from a character of the Borel of $G=GL_2(\mathbb Q_p)$, are always indecomposable, but they may nor be irreducible -- think of the case of the trivial character.

Another nice way to construct an example is with trees. You may know that $G$ acts transitively on the regular infinite tree $T$ of arity $p+1$ (i.e. every vertex has exactly $p+1$ neighbors -- this tree is the Bruhat-Tits tree of $G$). The stabilizer of a vertice is $ZK$ where $Z$ is the center and $K$ a compact maximal subgroup (= a conjugate of $GL_2(\mathbb Z_p)$). If you take for $W$ the set of functions with finite support on the set of vertices of the tree $T$, then $W$ is a representation of $G$ (since $G$ acts on $T$) which is smooth (stabilizers are intersection of finitely many $K_iZ$, with $K_i$ compact open), contains a $G$-morphism to the trivial representation (the linear form which to a function $f$ on $T$ attaches the sum of its value), but certainly does not contain the trivial representation as sub-representation, because $W$ has no function invariant by $G$ except 0 -- they would be of infinite support.

Small problem, $W$ is not admissible. But now take let $D: W \rightarrow W$ the operator which to a function $f$ attaches the function $g(x)=\sum f(y)$ or $y$ running mont the $p+1$ neighbors of $x$. And define $V = W/ (D-(p+1))W$. You can check easily that the trivial representation is still a quotient of $V$, and with some little geometric reasoning on the tree that I leave as an exercise, that the trivial representation is not a sub-representation of $V$, and that $V$ is admissible. Hence a second example of smooth-admissible non-semi-simple reducible representation which shows that the answer to your question is yes. (Actually this example is the dual of the first one).

It is indeed well-known that the category of smooth admissible representations of $G$ (and other reductive $p$-adic groups) is not semi-simple. The principal series, that is the representations induced from a character of the Borel of $G=GL_2(\mathbb Q_p)$, are always indecomposable, but they may nor be irreducible -- think of the case of the trivial character.

Another nice way to construct an example is with trees. You may know that $G$ acts transitively on the regular infinite tree $T$ of arity $p+1$ (i.e. every vertex has exactly $p+1$ neighbors -- this tree is the Bruhat-Tits tree of $G$). The stabilizer of a vertex is $ZK$ where $Z$ is the center and $K$ a compact maximal subgroup (= a conjugate of $GL_2(\mathbb Z_p)$). If you take for $W$ the set of functions with finite support on the set of vertices of the tree $T$, then $W$ is a representation of $G$ (since $G$ acts on $T$) which is smooth (stabilizers are intersection of finitely many $K_iZ$, with $K_i$ compact open), admits a non-trivial $G$-morphism to the trivial representation (the linear form which to a function $f$ on $T$ attaches the sum of its value), but certainly does not contain the trivial representation as sub-representation, because $W$ has no function invariant by $G$ except 0 -- they would be of infinite support.

Small problem, $W$ is not admissible. But now take let $D: W \rightarrow W$ the operator which to a function $f$ attaches the function $g(x)=\sum f(y)$ or $y$ running mont the $p+1$ neighbors of $x$. And define $V = W/ (D-(p+1))W$. You can check easily that the trivial representation is still a quotient of $V$, and with some little geometric reasoning on the tree that I leave as an exercise, that the trivial representation is not a sub-representation of $V$, and that $V$ is admissible. Hence a second example of smooth-admissible non-semi-simple reducible representation which shows that the answer to your question is yes. (Actually this example is the dual of the first one).

It is indeed well-known that the category of smooth admissible representations of $G$ (and other reductive $p$-adic groups) is not semi-simple. The principal series, that is the representations induced from a character of the Borel of $G=GL_2(\mathbb Q_p)$, are always indecomposable, but they may nor be irreducible -- think of the case of the trivial character.

Another nice way to construct an example is with trees. You may know that $G$ acts transitively on the regular infinite tree $T$ of parity $p+1$ (the Bruhat-Tits tree of $G$). The stabilizer of a vertice is $ZK$ where $Z$ is the center and $K$ a compact maximal subgroup (= a conjugate of $GL_2(\mathbb Z_p)$). If you take for $W$ the set of functions with finite support on the set of vertices of the tree $T$, then $W$ is a representation of $G$ (since $G$ acts on $T$) which is smooth (stabilizers are intersection of finitely many $K_iZ$, with $K_i$ compact open), contains a $G$-morphism to the trivial representation (the linear form which to a function $f$ on $T$ attaches the sum of its value), but certainly does not contain the trivial representation as sub-representation, because $W$ has no function invariant by $G$ except 0 -- they would be of infinite support.

Small problem, $W$ is not admissible. But now take let $D: W \rightarrow W$ the operator which to a function $f$ attaches the function $g(x)=\sum f(y)$ or $y$ running mont the $p+1$ neighbors of $x$. And define $V = W/ (D-(p+1))W$. You can check easily that the trivial representation is still a quotient of $V$, and with some little geometric reasoning on the tree that I leave as an exercise, that the trivial representation is not a sub-representation of $V$, and that $V$ is admissible. Hence a second example of smooth-admissible non-semi-simple reducible representation which shows that the answer to your question is yes. (Actually this example is the dual of the first one).

It is indeed well-known that the category of smooth admissible representations of $G$ (and other reductive $p$-adic groups) is not semi-simple. The principal series, that is the representations induced from a character of the Borel of $G=GL_2(\mathbb Q_p)$, are always indecomposable, but they may nor be irreducible -- think of the case of the trivial character.

Another nice way to construct an example is with trees. You may know that $G$ acts transitively on the regular infinite tree $T$ of arity $p+1$ (i.e. every vertex has exactly $p+1$ neighbors -- this tree is the Bruhat-Tits tree of $G$). The stabilizer of a vertice is $ZK$ where $Z$ is the center and $K$ a compact maximal subgroup (= a conjugate of $GL_2(\mathbb Z_p)$). If you take for $W$ the set of functions with finite support on the set of vertices of the tree $T$, then $W$ is a representation of $G$ (since $G$ acts on $T$) which is smooth (stabilizers are intersection of finitely many $K_iZ$, with $K_i$ compact open), contains a $G$-morphism to the trivial representation (the linear form which to a function $f$ on $T$ attaches the sum of its value), but certainly does not contain the trivial representation as sub-representation, because $W$ has no function invariant by $G$ except 0 -- they would be of infinite support.

Small problem, $W$ is not admissible. But now take let $D: W \rightarrow W$ the operator which to a function $f$ attaches the function $g(x)=\sum f(y)$ or $y$ running mont the $p+1$ neighbors of $x$. And define $V = W/ (D-(p+1))W$. You can check easily that the trivial representation is still a quotient of $V$, and with some little geometric reasoning on the tree that I leave as an exercise, that the trivial representation is not a sub-representation of $V$, and that $V$ is admissible. Hence a second example of smooth-admissible non-semi-simple reducible representation which shows that the answer to your question is yes. (Actually this example is the dual of the first one).