12
$\begingroup$

Let $V$ be a complex vector space of infinite dimension and let $(\pi,V)$ be a representation of the $p$-adic group $G:=GL_2(\mathbb{Q}_p)$. From representation theory, we know that if the representation $(\pi,V)$ is both smooth and irreducible, then $(\pi,V)$ is admissible. So now it is natural to think about the inverse question.

Does there exist a representation $(\pi,V)$ of $G$ which is smooth and admissible but not irreducible?

This question is not difficult. You can find two smooth and irreducible representations $(\pi_1,V_1)$ and $(\pi_2,V_2)$ of $G$ and give the example $(\pi_1 \oplus \pi_2, V_1 \oplus V_2)$. Let $\pi=\pi_1 \oplus \pi_2$ and $V=V_1 \oplus V_2$, then $(\pi,V)$ is the representation we want for the question.

But if we change a little on the question, does there exist a representation $(\pi,V)$ of $G$ which is smooth, admissible and indecomposable but not irreducible? I have no idea on this question since the above example does not work in this case.

Improve this question
$\endgroup$
5
  • $\begingroup$ www.ims.nus.edu.sg/preprints/2004-9.pdf 12.1 gives an example of complex representation of a p-adic group that is reducible but not completely reducible. Does this help? $\endgroup$
    – user43326
    Commented Sep 24, 2017 at 10:49
  • 1
    $\begingroup$ I guess you mean the direct sum, not the tensor product of $\pi_1$ and $\pi_2$. $\endgroup$
    – François Brunault
    Commented Sep 24, 2017 at 12:08
  • 1
    $\begingroup$ representations induced from characters of Borel subgroup is admissible, but not always direct sum of irreducible representations. $\endgroup$
    – Q-Zh
    Commented Sep 24, 2017 at 13:02
  • $\begingroup$ Thank you Francois. You are right. I made a mistake here. I will change it at once. $\endgroup$
    – JACK
    Commented Sep 24, 2017 at 18:11
  • $\begingroup$ The post also published at: math.stackexchange.com/questions/2442714/… $\endgroup$
    – I am not Paul Erdos
    Commented Sep 24, 2017 at 18:20

1 Answer 1

Reset to default
14
$\begingroup$

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).

Improve this answer
$\endgroup$
5
  • $\begingroup$ Thank you very much! But you know I'm a beginner of automorphic forms and representation theory. So I have a silly question: why the principal series which are induced from a character of the Borel subgroup of $G$ are always indecomposable? Could you briefly give a proof or give me some references to read? Thank you! $\endgroup$
    – JACK
    Commented Sep 25, 2017 at 7:02
  • 2
    $\begingroup$ A good reference for beginners is the monography by Bushnell and Henniart "The local Langlands conjecture for ${\rm GL}(2)$. They give a complete proof that the (admissible) induced representation from the trivial character of the Borel subgroup has length $2$ and is indecomposable. It has the trivial character as subrepresentation and the Steinberg representation as quotient. $\endgroup$
    – Paul Broussous
    Commented Sep 25, 2017 at 9:28
  • $\begingroup$ Paul's reference is good, or if you are ready to dip directly into the general theory (not only for $GL_2$, you can try the article of Cartier in Corvallis (ams.org/books/pspum/033.1). But this is why I also gave the second example with the tree. For this example, determining the structure of V is a series of really doable exercises. You need to prove that V has the trivial representation as quotient: done. That V has not the trivial representation as sub-object. Really it is not hard. And finally to prove that the kernel $V_0$ of the map $V$-to-trivial (that is the map... $\endgroup$
    – Joël
    Commented Sep 25, 2017 at 13:52
  • $\begingroup$ "sum of values") is irreducible. Still doable. The trick is to show that any non-zero sub-rep in $V$ has a representative in $W$ with at most 2 neighborhing points as its support. Anyway, like this you can get a complete and elementary prove that V is indecomposable, and has 2 Jordan-Holder factor, the trivial representation and $V_0$, which is called the Steinberg. $\endgroup$
    – Joël
    Commented Sep 25, 2017 at 14:04
  • $\begingroup$ Thanks to both of you, Paul and Joël. @Paul Broussous On the book "The local Langlands conjecture for $GL(2)$" you just mentioned, I found the chapter 3 is mainly about induced representations. But in the whole chapter, I did not find the corresponding proof. Maybe I missed something or I searched for the wrong chapter. Could you kindly tell me where can I find the theorem and proof? Many thanks! $\endgroup$
    – JACK
    Commented Sep 26, 2017 at 1:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .