3
$\begingroup$

I am reading part of Dipendra Prasad's paper found here: http://arxiv.org/pdf/1306.2729v1.pdf.

In it (in the middle of page 8) he writes that compactly induced representations are projective. Why is this true? In general, what objects are known to be projective in this category?

$\endgroup$

1 Answer 1

6
$\begingroup$

This follows from Frobenius reciprocity for compact induction. Let $G$ be the group of rational points of a reductive $p$-adic group and $K$ be a compact open subgroup of $G$. Then if $\lambda$ is any smooth representationf of $K$, we denote by ${\rm ind}_K^G \lambda$ the compactly induced representation. Then by Frobenius reciprocity, the functors ${\rm Hom}_G ({\rm ind}_K^G \lambda , -)$ and ${\rm Hom}_K (\lambda ,-)$ are isomorphic. It follows, by definition of projectivity, that ${\rm ind}_K^G \lambda$ is projective since $\lambda$ is projective as a $K$-module. Indeed any smooth representation of $K$ is semisimple.

The result remains true if $K$ is compact mod the center of G and $\lambda$ has a central character $\chi$ provided that you work with the category ${\rm S}_\chi (G)$ of smooth representations of $G$ with central character $\chi$.

It is also known that irreducible supercuspidal representations of $G$ with central character $\chi$ are projective (and injective) objects of the category $S_\chi (G)$. [Adler and Roche, Injectivity, Projectivity and Supercuspidal Representations J. London Math. Soc. (2004) 70 (2): 356-368] By the way the converse is true : if an object of $S_\chi (G)$ is projective and injective then it is supercuspidal.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.