MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let's take $D$ to be $n^2$-dimensional central algebra with an involution $*$ over a CM field $E=FE_0$ (where F is totally real, and $E_0$ is imaginary quadratic). Define $$G(R) = \{ x \in D \otimes _{\mathbb{Q}} R | x x^* \in R^{\times} \}$$ for $\mathbb{Q}$-algebras $R$. Complete this data to a Shimura datum (i.e. define $h: Res _{\mathbb{C} /\mathbb{R} } \mathbb{G} _m \rightarrow G _{\mathbb{R}}$ and $X$) just like in the book of Harris-Taylor, so that we get a Shimura variety $Sh _ K = G(\mathbb{Q}) \backslash G(\mathbb{A} _f) \times X / K$ of unitary type. Choose an irreducible representation $\xi$ of $G(\mathbb{A} _f)$ and form the associated sheaf $\mathcal{L} _{\xi}$ on $Sh _K$. Consider now the cohomology groups of $Sh _K$ and their decomposition: $$H^{n-1} _{et} (Sh_K, \mathcal{L} _{\xi}) = \oplus _{\pi } V _{\pi} \otimes \pi _f ^K$$ where $\pi$ runs over certain automorphic representations (I don't specify which) and where $V _{\pi} = Hom (\pi _f ^K, H^{n-1} _{et} (Sh_K, \mathcal{L} _{\xi}))$ (homomorphisms which are equvariant with respect to Hecke algebra).

Questions: What is known about the dimension of $V_{\pi}$ at the moment? Is it always equal to $n$? If not, what are the other possibilities for $dim V_{\pi}$? Fujiwara in his ICM 2006 talk ("Galois deformations and arithmetic geometry of Shimura varieties"), in footnote 31, says that probably it follows (for $\mathcal{L} _{\xi} = \bar{\mathbb{Q}} _l$ and for cuspidal locus) from the work of Ngo-Laumon on the fundamental lemma. Could someone provide either an explanation or a reference why is that?

Edit: additional question : what can be said of dimension of $V_{\pi}$ which appears in the cohomology group $H^{n-1} _{et} (Sh_K, \mathcal{L} _{\xi}) _{\mathfrak{m}}$ - i.e. localised at some "nice" (for example, non-Eisenstein) maximal ideal $\mathfrak{m}$ of Hecke algebra.

share|cite|improve this question
I am no expert at this sort of thing, but it seems to me that one issue that might stop the dimension of $V_\pi$ being $n$ is multiplicities. It seems to me that $V_\pi$ as you have defined it only depends on the isomorphism class of $\pi_f$, so if multiplicity one fails then $V_\pi$ is too big. In general you can completely "explicitly" compute $V_\pi$: it's a sum of some $g,K$-cohomology modules over the space of all auto reps -- the $(g,K)$-cohomology will be $n$-dimensional in most cases (perhaps there are issues with endoscopy if you're not careful to rule them out? not sure...)... – Kevin Buzzard Jun 16 '11 at 19:16
...but then you pick up some multiplicities if $\pi$ shows up with multiplicity greater than 1. – Kevin Buzzard Jun 16 '11 at 19:17
You're certainly right. In article of Kottwitz ("On $\lambda$ -adic representations...") he gets (for $\xi$ sufficiently regular and taking $\pi _f$ with some additional conditions) that $dim V _{\pi _f} = n a(\pi _f)$, where $a(\pi _f)$ is certain constant calculated by Matsushima's formula as you suggest. So I understand that in this situation, Fujiwara says that $a(\pi _f) =1$, but I would like to know exactly why, and what we can say more generally. – Przemyslaw Chojecki Jun 16 '11 at 22:07
I've not looked at this Fujiwara article, but let me just ask whether you're sure that he's asserting $a(\pi_f)=1$? Because in older approaches to the problem e.g by Taylor the strategy was to use the etale cohomology of unitary groups to produce the representation of dimension $a(\pi_f)n$ and then to use group theory/Lie algebras/Hodge-Tate theory to deduce the existence of the $n$-dimensional representation, thus simply constructing the correct representation whilst avoiding the issue of whether $a=1$. My impression was that $a>1$ was known to sometimes occur in this situation. On the... – Kevin Buzzard Jun 17 '11 at 7:28
...other hand, perhaps the article you're reading puts some assumptions on the unitary group to somehow force $a=1$. As I already said, unfortunately I'm not an expert in the technical details of this area. – Kevin Buzzard Jun 17 '11 at 7:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.