Let $K/\mathbf{Q}$ be an imaginary quadratic field, with $\sigma \in \mathrm{Gal}(K/\mathbf{Q})$ a generator. Suppose $\pi$ is a cuspidal automorphic representation of $GL_2 / K$ with central character $\omega_{\pi}$. If $\pi_{\infty}$ has Langlands parameter $z\to \mathrm{diag}(z^{1-k},\overline{z}^{1-k})$ for $k \geq 2$ an integer, and $\omega_{\pi}^{\sigma}=\omega_{\pi}$, then a well-known theorem of Taylor (proved in the early 90's) yields a compatible system of two-dimensional $\ell$-adic Galois representations of $\mathrm{Gal}(\overline{K}/K)$ attached to $\pi$. My question is simple: given all the recent work on the fundamental lemma and concomitant progress in the field of automorphic Galois representations, it is yet possible to prove this result without any Galois-invariance assumption on the central character $\omega_{\pi}$?


2 Answers 2


Various groups of people have thought about/are thinking about this. The natural source of the desired Galois reps. is a $U(2,2)$ Shimura variety. The problem is that the cohomology of this variety is not so easy to understand. The fundamental lemma certainly plays some role in controlling it, but I don't think that by itself it overcomes the key difficulties. (My own understanding of the issues is far from perfect, though.)

  • $\begingroup$ Ah, interesting! What, precisely, is the natural way of moving to this Shimura variety? The form $\pi \otimes \pi^{\sigma}$ is conjugate self-dual, so should transfer, but it's not obvious to me that you could easily recover $\rho_{\pi}$ from a tensor product like this. Or am I on the wrong track? $\endgroup$ Jan 4, 2011 at 19:22
  • $\begingroup$ If one constructs a quasi-split $GU(2,2)$ from the quad. imag. field $K$, then the hyperbolic $3$-manifolds that arise as congruence quotients for $\mathrm{GL}_2(K)$ appear in the Borel--Serre boundary of the Shimura varieties for $GU(2,2)$. $\endgroup$
    – Emerton
    Jan 4, 2011 at 19:32
  • $\begingroup$ Okay, but don't you get Eisenstein cohomology classes in this manner? Is it obvious that they're congruent modulo $\ell$-powers to cuspidal classes (which is what one would hope for in order to use pseudo-representations)? Thanks for your patience; I feel I'm missing something fundamental here. $\endgroup$ Jan 5, 2011 at 16:24
  • 2
    $\begingroup$ No, it is not obvious, and therein lies the difficulty. $\endgroup$
    – Emerton
    Jan 5, 2011 at 16:40

Taylor's theorem is about regular algebraic representations $\pi$. Most representations are not algebraic in any sense (cf. my answer to this question), hence they are not connected to any Galois representation (as far as we know).

Now for an algebraic $\pi$ the question arises where to look for the corresponding Galois representation. The difficulty with $GL_2/K$ is that the corresponding symmetric space is not an algebraic variety, hence there is no cohomology which would carry the relevant Galois representation. Taylor's idea was to pass to $GSp_4/\mathbb{Q}$ which is better in that regard, but the switch requires the condition on the central character. My feeling is that what is missing is a better (more general) construction yielding an object with a Galois action, not a better understanding of the fundamental lemma or Galois representations per se. This is not my field of expertise, so I better stop here.

  • $\begingroup$ My first paragraph was no criticism, I just wanted to emphasize the initial algebraic nature of $\pi$. $\endgroup$
    – GH from MO
    Jan 3, 2011 at 1:51

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