# Example of a non-odd motive appearing in cohomology of intermediate degree

I would like to know an example of a projective variety over a totally real field where a complex conjugation is not odd on some of its étale cohomology.

Edit: I am looking for the most interesting statement. Namely, is there an example of a connected projective variety $$X$$ of dimension $$>0$$ over a totally real field $$F$$ such that, for some $$i\ne0,\dim X$$, there is an irreducible constituent $$\rho$$ of the Galois representation $$H^{i}_{et}(X_{\overline{\mathbb{Q}}},\overline{\mathbb{Q}}_{\ell})$$ such that

• $$\dim_{\overline{\mathbb{Q}}_{\ell}}\rho\ge2$$,
• and for some complex conjugation $$c$$ in $$\text{Gal}(\overline{F}/F)$$, $$|\text{Trace}(\rho(c))|>1$$.

As posted in the comment, we know that it is expected to not arise from cohomological automorphic representations of $$\text{GL}_{n}$$, and it should have irregular Hodge–Tate weights.

• the following papers might be relevant: math.uchicago.edu/~fcale/papers/pst.pdf wwwf.imperial.ac.uk/~acaraian/papers/complexconjugation.pdf – vrz Feb 5 at 7:11
• Can you make precise the definition you are using of "odd"? The concept is normally only encountered for irreducible Galois representations (where it is the condition that the number of +1 and -1 eigenvalues for complex conj'n at each infinite place be equal if the dimension is even, or differ by 1 if the dimension is odd). E.g. would you consider the sum of 17 copies of the cyclotomic character to be odd? – David Loeffler Feb 5 at 9:04
• @DavidLoeffler Sorry, I have edited the question. I should've formulated my question more precisely. – GTA Feb 5 at 13:26
• Artin motives associated to even 2-dimensional Artin representations (which do exist) would seem to fit the requirements. – David Loeffler Feb 5 at 13:34
• @DavidLoeffler Yes! Sorry for changing the question again -- I would be extremely happy if there is an example of an even irreducible Artin representation appearing in some intermediate degree cohomology. – GTA Feb 5 at 15:03

Let $$A$$ be a principally-polarised abelian surface over $$\mathbf{Q}$$ which is "generic", i.e. $$End_{\overline{\mathbf{Q}}}(A) = \mathbf{Z}$$. Then the Galois action on $$H^1_{\mathrm{et}}(A_{\overline{\mathbf{Q}}}, \mathbf{Z}_p)$$ has to respect the polarisation, so we get a representation $$\rho: G_{\mathbf{Q}} \to \operatorname{GSp}_4(\mathbf{Z}_p)$$. It turns out that this representation has open image, and complex conjugation acts as a conjugate of $$c = (-1,-1,1,1)$$.
We can then consider the composite of $$\rho$$ with the 10-dimensional adjoint representation $$Ad^0$$ of $$\operatorname{GSp}_4$$ (whose underlying space is the Lie algebra $$\mathfrak{sp}_4$$). Since the image of $$\rho$$ is Zariski-dense in $$\operatorname{GSp}_4$$, $$Ad^0(\rho)$$ is irreducible; and if I'm not mistaken, $$Ad^0(c)$$ has four $$+1$$ eigenvalues and six $$-1$$ eigenvalues, so $$Ad^0(\rho)$$ is not odd.
Since $$\rho$$ appears in the cohomology of $$A$$, and $$Ad^0(\rho)$$ occurs as a direct summand of $$\rho^{\otimes m}(n)$$ for some $$m, n$$, it follows that some twist of $$\rho$$ appears in the cohomology of $$A \times \dots \times A$$. It should be easy to compute exactly which degree it shows up in, but it clearly can't be the top or bottom degree because $$A$$ is geometrically connected so those representations would be 1-dimensional.
(One can even show, using work of Ancona, that $$Ad^0(\rho)$$ is cut out inside $$A \times \dots \times A$$ by correspondences, so it corresponds to a pure motive.)