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.
 A: How about the following construction?
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.)
