Suppose that over some (algebraically closed) field $K$ of characteristic $p>0$ we have: numerical equivalence of cycles coincides with homological one with respect to ${\mathbb{Q}}\_{l}$ and ${\mathbb{Q}}\_{l'}$-adic etale cohomology for two (distinct) primes $l,l'\neq p$ (i.e. the Standard Conjecture D is fulfilled). Then it was proved in: 
Smirnov O., Graded associative algebras and Grothendieck standard conjectures, Invent. Math., 128 (1997), 201-206
that the Lefschetz standard conjecture holds also; hence the Kunneth decompositions of the motif of a smooth projective $P$ exists both with respect to $\mathbb{Q}\_l$-adic and with respect to $\mathbb{Q}\_{l'}$-adic etale cohomology. Are these two decompositions necessarily isomorphic? I suspect that that the answer is 'Yes' and the proof is easy, but I am not sure.


P.S. I don't understand why writing ${\mathbb{Q}}\_{l'}$ in my question leads to catostrophic appearance.



Could the Kunneth decomposition of a motif depend on the choice of $l$?
Let me develop YBL's answer a bit. (I wanted to make this a comment but it was too long...)


Consider a smooth variety $U$ over $\mathbb{F}\_p$ with function field $K$ such that your motive and its two Künneth decompositions extend over U. Take a $\mathbb{F}\_q$-rational point $x$ of $U$, and look at the specialization of everybody at $x$ (see André-Kahn, *Construction inconditionnelle de groupes de Galois motiviques*, section 3 for the definition of good reduction and specialization of motives). Then you get a motive over $\mathbb{F}\_q$ with two Künneth decompositions, but these have to be the same for the reason YBL gave : weights (more precisely, the projectors on the components of the Künneth decomposition are given in this case by rational polynomials in the graph of Frobenius that are independent of the Weil cohomology, see part III of Katz-Messing, *Some consequences of the Riemann hypothesis for varieties over finite fields*). So your two decompositions are the same on the specialization, but the specialization functor is faithful, so the two decompositions are the same on the original motive.


