In his Ph.D. thesis, Semenov shows that the motivic decomposition of a variety in general is not unique. He works in the category of Chow-Motives and not in a bigger category of motives.

Is there an example of two varieties having different Chow-Motive decomposition in the category of Chow motives, such that the respective summands split into finer/more summands in a bigger category, and so that these new decompositions coincide?

Maybe we can choose the same category but consider another equivalence relation for the Chow-Cycles instead. This would of course require some coarser relation than rational equivalence to be chosen in the first place.

Edit: Of course we could consider the algebraic closure of $k$, respectively its category of Chow-Motives, and interpret this as that certain "extended" category in question. In this category, two quadrics $X$, $Y$ will both become isomorphic to a sum of twisted Tate-Motives. If we choose them both to be Pfister quadrics of the same dimension, one being isotropic (i.e. hyperbolic), the other anisotropic, the decompositions won't be isomorphic over $k$. But this is a rather trivial example and somehow cheating, because one decomposition already contains only twisted Tate-Motives.

Isn't there more exciting stuff possible, maybe in Voevodsky's category?

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    $\begingroup$ The category of Chow motives is a full subcategory of Voevodsky's triangulated category and is moreover closed under taking direct summands so you cannot get anything new by considering Voevodsky's category. $\endgroup$ – ulrich Dec 14 '13 at 7:31
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    $\begingroup$ I suspect that Semenov works with motives with integral coefficients; conjecturally non-uniqueness does not happen for rational coefficients. Certainly, the category of cohomological motives with rational coefficients is a Krull-Schmidt one. $\endgroup$ – Mikhail Bondarko Dec 14 '13 at 8:23
  • $\begingroup$ Yes, Semenov works with integral coefficients. Sorry, i dont know anything about the category of cohomological motives. Might that be another term for something well known? $\endgroup$ – Jason Pioneer Dec 15 '13 at 2:50
  • $\begingroup$ Motives up to homological equivalence. $\endgroup$ – Mikhail Bondarko Dec 18 '13 at 9:26

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