Let $Q$ be a anisotropic quadric of dimension $d$ over $k$. We work in the category of effective Chow-Motives over $k$. Let $T$ be the Tate-Motive. For a motive $M$ we write $M(l)$ for its $l$-th Tate-Twist.

Assume we have a motivic decomposition $M(Q) = A \oplus B$. Assume further that over the algebraic closure of $k$, the motive $A$ decomposes as $T(a) \oplus T(b)$,while $a\leq b$. The motive $A$ is then called a binary summand of M(Q). The dimension of the motive $A$ is defined as $b-a$.

Question: Is it true that the dimension of every binary summand of a quadric is odd ? $*$

I think i once red it in a paper (of Vishik?), but i need the reference.

$*$ Only exception may be the case if its zero. For example the motive of a zero dimensional quadric splits in two trivial Tate-Motives and thus its motive has dimension zero.

  • $\begingroup$ I don't know much about motives, but shouldn't the motive of any smooth quadric surface $S$ be something like $T(0)\oplus \mathrm{Pic}(S)\oplus T(−2)$? Does this not give a counter-example, on taking $A=T(0)\oplus T(−2)$? $\endgroup$ – Daniel Loughran Aug 18 '14 at 17:22
  • $\begingroup$ The motive of a SPLIT quadric is does indeed have the form $T(0) \oplus T(1) \oplus T(1) \oplus T(2)$, while the $T(1)$ basically represent CH$^1(Q) = $Pic$(Q)$. This is a well known fact.But i look at a anisotropic projective quadric.Assume it has a decomposition into $A \oplus B$ over $k$. Then none of these two motives can contain a trivial Tate-Motive because this happens iff it is isotropic over $k$. Take for example a two dimensional Pfister-quadric. It has a decomposition into two Rost-Motives iff its anisotropic. But these split into $T(0) \oplus T(1)$ and $T(1) \oplus T(2)$! $\endgroup$ – nxir Aug 18 '14 at 18:23

The binary motive theorem states that a binary summand of the motive of a quadric has dimension $2^r-1$ for some $r$. This should be Theorem 6.1 in:

O. Izhboldin and A. Vishik. Quadratic forms with absolutely maximal splitting. Contemp. Math. 272 (2000), 103-125.

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