Bloch defines the motivic complexes $\mathbf{Z}(n)$ in his paper "Algebraic Cycles and Higher K-Theory" (1986).

Some references (that I currently am unable to track down) use $$\check{\mathbf{Z}}(n) := \mathbf{Z}(1)^{\otimes n} = \mathbf{G}_m[-1]^{\otimes n}$$ instead of $\mathbf{Z}(n)$.

What is the relation between $\check{\mathbf{Z}}(n)$ and ${\mathbf{Z}}(n)$?

  • 4
    $\begingroup$ It has nothing to do with Cech cohomology. It’s just notation for the tensor powers of the complex $\mathbf{Z}(1)$. $\endgroup$
    – user92332
    Feb 24, 2018 at 16:45
  • 1
    $\begingroup$ What tensor product are you using? I think they are equivalent if you consider the tensor product in the category of Voevodsky motives, but that of course is not the tensor product of sheaves of complexes. $\endgroup$ Feb 24, 2018 at 17:08
  • 1
    $\begingroup$ Surely the etale cohomology functor you are considering factors through the category of Galois representations - do you think one of them fails to correspond to the standard cyclotomic character? $\endgroup$
    – Will Sawin
    Feb 24, 2018 at 17:56

1 Answer 1


They are not quasi-isomorphic: your $\check{\mathbf Z}(n)$ is concentrated in a single degree. As Denis points out in the comments, your definition of $\check{\mathbf Z}(n)$ is wrong because you should use the tensor product of sheaves with transfers. In addition, you need to apply Suslin's $\mathbf A^1$-invariantification construction $C_*$ to the result.

Voevodsky's motivic complex $\mathbf Z(n)_V$ is $$\mathbf Z(n)_V = C_*(\mathbf G_m^{\otimes_{\mathrm{tr}}n})[-n],$$ where:

  • $\mathbf G_m$ is regarded as a presheaf with transfers, i.e., a presheaf on Voevodsky's category $\mathrm{Cor}_k$ of smooth separated $k$-schemes and finite correspondences. The transfers are given by norms of invertible functions.
  • $\otimes_{\mathrm{tr}}$ is the tensor product of sheaves with transfers, which is the Day convolution of the tensor product $X\otimes Y=X\times Y$ on $\mathrm{Cor}_k$.
  • $C_*(F)(X)$ is a chain complex concentrated in nonnegative degrees with $C_n(F)(X)=F(X\times \mathbf A^n)$.

$\mathbf Z(n)_V$ is quasi-isomorphic to Bloch's $\mathbf Z(n)$ as a complex of Zariski sheaves on smooth $k$-schemes. This combines several deep results of Voevodsky, Suslin, and Friedlander. A more or less self-contained proof is in Mazza–Voevodsky-Weibel's Lecture notes on motivic cohomology.

On the other hand, $\mathbf Z(n)_V$ is also a complex of étale sheaves, and if $m$ is prime to the characteristic, then $\mathbf Z/m(n)_V$ is quasi-isomorphic to $\mu_m^{\otimes n}$ as a complex of étale sheaves. This is Theorem 10.3 in the above book. Perhaps this answers your last question.

Edit: Actually the tensor product in the above formula for $\mathbf Z(n)_V$ must be derived, so it's not very explicit. The "official" definition is $$\mathbf Z(n)_V = C_*(\mathbf Z_{\mathrm{tr}}(\mathbf G_m^{\wedge n}))[-n],$$ where $\mathbf Z_{\mathrm{tr}}(\mathbf G_m^{\wedge n})$ means the quotient of the sheaf $\mathbf Z_{\mathrm{tr}}(\mathbf G_m^{\times n})$ sending $U$ to $\mathrm{Cor}_k(U,\mathbf G_m^{\times n})$ by the subsheaf generated by $\mathbf Z_{\mathrm{tr}}(\mathbf G_m^{\times n-i-1}\times\{1\}\times\mathbf G_m^{\times i})$. The relation with the other formula comes from the fact that $C_*(\mathbf Z_{\mathrm{tr}}(\mathbf G_m^{\wedge 1}))\to\mathbf G_m$ is a free resolution of $\mathbf G_m$ as a sheaf with transfers.

  • $\begingroup$ Great. I'm happy with this! $\endgroup$
    – user120812
    Feb 25, 2018 at 0:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy