TITLE: A question about the vanishing of motivic cohomology in negative Tate twist
QUESTION [9 upvotes]: Let $DM_{\text{gm}}$ be the category of Voevodsky´s geometric motives. Let $p,q\in \mathbb{Z}$ be integers with $p<0$.
Is it true that 
$$\text{Hom}_{DM_{\text{gm}}}(M_{\text{gm}}(X),\mathbb{Z}(p)[q])=0,$$
where $M(X)$ is the motive of a smooth scheme $X$ over a field $k$ and $\mathbb{Z}(p)$ is the Tate motive?

REPLY [5 votes]: Here's an easy way to see this (which is more or less an elaboration of Mikhail's answer):
Suppose $i > 0$ and $n\in \mathbb Z$. We have the Thom isomorphism
$$
H^n(X, \mathbb Z(-i)) \cong H^{n+2i}_X(\mathbb A^i_X, \mathbb Z(0)).
$$
Weight $0$ motivic cohomology is just Zariski cohomology, so
$$
H^n(X,\mathbb Z(0)) = \begin{cases} \mathbb Z^{\pi_0(X)} & \text{if }n=0,\\
0 & \text{otherwise.}\end{cases}
$$
Now the long exact sequence for cohomology with support shows that $H^*_X(\mathbb A^i_X,\mathbb Z(0))=0$.