I am in a situation where I need a result of the following form. Suppose $X$ is a smooth $k$-variety, $U$ is a dense open subvariety with complement $Z$ a smooth divisor. Let $\pi^X:X\rightarrow\text{Spec }k$ and $\pi^U:U\rightarrow\text{Spec }k$ be the structure morphisms. Let us work with Voevodsky motives $\text{DM}(k)$. Is it true that if $\pi^X_!\mathbf{1}_X(n)$ is effective for some integer $n$, then so is $\pi^U_!\mathbf{1}_U(n)$?
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2$\begingroup$ What about $X = \mathbb A^1, U = \mathbb G_m, n=1$? $\endgroup$– Will SawinCommented Sep 8, 2017 at 21:09
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1$\begingroup$ Don't we have $\pi^{\mathbb{A}^1}_!\mathbf{1}_{\mathbb{A}^1}\simeq \mathbf{1}_k(-1)[-2]$ and $\pi^{\mathbb{G}_m}_!\mathbf{1}_{\mathbb{G}_m}\simeq \mathbf{1}_k[-1]\oplus\mathbf{1}_k(-1)[-2]$? $\endgroup$– user114292Commented Sep 8, 2017 at 21:35
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$\begingroup$ So if you twist by (1), isn't the second one ineffective? $\endgroup$– Will SawinCommented Sep 9, 2017 at 11:10
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$\begingroup$ I suspect that $\pi^X_!1_X(n)$ is effective if and only if $i\ge \dim(X)$. The "if" direction here is easy, and you can probably prove the "only if" implication by looking at some cohomology. If this is true then the answer to you question is certainly positive. $\endgroup$– Mikhail BondarkoCommented Sep 9, 2017 at 14:40
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$\begingroup$ Note also that you cannot express the (Borel-Moore) motif of $U$ in terms of that of $X$ "after you get into motives over $\operatorname{Spec} k$". $\endgroup$– Mikhail BondarkoCommented Sep 15, 2017 at 7:56
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