Let $G=SL_2(q)$, and $\mu_{q+1}$ be the $q+1$ roots of unity in the algebraic closure $F=\overline{\mathbb{F}_q}$. Let $Y$ be the Drinfeld curve defined over $F$. In this case, Deligne-Lusztig induction for $\theta$ a character of $\mu_{q+1}$ is $$ R'(\theta)=[H^1_c(Y)\otimes_{K\mu_{q+1}} V_\theta]_G-[H^2_c(Y)\otimes_{K\mu_{q+1}} V_\theta]_G $$ where $K$ is some field, $V_\theta$ a $K\mu_{q+1}$-representation affording $\theta$, and the notation $[-]_G$ denotes the afforded character in the Grothendieck group $\mathscr{K}_0(KG)$ identified as the ring $\mathbb{Z}\operatorname{Irr}(G)$ of virtual characters of $G$, and $H^i_c(Y)$ is the $i$th $\ell$-adic cohomology group.
Since $Y$ is irreducible, its known that $[H^2_c(Y)]_{G\times\mu_{q+1}}=1_{G\times\mu_{q+1}}$. How does this imply that if $\theta$ is a nontrivial linear character of $\mu_{q+1}$, then $R'(\theta)=[H^1_c(Y)\otimes_{K\mu_{q+1}} V_\theta]_G$, a character of $G$? I don't understand why the second term $[H^2_c(Y)\otimes_{K\mu_{q+1}} V_\theta]_G$ disappears.