# RHom and Koszul complexes.

Let $A$ be a Koszul algebra over a field $k$ that is both left and right finite. One can consider the its Koszul complex $X$ as a dg $A^!$-$A$ bimodule. I want to show:

For all $i \in \mathbb{Z}$, $RHom_A(X, X \langle -i \rangle [i])$ has its homology concentrated in degree $0$.

So far here is what I have been trying: Since $A$ is Koszul, there is a graded projective resolution of $A_0$, call it $P_\bullet$ such that $AP_i ^i = P_i$. On the other hand as a right $A$ module $X$ is projective resolution of $A_0$. So I try and compute the total Hom space $Hom ^\bullet_A(P_\bullet, A_0\langle -i \rangle [i])$. I am having trouble seeing how this complex is exact everywhere except at $Hom^0_A(P_\bullet, A_0 \langle -i \rangle [i]) = Hom_A (P_i, A_0 \langle -i \rangle )$.

-
Is it clear to you that, since $P_j$ lives in (wrt Koszul grading) degree $\geq j$, we have $hom_A(P_j, A_0\langle -i\rangle)=0$ unless $i=j$? – Aaron May 14 '14 at 21:16
Yes, so for example its true that $Hom_A ( P_j, A_0 \langle -i \rangle ) = hom_A ( P_j , A_0 \langle -j \rangle )$ where the lower case is means we take only morphisms of degree zero, right? But how does this help, I feel like its telling me that the differentials are zero or something like that. – Anette May 14 '14 at 21:32
my usual notation is $Hom$ for ungraded (some people use different one), so you take morphisms in all degree, but the special nature of $A_0$ and $P_\bullet$ implies that $Hom_A(P_j,A_0\langle -i\rangle)=hom_A(P_j,A_0\langle -j\rangle)$. Now, just think about what $Hom_A^\bullet$ means, it takes morphisms of the complex $P_\bullet$ to a stalk complex (hence module) $A_0\langle-i\rangle$. So the $h$-th (homological) degree morphism is completely described by $Hom_A(P_h,A_0\langle-i\rangle[i]) = Hom_A(P_{h+i},A_0\langle -i\rangle)$. – Aaron May 14 '14 at 21:41
I am sorry for being so slow Aaron, really! So are you saying that the complex I should end up with should in homological degree zero be $\prod_{i \in \mathbb{Z}} hom_A ( P_i , \langle -i \rangle )$ and zeros in all other degrees? – Anette May 15 '14 at 0:52
How about this: let $C_\bullet := A_0\langle-i\rangle [i]$. Degree $j$ maps are built up from module hom $P_{x+j}$ to $C_x$. Becuase $hom_A(P_x,C_y)=0$ for all $y\neq i$, the degree $j$ maps from $P_\bullet$ to $C_\bullet$ are described by $hom_A(P_{j+i},C_i) = hom_A(P_{j+i},A_0\langle -i\rangle)=0$. – Aaron May 15 '14 at 11:03