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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 )$.

Anette
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  • 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 00:52
  • The conclusion is for each $i\geq 0$, you have $Hom_A^\bullet(P_\bullet,A_0\langle -i\rangle[i]) = Hom_A(P_i,A_0\langle -i\rangle) = Hom_A^0(P_\bullet, A_0\langle-i\rangle[i])$. So the complex is zero in all non-zero (homological) degree. – Aaron May 15 '14 at 10:32
  • Yes I want to believe this very badly, but I dont see why $Hom^j_A (P_\bullet, A_0\langle -i \rangle [i])$ for $j \neq 0$ is zero. It should just be $hom_A(P_{j+i}, A_0 \langle -j -i \rangle )$ as you point out, no? – Anette May 15 '14 at 10:41
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    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
  • Hey Aaron, Thanks for you comments and sorry it took my so long to respond. My problem was that actually all the hom's where supposed to be the graded category anyway, I was confusing myself. – Anette May 23 '14 at 22:39

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