# Formal DG-algebra

Let $\mathcal{C}$ be a nice $k$-linear abelian category (the example I have in mind is the category of coherent sheaves on a smooth projective variety over $\mathbb{C}$). Let $B \in D^{b}(\mathcal{C})$ such that the DG-algebra $\mathrm{RHom}(B,B)$ is formal (that is quasi-isomorphic, as a DG-algebra to its cohomology algebra).

I would like to know if there exists a complex $B^{\bullet}$ representing $B$ such that: $$\mathrm{Hom}^{\bullet}(B^{\bullet}, B^{\bullet}) = \mathrm{Ext}^{\bullet}(B,B),$$ where the sign equal really means equality (and not just quasi-isomorphism of DG algebras)?

Here $\mathrm{Hom}^{\bullet}(B^{\bullet}, B^{\bullet})$ is the DG-algebra defined by: _$\mathrm{Hom}^{k}(B^{\bullet}, B^{\bullet})$ is the vector space of sequences of maps $f^p : B^{p} \rightarrow B^{p+k}$,

_the multiplication map in $\mathrm{Hom}^{\bullet}(B^{\bullet}, B^{\bullet})$ is the composition of maps,

_the differential is the one inherited from that of $B^{\bullet}$.

Thanks a lot!

## 1 Answer

The following seems to be a counterexample. Let $\mathcal C$ be the category of $k[x]$-modules and $B=k=k[x]/(x)$.

The derived endomorphism DG-algebra of $B$ can be computed as $\operatorname{RHom}(B,B)=\operatorname{Hom}^\bullet(X^\bullet,X^\bullet)$, where $X^\bullet$ is the complex

$\cdots\rightarrow 0\rightarrow k[x]\stackrel{x}\longrightarrow k[x]\rightarrow 0\rightarrow \cdots$

Its cohomology $H^\bullet\operatorname{RHom}(B,B)=\operatorname{Ext}^\bullet(B,B)=\Lambda(t)=k\cdot 1\oplus k\cdot t$ is the exterior algebra on a degree $1$ generator and there is a unique quasi-isomorphism of $k$-DGAs $\Lambda(t)\rightarrow \operatorname{Hom}^\bullet(X^\bullet,X^\bullet)$ sending $t$ to the cochain map $X^\bullet\rightarrow X^\bullet[1]$ which is the identity in $k[x]$ in degree $0$ and trivial otherwise.

If there were a complex $B^\bullet$ with $\operatorname{Hom}^\bullet(B^\bullet,B^\bullet)=\Lambda (t)$ then $\operatorname{Hom}^0(B^\bullet,B^\bullet)$ would be $1$-dimensional as a $k$-vector space. This forces $B^\bullet$ to be globally $1$-dimensional over $k$. Since $B^\bullet$ is quasi-isomorphic to $B$, it must be $B^\bullet=B=k$, and in this case $\operatorname{Hom}^1(B,B)=0\ncong k$.

• Super! Thank you. Oct 24, 2015 at 13:26