# $Ext$-algebra generated by $Hom$ and $Ext^1$ as $A_\infty$-algebra?

In [Keller: A-infinity algebras in representation theory, Proposition 1(b)], Keller states that for an associative algebra the $Ext$-algebra of the simples is generated by $Ext^1(S,S)$ as an $A_\infty$-algebra.

Are there any other conditions on a module $M$ known to prove that $Ext^*(M,M)$ is generated by $Hom(M,M)$ and $Ext^1(M,M)$ as an $A_\infty$-algebra module $M$ over a finite dimensional associative algebra? E.g. conditions like $M$ generator of the derived category or something similar.

• It can't be true if $\operatorname{Ext}^1(M,M)=0$ but $\operatorname{Ext}^i(M,M)\neq0$ for some $i>1$, and it's easy to find examples where that happens. – Jeremy Rickard Aug 28 '15 at 10:07
• @JeremyRickard Thanks, I should have thought about that before posting. An example is obviously $k[1\to 2\to 3]/(ab)$. and $M=S(1)\oplus S(3)$. – Julian Kuelshammer Aug 28 '15 at 18:01
• This works for Koszul modules, iirc. – Mariano Suárez-Álvarez Aug 28 '15 at 18:15
• @MarianoSuárez-Alvarez Yes, you are right. But for Koszul modules, you also need only the algebra structure, not the $A_\infty$-structure, so it is quite restrictive. I want to know whether there are any other classes, where (at least sometimes) you need the $A_\infty$-structure as well. – Julian Kuelshammer Aug 28 '15 at 18:21
• For $N$-Koszul modules you do need also the higher products. The cup product on $Ext_A(k,k)$ with $A$ $3$-Koszul is in fact zero. (I wrote Koszul but I was think of $N$-Koszul :-/ ) – Mariano Suárez-Álvarez Aug 28 '15 at 18:22

Suppose $M=M_1 \oplus \dots \oplus M_r$ with $M_i$ indecomposable for $1 \leq i \leq r$. Let $\mathcal F (M_1, \dots, M_r)$ denote the category of modules which admit a filtration with subquotients in the $M_i$. The crucial assumption in Keller's proof is that $\mathcal F(M_1, \dots, M_r)$ is closed under syzygies.
With this assumption, any $n$-fold extension ($n>1$) $$\xi \in \operatorname{Ext}^n(M_u,M_v) \simeq \operatorname{Ext}^1(\Omega^{n-1}(M_u),M_v)$$ with $1 \leq u,v \leq r$ is the Yoneda splice of $n$ short-exact sequences $$0 \to X_{i+1} \to E_i \to X_i \to 0 \quad (0 \leq i \leq n-1)$$ where $X_0=M_u$, $X_n=M_v$, and each $X_i$ is an object in $\mathcal F(M_1, \dots, M_r)$ for $0 \leq i \leq n$.
As explained in the article you mentioned, the category $\mathcal F(M_1, \dots, M_r)$ is equivalent to a category of twisted stalks. A twisted stalk can be described as a pair $(B,\delta)$, where $B$ is a sequence of subfactors from $\{M_1, \dots, M_r\}$ and $\delta$ is an upper triangular matrix with entries from $\operatorname{Ext}^1(M,M)$.
If $X_i$ corresponds to a twisted stalk $(B,\delta)$ and $X_{i+1}$ corresponds to a twisted stalk $(B',\delta')$, then the short-exact sequence $0 \to X_{i+1} \to E_i \to X_i \to 0$ corresponds to a morphism of twisted complexes $(B,\delta) \to (B',\delta')[1]$. Such a morphism is a matrix with entries from $\operatorname{Ext}^1(M,M)$.
So each short-exact sequence $0 \to X_{i+1} \to E_i \to X_i \to 0$ can be described using only the $\operatorname{Ext}^1$ part of $\operatorname{Ext}^\ast(M,M)$. The higher multiplications enter the picture when we splice the short-exact sequences together. The rule for composing morphisms of twisted complexes involves $m_2$ and higher multiplications of the matrices involved. When splicing all the short-exact sequences together, we end up with an expression for $\xi \in \operatorname{Ext}^n(M_u,M_v)$ using only elements in $\operatorname{Ext}^1(M,M)$ and $m_2$ and higher multiplications. This shows that $\operatorname{Ext}^\ast(M,M)$ is generated in degree $0$ and $1$ as an $A_\infty$-algebra.
A class of examples where the assumption is satisfied is when $\{M_1, \dots, M_r\}$ are the standard modules over a quasi-hereditary algebra.