On the Lie algebra structure of $HH^1(A)$ of a finite-dimensional algebra $A$

TL;DR

This paper investigates the Lie algebra structure of the first Hochschild cohomology of finite-dimensional algebras, revealing conditions under which it is solvable or isomorphic to sl_2(k), based on properties of the Ext-quiver.

Contribution

It establishes new criteria linking the Ext-quiver properties of an algebra to the solvability and simplicity of its Hochschild cohomology Lie algebra.

Findings

If the Ext-quiver is a simple directed graph, then HH^1(A) is solvable.

If the Ext-quiver has no loops and at most two parallel arrows, and HH^1(A) is simple, then char(k) ≠ 2 and HH^1(A) ≅ sl_2(k).

Investigates symmetric algebras with a quiver having a vertex with a single loop.

Abstract

Let $A$ be a split finite-dimensional associative unital algebra over a field. The first main result of this note shows that if the Ext-quiver of $A$ is a simple directed graph, then $HH^1(A)$ is a solvable Lie algebra. The second main result shows that if the Ext-quiver of $A$ has no loops and at most two parallel arrows in any direction, and if $HH^1(A)$ is a simple Lie algebra, then char(k) is not equal to $2$ and $HH^1(A)\cong$ $sl_2(k)$. The third result investigates symmetric algebras with a quiver which has a vertex with a single loop.

Equations

$$H{H}^1(A;S{\otimes }_k{S}^{\vee })\cong {\mathrm{Ext}}_A^1(S,S)\cong {\mathrm{Hom}}_A(J(A)i/J(A{)}^{2}i,S)\cong {\mathrm{Hom}}_{A{\otimes }_k{A}^{\mathrm{op}}}(J(A)/J(A{)}^2,S{\otimes }_k{S}^{\vee }).$$

$${\dim }_k(H{H}^1(A))=e(A)-\ell (A)+1\le (\ell (A)-1{)}^2.$$

$$f=\sum _{i=-1}^{p-1}{\lambda }_i{f}_i$$

$$g=\sum _{i=t}^{p-2}{\mu }_i{f}_i$$

$$g=\sum _{i=0}^{p-2}{\mu }_i{f}_i$$

$$h=\sum _{i=s}^{p-2}{\nu }_i{f}_i$$

$$u=\sum _{i=r}^{p-2}{\tau }_i{f}_i$$

Full text

1. Introduction

Let $k$ be a field. Our first result is a sufficient criterion for $H\!H^{1}(A)$ to be a solvable Lie algebra, where $A$ is a split finite-dimensional $k$-algebra (where the term ‘algebra’ without any further specifications means an associative and unital algebra).

Theorem 1.1**.**

Let $A$ be a split finite-dimensional $k$-algebra. Suppose that the $\mathrm{Ext}$-quiver of $A$ is a simple directed graph. Then the derived Lie subalgebra of $H\!H^{1}(A)$ is nilpotent; in particular the Lie algebra $H\!H^{1}(A)$ is solvable.

The recent papers [4] and [8] contain comprehensive results regarding the solvability of $H\!H^{1}(A)$ of tame algebras and blocks, and [8] also contains a proof of Theorem 1.1 with different methods. We will prove Theorem 1.1 in Section 3 as part of the more precise Theorem 3.1, bounding the derived length of the Lie algebra $H\!H^{1}(A)$ and the nilpotency class of the derived Lie subalgbra of $H\!H^{1}(A)$ in terms of the Loewy length $\ell\!\ell(A)$ of $A$. The hypothesis on the quiver of $A$ is equivalent to requiring that $\mathrm{Ext}_{A}^{1}(S,S)=$ $\{0\}$ for any simple $A$-module $S$ and $\mathrm{dim}_{k}(\mathrm{Ext}^{1}_{A}(S,T))\leq 1$ for any two simple $A$-modules $S$, $T$. If in addition $A$ is monomial, then Theorem 1.1 follows from work of Strametz [10]. The hypotheses on $A$ are not necessary for the derived Lie subalgebra of $H\!H^{1}(A)$ to be nilpotent or for $H\!H^{1}(A)$ to be solvable; see [2, Theorem 1.1] or [8] for examples.

The Lie algebra structure of $H\!H^{1}(A)$ is invariant under derived equivalences, and for symmetric algebras, also invariant under stable equivalences of Morita type. Therefore, the conclusions of Theorem 1.1 remain true for any finite-dimensional $k$-algebra $B$ which is derived equivalent to an algebra $A$ satisfying the hypotheses of this theorem, or for a symmetric $k$-algebra $B$ which is stably equivalent of Morita type to a symmetric algebra $A$ satisfying the hypotheses of the theorem.

If we allow up to two parallel arrows in the same direction in the quiver of $A$ but no loops, then it is possible for $H\!H^{1}(A)$ to be simple as a Lie algebra. The only simple Lie algebra to arise in that case is $\mathfrak{sl}_{2}(k)$, with $\mathrm{char}(k)\neq$ $2$.

Theorem 1.2**.**

Let $A$ be a split finite-dimensional $k$-algebra. Suppose that $\mathrm{Ext}_{A}^{1}(S,S)=$ $\{0\}$ for any simple $A$-module $S$ and that $\mathrm{dim}_{k}(\mathrm{Ext}^{1}_{A}(S,T))\leq 2$ for any two simple $A$-modules $S$, $T$. If $H\!H^{1}(A)$ is not solvable, then $\mathrm{char}(k)\neq 2$ and $H\!H^{1}(A)/\mathrm{rad}(H\!H^{1}(A))$ is a direct product of finitely many copies of $\mathfrak{sl}_{2}(k)$. In particular, the following hold.

1. (i)

If $H\!H^{1}(A)$ is a simple Lie algebra, then $\mathrm{char}(k)\neq 2$, and $H\!H^{1}(A)\cong$ $\mathfrak{sl}_{2}(k)$. 2. (ii)

If $\mathrm{char}(k)=2$, then $H\!H^{1}(A)$ is a solvable Lie algebra.

This will be proved in Section 3; for monomial algebras this follows as before from Strametz [10]. An example of an algebra $A$ satisfying the hypotheses of this theorem is the Kronecker algebra, a $4$-dimensional $k$-algebra, with $\mathrm{char}(k)\neq 2$, given by the directed quiver with two vertices $e_{0}$, $e_{1}$ and two parallel arrows $\alpha$, $\beta$ from $e_{0}$ to $e_{1}$. This example is a special case of more general results on monomial algebras; see in particular [10, Corollary 4.17]. As in the case of the previous Theorem, the conclusions of Theorem 1.2 remain true for an algebra $B$ which is derived equivalent to an algebra $A$ satisfying the hypotheses of this theorem, or for a symmetric algebra $B$ which is stably equivalent of Morita type to a symmetric algebra $A$ satisfying the hypotheses of the theorem.

We have the following partial result for symmetric algebras whose quiver has a single loop at some vertex.

Theorem 1.3**.**

Suppose that $k$ is algebraically closed. Let $A$ be a finite-dimensional symmetric $k$-algebra, and let $S$ be a simple $A$-module. Suppose that $\mathrm{dim}_{k}(\mathrm{Ext}_{A}^{1}(S,S))=$ $1$ and that for any primitive idempotent $i$ in $A$ satisfying $iS\neq$ $\{0\}$ we have $J(iAi)^{2}=$ $iJ(A)^{2}i$. If $H\!H^{1}(A)$ is a simple Lie algebra, then $\mathrm{char}(k)=$ $p>2$ and $H\!H^{1}(A)$ is isomorphic to either $\mathfrak{sl}_{2}(k)$ or the Witt Lie algebra $W=$ $\mathrm{Der}(k[x]/(x^{p}))$.

This will be proved in Section 4, along with some general observations regarding the compatibility of Schur functors and the Lie algebra structure of $H\!H^{1}(A)$. Section 5 contains some examples.

2. On derivations and the radical

We start with a brief review of some basic terminology. Let $k$ be a field. The nilpotency class of a nilpotent Lie algebra ${\mathcal{L}}$ is the smallest positive integer $m$ such that ${\mathcal{L}}^{m}=\{0\}$, where ${\mathcal{L}}^{1}=$ ${\mathcal{L}}^{\prime}$ and ${\mathcal{L}}^{m+1}=$ $[{\mathcal{L}},{\mathcal{L}}^{m}]$ for $m\geq 1$. In addition, the derived length of a solvable Lie algebra is the smallest positive integer $n$ such that ${\mathcal{L}}^{(n)}=\{0\}$, where ${\mathcal{L}}^{(1)}=$ ${\mathcal{L}}^{\prime}$ and ${\mathcal{L}}^{(n+1)}=$ $[{\mathcal{L}}^{(n)},{\mathcal{L}}^{(n)}]$ for $n\geq 1$. A Lie algebra ${\mathcal{L}}$ is called strongly solvable if its derived subalgebra is nilpotent. A Lie algebra ${\mathcal{L}}$ of finite dimension $n$ is called completely solvable (also called supersolvable) if there exists a sequence of ideals ${\mathcal{L}}_{1}={\mathcal{L}}\supset{\mathcal{L}}_{2}\supset\cdots\supset{\mathcal{L}}_{n}\supset\{0\}$ such that $\mathrm{dim}_{k}({\mathcal{L}}_{i})=n+1-i$ for $1\leq i\leq n$.

Remark 2.1**.**

If $k$ is algebraically closed of characteristic zero, then the classes of strongly and completely solvable Lie algebras coincide with the class of solvable Lie algebras as a consequence of Lie’s theorem. Lie’s theorem does not hold in positive characteristic. If $k$ is algebraically closed of prime characteristic $p$, then by [3, Theorem 3], a finite-dimensional Lie algebra ${\mathcal{L}}$ over $k$ is strongly solvable if and only if ${\mathcal{L}}$ is completely solvable.

Let $A$ be a finite-dimensional $k$-algebra. We denote by $\ell(A)$ the number of isomorphism classes of simple $A$-modules. The Loewy length $\ell\!\ell(A)$ of $A$ is the smallest positive integer $m$ such that $J(A)^{m}=\{0\}$, where $J(A)$ denotes the Jacobson radical of $A$. We denote by $[A,A]$ the $k$-subspace of $A$ generated by the set of additive commutators $ab-ba$, where $a$, $b\in$ $A$. A derivation on $A$ is a $k$-linear map $f:A\to$ $A$ satisfying $f(ab)=$ $f(a)b+af(b)$ for all $a$, $b\in$ $A$. If $f$, $g$ are derivations on $A$, then so is $[f,g]=$ $f\circ g-g\circ f$, and the space $\mathrm{Der}(A)$ of derivations on $A$ becomes a Lie algebra in this way. If $c\in$ $A$, then the map $[c,-]$ defined by $[c,a]=$ $ca-ac$ is a derivation; any derivation of this form is called an inner derivation. The space $\mathrm{IDer}(A)$ of inner derivations is a Lie ideal in $\mathrm{Der}(A)$, and we have a canonical isomorphism $H\!H^{1}(A)\cong$ $\mathrm{Der}(A)/\mathrm{IDer}(A)$; see [12, Lemma 9.2.1]. It is easy to see that any derivation on $A$ preserves the subspace $[A,A]$, and that any inner derivation of $A$ preserves any ideal in $A$. A finite-dimensional $k$-algebra $A$ is called split if $\mathrm{End}_{A}(S)\cong$ $k$ for every simple $A$-module $S$. If $A$ is split, then by the Wedderburn-Malcev Theorem, $A$ has a separable subalgebra $E$ such that $A=$ $E\oplus J(A)$. Moreover, $E$ is unique up to conjugation by elements in the group $A^{\times}$ of invertible elements in $A$. A primitive decomposition $I$ of $1$ in $E$ remains a primitive decomposition of $1$ in $A$.

For convenience, we mention the following well-known descriptions of certain $\mathrm{Ext}^{1}$-spaces.

Lemma 2.2**.**

Let $A$ be a split finite-dimensional $k$-algebra, let $i$ be a primitive idempotent in $A$. Set $S=$ $Ai/J(A)i$ and $S^{\vee}=$ $iA/iJ(A)$. We have $k$-linear isomorphisms

[TABLE]

Lemma 2.3**.**

Let $A$ be a split finite-dimensional $k$-algebra. Let $i$ be a primitive idempotent in $A$, and set $S=$ $Ai/J(A)i$. We have $\mathrm{Ext}^{1}_{A}(S,S)=$ $\{0\}$ if and only if $iJ(A)i\subseteq$ $J(A)^{2}$.

Proof.

By Lemma 2.2, we have $\mathrm{Ext}^{1}_{A}(S,S)=$ $\{0\}$ if and only if $J(A)/J(A)^{2}$ has no simple bimodule summand isomorphic to $S\otimes_{k}S^{\vee}$. This is equivalent to $i\cdot(J(A)/J(A)^{2})\cdot i=$ $\{0\}$, hence to $iJ(A)i\subseteq$ $J(A)^{2}$ as stated. ∎

Lemma 2.4**.**

Let $A$ be a split finite-dimensional $k$-algebra, and let $E$ be a separable subalgebra of $A$ such that $A=$ $E\oplus J(A)$. Every class in $H\!H^{1}(A)$ has a representative $f\in$ $\mathrm{Der}(A)$ satisfying $E\subseteq$ $\mathrm{ker}(f)$.

Proof.

Let $f:A\to$ $A$ be a derivation. Since $E$ is separable, it follows that for any $E$-$E$-bimodule $M$ we have $H\!H^{1}(E;M)=$ $\{0\}$. In particular, the derivation $f|_{E}:E\to$ $A$ is inner; that is, there is an element $c\in$ $A$ such that $f(x)=$ $[c,x]$ for all $x\in$ $E$. Thus the derivation $f-[c,-]$ on $A$ vanishes on $E$ and represents the same class as $f$ in $HH^{1}(A)$. ∎

Lemma 2.5**.**

Let $A$ be a split finite-dimensional $k$-algebra, and let $E$ be a separable subalgebra of $A$ such that $A=$ $E\oplus J(A)$. Let $f:A\to$ $A$ be a derivation such that $E\subseteq$ $\mathrm{ker}(f)$. For any two idempotents $i$, $j$ in $E$ we have $f(iAj)\subseteq$ $iAj$ and $f(AiAj)\subseteq$ $AiAj$.

Proof.

Let $i$, $j$ be idempotents in $E$, and let $a$, $b\in$ $A$. We have $f(iaj)=$ $f(i^{2}aj)=$ $if(iaj)+$ $f(i)iaj=$ $if(iaj)$, since $i\in$ $E\subseteq$ $\mathrm{ker}(f)$. Thus $f(iaj)\in$ $iA$. A similar argument shows that $f(iaj)\in$ $Aj$, and hence $f(iaj)\in$ $iAj$. This shows the first statement. The second statement follows from this and the equality $f(biaj)=$ $f(b)iaj+bf(iaj)$. ∎

Lemma 2.6**.**

Let $A$ be a split finite-dimensional $k$-algebra such that $\mathrm{Ext}^{1}_{A}(S,S)=\{0\}$ for all simple $A$-modules $S$. Then for any derivation $f:A\to$ $A$ we have $f(J(A))\subseteq$ $J(A)$.

Proof.

Let $E$ be a separable subalgebra of $A$ such that $A=$ $E\oplus J(A)$. Let $I$ be a primitive decomposition of $1$ in $E$ (hence also in $A$). Note that if $i$, $j\in$ $I$ are not conjugate in $A^{\times}$, then $iAj\subseteq$ $J(A)$. The hypotheses on $A$ imply that $J(A)i/J(A)^{2}i$ has no summand isomorphic to $Ai/J(A)i$, and hence that $iJ(A)i\subseteq$ $J(A)^{2}$ for any $i\in$ $I$. Then $iJ(A)j\subseteq$ $J(A)^{2}$ for any two $i$, $j\in$ $I$ which are conjugate in $A^{\times}$. Let now $f:A\to$ $A$ be a derivation. As noted above, any inner derivation preserves $J(A)$. Thus, by Lemma 2.4, we may assume that $f|_{E}=0$. Since $J(A)=$ $\oplus_{i\in I}\ J(A)i$, it suffices to show that $f(J(A)i)\subseteq$ $J(A)i$, where $i\in$ $I$. If $j$ is conjugate to $i$, then $AjJ(A)i\subseteq$ $J(A)^{2}i$. Since $J(A)i=$ $\sum_{j\in I}\ AjJ(A)i$, it follows from Nakayama’s Lemma that $J(A)i=$ $\sum_{j}AjAi$, where $j$ runs over the subset $I^{\prime}$ of all $j$ in $I$ which are not conjugate to $i$. Now $f$ preserves the submodules $AjAi$ in this sum, thanks to Lemma 2.5. The result follows. ∎

The following observations are variations of the statements in [6, Proposition 3.5].

Proposition 2.7**.**

Let $A$ be a split finite-dimensional $k$-algebra, and let $E$ be a separable subalgebra of $A$ such that $A=$ $E\oplus J(A)$. For $m\geq 1$, denote by $D_{m}$ the subspace of $\mathrm{Der}(A)$ consisting of all derivations $f:A\to$ $A$ such that $E\subseteq$ $\mathrm{ker}(f)$ and such that $f(J(A))\subseteq$ $J(A)^{m}$. The following hold.

1. (i)

For any positive integers $m$, $n$ we have $[D_{m},D_{n}]\subseteq$ $D_{m+n-1}$. 2. (ii)

The space $D_{1}$ is a Lie subalgebra of $\mathrm{Der}(A)$, and for any positive integer $m$, the space $D_{m}$ is a Lie ideal in $D_{1}$. 3. (iii)

The space $D_{2}$ is a nilpotent ideal in $D_{1}$. More precisely, if $\ell\!\ell(A)\leq 2$, then $D_{2}=$ $\{0\}$, and if $\ell\!\ell(A)>2$, then the nilpotency class of $D_{2}$ is at most $\ell\!\ell(A)-2$.

Proof.

The space of derivations on $A$ which vanish on $E$ is easily seen to be closed under the Lie bracket on $\mathrm{Der}(A)$. Thus statement (i) follows from [6, Lemma 3.4]. Statement (ii) is an immediate consequence of (i). If $m\geq$ $\ell\!\ell(A)$, then $J(A)^{m}=$ $\{0\}$, and hence $D_{m}=$ $\{0\}$. Together with (i), this implies (iii). ∎

Proposition 2.8**.**

Let $A$ be a split finite-dimensional $k$-algebra, and let $E$ be a separable subalgebra of $A$ such that $A=$ $E\oplus J(A)$. For $m\geq 1$, denote by $D_{m}$ the subspace of $\mathrm{Der}(A)$ consisting of all derivations $f:A\to$ $A$ such that $E\subseteq$ $\mathrm{ker}(f)$ and such that $f(J(A))\subseteq$ $J(A)^{m}$. Suppose that every derivation $f$ on $A$ satisfies $f(J(A))\subseteq$ $J(A)$. Then the canonical algebra homomorphism $A\to$ $A/J(A)^{2}$ induces a Lie algebra homomorphism $\Phi:H\!H^{1}(A)\to$ $H\!H^{1}(A/J(A)^{2})$, and the following hold.

1. (i)

The canonical surjection $\mathrm{Der}(A)\to$ $H\!H^{1}(A)$ maps $D_{1}$ onto $H\!H^{1}(A)$. 2. (ii)

The canonical surjection $\mathrm{Der}(A)\to$ $H\!H^{1}(A)$ maps $D_{2}$ onto $\mathrm{ker}(\Phi)$; in particular, $\mathrm{ker}(\Phi)$ is a nilpotent ideal in the Lie algebra $H\!H^{1}(A)$. 3. (iii)

The Lie algebra $H\!H^{1}(A)$ is solvable if and only if $H\!H^{1}(A)/\mathrm{ker}(\Phi)$ is solvable. 4. (iv)

If the derived Lie algebra of $H\!H^{1}(A)$ is contained in $\mathrm{ker}(\Phi)$, then $H\!H^{1}(A)$ is nilpotent. 5. (v)

If the Lie algebra $H\!H^{1}(A)$ is simple, then $\Phi$ is injective.

Proof.

The hypotheses on $\mathrm{Der}(A)$ together with Lemma 2.4 imply that $H\!H^{1}(A)$ is equal to the image of the space $D_{1}$ in $H\!H^{1}(A)$, whence (i). The canonical surjection $\mathrm{Der}(A)\to$ $H\!H^{1}(A)$ clearly maps $D_{2}$ to $\mathrm{ker}(\Phi)$; we need to show the surjectivity of the induced map $D_{2}\to$ $\mathrm{ker}(\Phi)$. Note first that any inner derivation in $D_{1}$ is of the form $[c,-]$ for some $c$ which centralises $E$. Note further that the centraliser $C_{A}(E)$ of $E$ in $A$ is canonically isomorphic to $\mathrm{Hom}_{E\otimes_{k}E^{\mathrm{op}}}(E,A)$ (via the map sending an $E$-$E$-bimodule homomorphism $\alpha:E\to$ $A$ to $\alpha(1)$). Since $E$ is separable, hence projective as an $E$-$E$-bimodule, it follows that the functor $\mathrm{Hom}_{E\otimes_{k}E^{\mathrm{op}}}(E,-)$ is exact. In particular, the surjection $A\to$ $A/J(A)^{2}$ induces a surjection $C_{A}(E)\to$ $C_{A/J(A)^{2}}(E)$, where we identify $E$ with its image in $A/J(A)^{2}$. Let $f\in$ $D_{1}$ such that the class of $f$ is in $\mathrm{ker}(\Phi)$, or equivalently, such that the induced derivation, denoted $\bar{f}$, on $A/J(A)^{2}$ is inner. Then there is $c\in$ $A$ such that $\bar{f}=$ $[\bar{c},-]$, where $\bar{c}=$ $c+J(A)^{2}$ centralises the image of $E$ in $A/J(A)^{2}$. By the above, we may choose $c$ such that $c$ centralises $E$ in $A$. Then the derivation $f-[c,-]$ represents the same class as $f$, still belongs to $D_{1}$, and induces the zero map on $A/J(A)^{2}$. Thus $f-[c,-]$ belongs in fact to $D_{2}$, proving (ii). The remaining statements are immediate consequences of (ii). ∎

The next result includes the special case of Theorem 1.1 where $\ell\!\ell(A)\leq 2$.

Proposition 2.9**.**

Let $A$ be a split finite-dimensional $k$-algebra such that $J(A)^{2}=$ $\{0\}$. Suppose that for every simple $A$-module $S$ we have $\mathrm{Ext}^{1}_{A}(S,S)=$ $\{0\}$ and that for any two simple $A$-modules $S$, $T$ we have $\mathrm{dim}_{k}(\mathrm{Ext}_{A}^{1}(S,T))\leq 1$. Let $E$ be a separable subalgebra of $A$ such that $A=$ $E\oplus J(A)$. The following hold

1. (i)

If $A$ is basic and if $f$, $g$ are derivations on $A$ which vanish on $E$, then $[f,g]=$ [math]. 2. (ii)

The Lie algebra $H\!H^{1}(A)$ is abelian. 3. (iii)

Let $e(A)$ be the number of edges in the quiver of $A$. We have

[TABLE]

Proof.

In order to prove (i), suppose that $A$ is basic. Let $I$ be a primitive decomposition of $1$ in $A$ such that $E=$ $\prod_{i\in I}ki$. Let $f$ and $g$ be derivations on $A$ which vanish on $E$. Then $f$, $g$ are determined by their restrictions to $J(A)$. By Lemma 2.6, the derivations $f$, $g$ preserve $J(A)$. By the assumptions, each summand $iAj$ in the vector space decomposition $A=$ $\oplus_{i,j\in I}\ iAj$ has dimension at most one. By Lemma 2.5, any derivation on $A$ which vanishes on $E$ preserves this decomposition. Therefore, if $X$ is a basis of $J(A)$ consisting of elements of the subspaces $iAj$, $i$, $j\in$ $I$, which are nonzero, then $f|_{J(A)}:J(A)\to$ $J(A)$ is represented by a diagonal matrix. Similary for $g$. But then the restrictions of $f$ and $g$ to $J(A)$ commute. Since both $f$, $g$ vanish on $E$ , this implies that $[f,g]=0$, whence (i). If $A$ is basic, then clearly (i) and Lemma 2.4 together imply (ii). Since the hypotheses of the Lemma as well as the Lie algebra $H\!H^{1}(A)$ are invariant under Morita equivalences, statement (ii) follows for general $A$. In order to prove (iii), assume again that $A$ is basic. By the assumptions, $e(A)=$ $\mathrm{dim}_{k}(J(A))=$ $|X|$. One verifies that the extension to $A$ by zero on $I$ of any linear map on $J(A)$ which preserves the summands $iAj$ (with $i\neq j$), or equivalently, which preserves the one-dimensional spaces $kx$, where $x\in$ $X$, is in fact a derivation. By Lemma 2.4, any class in $H\!H^{1}(A)$ is represented by such a derivation. Thus the space of derivations on $A$ which vanish on $I$ is equal to $\mathrm{dim}_{k}(J(A))=e(A)$. Each $i\in$ $I$ contributes an inner derivation. The only $k$-linear combination of elements in $I$ which belongs to $Z(A)$ are the multiples of $1=$ $\sum_{i\in I}i$. Thus the space of inner derivations which annihilate $I$ has dimension $\ell(A)-1$, whence the first equality. Since there are at most $\ell(A)-1$ arrows starting at any given vertex, it follows that $e(A)\leq(\ell(A)-1)\ell(A)$, whence the inequality as stated. ∎

The above Proposition can also be proved as a consequence of more general work of Strametz [10], calculating the Lie algebra $H\!H^{1}(A)$ for $A$ a split finite-dimensional monomial algebra.

3. Proofs of Theorems 1.1 and 1.2

Theorem 1.1 is a part of the following slightly more precise result. Let $k$ be a field.

Theorem 3.1**.**

Let $A$ be a split finite-dimensional $k$-algebra. Suppose that for every simple $A$-module $S$ we have $\mathrm{Ext}^{1}_{A}(S,S)=$ $\{0\}$ and that for any two simple $A$-modules $S$, $T$ we have $\mathrm{dim}_{k}(\mathrm{Ext}_{A}^{1}(S,T))\leq 1$. Set ${\mathcal{L}}=$ $H\!H^{1}(A)$, regarded as a Lie algebra.

1. (i)

If $\ell\!\ell(A)\leq 2$ then ${\mathcal{L}}$ is abelian. 2. (ii)

If $\ell\!\ell(A)>2$, then the derived Lie algebra ${\mathcal{L}}^{\prime}=$ $[{\mathcal{L}},{\mathcal{L}}]$ is nilpotent of nilpotency class at most $\ell\!\ell(A)-2$. The derived length of $\mathcal{L}$ is at most $\log_{2}(\ell\!\ell(A)-1)+1$.

In particular, ${\mathcal{L}}$ is solvable, and if $k$ is algebraically closed, then ${\mathcal{L}}$ is completely solvable.

Proof.

If $\ell\!\ell(A)\leq$ $2$, then $J(A)^{2}=\{0\}$, and hence (i) follows from Proposition 2.9. Suppose that $\ell\!\ell(A)>2$. We may assume that $A$ is basic. Note that $A$ and $A/J(A)^{2}$ have the same $\mathrm{Ext}$-quiver, and hence we may apply Proposition 2.9 to the algebra $A/J(A)^{2}$; in particular, $H\!H^{1}(A/J(A)^{2})$ is abelian. Thus the kernel of the canonical Lie algebra homomorphism ${\mathcal{L}}=H\!H^{1}(A)\to$ $H\!H^{1}(A/J(A)^{2})$ contains ${\mathcal{L}}^{\prime}$. Proposition 2.8 implies that ${\mathcal{L}}^{\prime}$ is contained in the image of $D_{2}$, hence nilpotent of nilpotency class at most $\ell\!\ell(A)-2$ by Proposition 2.7. From the same proposition we have that if $f\in\mathcal{L}^{(n)}$, then $f(J(A))\subseteq J(A)^{2^{n-1}+1}$ for $n\geq 1$. Therefore the derived length is at most $\log_{2}(\ell\!\ell(A)-1)+1$. Since ${\mathcal{L}}^{\prime}$ is nilpotent, it follows that if $k$ is algebraically closed, then ${\mathcal{L}}$ is completely solvable. ∎

Proof of Theorem 1.2.

By Lemma 2.6, every derivation $f:A\to$ $A$ preserves $J(A)$, and hence sends $J(A)^{2}$ to $J(A)^{2}$. Thus the canonical map $A\to$ $A/J(A)^{2}$ induces a Lie algebra homomorphism $\varphi:\mathrm{Der}(A)\to$ $\mathrm{Der}(A/J(A)^{2})$ which in turn induces a Lie algebra homomorphism $\Phi:H\!H^{1}(A)\to$ $H\!H^{1}(A/J(A)^{2})$. By Proposition 2.8, $\mathrm{ker}(\Phi)$ is a nilpotent ideal. If $\mathrm{char}(k)=2$, then $H\!H^{1}(A/J(A)^{2})$ is solvable by [10, Corollary 4.12], and hence $H\!H^{1}(A)$ is solvable. Suppose now that $H\!H^{1}(A)$ is not solvable. Then, by the above, we have $\mathrm{char}(k)\neq 2$. Then, by [10, Corollary 4.11, Remark 4.16], the Lie algebra $H\!H^{1}(A/J(A)^{2})$ is a finite direct product of copies of $\mathfrak{sl}_{2}(k)$. Thus $H\!H^{1}(A)/\mathrm{ker}(\Phi)$ is a subalgebra of a finite direct product of copies of $\mathfrak{sl}_{2}(k)$, and hence $H\!H^{1}(A)/\mathrm{rad}(H\!H^{1}(A))$ is a subquotient of a finite direct product of copies of $\mathfrak{sl}_{2}(k)$. Since any proper Lie subalgebra of $\mathfrak{sl}_{2}(k)$ is solvable, it follows easily that the semisimple Lie algebra $H\!H^{1}(A)/\mathrm{rad}(H\!H^{1}(A))$ is a finite direct product of copies of $\mathfrak{sl}_{2}(k)$. ∎

4. Schur functors and proof of Theorem 1.3

The hypothesis $J(iAi)^{2}=$ $iJ(A)^{2}i$ in the statement of Theorem 1.3 means that for any primitive idempotent $j$ not conjugate to $i$ in $A$ we have $iAjAi\subseteq$ $J(iAi)^{2}$; that is, the image in $iAi$ of any path parallel to the loop at $i$ which is different from that loop is contained in $J(iAi)^{2}$. We start by collecting some elementary observations which will be used in the proof of Theorem 1.3. Let $k$ be a field.

Lemma 4.1**.**

Let $A$ be a $k$-algebra and $e$ an idempotent in $A$. Let $f:A\to$ $A$ be a derivation. The following hold.

1. (i)

We have $f(AeA)\subseteq$ $AeA$. 2. (ii)

We have $ef(e)e=$ [math]. 3. (iii)

We have $(1-e)f(e)(1-e)=$ [math]. 4. (iv)

We have $f(e)\in$ $eA(1-e)\oplus(1-e)Ae$. 5. (v)

We have $f(e)=[[f(e),e)],e]$; equivalently, the derivation $f-[[f(e),e],-]$ vanishes at $e$. 6. (vi)

If $f(e)=0$, then for any $a\in$ $A$ we have $f(eae)=$ $ef(a)e$; in particular, $f(eAe)\subseteq$ $eAe$ and $f$ induces a derivation on $eAe$. 7. (vii)

If $f(e)=0$ and if $f$ is an inner derivation on $A$, then $f$ restricts to an inner derivation on $eAe$.

Proof.

Let $a$, $b\in$ $A$. Then $aeb=$ $aeeb$, hence $f(aeb)=$ $aef(eb)+f(ae)eb\in$ $AeA$, implying the first statement. We have $f(e)=$ $f(e^{2})=$ $f(e)e+ef(e)$. Right multiplication of this equation by $e$ yields $f(e)e=$ $f(e)e+ef(e)e$, whence the second statement. Right and left multiplication of the same equation by $1-e$ yields the third statement. Statement (iv) follows from combining the statements (ii) and (iii). We have $[[f(e),e],e]=$ $[f(e)e-ef(e),e]$. Using that $ef(e)e=0$ this is equal to $f(e)e+ef(e)=f(e)$, since $f$ is a derivation. This shows (v). Suppose that $f(e)=0$. Let $a\in$ $A$. Then $f(eae)=$ $f(e)ae+ef(a)e+eaf(e)=$ $ef(a)e$, whence (vi). If in addition $f=$ $[c,-]$ for some $c\in$ $A$, then the hypothesis $f(e)=0$ implies that $ec=ce$, and hence (vi) implies that the restriction of $f$ to $eAe$ is equal to the inner derivation $[ce,-]$. This completes the proof of the Lemma. ∎

Proposition 4.2**.**

Let $A$ be a $k$-algebra, and let $e$ be an idempotent in $A$. For any derivation $f$ on $A$ satisfying $f(e)=0$ denote by $\varphi(f)$ the derivation on $eAe$ sending $eae$ to $ef(a)e$, for all $a\in$ $A$. The correspondence $f\mapsto\varphi(f)$ induces a Lie algebra homormophism $HH^{1}(A)\to$ $H\!H^{1}(eAe)$. If $A$ is an algebra over a field of prime characteristic $p$, then this map is a homomorphism of $p$-restricted Lie algebras.

Proof.

Let $f$ be an arbitrary derivation on $A$. By Lemma 4.1 (v), the derivation $f-[[f(e),e],-]$ vanishes at $e$. Thus every class in $H\!H^{1}(A)$ has a representative in $\mathrm{Der}(A)$ which vanishes at $e$. By Lemma 4.1 (vi), any derivation on $A$ which vanishes at $e$ restricts to a derivation on $eAe$, and by Lemma 4.1 (vii), this restriction sends inner derivations on $A$ to inner derivations on $eAe$, hence induces a map $H\!H^{1}(A)\to$ $H\!H^{1}(eAe)$. A trivial verification shows that if $f$, $g$ are two derivations on $A$ which vanish at $e$, the so does $[f,g]$, and an easy calculation shows that therefore the above map $H\!H^{1}(A)\to$ $H\!H^{1}(eAe)$ is a Lie algebra homomorphism. If $A$ is an algebra over a field of characteristic $p>0$, and if $f$ is a derivation on $A$ which vanishes at $e$, then the derivation $f^{p}$ vanishes on $e$ and the restriction to $eAe$ commutes with taking $p$-th powers by Lemma 4.1 (vi). This shows the last statement. ∎

We call the Lie algebra homomorphism $H\!H^{1}(A)\to$ $H\!H^{1}(eAe)$ in Proposition 4.2 the canonical Lie algebra homomorphism induced by the Schur functor given by multiplication with the idempotent $e$.

For $A$ a finite-dimensional $k$-algebra and $m$ a positive integer, denote by $H\!H^{1}_{(m)}(A)$ the subspace of $H\!H^{1}(A)$ of classes which have a a representative $f\in$ $\mathrm{Der}(A)$ satisfying $f(J(A))\subseteq$ $J(A)^{m}$.

Proposition 4.3**.**

Let $A$ be a split finite-dimensional $k$-algebra. Let $i$ be a primitive idempotent in $A$. Set $S=$ $Ai/J(A)i$. Suppose that $\mathrm{Ext}^{1}_{A}(S,S)=$ $\{0\}$. Then image of the canonical map $H\!H^{1}(A)\to$ $H\!H^{1}(iAi)$ is contained in $H\!H^{1}_{(1)}(iAi)$.

Proof.

By Lemma 2.3 we have $iJ(A)i=$ $iJ(A)^{2}i$. By Lemma 4.1 (v), any class in $H\!H^{1}(A)$ is represented by a derivation $f$ satisfying $f(i)=0$. Thus if $a\in$ $J(A)$, then $iai=$ $ibci$ for some $b$, $c\in$ $J(A)$, and hence $f(iai)=$ $if(b)ci+ibf(c)i\in$ $iJ(A)i$. ∎

Proposition 4.4**.**

Let $A$ be a split symmetric $k$-algebra. Let $i$ be a primitive idempotent in $A$. Set $S=$ $Ai/J(A)i$. Suppose that $\mathrm{Ext}^{1}_{A}(S,S)\neq$ $\{0\}$. Then the canonical Lie algebra homomorphism $H\!H^{1}(A)\to$ $H\!H^{1}(iAi)$ is nonzero.

Proof.

Set $S^{\vee}=$ $iA/iJ(A)$. Choose a maximal semisimple subalgebra $E$ of $A$. Since $\mathrm{Ext}^{1}_{A}(S,S)$ is nonzero, it follows from Lemma 2.2 that $J(A)/J(A)^{2}$ has a direct summand isomorphic to $S\otimes_{k}S^{\vee}$ as an $A$-$A$-bimodule. Since $A$ is symmetric, we have $\mathrm{soc}(A)\cong$ $A/J(A)$, and hence $\mathrm{soc}(A)$ has a bimodule summand isomorphic to $S\otimes_{k}S^{\vee}$. Thus there is a bimodule homomorphism $J(A)/J(A)^{2}\to$ $\mathrm{soc}(A)$ with image isomorphic to $S\otimes_{k}S^{\vee}$. Composing with the canonical map $J(A)\to$ $J(A)/J(A)^{2}$ yields a bimodule homomorphism $f:J(A)\to$ $\mathrm{soc}(A)$ with kernel containing $J(A)^{2}$ and with image isomorphic to $S\otimes_{k}S^{\vee}$. Extending $f$ by zero on $E$ yields a derivation $\hat{f}$ on $A$, by Lemma 2.4. Restricting $\hat{f}$ to $iJ(A)i$ sends $iJ(A)i$ to a nonzero subspace of $\mathrm{soc}(A)$ isomorphic to $iS\otimes_{k}S^{\vee}i$, hence onto $\mathrm{soc}(iAi)$. Thus the image of $\hat{f}$ under the canonical map $\mathrm{Der}(A)\to$ $\mathrm{Der}(iAi)$ from Proposition 4.2 is a nonzero derivation with kernel containing $ki+J(iAi)^{2}$ and image in $\mathrm{soc}(iAi)$. By [2, Corollary 3.2], the class in $H\!H^{1}(iAi)$ of this derivation is nonzero, whence the result. ∎

Proposition 4.5**.**

Let $p$ be an odd prime and suppose that $k$ is algebraically closed of characteristic $p$. Set $W=$ $\mathrm{Der}(k[x]/(x^{p}))$. For $-1\leq i\leq p-2$ let $f_{i}$ be the derivation of $k[x]/(x^{p})$ sending $x$ to $x^{i+1}$, where we identify $x$ with its image in $k[x]/(x^{p})$. Let $L$ be a simple Lie subalgebra of $W$. Then either $L=W$, or $L\cong$ $\mathfrak{sl}_{2}(k)$.

Proof.

Note that the subalgebra $S$ of $W$ spanned by the $f_{i}$ with $0\leq i\leq p-2$ is solvable. Thus $L$ is not contained in $S$. Note further that $\mathrm{dim}_{k}(L)\geq 3$. Therefore there exist derivations

[TABLE]

[TABLE]

belonging to $L$ with $\lambda_{-1}=1$, and $\mu_{t}=1$, where $t$ is an integer such that $0\leq t\leq p-2$. Choose $g$ such that $t$ is minimal with this property. But then $[f,g]$ belongs to $L$. Since $[f_{-1},f_{t}]=$ $(t+1)f_{t-1}$, the minimality of $t\geq 0$ forces $t=0$; that is we have

[TABLE]

and $\mu_{0}=1$. Since $\mathrm{dim}_{k}(L)\geq 3$, it follows that there is a third element $h$ in $L$ not in the span of $f$, $g$, and hence, after modifying $h$ by a linear combination of $f$ and $g$, we can choose $h$ such that

[TABLE]

for some $s$ such that $1\leq s\leq p-2$ and $\nu_{s}=1$. Choose $h$ such that $s$ is minimal with this property. Again by considering $[f,h]$, one sees that the minimality of $s$ forces $s=1$. If $L$ is $3$-dimensional, then $L\cong$ $\mathfrak{sl}_{2}(k)$, where we use that $k$ is algebraically closed. If $\mathrm{dim}_{k}(L)\geq 4$, then $L$ contains an element of the form

[TABLE]

with $2\leq r\leq p-2$ and $\tau_{r}=1$. But then applying $[f,-]$ and $[h,-]$ repeatedly to $u$ shows that $L$ contains a basis of $W$, hence $L=W$. ∎

Remark 4.6**.**

Note that if $\mathrm{char}(k)=p>2$, then the Witt Lie algebra $W$ contains indeed a subalgebra isomorphic to $\mathfrak{sl}_{2}(k)$. Let $\mathfrak{f},\mathfrak{e},\mathfrak{h}$ be elements of the basis of $\mathfrak{sl}_{2}(k)$ such that $[\mathfrak{e},\mathfrak{f}]=\mathfrak{h}$, $[\mathfrak{h},\mathfrak{f}]=-2\mathfrak{f}$, and $[\mathfrak{h},\mathfrak{e}]=2\mathfrak{e}$. Then we have a Lie algebra isomorphism $\mathfrak{sl}_{2}(k)\cong$ $\langle f_{-1},f_{0},f_{1}\rangle$ sending $\mathfrak{f}$ to $f_{-1}$, $\mathfrak{h}$ to $2f_{0}$, and $\mathfrak{e}$ to $-f_{1}$.

Proof of Theorem 1.3.

We use the notation and hypotheses of the notation in Theorem 1.3, and we assume that the Lie algebra $H\!H^{1}(A)$ is simple. We show that this forces $H\!H^{1}(A)$ to be a Lie subalgebra of the Witt Lie algebra $W$ with $\mathrm{char}(k)=p>2$, and then the result follows from Proposition 4.5.

Since $H\!H^{1}(A)$ is simple and since $\mathrm{Ext}_{A}^{1}(S,S)$ is nonzero, it follows from Proposition 4.4 that the canonical Lie algebra homomorphism $\Phi:H\!H^{1}(A)\to H\!H^{1}(iAi)$ from Proposition 4.2 is injective. By the assumptions, $iAi$ is a local algebra whose quiver has only one loop. Therefore $A\cong k[x]/(v)$ for some polynomial $v\in k[x]$ of degree at least $1$. Since $k$ is algebraically closed, $v$ is a product of powers of linear polynomials, say $\prod_{i}(x-\beta_{i})^{n_{i}}$, with pairwise distinct $\beta_{i}$ and positive integers $n_{i}$. Therefore $H\!H^{1}(iAi)$ is a direct product of the Lie algebras corresponding to these factors. It follows that $H\!H^{1}(A)$ is isomorphic to a Lie subalgebra of $H\!H^{1}(k[x]/((x-\beta)^{n}))$ for some positive integer $n$. After applying the automorphism $x\mapsto x+\beta$ of $k[x]$ we have that $H\!H^{1}(A)$ is isomorphic to a Lie subalgebra of $H\!H^{1}(k[x]/(x^{n}))$ for some positive integer $n$. If $\mathrm{char}(k)=p$ does not divide $n$ or if $\mathrm{char}(k)=0$, then the linear map sending $x$ to $1$ is not a derivation on $k[x]/(x^{n})$, and therefore $H\!H^{1}(k[x]/(x^{n}))$ is solvable in that case. Since Lie subalgebras of solvable Lie algebras are solvable, this contradicts the fact that $H\!H^{1}(A)$ is simple. Thus we have $\mathrm{char}(k)=p>0$ and $n=$ $pm$ for some positive integer $m$. Since $\mathrm{char}(k)=p$, it follows that the canonical surjection $k[x]/(x^{n})\to$ $k[x]/(x^{p})$ induces a Lie algebra homomorphism $H\!H^{1}(k[x]/(x^{n}))\to$ $W=$ $H\!H^{1}(k[x]/(x^{p}))$ with a nilpotent kernel. Thus $H\!H^{1}(A)$ is not containd in that kernel, and hence $H\!H^{1}(A)$ is isomorphic to a Lie subalgebra of $W$. The result follows. ∎

To conclude this section we note that although it is not clear which simple Lie algebras might occur as $H\!H^{1}(A)$ when $\mathrm{Ext}^{1}_{A}(S,S)=$ $\{0\}$ for all simple $A$-modules $S$, it easy to show that $H\!H^{\ast}(A)$ is not a simple graded Lie algebra (with respect to the Gerstenhaber bracket).

Proposition 4.7**.**

Let $A$ be a finite dimensional $k$-algebra, and assume that for every simple $A$-module $S$ we have $\mathrm{Ext}^{1}_{A}(S,S)=\{0\}$. Then $H\!H^{\ast}(A)$ is not a perfect graded Lie algebra. In particular, $H\!H^{\ast}$ is not simple.

Proof.

If $f\in C^{1}(A,A):=\mathrm{Hom}_{k}(A,A)$ and if $g\in C^{0}(A,A):=\mathrm{Hom}_{k}(k,A)$, then the Gerstenhaber bracket is given by $[f,g]=f(g)$, i.e. simply evaluating $f$ in $g$. Note that $1\in Z(A)=HH^{0}(A)$. By Lemma 2.4 and Lemma 2.6, $f$ preserves $J(A)$ and we may assume $E\subseteq\mathrm{ker}(f)$. Therefore the derived Lie subalgebra of $H\!H^{\ast}(A)$ does not contain $1_{A}$. ∎

Remark 4.8**.**

Lemma 4.1 and Proposition 4.2 hold for algebras over an arbitrary commutative ring instead of $k$.

5. Examples

Theorem 1.1 applies to certain blocks of symmetric groups.

Proposition 5.1**.**

Let $k$ be a field of prime characteristic $p$. Let $A$ be a defect $2$ block of a symmetric group algebra $kS_{n}$ or the principal block of $kS_{3p}$. Then $H\!H^{1}(A)$ is a solvable Lie algebra.

Proof.

From [9, Theorem 1] and from [7, Theorem 5.1] we have that the simple modules do not self-extend and the $\mathrm{Ext}^{1}$-space between two simple modules is at most one-dimensional. The statement follows from Theorem 1.1. ∎

Remark 5.2**.**

A conjecture by Kleshchev and Martin predicts that simple $kS_{n}$-modules in odd characteristic do not admit self-extensions.

Proposition 5.3**.**

Let $A$ be a tame symmetric algebra over a field $k$ with $3$ isomorphism classes of simple modules of type $3\mathcal{A}$ or $3\mathcal{K}$. Then $\mathrm{HH}^{1}(A)$ is a solvable Lie algebra.

Proof.

From the list at the end of Erdmann’s book [5] we have that the simple modules in these cases do not self-extend and that the $\mathrm{Ext}^{1}$-space between two simple modules is at most one-dimensional. The statement follows from Theorem 1.1. ∎

As mentioned in the introduction, the above Proposition is part of more general results on tame algebras in [4] and [8]. We note some other examples of algebras whose simple modules do not have nontrivial self-extensions.

Theorem 5.4** ([1, Theorem 3.4]).**

Let $G$ be a connected semisimple algebraic group defined and split over the field $\mathbb{F}_{p}$ with $p$ elements, and $k$ be an algebraic closure of $\mathbb{F}_{p}$. Assume $G$ is almost simple and simply connected and let $G(\mathbb{F}_{q})$ be the finite Chevalley group consisting of $\mathbb{F}_{q}$-rational points of G where $q=p^{r}$ for a non-negative integer $r$. Let $h$ be the Coxeter number of $G$. For $r\geq 2$ and $p\geq 3(h-1)$, we have $\mathrm{Ext}^{1}_{kG(\mathbb{F}_{q})}(S,S)=\{0\}$ for every simple $kG(\mathbb{F}_{q})$-module $S$.

Remark 5.5**.**

Let $G$ be a simple algebraic group over a field of characteristic $p>3$, not of type $A_{1},G_{2}$ and $F_{4}$. Proposition 1.4 in [11] implies that not having self-extensions does not allow to lift to characteristic zero certain simple modular representations. Therefore, for these cases the Lie structure of $H\!H^{1}$ plays a central role.

In the context of blocks with abelian defect groups one expects (by Broué’s abelian defect conjecture) every block of a finite group algebra with an abelian defect group $P$ to be derived equivalent to a twisted group algebra of the form $k_{\alpha}(P\rtimes E)$, where $E$ is the inertial quotient of the block and where $\alpha$ is a class in $H^{2}(E;k^{\times})$, inflated to $P\rtimes E$ via the canonical surjection $P\rtimes E\to$ $E$. Thus the following observation is relevant in cases where Broué’s abelian defect conjecture is known to hold (this includes blocks with cyclic and Klein four defect).

Proposition 5.6**.**

Let $k$ be a field of prime characteristic $p$. Let $P$ be a finite $p$-group and $E$ an abelian $p^{\prime}$-subgroup of $\mathrm{Aut}(P)$ such that $[P,E]=P$. Set $A=$ $k(P\rtimes E)$. Suppose that $k$ is large enough for $E$, or equivalently, that $A$ is split. For any simple $A$-module $S$ we have $\mathrm{Ext}^{1}_{A}(S,S)=$ $\{0\}$.

Proof.

Since $E$ is abelian, it follows that $\mathrm{dim}_{k}(S)=1$, and hence that $S\otimes_{k}-$ is a Morita equivalence. This Morita equivalence sends the trivial $A$-module $k$ to $S$, hence induces an isomorphism $\mathrm{Ext}_{A}^{1}(k,k)\cong$ $\mathrm{Ext}_{A}^{1}(S,S)$. It suffices therefore to show the statement for $k$ instead of $S$. That is, we need to show that $H^{1}(P\rtimes E;k)=$ $\{0\}$, or equivalently, that there is no nonzero group homomorphism from $P\rtimes E$ to the additive group $k$. Since $[P,E]=P$, it follows that every abelian quotient of $P\rtimes E$ is isomorphic to a quotient of $E$, hence has order prime to $p$. The result follows. ∎

Example 5.7**.**

If $B$ is a block of a finite group algebra over an algebraically closed field $k$ of characteristic $p>0$ with a nontrivial cyclic defect group $P$ and nontrivial inertial quotient $E$, then $H\!H^{1}(B)$ is a solvable Lie algebra, isomorphic to $H\!H^{1}(kP)^{E}$, where $E$ acts on $H\!H^{1}(kP)$ via the group action of $E$ on $P$. Indeed, $B$ is derived equivalent to the Nakayama algebra $k(P\rtimes E)$, which satisfies the hypotheses of Theorem 1.1 (thanks to the assumption $E\neq$ $1$, which implies $[P,E]=P$). Note that $kP$ is isomorphic to the truncated polynomial algebra $k[x]/(x^{p^{d}})$, where $p^{d}=$ $|P|$.

Acknowledgements. The second author has been supported by the projects, “Oberwolfach Leibniz Fellows”, DAAD Short-Term Grant (57378443) and 19880/GERM/15. He also thanks the Department of Mathematics of the University of Leicester.