heller functor of finite order

Question

Hi, let A be a finite dimensional selfinjective algebra. Assume mod A is periodic.That is for every finite dimensional module with no projective summand we have: $ \Omega^{n} (M) =M $ for some n ,where $ \Omega $ is the Hellerfunctor(giving the kernel of a projective cover of M ). When is it true that $ \Omega $ is of finite order,that is $ \Omega^{n} $ is isomorphic as a functor to the identity functor of the stable category of modules ?(Assume there is a common biggest period for all module if necassary)

Thanks for help

Answer 1

To the best of my knowledge, this is an open problem. There are no known examples where $\mbox{mod-}A$ is periodic (say, with a common period $n$ for all modules), but $\Omega^n$ is not isomorphic to the identity functor on the stable category. In fact, $\mbox{mod-}A$ being periodic may well imply that $A$ is a periodic algebra, meaning that $A$ has a periodic projective resolution as a module over its enveloping algebra. This is another open question. Examples where this happens include the preprojective algebras of Dynkin type and generalized Dynkin type, and their deformations [K. Erdmann, A. Skowronski. Trans. Amer. Math. Soc. 359 (2007), no. 6, 2625--2650], and all self-injective algebras of finite representation type [J. Pure Appl. Algebra 214 (2010), no. 6, 990–1000].

One class of examples where $\mbox{mod-}A$ is periodic, but it is not known (at least to me) whether $\Omega$ has finite order or not, is the so-called higher preprojective algebras studied by O. Iyama and S. Oppermann [arXiv:0912.3412v1]. Here, I believe the problem is connected to the question of whether the Nakayama automorphism has finite order.

Answer 2

Another example: If $A$ is a bialgebra over a field $k$ then $\Omega^n(k) = k$ implies $\Omega^n = id$.

The proof is rather simple: Choose a projective resolution $P \to k$ such that $\Omega^i(k)$ is the kernel of $P_{i-1} \to P_{i-2}$. Since $M$ is $k$-projective, $$P \otimes_k M\to k \otimes_k M \cong M$$ is a projective resolution of $M$ where $A$ acts diagonally on the tensor product (that's the place were the bialgebra structure is needed). If $f: M \to N$ is $A$-linear, then we obtain a commutative diagramm with exact rows $$\begin{array}{cccccccc} 0 & \to & M & \cong & k \otimes M & \to & P_{n-1} \otimes M & \to P_{n-2} \otimes M & \to \cdots \newline & & \scriptstyle f\; \displaystyle \downarrow & & \scriptstyle id \otimes f\;\displaystyle\downarrow & & \scriptstyle id \otimes f\; \displaystyle\downarrow & \scriptstyle id \otimes f\; \displaystyle \downarrow\newline 0 & \to & N & \cong & k \otimes N & \to & P_{n-1} \otimes N & \to P_{n-2} \otimes N & \to \cdots \newline \end{array}$$ Since up to a projective summand $\Omega^n(M)$ is just the kernel in a projective resolution, we have $$M = \Omega^n(M) \oplus(\text{projective}) \text{ and }f = \Omega^n(f) \oplus p,$$ i.e. $\Omega^n(M) = M$ and $\Omega^n(f) = f$ in the stable category. Thus $\Omega^n = id$.

Note: If $k$ is a commutative ring and $A$ projective as $k$-module then the same proof shows $\Omega^n = id$ on the stable category of finitely generated, $k$-projective $A$-modules.

Answer 3

I can show that the claim is true if $A$ is defined over $\mathbb F_{p^\infty}$, the algebraic closure of the field with $p$ elements. This proof will unfortunately not generalize to arbitrary algebraically closed fields (not even such of characteristic $p$), nor will it bound the exponent $n$ such that $\Omega^n \cong \rm id$ in any useful way. It does however show that it suffices to know that each simple module is periodic in order to conclude that $\Omega^n\cong \rm id$ for some $n$ (again, of course, only for algebras over $\mathbb F_{p^\infty}$).

Fix $n\in\mathbb N$ such that $\Omega^{n}(S) \cong S$ for any simple $A$-module $S$.

All but the last paragraph is the same as Theorem 2.5 in K. Erdmann and A. Skowronski. Periodic algebras. Trends in Representation Theory and Related Topics. European Math. Soc., Zurich, 2008. (you can view the relevant parts of this on google books).

First choose a projective cover $P$ of $_A A_A$ (i.e., $P$ is a projective $A$-$A$-bimodule). Then choose an $A$-$A$-bimodule $X$ as the kernel of the epimorphism from $P$ to $A$, i.e. we get a s.e.q. of $A$-$A$-bimdoules $$ 0 \longrightarrow X \longrightarrow P \longrightarrow A \longrightarrow 0 $$ As a sequence of left or right $A$-modules this sequence is split, hence $X$ is projective as a left and as a right $A$-module. This implies that tensoring the above sequence with any (left or right) $A$-module will again yield an exact sequence of left $A$-modules, with projective middle term (since $P\otimes_A M$ is projective for any $A$-module $M$). This shows that $-\otimes_A X$ is isomorphic to $\Omega(-)$ on the stable module category. Now $-\otimes_AX$ is a stable auto-equivalence of Morita-type, and so is $-\otimes_A X^{\otimes n}$. The latter sends simple modules to themselves, and therefore lifts to a Morita auto-equivalence, w.l.o.g. induced by the $A$-$A$-bimodule $Y$ (the fact that stable equivalences of Morita type which send simple modules to simple modules lift to Morita equivalences is usually attributed to Linckelmann). That is, $-\otimes_A Y \cong \Omega^n(-)$ on the stable module category. In particular $S\otimes_A Y \cong S$ for all simple $A$-modules $S$. But any Morita-autoequivalence which sends simple modules to themselves is induced by an automorphism $\alpha$ of $A$, i.e. $Y$ is isomorphic to the twisted $A$-$A$-bimodule $_{id} A_{\alpha}$.

Now comes the ugly part which doesn't work over arbitrary fields: $\alpha: A \longrightarrow A$ must have finite order, since every non-zero element in $\mathbb F_{p^\infty}$ is a root of unity and therefore every invertible matrix over $\mathbb F_{p^\infty}$ has finite order ( ${\rm Aut}(A) \leq {\rm GL}(A)$, and ${\rm GL}(A)$ is a torsion group). Hence there is some $m\in \mathbb N$ such that $Y^{\otimes m} \cong {_AA_A}$. But then $$\Omega^{m\cdot n}(-)\cong -\otimes_A Y^{\otimes m} \cong -\otimes_A {_{id} A _{\alpha^m}} \cong \rm id$$