# Periods in the trivial extension algebra of the incidence algebra of the divisor lattice

Definition of $C_L$ for people who like number theory:

Let $m$ be a number with prime factorisation $m=p_1^{n_1} ... p_r^{n_r}$ with $n_i>0$. Define $I_m$ to be the incidence algebra of the divisor lattice of $m$. Up to isomorphism this just depence on the $n_i$ and not on the primes so lets define with $L=[n_1+1,...,n_r+1]$, $I_L$ as the incidence algebra of the divisor lattice of a number $m$ with prime factorisation $m=p_1^{n_1} ... p_r^{n_r}$. Define $C_L$ as the trivial extension algebra of $I_L$. For the definition of trivial extension algebra see for example https://math.stackexchange.com/questions/229412/trivial-extension-of-an-algebra (together with a proof that this is always a Frobenius algebras) or the textbook on Frobenius algebras I by Skowronski and Yamagata.

Definition of $C_L$ for people who like Nakayama algebras:

For a natural number $n \geq 2$ denote by $A_n$ the hereditary Nakayama algebra with dimension $n$ given by quiver and relations over a field $K$ (that is the Nakayama algebra with Kupisch series $[n,n-1,...,2,1]$). For a list of integers $L=[n_1,...,n_r]$. Define the algebra $C_L$ as the trivial extension of the algebra $A_{n_1} \otimes_K \cdots \otimes_K A_{n_r}$.

Questions:

Is $C_L$ isomorphic to a known/studied algebra from the literature? Experiments with small lists suggest that the algebra is always periodic or at least all simple modules are periodic with the same period. (Recall that a module $M$ is called periodic with period $n$ when $n$ is the smallest integer such that $\Omega^n(M) \cong M$. An algebra is called periodic in case the regular module is periodic as a bimodule.) The period seems to behave in a strange way. What could it be (in case such a finite periodic exists)?

Here some examples, where after the list $L$ the number is the period of a simple module of $C_L$. Sadly even with a computer it takes long to do such calculations.

$L=[t]: 2t$ for $t \geq 1$ (this is well known)

$L=[2,2]: 5$

$L=[3,2]: 22$

$L=[4,2]:29$

$L=[5,2]: 12$ (this was surprising)

$L=[6,2]: 43$

$L=[4,3]:42$

$L=[3,3]: 8$

$L=[2,2,2]: 6$

$L=[3,2,2]: 26$

Guesses what the period might be for general $L$ are also welcome. I should say that it is not too good tested yet as even for those small lists it takes several minutes to do the calculations with a computer.

More general, one might ask what those periods are (in case they exist) for the trivial extension algebra of tensor products of hereditary path algebras of Dynkin type. For example the trivial extension of $k D_4 \otimes_K k A_2$ has the property that all simple modules have period equal to 6.

• I predict $L=[7,2]:50$ and $L=[8,2]:19$. – Jeremy Rickard Mar 14 '18 at 12:04
• @JeremyRickard $[7:2]$ is correct. Ill look at $[8:2]$ now. – Mare Mar 14 '18 at 13:34
• @JeremyRickard Your prediction for $[8,2]$ was also correct (and took the computer about an hour). Do you want to tell us your secret? – Mare Mar 14 '18 at 14:36
• I think I can show that for $[n,2]$ you get $21k+1$ if $n=3k$, $21k+8$ if $n=3k+1$, and $7k+5$ if $n=3k+2$. Also, for $[n^t]=[n,\dots,n]$ you get $(n-1)(t+1)+2$. For general $L$ I don’t know if there’s a nice explicit formula, but when I have time (probably later today) I’ll describe the method I used to get these results (by hand, no computer). – Jeremy Rickard Mar 14 '18 at 14:41
• It is very surprising to me that such complicated (and not very regular) numbers appear for this seemingly innocent problem. So partial results are very welcome. I also try to look more closely into this problem now for arbitrary hereditary algebras of Dynkin type. Luckily, we can assume that the quivers are always linear oriented (or any other orientation that might simplify things) when trying to prove periodicity of the algebras thanks to your old result that derived equivalent algebras have derived equivalent trivial extensions. – Mare Mar 14 '18 at 15:20

Calculating periods of simple modules for the trivial extension algebra $TA$ can be reduced to a calculation with $A$-modules (at least if $A$ has finite global dimension), which is much easier.

The starting point is a result of Happel, that for a finite-dimensional algebra $A$ with finite global dimension, the bounded derived category $D^b(\text{mod }A)$ is equivalent as a triangulated category to the stable module category $\underline{{\text{mod}}}\hat{A}$ of the repetitive algebra $\hat{A}$ of $A$, or equivalently the stable category $\underline{\text{mod}}_{\mathbb{Z} }TA$, of $\mathbb{Z}$-graded modules for the trivial extension algebra $TA=A\ltimes DA$, regarded as a graded algebra with $A$ in degree zero and $DA$ in degree one.

This equivalence sends an $A$-module $M$, regarded as an object of $D^b(\text{mod }A)$ concentrated in degree zero, to the graded $TA$-module with $M$ as degree zero component, and other degree components zero, with $DA$ acting as zero.

The shift functor in $D^b(\text{mod }A)$ is the usual "shift to the left" of complexes, and the shift functor in $\underline{\text{mod}}_{\mathbb{Z} }TA$ is the cosyzygy functor $N\mapsto\Omega^{-1} N$, which commutes with "forgetting the grading".

The "shift grading" functor $N\mapsto N(1)$ for $\underline{\text{mod}}_{\mathbb{Z} }TA$ corresponds to the functor $M\mapsto FM:=\mathbf{L}\nu M[1]$ for $D^b(\text{mod }A)$, where $\mathbf{L}\nu$ is the derived Nakayama functor (take a projective resolution and apply $-\otimes_ADA$).

So to calculate the period of an $A$-module $S$, considered as a $TA$ module, we can repeatedly apply $F$ to $S$, considered as an object of $D^b(\text{mod }A)$, until the first time we get a shift $S[p]$ of $S$, and then $p$ is the period of $S$.

Now let's apply this to the algebra $I_L=A_{n_1}\otimes_K\dots\otimes_K A_{n_r}$.

For the hereditary algebra $A_n$, let $X(n,1),X(n,2),\dots,X(n,n)$ be the simple modules, in the natural order where $X(n,1)$ is the injective simple and $X(n,n)$ is the projective simple, and let $X(n,0)$ be the $n$-dimensional projective-injective module.

The derived Nakayama functor for $A_n$ satisfies $\mathbf{L}\nu X(n,i)=X(n,i+1)[1]$ for $0<i<n$, $\mathbf{L}\nu X(n,0)=X(n,1)$, and $\mathbf{L}\nu X(n,n)=X(n,0)$.

Consider the $I_L$-modules $$X(i_1,\dots,i_r)=X(n_1,i_1)\otimes_K\dots\otimes_K X(n_r,i_r)$$ for $0\leq i_j\leq n_j$. These include all the simple modules.

Then the functor $M\mapsto FM=\mathbf{L}\nu M[1]$ described above satisfies $$FX(i_1,\dots,i_r)=X(i_1+1,\dots,i_r+1)[r+1-s],$$ where $s$ is the number of $i_j$ with $i_j\in\{0,n_j\}$ and where the $i_j$ are considered modulo $n_j+1$.

Iterating $F$, the first $l>0$ for which $F^lX(i_1,\dots,i_r)$ is a shift of $X(i_1,\dots,i_r)$ is $\operatorname{lcm}(n_1+1,\dots,n_r+1)$, and for each $j$, two in every $n_j+1$ of the modules in the $F$-orbit of $X(i_1,\dots,i_r)$ have $i_j\in\{0,n_j\}$, so the total shift of $F^lX(i_1,\dots,i_r)$, and hence the period of the corresponding module for the trivial extension algebra, is $$\operatorname{lcm}(n_1+1,\dots,n_r+1)\left(r+1-2\sum_{j=1}^r\frac{1}{n_j+1}\right).$$

Note that this only calculates the periods of simple modules (and shows that they all have the same period). Probably the trivial extension algebra itself has the same period, but I don't see how to prove that using this method.

• This looks beautiful. Thank you. I will look at it in detail until next week and then accept the answer. – Mare Mar 16 '18 at 20:20
• @Mare I've tidied this up, and added a couple of remarks. I've deleted the lengthy calculation for $L=[2,3]$ since this is now covered by the general case, although of course it's still available in the edit history if anybody's interested. – Jeremy Rickard Mar 17 '18 at 10:05