# Combinatorial inequality for dominant dimension

In the following I present a conjecture on Nakayama algebras that I have for nearly 2 years now. Since I was not able to solve it and it can be stated purely combinatorically, I thought it might be worth a try to post it on mathoverflow. A proof of conjecture 1 would have very nice consequences and applications that I present at the end. I will first present everything in elementary terms so that no knowledge of representation theory or homological algebra is needed.

Combinatorial definitions

An $$n$$-CNakayama algebra is a list of $$n$$ natural numbers $$[c_0,...,c_{n-1}]$$ (where we read the indices modulo $$n$$ so that $$c_i$$ is defined for all $$i \in \mathbb{Z}$$) satisfying $$c_i \geq 2$$, $$c_{n-1}-1=c_0=\min \{ c_0,c_1,...,c_{n-1} \}$$ and $$c_i -1 \leq c_{i+1}$$ for all $$i$$.

See also https://arxiv.org/pdf/1811.05846.pdf where CNakayama algebras are identified with periodic Dyck paths (two $$n$$-CNakayama algebras are identified in case their lists are cyclic shifts of eachother).

The defect $$\operatorname{Def(A)}$$ of an n-Cnakayama algebra A is by definition the cardinality of the set $$\{ i \in \{0,...,n-1 \} | c_{i-1} > c_i \}$$. A module is a tuple $$(i,k)$$ with $$i \in \mathbb{Z}$$ and $$k \in \{1,...,c_i \}$$ and we identify $$(i_1,k_1)$$ with $$(i_2,k_2)$$ in case $$i_1 \equiv i_2 \ mod \ n$$ and $$k_1=k_2$$. Formally, we also introduce modules $$(i,k)$$ with $$k \leq 0$$ that we call zero-modules and they are all identified. The index $$i$$ is called the top of a module.

A module $$(i,k)$$ is called projective in case $$k=c_i$$ and it is called projective-injective in case it is projective and additionally $$c_{i-1} \leq c_i$$.

Note that the defect simply counts how many projective modules there are that are not injective.

The first syzygy of a module $$M=(i,k)$$ is defined as $$\Omega^1(M):=(i+k,c_i-k)$$. The $$i$$-th syzygy module $$\Omega^i(M)$$ of $$M$$ is then defined inductively as $$\Omega^i(M):=\Omega^{1}(\Omega^{i-1}(M))$$. Set $$\Omega^0(M):=M$$. The i-th projective cover of a module $$M$$ is defined as the module $$(l,c_l)$$ when $$l$$ is the top of the module $$\Omega^l(M)$$. The dominant dimension $$\operatorname{domdim(M)}$$ of a module $$M$$ is defined as the smallest integer $$i$$ such that the $$i$$-th projective cover of $$M$$ is not projective-injective. By convention, the dominant dimension of a projective-injective modules and zero modules is infinite.

A module $$(i,k)$$ is called special in case $$c_{i-1} < c_i$$ and $$k \in \{c_{i-1},c_{i-1} +1 , ..., c_i -1 \}$$.

The dominant dimension $$\operatorname{domdim(A)}$$ of an $$n$$-CNakayama algeba A is defined as the minimum of all dominant dimensions of the special modules.

Conjecture 1: For any $$n$$-CNakayama algebra $$A$$, we have the inequality $$\operatorname{Def(A)} \cdot \operatorname{domdim(A)} \leq 2n-2.$$

Conjecture 2: In case an $$n$$-Cnakayama algebra $$A$$ has dominant dimension larger than or equal to $$n$$, we have $$\operatorname{Def(A)}=1$$.

Note that conjecture 1 implies conjecture 2 since the inequality gives $$\operatorname{domdim(A)} \leq \frac{2n-2}{\operatorname{Def(A)}} \leq n-1$$ in case $$\operatorname{Def(A)} \geq 2$$. Also note that an $$n$$-CNakayama algebra has $$\operatorname{Def(A)}=1$$ if and only if it is of the form $$[a,a,....a,a+1,...,a+1]$$.

$$\mathbf{Examples}$$

a) We look at the 6-CNakayama algebra [5,5,5,7,7,6]. It has defect equal to 2. The special modules are $$(3,6)$$ with dominant dimension 4 and $$(3,5)$$ with dominant dimension 4. Thus $$A$$ has dominant dimension 4 and the inequality $$2 *4 \leq 10$$ is true.

b) The $$n$$-CNakayama algebra $$[n,n+1,n+1,....,n+1]$$ has defect 1 and dominant dimension equal to $$2n-2$$, which shows that the inequality in conjecture 1 is optimal in case it is true.

$$\mathbf{Remarks}$$

A proof of conjecture 1 would have some nice applications. I name two here. First, it would substantially improve the main result on the dominant dimension of Nakayama algebras in https://www.sciencedirect.com/science/article/pii/S0021869317305999 .

A proof of conjecture 2 would be a needed key result to connect and complete two classification result that Dag Madsen and I work on.

The evidence for conjecture 1 (and thus also conjecture 2) is very good. Conjecture 1 is proven for $$n$$-CNakayama algebras with $$n \leq 12$$ and thus for over 2 million examples (here I mean 2 million different difference classes, see below).

$$\mathbf{More \ examples}$$

Say that two $$n$$-CNakayama algebra $$[c_0,c_1,...,c_{n-1}]$$ and $$[e_0,e_1,...,e_{n-1}]$$ are in the same difference class in case $$c_i \equiv e_i$$ mod $$n$$ for all $$i$$. Note that dominant dimension and defect are the same for two such lists in the same difference class. Thus the inequality has to be checked only for finitely many $$n$$-CNakyama algebras for a given $$n$$. Here are all (difference classes of) $$n$$-CNakayama algebras together with their dominant dimension for $$n=4$$:

 [ [ 2, 2, 2, 3 ], 4 ],
[ [ 2, 2, 3, 3 ], 1 ],
[ [ 2, 2, 4, 3 ], 1 ],
[ [ 2, 3, 2, 3 ], 2 ],
[ [ 2, 3, 3, 3 ], 2 ],
[ [ 2, 3, 4, 3 ], 1 ],
[ [ 2, 4, 3, 3 ], 1 ],
[ [ 2, 4, 4, 3 ], 1 ],
[ [ 2, 5, 4, 3 ], 1 ],
[ [ 3, 3, 3, 4 ], 5 ],
[ [ 3, 3, 4, 4 ], 3 ],
[ [ 3, 3, 5, 4 ], 2 ],
[ [ 3, 4, 3, 4 ], 1 ],
[ [ 3, 4, 4, 4 ], 1 ],
[ [ 3, 4, 5, 4 ], 1 ],
[ [ 3, 5, 4, 4 ], 1 ],
[ [ 3, 5, 5, 4 ], 1 ],
[ [ 3, 6, 5, 4 ], 1 ],
[ [ 4, 4, 4, 5 ], 2 ],
[ [ 4, 4, 5, 5 ], 4 ],
[ [ 4, 4, 6, 5 ], 1 ],
[ [ 4, 5, 4, 5 ], 2 ],
[ [ 4, 5, 5, 5 ], 6 ],
[ [ 4, 5, 6, 5 ], 1 ],
[ [ 4, 6, 5, 5 ], 1 ],
[ [ 4, 6, 6, 5 ], 2 ],
[ [ 4, 7, 6, 5 ], 1 ],
[ [ 5, 5, 5, 6 ], 1 ],
[ [ 5, 5, 6, 6 ], 2 ],
[ [ 5, 5, 7, 6 ], 1 ],
[ [ 5, 6, 5, 6 ], 1 ],
[ [ 5, 6, 6, 6 ], 3 ],
[ [ 5, 6, 7, 6 ], 1 ],
[ [ 5, 7, 6, 6 ], 1 ],
[ [ 5, 7, 7, 6 ], 1 ],
[ [ 5, 8, 7, 6 ], 1 ].


$$\mathbf{Extension \ of \ conjecture \ 1}$$

The following extension is suggested by a comment of Gjergji Zaimi. Define the super dominant dimenson $$\operatorname{sdomdim(A)}$$ of an $$n$$-CNakayama algebra as the sum of all dominant dimension of the special modules. A positive answer of the following question would be an extension of conjecture 1.

Question (by Gjergji Zaimi): Do we have $$\operatorname{sdomdim(A)} \leq 2n-2$$ for any $$n$$-CNakayama algebra?

The question has a positive answer for $$n \leq 12$$ and I try to look at it for higher $$n$$ with the computer.

Here are the values of the super dominant dimension for difference classes of $$n$$-CNakayama algebras $$n=4$$:

 [ [ 2, 2, 2, 3 ], 4 ]
[ [ 2, 2, 3, 3 ], 1 ]
[ [ 2, 2, 4, 3 ], 4 ]
[ [ 2, 3, 2, 3 ], 4 ]
[ [ 2, 3, 3, 3 ], 2 ]
[ [ 2, 3, 4, 3 ], 4 ]
[ [ 2, 4, 3, 3 ], 4 ]
[ [ 2, 4, 4, 3 ], 2 ]
[ [ 2, 5, 4, 3 ], 4 ]
[ [ 3, 3, 3, 4 ], 5 ]
[ [ 3, 3, 4, 4 ], 3 ]
[ [ 3, 3, 5, 4 ], 4 ]
[ [ 3, 4, 3, 4 ], 2 ]
[ [ 3, 4, 4, 4 ], 1 ]
[ [ 3, 4, 5, 4 ], 3 ]
[ [ 3, 5, 4, 4 ], 3 ]
[ [ 3, 5, 5, 4 ], 5 ]
[ [ 3, 6, 5, 4 ], 4 ]
[ [ 4, 4, 4, 5 ], 2 ]
[ [ 4, 4, 5, 5 ], 4 ]
[ [ 4, 4, 6, 5 ], 3 ]
[ [ 4, 5, 4, 5 ], 4 ]
[ [ 4, 5, 5, 5 ], 6 ]
[ [ 4, 5, 6, 5 ], 5 ]
[ [ 4, 6, 5, 5 ], 5 ]
[ [ 4, 6, 6, 5 ], 4 ]
[ [ 4, 7, 6, 5 ], 4 ]
[ [ 5, 5, 5, 6 ], 1 ]
[ [ 5, 5, 6, 6 ], 2 ]
[ [ 5, 5, 7, 6 ], 2 ]
[ [ 5, 6, 5, 6 ], 2 ]
[ [ 5, 6, 6, 6 ], 3 ]
[ [ 5, 6, 7, 6 ], 3 ]
[ [ 5, 7, 6, 6 ], 3 ]
[ [ 5, 7, 7, 6 ], 3 ]
[ [ 5, 8, 7, 6 ], 3 ]

• In your list, there is $c_1\ge c_0$ everywhere. Is that also a condition? What is wrong with [3,2,2,4]? – Wolfgang Jan 4 at 17:45
• @Wolfgang Thank you for your comment and the edit. One can assume without loss of generality that $c_0$ is the minimum of the $c_i$. I added that condition. – Mare Jan 4 at 17:48
• Maybe this is a silly question but is there an example where the sum of all dominant dimensions of all special modules exceeds $2n-2$? – Gjergji Zaimi Jan 5 at 21:49
• @GjergjiZaimi I think there should be such examples. I will search one with the computer now. Thanks for the interesting comment. – Mare Jan 5 at 21:52
• That's totally fine. Perhaps, for the sake of clarity it's worth pointing out that the number of special modules is the same as the defect of the algebra. Which is why the conjecture for the super dominant dimension implies conjecture 1 (in other words $\operatorname{sdomdim(A)}\geq \operatorname{def}(A)\operatorname{domdim}(A)$). – Gjergji Zaimi Jan 5 at 22:13

Here I will prove a bound that is much stronger in general. If $$b(A)$$ denotes the number of bounded projective-injective modules (a special kind of projective-injective module defined below) then we have $$\operatorname{sdomdim}(A)\le \operatorname{def}(A)+2b(A).$$ To see why this is much stronger notice that $$b(A)$$ is upper bounded by the total number of projective-injective modules which is simply $$n-\operatorname{def}(A)$$. So we get as a corollary $$\operatorname{sdomdim}(A)\le 2n-\operatorname{def}(A).$$ Notice that this implies your conjecture 2 immediately, and with a little more work also the stronger version of conjecture 1 (by a little more work, I mean we have to check the algebras of defect 1 by hand, but the conjectures are known in this case).

Summary of ideas:

We show that, starting from every special module, we can draw a "bouncing path" that ends in a projective but not injective module. The dominant dimension of the special module is bounded by twice the number of bounces in this path minus one. Adding these inequalities over all bouncing paths we get the desired bound.

Proof:

Let's call the set of all modules of the form $$(i,k)$$, with varying $$k$$, the $$i$$-th left interval. Every left interval contains a unique projective module. Similarly let's call the set of all modules of the form $$(k,i-k)$$, with varying $$k$$, the $$i$$-th right interval. Let's also denote the set of special modules as $$\mathcal L=\{L_1,L_2,\dots,L_t\}$$, the set of projective but not injective modules as $$\mathcal R=\{R_1,R_2,\dots,R_t\}$$, and finally the set of projective-injective modules as $$\{P_1,P_2,\dots,P_{n-t}\}$$, where $$t$$ is the defect of the algebra.

We can define a certain map on intervals:

• Every special module $$L=(i,k)$$ has a right interval associated to it, the $$(i+k)$$-th right interval, that we denote $$\phi(L)$$
• If $$I$$ is the $$i$$-th right interval we can define $$\phi(I)$$ to be the $$i$$-th left interval
• If the projective module, $$(i,c_i)$$ of the left interval $$I$$ is also injective, then define $$\phi(I)$$ to be the $$(i+c_i)$$-th right interval

We are now ready to describe a bijection between $$\mathcal L$$ and $$\mathcal R$$. For $$L_i$$ form the sequence of intervals $$\phi(L),\phi(\phi(L)),\cdots$$ until you stop (meaning you reach a left interval whose projective module is not injective). Call that projective module $$R_j=\pi(L_i)$$. Since we can define such sequences of intervals backwards in the opposite algebra, it is clear that $$\pi$$ is invertible, and thus a bijection. We can label such a chain of intervals as $$L_i\to\pi(L_i)$$.

Moreover if a module $$M$$ appears in an interval $$I$$ then $$\Omega(M)$$ appears in $$\phi(I)$$, therefore the dominant dimension of $$L_i$$ is upper bounded by one less than the number of intervals in the chain $$\phi(L_i),\phi^2(L_i),\dots$$. If there are $$r_i$$ projective-injective modules in this chain then there are $$2r_i+2$$ intervals, so we can say $$\operatorname{domdim}(L_i)\le 2r_i+1$$ Remark: the reason why we have an inequality rather than an equality is that it is possible for $$\Omega^k(L_i)$$ to have a projective cover that is not injective for $$k$$ smaller than $$2r_i+1$$.

It should be clear by construction that a projective injective module $$P$$ can appear in at most one such chain of intervals and if it does we call it bounded. Indeed, if you start from a bounded projective-injective module you start with the left and right intervals that meet at this module and keep applying $$\phi^{-1}$$ on the left of the chain and $$\phi$$ on the right of the chain, you will stop when you've hit some $$\phi(L_i)$$ on the left and the left interval of $$\pi(L_i)$$ on the right, and thus constructed the unique $$L_i\to \pi(L_i)$$ chain containing your bounded projective-injective module.

So we can finally write $$\operatorname{sdomdim}(A)=\sum_{i=1}^t \operatorname{domdim}(L_i)\le \sum_{i=1}^t(2r_i+1)=2b(A)+t$$ as desired.

• Thanks, I will read it now. Just one quick comment: For $def(A)=1$ it is known that $domdim(A) \leq 2n-2$ and thus also $sdomdim(A) \leq 2n-2$ for example by the result for Nakayama algebras in sciencedirect.com/science/article/pii/S0021869317305999 . – Mare Jan 7 at 21:26
• @Mare I edited to make the terminology changes, so I removed previous comments that are no longer relevant. – Gjergji Zaimi Jan 7 at 22:26
• I have to go to bed now and will read it again tomorrow and award the bounty. The proof looks a little like magic but it should work. The same proof by the way should also work for $n$-LNakayama algebras that are also defined as the lists with $[c_0,...,c_{n-1}]$ satisfying $c_i \geq 2$ for $i \neq n-1$, $c_{n-1}=1$ and $c_i -1 \leq c_{i+1}$ for all $i$ (they are in bijection with Dyck paths). So the inequality is true for general Nakayama algebras (although for LNakayama algebras the bound is not optimal). – Mare Jan 7 at 23:03