Specific projective dimension of a module over bound quiver

Question

Suppose $K$ is an algebraically closed field, and $A$ is the algebra presented by the quiver $$\require{AMScd} \begin{CD} 1 @>>> 2\\ @V{}VV @V{}VV \\ 3 @>>> 4 @>>> 5 \end{CD} $$ bound by $1\to 2\to 4 = 1\to3\to 4$ and $3\to4\to5 = 0$.

What is the projective dimension of the simple $A$-module $S(1)$?

Here is my work: the projective indecomposables have submodule lattices indicated by the following Alperin diagrams: $$\begin{array}{c@{}c@{}c@{}c@{}c} \newcommand{\kem}{\kern-1ex} \kem&\kem&\kem\kem1\kem&\kem&\kem \\ \kem&\kem\nearrow\kem&\kem&\kem\nwarrow\kem&\kem \\ 2\kem&\kem&\kem&\kem&\kem3 \\ \kem&\kem\nwarrow\kem&\kem&\kem\nearrow\kem&\kem \\ \kem&\kem&\kem\kem4\kem&\kem&\kem \end{array} \qquad \begin{array}{c} 2 \\ \uparrow \\ 4 \\ \uparrow \\ 5 \end{array} \qquad \begin{array}{c} 3 \\ \uparrow \\ 4 \\ \phantom{\uparrow} \\ \phantom{5}\end{array} \qquad \begin{array}{c} 4 \\ \uparrow \\ 5 \\ \phantom{\uparrow} \\ \phantom{5}\end{array} \qquad \begin{array}{c} 5 \\ \phantom{\uparrow} \\ \phantom{4} \\ \phantom{\uparrow} \\ \phantom{5}\end{array} $$

The projective cover of $S(1)$ is $P(1)$ and the kernel is the Heller operator of $S(1)$, namely $$\begin{array}{c@{}c@{}c@{}c@{}c} \newcommand{\kem}{\kern-1ex} 2\kem&\kem&\kem&\kem&\kem3 \\ &\kem\nwarrow\kem&\kem&\kem\nearrow\kem&\\ &\kem&\kem\kem4\kem&\kem& \end{array} $$ with $P(2) \oplus P(3)$ its projective cover.

Now the problem is to find the second Heller operator $\Omega^2(S(1))$. Its composition factors are clear: $S(4)$ and $S(5)$, but how do we know if it is $S(4) \oplus S(5)$ versus $P(4) = \begin{array}{c} 4 \\ \uparrow \\ 5 \end{array}$?

In the former case, the next projective cover is $P(5) = S(5)$ and the resolution terminates with projective dimension 3, but in the latter case the resolution terminates immediately with projective dimension 2.

The former case “uses” the 4 from $P(2)$ leaving $S(5)$ from $P(2)$ and $S(4)$ from $P(3)$. The latter case “uses” the 4 from $P(3)$ leaving the trailing $P(4)$ from $P(3)$. Neither of these is really the kernel, since it is some sort of “diagonal” submodule, but how do we know if the diagonal submodule is split or not?

Does the answer depend on the field?

Answer 1

As far as I know, field should not matter, unless your algebra has "weird relations", and in most case, you are pretty safe as long as characteristic of $k$ is not 2. I hope somebody can give some supplement answer here...

Back to your specific example, (say if we work over characteristic 0..) the (minimal) projective resolution of $S(1)$ is actually: $$ 0\to P(4)\to P(2)\oplus P(3) \to P(1) \to S(1) \to 0 $$ So the projective dimension is 2. Eplanation:

(1) Writing out explicitly what the maps are:

• first (surjective) map $ae_1 \mapsto a\overline{e_1}$, where $a\in A$, $e_1$ primitive idempotent and $\overline{e_1}=e_1+Rad(P(1))$.
• $\Omega(S(1))=ker(ae_1\mapsto a\overline{e_1}) = k-span\{\alpha,\beta,\gamma\alpha\}$, where $\alpha$ is ther arrow from 1 to 2, $\beta$ is the arrow from 1 to 3, and $\gamma$ arrow from 2 to 4.
• next map is projective cover of $\Omega(S(1))$, given by $ae_2 \mapsto a\alpha$ and $ae_3\mapsto a\beta$.
• So $\Omega^2(S(1))=k-span\{\delta-\gamma, \epsilon\gamma\}$, where $\delta$ is the arrow $3\to 4$ and $\epsilon$ is $4\to 5$. You can now construct an isomorphic to $P(4)$ via $ae_4\mapsto a(\delta-\gamma)$.

(2) After enough practice, you should be able to see that everything can be done combinatorially: $\Omega(S(1))$ is the 3-dimensional module with Loewy diagram being the one you have drawn out already. Well, you can see it comes from "tearing off the radical of $P(1)$ off from $P(1)$". The same principle applies: $\Omega^2(S(1))$ just come from "tearing off the part in $P(2)\oplus P(3)$ that isn't paste onto $M$ without destroying Loewy structure". Since both 2 and 3 can lie on top of 4, so you need to "tear off half of the 4 from $P(3)$, and half of the 4 from $P(5)$, and glue them back" (c.f. the map written explicitly above). Sorry this is vague, but this is how people who do computation in these algebras work, for real!

On a side notes, I think "Alperin diagrams" means the diagrams used by Alperin such as those in his book "Local representation theory" (I may be wrong), those are really filtration diagrams, which is different from what you have drawn here (Loewy diagram). Anyway, if you use "Loewy diagram", most people should know exactly what it is, less ambiguity.

EDIT-Correction: Thanks to Julian and Jack (the OP)! Yes, by "weird" I really mean situations like summands in a the relations reducing to zero; or in characteritic 2, minuses become pluses. When such situation does not occur, and since we are working over algebraically closed field, the homological algebras shouldn't change much. So this specific example is characteristic independent.