# Projective indecomposables versus general indecomposables

Given a finite dimensional algebra, what is the exact relation between the indecomposable projective modules, and a general indecomposable module? In the case of an oriented quiver without cycles for example, it is easy to find the simples, and the indecomposable projectives. What does this tell us about a general indecomposable module?

I guess the main question is, to understand the complete representation theory of a finite dimensional algebra (or let's say a quiver), what else do we need besides the simples and the projective indecomposables?

• You need to know all the other indecomposables, of which there is in general a whole lot. I really do not understand your question: have you tried finding all indecomposables in some simple example, like Dynkin diagrams of type $A$, say? As soon as you do that, you will answer this question yourself. – Mariano Suárez-Álvarez Jan 6 '13 at 10:08
• Finding all indecomposables for a general finite dimensional algebra is something we simply do not know how to do. As you say, finding the simples and the indecomposable projectives is trivial, so surely there is something that keeps people busy writing papers and books about this subject! :-) – Mariano Suárez-Álvarez Jan 6 '13 at 10:10
• Recover how? Unless you make the question concrete, it is impossible to answer it. For example, every module is a quotient of a projective, and if you know the indec. proj, you know all projectives... is this «recoving all modules from the indec. projectives»? – Mariano Suárez-Álvarez Jan 6 '13 at 10:49
• The simples and the indec. proj. modules are in general a minuscule part of the set of all indecomposable modules. For example, there are algebras of very small dimension which have exactly one simple module and one indecomposable projective, and whose set of modules contains copies of all modules of all other finite dimensional algebras (such algebras are called local wild algebras) – Mariano Suárez-Álvarez Jan 6 '13 at 10:51
• There are, though, classes of algebra for which an algorithm exists which will construct all indecomposables from the projectives; this is called the knitting algorithm. This works for a very special, small class of algebras, though. – Mariano Suárez-Álvarez Jan 6 '13 at 10:54

## 1 Answer

Let me elaborate a bit on Mariano's comments and since you specifically asked for the case of path algebras of quivers without oriented cycles I'll focus on that one.

What Mariano already mentioned in the comments is that the finite dimensional algebras can be split into three cases according to how complicated their representation theory is. I'm not going to give a precise definition here but roughly speaking one can say the following:

An algebra is called representation finite if there are only finitely many indecomposable modules up to isomorphism (e.g. for $A=k$ according to linear algebra $k$ is the only indecomposable module).

An algebra is called tame if the indecomposable modules in each dimension can be described by finitely many one-parameter families (in linear algebra you probably encountered Jordan normal form where matrices can be brought into a normal form $J(n,\lambda)$, where $n$ is the dimension and $\lambda$ is one-parameter. Although $k[X]$ is not finite dimensional, this is what you should compare it with.)

An algebra is called wild if its representation theory is at least as complicated as that of every other algebra (this can be made precise by functors $\operatorname{mod} B\to \operatorname{mod} A$ for every other algebra $B$ satisfying properties that you expect like preserving isomorphism or indecomposability, in fact it suffices to take the (though again infinite dimensional) algebra $k\langle X,Y\rangle$, the polynomial ring in two non-commuting variables).

A theorem by Drozd tells you that every finite dimensional algebra is either representation finite, tame or wild. (Some authors take tame to be not representation finite, some don't)

Now what happens if $A=k[Q]$ is the path algebra of a quiver (without oriented cycles of course). Then this was observed earlier by Gabriel for the representation finite case and independently to Nazarova and Donovan-Freislich for the tame case:

Theorem: $k[Q]$ is representation finite iff $Q$ is a simply-laced Dynkin diagram (with an arbitrary orientation of the edges). It is tame iff $Q$ is a simply-laced Euclidean diagram (also called extended Dynkin diagram, with an arbitrary orientation of the edges). And wild in all other cases.

Now we have a criterion to decide how many representations a given path algebra has. But I didn't give you an example of one yet. Instead I want to first give you an answer to another question, namely which dimension vectors do occur? Here a dimension vector of a representation $(M_i)_{i\in Q_0}$ is the vector $(\dim M_i)_{i\in Q_0}$ or equivalently $\dim \operatorname{Hom}(P_i,M)$. This question was answered by Kac in the 80s in the general case (for representation finite and tame path algebras, it was known before):

Theorem: Let $\mathfrak{g}$ be the Kac-Moody Lie algebra corresponding to the underlying graph of $Q$. Then there is a bijection between the positive roots of $\mathfrak{g}$ and the dimension vectors of $k[Q]$.

One can even say more about the number of representations of a given dimension vector. But I don't want to go into details here. Also this bijection can be enhanced algebraically via the so called Ringel-Hall algebra.

So we know (at least theoretically if we know enough representation theory of Lie algebras) which dimension vectors can occur. But we haven't constructed one yet, or even not given an example of a dimension vector. So as a last point let me explain the knitting algorithm that Mariano Suarez-Alvarez mentioned in the comments. This algorithm has the advantage that in contrast to the previous theorem it can also be done for non-path algebras and that it can be enhanced to really construct representations (I'll explain how to do it for dimension vectors, if you want to construct representations, you have to replace the knitting step by a similar pushout construction.)

Preparation: Write down all the projective modules and the indecomposable direct summands of their radicals in terms of dimension vectors.

Start: Write down all the simple projectives

Repeat the following two steps until you reach one of the halting conditions:

Insertion step: If the dimension vector you are currently studying occurs as the direct summand of a radical of a certain projective, write down this projective and an arrow from the module to the projective.

Knitting step: Add the dimension vectors that successors of you current dimension vector and substract your current dimension vector. Then put down the result and write arrows from all the successors of the current dimension vector to that new one.

Halting conditions: If you reach a negative dimension vector in the knitting step, the dimension vector you are currently studying should belong to an injective module. (Check that, otherwise there is some calculation error.) In that case see if there are other dimension vectors that you could "knit" further. If not, you have constructed all the dimension vectors of the indecomposable modules. The other condition is that if there is an entry of a dimension vector greater than $6$, then the other halting procedure will never be reached. Halt whenever you think you have constructed enough modules.

So let me give you three examples since I think it is difficult just working this out from these steps:

1) Let $Q=1\to 2$. Then the dimension vectors of the projectives are $(11)$ and $(01)$. Their radicals are $(01)$ and $0$, respectively. So we have to start with

$$(01)$$

Now according to the insertion step we have to insert $(11)$ and an arrow $(01)$ to $(11)$ (since $(01)$ occurs in the radical of $(11)$.) So we have the following:

$$\begin{array}{ccc}&&(11)\\&\nearrow\\(01)\end{array}$$

Now we have to do the knitting step so computing $(11)-(01)=(10)$. So we have the following:

$$\begin{array}{ccccc}&&(11)\\&\nearrow&&\searrow\\(01)&&&&(10)\end{array}$$

We cannot insert more and knitting further would result to $(10)-(11)=(0,-1)$, so we halt here and conclude that the path algebra of this quiver has $3$ indecomposable reprsentations (each of which is either simple or projective, so this is maybe not the best example).

2) Let $Q=1\to 2\to 3$ Then the projectives are $(111)$, $(011)$ and $(001)$. Their radicals $(011)$, $(001)$ and $0$ respectively. So start with:

$$(001)$$

Then insert

$$\begin{array}{ccc}&&(011)\\&\nearrow\\(001)\end{array}$$

Knit:

$$\begin{array}{ccccc}&&(011)\\&\nearrow&&\searrow\\(001)&&&&(010)\end{array}$$

Insert:

$$\begin{array}{ccccc}&&&&(111)\\&&&\nearrow\\&&(011)\\&\nearrow&&\searrow\\(001)&&&&(010)\end{array}$$

Knit:

$$\begin{array}{ccccc}&&&&(111)\\&&&\nearrow&&\searrow\\&&(011)&&&&(110)\\&\nearrow&&\searrow&&\nearrow\\(001)&&&&(010)\end{array}$$

Knit (we cannot insert anything else):

$$\begin{array}{ccccccc}&&&&(111)\\&&&\nearrow&&\searrow\\&&(011)&&&&(110)\\&\nearrow&&\searrow&&\nearrow&&\searrow\\(001)&&&&(010)&&&&(001)\end{array}$$

No further positive computation is possible so we stop here. So there are $6$ indecomposable representations and as you recognised some of them are neither simple nor projective.

3) The third one is $Q=1\stackrel{2}{\to} 2$, where the $2$ set above means that there are two arrows from $1$ to $2$. In the following diagrams, all arrows are supposed to be two arrows in fact. So the dimension vectors here are $(12)$ and $(01)$ with radicals $(01)\oplus (01)$ and $0$. So start with

$$(01)$$

Then insert (with two arrows, because the dimension vector appears twice in the radical:

$$\begin{array}{ccc}&&(12)\\&\nearrow\\(01)\end{array}$$

Now knit (you have to take the $(12)$ into account twice, think of it as being written twice with arrows $(01)\to (12)$):

$$\begin{array}{ccccc}&&(12)\\&\nearrow&&\searrow\\(01)&&&&(23)\end{array}$$

Knit again:

$$\begin{array}{ccccccc}&&(12)&&&&(34)\\&\nearrow&&\searrow&&\nearrow\\(01)&&&&(23)\end{array}$$

And so on, you will eventually reach an entry $6$ after some steps. If you want to practice it, doing $D_4$ next would be a good exercise and if you really have fun with it, try $E_8$. Also in fact this construction did not only give you the indecomposable representations, but also certain maps between them, the so called irreducible maps. The quiver you see is called Auslander-Reiten quiver.

As I said you can knit other algebras that are not path algebras at least if the quiver has no oriented cycles. If it has some, there are generalizations, you then have to cover the graph first and then knit in the cover and push down again. .

Let me finish by some literature to read something about that stuff: