Confusion in module structure of indecomposable injectives over quiver path algebra

Question

I got very confused while working out the explicit module structure on the indecomposable injective modules $I_i$ over a finite-dimensional quiver path algebra $kQ/I$. I know that the right injective indecomposable $I_i$, where $i$ is a vertex of $Q$, may be described as

$$I_i=\{\text{paths starting at }i \text{ modulo relations}\}^*,$$

where $^*$ denotes the $k$-dual module. But what does it mean? Does this mean that in this module a path $p$ corresponds to a linear map to $k$ denoted $p^*$ which takes $p$ to 1 and other paths to 0? If so, am I right in understanding that multiplication acts by "deletion", i.e. if $p^*\cdot q$ is $s^*$ such that in the quiver $qs=p$ (I am composing paths as maps, i.e. $q$ must start where $s$ begins)?

For example, consider a quiver $Q$ with 2 vertices and one edge $x: 1\to2$. I would say from the above description that in $I_i$ we have $x^*\cdot x=e_1^*$, since $(x^*\cdot x)(e_1)=x^*(x\cdot e_1)=1$?

By the way, I got lost in this because I was trying to understand the criteria for a quiver path algebra to be self-injective, for which I believe it is enough to have a permutation of vertices $\pi$ such that $P_i\cong I_{\pi(i)}$, but I also saw that it is enough to have $\text{soc}(P_i)\cong\text{top}(I_{\pi(i)})$. I would be thankful for any references on this topic!

Answer 1

There are several things to unpack here. Let's review the notions that already appear in your post, and we'll get to a nice description of injective modules at the end of this answer.

First, let's make our conventions explicit. From your post, I see you are considering right modules, and also that you are composing arrows in a quiver from right to left (that is, if the target of the arrow $x$ is the source of the arrow $y$, then $yx$ is their composition). Also, to simplify notations, let $A=kQ/I$.

Before looking at the injective modules, let's describe the indecomposable projective modules. For each vertex $i$, there is an idempotent $e_i$ in $A$ (corresponding to the path of length zero at $i$); the corresponding indecomposable projective right module is $e_iA$, which is the set of linear combinations of paths of $Q$ ending in $i$ (modulo $I$). Note that the indecomposable projective left module is $Ae_i$, which is the set of linear combinations of paths of $Q$ starting in $i$ (modulo $I$).

Now, to get injective right modules, we will use the duality functor $(?)^* = \operatorname{Hom}_k(?,k)$ from the category of left $A$-modules to the category of right $A$-modules. The duality functor sends projective modules to injective modules, and preserves indecomposability. Thus the indecomposable injective right module corresponding to the vertex $i$ is $$ I_i \cong (Ae_i)^* = \operatorname{Hom}_k(Ae_i,k). $$

The $A$-module structure of $\operatorname{Hom}_k(Ae_i,k)$ is defined as follows: for any $\phi\in \operatorname{Hom}_k(Ae_i,k)$ and any $a\in A$, $\phi\cdot a$ is the map sending $x$ to $\phi(a\cdot x)$.

In order to obtain a more explicit description of $I_i$, we turn to quiver representations. There is an equivalence of categories between the category of finite-dimensional right $A$-modules and the category of finite-dimensional representations of the opposite quiver $Q^{op}$ which satisfy the relations in $I$. This equivalence is fairly straightforward to describe on objects: given a right $A$-module $M$, the corresponding representation is defined by

• putting the vector space $Me_i$ at each vertex $i$, and
• putting the linear map $Me_j \to Me_i: m \mapsto ma$ at each arrow $a:i\to j$.

Let's apply this to the injective module $I_i$. For any vertex $j$, we have that $$ I_ie_j = \operatorname{Hom}_k(Ae_i, k)e_j \cong \operatorname{Hom}_k(e_jAe_i, k). $$

A basis of $I_ie_j$ is thus given by a dual basis of $e_jAe_i$, the space generated by paths from $i$ to $j$ in $Q$ (modulo $I$).

Finally (and perhaps most importantly in this discussion), for an arrow $a:h\to j$, the corresponding linear map $I_ie_j \to I_ie_h$ is dual to the map $\lambda_a:e_hAe_i \to e_jAe_i$ given by left multiplication by $a$. In particular, if you fix a basis of $e_hAe_i$ and $e_jAe_i$ and let $[\lambda_a]$ be the matrix of the map $\lambda_a$ in those bases, then the matrix of the dual map $I_ie_j \to I_ie_h$ in the dual bases is the transpose of $[\lambda_a]$.

In particular, if you know a representation $V$ isomorphic to the left projective $A$-module $Ae_i$, then to get a representation isomorphic to the right injective $A$-module $I_i$, all you need to do is take the transpose of all the matrices defining $V$.