Let Q be the quiver

enter image description here

bound by $αβ = 0$, $γδ = 0$. The indecomposable projective A-modules are given by enter image description here

where $A=KQ/I$. This an example in Assem-Simson-Skowronski book (Elements of the representation theory of the representation theory of finite dimensional algebras Part 1), page $80$. What I know is $P(i)=e_iA$, and I also know how to find $KQ$, but I do not know how to find $A=KQ/I$ and how could I get $P(2)$ and $P(3)$. So could you guys help me out?



Since you know that $P(i)=e_iA$, try to understand when this is nonzero. Let $p \in A$ be a path, then $e_i p \neq 0$ iff $p$ starts in $i$. This way you get a basis of this space. So, $P(1)=\langle e_1\rangle_K$, $P(2)=\langle e_2, \beta,\delta\rangle_K$, and $P(3)=\langle e_3, \alpha, \gamma, \alpha\delta, \gamma\beta\rangle_K$ (since $\alpha\beta$ and $\gamma\delta$ are relations).

The action of the arrows is just by concatenation (as usual in the path algebra).

The next step is to remember the equivalence between modules and representations: Whenever you have a module $M$, it is mapped to the representation given by the vector space $Me_j$ on each arrow, and the linear map $-\cdot a: Me_j\to Me_k$ for an arrow $a:j\to k$. In the example of $P(2)$ you get and in the example of $P(3)$ you get: (From this step, it is also easy to "see" the corresponding radicals.

As a final step, translate this to $K^r$ and matrices by using an explicit isomorphism to the claimed representations.

  • $\begingroup$ Could you explain more how I could get $P(2)$ and $P(3)$? Thanks, $\endgroup$ – Dan Sep 14 '14 at 20:04
  • 1
    $\begingroup$ @Dan I have expanded my answer. $\endgroup$ – Julian Kuelshammer Sep 18 '14 at 16:11
  • $\begingroup$ Thanks, you helped me to understand lots of things. $\endgroup$ – Dan Sep 19 '14 at 5:14

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