From a lower triangular matrix to its quiver representation I have a question about quivers:
Suppose we have an arbitrary lower triangular matrix algebra $A= \{\begin{pmatrix} 
a & 0 & 0\\
c & b & 0\\
e&d&a
\end{pmatrix}: a , b, c, d \in \mathbb C \}$. 
How can we find the related quiver to this algebra?
I know how to represent a finite quiver (A directed multigraph = a set of arrows and vertices with the set of arrows from $j$ to $i$ if $i\le j$) as a lower triangular matrix, but I don't know how to go from the other side?
Thanks!
 A: Your question is a bit confusing since you write "an arbitrary lower triangular matrix" but then give a very specific one. So let me give you a general receipe and illustrate it with this example.
A possible first step is to find out the Jacobson radical of the algebra. For that the characterisation that for a finite dimensional algebra it is a nilpotent ideal such that the quotient is semisimple is helpful. For lower triangular matrices, it is often the strictly lower triangular submatrices which work (since you also allow equalities between entries that is not always true). In the running example, it does work, it is easy to see that the lower triangular matrices $J$ form a nilpotent ideal and $A/J\cong \mathbb{C}\oplus \mathbb{C}$. 
The next step is not necessary here, but if in your decomposition of $A/J$, matrix rings over $\mathbb{C}$ occur, you would have to go to the basic Morita representative.
The number of factors of $A/J$ will be the number of vertices of the quiver. In this example, the decomposition $A/J\cong \mathbb{C}\oplus \mathbb{C}$ tells us, that the quiver has two vertices. Idempotent lifting tells us that for each factor there is an idempotent $e_i$ in $A$ whose residue class is the unit in the corresponding factor. In the example they are given by $a=1$, resp. $b=1$ and all the other parameters vanishing. 
The next step is to compute $J^2$ (just by multiplying the entries of $J$). In the example, it is given by the subset of matrices where only $e\neq 0$. The arrows of the quiver will then correspond to a basis of $J/J^2$. From where to where the arrows go, one has to choose specific bases of $e_iJ/J^2e_j$. In the example, one obtains one arrow $1\to 2$ and one arrow $2\to 1$. Thus, we have determined the quiver.
You have not asked for the relations, but in the example, the algebra is not isomorphic to a path algebra. By sending the idempotents and arrows to the corresponding elements of $A$, one obtains an algebra homomorphism $\mathbb{C}Q\to A$. In the example, the kernel will be the ideal generated by the path $2\to 1\to 2$. 
