Note: This question now has a [sister](https://mathoverflow.net/questions/279618/what-is-the-relation-between-coxeter-transformations-of-coxeter-systems-and-coxe) :-)


**The Coxeter transformation of a generalized Cartan matrix**:


In the paper
[The spectral radius of the Coxeter transformations for a generalized Cartan matrix](https://link.springer.com/article/10.1007/BF01450490), Claus Ringel defines a Coxeter transformation as follows, in several steps:


A *generalized Cartan matrix* of size $n$ is a matrix $A \in M^{n \times n}(\mathbb{Z})$ such that for all $i \neq j$ the following properties are satisfied:


* $A\_{ii} = 2$
* $A\_{ij} \leq 0$
* $A\_{ij} \neq 0 \Leftrightarrow A\_{ji} \neq 0$


He then goes on to define the reflection $R\_i: \mathbb{R}^n \to \mathbb{R}^n$ as the linear map (depending on $A$) which is given on the canonical basis by $e(j) \mapsto e(j) - A\_{ji}e(i)$. 


Now if $\pi: \{1, \dots, n\} \to \{1, \dots, n\}$ is any permutation, he calls the product $C(A, \pi) := R\_{\pi(n)} \cdots R\_{\pi(1)}$ a *Coxeter transformation* for $A$.


**The Coxeter transformation of a finite dimensional algebra**:


Let $k$ be a field and $A$ be a finite-dimensional $k$-algebra. Let $P(1), \dots, P(n)$ and $I(1), \dots, I(n)$ be the indecomposable projective $A$-modules up to isomorphism and the indecomposable injective $A$-modules up to isomorphism, respectively (in such a way that the orders correspond, i.e. $P(i) = Ae\_i$, $I(i) = D(e\_iA)$ with the same idempotent $e\_i$). Assume that the dimension vectors $\underline{\dim}(P(1)), \dots, \underline{\dim}(P(n))$ form a basis of $\mathbb{R}^n$. Then we can define the *Coxeter transformation* for $A$ (with respect to the ordering of the projective modules) by $\Phi\_{A}: \mathbb{R}^n \to \mathbb{R}^n, \ \underline{\dim}(P(i)) \mapsto -\underline{\dim}(I(i))$,


**Question**:


What precisely is the relation between these two notions of Coxeter transformations? How do properties of a generalized Cartan matrix and properties of a finite-dimensional algebra translate via this relation?



What is the relation between Coxeter transformations of generalized Cartan matrices and Coxeter transformations of finite-dimensional algebras?
Since I didn't want to think about permutations too much, here is an answer about relating Coxeter transformations of the form $C(A, \text{id})$ to Coxeter transformations of finite-dimensional path-algebras.


Let $A$ be a generalized Cartan matrix. Let $A\_+$ and $A\_-$ be defined by


$$\left( A\_+\right)\_{ij} =
\begin{cases}
A\_{ji}, \ i > j \\
1, \ i = j \ \ \ \ \ \ \ \ ; \\
0, \text{else}
\end{cases}
\ \ \ (A\_-)\_{ij} = 
\begin{cases}
A\_{ij}, \ i > j \\
1, \ i = j \\
0, \text{else}
\end{cases}$$


**Theorem 1:** We have $C(A, \text{id}) = -A\_{+}^{-1}A\_{-}^t$.


*Proof:* This is Theorem $2.9$ in this [paper](https://arxiv.org/pdf/math/0611201.pdf "ON THE PERIODICITY OF COXETER TRANSFORMATIONS AND THE NON-NEGATIVITY OF THEIR EULER FORMS") by Sefi Ladkani. Note that I changed the definition of $A\_+$ and $A\_-$ to match the definition of a Coxeter transformation I gave in my question. $\square$


Now let $H = kQ$ be the path algebra of the finite quiver $Q$ without oriented cycles. Let $S(1), \dots, S(n)$ be the simple modules, ordered in such a way that there is never an arrow in increasing direction, i.e. whenever $1 \leq i \leq j \leq n$ then there is no arrow $i \to j$ in $Q$. This can be achieved by labeling a sink of $Q$ with number $1$, a sink of the remaining quiver after killing $1$ with number $2$ and so on. The homological bilinear form $\left\langle - , -\right\rangle : \mathbb{Z}^n \times \mathbb{Z}^n \to \mathbb{Z}$ is given (since $H$ is hereditary) on dimension vectors by 
\begin{align\*}
\left\langle \underline{\dim} S(i), \underline{\dim} S(j)\right\rangle
& = \dim\_k \text{Hom}\_H(S(i), S(j)) - \dim\_k \text{Ext}\_H^1(S(i), S(j)) \\
& = \delta\_{ij} - \#\{\alpha: i \to j\}.
\end{align\*}
It is well known that the Coxeter transformation $\Phi\_H$ of $H$ is the unique Coxeter transformation of the homological bilinear form, in the sence of Ladkanis paper (i.e. $\left\langle x,y \right\rangle = - \left\langle y, \Phi\_{H}x \right\rangle$). Therefore, if we set $D$ the matrix with entries $(D)\_{ij} = \left\langle \underline{\dim}S(i), \underline{\dim}S(j)\right\rangle$ then we get $\Phi\_H = -D^{-1}D^t$. Furthermore, the matrix $A = D + D^t$ is a symmetric generalized Cartan matrix, and since $D$ has no nonzero entries at the upper triangle (remember the order of the simple modules), we get $A\_+ = D = A\_-$. We get the following:


**Theorem 2:** $\Phi\_H = C(A, \text{id})$.


*Proof:* By theorem $1$ we have $C(A, \text{id}) = -A\_+^{-1}A\_{-}^t = -D^{-1}D^t = \Phi\_H$. Compare also with Corollary $2.11$ in Ladkanis paper. $\square$


*Remark:* I'm pretty sure we can generalize this to every permutation. Also note that we can in this way represent every Coxeter transformation of a *symmetric* generalized Cartan matrix as the Coxeter transformation of a path algebra (since the numbers in the lower triangle show us exactly how many arrows we have to put between the indices).


**Added later**: There is still the question how properties of the algebra $H = kQ$ and properties of the associated Cartan matrix $A$ correspond. Remember that we have $A\_{ii} = 2$, $A\_{ij} = - \#\{\alpha: i \to j\}$ for $i > j$ and $A\_{ij} = A\_{ji}$ and that al arrows in $Q$ go in increasing direction. Let $q\_A$ be the quadratic form associated to $A$, i.e. $q\_A(x) = x^{t}Ax$ for $x \in \mathbb{Z}^n$. Let $q\_Q$ be the quadratic form of the quiver $Q$, i.e.
$$q\_Q(x) = \left\langle x, x \right\rangle = \sum\_{i = 1}^{n}x\_i^2 - \sum\_{\alpha \in Q\_1}x\_{s(\alpha)}x\_{t(\alpha)}.$$


**Theorem 3:** $q\_A = 2q\_Q$.


*Proof:* We have
\begin{align\*}
q\_A(x) & = x^tAx = \sum\_{i,j} x\_iA\_{ij}x\_j \\
& = \sum\_{i = 1}^{n}A\_{ii}x\_{i}^2 + \sum\_{i \neq j} A\_{ij}x\_ix\_j \\
& = \sum\_{i = 1}^{n}2x\_i^2 + \sum\_{i > j}2A\_{ij}x\_ix\_j \\
& = 2 \left( \sum\_{i = 1}^{n}x\_i^2 - \sum\_{\alpha \in Q\_1} x\_{s(\alpha)}x\_{t(\alpha)}\right) \\
& = 2q\_Q(x),
\end{align\*}
proving the claim. $\square$


Therefore, the quiver $Q$ is Dynkin (i.e. $H$ is representation finite), Euclidean (i.e. $H$ is tame) or wild (i.e. $H$ is wild) iff $A$ is positive definite, positive semidefinite (but not positive definite) or indefinite as a matrix.


