# Is it the Cartan, Coxeter or Gram matrix?

If you have a (numerical) Grothendieck group, it comes equipped with an Euler form and usually it has a Serre automorphism too. If the numerical Grothendieck group happens to be finitely generated (i.e. isomorphic to $\mathbb{Z}^n$ for some $n$), then choosing a basis you can express the Euler form (which is a bilinear form) and the Serre automorphism as a matrix in terms of this basis. In that case the terminology that is used in the literature is:

- Euler form $\langle-,-\rangle$
*Gram matrix**Cartan matrix*- Serre automorphism $\mathbb{S}$
*Coxeter matrix*

The Serre automorphism is sometimes called *Coxeter transformation* too.

If we denote the Cartan–Gram matrix for the Euler form by $A$ and the Coxeter matrix for the Serre automorphism by $C$, then there is a strong relationship between the two.

**Proposition** We have that
$$
C=-A^{-1}A^{\mathrm{t}}.
$$

*Proof.* Because $\langle-,-\rangle$ is nondegenerate, we have $\det A\neq 0$, and
$$
\begin{aligned}
\langle y,x\rangle
&=y^{\mathrm{t}}Ax \\
&=x^{\mathrm{t}}A^{\mathrm{t}}y \\
&=-x^{\mathrm{t}}A(-A^{-1}A^{\mathrm{t}})y \\
&=\left\langle x,(-A^{-1}A^{\mathrm{t}})y \right\rangle
\end{aligned}
$$
so we get the desired relationship by nondegeneracy. $\blacksquare$

To finish this blogpost, let's give the two easiest examples: curves and quivers.

**Example 1** The numerical Grothendieck group of a smooth projective curve $C$ is generated by $[k(p)]$ and $[\mathcal{O}_C]$. You can easily compute using Riemann–Roch that
$$
A=
\begin{pmatrix}
0 & -1 \\
1 & 1-g
\end{pmatrix}
$$
and
$$
C=
\begin{pmatrix}
1 & 2g-2 \\
0 & 1
\end{pmatrix},
$$
where $g$ is the genus of the curve.

**Example 2** The (numerical) Grothendieck group of a path algebra for an acyclic quiver has two obvious choices of basis: the simples, and the projectives. We will use the projectives here. Just note there is a strong connection between the Serre functor and Auslander–Reiten triangles, see e.g. theorem A of Reiten–Van den Bergh, Noetherian hereditary abelian categories.

If we take the Dynkin quiver $\mathrm{D}_4$ as an example, and we orient the arrows from left to right, then $$ A= \begin{pmatrix} 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$ and $$ C= \begin{pmatrix} 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & -1 & -1 & -1 \end{pmatrix}. $$