invariants in the case where $\R=\k$ is a result of Kac. The proof given in [@kraft-riedtmann §5.5] in the finite field case still work in the Frobenius $\k$-algebra case. It can actually be simplified as we now see. The proof reduces to the case where $Q'$ is obtained from $Q$ by inverting the orientation of one arrow $a:i\rightarrow j$ only. We then put

$$\H={\rm Hom}(\R^{\alpha(i)},\R^{\alpha(j)}), \hspace{1cm} \H'={\rm Hom}(\R^{\alpha(j)},\R^{\alpha(i)}),\hspace{1cm}\K=\bigoplus_{k\rightarrow l\in E\backslash\{a\}} {\rm Hom}_\R(\R^{\alpha(k)},\R^{\alpha(l)}).$$ We have $\rep^\alpha(Q,\R)=\K\oplus \H$ and $\rep^\alpha(Q',\R)=\K\oplus \H'$. The assertion (i) is then a consequence of the results of §\[Fouriergeneral\] with $X=\K$, $\M=\H$ and where we identified $\H'$ with $\H^\vee$ via the perfect pairing

$$\H'\times \H\rightarrow \R, \hspace{1cm} (f,g)\mapsto {\rm Tr}(g\circ f).$$ The above pairing is perfect as the multiplication $\R\times\R\rightarrow \R$ is perfect. Indeed, if $ab=0$ for all $a\in \R$, then $\lambda(ab)=0$ for all $a\in\R$ and so $\lambda$ vanishes on the ideal generated by $b$ which is impossible as the kernel of $\lambda$ does not contain non-trivial ideals of $\R$.

\(i) Assume that $Q'$ is a quiver obtained from $Q$ by changing the orientation of some arrows. Then

$$m_\alpha(\R Q)=m_\alpha(\R Q').$$

\(ii) We have

$$m_\alpha(\Pi_\R(Q))=m_\alpha(\R[\epsilon]Q).$$ \[finitecase\]

The assertion (ii) is a reformulation of Formula (\[mainfor\]). The assertion (i) 