Mutation of valued quivers

Question

Mutations of valued quivers are defined in cluster algebras II, Proposition 8.1 on page 28. I have a question about the number $c'$. For example, let $a = 2, b=1, c=1$ and consider the quiver $Q$: $1 \overset{a}{\to} 2 \overset{b}{\to} 3$ and there is an arrow $1 \overset{c}{\to} 3$ in this quiver $Q$.

Now we mutate at $2$. According the formula $ \pm \sqrt{c} \pm \sqrt{c'} = \sqrt{ab}$. In this example, it is $ -1 + \sqrt{c'} = \sqrt{2}$. But $c' = (\sqrt{2}+1)^2$ is not an integer. Where do I make a mistake? Thank you very much.

Answer 1

I don't think your diagram is coming from a skew-symmetrizable matrix. Let $B$ be a skew-symmetrizable matrix and assume $Q = \Gamma(B)$ so your quiver is the diagram of $B$ as defined in Definition 7.3 of the linked paper. Then we must have $$B = \begin{bmatrix}0 & x & y \\ \frac{-1}{x} &0 & z \\ \frac{-1}{y} & \frac{-2}{z} &0 \end{bmatrix}$$ with $x,y,z > 0$. There also must exist a diagonal matrix $D = \mathrm{diag}(r,s,t)$ with $r,s,t > 0$ so that $$DB = \begin{bmatrix}0 & rx & ry \\ \frac{-s}{x} &0 & sz \\ \frac{-t}{y} & \frac{-2t}{z} &0 \end{bmatrix}$$ is skew-symmetric. So, $rx = \frac{s}{x}$, $ry = \frac{t}{y}$, and $sz = \frac{2t}{z}$ or equivalently $x^2 = \frac{s}{r}$, $y^2 = \frac{t}{r}$, and $z^2 = \frac{2t}{s}$. But this leads to a contradiction since $\left(\frac{y}{x}\right)^2 = \frac{y^2}{x^2} = \frac{t}{s}$, but also $z^2 = 2\left(\frac{t}{s}\right)$.

All the variables above aren't completely necessary. Since all entries of $B$ are integers we see $x = y = 1$ and $z = 1$ or $z = 2$. Then then follows $r = s = t$ and the contradiction is $z^2 = 2$.