The Laurent Phenomenon

1 Matrix mutation

Definition 1.1
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1) An \(n\times n\) matrix \(B = (b_{i,j})\) with (say) rational entries is called skew-symmetrisable if there exists a diagonal matrix \(D = {\rm diag}(d_1, \ldots , d_n)\) with \(d_i \in \mathbb {Z}_{{\gt}0}\) such that \(DB\) is skew-symmetric, i.e.

\[ d_ib_{i,j} = -d_jb_{j,i}, \qquad \forall i,j \in [1,n]. \]

2) A mutation matrix is a skew-symmetrisable matrix with integer entries.

Definition 1.2
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Let \(B\) be an \(n \times n\) skew-symmetrisable matrix. The mutation of \(B\) in direction \(k \in [1,n]\) is the matrix \(\mu _k(B) = (b'_{i,j})\) defined by

\[ b_{i,j}' = \begin{cases} -b_{i,j} & \text{if } i = k \text{ or } j = k, \\ b_{i,j} + b_{i,k}b_{k,j} & \text{if } b_{i,k} {\gt}0 \text{ and } b_{k,j} {\gt} 0, \\ b_{i,j} - b_{i,k}b_{k,j} & \text{if } b_{i,k} {\lt}0 \text{ and } b_{k,j} {\lt} 0, \\ b_{i,j} & \text{otherwise.} \end{cases} \]
Lemma 1.3
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Let \(B\) be an \(n \times n\) skew-symmetrisable matrix, and fix \(k \in [1,n]\). Then the following hold:

  1. \(\mu _k(B)\) is skew-symmetrisable, with the same diagonal matrix \(D\).

  2. \(\mu _k(\mu _k(B)) = B\).

  3. If \(B\) is skew-symmetric, then \(\mu _k(B)\) is skew-symmetric.

Proof ▶

  1. Fix \(k \in [1,n]\), and let \(i,j \in [1,n]\) be arbitrary. By definition of \(\mu _k(B)\) and the skew-symmetry of \(B\), we have

    \[ b_{j,i}' = \begin{cases} -b_{j,i} & \text{if } i = k \text{ or } j = k, \\ b_{j,i} - b_{j,k}b_{k,i} & \text{if } b_{i,k} {\gt}0 \text{ and } b_{k,j} {\gt} 0, \\ b_{j,i} + b_{j,k}b_{k,i} & \text{if } b_{i,k} {\lt}0 \text{ and } b_{k,j} {\lt} 0, \\ b_{j,i} & \text{otherwise.} \end{cases} \]

    Suppose we are in the first case, i.e. \(i=k\) or \(j = k\). Then

    \[ d_i b_{i,j}' = d_i (-b_{i,j}) = - d_j (- b_{j,i}) = - d_j b_{j,i}'. \]

    Now suppose we are in the second case, i.e. \(b_{i,k} {\gt} 0\) and \(b_{k,j} {\gt} 0\). Then

    \begin{align*} d_i b_{i,j}’ & = d_i (b_{i,j} + b_{i,k}b_{k,j})\\ & = - d_j b_{j,i} - d_k b_{k,i}b_{k,j} \\ & = - d_j b_{j,i} - d_k b_{k,i}b_{k,j} \\ & = - d_j b_{j,i} + d_j b_{k,i}b_{j,k} \\ & = - d_j (b_{j,i} - b_{j,k}b_{k,i}) \\ & = - d_j b_{j,i}’. \end{align*}

    The remaining two cases are similar, and we omit them.

  2. This is a direct computation.

  3. This follows from 1.