Finding explicit unimodular transformations for Chern-Simons K-matrices

Question

An invertible, symmetric matrix with integer entries, $K$, that encodes the braiding and statistics of an Abelian topologically ordered state, is equivalent to another such matrix, $K'$, if there exists an integer unimodular matrix, $W$, such that $$W^T K W = K'$$ Suppose I have $K$ and $K'$ and want to find such a $W$. Does there exist a general procedure for finding this? I can't find a particularly clever way of doing it. However, it seems that perhaps people know how to do this since in -

http://arxiv.org/abs/1404.6256 and https://arxiv.org/abs/1310.5708

they find large $W$ matrices (even $10\times 10$). I would like to know if this can only be done generally in certain cases (in systems with no topological order i.e. det$|K| = 1$) or of there is a known procedure for finding such a transformation for any $K$.

See also - https://math.stackexchange.com/questions/1800488/existence-of-unimodular-congruence-transformation-for-symmetric-integer-matrice

Answer 1

Necessary and sufficient condition for conjugacy: see Sylvester's law of inertia. In your notation, the two symmetric matrices $K$ and $K'$ are conjugate by a transformation $W^TKW = K'$ if and only if they have the same number of positive and negative eigenvalues (no null eigenvalues since both are nonsingular). The eigenvalues themselves need not be identical though.

Necessary and sufficient condition for $W$ to be unimodular: $K$ need not be unimodular, but the conjugacy relation imposes $$ [\det(W)]^2 = \frac{\det(K')}{\det(K)} $$ so $W$ is unimodular if and only if $$ \det(K) = \det(K') $$

Finding transformation $W$: Diagonalize $K$ and $K'$ as $$ K = U^T Q U,\;\;\;\;,U^TU = UU^T = I\;\;\;\;\;Q=Q^T=\text{Diag}(\lambda_K) $$ $$ K' = {U'}^T{Q'}{U'}, \;\;\;\;\;{U'}^T{U'} = {U'}{U'}^T = I\;\;\;\;\;{Q'}={Q'}^T= \text{Diag}(\lambda_{K'}) $$ where the diagonal matrices $Q$, $Q'$ list the eigenvalues $\lambda_K$, $\lambda_{K'}$ in order, say from smallest to largest. Further define $S = \text{Diag}(\sqrt{|\lambda_K|})$, ${S'} = \text{Diag}(\sqrt{|\lambda_{K'}|})$, and rewrite $$ Q = S^T D S,\;\;\;\;\; Q' = {S'}^T D {S'} $$ where $$D = \text{Diag}(\text{sign}(\lambda_K)) = \text{Diag}(\text{sign}(\lambda_{K'}))$$ is now a diagonal unimodular matrix, $[\det(D)]^2 = 1$. Substituting everything into the conjugacy relation brings it to the form $$ (S U W)^T D (S U W) = ({S'}{U'})^T D ({S'}{U'}) $$ and allows the identification, for instance, $$ S U W = {S'}{U'} $$ which eventually gives $$ W = U^T S^{-1}{S'}{U'} $$ If $\det(K) = \det(K')$ then also $\det(S) = \det(S')$ and so $\det(W) = \pm 1$, depending on $\det(U)$, $\det(U')$.

Note however that $W$ is not unique. Any ${\bar W} = U^T S^{-1}V{S'}{U'}$ with $V$ such that $V^TDV = D$ (for example $[V, D] = [V^T, D] = 0$ and $V^TV = I$) works just as well.