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Suppose that we have a matrix $A$ of a quadratic form $Q_A$ of signature $(n,1)$ and a matrix $B$ of a quadratic form $Q_B$ which also has signature $(n,1)$. Let $O(Q_A)$ be the orthogonal group that preserves the form $Q_A$ and let $O(Q_B)$ be the orthogonal group that preserves the form $Q_B$. Then $O(Q_A)$ and $O(Q_B)$ are isomorphic since $Q_A$ and $Q_B$ have the same signature. Let $G$ be a discrete subgroup of $O(Q_A)$, and suppose that $G$ is given by generators and relations. Does anyone know how to obtain a faithful representation of $G$ into $O(Q_B)$? Equivalently, how does one find generators of $G$ in $O(Q_B)$ that satisfy the defining relations for $G$?

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The isomorphism $O(Q_A)\to O(Q_B)$ is given by conjugation via a matrix $M$. Thus, you just conjugate generators of $G$ by the matrix $M$. Is this what you wanted to know? – Misha May 25 '13 at 22:26
Thanks Misha, yes, this is what I guessed. I'm working explicitly and got stuck in the computations. Will try again. Many thanks. – Lisa Carbone May 26 '13 at 8:49

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