“…Then by Lemma 4.1(b) there exists g ∈ SL(2, R) such that g t P g is the matrix of the form ± √ − det P diag(1, −1). Since det Q = det P − det S ≤ 0, and thanks to (16) and (17) above, it follows that the sum of the entries of g t P g cannot be zero, so by Lemma 4.2 there exists a matrix k ∈ SO(1, 1) such that for Q ′ = k t (g t Qg)k = (q ′ ij ), where either q ′ 11 or q ′ 22 vanishes. Since the subgroup SO(1, 1) ⊂ SL(2, R) preserves P ′ = g t P g, it follows from (17) that either q ′ 11 √ − det P = 0, or q ′…”