An efficient algorithm to find a linear transformation between two ternary quadratic forms - Computer Science Stack Exchange most recent 30 from cs.stackexchange.com 2019-08-22T07:14:56Z https://cs.stackexchange.com/feeds/question/103171 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://cs.stackexchange.com/q/103171 2 An efficient algorithm to find a linear transformation between two ternary quadratic forms Dima Koshelev https://cs.stackexchange.com/users/99315 2019-01-21T16:35:44Z 2019-01-22T21:40:28Z <p>Let <span class="math-container">$\mathbb{F}_p$</span> be a prime finite field for <span class="math-container">$p &gt; 2$</span>. Consider two ternary quadratic forms <span class="math-container">$$Q_1\!: x^2 - a_1(t)y^2 - b_1(t)z^2,\\ Q_2\!: x^2 - a_2(t)y^2 - b_2(t)z^2$$</span> over the field <span class="math-container">$\mathbb{F}_p(t)$</span> of rational functions with coefficients from <span class="math-container">$\mathbb{F}_p$</span>. For simplicity let <span class="math-container">$a_1, a_2, b_1, b_2 \in \mathbb{F}_p[t]$</span> are polynomials without multiple roots and <span class="math-container">$a_1$</span>, <span class="math-container">$b_1$</span> (respectively <span class="math-container">$a_2$</span>, <span class="math-container">$b_2$</span>) have no common roots. </p> <p>Is there an efficient algorithm to find a linear transformation (over <span class="math-container">$\mathbb{F}_p(t)$</span>) between <span class="math-container">$Q_1$</span> and <span class="math-container">$Q_2$</span> if it exists?</p> <p>There is the theory that relates such forms and quaternion algebras (see, for example, <span class="math-container">$\S$</span>1.4 in Book of Gille, Szamuely - Central Simple Algebras and Galois Cohomology). For example, for any non-zero polynomial <span class="math-container">$f \in \mathbb{F}_p[t]$</span> and <span class="math-container">$p &gt; 2$</span> the following quadratic forms are isomorphic: <span class="math-container">$$Q_1\!: x^2 - y^2 - f(t)z^2,\\ Q_2\!: x^2 - y^2 - z^2$$</span> This is true, because <span class="math-container">$Q_1$</span> can be reduced to the quadratic form <span class="math-container">$Q_3\!: x^\prime y^\prime-(z^\prime)^2$</span> by the transformation <span class="math-container">$$x := x^\prime+\frac{y^\prime}{4f},\qquad y := x^\prime-\frac{y^\prime}{4f},\qquad z := \frac{z^\prime}{f}.$$</span> It is well known that any two conics (including <span class="math-container">$Q_2$</span>, <span class="math-container">$Q_3$</span>) over a finite field are isomorphic.</p>