Quadratic forms over symmetric $3\times3$ matrices - MathOverflow most recent 30 from http://mathoverflow.net2013-05-27T02:57:18Zhttp://mathoverflow.net/feeds/question/98197http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/98197/quadratic-forms-over-symmetric-3-times3-matricesQuadratic forms over symmetric $3\times3$ matricesDenis Serre2012-05-28T15:34:56Z2012-05-28T15:34:56Z
<p>This question is motivated by a technique called <em>compensated compactness</em> (CC), which was elaborated by L. Tartar and F. Murat for the analysis of PDEs. The foreword of CC is a problem about quadratic forms: given a cone $\Lambda$ in some real vector space $V$, <strong>determine</strong> all the quadratic forms $q$ on $V$ such that $q\ge0$ over $\Lambda$. In practice, $\Lambda$ is defined by homogeneous polynomial equation(s).</p>
<p>There is no general way to solve this problem, but the following case could be important:</p>
<blockquote>
<p>$V=Sym_3$ is the space of real symmetric matrices $S$, and $\Lambda$ is defined by $\det S=0$.</p>
</blockquote>
<p>Some elementary remarks:</p>
<ul>
<li>The set of such quadratic forms is itself a cone ${\mathcal K}$, but a closed convex one. We are really interested in the <em>extremal lines</em> of ${\mathcal K}$.</li>
<li>This cone is congruence-invariant: if $q\in{\mathcal K}$, and if $P\in GL_3({\mathbb R})$, then $q_P:S\mapsto q(P^TSP)$ is an element of ${\mathcal K}$.</li>
<li>The form $$q_0(S)=({\rm Tr} S)^2-4(s_{11}s_{22}-s_{12}^2+s_{33}s_{22}-s_{23}^2+s_{11}s_{33}-s_{13}^2)$$ is a non-trivial element of ${\mathcal K}$. In terms of the spectrum of $S$, it writes as $\sigma_1^2-4\sigma_2$, where $\sigma_j$ is the $j$-th elementary symmetric polynomial. It is non-negative over the real triplets $(\lambda,\mu,\nu)$ such that $\lambda\mu\nu=0$.</li>
</ul>