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  "body": "\n## Definition\n\nLet $m, n \\ge 1$ and let $T(\\xi) \\in \\R^{\\tau}$ denote the vector of all [[minors|0BT9BYWVRY]] of all orders $1, \\dots, \\min(m, n)$ of a matrix $\\xi \\in \\R^{m \\times n}$.\nA function $f \\colon \\R^{m \\times n} \\to (-\\infty, +\\infty]$ is *polyconvex* if there is a [[convex function|07GR3TNW4H]] $g \\colon \\R^{\\tau} \\to (-\\infty, +\\infty]$ with\n$$f(\\xi) = g\\big(T(\\xi)\\big) \\quad \\text{for all } \\xi \\in \\R^{m \\times n}.$$\n\n## Remarks\n\nFor $m = n = 2$, $T(\\xi) = (\\xi, \\det \\xi) \\in \\R^5$.\nFor $m = n = 3$, $T(\\xi) = (\\xi, \\operatorname{cof} \\xi, \\det \\xi) \\in \\R^{19}$, with $\\operatorname{cof} \\xi$ the cofactor matrix.\n\nExample: for $m = n = 2$, $f(\\xi) = \\det \\xi$ is polyconvex, with $g$ linear, although $f$ is not convex.\n\nExample: $f(\\xi) = \\abs{\\xi}^2 + h(\\det \\xi)$ for $3 \\times 3$ matrices, with $h(d) = d^{-1}$ for $d \u003e 0$ and $h(d) = +\\infty$ for $d \\le 0$.\nIt is polyconvex, since $h$ is convex on $\\R$ with values in $(-\\infty, +\\infty]$.\n\nNon-example: for $m = n = 2$, $f(\\xi) = -(\\det \\xi)^2$ is not polyconvex.\nAlong $\\xi(t) = I + t \\, e_1 \\otimes e_1$ it equals $-(1 + t)^2$, which is not convex.\nSo $f$ is not [[rank-one convex|GK4P8G0PB2]], while every finite polyconvex function is.\n\nFor real-valued $f$, polyconvexity implies [[quasiconvexity|DVJ17KVWXP]], and the converse fails when $m, n \\ge 2$.\n",
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