Is $\ln \det F$ a rank one convex function of $F$ 
Let $F=(F_{ij})$ be a matrix where $i,j \in \{1,2,...,d\}$. $F_{ij} \in \mathbb{R}$. And $F$ is taken to be invertible. Is $\ln \det F$ a convex function of $F$? Or is it a rank-one convex function of $F$?

Since I thought it is rank-one convex, but I now find ${\partial \ln \det F \over \partial F_{i\alpha } \partial F_{j \beta} }\xi_i \xi_j \eta_\alpha \eta_\beta  = (\xi F^{-1} \eta)^2$ may not be greater than $c |\xi|^2 |\eta|^2$. Is there some error in my calculation?
And if we suppose $F$ to be positive-definite, could we say $\ln \det F$ is a convex function of $F$?
 A: Let $S$ (resp.) $S^*$ be the set of symmetric (resp. symmetric invertible) matrices and let $C$ be an open  convex subset of $S^*$. Put $f:A\in C\rightarrow \log(|\det(A)|)$.
Proposition. $f$ is a concave function iff $C\subset S^+$, the set of symmetric $>0$ matrices or $C\subset S^-$, the set of symmetric $<0$ matrices.
Proof. $Df_A:H\in S\rightarrow \operatorname{tr}(HA^{-1})$ and $D^2f_A:(H,K)\in S^2\rightarrow -\operatorname{tr}(HA^{-1}KA^{-1})$. 
Then a necessary condition is, for every $H\in S\setminus 0$, $\operatorname{tr}( (HA^{-1})^2)\geq 0$. We may assume that $A^{-1}=\operatorname{diag}((\lambda_i))$ where $\lambda_i\not=0$; to prove one of both inclusions, it suffices to show that $\lambda_1\lambda_2>0$. Consider $H=\operatorname{diag}(\begin{pmatrix}0&1\\1&0\end{pmatrix},0_{n-2,n-2})$; thus $\operatorname{tr}((HA^{-1})^2)=2\lambda_1\lambda_2$ and we are done.
It remains to see why $f$ is strictly concave on $S^+$ and $S^-$; that comes from the fact that, if $A\in S^+$ or $S^-$ and $H\in S\setminus 0$, then the eigenvalues of $HA^{-1}$ are real and not all $0$; that implies $\operatorname{tr}((HA^{-1})^2)>0$.
EDIT (second version). Answer to sam.  Let $C\subset GL_d(\mathbb{R})$ be an open rank-one convex set. The condition for $f$ to be rank-one concave is: for every $A\in C$ and for every vectors $u,v$, $D^2f_A(uv^T,uv^T)=tr((uv^TA^{-1})^2)\geq 0$. Note that $rank(uv^TA^{-1})\leq 1$, that implies $spectrum(uv^TA^{-1})=\{tr(uv^TA^{-1}),0,\cdots,0\}$ and $tr((uv^TA^{-1})^2)=(tr(uv^TA^{-1}))^2\geq 0$.
Note that $GL_d^+(\mathbb{R})$ and $GL_d^-(\mathbb{R})$ are open rank-one convex sets (since the function $t\rightarrow \det(A+tuv^T)$ has at most one signum change). Finally, if $C=GL_d^+(\mathbb{R})$ or $C=GL_d^-(\mathbb{R})$, then $f$ is rank-one concave. 
