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Let $f:\mathbb{R}^n\to \mathbb{R}^n$ be a $C^2$-function and let $H=\left(\frac{\partial^2f}{\partial x_i \partial x_j}\right)_{1\le i,j\le n}$ be its Hessian matrix. Suppose I know that $ \det H(x_1,\ldots,x_n)\ge 0$ for all $x=(x_1,\ldots,x_n)$.

Does this have any geometric meaning for $f$?

e.g., when $n=1$, this means that $f$ is convex. This no longer holds for $n\ge 2$, but when $f$ is convex the Hessian determinant is certainly positive, so perhaps one could wonder if a weaker property holds.

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    @trony: I've read your post again, but I still didn't see much meaning from$\det H$. There are many situations in which \det H>0. I don't think \det H>0 can imply anything. In stead, checking the eigenvalues of $H$ might be meaningful.2011-09-07

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The determinant is the product of the eigenvalues. In two dimensions, the product being positive means that the two eigenvalues have the same sign, so $f$ is either concave or convex, and if the first derivative vanishes the point is an extremum. In higher dimensions, there's no such conclusion, since you could have any number of both positive and negative eigenvalues and have the determinant come out positive or negative.