Hessian-Matrix positive definite $\iff$ $a$ local minimum? It is commonly known that if $f$ is twice differentiable, $\nabla f(a) = 0$ and  $H_f(a)$ positive definite, $a$ is a local minimum.
So, in short: $H_f(a)$ positive definite $ \implies $ $a$ local minimum
But what about " $\impliedby $ " ? What can I say about the Hessian-Matrix, when I know $a$ is a local minimum?
Do cases exist, where $a$ is a local minimum and the Hessian-Matrix is not positive definite ? 
 A: Indeed such cases exist.  For example, consider the function $f$ on $\Bbb R^n$ given
$f(x_1, x_2, \ldots, x_n) = \sum_1^n x_i^{2m}, \tag{1}$
where $m \ge 2$.  We have
$\nabla f(x_1, x_2, \ldots, x_n) = 2m(x_1^{2m - 1}, x_2^{2m -1}, \ldots, x_n^{2m -1}); \tag{2}$
the only critical point of $f$ occurs at $0$, and it is evidently a global minimum.  But the matrix of second derivatives of $f$ is
$H(f) = 2m(2m - 1)\text{diag}(x_1^{2m - 2}, x_2^{2m -2}, \ldots, x_n^{2m - 2}), \tag{3}$
that is, the diagonal matrix whose $ii$ entry is $2m(2m -1)x_i^{m - 2}$; clearly
$H(f)(0) = 0. \tag{4}$
Thus $f(x_1, x_2, \ldots, x_n)$ provides an example of a function with a unique global minimum at $0$, but a vanishing Hessian there.  This is a case such as the OP sought in his question; evidently, there are many more such examples; constructing such is not too hard.
I think the best you can say is that $H(f)(a)$ must be positive semi-definite; it cannot have any negative eigenvalues, lest $f$ take on a value less that $f(a)$ in neighborhoods of $a$; it may, however, have eigenvalues equal to $0$, as the above example illustrates.
