If the hessian evaluated at a critical point is positive (negative) definite, then we can conclude that it's a local minimum (maximum) there. If the hessian is indefinite (both negative and positive eigenvalues), then it's a saddle point.
What happens if the Hessian is positive SEMI-definite? According to https://en.wikipedia.org/wiki/Second_partial_derivative_test, it seems it is inconclusive whether the point is a max/min/saddle point, but I am wondering why this is the case. If the hessian was globally positive semi-definite, then the function is convex, and hence a local minima exists. But in the current local case, why can't we say that if the Hessian at a critical point is PSD, then it's a local minima?
This is the same as the case in one dimensional optimization of the second derivative being 0. Namely, 0 is not a local minimum of the function $x^3$, and an analogous multivariate example to work out is the function $f(x,y)=x^3+y^2$.