If the torus is obtained from the unit square by identifying opposite edges, then you can use any constant (nonzero) vector field on the square.
The question is :
Prove that an orientable compact surface in R^3 has a differentiable vector field without singular points iff S is homeomorphic to a torus
I have proved one direction by using Poincare's theorem ,but I can't prove that if S is homemorphic to a torus then it has a vector field without singular points,please help.