huyichen said:"if a vector field has only nondegenerate zeros then the number of zeros is bounded"
With no idea how to show that without using Poincare-Hopf Theorem. Any proof possible without using any concept from algebraic topology?
Nondegenerate refers to the behavior of the vector field at each zero point. A nondegenerate zero indicates that the vector field is not changing direction or becoming zero at that point, but rather has a distinct, nonzero direction.
The number of nondegenerate zeros of a vector field is bounded by the dimension of the space in which it is defined. For example, a vector field in two-dimensional space can have at most two nondegenerate zeros, while a vector field in three-dimensional space can have at most three nondegenerate zeros.
No, a vector field cannot have an infinite number of nondegenerate zeros. As mentioned before, the number of nondegenerate zeros is bounded by the dimension of the space in which the vector field is defined. Therefore, a vector field can have at most a finite number of nondegenerate zeros.
Nondegenerate zeros play a crucial role in determining the behavior of a vector field. They represent points where the vector field is not changing direction or becoming zero, and as a result, they can act as attractors, repellers, or saddle points for the surrounding trajectories of the field.
The number of nondegenerate zeros of a vector field is related to its divergence through the Poincaré-Hopf index theorem. This theorem states that the sum of the indices of all nondegenerate zeros of a vector field is equal to the Euler characteristic of the space in which the field is defined. Therefore, the number of nondegenerate zeros can give insight into the overall behavior of the vector field and its divergence.