Index Theory: Can a closed curve around a single unstable fixed point have index $0$?

Question

I know that a closed curve containing zero fixed points has index $0$. Is the converse also always true?

Answer 1

In index theory, a closed curve around a fixed point is $1$ if the fixed point is a node, spiral, or center and $-1$ if it is a saddle and a curve around multiple fixed points in the sum of the indices.

To see that a saddle is $-1$, consider a vector field $\mathbf{V}(\theta)$. Then $$ \Theta(\theta) = \tan^{-1}\frac{\mathbf{V}_2(\theta)}{\mathbf{V}_1(\theta)} $$ With a saddle, $\mathbf{V} = (x,-y)$ or $\mathbf{V} = (x,y)$ so $\Theta = -\theta$ and $\Theta = 0.5\pi - \theta$, respectively. Thus, a saddle has index $-1$.

Answer 2

The converse is not true in general. The index of a curve which encloses multiple fixed points is equal to the sum of the indices of curves which would enclose each fixed point on its own.

For example, if the curve encloses a stable fixed point and a saddle point, it will have $\text{index} = -1 + 1 = 0$, yet will enclose two fixed points.