zero of vector field with index 0

I’m currently studying vector fields on surfaces in the $\mathbb R^3$ and I currently I am doing some reading on the index of zeros of vector fields, which got me wondering: Is it possible to find a vector field, which has a zero that has index 0? Or is this not possible?
The definition of the index I am using is the number of rotations a vector does while making one round around the zero on a small circle with the zero in the middle. So if the index of the zero is 0, this would mean the field would rotate clockwise and counter-clockwise and intentionally I think this would mean the vector field is not contiuous. Is that true?

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Completely possible. Consider
$$F(x, y) = [x^2, 0].$$

That’s got a zero at the origin whose index is zero. (Note, too, that this field is continuous, differentiable, etc.)

The zero at the origin of the field above is not isolated, however — it’s zero everywhere on the $y$ axis. A more “typical” example is
$$G(x, y) = (x^2 + y^2) [1, 0]$$
which is nonzero everywhere except the origin, but has index zero there.

Another example is provided by the vector field $V = (V_x, V_y)^T = (x^2, -y)^T$, which has one zero at $(0, 0)$. It is amusing to trace the rotation of $V$ as we traverse a small circle of radius $\epsilon$ centered at $0$. Following the path $\epsilon (\cos \theta, \sin \theta)$, where $\theta$ is taken counter-clockwise increasing, we see that $V = (\epsilon^2, 0)$ at $\theta = 0$, and rotates (while admittedly also changing in magnitude) in a clockwise direction to reach $V = (0, -\epsilon)$ at $\theta = \pi / 2$; then as $\theta$ continues to increase from $\pi / 2$ to $3 \pi / 2$, $V$ rotates in a counter-clockwise direction to reach a value of $(0, \epsilon)$ when $\theta = 3\pi / 2$, passing though the value $(\epsilon^2, 0)$ (again) as $\theta$ passes through $\pi$; then as $3 \pi / 2 \to \theta \to 2 \pi$, $V$ again rotates in a clockwise direction, returning finally to $V = (\epsilon^2, 0)$ when $\theta$ reaches $2 \pi$. The rotational direction of $V$ reverses itself at $\theta = \pi / 2, 3\pi / 2$; as such, $V$ never completes a full revolution as the $\epsilon$-circle is traversed; indeed, it is easy to see that the argument of $V$, if measured in the same coordinate system as $\theta$, never attains the value $\pi$. The index of $V$ at $(0, 0)$ is thus $0$.

It seems to me it is edifying to try and construct a phase portrait of $V$ near $(0, 0)$, either with pencil and paper, a computer graphics system, or simply in one’s imagination (where true phase portraits truly exist! ;-)). What is seen is that for $x \ge 0$, $V$ looks like a saddle, while for $x \le 0$, it has the appearance of a node. A related situation occurs in my answer to this question: Bifurcation and homoclinic orbits., where a zero with the same phase portrait occurs in the context of a bifurcation of a one-parameter family of ODEs in $\Bbb R^2$. There, a saddle and a node meet one another, and at the moment they “collide”, if you will, the exact vector field $V$ discussed here “occurs.” The index of the saddle is negative one; that of the node is one; and it is no accident that $1 + (-1) = 0$, the index of $V$, but I leave further unraveling of this riddle to my readers, who might like this wikipedia page.

Hope this helps. Cheerio,

and as always,

FIat Lux!!!