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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?

dinosaur
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2 Answers2

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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.

John Hughes
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  • So I guess my intuition was wrong with that. Is there a more analytic approach to calculating the index of a zero? I feel like counting the rotations along a circle is a rather inaccurate approach. – dinosaur Dec 20 '13 at 19:43
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    Actually, counting rotations around a small circle is rock solid. That's in part because your answers have to be multiples of $2\pi$. If you added up all the angle changes and got $710$ degrees, you could be pretty sure the real answer was $720$ degrees, for example. (Many computer programs that do such computations use exactly this idea.) – John Hughes Dec 20 '13 at 19:51
  • okay, I think I have to study this a bit more. Thank you! – dinosaur Dec 20 '13 at 19:55
  • Index is something that is well-defined when the zero of the vector field is isolated. In terms of the naive definition of index given here, one can't keep track of the "turning" of the vector field when the vector field vanishes. In the "example" given above one can't keep track of the `turning' when $x=0$. (When someone invokes cliches like "rock solid", one should immediately be suspicious). – Chris Judge May 30 '17 at 12:29
  • Chris is right: I should have said that the definition is rock solid for generic vector fields. (And the example I gave was by no means generic). I'll edit slightly. – John Hughes May 30 '17 at 12:45
  • The latter example is a good one. – Chris Judge May 30 '17 at 14:23
  • The first example is the gradient of the function $(x,y) \mapsto x^3/3$. The gradient of $x^3 -\epsilon xy^2$ is a small perturbation of the first example and has index -3. Without assuming isolated zeros, the index is not stable. – Chris Judge May 30 '17 at 14:44
  • Yep. And non-isolated zeroes are not "generic", in the sense that an arbitrarily small perturbation can make them isolated. (I admit that "generic" is a bit vague, but then again, I also suspect that OP doesn't know all about Sard's Theorem, so trying to be a lot more precise probably isn't useful.) – John Hughes May 30 '17 at 14:49
  • @ dinosaur. There is a more analytic way to describe the index. If the vector field is $F= (F_1,F_2)$, then the exterior derivative of the function $\arctan(F_2/F_1)$ defines a 1-form on a punctured disk neighborhood of the isolated zero. The index of the zero is $1/2\pi$ times the integral of this (closed) 1-form over any loop that generates the fundamental group of the punctured disk neighborhood and that has the standard counterclockwise orientation. See section 36.6 of the English edition of Arnold's book "Ordinary Differential Equations". – Chris Judge May 31 '17 at 19:21
  • That works fine if the "direction" (i.e., that "arctan" expression, suitably extended to handle cases where $F_1 = 0$) is a differentiable function of position, but when it's merely continuous, the "turning number" (or, better, the (algebraic) degree of the induced map from $S^1$ to $S^1$) definition still works, while the exterior derivative doesn't do anything for you at all. – John Hughes May 31 '17 at 19:43
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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!!!

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Robert Lewis
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