Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I found a term "winding number of vector field with respect to another vector field" in a paper without definition. Because my paper I am reading is talking about the surface, so I don't know if I can use the definition of winding number of vector field like the case in plane? And for the case of surface, why we need to add another vector field for respect? Thanks for any help!

share|cite|improve this question
Do you know anything about the zeros of these vector fields? These can make the discussion a little tricky. – Sam Lisi Mar 26 '13 at 20:55
what paper would that be? – Will Jagy Mar 26 '13 at 21:21

Think first of the winding of a vector field $v$ on the plane. In order to define this, you take $v/ |v|$ and look at what this does as a map to the circle. In doing this, you are secretly identifying a vector field on the plane that doesn't wind at all (eg. for which the map to $S^1$ is constant). In the case of the plane, this is just a constant vector field, which is a well-defined notion.

Now, suppose you have two vector fields on a surface $\Sigma$, call them $v_1$ and $v_2$. Let's make the assumption that the two vector fields vanish at the same set of points. Then, we can write $v_1 = A v_2$, where at each point $z\in \Sigma$, $A(z)$ is a linear map from $T_z \Sigma \to T_z\Sigma$ and has positive determinant in an oriented basis (its existence and non-degeneracy use the assumption that the two vector fields vanish together). Now, by a polar decomposition, you can write this as some positive definite matrix times a unitary complex matrix, i.e. times a rotation matrix. You can identify rotation matrices with $S^1$, and this gives you the map you were looking for.

Here's an illustration of why we want a reference vector field. Consider the surface to be the annulus in the complex plane with coordinates $z = x+iy$ given by $1/2 \le |z| \le 2$. Now look at the radial vector field $x\partial_x + y \partial_y$. If you look at the winding number of this with respect to the vector field $\partial_x$, you get $1$ (maybe $-1$, I haven't checked the sign). If, however, you change to polar coordinates, your annulus looks like some $I \times S^1$ for $I$ an interval, with coordinates $s+it$, with $t$ an angular variable in $S^1$. Then, it is natural to take a reference vector field by $\partial_t$. It's an easy exercise to check that my vector field is a multiple of $\partial_s$, and thus has winding number $0$ with respect to this vector field.

Another way of saying what I am saying: on an oriented surface, a non-vanishing vector field gives a trivialisation of the tangent bundle since the tangent bundle is a complex line bundle. The reference vector field is just a choice of trivialisation.

In what I have said, some care needs to be taken when the reference vector field vanishes. In that case, I think that you want to cut out a small neighbourhood of the zero of the reference vector field, and study the winding number as above on the complement. You then probably want to look at the index of these vector fields at that point. I haven't thought about this aspect enough.

share|cite|improve this answer
@ Sam Lisi Thanks for your answer. Is the discussion above the definition of winding number of vector field with respect to another vector filed? By my understanding, it's more general than the definition of index of vector field at some isolated zero, is it true? – mapping Mar 27 '13 at 9:05
It's not more general... what I described above really uses the fact that the structure group of the tangent bundle to an oriented surface can be reduced to U(1). This is not true for a general manifold. In some sense, what I describe can be generalized to a discussion of Maslov classes of Lagrangians in symplectic manifolds... but that's much fancier than what you are interested in. – Sam Lisi Mar 27 '13 at 12:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.