Ngô Quốc Anh

July 15, 2010

The winding number is a special case of the Brouwer degree

Filed under: Giải tích 7 (MA4247) — Ngô Quốc Anh @ 18:25

In the classical complex analysis, the winding number of a plane curve \Gamma with respect to a point a \notin \mathbb C \setminus \Gamma is defined by

\displaystyle w(\Gamma ,a) = \frac{1}{{2\pi i}}\int_\Gamma {\frac{{dz}}{{z - a}}} .

Let \Omega \subset \mathbb R^n be open and bounded. Let f : \Omega \to \mathbb R^n and p \notin f(\partial \Omega). Notation J_f(a) denotes the Jacobian of f evaluated at a. It is well-known that the Brouwer Degree Theory, usually denoted by \deg, is constructed for continuous function, C^0(\overline\Omega)-class, via the following steps

For the C^1(\overline\Omega)-class: We assume J_f(f^{-1}(p)) \ne 0 then

\displaystyle\deg (f,\Omega ,p) = \begin{cases} \sum\limits_{x \in {f^{ - 1}}(p)} {{\rm sgn} {J_f}(x)} , & {f^{ - 1}}(p) \ne \emptyset ,\\ 0, & {f^{ - 1}}(p) = \emptyset.\end{cases}

For the C^2(\overline\Omega)-class: In case we want to remove the condition J_f(f^{-1}(p)) \ne 0, we then define

\displaystyle\deg (f,\Omega ,p) = \deg (f,\Omega ,p')

where p' is any regular value of f sufficiently closed to p in the sense that

\displaystyle\left\| {p - p'} \right\| < {\rm dist}(p,f(\partial \Omega )).

For the C^0(\overline\Omega)-class: In this case, we define

\displaystyle\deg (f,\Omega ,p) = \deg (g,\Omega ,p)

where g \in C^2(\overline\Omega) is sufficiently closed to f in the sense that

\displaystyle\left\| f-g \right\| < {\rm dist}(p,f(\partial \Omega )).

Now we prove the following fact

Theorem.  Let B(0,1)\subset \mathbb C be the unit ball and \Gamma = \partial B(0,1). Assume f : \overline{B(0,1)} \to \mathbb C is a C^1 function and a \notin f(\Gamma). Then

\displaystyle\deg (f,B(0,1),a) = \frac{1}{{2\pi i}}\int_{f(\Gamma )} {\frac{{dz}}{{z - a}}} .

Proof. It is sufficient to prove in the case when

a \notin f(\text{critical points of } f).

Let us assume

f^{-1}(a)=\{z_1,z_2,...,z_k\}.

Then we need to show that

\displaystyle\deg (f,B(0,1),a) = \sum\limits_{i = 1}^k {\text{sgn} {J_f}({z_i})} .

Take \varepsilon>0 small enough such that the \overline V_i‘s are disjoint, where V_i=B(z_i, \varepsilon). It is clear that

\displaystyle\text{sgn} {J_f}(z) = \text{sgn} {J_f}({z_i}), \quad \forall z \in {V_i} \subset {\overline V _i} \subset B(0,1)

and the restriction of f to \overline V_i is a homeomorphism for all k.

Put S_i =\partial V_i then f(S_i) is a Jordan curve such that a lies in its interior region. Besides, f(S_i) has the same orientation as S_i if J_f(z_i)>0 and the opposite orientation if J_f(z_i)<0.

Now set

\displaystyle U = \overline {B(0,1)} \backslash \bigcup\limits_{i = 1}^k {{V_i}} .

Then |f(z)-a|>\alpha in U for some \alpha>0. We can divide U into small rectangles R_j such that |f(z)-f(w)|<\alpha on each latex R_j. Since the image f(\partial(R_j \cap V)) does not wind around a, we have

\displaystyle w(f(\partial ({R_j} \cap V)),a) = 0

and summing over all R_j yields

\displaystyle\int_{f(\Gamma )} {\frac{{dz}}{{z - a}}} = \sum\limits_{i = 1}^k {\int_{f({S_i})} {\frac{{dz}}{{z - a}}} } .

Since the orientation of f(S_i) is determined by J_f(z_i), f(S_j) winds exactly once around a. Thus, we obtain

\displaystyle {\int_{f({S_i})} {\frac{{dz}}{{z - a}} = \text{sgn} {J_f}({z_i})} }.

The proof now easily follows.

3 Comments »

  1. Sweet! I will might use this in my thesis🙂
    But it seems that one needs Jordan curve Theorem in this proof. Also I haven’t found any good explanations for that “Jacobian sign and orientation connection”.

    Any way, congratulations for your Ph.D!
    I too self-study GR but on elementary level: I just finished Woodhouse’s book General Relativity.

    Comment by JL — February 26, 2013 @ 12:49

    • I forgot to mention, there is a minor typo on the last line: right hand side should state 2\pi i J_f(z_i).

      Comment by JL — February 26, 2013 @ 18:17

    • Thanks JL, I wish you all the best.

      Comment by Ngô Quốc Anh — February 26, 2013 @ 21:35


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