# 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.

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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