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Let $B(x_0,y_0)$ be an open unit disk. Assume that $F(x,y)=(f_1 (x,y),f_2 (x,y)):\overline{B(x_0,y_0)}\rightarrow \mathbb{R}^2$ is a diffeomorphism. Assume also that $F(x,y) \ne (s_0,t_0) $, for all $(x,y)\in \partial B(x_0,y_0)$.

What is the elementary way to evaluate the following result?

$$\frac{1}{2\pi}\int_{\partial B(x_0,y_0)}\frac{(f_1-s_0) df_2-(f_2-t_0)df_1}{(f_1-s_0) ^2+(f_2-t_0) ^2}=sign \det F'(x_0,y_0)=±1. $$

Can that be shown by using Green's theorem and then change of variables formula? Since $F$ is a diffeomorphism, due to inverse function theorem $\det F'(x,y) \ne 0$, for all $(x,y) \in B(x_0,y_0)$, and by Jordan curve theorem $F(\partial B(x_0,y_0))$ is a Jordan curve. Therefore inner domain is simply connected and we can use Green's theorem. But I'm not sure how to calculate that integral.

Hulkster
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  • As you're probably aware, if $$ \omega = \frac{x_{1}, dx_{2} - x_{2}, dx_{1}}{2\pi(x_{1}^{2} + x_{2}^{2})}, $$ (formally, $d\theta/(2\pi)$ on $\mathbf{R}^{2}\setminus{(0,0)}$), then your integral is $$ \int_{\partial B(x_{0}, y_{0})} F^{*}\omega = \int_{F(\partial B(x_{0}, y_{0}))} \omega, $$ a winding number, which can be $0$ (by Green's theorem, for example) if $F(\partial B)$ doesn't wind around the origin. In your integrand, do you mean "$f_{i} - f_{i}(x_{0},y_{0})$" instead of "$f_{i}$" (and "$F(x,y) \neq (0,0)$" is spurious)? – Andrew D. Hwang Feb 04 '15 at 17:32
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    Thanks! It was a typo. But what is the elementary/naked way to show that pullback substitution? One needs change of variables formula? – Hulkster Feb 04 '15 at 17:45
  • An hour ago it was just false in certain cases; now it doesn't make sense anymore. – Christian Blatter Feb 04 '15 at 18:53
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    @Blatter: Thank you for your laconic comment. But if you have same relevant advices, please tell me. I belive that you do know what I'm trying to ask; winding number of a diffeomorphism with respect to a unit circle. – Hulkster Feb 04 '15 at 19:24

1 Answers1

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Assume that $B$ is a disk with center ${\bf z}_0$ in the $(x,y)$-plane, that $$f:\quad \bar B\to{\Bbb R}^2, \qquad (x,y)\to(u,v):=\bigl(f_1(x,y),f_2(x,y)\bigr)$$ is a diffeomorphism, and that ${\bf 0}\notin f(\partial B)$. Then $$N:={1\over2\pi}\int_{\partial B}{f_1\>df_2-f_2\>df_1\over f_1^2+f_2^2}= \left\{\eqalign{{\rm sgn}\bigl(J_f({\bf z}_0)\bigr)&\qquad\bigl({\bf 0}\in f(B)\bigr)\cr 0\qquad&\qquad\bigl({\bf 0}\notin f(B)\bigr) \cr}\right.$$ Proof. Consider the vector field $${\bf A}(u,v):=\left({-v\over u^2+v^2}, \>{u\over u^2+v^2}\right)=\nabla\arg(u,v)$$ in the punctured $(u,v)$-plane. I claim that $$N={1\over2\pi}\int_{f(\partial B)}{\bf A}\cdot d{\bf w}\ .\tag{1}$$ Subproof. Let $$t\mapsto\bigl(x(t),y(t)\bigr)\qquad(0\leq t\leq T)$$ be a parametrization of $\partial B$. Then $$t\mapsto {\bf w}(t):=\bigl(f_1\bigl(x(t),y(t)\bigr), f_2\bigl(x(t),y(t)\bigr)\bigr)\qquad (0\leq t\leq T)$$ is a parametrization of $f(\partial B)$. It follows that $$\eqalign{&\int_{f(\partial B)}{\bf A}\cdot d{\bf w} =\int_0^T\left({-f_2\over f_1^2+f_2^2}\bigg|_{{\bf w}(t)}\dot u(t)+ {f_1\over f_1^2+f_2^2}\bigg|_{{\bf w}(t)}\dot v(t) \right)\ dt\cr &=\int_0^T \left({-f_2\over f_1^2+f_2^2}\bigl(f_{1.1}\dot x(t)+f_{1.2}\dot y(t)\bigr)+ {f_1\over f_1^2+f_2^2}\bigl(f_{2.1}\dot x(t)+f_{2.2}\dot y(t)\bigr)\right)\>dt \cr &=\int_{\partial B}{f_1\>df_2-f_2\>df_1\over f_1^2+f_2^2}\quad.\qquad\square\cr}$$ Note that $${\rm curl}\>{\bf A}(u,v)\equiv0\ .$$ When ${\bf 0}\notin f(B)$ we can apply Green's theorem in $(1)$ and obtain $$N={1\over2\pi}\int_{f(\partial B)}{\bf A}\cdot d{\bf w}={1\over2\pi}\int_{f(B)}{\rm curl}\>{\bf A}(u,v)\>{\rm d}(u,v)=0\ .$$ When ${\bf 0}=f({\bf p})\in f(B)$ we introduce $B':=B\setminus B_\epsilon({\bf p})$ and apply Green's theorem to $f(B')$. We then get $$N={1\over2\pi}\int_{f(\partial B)}{\bf A}\cdot d{\bf w}={1\over2\pi}\int_{f(\partial B')}{\bf A}\cdot d{\bf w}+{1\over2\pi}\int_{f(\partial B_\epsilon)}{\bf A}\cdot d{\bf w}\ .$$ Since the first term on the right hand side vanishes we have $$N={1\over2\pi}\int_{f(\partial B)}{\bf A}\cdot d{\bf w}={1\over2\pi}\int_{f(\partial B_\epsilon)}{\bf A}\cdot d{\bf w}\ .$$ Now $f(\partial B_\epsilon)$ is a tiny ellipse around ${\bf 0}$, and a suitable limiting argument should then prove that $N= {\rm sgn}\bigl(J_f({\bf z}_0)\bigr)$ in this case.

Christian Blatter
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  • (+1) Incidentally, I'm guessing the new notation means $F(x_{0}, y_{0}) = (s_{0}, t_{0})$, so the condition "$F(x, y) \neq (s_{0}, t_{0})$ on the boundary" is automatic, and the boundary winds $\pm1$ times. Particularly, the question is in fact sensible, and you've answered it. :) – Andrew D. Hwang Feb 04 '15 at 21:45
  • @Blatter. Can you clarify your methods?

    Since circle $\partial B({\bf z}0)$ is positively oriented, we know that the orientation of Jordan curve $f(\partial B({\bf z}_0))$ is ${\rm sgn}\bigl(J_f({\bf z}_0)\bigr)$, and if ${\bf 0}\in f(B)$, then the orientation of $f(\partial B\epsilon ({\bf 0}))\subset f( B({\bf z_0}))$ is also ${\rm sgn}\bigl(J_f({\bf z}_0)\bigr)$. If the vector field $\bf{A}$ is trivial, i.e. $${\bf A}(x,y)=\left({-y\over x^2+y^2}, >{x\over x^2+y^2}\right)$$

     – Hulkster Feb 06 '15 at 01:06
  • it's easy to conclude from Green's theorem that $${1\over2\pi}\int_{f(\partial B_\epsilon (\bf 0))}{\bf A}(x,y)\cdot (dx,dy)={\rm sgn}\bigl(J_f({\bf z}0)\bigr).$$ But, if the vector field $\bf{A}$ is not trivial, as $${\bf A}(f_1,f_2)=\left({-f_2 (x,y)\over f_1^2 (x,y)+f_2^2 (x,y)}, >{f_1 (x,y)\over f_1^2(x,y)+f_2^2(x,y)}\right)$$ How can you calculate the integral $${1\over2\pi}\int{f(\partial B_\epsilon (\bf 0))}{\bf A}(f_1,f_2)\cdot (df_1,df_2)=\space ????.$$ Sorry if I stress this too much, but I really don't understand, and that calculation is at the heart of my original question. – Hulkster Feb 06 '15 at 01:07