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.

  • $\begingroup$ 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)? $\endgroup$ – Andrew D. Hwang Feb 4 '15 at 17:32
  • 1
    $\begingroup$ Thanks! It was a typo. But what is the elementary/naked way to show that pullback substitution? One needs change of variables formula? $\endgroup$ – Hulkster Feb 4 '15 at 17:45
  • $\begingroup$ An hour ago it was just false in certain cases; now it doesn't make sense anymore. $\endgroup$ – Christian Blatter Feb 4 '15 at 18:53
  • 1
    $\begingroup$ @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. $\endgroup$ – Hulkster Feb 4 '15 at 19:24

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.

  • $\begingroup$ (+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. :) $\endgroup$ – Andrew D. Hwang Feb 4 '15 at 21:45
  • $\begingroup$ @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. $${\bf A}(x,y)=\left({-y\over x^2+y^2}, \>{x\over x^2+y^2}\right)$$ $\endgroup$ – Hulkster Feb 6 '15 at 1:06
  • $\begingroup$ 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. $\endgroup$ – Hulkster Feb 6 '15 at 1:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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