Let $f: K_1(0) \rightarrow \mathbb{R}^2$ be continuously differentiable, $\{z_1,..,z_m\}=f^{-1}(a)$ with a regular $a \in \mathbb{R}^2$. We choose $\epsilon$ small enough so that $f\vert_{\overline{U}_\epsilon(z_i)}$ are homeomorphisms.

Consider $\partial U_\epsilon(z_i)=:S_i$. Also view $S_i$ as the path surrounding the $\epsilon$-circle in the mathematically positive direction. Then $f(S_i)$ is a Jordan curve with $a$ on the inside of $f(S_i)$, since $f$ is a homeomorphism on $S_i$.

$f(S_i)$ has the same orientation as $S_i$, if $det ~J_f(z_i) > 0$ and the inverse orientation, if $det ~J_f(z_i)<0$.

Why does the determinant of the Jacobian decide the orientation of $f(S_i)$?

  • $\begingroup$ Two remarks: 1) Although you work in $\mathbb{C}$ it seems that you really mean $\mathbb{R}^2$. 2) You can only assume that $f$ is homeomorphic in a neighborhood of $z_i$ if $J_f(z_i)$ is non-singular. $\endgroup$ – Paul Frost Jul 20 '18 at 9:14
  • $\begingroup$ ad 2) We only look at regular a, yes. I added it to the question. ad 1) What difference does it make in this case? $\endgroup$ – akwa Jul 20 '18 at 9:36
  • $\begingroup$ Not a real difference, but using $\mathbb{C}$ can be misleading. I guess many people who see the notation $f : K_1^{\mathbb{C}}(0) \to \mathbb{C}$ expect that $f$ is complex differentiable but read that it is continuously differentiable in the sense of real analysis, and they will ask why you didn't use $\mathbb{R}^2$ just to avoid "stumbling over it". $\endgroup$ – Paul Frost Jul 20 '18 at 9:51
  • $\begingroup$ How do you define the orientation of a Jordan curve $C \subset \mathbb{R}^2$ parametrized by a path $u : [0,1] \to \mathbb{R}^2$? Via winding number? $\endgroup$ – Paul Frost Jul 20 '18 at 10:09
  • $\begingroup$ Ok, I understand, thanks! yes, exactly. $\endgroup$ – akwa Jul 20 '18 at 10:21

We shall use the fact that the winding number is homotopy invariant.

We only consider $S = S_1$. Let $L : \mathbb{R}^2 \to \mathbb{R}^2$ denote the linear isomorphism of $\mathbb{R}^2$ corresponding to $J_f(z_1)$; it is characterized by $\lim_{u \to 0}\frac{\lVert f(z_1 + u) - f(z_1) - L(u) \rVert}{\lVert u \rVert} = 0$. Let $r = \min\{ \lVert L(z)\rVert \mid \lVert z \rVert = 1 \}$. We have $r > 0$ because $L$ is an isomorphism. Then

$$\frac{\lVert f(z_1 + u) - f(z_1) \rVert}{\lVert u \rVert} \ge \frac{\lVert L(u) \rVert}{\lVert u \rVert} - \frac{\lVert f(z_1 + u) - f(z_1) - L(u) \rVert}{\lVert u \rVert} = \\ \lVert L(\frac{u}{\lVert u \rVert}) \rVert- \frac{\lVert f(z_1 + u) - f(z_1) - L(u) \rVert}{\lVert u \rVert} > \frac{r}{2}$$

for $0 < \lVert u \rVert \le \mu$ (we may of course assume that $\mu \le \epsilon_1$). This shows that there exists $0 < \rho \le \mu$ such that

$$(\ast) \phantom{xx} \lVert f(z_1 + u) - f(z_1) - L(u) \rVert < \lVert f(z_1 + u) - f(z_1) \rVert$$

for $0 < \lVert u \rVert \le \rho$ because otherwise we would obtain a sequence $(u_n)$ such that $0 < \lVert u_n \rVert \le \mu$ and $u_n \to 0$ and

$$\frac{\lVert f(z_1 + u_n) - f(z_1) - L(u_n) \rVert}{\lVert u_n \rVert} \ge \frac{\lVert f(z_1 + u_n) - f(z_1) \rVert}{\lVert u_n \rVert} > \frac{r}{2} .$$

A geometric consequence of $(\ast)$ is

$$(\ast\ast) \phantom{xx} \text{The line segment connecting } f(z_1 + u) \text{ and } f(z_1) + L(u) \text{ does not contain } \\ a = f(z_1)$$

for $0 < \lVert u \rVert \le \rho$.

Now let $u : [0,1] \to K_1(0)$ be the closed path $u(t) = z_1 + \epsilon_1e^{2\pi i t}$ (here $\mathbb{C}$ enters again, but only for convenient notation). It is homotopic in $K_1(0) \backslash \{ z_1 \}$ to the path $v(t) = z_1 + \rho e^{2\pi i t}$. Hence $f \circ u$ and $f \circ v$ are homotopic in $\mathbb{R}^2 \backslash \{ a \}$ and therefore have the same winding number. We claim that the paths $f \circ v$ and $w(t) = a + L(\rho e^{2\pi i t})$ are homotopic in $\mathbb{R}^2\backslash \{ a \}$. Define $H : [0,1] \times [0,1] \to \mathbb{R}^2\backslash \{ a \}, H(t,s) = (1-s)f(v(t)) + sw(t)$. By $(\ast \ast)$ this is well-defined (note $f(v(t)) = f(z_1 + \rho e^{2\pi i t}), w(t) = f(z_1) + L(\rho e^{2\pi i t})$). We see that $H_0 = f \circ v, H_1 = w$ and all $H_t$ are closed paths (i.e. $H_t(0) = H_t(1)$).

It therefore suffices to determine the winding number of $w$ which can be safely left as an exercise. Note that the winding number of $w$ around $a$ is the same as the winding number of $\omega(t) = L(\rho e^{2\pi i t})$ around $0$ and the latter has the same winding number as $\sigma(t) = L(e^{2\pi i t})$ (use again a homotopy).

The "philosophy" behind this proof is this: The differentiable maps are precisely those which can be nicely approximated by affine maps around each point. This implies that each sufficiently small circle around a point in which $f$ has a non-singular Jacobian is mapped to a Jordan curve which can be nicely approximated by the image of the circle under an affine map whose linear part is the derivative of $f$.

Remark concerning the winding number of the path $\sigma$:

The corresponding integral can be computed explicitly although it is somewhat tedious and requires some knowledge about the integration of trigonometric expressions of the form $\frac{1}{a \cdot sin^2t + b \cdot sint \cdot cost + c \cdot cos^2t}$. You will find that the the winding number of $\sigma$ is $\frac{detL}{\lvert detL \rvert}$.

An alternative approach is to use once more the homotopy invariance of the winding number and to construct a homotopy from $\sigma$ to one of the paths $a(t) = e^{2 \pi it }$ (if $detL > 0$) or $b(t) = e^{-2 \pi it }$ (if $detL < 0$).

Let $e_1 = (1,0), e_2 = (0,1)$ and let $R(\alpha)$ denote the rotation by the angle $\alpha$. There exists $\varphi$ such that $R(\varphi)(L(e_1)) = re_1$ with $r = \lVert L(e_1) \rVert > 0$. Then $\sigma$ is homotopic to $\sigma_1(t) = (R(\varphi)L)(e^{2\pi i t})$ (via $h(t,s) = (R(s\varphi)L)(e^{2\pi i t})$). Rotating is essential to avoid that $h(t,s) = 0$ for some $s$. $L_1 = R(\varphi)L$ is a linear isomorphism having the same determinant as $L$. We have $L_1(e_1) = re_1 = (r,0)$ and $L_1(e_2) = (w_1,w_2)$; this shows $det L = det L_1 = rw_2$. Let $\epsilon = sign(det L)$. Then $\sigma_1$ is homotopic to $a_\epsilon(t) = e^{\epsilon 2 \pi it }$ (via $g(t,s) = (1-s)\sigma_1(t) + sa_\epsilon(t)$; note that $g(t,s) \ne 0$ for all $t,s$).

  • $\begingroup$ Thank you very much! I have a few questions. 1) How exactly do we get (**) ? Everything up until then is clear. 2) I'm not sure how to construct the homotopy in the end for the two different cases. Could it be $h(s,t)=s e^{2\pi i t}+ (1-s)L(e^{2 \pi i t})$, and analogously for b(t)? I believe that would get us the results 1 and -1 for the winding number, which is what we want. $\endgroup$ – akwa Jul 23 '18 at 16:32
  • $\begingroup$ Ad $(**)$: By $(*)$ the distance from $f(z_1 +u)$ to $f(z_1) + L(u)$ is smaller than the distance from $f(z_1 + u)$ to $f(z_1 )= a$, therefore $a$ cannot be contained in the line segment. Concerning the homotopy I shall supplement my answer. Your "linear" homotopy $h(t,s)$ is in general unusable because $h(t,s) = 0$ is possible (e.g. for $L(x) = -x$ which has determinant $+1$) $\endgroup$ – Paul Frost Jul 23 '18 at 22:26
  • $\begingroup$ I understand. Right, I hadn't checked whether h(t,s)=0 yet. Thank you! $\endgroup$ – akwa Jul 24 '18 at 10:20
  • $\begingroup$ Note that the homotopy $g(t,s)$ doesn't meet $0$ because for $t = 0, \pi$ the points $\sigma_1(t), a_\epsilon(t)$ are both real and lie on the same side of $0$, and for all other $t$ they lie in the same open halfspace above or below $\mathbb{R} \times \{ 0 \}$. $\endgroup$ – Paul Frost Jul 24 '18 at 12:38

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