Vector fields in the plane

Give the details for the claim that the degree of $vm_{x,r}$ does not depend on the choice of radius $r$ as long as we are inside a disk where $v$ only vanishes at $x$.

Setup: a vector field on $X \subset \mathbb {R}^{2}$ is a continuous function $v : X \rightarrow \mathbb {R}^{2}$. For a circle of radius $r$ about a point $x$, let $m_{x,r}(z) = x + rz$, giving a map $vm_{x,r} : S^1 \rightarrow \mathbb {R}^{2}$; if $v$ does not vanish on this circle, $\deg (vm_{x,r})$ is defined as usual (via the normalized map $z \mapsto vm_{x,r}(z)/\left|vm_{x,r}(z)\right|$ into $S^1$). A point $x$ where $v$ vanishes is called a singularity; it is isolated if there is a deleted neighborhood of $x$ where $v$ is nonzero. For an isolated singularity $x$, the index $i(x)$ is the degree of $vm_{x,r}$ for a small enough $r$ that the only zero inside the disk of radius $r$ about $x$ is $x$ itself.

Give an explicit construction of the homotopy of $vm_{2,1}$ to $z$ described in the computation of the index of the vector field $v(z) = z(z-2)$ at its singularity $2$.

Recall the analogous computation at the singularity $0$: $z - 2$ extends to a map of the unit disk into $\mathbb {R}^{2} \setminus \left\{ 0\right\}$, so $H(z,t) = tz - 2$ is a homotopy into $\mathbb {C} \setminus \left\{ 0\right\}$ between $z-2$ and the constant map $-2$; then $V(z,t) = zH(z,t)$, normalized, gives a homotopy between $v(z)$ and $-z$ as maps $S^1 \rightarrow S^1$, and $-z$ is homotopic to $z$ via $k(z,t) = e^{\pi i t} z$. For the singularity at $2$, restrict to the circle of radius $1$ about $2$ and look at $vm_{2,1}(z) = (2+z)z$; homotope the $(2+z)$ term to $2$ to get a homotopy of this to $2z$, which normalizes to a homotopy of $vm_{2,1}(z)$ to $z$.

Each part of the figure below depicts the vector field on a circle about an isolated singularity. Determine the index of each of the singularities.

Vector fields for Exercise 3.6.3.

Give an example of a vector field in the plane which has two singularities of index $1$ and one singularity of index $-1$.

Show that $f : S^1 \rightarrow S^1$, $f(z) = c$ (a constant), is homotopic to $g(z) = 1$.

Show that, if $f : S^1 \rightarrow \mathbb {C} \setminus \left\{ 0\right\}$ satisfies $f(z) = f_1(z) f_2(z)$ (complex multiplication of $f_1$ and $f_2$) and $f_2$ is homotopic to $f_3$, then $f$ is homotopic to $g(z) = f_1(z) f_3(z)$.

Show that, if $f(z) : S^1 \rightarrow \mathbb {C} \setminus \left\{ 0\right\}$ extends over the unit disk to $F : D^2 \rightarrow \mathbb {C} \setminus \left\{ 0\right\}$, then $f(z)$ is homotopic to $g(z) = 1$.

Give an explicit homotopy between $f(z) = 2z - 5$ and $g(z) = 1$ as maps from $S^1$ to $\mathbb {C} \setminus \left\{ 0\right\}$.

Suppose $v(z) = v_1(z) v_2(z)$ (complex multiplication) is a vector field with an isolated singularity at $x$. Suppose $v_1(x) = 0$, $v_2(x) \neq 0$, and the only singularity of $v$ inside $B(x,r)$ is at $x$. Show that the index of $v$ at $x$ as computed on the circle of radius $r$ about $x$ is the same as the index of $v_1$ as computed on the same circle.

Suppose $v(z) = (z-x_1) \cdots (z-x_n)$ is given as a factored polynomial. Show that $v$ has isolated singularities $x_1, \ldots , x_n$ and that each has index $1$, so the index of the vector field is $n$.

Verify that $r : S^1 \times I \rightarrow D^2$, $r(z,t) = tz$, is a quotient map, and that the induced map $\bar{K}' : D^2 \rightarrow \mathbb {R}^{2} \setminus \left\{ 0\right\}$ gives an extension of $f$, in the context of the proof of the following lemma.

Lemma 3.6.1: Let $f : S^1 \rightarrow \mathbb {R}^{2} \setminus \left\{ 0\right\}$ be a continuous map whose composition with $u(z) = z/\left|z\right|$ has degree $0$. Then $f$ extends to a continuous map from $D^2$ to $\mathbb {R}^{2} \setminus \left\{ 0\right\}$.

In the proof, a homotopy $K : I \times I \rightarrow (\mathbb {R}^{2} \setminus \left\{ 0\right\} )$ between $fp$ (where $p(t) = e^{2\pi i t}$) and a constant map induces, since $S^1$ is a quotient of $I$ via $p$, a homotopy $\bar{K} : S^1 \times I \rightarrow \mathbb {R}^{2} \setminus \left\{ 0\right\}$ with $\bar{K}(z,1) = f(z)$.

Verify the claim made in the proof of the following proposition, for the base case $k=1$: that the degree of $vm_{x,r}$ is the same as the degree of $vm_{z_1, r_1}$, using a homotopy between the map $m_{x,r}$ and the map $m_{z_1, r_1}$ onto a small circle around the singularity $z_1$.

Proposition 3.6.2: Suppose $v$ is a vector field defined on the disk $B(x,r)$ centered at $x$ which has isolated singularities at $z_1, \ldots , z_k$. The degree of the map $vm_{x,r}$ is the sum of the indices of the singularities; that is, the index of the vector field may be computed by computing the degree using the vector field on the boundary circle.