Vector-field index theory

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.

Suppose $S_1, S_2$ are equivalent differential structures on $M$. Show that $f : M \rightarrow \mathbb {R}$ is differentiable in terms of $S_1$ iff it is differentiable in terms of $S_2$. Thus equivalent differential structures have exactly the same differentiable maps to $\mathbb {R}$.

Draw a picture to show the effect on the vector field $v(z) = z^2$ on the unit circle when we reflect $v(z)$ through the line perpendicular to the line from the origin to $z$.

The vector field v(z) = z^2.

Your picture should show the images of the vectors pictured in the figure above. Verify from your picture that the degree of the new vector field is $0$.

An orthogonal linear transformation from $S^1$ to $S^1$ is the composition of the reflection $r(z) = \bar{z}$ with multiplication by a number in $S^1$ (i.e. $g(z) = az$, where $a \in S^1$) or a map of the form of $g$ itself. Show that the degree of $r$ is $-1$. Show that the degree of an orthogonal linear transformation is $\pm 1$. Determine the inverses of $g$ and $r$, and show that they have the same degree as the original map $g$ or $r$.

Prove that any continuous map $f : S^1 \rightarrow S^1$ is homotopic to $z^n$ for some $n$.

Show that, if $f : S^2 \rightarrow \mathbb {R}^{3}$ is continuous, then there must be some $x \in S^2$ so that $f(x) = kx$, where $k \in \mathbb {R}$ and $x$ is thought of as a point in $\mathbb {R}^{3}$.

Give an example of a vector field on the torus $T$ with no singularities.

Recall the differential structure on the projective plane $P$ induced via the covering map $p : S^2 \rightarrow P$: there is a cover $\left\{ U_i\right\}$ of $P$ so that $p^{-1}(U_i) = U_{i1} \cup U_{i2}$, $p \vert U_{ij} : U_{ij} \rightarrow U_i$ a homeomorphism with inverse $q_{ij}$, and $T(x) = -x$ a diffeomorphism of $S^2$ interchanging $U_{i1}$ with $U_{i2}$; taking diffeomorphisms $h_{i1} : U_{i1} \rightarrow \bar{U}_i$ together with $h_{i2} = h_{i1} T$, the differential structure on $P$ uses the maps $\bar{h}_i : U_i \rightarrow \bar{U}_i$, $\bar{h}_i = h_{i1} q_{i1} = h_{i2} q_{i2}$.

Prove Theorem 3.7.5 as outlined below.

Theorem 3.7.5: The index of a vector field on a compact connected surface is given by the Euler characteristic.

Outline: it was already shown directly that this holds for $S$ (Theorem 3.7.1, index $2$), $T$ (Theorem 3.7.2, index $0$), and $P$ (Theorem 3.7.3, index $1$), each matching the Euler characteristic of the surface. Lemma 3.7.4: Suppose $M, N$ are surfaces so that the index of any vector field on each is given by the Euler characteristic. Then the same is true of $M \# N$. (Sketch of Lemma 3.7.4: form $M \# N$ by removing a disk from each of $M, N$ and gluing the boundary circles; a vector field $v$ on $M \# N$ determines vector fields $v_M, v_N$ on $M, N$ by extending over the removed disks via radial damping, contributing one new singularity each. If $d_M, d_N$ denote the degrees of $v_M, v_N$ on their boundary circles, then $I(v_M) + d_M = \chi (M)$, $I(v_N) + d_N = \chi (N)$, and $d_M + d_N = 2$ since the vector fields on the two disks are related on their common boundary the same way as for the sphere; hence $I(v) = I(v_M) + I(v_N) = \chi (M) + \chi (N) - 2 = \chi (M \# N)$.) Since every compact connected surface is homeomorphic to a connected sum of copies of $S, T, P$, induction using Lemma 3.7.4 together with the base cases for $S, T, P$ proves the theorem.

Give an alternate proof of Theorem 3.7.5 (the index of a vector field on a compact connected surface is given by the Euler characteristic) for the surface $T^{(k)}$ as being formed from two regions in the plane with $k$ holes in each, by identifying points on the boundary circles. Such a description arises from thinking of $T^{(k)}$ as sitting in $\mathbb {R}^{3}$ symmetrically with respect to the $xz$-plane and then slicing it by that plane and flattening out the two halves.

Corresponding vector fields from T^{(3)}.

Give an analysis of how tangent vectors on identified circles correspond, similar to the analysis for $T$ (Theorem 3.7.2, obtained by cutting the torus into two annuli along its equator circles) to relate the indices of the vector fields on the two halves. The figure above depicts this decomposition in the case of $T^{(3)}$.

Suppose $M$ is a compact, connected oriented surface with a vector field of index $-2$. Identify $M$.

Show that there are vector fields on $D^2$ with a prescribed integer as index, and so Theorem 3.7.5 (the index of a vector field on a compact connected surface is given by the Euler characteristic) does not apply to surfaces with boundary without some additional hypothesis.

By adding disks to a surface with boundary, find a formula for the index of a vector field on a surface with boundary where the vector field is required to be an exterior normal vector field on each boundary circle.