Vector Field Index Euler Characteristic and Poincare Hopf — Examples

By [F1] the only nonzero rational Betti numbers of $S^{2m}$ are in degrees $0$ and $2m$, both equal to $1$, so the alternating sum of the definition gives $\chi (S^{2m})=(-1{)}^0\cdot 1+(-1{)}^{2m}\cdot 1=2\mathrm{\ne }0$.

If a smooth field on $S^{2m}$ were nowhere zero, [F2] would force $\chi (S^{2m})=0$, contradicting step 1.1; hence every smooth vector field on an even-dimensional sphere has at least one zero.

The map $z\mapsto iz$ is $\mathbb{R}$-linear, hence smooth, with $\mathrm{\mid }iz\mathrm{\mid }=\mathrm{\mid }z\mathrm{\mid }=1$ on the sphere, so $X$ is a smooth nowhere-zero map of the sphere to itself.

Put $d=2m$. The $2d$ hemispheres $U_k^s=\{x\in S^{d-1}:s{x}_k>0\}$, $1\le k\le d$, $s\in \{-1,1\}$, have charts ${\phi }_k^s$ deleting coordinate $k$, with image the open unit ball and inverse inserting $x_k=s\sqrt{1-\mathrm{\mid }u{\mathrm{\mid }}^2}$. They cover the sphere and have smooth transitions. Their induced bundle charts define a topology by pulling back Euclidean open sets; [F3] makes the definitions agree on overlaps. The bundle is Hausdorff: distinct base points are separated by inverse images of disjoint base neighbourhoods, and vectors over the same point are separated in one bundle chart. The inverse images of rational balls in these finitely many bundle charts give an explicitly countable basis. Thus [F3] supplies a smooth tangent-bundle structure without choice. Every other smooth base chart has compatible induced charts, so this is the canonical structure.

For each $z$, the smooth curve $t\mapsto (z+tiz)\mathrm{/}\mathrm{\mid }z+tiz\mathrm{\mid }$ lies in the sphere and has velocity $iz$ at zero by [F1], so [F4] makes $X(z)$ a tangent vector. In a hemisphere chart, its bundle coordinates are $(u,D{\phi }_k^s(X(({\phi }_k^s{)}^{-1}(u))))$: the fibre part simply deletes coordinate $k$ from $i({\phi }_k^s{)}^{-1}(u)$, hence is smooth. Therefore $z\mapsto (z,iz)$ is a smooth section of the canonical bundle constructed in step 1.2 and is nowhere zero by step 1.1. This proves the unconditional field claim, consistent with $\chi (S^{2m-1})=0$ by [F2] and, under AC, with the converse of Poincare-Hopf.

The fields $X_1,X_2,X_3$ are linear, so their derivatives at every point are the matrices $I$, $-I$ and $\mathrm{diag}(1,-1)$; each matrix is invertible, and $X_i(u)=A_{i}u=0$ has the unique solution $u=0$ since $A_i$ is invertible, so the origin is the only zero of each field and it is nondegenerate.

The determinants are $\det I=1$, $\det (-I)=(-1{)}^2=1$ and $\det \mathrm{diag}(1,-1)=-1$, so [F2] gives the displayed indices $+1,+1,-1$; in particular a source and a sink on a surface both contribute $+1$, and a saddle contributes $-1$.

The field $X$ is linear with $D{X}_u=I_n$, invertible at every point, so its only zero is the centre $0$ and that zero is nondegenerate; by [F2] its index is $\mathrm{sign}\det I_n=+1$, so the index sum is $+1$.

On the boundary sphere the outward normal is the radial vector $u$, so $⟨X(u),u⟩=\mathrm{\mid }u{\mathrm{\mid }}^2=1>0$ and the field is strictly outward by [F1]; by [F3] the index sum equals $\chi (D^n)$, and the value $+1$ computed in step 1.1 matches $\chi (D^n)=1$.

The field $X(u)=-u$ is linear with derivative $-I_n$ everywhere, so its only zero is the centre $0$ and it is nondegenerate there with index $(-1{)}^n=-1$ because $n$ is odd; on the boundary sphere $⟨X(u),u⟩=-1<0$, so $X$ is nonzero but strictly inward.

On the other hand $\chi (D^n)=1$ by [F3], so the index sum $-1$ differs from $\chi (D^n)=1$; therefore the conclusion of the boundary form of Poincare-Hopf fails for this field even though it is nonzero on the boundary, and the outwardness hypothesis of Poincare-Hopf with outward-pointing boundary is load bearing.

The interval $[0,1]$ has dimension one, which is odd, but by [F1] its Euler characteristic is $\chi ([0,1])=1\mathrm{\ne }0$; thus the conclusion of the odd-dimensional vanishing fails as soon as the closedness hypothesis is dropped.

The standard outward field $X(t)=t-\frac{1}{2}$ has the single nondegenerate zero $t=\frac{1}{2}$ of index $+1$ by [F2], so its index sum is $1=\chi ([0,1])$ and the outward-boundary formula remains true here; the interval therefore refutes only the closedness hypothesis of the odd-dimensional corollary, not the boundary form itself.