Manifolds with Boundary Collars and Orientations — Examples

For $n\ge 1$, the identity map is a global boundary chart on ${\mathbb{H}}^n$; for $n=0$, the unique map identifies the point with ${\mathbb{R}}^0$. These charts give the asserted smooth manifolds with boundary.

For $n\ge 1$, [L1] and [L2] identify the intrinsic interior with $\{x^n>0\}$ and the intrinsic boundary with $\{x^n=0\}$. The identity chart identifies every tangent space with ${\mathbb{R}}^n$, and [L3] makes the inward vectors at the face precisely those with positive last component. For $n=0$, [L1] makes the unique point interior and the boundary empty.

Let $p\in S^{n-1}$ and choose $k$ with $p_k\mathrm{\ne }0$. The map $H(x)=(x_1,\dots ,\widehat{x_k},\dots ,x_n,\rho (x))$ has invertible derivative at $p$, since $\mathrm{\partial }\rho \mathrm{/}\mathrm{\partial }{x}_k(p)=-2p_k\mathrm{\ne }0$. By [L1], $H$ is a diffeomorphism near $p$. Because $B^n=\{\rho \ge 0\}$, its restriction is a half-space chart near $p$; ordinary Euclidean charts cover $\{\mathrm{\mid }x\mathrm{\mid }<1\}$. Every transition between these charts is the restriction of a composition of the corresponding Euclidean diffeomorphisms and their inverses, so the covering atlas is compatible. Thus [L2] makes $B^n$ a smooth manifold with boundary, and the chart calculation identifies its boundary with $S^{n-1}$.

By construction, $\rho \ge 0$ on $B^n$, ${\rho }^{-1}(0)=S^{n-1}=\mathrm{\partial }{B}^n$, and $d{\rho }_x(v)=-2⟨x,v⟩$ is nonzero when $\mathrm{\mid }x\mathrm{\mid }=1$. Hence $\rho$ satisfies [L3].

Along $\{1\}\times S^1$, the outward vector is $+{\mathrm{\partial }}_t$, whereas along $\{0\}\times S^1$ it is $-{\mathrm{\partial }}_t$.

If $v$ is a positive tangent vector to $S^1$, then $({\mathrm{\partial }}_t,v)$ is positive by [L1]. Thus [L2] makes $v$ positive on $\{1\}\times S^1$, while it makes $-v$ positive on $\{0\}\times S^1$, because $(-{\mathrm{\partial }}_t,-v)$ has the same orientation as $({\mathrm{\partial }}_t,v)$. The two boundary-circle orientations are opposite.

The vectors $+{\mathrm{\partial }}_x$ at $b$ and $-{\mathrm{\partial }}_x$ at $a$ point out of the interval.

At $b$, the outward vector $+{\mathrm{\partial }}_x$ is positive, so [L1] gives the point $b$ the positive orientation. At $a$, the outward vector $-{\mathrm{\partial }}_x$ is negative, so [L1] gives $a$ the negative orientation; [L2] makes these conclusions independent of the particular outward vectors. Hence $\mathrm{\partial }[a,b]=\{b\}-\{a\}$.

The map $c$ is smooth and satisfies $c(u,0)=u$. Since $1-t>0$, its inverse on its image is $x⟼(\frac{x}{\mathrm{\mid }x\mathrm{\mid }},1-\mathrm{\mid }x\mathrm{\mid }),$ which is smooth; hence $c$ is a smooth embedding.

Its image is $\{x\in B^n:\mathrm{\mid }x\mathrm{\mid }>1-\epsilon \}$, which is open in $B^n$ and contains $S^{n-1}=\mathrm{\partial }{B}^n$ by [L1]. Consequently $c$ satisfies every clause of [L2] and is a collar.

For $x\mathrm{\ne }0$, write $x=ru$ with $0<r\le 1$ and define $F([x,\pm ])=(\sin (\pi r\mathrm{/}2)u,\pm \cos (\pi r\mathrm{/}2))$, while $F([0,\pm ])=(0,\dots ,0,\pm 1)$. At $r=1$ both formulas give $(u,0)$, so [L1] makes $F$ well defined on the double.

On either disk the first component is $\frac{\sin (\pi \mathrm{\mid }x\mathrm{\mid }\mathrm{/}2)}{\mathrm{\mid }x\mathrm{\mid }}x$, with its value at $x=0$ defined by the smooth even power series, and the last component is $\pm \cos (\pi \mathrm{\mid }x\mathrm{\mid }\mathrm{/}2)$, also a smooth function of $\mathrm{\mid }x{\mathrm{\mid }}^2$. Thus the restrictions are smooth at the two poles.

In the signed seam coordinate $s$, where $x=(1-\mathrm{\mid }s\mathrm{\mid })u$ and the sign of $s$ records the label, [L2] and [L3] rewrite the map as $F(u,s)=(\cos (\pi s\mathrm{/}2)u,\sin (\pi s\mathrm{/}2))$. This is a smooth local diffeomorphism across $s=0$. The formula is bijective because the last coordinate selects the hemisphere and its absolute value determines $r$; its inverse is smooth in the pole charts and in these seam charts. Hence $F$ is a diffeomorphism from the labelled double to $S^n$.

Suppose $B$ had an orientation. Pulling its determinant rays back by the local diffeomorphism $q$ would orient the connected strip $\mathbb{R}\times [-1,1]$. Relative to the standard ray of $({\mathrm{\partial }}_t,{\mathrm{\partial }}_s)$, this continuous choice has one constant sign on the strip.

Since $q\circ g=q$, the pulled-back orientation would have to be invariant under $g$. But $Dg=\mathrm{diag}(1,-1)$ has determinant $-1$ and reverses every determinant ray, contradicting step 1.1. Hence $B$ is nonorientable by [L2].

The two boundary lines of the strip are exchanged by $g$, so their quotient is one component. The map $t<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>2\mapsto q(t,1)$ is a smooth bijection $\mathbb{R}\mathrm{/}2\mathbb{Z}\to \mathrm{\partial }B$ with smooth inverse in the quotient charts. It identifies $\mathrm{\partial }B$ with a circle, whose positive $t$-direction supplies an orientation under [L1]. This orientation is chosen on the boundary itself and is not induced from the nonorientable band.

At $(p_1,p_2)$, the ordered tangent-space splitting is $T_{p_1}{S}^1\oplus T_{p_2}{S}^1$. By [L1], the product ray is generated by placing a positive vector of the first factor before a positive vector of the second.

The angular product chart has ordered frame $({\mathrm{\partial }}_{{\theta }_1},{\mathrm{\partial }}_{{\theta }_2})$, whose entries are positive in their respective circle factors. Step 1.1 and [L2] therefore identify its ray with the product orientation, which is the standard torus orientation for these chosen angular directions and factor order.

The parametrization $[-1,1]\to S$, $s\mapsto (0,s)$, is a diffeomorphism onto $S$ with its subspace topology, and its derivative $v\mapsto (0,v)$ is injective. Thus the inclusion is a smooth embedding, so [L1] makes $S$ an embedded submanifold with boundary $\mathrm{\partial }S=\{(0,-1),(0,1)\}$.

Since $\mathrm{\partial }M=\{0\}\times \mathbb{R}$, one has $S\cap \mathrm{\partial }M=S$, which is not the two-point set $\mathrm{\partial }S$. The equality required by [L2] therefore fails, so $S$ is not neat (independently of the transversality condition).

In the identity boundary chart on $M$, the vector $X_0={\mathrm{\partial }}_x$ has coordinate $+1$, so it is inward at $0$ by [L1].

Any integral curve through $0$ must satisfy ${\gamma }^{\mathrm{\prime }}(t)=1$ and $\gamma (0)=0$, hence $\gamma (t)=t$. It lies in $M$ for $t\ge 0$ but not for any $t<0$. Therefore no flow through $0$ can be defined for negative time while remaining in the half-line.

At $u\in S^{n-1}$, the radial vector $u$ points outward from $B^n$.

If $n=1$, then $B^1=[-1,1]$ and [L3] gives the positive determinant ray at $1$ and the negative determinant ray at $-1$, exactly as the outward vectors $1$ and $-1$ require under [L1]. If $n\ge 2$, a tangent basis $(v_1,\dots ,v_{n-1})$ is positive precisely when $(u,v_1,\dots ,v_{n-1})$ is positive in the standard orientation of ${\mathbb{R}}^n$. Thus in every case the boundary orientation is the standard hypersurface orientation cooriented by the radial outward normal.