Riemannian Metrics, Killing Fields, Morse Theory, etc.

Lee Smooth Manifolds

First, a warmup: (Bundle Characterization Lemma) Let $\pi :E\to M$ and ${\pi }^{\prime }:E^{\prime }\to M$ be smooth vector bundles over a smooth manifold $M$ with or without boundary. A map $\mathcal{F}:\Gamma (E)\to \Gamma (E^{\prime })$ is $C^{\infty }(M)$-linear iff there is a smooth bundle homomorphism $F:E\to E^{\prime }$ such that $\mathcal{F}(\sigma )=F\circ \sigma$ for all $\sigma \in \Gamma (E)$.

Proof) The converse is trivial, so we focus on the forward direction. A natural first step is to show $\mathcal{F}$ acts pointwise on tangent vectors. Take an arbitrary point $p\in M$; let $E$ be rank $k$ and $E^{\prime }$ be rank $k^{\prime }$.

First, assume $\sigma \in \Gamma (E)$ vanishes in some neighborhood $U$ of $p$. Taking a bump function $\psi$ supported in $U$ and equal to $1$ in some neighborhood of $p$, $\mathcal{F}(\sigma )(p)=\psi (p)\mathcal{F}(\sigma )(p)=\mathcal{F}(\psi \sigma )(p)=0$, and thus $\mathcal{F}$ acts locally. We must go one step further, assume $\sigma \in \Gamma (E)$ vanishes at $p$. Taking a neighborhood $U$ of $p$ and smooth local frame $({\sigma }_i)$ defined on some neighborhood of $\bar{U}$, we can represent $\sigma {|}_{\bar{U}}$ as $\sigma {|}_{\bar{U}}=f^i{\sigma }_i$, where the coefficients $f^i$ vanish at $p$. As all the functions $f^i$ and ${\sigma }_i$ are defined on a closed set, they can be smoothly extended to functions $g^i$ and ${\lambda }_i$ defined on the entirety of $M$. Then, The vector field $g^i{\lambda }_i$ agrees with $\sigma$ in some neighborhood of $p$, and satisfies $\mathcal{F}(g^i{\lambda }_i)(p)=g^i(p)\mathcal{F}({\lambda }_i)(p)=0$, forcing $\mathcal{F}(\sigma )(p)=0$, as desired. Thus, $\mathcal{F}$ acts pointwise on vectors.

Let $F$ be the function such that, for any $v\in E_p$, $F(v)=\mathcal{F}({\sigma }_v)(p)$, where ${\sigma }_v$ is any smooth section such that ${\sigma }_v(p)=v$. $F$ is trivially linear on each fiber and covers ${\text{Id}}_M$; it suffices to establish smoothness. Let $U$ be a neighborhood of $p$ on which there is a smooth local frame ${\sigma }_i$ of $E$ and ${\sigma }_i^{\prime }$ of $E^{\prime }$ defined on some neighborhood of $\bar{U}$; obtain global smooth sections ${\lambda }_i$ by extending ${\sigma }_i{|}_{\bar{U}}$ to $M$. Then, $\mathcal{F}({\lambda }_i)$ is a smooth section of $E^{\prime }$, and thus there are smooth functions $a_i^j\in C^{\infty }(U)$ such that $\mathcal{F}({\lambda }_i){|}_U=a_i^j{\sigma }_j^{\prime }$. In the coordinates induced by ${\sigma }_i$ and ${\sigma }_i^{\prime }$, $F$ is simply matrix multiplication by $a_i^j$; as every component is smooth, $F$ is smooth on $U$. As $p$ is arbitrary, $F$ is smooth.

Problem 1 (E13-1). If $(M,g)$ is a Riemannian $n$-manifold with or without boundary, show that the unit tangent bundle $UM=\{(x,v)\in TM:|v{|}_g=1\}$ is a smooth fiber bundle over $M$ with model fiber ${\mathbb{S}}^{n-1}$.

Proof) We first slightly generalize the smooth manifold chart lemma. Given any set $S$ and fixed $n$, suppose we are given a cover $U_{\alpha }$ with bijective maps ${\tau }_{\alpha }:U_{\alpha }\to M_{\alpha }$ where $M_{\alpha }$ is a $n$-manifold with or without boundary. If we have that: (a) The cover $(U_{\alpha },{\tau }_{\alpha })$ is such that for any $\alpha$ and $\beta$, ${\tau }_{\alpha }(U_{\alpha }\cap U_{\beta })$ and ${\tau }_{\beta }(U_{\alpha }\cap U_{\beta })$ are open and ${\tau }_{\beta }\circ {\tau }_{\alpha }^{-1}$ is a diffeomorphism from ${\tau }_{\alpha }(U_{\alpha }\cap U_{\beta })$ to ${\tau }_{\beta }(U_{\alpha }\cap U_{\beta })$, (b) The cover $U_{\alpha }$ has a countable subcover (c) For any $p,q\in S$, there is either one $U_{\alpha }$ such that $p,q\in U_{\alpha }$ or $p\in U_{\alpha }$, $q\in U_{\beta }$ such that $U_{\alpha }\cap U_{\beta }=\varnothing$. Then there is a unique topology and smooth structure on $S$ such that each of the maps ${\tau }_{\alpha }$ are a diffeomorphism. This easily follows from the original smooth manifold chart lemma by declaring $({\tau }_{\alpha }^{-1}(U),\varphi \circ {\tau }_{\alpha })$ to be smooth charts for any $\alpha$ and smooth chart $(U,\varphi )$ of $M_{\alpha }$.

For any $p\in M$, there is a neighborhood $U$ such that there is a local frame $({\sigma }_1,\dots ,{\sigma }_n)$ on $U$; we can use Gram-Schmidt to make this frame orthonormal with respect to $g$. Using the coordinates induced by this orthonormal frame, we can see $UM\cap {\pi }^{-1}(U)$ is diffeomorphic to $U\times {\mathbb{S}}^{n-1}$. Taking all such $U$s and diffeomorphisms gives us a structure satisfying the general smooth manifold chart lemma above. Giving $UM$ the corresponding topology and smooth structure, we can clearly see it is a fiber bundle over $M$ with model fiber ${\mathbb{S}}^{n-1}$.

Problem 2 (E13-3). Let $M$ be a smooth manifold. (a) If there exists a global nonvanishing vector field on $M$, then there exists a global smooth nonvanishing vector field. (b) If there exists a linearly independent $k$-tuple of vector fields on $M$, then there exists such a $k$-tuple of smooth vector fields.

Proof) (a) Let $g$ be an arbitrary Riemannian metric on $M$. It suffices to show that, for any continuous positive function $\delta :M\to \mathbb{R}$ and arbitrary vector field $V\in \Gamma (TM)$, there exists some $W\in \mathfrak{X}(M)$ such that, for any $p\in M$, $|V_p-W_p{|}_g<\delta (p)$. We can then apply this lemma to the nonvanishing vector field with $\delta$ as its $g$-norm to complete the proof.

For any $p\in M$, take a neighborhood $U$ on which there exists an orthonormal frame $({\sigma }_1,\dots ,{\sigma }_n)$; let $V=v^i{\sigma }_i$, where the functions $v^i$ are continuous. There exists a neighborhood $V_1\subset U$ of $p$ such that, for any $q\in V_1$, $\sum (v^i(q)-v^i(p){)}^2<(\delta (p)/2{)}^2$, and a neighborhood $V_2\subset U$ of $p$ such that $\delta (q)>\delta (p)/2$ for all $q\in V_2$. Taking $U^{\prime }=V_1\cap V_2$, we can see that if we define the smooth local vector field $W$ on $U^{\prime }$ such that $W(q)=v^i(p){\sigma }_i(q)$, for any $q\in U^{\prime }$, $|W(q)-V(q){|}_g=\sqrt{\sum (v^i(q)-v^i(p){)}^2}<\delta (p)/2<\delta (q)$. Taking such neighborhoods and local vector fields for all points $p$ and linking them via partitions of unity gives us the desired smooth vector field $W$ over $M$ that is $\delta$-close to $V$.

(b) Let $(V_1,\dots ,V_k)$ denote the $k$-tuple of independent vector fields. Fix an arbitrary Riemannian metric $g$, and let $\delta (p)$ be

$$\delta (p)=\inf \left\{\sqrt{\sum _{i=1}^k||V_i(p)-v_i|{|}_g^2}|(v_1,\dots ,v_k)\text{ is linearly dependent}\right\}$$

$\delta$ is clearly positive and continuous (take a local orthonormal frame); take smooth vector fields $(W_1,\dots ,W_k)$ such that $W_i$ is $\delta (p)/\sqrt{k}$-close to $V_i$ (by the logic in (a)). As $(W_1,\dots ,W_k)$ is smooth and linearly independent, we are done.

Problem 3 (E13-5). Suppose $(M,g)$ is a Riemannian manifold. A smooth curve $\gamma :J\to M$ is unit-speed if $|{\gamma }^{\prime }(t){|}_g\equiv 1$. Prove that every smooth curve with nowhere-vanishing velocity has a unit-speed reparametrization.

Proof) Assume $J$ is an open interval. Fix a point $p\in J$ and define a function $\tau :J\to \mathbb{R}$ by

$$\tau (q)={\int }_p^q|{\gamma }^{\prime }(t){|}_{g}dt$$

As $|{\gamma }^{\prime }(t){|}_g$ is a smooth function of $t$ and ${\tau }^{\prime }(q)=|{\gamma }^{\prime }(q){|}_g>0$, $\tau$ is an injective local diffeomorphism with an open image, and thus is a diffeomorphism with some open interval $I$. Let $\tilde{\gamma }:I\to M$ defined by $\tilde{\gamma }=\gamma \circ {\tau }^{-1}$ be a reparametrization of $\gamma$. For any $q\in I$,

$$|{\tilde{\gamma }}^{\prime }(q){|}_g=|{\gamma }^{\prime }({\tau }^{-1}(q))({\tau }^{-1}{)}^{\prime }(q){|}_g=\frac{|{\gamma }^{\prime }({\tau }^{-1}(q)){|}_g}{|{\gamma }^{\prime }({\tau }^{-1}(q)){|}_g}=1$$

and we are done.

Remark: This implies every Riemannian 1-manifold is flat as any chart can transformed, via diffeomorphism, to a unit-speed curve whose corresponding coordinate frame is (trivially) commuting and orthonormal. Remark 2: In the case that $J$ is not an open interval, simply extend it to a nonvanishing-velocity curve on an open interval, find the unit-speed reparametrization, and restrict the reparametrization.

Problem 4 (E13-9). Show that the image of $F:{\mathbb{R}}^2\to {\mathbb{R}}^3$ given by $F(u,v)=(u\cos v,u\sin v,v)$ (the helicoid $H$) is locally isometric but not globally isometric to the catenoid $C$, the surface of revolution generated by the curve $\gamma (t)=(\cosh t,t)$.

Proof) Note that the helicoid is homeomorphic to ${\mathbb{R}}^2$ and thus has a trivial fundamental group, whereas the catenoid is homeomorphic to $\mathbb{R}\times \mathbb{S}$, and thus ${\pi }_1(C)\cong \mathbb{Z}$. Therefore, there can be no homeomorphism between $C$ and $H$, meaning there cannot be a global isometry.

We now show $H$ is locally isometric to $C$. Recall that the pullback metric on $H$ with its global parametrization $F$ is $d{u}^2+(u^2+1)d{v}^2$, and that the function $f:{\mathbb{R}}^2\to {\mathbb{R}}^2$ defined by $f(x,y)=(\sinh x,y)$ is a global diffeomorphism. Take any $p\in H$ and sufficiently small neighborhood $U$ of $F^{-1}(p)$ such that taking $V=f^{-1}(U)$ and defining $\tau (\theta ,t)=(\cos \theta \cosh t,\sin \theta \cosh t,t)$ gives a local parametrization of $C$. The pullback metric on $U$ is then $(\cosh t{)}^2(d{\theta }^2+d{t}^2)$, which pulls back via $f$ into $d{u}^2+(u^2+1)d{v}^2$, as desired. Thus, the map $\tau \circ f^{-1}\circ F^{-1}$ is an isometry from some neighborhood of $p$ to some open set of $C$, completing the proof.

Remark: Recall that “locally isometric” is a non-symmetric property.

Problem 5 (E13-13). Let $(M,g)$ be a Riemannian manifold. A smooth vector field $V$ on $M$ is called a Killing vector field for $g$ if its flow acts by isometries of $g$. (a) Show that the set of all Killing vector fields is a Lie subalgebra of $\mathfrak{X}(M)$. (b) Show that a smooth vector field $V$ on $M$ is a Killing vector field iff it satisfies the following equation in each smooth local coordinate chart:

$$V^k\frac{\partial g_{ij}}{\partial x^k}+g_{jk}\frac{\partial V^k}{\partial x^i}+g_{ik}\frac{\partial V^k}{\partial x^j}=0$$

Proof) (a) Note that $V\in \mathfrak{X}(M)$ is a Killing vector field iff the tensor field $g$ is invariant under the flow of $V$; this implies that $V$ is a Killing vector field iff ${\mathcal{L}}_{V}g=0$. This implies the Killing vector fields are closed under scalar multiplication and addition as the map $V\mapsto {\mathcal{L}}_{V}g$ is $\mathbb{R}$-linear. It suffices to show that, if $V,W\in \mathfrak{X}(M)$ satisfies ${\mathcal{L}}_{V}g={\mathcal{L}}_{W}g=0$, then ${\mathcal{L}}_{[V,W]}g=0$. Note that

$${\mathcal{L}}_{[V,W]}(d{x}^p\otimes d{x}^q)=d({\mathcal{L}}_{[V,W]}{x}^p)\otimes d{x}^q+d{x}^p\otimes d({\mathcal{L}}_{[V,W]}{x}^q)$$

Furthermore,

$$({\mathcal{L}}_V{\mathcal{L}}_W-{\mathcal{L}}_W{\mathcal{L}}_V)(d{x}^p\otimes d{x}^q)=({\mathcal{L}}_V{\mathcal{L}}_{W}d{x}^p-{\mathcal{L}}_W{\mathcal{L}}_{V}d{x}^p)\otimes d{x}^q+d{x}^p\otimes ({\mathcal{L}}_V{\mathcal{L}}_{W}d{x}^q-{\mathcal{L}}_W{\mathcal{L}}_{V}d{x}^q)$$ $$=(d(VW{x}^p)-d(WV{x}^p))\otimes d{x}^q+d{x}^p\otimes (d(VW{x}^q)-d(WV{x}^q))$$

These two sides are equivalent, and thus ${\mathcal{L}}_{[V,W]}(d{x}^p\otimes d{x}^q)=({\mathcal{L}}_V{\mathcal{L}}_W-{\mathcal{L}}_W{\mathcal{L}}_V)(d{x}^p\otimes d{x}^q)$. A general consequence of this is that, for any tensor field $A$, ${\mathcal{L}}_{[V,W]}A={\mathcal{L}}_V{\mathcal{L}}_{W}A-{\mathcal{L}}_W{\mathcal{L}}_{V}A$. This identity is clearly sufficient for (a).

(b) $V$ is a Killing vector field iff $0={\mathcal{L}}_{V}g=({\mathcal{L}}_V{g}_{ij})d{x}^i\otimes d{x}^j+g_{ij}({\mathcal{L}}_{V}d{x}^i\otimes d{x}^j+d{x}^i\otimes {\mathcal{L}}_{V}d{x}^j)$. Isolating the $d{x}^i\otimes d{x}^j$ component gives

$$V^k\frac{\partial g_{ij}}{\partial x^k}+g_{kj}\frac{\partial V^k}{\partial x^i}+g_{ik}\frac{\partial V^k}{\partial x^j}=0$$

We are done.

Problem 6 (E13-14). Let $K\subset \mathfrak{X}({\mathbb{R}}^n)$ be the Lie algebra of Killing vector fields with respect to the Euclidean metric, and let $K_0\subset K$ be the subspace of such fields vanishing at the origin. (a) Show that the map

$$V\to \left(\frac{\partial V^i}{\partial x^j}(0)\right)$$

is an injective linear map from $K_0$ to $\mathfrak{o}(n)$ (the Lie algebra of skew-symmetric matrices). (b) Show that the following vector fields form a basis for $K$:

$$\frac{\partial }{\partial x^i},1\le i\le n;x^i\frac{\partial }{\partial x^j}-x^j\frac{\partial }{\partial x^i},1\le i<j\le n$$

Proof) (a) This map must land in $\mathfrak{o}(n)$ as, by 5(b), in canonical coordinates any Killing vector field $V\in K_0$ must satisfy

$$\frac{\partial V^i}{\partial x^j}+\frac{\partial V^j}{\partial x^i}=0$$

for any $i,j$ at the origin at which it vanishes, forcing the Jacobian to be skew-symmetric at the origin. The map is also clearly linear; it suffices to show injectivity. We begin with the following general Lemma:

Lemma) Let $(M,g)$ be a complete Riemannian manifold and $V$ be a Killing vector field with respect to $g$. The flow $\theta$ of $V$ must then be complete.

FTSOC, assume $p\in M$ is such that its integral curve $\gamma$ is not defined for all $\mathbb{R}$; let its domain be $(a,b)$ and assume, WLOG, that $b<\infty$. For any $t\in (a,b)$, note that

$$|{\gamma }^{\prime }(t){|}_g=|V_{\theta (t,p)}{|}_g=|(d{\theta }_t{)}_p{V}_p{|}_g=|V_p{|}_g$$

as $V$ is clearly invariant under its own flow and ${\theta }_t$ is an isometry. Therefore, for any $t_1,t_2\in (a,b)$, the Riemannian distance between $\gamma (t_1)$ and $\gamma (t_2)$ is bounded by

$$d(\gamma (t_1),\gamma (t_2))\le \left|{\int }_{t_1}^{t_2}|{\gamma }^{\prime }(t){|}_g\right|=|V_p{|}_g|t_2-t_1|$$

Thus, under the Riemannian metric the sequence $\gamma (b-1/n)$ is Cauchy and converges by completeness, implying ${\lim }_{t\to b-}\gamma (t)$ exists; we call this point $q\in M$. By the Fundamental ODE Theorem, there exists some neighborhood $U$ of $q$ and $\epsilon >0$ such that for any $u\in U$, an integral curve is defined on $(-\epsilon ,\epsilon )$. Take some $t^{\prime }\in (a,b)$ such that $b-t^{\prime }<\epsilon /2$ and $\gamma (t^{\prime })\in U$; then, the integral curve $\gamma$ can be extended, contradicting maximality.

As ${\mathbb{R}}^n$ is complete with the Euclidean metric, the lemma implies any Killing vector field in $K$ has complete flow. Therefore, for any $t\in \mathbb{R}$, ${\theta }_t$ is an isometry from ${\mathbb{R}}^n$ to itself. Now, assume that $V$ is a Killing vector field such that $V(0)=0$ and such that its Jacobian also vanishes at the origin. One can compute

$$\frac{d}{dt}(d({\theta }_t{)}_0{)}_{ij}={\frac{d}{dt}\left(\frac{\partial }{\partial x^j}{\theta }^i(t,x)\right)|}_{x=0}={\frac{\partial }{\partial x^j}\frac{\partial }{\partial t}{\theta }^i(t,x)|}_{x=0}={\frac{\partial }{\partial x^j}{V}_x^i|}_{x=0}=0$$

Therefore, the matrix $d({\theta }_t{)}_0$ is invariant with time. In other words, for any $t\in \mathbb{R}$, ${\theta }_t$ is an isometry from ${\mathbb{R}}^n$ to itself such that ${\theta }_t(0)=0$ and $d({\theta }_t{)}_0=d({\theta }_0{)}_0=I_n$; the only such isometry is the identity map. Therefore, $V$ must simply be the zero vector field, making the map injective as desired.

(b) It suffices to show for $i<j$, $(x^i\partial /\partial x^j)-(x^j\partial /\partial x^i)$ is a Killing vector field; the rest is trivial. This follows from the fact that 5(b) can be straightforwardly strengthened from “each smooth local coordinate chart” to “some local coordinates around each point” as its proof is simply the direct Lie derivative computation. As 5(b) is satisfied in the canonical global coordinates of ${\mathbb{R}}^n$, we are done.

Problem 7 (E13-17). Let $M$ be a connected noncompact smooth manifold and let $g$ be a Riemannian metric on $M$. Prove that there exists a positive function $h\in C^{\infty }(M)$ such that the Riemannian metric $\tilde{g}=hg$ is complete. Use this to prove that every connected smooth manifold admits a complete Riemannian metric.

Proof) Let $f$ be a positive smooth exhaustion function on $M$. It suffices to show $h$ can be chosen such that $f$ is bounded on $\tilde{g}$-bounded sets; this is because any $\tilde{g}$-Cauchy sequence is bounded in $\tilde{g}$ and must then be bounded in $f$. As it is then contained in some compact set of $M$, it must converge, proving completeness.

We sequentially reduce the goal of the problem until it becomes straightforward. Intuitively, we must modify $g$ such that $f$ “grows slower” than $\tilde{g}$; more precisely, it is sufficient to find $h$ such that for any $p,q\in M$, $d_{\tilde{g}}(p,q)\ge f(p)-f(q)$. As the LHS is a path integral, it is natural to represent the RHS as a path integral as well; in particular, for any piecewise smooth path segment $\gamma :[a,b]\to M$ such that $\gamma (a)=q$ and $\gamma (b)=p$, $f(p)-f(q)={\int }_a^{b}df\circ {\gamma }^{\prime }(t)dt$.

We can modestly strengthen the goal to finding $\tilde{g}$ such that, for any piecewise smooth path segment $\gamma :[a,b]\to M$, the following inequality holds:

$${\int }_a^b⟨d{f}^{♯}(\gamma (t)),{\gamma }^{\prime }(t){⟩}_{\tilde{g}}dt={\int }_a^{b}df\circ {\gamma }^{\prime }(t)dt\le {\int }_a^b|{\gamma }^{\prime }(t){|}_{\tilde{g}}dt$$

By Cauchy-Schwarz, $⟨d{f}^{♯},{\gamma }^{\prime }(t){⟩}_{\tilde{g}}\le |d{f}^{♯}{|}_{\tilde{g}}|{\gamma }^{\prime }(t){|}_{\tilde{g}}$. Therefore, we can further strengthen our goal to finding $\tilde{g}$ such that, for any $x\in M$, $|d{f}^{♯}(x){|}_{\tilde{g}}\le 1$. Then, for any piecewise smooth curve segment $\gamma :[a,b]\to M$ such that $\gamma (a)=q$ and $\gamma (b)=p$, we have that

$$f(p)-f(q)={\int }_a^{b}df\circ {\gamma }^{\prime }(t)dt={\int }_a^b⟨d{f}^{♯}(\gamma (t)),{\gamma }^{\prime }(t){⟩}_{\tilde{g}}dt\le {\int }_a^b|d{f}^{♯}(\gamma (t)){|}_{\tilde{g}}|{\gamma }^{\prime }(t){|}_{\tilde{g}}dt\le {\int }_a^b|{\gamma }^{\prime }(t){|}_{\tilde{g}}dt$$

as this holds for any arbitrary piecewise smooth curve segment, this clearly implies that $f$ is bounded on any $\tilde{g}$-bounded set.

We now find such an $h$, taking note that the musical isomorphisms are metric-dependent. If we scale $g$ as $\tilde{g}=hg$ and write ${\nabla }^{g}f=d{f}^{{♯}_g}$ and ${\nabla }^{\tilde{g}}f=d{f}^{{♯}_{\tilde{g}}}$, for any $v\in T_{p}M$, $⟨{\nabla }^{g}f(p),v{⟩}_g=df(v)=⟨{\nabla }^{\tilde{g}}f(p),v{⟩}_{\tilde{g}}=h(p)⟨{\nabla }^{\tilde{g}}f(p),v{⟩}_g$, implying that ${\nabla }^{\tilde{g}}f={\nabla }^{g}f/h$. Therefore, the norm $|{\nabla }^{g}f(x){|}_g$ becomes scaled as $|{\nabla }^{\tilde{g}}f(x){|}_{\tilde{g}}=|{\nabla }^{g}f(x){|}_{\tilde{g}}/h=|{\nabla }^{g}f(x){|}_g/\sqrt{h(x)}$. In other words, we hope to find a smooth function $h\in C^{\infty }(M)$ such that for any $x\in M$, $h(x)\ge |{\nabla }^{g}f(x){|}_g^2$. The smooth function $h(x)=|{\nabla }^{g}f(x){|}_g^2+1$ suffices.

The second part of the problem follows by letting $g$ be an arbitrary Riemannian metric and augmenting it to be complete.

Remark: If not specified $d{f}^{♯}$ refers to $d{f}^{{♯}_{\tilde{g}}}$.

Problem 8 (E13-24). Let $M$ be a compact smooth $n$-manifold, and suppose $f$ is a smooth real-valued function on $M$ that has only finitely many critical points $\{p_1,\dots ,p_k\}$, with corresponding critical values $\{c_1,\dots ,c_k\}$ labeled so that $c_1\le \cdots \le c_k$. For any $a<b\in \mathbb{R}$, define $M_a=f^{-1}(a)$, $M_{[a,b]}=f^{-1}([a,b])$, and $M_{(a,b)}=f^{-1}((a,b))$. If $a$ and $b$ are regular values, note that $M_a$ and $M_b$ are embedded hypersurfaces in $M$, $M_{(a,b)}$ is an open submanifold, and $M_{[a,b]}$ is a regular domain. (a) Choose a Riemannian metric $g$ on $M$, let $X$ be the vector field $X=\mathrm{grad}f/|\mathrm{grad}f{|}_g^2$ on $M\setminus \{p_1,\dots ,p_k\}$, and let $\theta$ denote the flow of $X$. Show that $f({\theta }_t(p))=f(p)+t$ whenever ${\theta }_t(p)$ is defined. (b) Let $[a,b]\subseteq \mathbb{R}$ be a compact interval containing no critical values of $f$. Show that $\theta$ restricts to a diffeomorphism from $[0,b-a]\times M_a$ to $M_{[a,b]}$.

Proof) (a) follows from the straightforward computation

$${\frac{d}{dt}f({\theta }_t(p))|}_{t_0}=df(X(\theta (t_0,p)))=\frac{⟨\mathrm{grad}f(\theta (t_0,p)),\mathrm{grad}f(\theta (t_0,p)){⟩}_g}{|\mathrm{grad}f(\theta (t_0,p)){|}_g^2}=1$$

For (b), note that for any $a<b$ such that $[a,b]$ has no critical points, the set $O(a,b)\subset \mathbb{R}\times M$ defined by $O(a,b)=\{(t,x):f(x),t+f(x)\in [a,b]\}$ is a subset of the flow domain $\mathcal{D}$. This is because, for any $p\in M$, there exists some neighborhood $U$ and constant $\epsilon >0$ such that $(-\epsilon ,\epsilon )\times U\subset \mathcal{D}$. As $M_{[a,b]}$ is compact, finitely many such neighborhoods cover it, meaning there is some constant $ϵ>0$ such that $(-ϵ,ϵ)\times M_{[a,b]}\subset \mathcal{D}$. Thus, an integral curve cannot end inside $M_{[a,b]}$, forcing $O(a,b)\subset \mathcal{D}$.

To avoid regular domain bookkeeping (which is frustrating), we simply construct the diffeomorphism and its inverse by hand. Define $\Phi :[0,b-a]\times M_a\to M_{[a,b]}$ by $\Phi (t,p)=\theta (t,p)$ with inverse ${\Phi }^{-1}:M_{[a,b]}\to [0,b-a]\times M_a$ defined by ${\Phi }^{-1}(p)=(f(p)-a,\theta (a-f(p),p))$. Both are smooth, finishing the proof.

Problem 9 (E13-25). Suppose $M$ is a smooth manifold that admits a proper smooth function $f:M\to \mathbb{R}$ with no critical points. Show that $M$ is diffeomorphic to $N\times \mathbb{R}$ for some compact smooth manifold $N$.

Proof) Take some Riemannian metric $g$ on $M$ and consider the vector field $X=\mathrm{grad}f/|\mathrm{grad}f{|}_g^2$. As in Problem 8, the flow $\theta$ of $X$ satisfies $f(\theta (t,p))=f(p)+t$. The flow of $X$ must be complete, as if, say, an integral curve $\gamma$ starting at $p$ is defined on some interval $(a,b)$ with $b<\infty$, the set $f^{-1}([f(p)+b-1,f(p)+b])$ is compact and thus has some $\epsilon >0$ such that $(-\epsilon ,\epsilon )\times f^{-1}([f(p)+b-1,f(p)+b])\subset \mathcal{D}$; thus $\gamma$ can be extended, an immediate contradiction.

Consider the level set $N=f^{-1}(0)$ which is a compact hypersurface of $M$. The restriction of $\theta$ to $\mathbb{R}\times N$ (the flowout of $N$) is a bijective immersion and thus a diffeomorphism. We are done.