Vector Fields Flows and Lie Derivatives

In the induced tangent-bundle chart over $U$, the coordinate fields $\mathrm{\partial }\mathrm{/}\mathrm{\partial }{x}^1,\dots ,\mathrm{\partial }\mathrm{/}\mathrm{\partial }{x}^n$ form a local frame of $TM{\mathrm{\mid }}_U$, so [L1] says that $X$ is smooth exactly when it can be written with smooth coefficient functions $X^1,\dots ,X^n$ in that frame.

If one changes charts, [L2] expresses the new coefficients as linear combinations of the old ones with smooth Jacobian entries. Hence the criterion from step 1.1 is independent of the chosen chart.

Therefore a vector field is smooth exactly when its coordinate components in a chart are smooth.

For each point $p\in M$, [L1] gives $X_p(fg)=f(p)X_p(g)+g(p)X_p(f)$. By the definition of the action on functions, this is exactly $(X(fg))(p)=f(p)(Xg)(p)+g(p)(Xf)(p)$.

To see that $Xf$ is smooth, write locally $X={\sum }_i{X}^i\mathrm{\partial }\mathrm{/}\mathrm{\partial }{x}^i$ using [L2]. Then $Xf={\sum }_i{X}^i{\mathrm{\partial }}_i(f\circ x^{-1})\circ x$, a sum of products of smooth functions.

Since the equality in step 1.1 holds for every $p$, one has $X(fg)=fXg+gXf$ as functions on $M$. The map $f\mapsto Xf$ is $\mathbb{R}$-linear for the same pointwise reason.

Therefore $X$ acts on $C^{\mathrm{\infty }}(M)$ as an $\mathbb{R}$-linear derivation.

The forward map is well defined by [L1]: every smooth vector field $X$ yields an $\mathbb{R}$-linear derivation $f\mapsto Xf$.

Fix $p\in M$. If global smooth functions $f$ and $g$ agree on a neighbourhood of $p$, choose an open set $W$ on which $f=g$ and use [L5] to choose $\chi :M\to [0,1]$ that is $1$ on a neighbourhood of $p$ and has support contained in $W$. Then $\chi (f-g)=0$, so evaluating the Leibniz rule for $D$ at $p$ gives $0=D(\chi (f-g))(p)=\chi (p)D(f-g)(p)+(f-g)(p)D(\chi )(p)=D(f)(p)-D(g)(p)\mathrm{.}$ Thus $f\mapsto D(f)(p)$ depends only on the germ of $f$ at $p$, and it defines a derivation $D_p:C_p^{\mathrm{\infty }}(M)\to \mathbb{R}$. By [L2], there is a unique tangent vector $X_p\in T_{p}M$ with $X_p([f])=D_p([f])$.

Let $p\in M$, choose a chart $(U,x^1,\dots ,x^n)$ around $p$, and use [L5] again to choose $\chi :M\to [0,1]$ that is $1$ on a neighbourhood $V$ of $p$ and has support contained in $U$. For each $i$, let ${\tilde{x}}^i$ be the global smooth function that equals $\chi x^i$ on $U$ and $0$ outside $U$. Then for every $q\in V$, the germs of ${\tilde{x}}^i$ and $x^i$ agree at $q$, so [L3] writes $X_q={\sum }_{i}D({\tilde{x}}^i)(q)\frac{\mathrm{\partial }}{\mathrm{\partial }{x}^i}{\mid }_q\mathrm{.}$ Each coefficient function $D({\tilde{x}}^i){\mathrm{\mid }}_V$ is smooth, because $D({\tilde{x}}^i)$ is a global smooth function. Hence [L4] makes $X$ smooth on $V$.

Since every point has a neighbourhood $V$ on which step 2.1 makes $X$ smooth, the pointwise-defined tangent vectors $X_p$ form a global smooth vector field $X$ on $M$.

By construction, $Xf=D(f)$ for every smooth function $f$, so the map from smooth vector fields to derivations is surjective. If two smooth vector fields induce the same derivation, then their values at each point agree on every smooth function, hence are equal by [L2]; thus the map is injective.

Therefore smooth vector fields and $\mathbb{R}$-linear derivations of $C^{\mathrm{\infty }}(M)$ are in bijection.

Assume $X$ and $Y$ are $F$-related. For any $p\in M$ and $f\in C^{\mathrm{\infty }}(N)$, [L1] and [L2] give $(X(f\circ F))(p)=X_p(f\circ F)=d{F}_p(X_p)(f)=Y_{F(p)}(f)=((Yf)\circ F)(p)$.

Conversely, assume $X(f\circ F)=(Yf)\circ F$ for every smooth $f$ on $N$. Fix $p\in M$, and let $[g]\in C_{F(p)}^{\mathrm{\infty }}(N)$ be a smooth germ. Choose an open neighbourhood $W$ of $F(p)$ on which $g$ is represented by a smooth function, and use [L3] to choose $\chi :N\to [0,1]$ that is $1$ on a neighbourhood of $F(p)$ and has support contained in $W$. Let $h$ be the global smooth function obtained by extending $\chi g$ by $0$ outside $W$. Then $[h]=[g]$ at $F(p)$. The hypothesis applied to $h$ gives $X_p(h\circ F)=Y_{F(p)}(h)$, and [L1] identifies the left-hand side with $d{F}_p(X_p)([h])$. Thus $d{F}_p(X_p)([g])=Y_{F(p)}([g])$ for every germ $[g]$, so $d{F}_p(X_p)=Y_{F(p)}$.

Therefore $X$ and $Y$ are $F$-related exactly when the derivation intertwining law holds for every smooth target function.

By [L1], every point of $S$ has a slice chart $(U_{\alpha },x_{\alpha })$ in which $S\cap U_{\alpha }$ is given by $x_{\alpha }^{k+1}=\cdots =x_{\alpha }^n=0$. On that slice, $Y$ has smooth coordinate components, so extending those coefficient functions constantly in the normal coordinates defines a smooth vector field ${\tilde{Y}}_{\alpha }$ on $U_{\alpha }$.

The open sets $U_{\alpha }$ cover $S$. Choose a smaller open neighbourhood $U\subseteq {\bigcup }_{\alpha }{U}_{\alpha }$ of $S$, and by [L2] choose a partition of unity $({\rho }_{\alpha })$ on $U$ subordinate to $(U_{\alpha }\cap U)$. Then $\tilde{Y}:={\sum }_{\alpha }{\rho }_{\alpha }{\tilde{Y}}_{\alpha }$ is a smooth vector field on $U$, and on $S$ the coefficients sum to those of $Y$, so $\tilde{Y}{\mathrm{\mid }}_S=Y$.

Assume now that $S$ is closed. By [L4], $S$ has an open tubular neighbourhood $V$, and step 2.1 gives a smooth extension $\tilde{Y}$ on some neighbourhood $U$ of $S$. Replace $U$ by $U\cap V$, which is still an open neighbourhood of $S$.

Because $S$ is closed in the open set $U$, [L3] gives a smooth function $\chi :M\to \mathbb{R}$ with $\chi =1$ on $S$ and $\mathrm{supp}\chi \subseteq U$. Define $\widehat{Y}:=\chi \tilde{Y}$ on $U$ and $\widehat{Y}:=0$ on $M\setminus \mathrm{supp}\chi$. This is a smooth global vector field and restricts to $Y$ on $S$.

Therefore every smooth vector field along an embedded submanifold extends to a neighbourhood, and to all of $M$ when the submanifold is closed.

In a slice chart $(U,x^1,\dots ,x^n)$ for $S$, the submanifold is given by $x^{k+1}=\cdots =x^n=0$. By [L2], write $X={\sum }_i{X}^i\mathrm{\partial }\mathrm{/}\mathrm{\partial }{x}^i$ with smooth coefficients on $U$.

Tangency means that at each point of $S\cap U$ the normal components vanish: $X^{k+1}=\cdots =X^n=0$ on $S\cap U$. Hence on $S\cap U$ the field is $X{\mathrm{\mid }}_{S\cap U}={\sum }_{i=1}^k(X^i{\mathrm{\mid }}_{S\cap U})\frac{\mathrm{\partial }}{\mathrm{\partial }{x}^i},$ whose coefficients are smooth on the slice.

The local expressions from step 2.1 define a smooth section of $TS$ in each restricted slice chart, and these local sections agree on overlaps because they are all restrictions of $X$. By [L3], they therefore glue to a smooth vector field on $S$.

Therefore a smooth ambient vector field tangent to an embedded submanifold restricts to a smooth vector field on that submanifold.

By [L1], both $X$ and $Y$ are $\mathbb{R}$-linear derivations, so their commutator is automatically $\mathbb{R}$-linear. It remains to prove the Leibniz rule.

Expand $X(Y(fg))$ using [L1] twice: $X(Y(fg))=X(fYg+gYf)=XfYg+fX(Yg)+XgYf+gX(Yf)\mathrm{.}$

Similarly, $Y(X(fg))=YfXg+fY(Xg)+YgXf+gY(Xf)\mathrm{.}$

Subtracting step 1.3 from step 1.2 cancels the mixed first-order products, leaving $[X,Y](fg)=f[X,Y]g+g[X,Y]f\mathrm{.}$

Therefore $[X,Y]$ is an $\mathbb{R}$-linear derivation of $C^{\mathrm{\infty }}(M)$.

Fix $p\in U$. By [L4], choose a smooth function $\chi :M\to [0,1]$ that is $1$ on a neighbourhood $W$ of $p$ and has support contained in $U$. For each $j$, let ${\tilde{x}}^j$ be the global smooth function that equals $\chi x^j$ on $U$ and $0$ outside $U$. Then ${\tilde{x}}^j=x^j$ on $W$, and because $\chi$ is constant on $W$, one also has $X({\tilde{x}}^j)=X^j$ and $Y({\tilde{x}}^j)=Y^j$ on $W$.

By [L1] and [L2], $[X,Y]$ is induced by a unique smooth vector field on $M$. Evaluating it on ${\tilde{x}}^j$ at $p$ and using step 1.1 gives $[X,Y]({\tilde{x}}^j)(p)=X(Y^j)(p)-Y(X^j)(p)={\sum }_i{X}^i(p){\mathrm{\partial }}_i{Y}^j(p)-{\sum }_i{Y}^i(p){\mathrm{\partial }}_i{X}^j(p)\mathrm{.}$

Because ${\tilde{x}}^j$ and $x^j$ agree near $p$, the $j$-th coordinate coefficient of $[X,Y]$ at $p$ is exactly $[X,Y]({\tilde{x}}^j)(p)$. Since $p$ was arbitrary, step 2.1 and [L3] give the displayed coordinate formula for $[X,Y]$ on $U$.

By [L1] and [L2], the commutator $[X,Y]$ is again a smooth vector field, so the bracket closes on $\mathfrak{X}(M)$.

Bilinearity and antisymmetry follow from the corresponding identities for commutators of $\mathbb{R}$-linear endomorphisms of $C^{\mathrm{\infty }}(M)$: $[aX+bY,Z]=a[X,Z]+b[Y,Z],[X,Y]=-[Y,X]\mathrm{.}$

The operator commutator satisfies the Jacobi identity $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$ on $C^{\mathrm{\infty }}(M)$, again by direct expansion in the endomorphism algebra.

Steps 1.1-1.3 are exactly the Lie-algebra axioms, so smooth vector fields form a Lie algebra under the Lie bracket.

For any test function $h$, $[X,fY](h)=X(fYh)-fY(Xh)=f[X,Y](h)+(Xf)Yh$ by one application of the Leibniz rule from [L1]. Since this holds for every $h$, one has $[X,fY]=f[X,Y]+(Xf)Y$.

Apply step 1.1 with $fX$ in place of $X$ and use [L1] once more: $[fX,gY]=g[fX,Y]+(fXg)Y=g(f[X,Y]-(Yf)X)+f(Xg)Y\mathrm{.}$ Collecting terms gives $[fX,gY]=fg[X,Y]+f(Xg)Y-g(Yf)X\mathrm{.}$

Therefore the Lie bracket satisfies the displayed Leibniz rules with function multiples.

For every $f\in C^{\mathrm{\infty }}(N)$, [L1] gives $X_2(f\circ F)=(Y_{2}f)\circ F\text{and}{X}_1(f\circ F)=(Y_{1}f)\circ F\mathrm{.}$

Apply $X_1$ to the first identity and $X_2$ to the second. Using [L1] again on the resulting target functions yields $X_1{X}_2(f\circ F)=(Y_1{Y}_{2}f)\circ F,X_2{X}_1(f\circ F)=(Y_2{Y}_{1}f)\circ F\mathrm{.}$

Subtracting the identities in step 2.1 gives $[X_1,X_2](f\circ F)=([Y_1,Y_2]f)\circ F$ for every smooth $f$ on $N$. By [L2], [L3], and [L1], this is exactly the statement that $[X_1,X_2]$ and $[Y_1,Y_2]$ are $F$-related.

Therefore related vector fields have related Lie brackets.

By [L1], the vector fields $X$ and $F_{\ast }X$ are $F$-related, and likewise $Y$ and $F_{\ast }Y$ are $F$-related.

Applying [L2] to step 1.1 shows that $[X,Y]$ is $F$-related to $[F_{\ast }X,F_{\ast }Y]$. By the definition of pushforward in [L1], this means exactly that $F_{\ast }[X,Y]=[F_{\ast }X,F_{\ast }Y]$.

In the chosen chart, the coefficient functions of $\mathrm{\partial }\mathrm{/}\mathrm{\partial }{x}^i$ and $\mathrm{\partial }\mathrm{/}\mathrm{\partial }{x}^j$ are constants: each is either $0$ or $1$.

Substituting those constant coefficients into the formula of [L1] makes every derivative term vanish, so each coefficient of the bracket is zero.

Therefore the coordinate vector fields commute.

Choose a chart $(V,x)$ around $p$. By [L1], $x:V\to x(V)\subseteq {\mathbb{R}}^n$ is a diffeomorphism onto an open set, and by [L2] the vector field $X{\mathrm{\mid }}_V$ corresponds to a smooth Euclidean vector field $\tilde{X}$ on $x(V)$.

Apply [L3] to $\tilde{X}$ at the point $x(p)$. This gives $h>0$, an open neighbourhood $W\subseteq x(V)$ of $x(p)$, and a smooth map $\tilde{\mathrm{\Phi }}:(-h,h)\times W\to x(V)$ whose time slices are the unique integral curves of $\tilde{X}$.

Set $U:=x^{-1}(W)$ and define $\mathrm{\Phi }(t,q):=x^{-1}(\tilde{\mathrm{\Phi }}(t,x(q)))$. Because $x$ and $x^{-1}$ are smooth by [L1], $\mathrm{\Phi }$ is smooth. Each curve $t\mapsto \mathrm{\Phi }(t,q)$ is an integral curve of $X$ and satisfies $\mathrm{\Phi }(0,q)=q$.

If another curve in $M$ through $q\in U$ solved the same initial-value problem, its coordinate expression under $x$ would solve the Euclidean problem for $\tilde{X}$ with the same initial value. Uniqueness in [L3] then forces the two curves to agree.

Therefore $X$ has unique local integral curves depending smoothly on the initial point.

By [L1], there exists at least one integral curve of $X$ through $p$ on some open interval about $0$. If two such curves are defined on overlapping intervals, [L1] forces them to agree on the overlap because they solve the same initial-value problem at any common time.

Let $I_p$ be the union of all intervals carrying an integral curve through $p$, and define ${\gamma }_p(t)$ by any one of those curves. Step 1.1 shows this is well defined on $I_p$. Because all the intervals contain $0$ and pairwise overlap along the common trajectory, their union is again an interval.

The map ${\gamma }_p:I_p\to M$ is an integral curve, since near each $t\in I_p$ it coincides with one of the local curves from which it was assembled. If it extended to a larger interval, that larger curve would belong to the family defining $I_p$, contradicting the definition of the union.

Therefore ${\gamma }_p$ is the unique maximal integral curve of $X$ through $p$.

By [L1], for each $p\in M$ there is a unique maximal integral curve ${\gamma }_p:I_p\to M$ through $p$. Therefore the set $\mathcal{D}$ and the map $\mathrm{\Phi }(t,p)={\gamma }_p(t)$ are well defined, each fibre ${\mathcal{D}}_p=I_p$ is an open interval containing $0$, and $\mathrm{\Phi }(0,p)=p$.

Fix $p\in M$ and $s\in I_p$, and put $q:=\mathrm{\Phi }(s,p)={\gamma }_p(s)$. Then the translated curve $t\mapsto {\gamma }_p(t+s)$ is an integral curve through $q$ on the interval $I_p-s:=\{t:t+s\in I_p\}$. By uniqueness of maximal integral curves, it agrees with ${\gamma }_q$ on their common domain, so $\mathrm{\Phi }(t,\mathrm{\Phi }(s,p))=\mathrm{\Phi }(t+s,p)$ whenever both sides are defined. The same translation argument applied to ${\gamma }_q$ shows $I_q=I_p-s$.

Let $W\subseteq \mathcal{D}$ be the set of all $(t,p)\in \mathcal{D}$ such that $\mathrm{\Phi }$ is defined and smooth on some product neighbourhood $J\times U\subseteq \mathcal{D}$ of $(t,p)$. By [L2], every $(0,p)$ lies in $W$. Suppose $W\mathrm{\ne }\mathcal{D}$. Choose $(\tau ,p_0)\in \mathcal{D}\setminus W$; replacing $\tau$ by $-\tau$ if needed, we may assume $\tau >0$. Let $t_0:=\inf \{t\in \mathbb{R}:(t,p_0)\mathrm{\notin }W\}\mathrm{.}$ Then $t_0\in I_{p_0}$, because $(0,p_0)\in W$ and $I_{p_0}$ is an open interval containing $\tau$. Put $q_0:=\mathrm{\Phi }(t_0,p_0)$. Applying [L2] at $q_0$ gives $\epsilon >0$ and an open neighbourhood $U_0$ of $q_0$ such that the local flow is smooth on $(-\epsilon ,\epsilon )\times U_0$. Choose $t_1<t_0$ with $t_1+\epsilon >t_0$ and $\mathrm{\Phi }(t_1,p_0)\in U_0$. Because $(t_1,p_0)\in W$, there is a product neighbourhood $(t_1-\delta ,t_1+\delta )\times U_1\subseteq W$; shrinking $U_1$ if necessary, we may assume $\mathrm{\Phi }(\{t_1\}\times U_1)\subseteq U_0$.