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The chain rule shows that $\alpha {\varphi }_t{\alpha }^{-1}$ is the local flow of the pushforward of a vector field ${\alpha }_{\ast }X$. If this equals $X$, uniqueness of integral curves of a vector field gives commutation. Conversely, differentiate the commuting identity at $t=0$ to obtain $d{\alpha }_p{X}_p=X_{\alpha (p)}$. All identities hold on their common domains; the vector field need not be complete.

A smooth vector field on a smooth manifold $M$ is a smooth section $X:M\to TM$ of its tangent bundle. It acts on smooth functions by $Xh(p)=d{h}_p(X_p)$. In a manifold chart, $X={\sum }_i{X}^i{\partial }_i$. The Lie bracket of vector fields is the commutator of their actions on functions:Expanding in coordinates makes all second derivatives cancel and givesThus the commutator is again a smooth vector field, not a second-order operator. The coordinate expression represents the intrinsic derivation on functions, so agrees on overlaps.

For a diffeomorphism $f:M\to N$, its pushforward of a vector field isFor every smooth function $h$ on $N$, the chain rule gives $((f_{\ast }X)h)\circ f=X(h\circ f)$. Applying this twice yieldsSmooth functions determine a tangent vector, and $f$ is surjective, so

On a Lie group, a left-invariant vector field satisfies $(L_g{)}_{\ast }X=X$ for every left translation $L_g(h)=gh$. Given $\xi \in T_{e}G$, setThis field is smooth by smoothness of multiplication. The equality $L_a{L}_g=L_{ag}$ proves left invariance. Conversely, left invariance forces $X(g)=(d{L}_g{)}_{e}X(e)$, so $X_{\xi }$ is the unique left-invariant field with value $\xi$ at $e$. This is a linear isomorphism between $T_{e}G$ and the space of left-invariant vector fields.

The pushforward identity applied to $L_g$ shows that $[X_{\xi },X_{\eta }]$ is again left invariant. Consequentlydefines a Lie bracket on $T_{e}G$. Bilinearity and antisymmetry follow from the operator commutator. Its Jacobi identity follows by expanding the three nested commutators: the six triple compositions cancel in pairs. These properties transfer through the linear isomorphism, making $T_{e}G$ the Lie algebra of $G$.

For the general linear group $G=\mathrm{GL}(n,\mathbb{R})$, use matrix-entry coordinates $x_{pq}$. It is an open subset of $M_n(\mathbb{R})$, so every tangent space is naturally $M_n(\mathbb{R})$. Since $L_g(h)=gh$, its differential sends $A$ to $gA$, and henceFor $A,B\in M_n(\mathbb{R})$,and reversing $A,B$ gives $(gBA{)}_{pq}$. Thus $[X_A,X_B]=X_{AB-BA}$; evaluating at the identity gives the matrix commutator

Let $\overline{\mathbf{r}}=a^2\mathbf{r}/r^2$ and $\hat{\mathbf{r}}=\mathbf{r}/r$. Its Jacobian isThe chain rule therefore yieldsTaking its cross product with $\mathbf{r}$ kills the radial term and proves that spherical inversion preserves angular derivatives:This is also the invariance of the differential expression underlying the angular momentum operator.

A scalar spherical harmonic $Y$ depends only on direction, so $Y(\hat{\mathbf{r}})=Y(\widehat{\overline{\mathbf{r}}})$ and its gradient is tangential. Thus $\hat{\mathbf{r}}\cdot {\nabla }_{\overline{\mathbf{r}}}Y=0$, andEach of the three normalized vector spherical harmonic families is consequently unchanged as an angular basis function:The normalization requires $n\ge 1$ for $\mathbf{B}$ and $\mathbf{C}$; at $n=0$ the gradient vanishes and only $\mathbf{P}$ remains.

This substitution into the basis functions is distinct from the pushforward of a vector field of a physical vector field by inversion. Under that Jacobian map the radial vector picks up $-a^2/r^2$ and the two tangential vectors pick up $+a^2/r^2$. Likewise, the weighted Kelvin transform of a scalar includes the extra factor $a/r$. Neither operation changes the angular differential identity just proved.

The local flow of $X$ is the smooth map $\Phi :D\to M$, where $D\subset \mathbb{R}\times M$ is the maximal open domain of initial-point/time pairs, whose curve $t\mapsto \Phi (t,q)={\varphi }_t(q)$ is the maximal integral curve of a vector field through $q$, with ${\varphi }_0(q)=q$. Smooth dependence in the ordinary differential equation theorem gives joint smoothness and openness of $D$. Each time slice is a diffeomorphism between its domain and image, with inverse ${\varphi }_{-t}$.

Here is the group law, including its domain qualification. Fix $s,q$ where the relevant trajectories exist. Both curvessolve the same autonomous ordinary differential equation and have value ${\varphi }_s(q)$ at $t=0$. The uniqueness from the first part therefore giveson their common interval. These are local identities; neither the vector field nor its local flow has been assumed complete.

For the commuting local flows ${\varphi }_t$ and ${\psi }_s$, differentiate ${\varphi }_t({\psi }_s(q))={\psi }_s({\varphi }_t(q))$ with respect to $s$ at zero. The chain rule givesThus the pushforward of a vector field $Y$ by the $X$ flow equals $Y$. Interchanging the two fields gives the analogous identity for $X$ under ${\psi }_s$.

Now let ${\mathcal{D}}_q=\mathrm{span}(X_q,Y_q)$. Pointwise linear independence makes this a rank-two smooth distribution. The identity just proved implies $[X,Y]=0$: in coordinates, differentiating $D_q{\varphi }_t{Y}_q=Y_{{\varphi }_t(q)}$ at $t=0$ gives $DX(q)Y_q=DY(q)X_q$, precisely the vanishing Lie bracket of vector fields. The bracket of two local sections is also in $\mathcal{D}$, sinceand the analogous terms for $[fX,gX]$ and $[fY,gY]$ remain in the same span. Therefore $\mathcal{D}$ is an involutive distribution.

The Frobenius theorem says that a smooth constant-rank involutive distribution has local coordinates $(u^1,\dots ,u^n)$ in which it is spanned by ${\partial }_{u^1},{\partial }_{u^2}$, and has unique maximal connected integral manifolds through its points. These are the leaves of a regular foliation. Let $S$ be the maximal leaf through $p$. The global confinement to an immersed leaf gives the answer: $S$ is a two-dimensional immersed integral manifold and every smooth $\mathcal{D}$-tangent curve starting at $p$ stays in $S$.

For the curve assertion, in each Frobenius chart the transverse coordinate functions $u^3,\dots ,u^n$ have zero derivative along such a curve. Hence its segment in that chart stays in one plaque. On any compact subinterval of the curve's parameter interval, a finite chain of overlapping such segments connects the initial plaque to the last; all belong to the same maximal leaf. Exhausting the parameter interval proves the assertion for every time.

The two commuting local flows also give the leaf parametrization directly near $p$:Its partial derivatives are ${\partial }_{s}F=X_F$ and ${\partial }_{t}F=Y_F$, by the pushforward identity. Its differential has rank two. The constant rank theorem makes a sufficiently small image an embedded local plaque with tangent plane $\mathcal{D}$. This is a concrete local construction of the integral manifold.

There is a global distinction in the word “submanifold”. If it is required to mean an embedded submanifold, the assertion for all times is not valid in general. On the three-dimensional torus ${\mathbb{R}}^3/{\mathbb{Z}}^3$, takeThe fields are independent and commute, and their dense immersed cylinder in a three-dimensional torus through zero isIt is an injectively immersed cylinder $\mathbb{R}\times (\mathbb{R}/\mathbb{Z})$ and is dense in the torus. Indeed, fixing $x\mathrm{mod}1$ and varying $s$ by integers makes $\alpha s\mathrm{mod}1$ dense by an irrational rotation of the circle; $t$ is unrestricted. Every point of this leaf is reached by a tangent curve $r\mapsto (rs,rt,\alpha rs)\mathrm{mod}{\mathbb{Z}}^3$. An embedded surface containing it would, in a submanifold chart near one of its points, be a closed coordinate plane containing a dense subset of the ambient open chart, an impossibility. Thus the global conclusion uses an immersed leaf; the embedded conclusion is local, for curve segments remaining in a suitable chart.

A smooth vector field is a smooth section of a vector bundle $X:M\to TM$ of the tangent bundle, so $X_p\in T_{p}M$. In a manifold chart, $X=X^i{\partial }_i$ with smooth coefficients. It acts on a smooth function by $Xf=X^i{\partial }_{i}f$. For a diffeomorphism $\alpha :M\to N$, the pushforward of a vector field isThe Lie bracket of vector fields is the commutator of their actions on smooth functions:The second derivatives of $f$ cancel, so this is again a vector field. The corresponding Jacobi identity is $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$.

A local flow of $X$ is a smooth map $(t,p)\mapsto {\varphi }_t(p)$ on an open neighbourhood of $\{0\}\times M$ satisfying ${\varphi }_0(p)=p$ and ${\partial }_t{\varphi }_t(p)=X_{{\varphi }_t(p)}$. On a sufficiently small neighbourhood and time interval, each ${\varphi }_t$ is a diffeomorphism onto its image, with inverse ${\varphi }_{-t}$. Uniqueness of integral curves of a vector field gives ${\varphi }_{s+t}={\varphi }_s\circ {\varphi }_t$ wherever both sides are defined. These are local statements; no assumption of a complete vector field is required.

A smooth tensor field of type $(r,s)$ is a smooth section of $(TM{)}^{\otimes r}\otimes (T^{\ast }M{)}^{\otimes s}$, where the factors are the tangent bundle and cotangent bundle. The flow definition of the Lie derivative of a tensor field isHere pullback applies $d{\varphi }_t^{-1}$ to each vector factor and the dual of $d{\varphi }_t$ to each covector factor, evaluating $T$ at ${\varphi }_t(p)$. Thus all tensors being differentiated lie in the same fibre over $p$.

For a smooth function, ${\varphi }_t^{\ast }f=f\circ {\varphi }_t$, and the chain rule givesFor a vector field $Y$, in local coordinates the expansions ${\varphi }_t(x)=x+tX(x)+O(t^2)$ and $d{\varphi }_t(x)=I+td{X}_x+O(t^2)$ giveConsequentlyPullback preserves tensor products and tensor contractions, so differentiation makes this definition a tensor derivation. It therefore agrees on every tensor field with the Lie derivative of a tensor field determined by these two formulas.

Now put ${\psi }_t=\alpha \circ {\varphi }_t\circ {\alpha }^{-1}$. The chain rule givesThus ${\psi }_t$ is the local flow of ${\alpha }_{\ast }X$. In the case $M=N$, if ${\alpha }_{\ast }X=X$, uniqueness of integral curves of a vector field makes ${\psi }_t={\varphi }_t$ on their common domains, or $\alpha \circ {\varphi }_t={\varphi }_t\circ \alpha$. Conversely, differentiating this commuting identity at zero gives $d{\alpha }_p{X}_p=X_{\alpha (p)}$, hence ${\alpha }_{\ast }X=X$. This proves that diffeomorphism invariance of a vector field is equivalent to commuting with its local flow, with every identity understood on the domain where its compositions exist.

At $p\in \mathcal{Q}$, the differential of a smooth map defines the pushforward of a vector field byThis is a tangent vector at $\psi (p)$, and varies smoothly as a section of the pullback tangent bundle. In local coordinates its components are $X^i(p){\partial }_i{\psi }^a(p)$.

For a general smooth map between manifolds, this is a vector field along a map, rather than an intrinsic vector field on all of $N$. A vector field $Y$ on the image can be defined only if these vectors agree whenever two points have the same image, and the resulting field must also be smooth; a projectable vector field must meet that requirement. For a diffeomorphism, there is no ambiguity and $Y_q=d{\psi }_{{\psi }^{-1}(q)}{X}_{{\psi }^{-1}(q)}$. For example, $\psi (x)=x^2$ on $\mathbb{R}$ and $X={\partial }_x$ give $2x{\partial }_y$, with opposite nonzero values at the two preimages of $y>0$. Thus the source's notation must be read with this qualification.