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Every smooth left-invariant vector field on a Lie group is a complete vector field. Translate a local integral curve of a vector field through the identity to every other point. The same positive local existence interval works at every initial point, so no maximal integral curve of a vector field can have a finite endpoint. Uniqueness then makes the curve through the identity a one-parameter subgroup.

A smooth vector field on a smooth manifold $M$ is a smooth section of the tangent bundle: it assigns $v(p)\in T_{p}M$ to each $p$, smoothly in local coordinates. In a coordinate chart it has the form $v={\sum }_i{a}_i\partial /\partial x_i$, with smooth coefficients $a_i$. Equivalently, it acts on smooth functions as a derivation, $v(fh)=v(f)h+fv(h)$.

For a Lie group $G$, write $L_a(h)=ah$ for Left translation on a Lie group. A left-invariant vector field satisfiesThus it is determined by its value at the identity. Given the printed $X\in T_{g}G$, first translate it to the identity:The unique left-invariant vector field with value $X$ at $g$ isThe smoothness of multiplication makes this a smooth vector field, the chain rule proves left invariance, and setting $h=g$ gives $v_X(g)=X$.

We prove the completeness of left-invariant vector fields: this left-invariant vector field is a complete vector field. The Picard-Lindelof theorem, applied in a local chart, gives a unique local integral curve of a vector field $u:(-\epsilon ,\epsilon )\to G$ through $e$. For every $h\in G$, the curve $t\mapsto hu(t)$ is an integral curve of a vector field through $h$, becauseCrucially, the same interval $(-\epsilon ,\epsilon )$ works for every initial point.

Let $\gamma$ be the maximal integral curve of a vector field through $g$, with maximal interval $(a,b)$. If $b<\infty$, choose $t_0\in (a,b)$ with $b-t_0<\epsilon /2$. The curveexists on $(t_0-\epsilon ,t_0+\epsilon )$ and agrees with $\gamma$ on the overlap by uniqueness of the local ordinary differential equation. It extends $\gamma$ past $b$, a contradiction. The same argument at a finite $a$ excludes that possibility. Thus $(a,b)=\mathbb{R}$, and uniqueness on overlapping intervals gives uniqueness on all of $\mathbb{R}$.

After establishing completeness, uniqueness also gives the one-parameter subgroup law for the global curve through $e$:Both sides, as curves in $t$, are integral curves of a vector field through $u(s)$ at $t=0$. Defining the Exponential map of a Lie group by $u(t)=\exp (t\xi )$, the answer is

The identity component $G^{\circ }$ is an open normal subgroup, and every open identity neighbourhood generates it. Let $G^{\circ }$ be the connected component of $e$. The product of connected spaces is connected, so the image of $G^{\circ }\times G^{\circ }$ under multiplication is connected and contains $e$. It is therefore contained in $G^{\circ }$. Inversion has the same property. Thus $G^{\circ }$ is a subgroup.

A smooth manifold is locally connected. In particular, a coordinate neighbourhood of $e$ can be chosen homeomorphic to an open ball, so there is a connected open neighbourhood $O$ of $e$ contained in $G^{\circ }$. Its translates $hO$, $h\in G^{\circ }$, are open and lie in $G^{\circ }$, and cover $G^{\circ }$. Therefore the identity component of a Lie group is open in $G$. Conjugation by any $a\in G$ is a homeomorphism fixing $e$, so it maps $G^{\circ }$ into itself; conjugation by $a^{-1}$ gives the reverse inclusion. Hence $G^{\circ }$ is a normal subgroup.

If $U$ is an open neighbourhood of $e$ in $G^{\circ }$, let $H=⟨U⟩$, allowing inverses in the meaning of generated subgroup. For each $h\in H$, $hU$ is open and lies in $H$, so $H={\bigcup }_{h\in H}hU$ is open in $G^{\circ }$. Every other left coset of $H$ is also open. Thus $H$ is both open and closed in the connected space $G^{\circ }$. It is nonempty, so $H=G^{\circ }$.

A quadratic form on $V={\mathbb{R}}^n$ is a function $Q(v)=B(v,v)$ for a symmetric bilinear form $B$. Equivalently, $Q(tv)=t^{2}Q(v)$ and the polarizationis a bilinear map. In coordinates there is a unique real symmetric matrix $A$ with $Q(v)=v^{T}Av$. No assumption of nondegeneracy or positive definiteness is needed.

An element $g\in {\mathrm{GL}}_n(\mathbb{R})$ stabilizes $Q$ when $Q(gv)=Q(v)$ for every $v$. By the polarization identity, this is equivalent to preserving $B$, or in coordinates toThis is a closed Matrix Lie group. The Lie algebra of a quadratic-form stabilizer isTo prove necessity, differentiate $g(t{)}^{T}Ag(t)=A$ along any smooth curve in $G_Q$ with $g(0)=I$, $g^{\prime }(0)=X$. The derivative at zero is $X^{T}A+AX$.

To prove sufficiency, suppose $X^{T}A+AX=0$ and consider the matrix exponential $g(t)=e^{tX}$. ThenIts value at zero is $A$, so $e^{tX}\in G_Q$ for every real $t$. This curve has derivative $X$ at zero, proving the claimed tangent space description even for a degenerate quadratic form.

Equivalently, the condition is $B(Xu,v)+B(u,Xv)=0$ for all $u,v$. For a nondegenerate quadratic form it is the corresponding Special orthogonal Lie algebra. To see the degenerate case explicitly, choose a basis withWriting $X=\left(\begin{array}{cc}P& R\\ S& T\end{array}\right)$, the condition becomeswhile $S,T$ are arbitrary. Thus the nullspace of the quadratic form is invariant, but arbitrary infinitesimal maps into it are allowed. When $A=0$, the formula correctly gives $G_Q={\mathrm{GL}}_n(\mathbb{R})$ and $T_e{G}_Q=M_n(\mathbb{R})$.

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.

For a smooth vector field $X$, solve the initial-value problem $d{\varphi }_t(p)/dt=X({\varphi }_t(p))$, ${\varphi }_0(p)=p$. Local existence, uniqueness and smooth dependence for ordinary differential equations give a local flow. Uniqueness yields ${\varphi }_{t+s}(p)={\varphi }_t({\varphi }_s(p))$ whenever both sides are defined, and ${\varphi }_{-t}$ is the inverse of ${\varphi }_t$ on the corresponding domains. Thus each local flow map is a diffeomorphism between open subsets.

A group of diffeomorphisms of the entire manifold for every real $t$ requires a complete vector field. This follows, for example, on a compact manifold without boundary: local existence times can be made uniform on a finite cover, and repeated continuation prevents finite-time escape. Without completeness, the source's group is only local. The vector field $x^2{\partial }_x$ on $\mathbb{R}$ has flow $x\mapsto x/(1-tx)$ and blows up at $t=1/x$ for $x>0$, so smoothness alone cannot imply a global one-parameter group.