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Differentiate the displayed homogeneity relation in $\lambda$ at $1$. On every regular point of $H=0$, the radial vector field $x$ is therefore perpendicular to the normal gradient and tangent to the level surface. Its flow $x(t)=e^{t}x(0)$ preserves that zero set. At a singular vertex this flow formulation remains meaningful even if there is no tangent plane.

On a fiberwise star-shaped neighborhood, let $H_s$ scale fibers by $s$ and let $E$ be the radial vector field. For a closed differential form $\delta$ with zero pullback to the zero section, the homotopy primitive $\eta ={\int }_0^1{s}^{-1}{H}_s^{\ast }({\iota }_E\delta )ds$ satisfies $d\eta =\delta$. For a two-form, it vanishes as an ambient covector along that section, providing the fixed-submanifold condition in the relative Moser theorem.

Two closed symplectic forms agreeing as ambient bilinear forms along a compact embedded submanifold are equivalent near it by a diffeomorphism fixing it pointwise. Shrink until their affine interpolation is nondegenerate. A relative homotopy primitive $\eta$ for their difference vanishes as an ambient covector on the submanifold. Solving ${\iota }_{Z_t}{\omega }_t=-\eta$ gives a vector field vanishing there; its local flow fixes the submanifold and makes ${\omega }_t$ constant under pullback.

For a smooth function $f$ on a Riemannian manifold, its Riemannian gradient is the unique vector field satisfying $g({\nabla }_{g}f,V)=df(V)$ for every vector field $V$. It depends on the Riemannian metric, and its negative generates the downward gradient flow.

A smooth local basis of vector fields whose inner products under a Riemannian metric are the identity matrix. Tracing the Riemannian Hessian in such a frame gives the Laplace-Beltrami operator, with the chosen sign.

The right-invariant vector field associated with a Lie algebra element $\xi$ is obtained by right translation of its value at the identity. Its flow is $g\mapsto \exp (t\xi )g$, hence consists of left translations on a Lie group. Right invariance follows because left and right multiplication commute.

For a smooth function $f$ and vector field $X$,on every tensor field, with $X\otimes df$ interpreted as the endomorphism $Y\mapsto Y(f)X$. Both sides vanish on functions; on vector fields this follows from $[fX,Y]=f[X,Y]-Y(f)X$. Both are contraction-compatible tensor derivations, so agreement on functions and vector fields proves equality on all tensor types. In particular ${\mathcal{L}}_{fX}\omega =f{\mathcal{L}}_X\omega +\omega (X)df$.

The vector field $S=t{\partial }_t+x\cdot \nabla$ generates simultaneous spacetime dilation. For the flat d'Alembert operator, $[\square ,S]=2\square$. Consequently it commutes with the homogeneous wave equation at the level of its solution set, although its operator commutator is nonzero.

A smooth positive Hermitian bundle metric admits local orthonormal frames by the smooth Gram-Schmidt process. In such a frame, metric compatibility gives $A_{ji}(V)+\overline{A_{ij}(V)}=0$ for real vector fields $V$, so the connection matrix is skew-Hermitian. The orthonormal frame is smooth; a holomorphic orthonormal frame is not generally available.

The vector fields ${\partial }_t$ and ${\partial }_i$ generate time and space translations and commute with the flat d'Alembert operator. They form the unweighted part of the commutation vector fields for the wave equation.

To deduce $D{u}_t=I+tDF+o(t)$ from $u_t(x)=x+tF(x)+o(t)$, the remainder must be small in a topology controlling its spatial derivatives, such as $C^1$ locally. Small displacement alone is insufficient: $u_t(x,y,z)=(x+t^2\sin (x/t^2),y,z)$ differs uniformly from the identity by $o(t)$, but has Jacobian determinant $2$ at $x=0$ for nonzero $t$. A smooth flow map generated by a smooth vector field has the required derivative control.

The vector field ${\Omega }_{ij}=x_i{\partial }_j-x_j{\partial }_i$ is an infinitesimal spatial rotation. It commutes with the flat d'Alembert operator and supplies angular derivatives in the Klainerman-Sobolev inequality.

For an upward-oriented band of the circular elliptic paraboloid between radii $r_0$ and $R$, the vector field $B=(-y^3,x^3,z^3)$ has vertical curl $3(x^2+y^2)$. Its surface integral is $3\pi (R^4-r_0^4)/2$. Stokes theorem gives the same value from the counterclockwise outer circle minus the counterclockwise inner circle. Equivalently the induced inner boundary direction is clockwise. Forgetting the inner boundary or its reversed orientation changes the answer.

Near a point where a smooth vector field is nonzero, there are coordinates in which it is $\partial /\partial x_1$. Flow a transverse coordinate slice by the local flow; the inverse function theorem makes the resulting map a coordinate chart. This is the first step of the coordinate proof of the Frobenius theorem.

The surface divergence of an ambient vector field is the tangential trace ${\nabla }_s\cdot \mathbf{v}=({\delta }_{ij}-n_i{n}_j){\partial }_j{v}_i$. For a tangent vector field it is the intrinsic divergence. For $\mathbf{v}=\mathbf{n}$ it gives the sum of the two principal curvatures; a sphere with outward normal has ${\nabla }_s\cdot \mathbf{n}=2/a$.

A vector field on a symplectic manifold is symplectic when its flow preserves the symplectic form. Infinitesimally this is ${\mathcal{L}}_X\omega =0$. By Cartan's magic formula it is equivalent to $d({\iota }_X\omega )=0$. A Hamiltonian vector field has an exact contraction with the form, so is symplectic. On the standard two-dimensional torus, ${\partial }_x$ contracts to the closed but nonexact form $dy$, providing a symplectic non-Hamiltonian example.