Surface covariant derivative

For a tangent vector field $V$ along a curve on an embedded surface, the surface covariant derivative is the tangential projection of its ordinary derivative:It is the covariant derivative induced by the surface metric.

The orthogonal projector onto the tangent plane of a surface with unit normal vector $\mathbf{n}$. It removes the normal component of a vector and defines the surface gradient ${\nabla }_s=P\nabla$.

On a sphere of radius $a$, ${\nabla }_s\mathbf{n}=P/a$, ${\nabla }_s\cdot \mathbf{n}=2/a$, and ${\Delta }_s\mathbf{n}=-2\mathbf{n}/a^2$. For a constant vector $\mathbf{U}$, ${\nabla }_s\cdot (P\mathbf{U})=-2\mathbf{U}\cdot \mathbf{n}/a$. These identities simplify dipolar surface diffusion and interfacial transport.

For an embedded surface with unit normal vector $\mathbf{n}$, the surface gradient is ${\nabla }_{s}f=(\mathbf{I}-\mathbf{n}{\mathbf{n}}^T)\nabla f$. It projects the ambient gradient onto the tangent space and is independent of how $f$ is extended normally off the surface.

Let ${\nabla }_{\tan }F=\nabla F-{\mathbf{e}}_r{\partial }_{r}F$ be the tangential surface gradient on a sphere of radius $r>0$. For smooth scalar fields, set $\mathbf{f}=R{\mathbf{e}}_r+{\nabla }_{\tan }G+{\mathbf{e}}_r\times {\nabla }_{\tan }F$. Its radial curl detects only $F$:To prove this, use the radial curl formula $(r\sin \theta {)}^{-1}[{\partial }_{\theta }(\sin \theta f_{\varphi })-{\partial }_{\varphi }{f}_{\theta }]$. The radial component $R$ never enters it. The $G$ terms cancel as $G_{\varphi \theta }-G_{\theta \varphi }=0$ by symmetry of second derivatives, while the $F$ terms give the unit-sphere Laplace-Beltrami operator. Taking the dot product with the position vector $\mathbf{r}=r{\mathbf{e}}_r$ gives the displayed $r^{-1}$ version. This tangential gradient decomposition does not assert that the whole vector field is solenoidal.

On the graph of a function $z=f(x,y)$, assume $f$ is differentiable, its planar gradient $p$ is nonzero, and speed $U>0$ is measured by the three-dimensional Euclidean norm. The upward unit normal vector is $n=(-p,1)/\sqrt{1+|p{|}^2}$. Projecting the downward vertical vector $-e_z$ onto the tangent plane gives $-(p,|p{|}^2)/(1+|p{|}^2)$. By the Cauchy-Schwarz inequality, normalizing this projected vector produces the tangent direction with greatest height decrease at the prescribed speed, giving the displayed velocity. Its vertical component is $-U|p|/\sqrt{1+|p{|}^2}$, and its horizontal direction is parallel to $-\nabla f$. Consequently its planar paths agree with negative gradient flow up to a change of time, although the speeds differ. Where $p=0$, every tangent direction has zero first-order height change and this normalized formula is undefined.

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 ${\nabla }_s\cdot \mathbf{n}=2/a$.

For a smooth tangent vector field $v$ on a compact orientable surface with smooth boundary and outward conormal of a surface boundary, the integral of its surface divergence is its boundary flux. Extend the unit normal vector to a unit field $m$. The curl of a cross product gives $m\cdot \nabla \times (m\times v)=\nabla \cdot v-m_i{m}_j{\partial }_j{v}_i$ on the surface: $m\cdot v=0$ there and $m\cdot {\partial }_{j}m=0$ by unit length. Apply Stokes theorem and use $(m\times v)\cdot d\mathbf{r}=v\cdot (d\mathbf{r}\times m)$ to obtain the formula.