TITLE: Calculus on surfaces and chain rule
QUESTION [5 upvotes]: Define the surface gradient operator on any surface $S$ as
$$\nabla_S f = \nabla f - \nabla f \cdot \nu_S \nu_S$$
where $\nu_S$ is the outward unit normal on $S$.
Let $T:S_1 \to S_2$ be a $C^2$ diffeomorphism between hypersurfaces.
Consider $v:S_2 \to \mathbb{R}$ a function. Is is true that
$$\nabla_{S_1}(v \circ T(x)) = (\mathbf{D}T)^T(\nabla_{S_2}v)|_{T(x)}?$$
i.e. does the chain rule hold? Can someone prove it?
Here $\mathbf{D}T$ denotes the Jacobian of $T$ wrt. the orthogonal basis of tangent space.

REPLY [2 votes]: Yes , it is indeed true. Observe the relation with the total derivatives 
$ v'(x)(y) = \nabla v(x).y $ and $ T'(x)(y) = DT(x)y $. Now you have chain rule as $$ (v\circ T)'(x) = (v'(T(x))\circ T'(x) $$ Thus for all $ y $ we have $$ \nabla (v\circ T)(x).y = (v\circ T)'(x)(y) = v'(T(x))(T'(x)(y)) = \nabla v(T(x)).DT(x)y \\= (DT)^T\nabla v(T(x)).y $$ Thus $ \nabla(v\circ T)(x) = (DT)^T\nabla v(T(x)) $. Now we observe that $ DT \nu_{S_1} = \nu_{S_2} $, hence we have $$ \nabla(v\circ T(x)).\nu_{S_1} \nu_{S_1} = (DT)^T\nabla v(T(x)).\nu_{S_1}\nu_{S_1} = (\nabla v(T(x).DT\nu_{S_1})\nu_{S_1} = \nabla v(T(x)).\nu_{S_2} DT^T\nu_{S_2} $$ Now as $ \nabla v(T(x)).\nu_{S_2} $ is scalar hence $\nabla(v\circ T(x)).\nu_{S_1}\nu_{S_1} = (DT)^T(\nabla v(T(x)).\nu_{S_2}\nu_{S_2}) $ Thus you find combining all 
$$ \nabla_{S_1}(v\circ T(x)) = (DT)^T(\nabla_{S_2}v(T(x))) $$