# tangent space duality of a surface

let us describe how the euclidean duality is carried to the tangent bundle of a surface.

We know how euclidean duality is: For each euclidean space, $\mathbb{R}^n$, the coordinated (rectangular)  functions: $x^i(p)=p^i$ are linear, so their derivatives, $Jx^i$, are equal themselves i.e the gradients obey $Jx^i=x^i$, they are dubbed $dx^i$ and they satisfy duality:

$dx^i(e_k)={\delta^i}_k$

Now, if $\Phi:\Omega\to\mathbb{R}^3$  is a parameterization of the surface, $\Sigma\subset\mathbb{R}^3$, and $f:\Sigma\to\mathbb{R}$ is a “measure” then, doing calculus in the surface means do calculus to $f\circ\Phi$. Let $g=f\circ\Phi$.

Let us name $\xi^1,\xi^2,\xi^3$ the coordinated (rectangular) functions on $\mathbb{R}^3$.

So by the chain rule we have: $Jg=Jf\cdot J\Phi$ that is, in terms of gradients:

${\rm grad}(g)={\rm grad}(f)\!\cdot\!J\Phi$

or

$[\begin{array}{cc}\frac{\partial g}{\partial x^1}&\frac{\partial g}{\partial x^2}\end{array}]=[\begin{array}{ccc}\frac{\partial f}{\partial\xi^1}&\frac{\partial f}{\partial\xi^2}&\frac{\partial f}{\partial\xi^3}\end{array}]\!\cdot\!\left(\begin{array}{cc}\frac{\partial \xi^1}{\partial x^1}&\frac{\partial \xi^1}{\partial x^2}\\\frac{\partial \xi^2}{\partial x^1}&\frac{\partial \xi^2}{\partial x^2}\\\frac{\partial \xi^3}{\partial x^1}&\frac{\partial \xi^3}{\partial x^2}\end{array}\right)$

So for the functions $u^i=x^i\circ\Phi^{-1}$ we get $x^i=u^i\circ\Phi$ and by the same rule just above

$dx^i=du^i\!\cdot\!J\Phi$

where evaluating at the basis $e_1,e_2$ of $\mathbb{R}^2$ give

$du^i(J\Phi e_k)=dx^i(e_k)={\delta^i}_k$

but since it is known that the $J\Phi e_1,J\Phi e_2$ generate $T_p\Sigma$, then both: $du^1,du^2$ generate $T_p\Sigma^*$, the co-tangent space at $p\in\Sigma$,… wanna see a picture?