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?

 

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Filed under cucei math, differential geometry, fiber bundle, multilinear algebra

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