# Tag Archives: geometría diferencial de superficies

## four geodesics in the band on the right of this post you can see 4 (approximately) geodesics in the mobius strip, one is transversal to the core curve and the others (three)  begin at that transversal… One almost can see that all geodesics in the surface have curvature but without 3d torsion… would it be a theorem?

Rendered with Mathematica-6.

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Filed under fiber bundle, multilinear algebra, topology

## a GPS for a surface to do calculus on it ## 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?