Monthly Archives: March 2010

Moxi: the cartesian product of the mobius-strip and an interval


MoxI is a solid Klein bottle, the genus one nonorientable handlebody. Here we see it with some typical curves on

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look a rough view

another earlier

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geodesics in the band with better numerical resolution


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perturbated core curve


the core curve of the mobius strip perturbated by 0.1\cos(11\ t) 

But if we use 0.4\cos(12.5t)\sin(3t)\cos(t)

 
 

 

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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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a GPS for a surface to do calculus on it


 

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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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torus in chains


more than thousand googolplex parole

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