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Suppose we have a cone and we wish to parallel transport a vector $w=(0,1,0)$ from along the curve $\alpha(s)=(\sqrt{2}/2 \cos(v\sqrt{2}),\sqrt{2}/2 \sin(v\sqrt{2}),\sqrt{2}/2)$ from $p=\alpha(0)$ to $q=\alpha(\frac{\pi\sqrt{2}}{4})$

I take polar coordinates on the plane and construct the isometry $F:(\rho,\theta)\in (0,\infty)\times(0,\sqrt{2}\pi) \longrightarrow F(\rho,\theta)=\left(\frac{\sqrt{2}}{2} \rho\cos(\theta\sqrt{2}),\frac{\sqrt{2}}{2} \rho\sin(\theta\sqrt{2}),\rho\frac{\sqrt{2}}{2}\right)$.

$F$ is a local isometry between the plane and the cone, so it is equivalent if we do parallel transport on the plane or on the cone. See this image from Do Carmo's Differential Geometry of Curves and Surfaces. enter image description here

So now I know the angle between the parallel transport of $w$ and the $\alpha'(\frac{\pi\sqrt{2}}{4})$.

But how do I finish? How do I get the parallel transport of $w$?

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You compute $q$, and $v = \alpha'(\frac{\pi\sqrt{2}}{4})$, and let $v' = v/\|v\|$, then compute a second unit vector $u'$ perpendicular to $v'$, lying in the tangent plane to the cone at $q$.

Now you take $\cos \theta v' + \sin \theta u'$ (where $\theta$ is your computed angle between the transported $w$ and $v'$).

In other words, you take the $t(s)$, $n(s)$ basis for the plane from the right hand diagram and map it to the $v', u'$ basis for the cone at $q$. (Here $n(s)$ denotes a vector in the plane perpendicular to $t(s)$.)

Yes, you have to worry about the orientation of $u'$; there are two choices for $u'$ (or, if you like, for $n(s)$.

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  • $\begingroup$ If I understand correctly. We construct an orthonormal basis on the plane that "constains" $t(s)$ and an orthonormal basis on the tangent plane at $q$ that "contains" $v$. Then we construct an orthonormal transformation $T$ and compute the image of the parallel transport on the plane by $T$. $\endgroup$ – user203327 May 19 '15 at 10:34
  • $\begingroup$ Exactly right. And the only tricky bit is orientation, but since this is in $R^3$, it's not too hard to get that right (we hope). Perhaps an alternative way to think of this is to say that your map $f$ from the cone $C$ to the plane can be extended to a map $F$ of a neighborhood of $C$ (away from the vertex!) to a thickened plane, sending the normal vector $n$ at each point to the $+z$ direction. You then want dF(q), applied to $u', v', n$ to yield a positively oriented basis for 3-space. $\endgroup$ – John Hughes May 19 '15 at 11:19

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