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Say we have 2 surfaces $M$ and $\hat M$ that intersect perpendicularly --> $\left<n,\hat n\right> = 0$ along the curve of the intersection intersection, where $n$ is the unit normal to $M$ and $\hat n$ is the unit normal to $\hat M$.

Assume the intersection of $M$ and $\hat M$ is the image of a unit speed curve $\gamma$ that is a geodesic in both $M$ and $\hat M$.

How can we show that $\gamma$ is a straight line?


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up vote 3 down vote accepted

Assuming we're in $3$-space (so that "the unit normal to a surface" makes sense):

For each point $\gamma(t)$ on the curve, consider $\gamma''(t)$, the curvature of $\gamma$. Since $\gamma$ is a geodesic on $M$, $\gamma''(t)$ must have no component along $M$, so $\gamma''(t)$ is parallel to $n$. But similarly, $\gamma''(t)$ is parallel to $\hat n$, while $n\cdot \hat n=0$. Therefore $\gamma''(t) = 0$ for all values of $t$, so $\gamma$ is a straight line.

In particular, you don't need $M$ and $\hat M$ to intersect perpendicularly, just transversely, for any shared geodesic to be a straight line.

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Is gamma''(t) the curvature? I don't know about this definition, can you please explain? Thanks – mary Oct 8 '12 at 21:29
Since $\gamma$ is unit speed, the acceleration vector $\gamma''(t)$ is always orthogonal to the tangent vector $\gamma'(t)$. Hence it measures, not just how fast the velocity is changing, but how fast the curve is turning, and in what direction. I added a link in the answer proper. – Owen Biesel Oct 8 '12 at 21:54
I think it's only true in the hypersurface in $ R^3 $ . – Illuminata Mar 15 at 6:00

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