Extrapolation on the p-norm sphere using the exponential map - MathOverflow most recent 30 from http://mathoverflow.net2013-05-25T14:20:00Zhttp://mathoverflow.net/feeds/question/90440http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/90440/extrapolation-on-the-p-norm-sphere-using-the-exponential-mapExtrapolation on the p-norm sphere using the exponential mapAntoine Levitt2012-03-07T10:06:02Z2012-03-07T10:35:25Z
<p>Hi,</p>
<p>In order to follow a branch of solutions to an implicit equation on the manifold $M = \lbrace x \in \mathbb R^n, \|x\|_p = 1 \rbrace $, I'm interested in the following problem. Given two points $a$ and $b$ in $M$, is there a natural way to extrapolate them to a third point $c$, as in the flat case $c = 2b - a$ ? </p>
<p>In particular, I'm trying the following strategy. Find v such that $\exp_b(-v) = a$, and compute $c$ as $\exp_b(v) = a$, where $\exp_b$ is the exponential map on $T_bM$. This should be a well-posed problem for $b$ and $a$ close enough, but my knowledge of differential geometry is very limited, and I do not know if it possible to solve this problem in closed form for general $p$. I expect it only makes sense for $p \geq 2$, but I'd be happy to be proved wrong. I have solved the case $p=2$ using the formula from <a href="http://www.math.duke.edu/~bryant/267/Day12Exercises.pdf" rel="nofollow">http://www.math.duke.edu/~bryant/267/Day12Exercises.pdf</a> (it reduces to simple geometry)</p>
<p>If the inverse problem (logarithm map) is too hard to solve, I'm still interested in the direct problem (exponential map), because if I have an explicit derivative of the curve available, I can use that as $v$. Of course, a trivial strategy is to extrapolate linearly the curve and then project back, but I feel that's clumsy.</p>