Suppose I have a normed vectorspace $(X,\|.\|)$ and a (differential) path $\gamma:[0,1]\rightarrow X$. Can the Length of the curve be defined as $L(\gamma)=\int_0^1\|\gamma'(t)\|\text{d}t$ Or do other modifications have to be applied? In the wikipedia articles I found, all proofs were done via the euclidean norm, with no notion of arbitrary norms. The only thing I found that was more in general was the article about riemann submanifolds, but I am not happy with using something we have yet to learn.
Length of a curve in normed spaces
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1Definition is fine as stated. – 2012-12-25
1 Answers
The definition that you mention is fine, although I suspect you might want an affine vector space instead of a simple vector space. This just means you can have vectors based at any point and not just the origin. Of course it's fine with a vector space, but might seem a little unnatural.
If you are to progress your theory then you need to find an arc-length parameter, i.e. a parameter $s$ for which $||d\gamma/ds|| = 1$ for all $s$. After that, you will need to think about what curvature means.
You would be surprised about how varied the geometries can be. There is a geometry based on area instead of length. In this geometry you look for an arc-"length" parameter $s$ for which
$ \det\left( \frac{d\gamma}{ds},\frac{d^2\gamma}{ds^2}\right) = 1$
for all $s$. Notice that $\det$ measures the oriented area spanned by $d\gamma/ds$ and $d^2\gamma/ds^2$. It turns out that this arc-"length" parameter is given by:
$s(t) = \int \det\left(\dot{\gamma},\ddot{\gamma}\right)^{1/3} \, dt \, . $
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1@CBenni No worries. PM me and we can continue the discussion. Merry Christmas! – 2012-12-25