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.

This is fine, but you'll most likely want an affine space, not just a vector space. https://en.wikipedia.org/wiki/Affine_space – Fly by Night Dec 25 '12 at 15:02

1Definition is fine as stated. – Dec 25 '12 at 15:06
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 arclength 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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1Thank you ;) I am basically just looking to calculate the length of a path in $(\mathbb{R}^2,\.\)$ where $\.\$ is any norm. I kept the question general in sake of meeting this pages requirements ;) – CBenni Dec 25 '12 at 15:49

1@CBenni It was a very nice question, and I'm pleased to see you're thinking in general terms. If I were you then I'd chose a favourite norm and run with it. You'd be amazed by where it takes you. Like I said: arclength is the first thing to generalise, and then curvature. After that you can talk about vertices and inflections. In the area case the vertices relate to unusuallyhigh contact with ellipses and hyperbolae, while the inflections relate to contact with parabolae. (In the Euclidean case it's contact with circles and lines.) *Be bold*: you have no idea where your ideas will lead you. – Fly by Night Dec 25 '12 at 20:56

1Basically I am already amazed to where I got. This question is connected to http://math.stackexchange.com/questions/254620/piinarbitrarymetricspaces which is something that was created in a boring reading and is slowly taking shape. Thanks to you, I finally was able to find a quasinorm for which I can calculate $\pi$ with mathematica, and therefore I could find a norm with $\pi\approx 42$ :D – CBenni Dec 25 '12 at 21:11

@CBenni Be very careful if you use the "area norm". This is only valid for curves without Euclidean inflections. In other words, we need $\dot{\gamma}$ and $\ddot{\gamma}$ to be linearly independent along the curve. (Where $\dot{\gamma}$ is differentiation of $\gamma$ with respect to a parameter *of your choice*, $t$.) If they are dependent, then we need to delete the bad points and consider the arcs. – Fly by Night Dec 25 '12 at 21:16

I will have to look into that a bit more closely, its too late in the evening to do that now ^.^ – CBenni Dec 25 '12 at 21:18

1@CBenni No worries. PM me and we can continue the discussion. Merry Christmas! – Fly by Night Dec 25 '12 at 21:20