Most of the special properties can be traced back to its special relationship with the inner product. And inner products have somewhat more elementary properties, so in that sense it explains the special position of the euclidean norm.

This has nothing to do with the coordinates by the way. If you want a different norm you'll first have to figure out an alternative to the bilinearity that gives the inner product its special properties.

Though bilinearity is pretty special itself, given the link between the tensor space and the linear algebra equivalent of currying.

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srean a day ago | parent [-]

> This has nothing to do with the coordinates by the way.

I think it does. Both decompose along orthogonal directions. See my comment here https://news.ycombinator.com/item?id=45248881

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shiandow 19 hours ago | parent | next [-]

I mean they decompose in orthogonal components for all Lp norms I think? Is there a norm for which (x,0) is not the closest point to (x,y) on the x-axis?

	▲	srean 5 hours ago \| parent [-]
		Not quite sure what you mean but now I agree with your previous point that the coordinate system is irrelevant because these normal derived Lp metrics can be specified in a coordinate free way just by specifying their unit ball set

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JadeNB a day ago | parent | prev [-]

I think that's arguably an a posteriori explanation: you can find orthogonal coordinates with respect to which the L^2 norm has a nice form, but you can also single out the L^2 norm in various ways (for example, by its large symmetry group, or the fact that it obeys the parallelogram law—or even just the fact that "orthogonal" makes sense!) without ever directly referencing coordinates.

	▲	a day ago \| parent [-]
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