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Most, if not all, of the notions of derivative that I have so far seen have the property that they are locally defined -- meaning that the derivative of a map-type object at a point depends on the map only in an arbitrarily small neighborhood of said point.

One could define some sort of non-local "differential" operator in real analysis for example by convolving a derivative with a function having non-trivial support. I use quotes here, because the operator involves integration as well as differentiation.

A less explicit but more apropos example would be an operator $N$ taking a manifold morphism $\phi$ to a vector bundle morphism $N\phi$, but which does not respect the restriction and gluing concepts used in sheaf theory (which I understand to be an abstract way to say that something is locally defined). The tangent map operator on manifold morphisms is a similar type of object, except that it can be shown to be locally defined. Thus $N\phi$ would "taste" like a derivative in that it has the same form as $T\phi$, but is not locally defined.

  1. Could an operator such as $N$ qualify as a differential operator (equivalently, would $N\phi$ qualify as a derivative of $\phi$)?
  2. Generally, what is the essential quality defining a "derivative"? My guess is that an answer would live in the realm of sheaf theory.
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A keyword you might search for is pseudo-differential operators, which is usually attached to well-behaved operators in that spirit... –  Mariano Suárez-Alvarez Jan 30 '12 at 0:16
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Yes, pseudo-differential operators are defined by convolution and are somehow not local but very close to being local. –  Tom Goodwillie Jan 30 '12 at 0:29
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As for «what's the essential quality of a differential operator», one answer is Peetre's theorem. (Wikipedia's write-up on this is particularly obfuscated, so Googling a bit might be helpful) –  Mariano Suárez-Alvarez Jan 30 '12 at 0:33
    
A notion of derivative applies for functions on many finite sets in $\mathbb{R}^n$ (say). Though the Banach algebra of bounded functions on a finite set turns out to carry no nonzero derivations, one can pick a suitable function space for which the interpolation problem has a unique solution and differentiate the interpolant. This is intrinsically non-local. –  Steve Huntsman Jan 30 '12 at 3:15
    
Re: my comment above, see mathoverflow.net/questions/32667/… –  Steve Huntsman Jan 30 '12 at 3:17
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