Timeline for Different ways of thinking about the derivative
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
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| May 20, 2010 at 20:00 | comment | added | KConrad | Since not all local rings contain copies of their residue field, the concept of a derivation is not more geometric than symbolic. I said the description of derivations is close to Thurston's symbolic item simply because it is a very algebraic way of talking about derivatives. | |
| May 20, 2010 at 19:58 | comment | added | KConrad | In the geometric setting, A/M is the real numbers and K is always the set of constant functions on a nbd. of the point of interest. | |
| May 20, 2010 at 19:57 | comment | added | KConrad | Oh, I should say that the assumption that the residue field has an embedding back into the local ring that is a section to the map A --> A/M could be said in a more down-to-earth way: there is a subfield of A which is a set of representatives for the residue field A/M. That is, there's some field K inside A such that the map A --> A/M is onto when we restrict it to K. If that's the case, the map K --> A/M is onto and it's automatically one-to-one since any ring hom. out of a field is one-to-one. Thus K is isom. to A/M so the inverse map A/M --> K is a section to A --> A/M. | |
| May 20, 2010 at 19:41 | comment | added | KConrad | The f-values are in F = A/M, so in the last expression things only matter mod M. Since a = c mod M and b = d mod M, cf(b) + df(a) = af(b) + bf(a). Voila: f(ab) = af(b) + bf(a). Thus, if we have an embedding of the residue field F = A/M back into A which is a section to the canonical reduction map A ---> A/M, any F-linear map M/M^2 ---> F is canonically associated to an A-linear derivation A ---> F. The "other natural condition" that you were missing is precisely this section of the residue field map A --> A/M (loosely, that A/M lives inside A). That's all! | |
| May 20, 2010 at 19:38 | comment | added | KConrad | So now we have a canonical function f : A ---> F where f(c+m) = f(m) for c in F (as subset of A) and m in M. It is A-linear and it is 0 on F and M^2. For a and b in A, write a = c + m and b = d + n where c, d are in F and m, n are in A. Then f(a) = f(m) and f(b) = f(n). Now for the product rule! We have f(ab) = f((c+m)(d+n)) = f(cd+cn+dm + mn) = f(cd) + f(cn) + f(dm) + f(mn). Since f is 0 on F and M^2, this is f(cn) + f(dm) = cf(n) + df(m) = cf(b) + df(a). More next... | |
| May 20, 2010 at 19:32 | comment | added | KConrad | This direct sum decomposition is obvious geometrically: h(x) = h(a) + (h(x) - h(a)). OK, now let f : M/M^2 --> F be an F-linear map (note since F = A/M that M/M^2 is naturally an A/M-vector space). Saying it's F-linear is the same as saying it's A-linear since M/M^2 and F = A/M are both A-modules and in both cases M multiplies on them like 0. Since M is an A-module (it's an ideal!), an A-linear map f : M/M^2 --> F can be pulled back to an A-linear map M --> F which is 0 on M^2. Let us extend to a map A ---> F by just declaring it to act on A = F + M as 0 on F-part. More next.. | |
| May 20, 2010 at 19:28 | comment | added | KConrad | Let us make an assumption that is true in setting of manifolds but is not always true: the field F = A/M naturally lies inside the ring A. More precisely, suppose there is a homomorphism F ---> A such that the composite F --> A --> A/M is the identity. (For rings of functions at a point on a manifold this is very natural since the real numbers sit inside A as constant functions naturally. But not every local ring naturally contains its residue field, such as a local ring of char. 0 whose residue field has characteristic p.) We get a natural direct sum decomposition A = F (+) M. More next... | |
| May 20, 2010 at 19:23 | comment | added | KConrad | The space of derivations at a point can be identified with the dual space of M/M^2 (M being the maximal ideal of functions vanishing at the point), and from the special features of the manifold setting the mere linearity can indeed be pulled back to recover a derivation. Here's the setup. Let A be a local ring with maximal ideal M and set F = A/M. (Think A = local ring of smooth functions at point P on manifold, M = elements of A vanishing at P, F = real numbers). We want to show any F-linear map f : M/M^2 --> F can be pulled back naturally to a derivation A ---> F. More in next comment. | |
| May 20, 2010 at 18:14 | comment | added | Vectornaut | To me, this way of thinking seems more geometric than symbolic, because if R is the ring of smooth functions on a manifold, the product rule guarantees that derivations are local: if two functions agree on an open neighborhood of x, their derivations agree at x. Ever since I found that out, I've been wondering whether you can derive the product rule from linearity, locality, and some other natural condition (I don't think the first two are enough). Thoughts? | |
| May 18, 2010 at 20:48 | comment | added | KConrad | I don't know about that. | |
| May 18, 2010 at 20:34 | comment | added | Dan Piponi | It makes sense to talk of derivations on semirings. (In fact, my answer is about an example.) Is there much of a developed theory of calculus on semirings. In particular, mutually recursively defined types look remarkably like varieties over semirings meaning we can mimic the construction of tangent bundles over semirings. Is there any developed theory for this? Maybe this should be a proper question. | |
| May 17, 2010 at 21:58 | history | answered | KConrad | CC BY-SA 2.5 |