Stanford MATH 396 - Globalization vua Bump Functions

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Math 396. Globalization via bump functions1. MotivationIn the text (as with most books on differential geometry), all notions that ought to be local (suchas tangent vectors, connections, curvature, etc.) are initially defined in terms of objects that arefar too “global” such as global functions, global vector fields, and so on. A s neaky device usingbump functions enables one to show that such global definitions are actually equivalent to the localones that we shall use.The aim of this handout is to explain how this equivalence works in the case of tangent spacesat a point: we gave a manifestly local definition in class, but the text gives the usual “global”definition that is found in virtually all introductory books on differential geometry. There are someuseful applications of the ability to formulate local notions in terms of global structures, but it is abit unnatural to do so in the foundations of the theory. Moreover, such a global point of view is notavailable in the foundations of the theory of real-analytic (or complex-analytic manifolds), due tothe lack of interesting analytic bump functions. We will leave it to the interested reader to adapt themethod below to verify that other later notions that we define locally are equivalent to the “global”definitions found in the course text or in other books. The local definitions are always adequate forall proofs, so for our purposes it doesn’t really matter than this equivalence holds (and in a senseit is better to not use it early in the development of the theory, in order that one develops goodhabits of mind that are necessary for the theories of real-analytic and complex-analytic manifolds).2. Another perspective on tangent vectorsLet X be a Cp-premanifold with corners, 0 ≤ p ≤ ∞, and Oxthe local ring at x ∈ X. Let mxbethe kernel of the evaluation map ex: Ox R defined by f 7→ f(x). By the very definition of howwe add and multiply functions, it is clear that exis a ring homomorphism such that ex(c) = c forall c ∈ R (so in particular, exis surjective). To be a bit pedantic, here are the details. Let (U, f)and (V, g) be representatives for germs in Ox. By definition of the R-algebra structure on Ox, forany a, b ∈ R, a[(U, f)] + b[(V, g)] is represented by (U ∩ V, a · f|U∩V+ b · g|U∩V). By the definitionof the R-algebra structure on the set of R-valued functions on open sets in X (such as U ∩V ), thevalue of a · f|U∩V+ b · g|U∩Vat x ∈ X is a · f(x) + b · g(x) = a · ex([(U, f)]) + b · ex([(V, g)]). Thissays exactly that for s, s0∈ Oxand a, a0∈ R, ex(as + a0s0) = aex(s) + a0ex(s0). This shows that exis R-linear. The same argument proves ex(c) = c for c ∈ R and ex(ss0) = ex(s)ex(s0), so e is a mapof R-algebras.Somewhat more amusing is the fact that such properties uniquely characterize the evaluationmap ex; that is, if we focus on the R-algebra Oxand discard the space X and point x that gaverise to it, we can still define the map ex. To make sense of this, the key is the following specialproperty of the local ring at x:Lemma 2.1. A germ f ∈ Oxhas a multiplicative inverse in Ox(i.e., fg = 1 in Oxfor someg ∈ Ox) if and only if f 6∈ mx. In other words, the kernel mxof exis precisely the set of f ∈ Oxthat do not have a multiplicative inverse.We warn the reader that if we choose a representative for a germ, to say that the germ hasa multiplicative invers e in the ring of germs does not imply that the represe ntative function isnowhere zero on its domain; all one can infer is that it has to be non-vanishing near x.Proof. If f ∈ Oxsatisfies fh = 1 for some h ∈ Oxthen applying exgives ex(f)ex(h) = ex(fh) =ex(1) = 1 in R, so ex(f) 6= 0. That is, if f ∈ Oxhas a multiplicative inverse then f 6∈ mx.12Conversely, if f 6∈ mxthen we want to show that f has a multiplicative inverse. In terms ofa representative function on an open, it suffices to show that if U is an open set around x andf ∈ O (U) satisfies f(x) 6= 0 then on some small open U0⊆ U around x there exists h ∈ O(U0) thatis a multiplicative inverse to f|U0. To prove this we may shrink U around x, and so by continuityof f we may assume that f is non-vanishing on U. Thus, there is a reciprocal 1/f as an R-valuedfunction on U and we just have to check that it lies in O (U). This is a local problem on U, and so byworking in local Cp-charts we can shift the problem over to one for Cp-functions on open domainsin sectors in Rn: if such a function is non-vanishing then its pointwise reciprocal is also Cp. Thisfollows from the stability of the Cpproperty under composition and the fact that x 7→ 1/x is a C∞map from R×to R×. (Or one could use Whitney’s extension theorem to reduce the problem tothe case of non-vanishing Cpfunctions on open sets in Rn, at least for the non-trivial case p > 0,but that would be overkill for the present circumstances.) Now we can prove the uniqueness for ex:Theorem 2.2. Let e : Ox→ R be a ring homomorphism such that e(c) = c for all c ∈ R (i.e., eis a map of R-algebras). The map e must equal ex.The significance is this: the R-algebra map exis intrinsic to the R-algebra Oxwithout needingto mention the geometric data such as X or x that gave rise to Ox.Proof. Since e(c) = c for all c ∈ R, certainly e is surjective. We next check that e kills mx. If f ∈ mxis not killed by e, so e(f) = c ∈ R is nonzero, then e(f −c) = e(f)−e(c) = c−c = 0. However, sincef ∈ mxand c 6∈ mx(as ex(c) = c 6= 0), we must have f −c 6∈ mxbecause mx= ker exis an R-linearsubspace of Ox. Thus, by the lemma it follows that f −c ∈ Oxhas a multiplicative inverse, say h.Applying the ring homomorphism e to the identity (f −c)h = 1 in Oxgives e(f −c)e(h) = e(1) = 1in R, a contradiction since e(f −c) = 0. Thus, e(f) must have been zero after all. This shows thate kills mx.It follows that e and exagree on mx⊆ Ox(both vanish) and they agree on R ⊆ Ox(both sendc to c for all c ∈ R). Hence, to conclude e = exit suffices to show R ⊕mx= Ox. That is, we wantthat each f ∈ Oxis uniquely expressible as f = c + f0with c ∈ R and f0∈ mx. The existencefollows from the definition of mxand the identity f = f(x) + (f −f(x)). As for uniqueness, we justneed R ∩ mx= 0, and this is clear. In the C∞case, we can prove that not only the evaluation map at x but even the notion oftangent vector at x is intrinsic to the R-algebra Oxand does not need …


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Stanford MATH 396 - Globalization vua Bump Functions

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