A manifold on the real commutative Banach algebra $C([0,1];\mathbb R)$ that cannot be embedded in the finite-dimensional Euclidean space $C([0,1];\mathbb R^n)

See discussions, stats, and author profiles for this publication at: https://www.researchgate.net/publication/335568846 A manifold on the real commutative Banach algebra $C([0,1];\mathbb R)$ that cannot be embedded in the finite-dimensional Euclidean space $C([0,1];\mathbb R^n) Preprint · September 2019 DOI: 10.13140/RG.2.2.17114.90562 CITATIONS 0 READS 13 1 author: Hiroki Yagisita Kyoto Sangyo University 27 PUBLICATIONS227 CITATIONS SEE PROFILE All content following this page was uploaded by Hiroki Yagisita on 02 September 2019. The user has requested enhancement of the downloaded file. arXiv:submit/2826675 [math.GT] 2 Sep 2019 A manifold on the real commutative Banach algebra C([0,1]; R)that cannot be embedded in the finite-dimensional Euclidean space C([0,1]; Rn) Hiroki Yagisita (Kyoto Sangyo University) Abstract: C([0,1]; R) and R2( = C({0,1};R) ) are simple examples of a real commutative Banach algebra. C([0,1]; C) and C2( = C({0,1};C) ) are simple examples of a commutative C∗-algebra. Here, we consider C([0,1]; R), which is the set of all real-valued continuous functions on the bounded closed interval [0,1]. The interval is a contractible compact Hausdorff space. The direct product space (C([0,1]; R))n( = C([0,1]; Rn) ) is a real Banach space and a free C([0,1]; R)-module. A C1-mapping from an open set of C([0,1]; Rn) to C([0,1]; R) is said to be C([0,1]; R)-smooth, if its Frechet derivatives are C([0,1]; R)-linear. Then, we can define a concept of an n-dimensional smooth C([0,1]; R)-manifold. In this memo, we see existence of a connected metrizable 1-dimensional smooth C([0,1]; R)-manifold that cannot be embedded in C([0,1]; Rn). In Section 1, as a primitive observation, we see that embeddability in the Cartesian space C([0,1]; Rn) as a smooth C([0,1]; R)-submanifold implies existence of an analog of a bump function. Then, in Section 2, we construct an example where no such analog exists. Keywords: Serre-Swan theorem, complex vector bundle, locally trivial fiber space, partition of unity, Gelfand representation, von Neumann algebra, Radon measure, vector sheaf. 1 1 Primitive obstruction Here, we make a primitive observation about a bump function and a separation function. Proposition 1 (Embedding condition) : Let Mbe a C([0,1]; R)-smooth manifold. Suppose that there exist n∈N and an embedding of Min C([0,1]; Rn) as a smooth C([0,1]; R)-submanifold. Then, the followings hold. (1) Let Ube an open set of M. Let a∈U. Then, there exist open sets Vand Wof Mand a C([0,1]; R)-smooth mapping ffrom Mto C([0,1]; R) such that the followings hold. a∈W, W⊂V, V ⊂U hold. For any x∈Mand t∈[0,1], 0≤(f(x))(t)≤1 holds. For any y∈Wand t∈[0,1], (f(y))(t) = 1 holds. Let f.supp (f) := {z∈M| ∀ t∈[0,1] : (f(z))(t)6= 0 }. Then, f.supp (f)⊂V holds. (2) Let a, b ∈Mand a6=b. Then, there exist a C([0,1]; R)-smooth mapping ffrom Mto C([0,1]; R) and t0∈[0,1] such that (f(a))(t0)6= (f(b))(t0) holds. Remark : In general, f.supp (f)⊂supp (f) holds. Usually, f.supp (f)6= supp (f) holds. 2 Proof : (1) Let ρbe a usual bump C∞-function at 0 on Rn. Let εbe a small positive number. Then, as we put (f(x))(t) := ρ(x−a)(t) ε for x∈M(⊂C([0,1]; Rn) ) and t∈[0,1], it follows. (2) Let U:= M\ {b}. Then, from (1), it follows.  2 Simple example For convenience of description, we consider [−1,+1] instead of [0,1]. Theorem 2 (Strange cylinder) : There exists a connected metrizable 1-dimensional C([−1,+1]; R)-smooth manifold Msuch that for any n∈N, there exists no embedding of Min C([−1,+1]; Rn) as a smooth C([−1,+1]; R)-submanifold. Proof : We define an equivalence relation ∼of C([−1,+1]; R) as follows. Let (x, y)∈(C([−1,+1]; R))2. Then, x∼yholds, if and only if there exists n∈Zsuch that for any t∈[−1,+1], y(t) = x(t) + nt holds. So, let M:= (C([−1,+1]; R))/∼. Let a(t) := 0 (−1≤t≤0), t(0 ≤t≤+1). Then, 0 6∼ aholds. That is, [a]6= [0] holds. Now, we show that there exists no bump function for ([a], M \ {[0]}). To that end, we show that no C([−1,+1]; R)-smooth function on Mseparates [0] and [a]. That is, we show that for any C([−1,+1]; R)-smooth mapping ffrom Mto C([−1,+1]; R), f([0]) = f([a]) holds. Let fbe a C([−1,+1]; R)-smooth map from Mto C([−1,+1]; R). Let gs:= (1 −s)0 + sa (s∈[0,1]). Then, for any s1, s2∈[0,1] and t∈[−1,0], gs1(t) = gs2(t) 3 holds. Hence, because fis C([−1,+1]; R)-smooth, for any t∈[−1,0], the map s∈[0,1] 7→ (f[gs])(t)∈Ris locally constant and, so, (f([0]))(t) = (f([g0]))(t) = (f([g1]))(t) = (f([a]))(t) holds. On the other hand, let hs:= (1 −s)t+sa (s∈[0,1]). Then, for any s1, s2∈[0,1] and t∈[0,+1], hs1(t) = hs2(t) holds. Hence, in virtue of 0∼t, for any t∈[0,+1], (f([0]))(t) = (f([t]))(t) = (f([h0]))(t) = (f([h1]))(t) = (f([a]))(t) holds. Therefore, f([0]) = f([a]) holds. That is, no C([−1,+1]; R)-smooth function on Mseparates [0] and [a]. So, there exists no bump function for ([a], M \ {[0]}).  Problem 3 : We define an equivalence relation ∼of Mas follows. Let ([x],[y]) ∈M2. Then, [x]∼[y] holds, if and only if there exists m∈Zsuch that [y] = [x] + m[a] holds. So, let N:= M/ ∼. Then, do there exist k∈Nand an embedding of Nin C([−1,+1]; Rk) as a smooth C([−1,+1]; R)-submanifold ? Comment : Kasuya proposed a candidate for a compact C2-manifold Nsuch that for any C-manifolds M1and M2,Ncan not be embedded in M1×M2 as a C2-manifold. The construction of our example was inspired by this proposal. — Problem 4 :L∞([0,1]; C) is a commutative von Neumann algebra. Consider a L∞([0,1]; R)-manifold and a L∞([0,1]; C)-manifold. — 4 Oral : If there is not much previous research, it will be either Case 1: It has not been considered seriously. or Case 2: It is difficult to find related interesting problems. Or, even if you find a problem, it is difficult to get even a “partial result”. By the way, manifolds on a (real or complex) commutative Banach algebra are very natural objects. Nevertheless, even an ordinary mathematician like me can get some “slight partial results” with just a few thoughts, and is likely to hit the case 1. Now, it is great to create a novel breakthrough for a subject that has been studied extensively, but this is difficult. If you are a mediocre mathematician like me, you will likely be able to do a much better study on a subject that has not yet been considered. But, there is one problem here. Because the research field has not been established, even if a result is obtained, there are not many people who immediately evaluate it. Therefore, it is not recommended that people who want to acquire academic posts in the future do their main research. 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Yagisita, A remark on locally direct product subsets in a topological Cartesian space, arXiv.org. 6 View publication statsView publication stats