By Chern S.S., Li P., Cheng S.Y., Tian G. (eds.)

Those chosen papers of S.S. Chern talk about themes akin to fundamental geometry in Klein areas, a theorem on orientable surfaces in 4-dimensional area, and transgression in linked bundles

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Let (y0 , . . , yL ) be in YL (πs ). As πu is onto, we may ﬁnd z in Z such that πu (z) = πs (y0 ). Then (y0 , . . , yL , z) is in ΣL,0 (π) and its image under ρL, is (y0 , . . , yL ). Next, we check that ρL, is u-resolving. Suppose that (y0 , . . , yL , z0 ) and (y0 , . . , yL , z0 ) are unstably equivalent and have the same image under ρL, . The ﬁrst fact implies, in particular, that z0 and z0 are unstably equivalent. The second fact just means that (y0 , . . , yL ) = (y0 , . . , yL ). Since the points are in ΣL,0 , we also have πu (z0 ) = πs (y0 ) = πs (y0 ) = πu (z0 ).

If π is an s/u-bijective pair for (X, ϕ), then for all L, M ≥ 0, (ΣL,M (π), σ) is a Smale space. In fact, we can say more. 6. If π is an s/u-bijective pair for (X, ϕ), then for all L, M ≥ 0, (ΣL,M (π), σ) is a shift of ﬁnite type. Proof. We begin with the case L = M = 0. The map ρu : (Σ(π), σ) → (Y, ψ) is u-bijective. 12. On the other hand, by deﬁnition, the unstable sets of (Y, ψ) are totally disconnected. A similar argument using Z in stead of Y shows that the stable sets of Σ(π) are totally disconnected.

Each singleton {+∞} and {n}, n ∈ Z is in CO s (Σ, σ). Now, we have some notational diﬃculties because our space carries an obvious order structure and we would like to look at intervals, such as [−∞, n] = {a | −∞ ≤ a ≤ n}. Unfortunately, as we are in a Smale space, the bracket has another meaning. We use [, ] in the order sense only. Moreover, each interval [−∞, n], n ∈ Z is also in CO s (Σ, σ). Notice that, for n ≥ 1, the Smale bracket of n with +∞ is +∞ and so {n} ∼ {+∞}. It follows that in the group Ds (Σ, σ), < [−∞, n] >=< [−∞, m] >, for every m, n ∈ Z and < {n} >=< {+∞} >, for every n in Z.