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Smooth Structures on Spheres

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Smooth Structures on Spheres Math 120B Discussion Session Week 4 Notes April 25, 2019 Today we'll discuss some basic notions about manifolds. First, we'll construct an atlas for the n n+1 unit sphere S ⊂ R , and then we'll obtain some manifolds as level sets of smooth functions. Smooth structures on spheres Recall that a smooth manifold consists of two pieces of data: a metric space1 M, and an atlas, con- sisting of coordinate charts (U; φ) on M. To qualify as an atlas, this collection of charts must cover M, must have smooth transition functions, and must be maximal with respect to this property. It is common to refer to the atlas as a smooth structure on M. Because atlases have this property of being maximal, we can determine a smooth structure on M by covering M with coordinate charts whose pairwise transition maps are smooth. We simply declare that A contains these coordinate charts, and then we're obligated to toss in any other coordinate charts which are compatible with these. Let's see an example of this when M = Sn. Just as in the n = 2 case, we may use stereographic projection to cover Sn with just two charts. We have n 2 2 n+1 S = f(x1; : : : ; xn+1)jx1 + ··· + xn+1 = 1g ⊂ R : n n Now consider open sets U± = S n f(0;:::; 0; ∓1)g and define φ± : U± ! R by x1 xn φ±(x1; : : : ; xn+1) = ; ··· ; : 1 ± xn+1 1 ± xn+1 Notice that φ+ omits the south pole, while φ− omits the north pole, and that the image of each of n these charts is R . By now you've probably computed the transition map −1 n n φ− ◦ φ+ : R n 0 ! R n 0 at least fifteen times and found it to be smooth. This tells us that the coordinate charts (U+; φ+) and (U−; φ−) can live in the same atlas A, but additionally they determine an atlas, since they cover M = Sn. As above, we simply add to A all the coordinate charts which are compatible with (U+; φ+) and (U−; φ−). Recall that smooth manifolds (M; A) and (N; B) are diffeomorphic if there is a homeomorphism f : M ! N with the property that every composition ◦ f ◦ φ−1 is smooth, where (U; φ) 2 A and (V; ) 2 B. This leads to an interesting question. Can we choose an atlas on Sn which is not diffeomorphic to the atlas A constructed above? Perhaps we can start with a different cover of Sn by coordinate charts and end up with a totally new smooth structure. Such an atlas, not diffeo- morphic to the one we've constructed, is known as an exotic smooth structure on Sn. 1This is our textbook's convention. A more common approach is to demand that M be a topological manifold, which means that it's a Hausdorff, second countable, locally Euclidean space. Avoiding the definitions of these terms is nice for now, but you may care about topological manifolds at some point. 1 Figure 1: A family of smooth manifolds, along with a singular space. The smooth Poincar´econjecture asserts that no exotic structures exist on Sn. This means that if we have any atlas A~ on Sn, then (Sn; A) and (Sn; A~) are diffeomorphic. The smooth Poincar´e conjecture is known to be true for some small dimensions, but in 1956, Milnor constructed exotic smooth structures on S7. In fact, Kervaire and Milnor went on to show that S7 admits exactly 28 distinct smooth structures. Here are some other values computed by Kervaire and Milnor: n 1 2 3 4 5 6 7 8 9 10 11 12 smooth structures on Sn 1 1 1 ?? 1 1 28 2 8 6 992 1 Unfortunately we can't get into the construction of exotic structures here, but notice that the smooth Poincar´econjecture remains open for n = 4. Hypersurfaces as manifolds In lecture you showed that a good way to produce a differentiable n-manifold is as a hypersurface n+1 of a smooth function f : R ! R. We're not going to reprove this, but we'll state it in a (slightly) more general form before giving some examples and discussing why this is a reasonable construction. m k Fact. Let F : R ! R be a smooth function, for some 1 ≤ k ≤ m, and let JF (p) be the Jacobian matrix 0 1 1 1 @F ··· @F @x1 @xm B . C JF = B . .. C @ k k A @F ··· @F @x1 @xm m k of F , evaluated at p 2 R . Fix c 2 R and set m Mc = fp 2 R jf(p) = cg: m If rank(JF (p)) = k for all p 2 Mc, then Mc ⊂ R is a smooth manifold of dimension m − k. 2 2 Figure 2: Some level curves of a smooth function h: T ! R, two of which are submanifolds. 3 2 2 2 Example 1. Consider the function F : R ! R defined by F (x; y; z) = x + y − z . The Jacobian matrix JF (x; y; z) = 2x 2y −2z will have full rank whenever (x; y; z) 6= (0; 0; 0). So for any constant c 6= 0 = F (0; 0; 0), F −1(c) is a smooth 2-manifold. If c < 0, then Mc is a hyperboloid of two sheets, and in fact we can show that for any c0; c1 < 0, the manifolds Mc0 and Mc1 are diffeomorphic. On the other hand, when c > 0, Mc is a hyperboloid of one sheet, and again we have the property that all such Mc are diffeomorphic. What happens when c = 0? Then M0 is the singular set f(x; y; z)jx2 + y2 = z2g; 3 which is a (double) cone in R . As we can see in Figure1, our family of manifolds Mc, with c < 0, degenerates into this cone as c approaches 0 and comes out on the other side (when c > 0) as a different family of smooth manifolds, topologically distinct from the previous family. The tangent space to a smooth manifold hasn't yet been discussed in lecture, but for hypersur- faces in Euclidean space, we can make sense of this in the same way we approached the tangent 3 N space to a surface in R . Namely, the tangent space to M ⊂ R can be thought of as all the possible velocity vectors of curves in M: TpM = fγ_ (0)jγ :(−, ) ! M has γ(0) = pg: m k If we realize M as a regular hypersurface of a smooth function F : R ! R , then TpM ought to be a vector space of dimension m − k. How can we realize this? Let γ :(−, ) ! M be a curve in k M with γ(0) = p. Because M is a regular hypersurface of F , the function F ◦ γ :(−, ) ! R is constant, so differentiating gives k JF (γ(t)) · γ_ (t) = 0 2 R ; for every t 2 (−, ). In particular, this holds when t = 0. Soγ _ (0) is in the kernel of JF (p), and in fact we can write TpM = ker JF (p): Because JF (p) has rank k, its kernel has dimension m − k | precisely the dimension we expected for TpM. We can think of the regularity condition rank(JF (p)) = k as demanding that the tangent space to F −1(c) be reasonable. 3 Digression. Because we haven't yet defined the tangent space to an abstract manifold, we can't yet speak very rigorously about what it means for a map F : M ! N of manifolds to have full rank. But if we could, then the above fact would continue to hold. In particular, if c 2 N is a regular value of a smooth map F : M ! N of manifolds, then F −1(c) ⊂ M is a smooth submanifold of dimension dim M − dim N. Just as before, these smooth submanifolds vary in continuous families, with changes in their topology occurring whenever c passes through a critical point. For instance, 2 consider the height function h: T ! R on the torus, depicted in Figure2. Can you identify the critical points of this function? Which preimages h−1(c) are smooth submanifolds, and how does the topology of the preimages change as c passes through critical points? A manifold of matrices The last example of a smooth manifold that we'll consider is a bit more abstract. Let M(n; R) n2 denote the collection of n × n matrices with real entries. We can identify M(n; R) with R entry- wise. Now let SL(n; R) denote the special linear group | that is, n × n matrices with determinant 1. Then SL(n; R) is the preimage of 1 under the map f : M(N; R) ! R : A 7! det A: This map is certainly smooth, since det A can be written as a polynomial in the entries of A. The question, then, is whether the Jacobian matrix Jf (A) has full rank whenever A 2 SL(n; R). Let's consider the case n = 2. Then we may write a b A = and f(A) = ad − bc: c d So the Jacobian matrix Jf (A) is given by Jf (A) = d −c −b a : As long as A is not the zero matrix, this will have full rank. In particular, Jf (A) has full rank whenever A 2 SL(2; R), so SL(2; R) is a smooth manifold of dimension 3. For a general n we can similarly check that Jf (A) has full rank whenever A 6= 0, so SL(n; R) is a smooth submanifold of 2 M(n; R) of dimension n − 1.
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