Math 15B: Solutions 1 Due Thursday, January 18, 018 (1) Let π : X X be a covering space. Let Φ be a smooth structure on X. Prove that there is a smooth structure Φ on X so that π : ( X, Φ) (X, Φ) is an immersion. Solution. Note that X is Hausdorff: Any two points not in the same fiber of π can be separated by preimages of open sets in X, and any two points in the same fiber can be separated due the covering space condition. For X to be second-countable it is necessary and sufficient that the fibers of π are at most countable. So we assume that this is the case. Now we need to construct a compatible system of charts for X. Let y X and x := π(y) X. Let ϕ : V U R n be a chart around x belonging to ϕ. Since π is a covering map, there an open subset W X containing x and such that π 1 (W ) = i I W i for some index set I, where each W i is an open subset of X which is mapped homeomorphically mapped onto W by π. let j I be the unique index for which y W j. Then ψ y,ϕ,w := ϕ V W π π 1 (V ) W j : π 1 (V ) W j ϕ(v W ) is a homeomorphism onto an open subset of R n, that is a chart for X around y. topological manifold. Thus X is a We claim this system of charts even defines a smooth structure. Suppose ψ := ψ y,ϕ,w and ψ := ψ y,ϕ,w are two such charts, where ϕ : V U and ϕ : V U are charts for X. Let Ṽ and Ṽ denote the domains of ϕ and ϕ respectively. Then by construction π V has a continuous inverse, and ψ 1 = π 1 Ṽ ϕ 1. Thus ψ ψ 1 = ϕ π π Ṽ 1 ϕ 1 = ϕ ϕ 1. This map is a diffeomorphism on the open subset of U where it is defined, since the charts ϕ and ϕ are compatible. This shows that the charts we constructed for X are compatible and thus can be extended to a smooth structure. () Prove that the following are submanifolds of the space of n n matrices, Mat n,n (R) = R n. Compute their dimensions. 1. GL n (R). SL n (R) 3. SO(n). Solution. 1. GL n (R) is the set of matrices with nonzero determinant. The determinant function det : Mat n,n (R) R is a polynomial in the matrix entries and is therefore smooth and in particular continuous. Thus GL n (R) = det 1 (R \ {0}) is an open subset of Mat n,n (R) as the preimage of an open set under a continuos function. As an open subset of a manifold it is a submanifold.. SL n (R) is the set of matrices with determinant one, that is the set det 1 ({1}). If we can show that 1 is a regular value of the determinant, it follows that SL n (R) is a submanifold of Mat n,n (R) by the implicit function thorem. So let A be any matrix with det A = 1. We 1
want to show that D A det is surjective. Let M i,j denote the (i, j)-th minor of A, i.e. the determinant of the Matrix obtained by deleting the i-th row and j-th column from A. By the Laplace formula for the first row of A, we have M 1,j 0 for some 1 j n, since otherwise we would have det A = 0. Let B t be the matrix obtained by replacing the first row of A with a vector whose only nonzero entry is ( 1) j t in position j. Then det B t = tm 1,j. Since the determinant is linear in rows, we have det(a + B t ) = det A + det B t = det A + tm 1,j. Thus the derivative of the function t det(a + B t ) at t = 0 is M 1,j which is nonzero. This shows that D A det is surjective by the chain rule. 3. Since SO(n) is a connected component of O(n), it is enough to show that the latter is a submanifold of the space of matrices. Consider the space Sym n (R) of symmetric n n matrices. Any such matrix is uniquely determined by its entries on and above the diagonal, and there are no restrictions for what these entries can be. Thus Sym n (R) is a real vector space of dimension n(n + 1)/, and as such has a canonical smooth structure (using the identification with euclidean space coming from any choice of basis). Consider the map Φ : Mat n,n Sym n (R), A A T A. By definition of the orthogonal group, O(n) = Φ 1 ({Id n }). So by the implicit function theorem, it is enough to show that Id n is a regular value for Φ. Let A O(n). We claim that D A Φ is surjective. Indeed for B Mat n,n we have Φ(A + tb) = (A + tb) T (A + tb) = A T A + t(a T B + B T A) + t B T B. Thus D A Φ B = A T B + B T A. Now let C Sym n (R). Then D A Φ AC = AT (AC) + (AC)T A = AT AC + C T A T A = C, using that A O(n). Since C Sym n (R) was arbitrary, this shows surjectivity of D A Φ. (3) Do the following problems from Hirsch, Differential Topology : Chapter 1, section 1, # 1,, 3 Chapter 1, section, # 10, 1, 13 Chapter 1, Section 1, Exercise 1: Solution. The Grassmannian G n,k is defined as the set of k-planes of R n. There are several equivalent ways to define a topology on the Grassmannian, for example one can introduce a metric by d(e, E ) = max{d(x, y) x E, y E, x = y = 1} or one cold require that the open subsets are exactly the sets of planes that intersect a given open subset of R n. We are the topology as the one coming from a chosen system of charts as in the gluing lemma (since the set is already given, the cocycle condition automatically holds). The verification that this coincides with other possible definitions is omitted. For every k-plane E we define an open subset U E G n,k and a chart ϕ E : U E Hom(E, E ) = R k(n k), where E denote the orthogonal complement of E, as follows: Let π E, π E denote the
orthogonal projections on E and E respectively. We define U E as the open subset of k-planes F with F E = {0}. This implies that for every F U E the kernel of π E F is trivial. Thus π E F is invertible and we can set ϕ E (F ) := π E (π E F ) 1. We claim that ϕ is a bijection. Indeed, we can decompose R n orthogonally as R n = E E. This identifies U E with the set of linear subspaces of E E on which the projection to the first factor is an isomorphism, which is exactly the set of graphs of linear maps E E. Then ϕ E is just the (inverse of the) correspondence between a function and its graph and thus bijective. Now suppose that E is another k-plane. We want to show that the two charts are compatible. For this choose a basis B = (b 1,..., b n ) on R n such that E = b 1,..., b k and e = b k+1,..., b n. This data gives a canonical identification Hom(E, E ) Mat n k,k (R). Now let C = (c 1,..., c k ) be some basis of E, D = (d 1,..., d n k ) a basis of E and P B,C and P B,D be the matrices representing π E and π E in the chosen bases respectively. Then ϕ E(U E U E ) Hom(E, E ) is exactly the set of maps α such that π E maps the graph of α bijectively onto E. Under the identification with Mat n k,k, this is the set of matrices A, such that ( )) Id k det (P B,C 0. A Since Matrix multiplication and determinant are continuous, this specifies an open subset in the space of matrices. On this open subset, the map ϕ E ϕ 1 E is explicitly given by ( )) 1 Id k A P B,D A (P B,C. A This is a rational function in the coordinates of A and therefore smooth wherever it is defined. Now by the gluing Lemma, G n,k is a topological manifold. As we have shown, the given charts even specify a smooth (even analytic) structure. Chapter 1, Section 1, Exercise /3: Solution. Consider CP n = C n+1 \{0}/C with the quotient topology. On CP n we have homogeneous coordinates [z 0,..., z n ]. Let U i = {[z 0,..., z n ] z i 0}, which is open, since its preimage under the quotient map is, and ϕ i : U i C n, [z 0,..., z n ] (z 0 /z i,..., ẑi/z i,..., z n /z i ), where the hat denotes omission. Since composition of ϕ i with the quotient map gives a continuous map from an open subset of C n+1 \ {0}, it follows that the ϕ i are continuous by the universal property of the quotient topology. Moreover, ϕ i has an inverse given by (w 1,..., w n ) [w 1,..., w i, 1, w i+1,..., w n ], which is continuous since it factors through a continuous map to C m+1 \ {0}. This shows that the ϕ i are homeomorphisms and thus CP n is a topological manifold. 3
For i < j, the transition function ϕ ij = ϕ j ϕ 1 i is given by (w 1,..., w n ) (w 1,..., w i, 1, w i+1..., w j 1, w j+1..., w n ) 1 w j, defined wherever w j is nonzero. Since the field operations on the complex numbers are rational functions in the natural real coordinates, this is clearly a smooth analytic function. The solution works verbatim with the complex numbers replaced by the quaternions, with the caution that it is only a skew-field. Chapter 1, Section, Exercise 10.b): Show that SO(3) = RP 3. Solution. We will to show that there exists a surjective homomorphism S 3 SO(3) which is a local diffeomorphism and that all preimages are a set of antipodal points on the sphere. Since RP 3 = S 3 /{±1}, this gives the desired result. To construct this map, consider the quartenions H = {w + xi + yj + zk w, x, y, z R}. The space of nonzero quaternions is naturally a Lie-group with respect to quaternionic multiplication. Quaternionic multiplication fulfills q 1 q = q 1 q and thus restricts to a natural group structure on the unit quaternions, which we identify with the 3-sphere S 3 H. We also identify R 3 with the space of totally imaginary quaternions via (x, y, z) xi + yj + zk. Now it is a straightforward to check that for r, q H with r totally imaginary, also qrq, where the overline denotes conjugation. Under the above identifications, this defines an action S 3 R 3 R 3, (q, r) qrq and one can check, which corresponds to a group homomorphism Φ : H \ {0} GL(3). By inspection one sees that Φ is given by a polynomials, so in particular it is smooth. For a unit quaternion q, we have qrq = r, so S 3 acts by isometries under this action, in particular Φ(S 3 ) is contained in O(3), and since the sphere is connected even in SO(3). Let Ψ : S 3 SO(3) denote the restriction. We claim it is a local diffeomorphism. By homogeneity and since both manifolds are 3-dimensional, it is enough to show that the map of tangent spaces D 1 Ψ : T 1 S 3 T Idn SO(3) is bijective. Since Ψ is a restriction of Φ, it is enough to show that T 1 Φ has trivial kernel on T 1 S 3 T 1 H. Note that T 1 S 3 is canonically identified with the totally imaginary quaternions. So let s 0 be totally imaginary and consider the map ϕ s (t) = Φ(1 + st) with ϕ s (t)(r) = (1 + st)r(1 + st) = r + t(qr + rs) + t srs = r + t(sr rs) t srs Thus D 1 Φ s is the linear map r sr rs. Taking r to be any totally imaginary quaternion that does not commute with s (e.g. i if s is not a real multiple of i and j otherwise), shows that this is nonzero. Thus we have shown that Ψ is a local diffeomorphism. Since both S 3 is compact, its image in SO(3) is closed, but since Ψ is a local diffeomorphism also open, thus equal to SO(3), which is connected. 4
It remains to determine the kernel of Ψ. That is those unit quaternions q such that for every totally imaginary quaternion r we have qrq = r, or equivalently qr = rq. By taking i and j for r, one sees that this implies q is totally real. Thus q = {±1}. 5