Math 05B Homework Edward Burkard 0.0. Lecture Problems. 1. Show that a differentiable function is continuous. 0. Kaiser Wilhelm Problems Let f : R n R m be a differentiable function. Then for any ɛ > 0 there is a δ > 0 such that if h < δ, then f(x + h) f(x) L(h) < ɛ. h So f(x + h) f(x) L(h) < ɛ h, and f(x + h) f(x) L(h) f(x + h) f(x) L(h). Thus: f(x + h) f(x) L(h) < ɛ h f(x + h) f(x) < ɛ h + L(h) Let k = ɛ + Df(x) and δ = ɛ k, then we get continuity. < ɛ h + h Df(x) = (ɛ + Df(x) ) h. Let A R n and B R n be closed sets such that ζ A and η B. Define Give F (A, B) the uniform metric: F (A, B) := {ϕ : A B ϕ(ζ) = η}. d(ϕ, ψ) = sup { ϕ(x) ψ(x) }. x A Show that (F (A, B), d) is a complete metric space. Let {ϕ n } n I be a Cauchy sequence in F (A, B). Thus ɛ > 0 N N such that m, n > N we have d(ϕ m, ϕ n ) < ɛ. Equivalently stated we have sup x A { ϕ m (x) ϕ n (x) } < ɛ. If we fix a A then {ϕ n (a)} is a Cauchy sequence in B, a complete metric space (as B is a closed subset of R n which is a complete metric space). Thus ϕ n (a) ϕ(a). Since this is true for all a A we get that ϕ n ϕ F (A, B) 3. Use Invariance of Domain to show that local dimension is well-defined and locally constant (and hence constant if M is connected). Let M be a topological manifold and let x M. Suppose that m and n are local dimensions for some charts containing x. Let (U, ϕ) be the chart where ϕ : U ϕ(u) R m and let (V, ψ) be the chart where ψ : V ψ(v ) R n. Since every manifold is locally (path) connected choose C U V to be the open connected set containing x. Since ϕ and ψ are homeomorphisms, we have that ϕ : C ϕ(c) and ψ : C ψ(c) are homeomorphisms, and, moreover, ϕ(c) and ψ(c) are connected. Note the fact that an open connected subset of R k is homeomorphic to R k. Let h m : ϕ(c) R m and h n : ψ(c) R n be the corresponding homeomorphisms. Then: R n h 1 n ψ(c) ψ 1 C ϕ ϕ(c) hm R m is a homeomorphism. So by the next exercise, which uses Invariance of Domain, we have that n = m. Since these charts were arbitrary we get that local dimension is well defined. Since local dimension is locally constant we have a continuous function d : M N (the dimension function ). If we assume that M is connected then d(m) must be connected and hence d is a constant map. 4. Use Invariance of Domain to show R n is not homeomorphic to R m if m < n. 1
View R m as a subspace of R n. Given any nonempty open subset U of R m, it is not open in R n. (Let U R m be open and let 0 R. When viewed in R n it is of the form: U {0} {0}. This is clearly not }{{} n m times open.) Suppose h : R n R m is a homeomorphism. Define i : R m R n by (x 1,..., x m ) (x 1,..., x m, 0,..., 0). Then i : R m R n is an embedding. Thus i h : R n R n is 1-1 and continuous since i and h are individually. Let V be an open set of R n. Then by Invariance of Domain we get that i h(v ) is open in R n, however this is not possible since h(u) is open in R m and i h(v ) = h(v ) {0} {0}. Thus i h(v ) is not open }{{} n m times in R n which is a contradiction. h is not a homeomorphism, thus R n is not homeomorphic to R m. 5. Show that Φ : M N is smooth at x M iff ψ Φ ϕ 1 is smooth at x for some chart (U, ϕ) with x U and some chart (V, ψ) with Φ(x) V. (= ) Clearly if ψ Φ ϕ 1 is smooth at ϕ(x) for all charts (U, ϕ) with x U and all charts (V, ψ) with Φ(x) V, then it is smooth with respect to any choice of (U, ϕ) and (V, ψ). ( =) Let M = {(U ξ, ϕ ξ )} be the atlas on M and N = {(V η, ψ η )} be the atlas on N. Let (U, ϕ), x U, be a chart on M and (V, ψ), Φ(x) V, a chart on N such that ψ Φ ϕ 1 is smooth at ϕ(x). Let (U ξ, ϕ ξ ) be any chart in M and (V η, ψ η ) be any chart in N such that x U ξ and Φ(x) V η. If we have the case where Φ(U) V, we can replace U by E where E U is an open neighborhood of x such that Φ(E) V. Similarly we can create E ξ such that Φ(E ξ ) V η. Then ψ η Φ ϕ 1 ξ : ϕ ξ (U U ξ ) ψ(v V η ) is given by ϕ ξ (U U ξ ) ϕ ϕ 1 ξ φ(u U ξ ) ψ Φ ϕ 1 ψ(v V η ) ψη ψ 1 ψ η (V V η ) (or quickly viewed as: ψ η ψ 1 (ψ Φ ϕ 1 ) ϕ ϕ 1 ξ = ψ η Φ ϕ 1 ξ ) So we have that ψ η Φ ϕ 1 ξ is smooth at ϕ ξ (x) as it is just a composition of smooth functions. 6. If c 1, c : R R n satisfy d dt c 1(t) t=0 = d dt c (t) t=0, show d dt (f c 1)(t) t=0 = d dt (f c )(t) t=0 f C (U p, R). Let f C (U p, R), and suppose c 1, c : R R n satisfy d dt c 1(t) t=0 = d dt c (t) t=0. We need c 1 (0) = c (0) because if not we have the following counterexample: Suppose c 1 (t) = (cos t, sin t) and c (t) = (cos t +, sin t + ) with f(x, y) = x + y. Then c 1(t) = ( sin t, cos t) = c (t), but f c 1 (t) = sin t + cos t = 1 and f c (t) = So (f c 1 ) (0) = 0 and cos t + 4 cos t + 4 + sin t + 4 sin t + 4 = 4 cos t + 4 sin t + 8. (f c ) (0) = 1 (4 cos t 4 sin t) 1 4 cos t + 4 sin t + 8 t=0 = 1 (4) 1 4 + 8 = 1 0. Then: d dt (f c 1)(t) t=0 = c 1(0)f (c 1 (0)) = c (0)f (c 1 (0)) = c (0)f (c (0)) = d dt (f c )(t) t=0 Thus since f was arbitrary, we are done. 7. Prove part of the Whitney Embedding Theorem.
3 As M is a compact manifold, we can cover M by a finite number of charts (U i, φ i ), i = 1,..., k. We can also assume that there are sets V i covering M such that V i U i for each i. By theorem 10.6 of Bredon there are smooth functions λ i : M R which are 1 on V i and have support in U i. Now define ψ i (p) = λ i (p)φ i (p) for p U i and 0 otherwise. Then each ψ i is smooth as it is a product of smooth functions. Define θ : M (R n ) k R k by and note that θ(p) = (ψ 1 (p),..., ψ k (p), λ 1 (p),..., λ k (p)) Dθ = Dψ 1 Dψ k Dλ 1 Dλ k. Our first claim is that θ is an immersion. Look at a point p M which must lie in some V i, say p V j. Since λ j = 1 near p, ψ j coincides with φ j near p. Thus Dψ j = Dφ j near p, and Dφ j is one-to-one near p as it is a chart. Thus it follows that Dθ is one-to-one at p, and hence since p was arbitrary, θ is an immersion. Now we claim that θ is one-to-one. Suppose that θ(p) = θ(q). Then λ i (p) = λ i (q) for each i. Now p V i for some i and, for that i we have λ i (p) = 1. Thus we have φ i (p) = 1 λ i (p) ψ i(p) = 1 λ i (q) ψ i(q) = φ i (q) which forces that p = q since φ i is one-to-one as it is a chart. Since we have that M is compact, (R n ) k R k is Hausdorff, and θ is one-to-one, onto its image, and continuous (being an immersion), it is a homeomorphism onto its image. So θ is an embedding of M n into R N for some large N. 8. Prove the path-lifting lemma: Let p : E B be a covering map, let p(e 0 ) = b 0. Any path F : [0, 1] B beginning at b 0 has a unique lifting to a path f in E beginning at e 0. Existence: Cover B by open sets U which are evenly covered by p. Pick a partition P = {0 = s 0 < s 1 < < s n = 1} of [0, 1] such that for each i the set f([s i, s i+1 ]) lies in such an open set U by using the Lebesgue number lemma. Define f as follows: First define f(0) = e 0. Then, supposing that f(s) is defined for 0 s s i, define f on [s i, s i+1 ] as follows: The set f([s i, s i+1 ]) lies in some open set U that is evenly covered by p. Let {V α } be a partition of p 1 (U) into slices (i.e. for each p Vα : V α U is a homeomorphism). Now f(s i ) lies in one of these sets, say V 0. Define f(s) for s [s i, s i+1 ] by f(s) = ( ) 1 p V0 (f(s)). Since p V0 is a homeomorphism, f will be continuous on [s i, s i+1 ]. In this way we can define f on all of the subintervals of [0, 1] created by the partition, and since f is uniquely defined on the overlap of each subinterval, by the pasting lemma, f is continuous. Uniqueness: Suppose we have another lifting f of f beginning at e0. Then we must have f(s0 ) = e 0 = f(s 0 ). Suppose that f(s) = f(s) for all s [0, si ]. Let V 0 be defined as above, then for s [s i, s i+1 ], f(s) is defined to be ( p V0 ) 1 (f(s)). As f is a lifting of f is must be that f([si, s i+1 ]) p 1 (U) = V α. Recall that the V α s are open and disjoint, and because f([si, s i+1 ]) is connected, it lies in one, and only one V α. Since f(s i ) = f(s i ) V 0, it must be that f([si, s i+1 ]) V 0. Now it only remains to show that f = f on [si, s i+1 ]. Let s [s i, s i+1 ], then p 1 (s) consists of one point in each V α. Call the point in V 0, y. Then f(s) = y. But, by the same reasoning, f(s) = y, so they are equal on the entire interval [si, s i+1 ]. Thus by induction we have that f = f on [0, 1]. 0.1. Problem Set 1. Problem 1. Recall that a subset C of R n is called convex iff for a, b C the entire line segment ab between a and b lies in C. Let C R n be compact and convex. Show that any C 1 map is Lipshitz. f : C R n Let p, q C. Theorem, i.e.: Let x be the point on the line segment joining x and y given by the Mean Value
4 f(x) f(y) = n f i=1 x i ( x)(x i y i ) < by compactness. Thus: n ( { }) f(x) f(y) f ( x)(x i y i ) f x i Let M = max ( x) x i Thus f(x) f(y) K x y f is Lipshitz. i=1 n M x i y i = M i=1 n x i y i i=1 Mn x y = K x y Problem. Let C R n be convex and let f : C R n be C 1. Suppose that for some x 0 C and λ > 0 we have and λ = min v R n, v =1 df x 0 (v) Show that for any R with B(x 0, λr) C. Hint: Given y B(f(x 0 ), R) let df x df x0 < 1 λ for all x C. B(f(x 0 ), R) f(b(x 0, λr)) ϕ y : R n R n be ϕ y (x) = x + df y 1 0 (y f(x)) and argue that ϕ y is a contraction mapping of C with Lipshitz constant 1. Notice that ϕy (x) = x is equivalent to f(x) = y. Let y B(f(x 0 ), R) and let ϕ y : R n R n be given by Then ϕ y (x) = x + df y 1 0 (y f(x)). dϕ y (x) = 1 + df 1 y 0 ( df x ) = df y 1 0 df y0 + df y 1 0 df x = df y 1 0 (df y0 df x ) Thus we have dϕ y (x) = df y 1 0 df y0 df x λ 1 λ = 1 ( ). Aside: Since C is convex, let x, y C and define l(t) = (1 t)x + ty, t [0, 1], to be the line in C connecting x and y. Define g(t) = f(l(t)). So g (t) = f (l(t))l (t). Rewriting in terms of the differential: dg t = df l(t) ( x + y) Now take the norm dg t = df l(t) x + y M y x By the mean value theorem there is a t [0, 1] such that Thus f(y) f(x) = dg t and so g(1) g(0) = dg t (1 0) dg t = f(y) f(x) M y x
5 Now apply (*) to the above equation, since dϕ y x < 1 we get ϕ y (x 0 ) ϕ y (x) 1 x 0 x for x, x 0 B(f(x 0 ), R). Thus ϕ y is a contraction mapping with Lipshitz constant 1. Define B = B(x 0, λr). We want to show that ϕ y : B B. ϕ y (x) x 0 = ϕ y (x) ϕ y (x 0 ) + ϕ y (x 0 ) x 0 ϕ y (x) ϕ y (x 0 ) + ϕ y (x 0 ) x 0 = λr df 1 y 0 y f(x 0 ) + 1 x x 0 1 λr + λ R Thus x B(x 0, λr). So ϕ y : B B is a contraction mapping and B is a complete metric space, thus by the contraction mapping principle we have a unique x B such that ϕ y (x) = x. Thus by the definition of ϕ y we have that f(x) = y, which means that y f(b(x 0, λr) (since y = f(x) and x B(x 0, λr). B(x 0, R) f(b(x 0, λr). 0.. Problem Set. Problem 1. a. Show that stereographic projection gives a smooth structure on S n. b. Show that orthogonal projection from hemispheres to the corresponding coordinate subspaces gives a smooth structure on S n. c. Show that the structures in parts a and b are the same. Too Lazy at the Moment. a. Let p = (0, 0,..., 0, 1) be the north pole of S n. Consider the map ϕ : S n \{p} R n given by: 1 ϕ(x) = ϕ(x 1,..., x n+1 ) = (x 1,..., x n ). 1 x n+1 This map is the Stereographic Projection. It has a well known inverse ψ : R n S n \ {p} given by ψ(y) = ψ(y 1,..., y n ) = 1 + y (y 1,..., y n, y 1 ). Clearly these are both continuous as they are just quotients of polynomials. To see that they are bijective I show that their composition each way is the identity: ( ) y1 ϕ(ψ(y)) = ϕ(ψ(y 1,...y n )) = ϕ 1 + y,..., y n 1 + y, y 1 ( ) 1 y1 = y +1 1 + y,..., y n 1 + y = y ψ(ϕ(x)) = ψ(ϕ(x 1,..., x n+1 )) = = 1 + x 1 + +x n (1 x n+1) x 1 1 x n+1,..., 1 x n+1+x n+1 +x 1 + +x n (1 x n+1) ( = (1 x n+1 ) = x x 1 1 x n+1,..., This gives a homeomorphism between R n and S n \ {p}. Similarly we have the maps ϕ x n 1 x n+1, x 1 + +x n (1 x n+1) 1 ( x 1 x n,...,, x ) n+1(1 x n+1 ) 1 x n+1 1 x n+1 (1 x n+1 ) x n 1 x n+1, x n+1 1 x n+1 )
6 Problem. Show that p : R S 1 p(t) = (cos t, sin t) is a cover. Continuous) Clearly p is continuous as its components are continuous functions. Onto) Given any point (x, y) S 1 there is a unique angle 0 θ < π such that (cos θ, sin θ) = (x, y). Choose t = θ. Evenly Covers) Let q = (x, y) S 1 and suppose that p(c) = q where 0 c < π. Let a, b be such that c π < a < c < b < π and p(a) and p(b) form the end points of an interval on the circle containing q. Let G = (a, b), then q p(g) = (p(a), p(b)) and p 1 ( (p(a), p(b))) = (a + kπ, b + kπ). Thus p evenly covers S 1. Therefore p is a covering map. Problem 3. Let p : M N be a covering map. Show that a smooth structure on N determines a unique smooth structure on M so that p is smooth and dp is an isomorphism at every point. k Z The first thing to note here is that p is locally a homeomorphism since for each x M there is a neighborhood U of x with p 1 (U) = V x a disjoint union of open sets and p Vx : V x U is a homeomorphism. Let N = {(U α, ϕ α )} be the smooth structure and use p to pull the smooth structure of N back to M. For any x U α, use p to find a neighborhood U x of x such that p 1 (U x U α ) = V xα and p Vxα is a homeomorphism. Define ψ xα := ϕ α p Vxα, and claim that M = {(V xα, ψ xα )} is a smooth structure for M. Let (V 1, ψ 1 ) and (V, ψ ) be two overlapping charts in M. We want to show: ψ ψ1 1 : ψ 1 (V 1 V ) ψ (V 1 V ) is a diffeomorphism. Note that given any y V 1 V we have p(y) = x p(v 1 V ), so we may choose, for V 1 = U 1α and V = U β, α = β. So ψ ψ1 1 1(V 1 V )) = ψ (V 1 V ) ψ i are diffeomorphisms = ϕ p V (V 1 V ) definition of ψ = ϕ (p(u 1α U α )) α = β = ϕ (U 1 U ) definition of p = ϕ 1 (U 1 U ) smooth structure of N = ψ 1 ψ 1 (V 1 V )) same argument but in reverse Problem 4. Let a group G act on the topological space X. The action is called properly discontinuous if and only if any point x X has a neighborhood U x so that U x g(u x ) = for all g G, g e (the identity of the group). a. If G acts properly discontinuously on X, show that the quotient map π : X X/G is a cover. b. Let X and G be as in part a. If X is a smooth manifold and the action is by diffeomorphisms, show that X/G is also a smooth manifold, and that there is a unique smooth structure on X/G so that π is smooth and dπ is an isomorphism at every point. a. We need to show that π is a covering map. Continuous) We get this for free since quotient maps are automatically continuous by definition.
7 Onto) This also comes for free since quotient maps are defined to be onto. Evenly Covers) Let x X/G. Choose any single point y π 1 (x) and let U y be an appropriate neighborhood of y such that U y g(u y ) = for all g G, g e. Choose U x := π(u y ) to be the open neighborhood of x. Then π 1 (U x ) = g G g(u y ) is a disjoint union of subset all homeomorphic to U x. b. Problem 5. Show that each of the following actions are properly discontinuous. In Parts a d the quotient is homeomorphic to a space that you are already familiar with. Identify the space. In Parts e and f the quotient is called a Lens Space. a. Z acts on R via Z R R b. Z Z acts on R R via (n, t) t + πn. (Z Z) (R R) R R (m, n, s, t) (s + πm, t + πn). c. Z acts on S n via the antipodal map. d. View S 1 as the unit sphere in C. For any k Z with k, Z k acts on S 1 via Z k S 1 S 1 ([n], z) e i(n π k ) z. e. View S 3 as the unit spere in C C. For any k Z with k, Z k acts on S 3 via Z k S 3 S 3 ([n], z 1, z ) (e i(n π k ) z 1, e i(n π k ) z ). The quotient is called a Lens Space. f. Generalize Part e to any odd dimesional sphere. a. This is a circle. This is because the action breaks the line at every nπ, n Z into closed intervals, and then identifies all of those intervals, then identifies the endpoints of the remaining interval, creating a circle. Let x R consider the open set U x := (x π, x + π ). Then for n Z \ {0}, g(u x ) = g(x π, x + π (4n 1)π ) = (x +, x + (4n+1)π ). So clearly U x g(u x ) =. Thus this is a properly discontinuous action. b. This is a torus. This is very similar to the previous action except that it creates π π boxes which are identified of which the parallel edges are identified, which is a torus. Let (x, y) R R consider the open set U (x,y) := (x π, x + π ) (y π, y + π ). Then for (m, n) (Z Z) \ {(0, 0)}, g(u (x,y) ) = g(x π, x + π ) (y π, y + π (4m 1)π ) = (x +, x + (4m+1)π ) (y + (4n 1)π, y + (4n+1)π ). So clearly U x g(u x ) =. Thus this is a properly discontinuous action. c. This is RP n. This action identifies the antipodal points of the sphere{ which is the definition of RP n. p if [m] = 0 We can completely write down this action (Z = {0, 1}): ([m], p) p if [m] = 1. Let p S n. Treat p as the north pole of the sphere and let U p be the upper open hemisphere of S n. Then if n = 1 we have [m](u p ) = L p := { p p U p } (this is the open lower hemisphere of S n relative to the north pole p). Thus we have U p L p = and thus the action is properly discontinuous. d. This is a circle. e.
8 f. The generalization is as follows: View S n+1 as the unit spere in n+1 i=1 C. For any k Z with k, Z k acts on S n+1 via Z k S n+1 S n+1 ([m], z 1, z,..., z n+1 ) (e i(m π k ) z 1, e i(m π k ) z,..., e i(m π k ) z n+1 ). 0.3. Problem Set 3. Problem 1. Identify R with C. Recall that a differentiable map f : R R is Complex Analytic iff when we view f as a map from C to C its partial derivatives are all complex linear. Let the component functions of f be u and v, i.e. f (u, v) or in complex notation f = u + iv. Show that f is Complex Analytic iff it satisfies the Cauchy-Riemann equations u x = v y v x = u y. Hint: the standard matrix representation of df is of course u v x y v x u y A map L : C C is complex linear iff L(z) = ω 0 z for some fixed complex number ω 0. Find necessary conditions on ( a ) b c d that make the corresponding map complex linear. (= ) Suppose that f(z) = u(z) + iv(z) is a complex analytic function. View z as z = x + iy. Recall the complex analysis definition of complex analytic: f is complex analytic if it posses a derivative wherever it is defined. The derivative of f is given by. f f(z + h) f(z) (z) = lim. h 0 h Since the derivative is defined, the limit of the difference quotient is the same no matter how h approaches 0. First choose only real values for h, then y is held constant and f (z) = f x = u x + i v x. Second choose only imaginary values for h, say ik. Then f f(z + k) f(z) (z) = lim = i f k 0 ik y = i u y + v y. Thus we have f x = i f y. Now equating the real and imaginary parts we get the system of equations: { u x = v y v x = u y AKA the Cauchy-Riemann equations. ( =) Suppose that f satisfies the Cauchy-Riemann equations. Then we both f x that f is differentiable and hence complex analytic. and f y exist which means
( ) ( ) a b x Define L(z) = where z = x + iy and let ω c d y 0 = α + iβ. Then ( ) ( ) ( ) a b x ax + by =. c d y cx + dy We want that: (ax + by) + (cx + dy)i = ω 0 (x + iy) = (α + iβ)(x + iy) = (xα yβ) + i(xβ + yα) So we have that a = α = d and b = β = c and thus the conditions on the matrix are that a = d and b = c. 9 0.4. Problem Set 4. Problem 1. Let M be a manifold, U M an open subset, K U a set that is closed in M, and let f : U R be smooth. Prove that the restriction of f to K extends to a smooth function f : M R. We have two disjoint closed sets: K and M \ U. Use Smooth Urysohn s lemma to create a smooth function g : M R with g(k) 1 and g(m \ U) 0. Define the function ˆf : M R by ˆf(x) = { f x U 0 otherwise and define f : M R by f(x) = g(x) ˆf(x). Now notice that, since g 1 on K we have that f = f K. Note that ˆf is smooth on U since it is just f K there, and it is smooth on M \ U since it is constant there, however we do not know if ˆf is smooth on all of M. Since g(x) smoothly approaches 0 as d(x, M \ U) 0 (d is the distance function on M) we have that f is smooth on M since it is the product of two smooth functions. The fact that it is smooth is a result of the fact that g smooths out the transition of ˆf from U to M \ U. Problem. Let f be in C ([0, 1] [0, 1]) and let f : U R be a smooth extention of f to a neighborhood U of [0, 1] [0, 1]. Prove that d f (0,0) depends only on f and not on the choice of f. Let f and f be defined as above. Let points in U be represented by (x, y). Since d f (0,0) = f f x y (0,0) (0,0) it suffices to check that f and f depend only on f. Observe the following: (0,0) (0,0) x This implies that we can think of only using t [0, 1], so: y f f(t, 0) f(0, 0) = lim x t 0 t (0,0) = lim t 0 + f(t, 0) f(0, 0) t f(t, 0) f(0, 0) f(t, 0) f(0, 0) lim = lim t 0 + t t 0 t = f x (0,0) which clearly only depends on f. We can similarly show that f depends only on f and hence that (0,0) d f (0,0) depends only on f. y Problem 3. Recall that the set of invertible n n matrices is a group under matrix multiplication. It is called the General Linear Group, and denoted by Gl n.
10 a. Give a brief explanation of why Gl n is a Lie Group. The orthogonal group, O n is the subgroup of Gl n consisting of those martices B so that B T B = BB T = I, (the identity matrix) In this exercise you will argue that O n is a Lie subgroup of Gl n. b. Show that the map Φ : Gl n Gl n defined by is smooth. Φ(B) = B T B Hint. Its coordinate functions are polynomials. c. Argue that T I Gl n can be identified with the vector space M n of all n n matrices. d. Show that relative to the identification T I Gl n = M n, the differential is (dφ) I : T I Gl n T I Gl n (dφ) I (A) = A T + A. e. Show that Φ has constant rank n(n+1). f. Conclude that the orthogonal group O n Gl n is a Lie subgroup of Gl n of dimension n(n 1). a. For Gl n to be a Lie Group we need its group operation to be smooth. Indeed its group operation, matrix multiplication, is smooth as its coordinate functions are just polynomials which are clearly smooth. b. As the hint suggests, the map Φ(B) = B T B is just matrix multiplication which has smooth coordinate functions, making Φ smooth. c. Consider the map det : M n R (the determinant map). It is smooth as it is just a polynomial. Thus det 1 (R \ {0}) = Gl n is open in M n. Thus T A Gl n = T A M n for all A GL n, in particular T I Gl n = T I M n. There is a natural identification of M n with R n where I M n gets identified with e R n. Hence we also have T I M n = T e R n = R n = M n. So since T I Gl n = T I M n, we have T I Gl n = Mn. d. Let α : [ η, η] Gl n be given by α(t) = I + ta for some A Gl n. Then α(0) = I and α (0) = A. So dφ I (A) = d dt (Φ α)(t) t=0 = d dt Φ(I + ta) t=0 = d dt (I + ta T )(I + ta) t=0 = d dt (I + ta T + ta + t A T A) t=0 = (A T + A + ta T A) t=0 = A T + A e. First we note that Φ actually maps Gl n into S n the set of symmetric n n matrices. So anything in the image of Φ has n j=1 j = n(n+1) independent entries as every matrix is equal to its transpose. It follows that the rank of Φ is n(n+1). f. O n = Φ 1 (I) and hence O n has dimension n n(n+1) = n(n 1). Problem 5. If M is a compact n-manifold without boundary, show it admits a smooth immersion into R n. Problem 6. Prove that a compact n-manifold cannot be diffeomorphic to any subset of R n.
11 Let M be a compact n-manifold and let X R n. Suppose that f : M X R n is a diffeomorphism. If M is empty, then it is diffeomorphic to R n. So suppose that M is nonempty. Since M is open in M, we have, by Smooth Invariance of Domain, that X is open in R n. As f is a diffeomorphism X is compact and since R n is Hausdorff X is also closed. So we have that X is both open and closed in R n, and hence X is either all of R n (this also takes care of the case where X could only be a connected component of R n as it R n only has one connected component) or the empty set. Clearly that the empty set is ruled out since M is nonempty. Since M is compact, X is also compact as it is the continuous image of a compact set. However if X is compact it cannot be all of R n as required above. Hence we have a contradiction. M is not diffeomorphic to any subset of R n. Problem 7. A Riemannian metric on smooth manifold is an assignment of an inner product on each tangent space g p : T p M T p M R. These inner products are required to vary smoothly in the sense that if X and Y are smooth vector fields then the function M R p g p (X p, Y p ) is required to be smooth. Show that every smooth manifold admits a Riemannian metric. Claim: The usual inner product on R n is a Riemannian metric. proof: Given any point p R n, T p R n = R n and hence the inner product is defined on the tangent spaces. Let X = (x 1,..., x n ) and Y = (y 1,..., y n ) be smooth vector fields on R n (note that this means that each component function of the vector field must be smooth). Then X, Y = n i=1 x iy i is clearly smooth as sums and products of smooth functions are smooth. Now embed M into R k for some k N (Whitney embedding theorem). Let the embedding be ι : M R k. For each p M dι p : T p M T ι(p) R k = R k is an isomorphism onto some subspace of T ι(p) R k = R k and hence we may restrict the inner product from R k onto this subspace, yielding an inner product on the tangent spaces of the manifold. Since this inner product is a Riemannian metric, we have just shown that there is a Riemannian metric on M. 0.5. Problem Set 5. Problem 1. Prove that homotopy groups are homotopy invariants. (Hint: Read the bottom of page 131 and the top of page 13.) Problem. Assuming that π n (S n, p) = Z for all p S n, prove a. There is no retraction r : D n+1 S n. b. (Brouwer s Fixed Point Theorem) Any continuous map D n+1 D n+1 has a fixed point. c. S n is not contractible. a. First a lemma: Lemma. If r : X A is a retraction, then the homotopy group homomorphism induced by the inclusion map j : A X is injective. proof: Consider the composition r j : A A which is the identity on A. Then the induced homomorphism is (r j) = r j : π n (A, a 0 ) π n (A, a 0 ) which is the identity automorphism of the group. Since this is a bijection, set theory tells us that j : π n (A, a 0 ) π n (X, a 0 ) is necessarily injective. Assume that there is a retract r : D n+1 S n. Note that since D n+1 is convex its n th homotopy group is trivial (i.e. π n (D n+1, b) = {e}). Now let j : S n D n+1 be the inclusion map. Then the above lemma tells us that j : π n (S n, p) π n (D n+1, p) is injective. However π n (S n, p) = Z and π n (D n+1, p) = {e}, so it is impossible that j is an injection, which is a contradiction. There is no retraction from D n+1 to S n.
1 b. c. This is obvious since a contractible space has the homotopy type of a point (i.e. if a space X is contractible, π n (X, p) = {e} for all n) and π n (S n, p) = Z which is a contradiction if it is assumed that S n is contractible. 0.6. Sard s Theorem Handout. Problem 3. Show that there is no surjective differentiable map R n R n+1. Suppose that f : R n R n+1 is a surjective differentiable map. Then by Sard s theorem the set of regular values of f are dense in R n+1 (i.e. Df(y) is onto for each regular value y). But this means we have an onto linear map between R n and R n+1 at each regular point, which is impossible. Thus f cannot be onto and, moreover, f(r n ) has measure zero. Problem 4. Let M n be a compact manifold, f : M n R n+1 differentiable, and 0 / f(m). Show that there exists a line through the origin of R n+1, which only meets finitely many points of f(m n ). Had I known how to do this problem months ago, I would have PhD passed the qual... :( Since f(m n ) R n+1 \ {0} we can consider the map g : f(m n ) S n given by g(x) = x x. Also consider the map h : S n RP n which is the usual projection map (i.e. think of RP n as lines through the origin of R n+1 ). Now consider the composition: M n f f(m n ) g S n h RP n It suffices to show that for some p RP n the preimage (g h) 1 (p) is finite as each (h g) 1 (p) is a line through the origin of R n+1. Note that both g and h are smooth and so the composition h g is smooth as well. Also note that f(m n ) is compact since f is continuous. There are two possible cases: (a) If h g is not onto, then there obviously exists a p RP n such that (h g) 1 (p) = which is clearly finite. (b) If h g is onto, then by Sard s theorem the set of regular values of h g is dense in RP n and thus by the following theorem: If Φ : M m N n is a smooth map between smooth manifolds with m n and y N is a regular value, then Φ 1 (y) is a smooth submanifold of M of dimension m n. we get that there is a regular value y RP n such that (h g) 1 (y) is a smooth submanifold of dimension n n = 0 and thus is discrete. Thus (h g) 1 (y) is closed since it is a discrete set, and therefore compact since it is closed in a compact space. Now since a compact discrete space is finite (see proof of this below) we get that (h g) 1 (y) is finite and thus we have proved our desired conclusion. Let X be a compact disctete space. For each x X there is an open set U x containing x such that U x (X \ {x} = which is possible since X is discrete. The collection C = {U x } x X forms an open cover of X. But since X is compact C has a finite subcover. But since each U x is uniquely determined by x and there are finitely many U x s in the cover, hence X must be finite. Problem 8. Let M k R n+1 be a compact submanifold and n k. Show that, for the projection π : R n+1 H n onto a suitable hyperplane H of R n+1, the restriction π M : M H is an immersion. Hint: consider the (k 1)-dimensional manifold P T M, whose elements are the 1-dimensional subspaces of the tangent spaces of M, and study the canonical map P T M RP n. Problem 9. Let M k R n+1 be a compact submanifold and n k + 1. Show that, for the projection π : R n+1 H n onto a suitable hyperplane of R n+1, the restriction π M : M H is an embedding.
0.7. Homotopy Handout. Problem 3. If two maps f and g from X to S p satisfy f(x) g(x) < for all x, prove that f is homotopic to g, the homotopy being smooth if f and g are smooth. Given any two points in R p+1 on S p that are not antipodal we can connect them by the straight line l(t) = tp + (1 t)q, t [0, 1]. Now normalize this line so that it lies entirely on S p : L(t) = tp + (1 t)q tp + (1 t)q. Notice that this map is only defined if p is not antipodal to q, because if they were we would have tp + (1 t)q = tp + (1 t)( p) = tp + tp p = tp p which is zero when t = 1 which would make L ( 1 ) undefined. Now consider two maps f, g : X S p such that f(x) g(x) x X (this is equivalent to f(x) g(x) < x X). Then define the map H(x, t) = tf(x) + (1 t)g(x) tf(x) + (1 t)g(x). This is clearly continuous since if we fix a t value we get a combination of continuous maps that remains continuous, and if we fix an x value we get an arc on the sphere. Now notice that and H(x, 0) = H(x, 1) = 0f(x) + (1 0)g(x) 0f(x) + (1 0)g(x) = g(x) g(x) = g(x) 1f(x) + (1 1)g(x) 1f(x) + (1 1)g(x) = f(x) f(x) = f(x) so H is a homotopy between f and g. Notice that this map would remain smooth if both f and g are smooth because then H would just be a combination of smooth maps. 13 Problem 4. If X is compact, show that every continuous map X S p can be uniformly approximated by a smooth map. If two maps X S p are continuously homotopic, show that they are smoothly homotopic. Let f : X S p be any continuous map. Then by the Stone-Weierstrass Theorem we can uniformly approximate f by polynomials (in each component). But we know that polynomials are smooth functions, so this approximation is smooth. Now if two maps f and g are continuously homotopic (i.e. there is a homotopy H : X I S p ), fix t for H and uniformly approximate H(x, t ) with a smooth function. Since these two functions are arbitrarily close, by the previous exercise, they are homotopic. Call this new homotopy Z : (X I) J S p. Then the composition of Z and H yields a new, smooth homotopy between f and g. Problem 5. If m < p, show that every map M m S p is homotopic to a constant. First note that S p \ n = R p for any n S p ; then since R p is contractible we have that S p \ {p} is also contractible. Also note that f cannot be onto (see exercise 0.6.3 above). Let f : M S p be continuous and let f(m); let c : M { } be a constant map; lastly let d : S p \ {n} { } (n / f(m) which is possible since f is not onto) be the null-homotopy from S p \ {n} to { }. Thus d f : M { } is a null-homotopy of f with c. 0.8. Hopf Fibration Problem Set. To define the Hopf fibration h : S 3 S we think of S 3 are the unit sphere in C C and S as the unit sphere in C R. With respect to these coordinates, the formula for h is h : (a, c) (a c, a c ). Problem 1. Show that the image of h is indeed contained in S.
14 What we need to show is that all of the image points have length 1. Rewrite a = w+xi and c = y +zi. Then a = w + x, c = y + z, a c = (wy + xz) + (xy wz)i, and āc = (wy + xz) (xy wz)i. So: (a c, a c ) = (a c) (āc) + ( a c ) = 4w y + 4w z + 4x y + 4x z + w 4 + w x w y w z + x 4 x y x z + y 4 + y z + z 4 = w 4 + w x + w y + w z + x 4 + x y + x z + y 4 + y z + z 4 = (w + x + y + z ) = (a, c) 4 = 1 4 = 1 Problem. Identify S 1 with the unit circle in C. Consider the S 1 -action on S 3, H : S 1 S 3 S 3, H : (ω; (a, c)) (ωa, ωc), where the multiplication takes place in C. Show that the orbits of this circle action coincide with the fibers of h. (The fibers of a map are the preimages of points.) Problem 3. Recall that U() is the group of ( )- unitary matrices with complex entries. That is it is the group of ( )-matrices with complex entries so that AA T = A T A = I. There is a natural action of U() on C by matrix multiplication: U() C C (( ) ( )) ( α β a α β, δ δ c δ δ ) ( a c Since this action is by isometries, it preserves the unit 3-sphere, S 3. a. Use the fact that the U() action is complex linear to conclude that is commutes with the Hopf action. That is ) ( ωa A ωc ( a = ωa c Conclude that multiplication by an element A of U() induces a homeomorphism  : S S so that S 3 A S 3 h h S  S commutes. Here the top map is multiplication by A, and the maps on the left and right are the Hopf fibration. Because of this, U() is said to act by symmetries of the Hopf fibration. b. By considering the actions of the circle subgroups {( ) ω 0 ω = 1} 0 1 and {( cos θ sin θ sin θ cos θ ). ) θ [0, π)} of U(), show that h : S 3 S is onto. c. Show that any orientation preserving isometry of S lifts to a symmetry of the Hopf fibration. d. Show that h is a quotient map. (Hint: h is actually a closed map.) ).
15 Problem 4. a. Show that if a 0 and c 0 then the tangent space to S 3 at (a, c) is { ( a T (a,c) S 3 = span (ia, ic), a c, c ) ( c a, i a )} c c, i a c a. b. If a = 0 show that the tangent space to S 3 at (0, c) is T (0,c) S 3 = span{(0, ic), (1, 0), (i, 0)}. c. If c = 0 show that the tangent space to S 3 at (a, 0) is T (a,0) S 3 = span{(ia, 0), (0, 1), (0, i)}. Problem 5. Show that h is a submersion. Hint: Use exercise 4 to describe the tangent space of S 3. Notice (ia, ic) is tangent to the orbits ( of the Hopf action, H, and hence to the fibers of h. Therefore dh (a,c) (ia, ic) = 0. a To compute dh (a,c) a c, c ) c a, find a curve γ in S 3 such that γ(0) = (a, c), ( a γ (0) = a c, c ) c a, and use ( a dh (a,c) a c, c ) c a = (h γ) (0). To find γ use the fact that if p, q S n are perpindicular, then is a curve in S n with γ p,q (t) := (cos t)p + (sin t)q γ p,q (0) = p, and γ p,q(0) = q. Problem 6. Recall that RP n is S n /Z where the Z action is by the antipodal map. Let p : S n RP n be the corresponding quotient map. Show that RP n+1 is homeomorphic to the mapping cone of p. Problem 7. Recall that CP n is the space of complex lines through the origin of C n+1. That is CP n (C n+1 \ {0})/ where v, w C n+1 \ {0} are equivalent (v w) iff v = z w for some complex number z. Prove that CP is homeomorphic to the mapping cone of the Hopf fibration h 1 : S 3 S. Hint: The Hopf circle action on the 5-sphere is S 1 S 5 S 5 (ω, z 1, z, z 3 ) (ωz 1, ωz, ωz 3 ). Show that CP can also be viewed as the quotient of this circle action. Let the quotient map be called h : S 5 CP. This map is also called a Hopf fibration. Map D 4 C C S 5 by ι : (z 1, z ) (z 1, z, 1 z 1 z ) and to CP by h ι(z 1, z ) = [z 1, z, 1 z 1 z ]. Show that h ι is onto CP. Show that h ι is 1-1 on the interior of D 4. What is the behavior of h ι on the boundary of D 4? Problem 8. What is the symmetry group of h : S 5 CP?
16 Problem 9. Can you generalize problem 7 and view CP 3 as a mapping cone? 0.9. Homogeneity Problem Set. Definition. A Lie group, G, is simultaneously a smooth manifold and a group. The structures are compatable in the sense that the group operations and are smooth. M : G G G M(g, h) = gh ι : G G ι(g) = g 1 Problem 1. Show that a Lie group is smoothly homogeneous in the sense that given any g, h G there is a diffeomorphism Φ : G G so that Φ(g) = h. Hint: Given g G consider the map L g : G G L g (h) = gh. Problem. Show that all Lie groups are parallelizable. Hint. Let e G be the identity element. Move T e G around using the differentials of the L g s from Problem 1. Problem 3. Let M be a connected smooth manifold. Show that any two (distinct) points p, q M can be connected by an embedded smooth curve. Hint. Suppose γ : [0, 1] M is any smooth embedded curve from p to q. Cover γ[0, 1] by a finite number of coordinate charts and use the fact that the result holds on R n. Problem 4. Show that a connected smooth manifold is smoothly homogeneous in the sense that given any p, q M there is a diffeomorphism Φ : M M so that Φ(p) = q. Hint. Use problem 3 to connect p and q by a smooth embedded curve, γ. Argue that the tangent field to γ can be extended to a smooth vector field X with compact support. Use the fact (see e.g. Theorem 4.1.11 in Conlon) that a vector field with compact support is complete in the sense that it generates a global flow. (A vector field has compact support iff it vanishes outside of a compact set.)
17. Differentiable Manifolds.. Differentiable Manifolds. Problem 4. Let X be the graph of the real valued function θ(x) = x of a real variable x. Define a function structure on X by f F (U) f is the restriction to U of a C functional structure on some open set V in the plane with U = V X. Show that X with this structure is not diffeomorphic to the real line with the usual C structure..5. Tangent Vectors and Differentials. Problem 1. If φ : R m R n is a linear map and we identify T p (R k ) with R k by identifying basis vector, show that φ becomes φ. ( k ) φ (v) = φ a i x i=1 i k ( ) = a i φ (since φ is linear) x i=1 i k ( ) = a i φ x i=1 i ( k ) = φ a i (since φ is linear) x i = φ(v) i=1 x i with the ith standard Problem. If the curve φ : R R n is an embedding then show that φ ( d dt ) coincides with the classical notion of the tangent vector to the curve φ under the identification of the tangent space to a euclidean space with the euclidean space. Since φ is an embedding, then φ is a smooth curve. Thus ( ) d φ (f) = d dt dt (f φ) = D φ(f). Problem 3. For a smooth function f defined on a neighborhood of a point p R n, the gradient f = grad f of f is the vector ( f,..., f ). x 1 x n For a vector v R n show that the directional derivative D v, denoted by D γv in Example 5.4, satisfies the equation D v f = f, v, the standard inner product of f with v in R n. Fix a point p R n and a vector v R n. Define l v (t) = p + tv to be the line passing through p with direction v. Then we have the tangent vector D lv : where a i = dx i dt D lv (f) = n i=1 a i f x i = d(p i + tv i ) t=0 dt = v i t=0
18 where l v (t) = (p 1 + tv 1,..., p n + tv n ). Thus: as desired. D lv (f) = = n i=1 n i=1 a i f x i v i f x i f f f = v 1 + v + + v n x 1 x x ( n f = (v 1, v,..., v n ),, f,..., f ) x 1 x x n = v, f = f, v Problem 4. If M m R n is a smoothly embedded manifold and f is a smooth real valued function defined on a neighborhood of p M m in R n and which is constant on M, show that f is perpendicular to T p (M) at p. Let p M. Let c : R M be a smooth function such that c(0) = p. Then the tangent vector to c at p is given by D c (g) = d dt g(c(t)) t=0 = v p where g : U R is any smooth function defined on a neighborhood of p. Note that f defined in the problem is one of these smooth functions when its domain is restricted to M. Then we note that f(c(t)) = k a constant, and so D c (f) = 0. Thus by the previous exercise 0 = D c (f) = f, v p So since c was arbitrary, this is true for all tangent vectors v p T p M..6. Sard s Theorem and Regular Values. Problem. For the map φ(x, y) = x y of the plane to the line, what are the regular values? We want the points in R such that Df 0. Compute Df: Df(x, y) = (x y) Notice that Df(x, y) 0 iff x = y = 0. So the regular points of f are {(x, y) R (x, y) (0, 0)}. Thus the regular values of f are R \ {0}. Problem 3. For the map φ(x, y) = sin (x + y ) of the plane to the line, what are the regular values? We want the points in R such that Df 0. Compute Df: Df(x, y) = (x cos (x + y ) y cos (x + y )) Notice that Df(x, y) 0 iff x = y = 0 or x + y = (n+1)π with n N 0. So the regular points of f are {(x, y) R (x, y) (0, 0) or x + y = (n+1)π, n N 0 }. Thus the regular values of f are R \ { 1, 0, 1}..7. Local Properties of Immersions and Submersions. Problem 1. Consider the real valued function f(x, y, z) = ( (x + y ) 1 ) + z on R 3 \ {(0, 0, z)}. Show that 1 is a regular value of f. Identify the manifold M = f 1 (1). M is a torus.
19 Problem. Show that the manifold M of Problem 1 is transverse to the surface Identify the manifold M N. N = {(x, y, z) R 3 x + y = 4}. M N is two circles of radius at the top and bottom of the torus. Problem 3. Show that the manifold M of Problem 1 is not transverse to the surface Is M N a manifold? N = {(x, y, z) R 3 x + y = 1}. Problem 4. Show that the manifold M of Problem 1 is not transverse to the plane Is M N a manifold? N = {(x, y, z) R 3 x = 1}..8. Vector Fields and Flows. Problem 1. On the -sphere, consider the flow θ(t, (x, y, z)) = (x, y cos (t) z sin (t), y sin (t) + z cos (t)). Find the vector field on S induced by this flow. dθ = (0, y sin (t) z cos (t), y cos (t) z sin (t)) dt ( ) d θ = dθ dt dt = (0, y sin (0) z cos (0), y cos (0) z sin (0)) = (0, z, y) (0,(x,y,z)) (0,(x,y,z)) So our vector field is given by: ξ(x, y, x) = (0, z, y). Problem. Consider the vector field ξ(x) = x on R. Show that ξ is the tangent field to a flow, and find the flow. (Hint: In classical notation, this vector field corresponds to the initial value problem dy dt = y, y(0) = x.) dy dt = y is a linear first order initial value problem and has general solution y = cet with initial value y(0) = x. Thus using the initial value we get: and thus the flow is θ(t, x) = xe t. For verification: x = ce 0 = c Evaluate at t = 0: as it should be. dθ dt = xet dθ dt = e 0 x = x = ξ(x) t=0 Problem 3. Show that the vector field ξ(x) = x is not the tangent field of any (global) flow.
0 Following the hint from the previous problem let dy dt = y, y(0) = x. Solve the equation: dy dt = y y dy = dt y dy = dt y 1 = t + c y 1 = c t y = 1 c t Now using the initial value y(0) = x: y(0) = 1 c 0 = 1 c = x and thus c = 1 x, so: y = 1 1 x t = x 1 xt. Therefore our flow is: θ(t, x) = x 1 xt. Consider the flow at the point (t, x) = (1, 1). θ(1, 1) is undefined thus ξ is not the tangent field of any global flow. Problem 4. If X and Y are vector fields on M then XY makes sense as an operator on smooth real valued functions on M. Show that [X, Y ] = XY Y X is a vector field. (This is called the Lie bracket of X and Y. Sometimes it is defined with the opposite sign.) Also show that XY itself is not a vector field. R-linear) Since [X, Y ] = [Y, X] it suffices to show linearity in one component. Let X, Y, Z be any smooth vector fields on M and let a, b R. Then: [ax + by, Z] = ax + by (Z) Z(aX + by ) = ax(z) + by (Z) Z(aX) Z(bY ) = ax(z) + by (Z) az(x) bz(y ) = ax(z) az(x) + by (Z) bz(y ) = a[x, Z] + b[y, Z] Problem 5. Show that the Klein bottle has an everywhere nonzero vector field. Describe the resulting flow.
1 View the Klein bottle K as the identification space of the space X := [0, 1] [0, 1]. Consider the vector field ξ(x, y) = (1, 0) on X. This will yield a vector field on K. The flow lines arising from this vector field are loops around K..11. Tubular Neighborhoods and Approximations. Problem 1. A probability vector in R n whose coordinates are all nonnegative and add to 1. A stochastic matrix is an n n matrix whose columns are probability vectors. Use Corollary 11.1 to show that every stochastic matrix A has a fixed probability vector under the mapping v Av. Let P n := {(v 1,..., v n ) n i=1 v i = 1 and v i 0} be the set of n-dimensional probability vectors. First oberve the n = 1 case: The only 1 1 stochastic matrix is A = (1) and since P 1 = {1} it is clear that A has a fixed probability vector. Now for n : Notice that P n is homeomorphic to D n 1. So let A be a stochastic matrix and let ϕ : P n D n 1 be the aforementioned homeomorphism, then ϕ A ϕ 1 : D n 1 D n 1 is a continuous map (A is continuous as it is linear), and hence by Brouwer s Fixed Point Theorem has a fixed point, call it η. Since ϕ is a homeomorphism we can write η = ϕ(ξ). So
Hence ξ is a fixed point of A. ϕ A ϕ 1 (η) = η ϕ A ϕ 1 (ϕ(ξ)) = ϕ(ξ) ϕ A(ξ) = ϕ(ξ) ϕ 1 ϕ A(ξ) = ϕ 1 ϕ(ξ) A(ξ) = ξ Problem. Use Corollary 11.1 to show that every n n matrix with positive real entries has a positive real eigenvalue. 3. Fundamental Group 3.. The Fundamental Group. Problem 1. Let G be a topological group with unity element e. For loops f, g : (S 1, p) (G, e) define a loop f g(t) = f(t)g(t) by the pointwise product in G. Show that f g f g rel p. Let f and g be loops as defined above. Then Problem. Let G be a topological group with unity element e. Show that π 1 (G, e) is abelian. (Hint: Use Problem 1 and the idea of Problem 1 to show that f g g f.) Problem 3. If K is the Klein bottle, show that π 1 (K ) is generated by two elements, say α and β obtained from the longitudinal and latitudinal loops. Also show that there is the relation (with proper assignment of α and β) αβα 1 = β 1. (You are not asked to show that this is the only relation, but, in fact, it is.) (Hint: Use the fact that a smooth loop must miss a point.) Since we cannot have a smooth onto map from S 1 to K (by Sard s Theorem), let be the missed point. Let p K and let l be a loop in K based at p.