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1 Math 7350 Geometry of Manifolds Dr. Vaughn Climenhaga, PGH 651A Selected HW solutions Spring 2015 HW 1, #1. (Lee, Problem 1-4). Locally finite covers Let M be a topological manifold, and let U be an open cover of M. (a) Suppose each set in U intersects only finitely many others. Show that U is locally finite that is, every point of M has a neighbourhood that intersects at most finitely many of the sets in U. Solution. Given x M, there is U U such that x U, because U covers M. Now U is a neighbourhood of x, and by the hypothesis, U intersects only finitely many of the sets in U. Thus U is locally finite. (b) Give an example showing that the converse may be false. Solution. Take any infinite but locally finite cover U (for example, U = {(n 1, n + 1) n Z} with M = R), and define a new cover V = U {M}. Then V is still locally finite but it has a set (M) that intersects infinitely other cover elements. (c) Show that the converse is true if the elements of U are precompact (have compact closures). Solution. Suppose U is locally finite. Thus for every x M there is a neighbourhood V x such that V x intersects only finitely many of the sets in U. Let U U be precompact. Then U is compact and is covered by the open sets V x, so it is covered by finitely many of them, say V x1,..., V xn. Any element of U that intersects U must intersect one of the V xi. Since each of these only intersects finitely many elements of U, the same is true of U, and hence of U.

2 Math 7350 Selected HW solutions Page 2 of 30 HW 1, #2. (Lee, Problem 1-6). Distinct smooth structures Let M be a nonempty topological manifold of dimension n 1. If M has a smooth structure, show that it has uncountably many distinct ones. [Hint: first show that for any s > 0, F s (x) = x s 1 x defines a homeomorphism from the unit ball in R n to itself, which is a diffeomorphism if and only if s = 1.] Solution. First we prove the hint. Given x R n \ 0 write ˆx = x/ x for the normalisation of x. Note that given s > 0 and x B n (the unit ball in R n ) we have F s (x) = x sˆx, so in particular F s (x) < 1 whenever x < 1, and we have F s (B n ) B n. Moreover, given s, t > 0 we have F s F t (x) = F s ( x tˆx) = x stˆx = F st (x). In particular, F s F 1/s is the identity, so F s is a bijection from B n to itself. Note that x x s = ( i (xi ) 2 ) s/2 is a continuous function B n R (as a composition of continuous functions), and x ˆx is continuous on B n \ 0, so F s is continuous (in fact smooth) on B n \ 0. Moreover, F s (x) 0 as x 0, so F s is continuous on B n. Its inverse F 1/s is continuous as well, so F s is a homeomorphism. If s 1 then F s is a diffeomorphism (the identity). If s < 1 then F s is not differentiable at 0. If s > 1 then Fs 1 = F 1/s is not C 1 at 0. Thus F s is not a diffeomorphism for any s 1. Now we use the hint to prove the result. Let A be any smooth atlas on M. Fix p M and (U, ϕ) A such that p U. Since ϕ(u) R n is open there is r > 0 such that B(ϕ(p), r) ϕ(u). Let V = ϕ 1 (B(ϕ(p), r)) U, and define ψ : U R n by ψ(q) = (ϕ(q) ϕ(p))/r. Note that ψ(p) = 0 and ψ(v ) = B n. Let A = A {(V, ψ)} and note that A is smooth because every transition map involving (V, ψ) is the composition of a linear map and a transition map involving ϕ. Given (W, θ) A, let W = W \ {p} and θ = θ W. Let A be the atlas obtained from A by replacing every (W, θ) (except (V, ψ)) with (W, θ ); this is again a smooth atlas.

3 Math 7350 Selected HW solutions Page 3 of 30 Given s > 0, let A s be the atlas obtained from A by replacing (V, ψ) with (V, F s ψ). Note that this is an atlas because F s is a homeomorphism from B n = ψ(v ) to itself. It is a smooth atlas because every transition map is the composition of F s (away from 0) with a transition map from A. Thus it defines a unique smooth structure. Finally, if A s and A t define the same smooth structure, then the transition map between (V, F s ψ) and (V, F t ψ) must be a diffeomorphism. This transition map is given by (F s ψ) (F t ψ) 1 = F s F 1 t = F s/t, so it is a diffeomorphism if and only if s = t. Thus the smooth structures induced by A s are distinct for s > 0, so there are uncountably many distinct smooth structures on M. HW 1, #4. (Lee, Problem 1-9). Complex projective n-space Complex projective n-space, denoted by CP n, is the set of all 1-dimensional complex-linear subspaces of C n+1, with the quotient topology inherited from the natural projection π : C n+1 \ {0} CP n. Show that CP n is a compact 2n-dimensional topological manifold, and show how to give it a smooth structure analogous to the one we constructed for RP n. Note that we identify C n+1 with R 2n+2 via the correspondence (x 1 + iy 1,..., x n+1 + iy n+1 ) (x 1, y 1,..., x n+1, y n+1 ). Solution. Given 1 j n + 1, let Ũj = {(z 1,..., z n+1 ) C n+1 z j 0}. Define a map ϕ j : Ũj C n by ( z ϕ j (z 1,..., z n+1 1 ) = z,..., zj 1 j z j ), zj+1,..., zn+1. z j z j Then ϕ j is continuous and is constant on fibres of π, so there is a unique continuous map ϕ j : U j := π(ũj) C n such that the following

4 Math 7350 Selected HW solutions Page 4 of 30 diagram commutes. C n+1 \ 0 Ũ j ϕ j C n π π CP n U j ϕ j Moreover, ϕ j is a bijection since its inverse is given by (w 1,..., w n ) [w 1,..., w j 1, 1, w j+1,..., w n ], and ϕ 1 j is continuous since it is the composition of continuous maps. Thus ϕ j is a homeomorphism, and since we identify C n with R 2n, we have shown that CP n is locally Euclidean. To show CP n is a topological manifold it remains to show that it is second countable and Hausdorff. The first of these follows since it is a quotient of a second countable space. For Hausdorff, note that if [z 1 ], [z 2 ] U j for some j, then they can be separated by disjoint open sets since the same is true of ϕ j (z 1 ), ϕ j (z 2 ) C n. So we consider the case where there is no U j containing both [z 1 ] and [z 2 ]. Given j k let A j,k = {[z] z j > z k } CP n ; this is open since its preimage under π is open in C n+1 \ 0. By the assumption on z 1, z 2 there are j k such that [z 1 ] U j and [z 2 ] U k, but z j 1 = z2 k = 0. Thus z 1 A j,k and z 2 A k,j. Clearly A j,k A k,j = so this suffices. We have shown that CP n is a topological manifold of dimension 2n. To show that it is compact we let S 2n+1 C n+1 be the unit sphere in R 2n+2 (with the natural identification between R 2n+2 and C n+1 ), and define τ : C n+1 S 2n+1 by τ(z) = z/ z, where z = ( j zj 2 ) 1/2. Writing ˆπ for the restriction of π to S 2n+1, we see that π = ˆπ τ. In particular, CP n is a quotient space of S 2n+1 via the map ˆτ. Because S 2n+1 is compact, it follows that CP n is as well. Finally, we check that A = {(U j, ϕ j ) j = 1,..., n + 1} is a smooth atlas on CP n. If j < k, the transition map ϕ k ϕ 1 j : ϕ j (U j U k ) C n is given by ϕ k ϕ 1 j (w 1,..., w n ) = ϕ k [w 1,..., w k 1, 1, w k+1,..., w n ] ( w 1 = w,..., wj 1, wj+1,..., wk 1 1, j w j w j w j w, wk+1 j w j,..., wn+1 w j ).

5 Math 7350 Selected HW solutions Page 5 of 30 Thus every (complex) coordinate function of ϕ k ϕ 1 j w wl for some l. Writing w j = x j + iy j, we see that w j is of the form w l w j = xl + iy l x j + iy j = (xl + iy l )(x j iy j ) (x j ) 2 + (y j ) 2 = xl x j + y l y j + i(y l x j y j x l ) (x j ) 2 + (y j ) 2, so every (real) coordinate function of ϕ k ϕ 1 j is of the form (x, y) xl x j + y l y j (x j ) 2 + (y j ) 2 or (x, y) yl x j y j x l (x j ) 2 + (y j ) 2 for some l. These are smooth as long as (x j, y j ) (0, 0), which holds on ϕ j (U j ), so ϕ k ϕ 1 j is smooth as a map on R 2n. HW 2, # 1. More on Grassmanians Let V be a n-dimensional real vector space and recall that given an integer 1 k n, G k (V ) is the Grassman manifold whose elements are all the k-dimensional subspaces of V. (a) We have seen that G k (V ) is a smooth manifold for each k. Prove that it is compact. Solution. Fix a basis for V, so we work with G k (R n ). Each S G k (R n ) is determined by a set of k orthonormal vectors in R n. Writing these as columns of an n k matrix gives A M n k (R) with A t A = I k. Let X be the set of such matrices. Then X is a closed and bounded subset of R nk, hence X is compact. Let π : X G k (R n ) be the map that assigns to each A X the subspace spanned by its columns. Then π is surjective. We claim that π is continuous. To this end we look at its coordinate representation. Recall that a chart for G k (R n ) is given by fixing a decomposition R n = P Q, where dim P = k, and using the bijection between U Q = {S G k (R n ) S Q = {0}} and the set of linear functions from P to Q. Let R P be projection onto P along Q, and R Q the projection onto Q along P. Then the linear map L: P Q associated to π(a) G k (R n ) is uniquely determined by the condition that L(R P v) = R Q v for every column v of the matrix A. Now it is easy to

6 Math 7350 Selected HW solutions Page 6 of 30 see that varying A continuously means varying the columns v continuously, which in turn varies the map L continuously. Thus π is continuous. (b) Prove that G k (V ) and G n k (V ) are diffeomorphic. Solution. Fix an inner product on V and let F : G k (V ) G n k (V ) be the map that takes a subspace S to its orthogonal complement S. Clearly F is a bijection, we claim that it is smooth. (This is enough since its inverse is of the same form.) To this end, let V = P Q be an orthogonal decomposition of V and let U Q be as above. Note that if S U Q then V = Q S, so we have V = Q S, hence F (S) = S U Q. Thus we write F in local coordinates given by the charts U Q, U Q. Let L: P Q be linear and let S G k (V ) be its graph. We want to find ˆF (L): P Q. Given v P, the image ˆF (L)(v) Q is determined by the condition that v + ˆF (L)v S ; that is, v + ˆF (L)v, w + Lw = 0 for all w P. Expanding the inner product gives v, w + v, Lw + ˆF (L)v, w + ˆF (L)v, Lw = 0. The first term always vanishes because v P and w P. Similarly for the last term (with Q and Q). Thus we have ˆF (L)v, w = v, Lw for all v P and w P. This means that once we fix bases, ˆF (L) is represented by the matrix A t, where A is the matrix representing L. Thus ˆF maps A to A t and is smooth. HW 2, # 3. (Lee, Problem 2-9). Let p be a non-zero polynomial in one variable with complex coefficients. Show that there is a unique continuous map p: CP 1 CP 1 such that the following diagram commutes,

7 Math 7350 Selected HW solutions Page 7 of 30 where G is as in the previous problem. C G CP 1 p p C G CP 1 Show that the map p is smooth. Is it a diffeomorphism? Solution. Recall that G(z) = [z, 1] = {(w 1, w 2 ) C 2 w 1 = zw 2 }. In order for the diagram to commute, we must have p([z, 1]) = p(g(z)) = G(p(z)) = [p(z), 1] for all z C. Whenever w 2 0 we have [w 1, w 2 ] = [ w1, 1] and so w [ ( ) ] 2 w p([w 1, w 2 1 ]) = p, 1. This defines p on CP 1 \ {[1, 0]}. Note that p is smooth on this set because its coordinate representation is p, which is a polynomial and hence analytic. w 2 In order to extend p to all of CP 1, we need to use a chart containing [1, 0]. Such a chart is given by the inverse of the map F : C CP 1 defined by F (z) = [1, z]. To find the coordinate representation of p with respect to the chart F 1, we write p(z) = a 0 + a 1 z + + a n z n for some a 0,..., a n C, and observe that given z C \ {0} for which p(1/z) 0, we have ( ) ]) 1 F 1 p F (z) = F 1 ( p([1, z])) = F ([p 1, 1 z 1 = 1/p(1/z) = a 0 + a 1 z a n z n z n =. a 0 z n + a 1 z n a n If p(z) a 0 is a constant polynomial, then we have p([0, 1]) = [a 0, 1], otherwise we have p([0, 1]) = [0, 1]. In the first case p is a constant function and therefore smooth. In the second, there is a neighbourhood

8 Math 7350 Selected HW solutions Page 8 of 30 U C of 0 such that p(1/z) 0 for all z U \ {0}, and in particular, the formula above shows that p is smooth on this neighbourhood, hence smooth on CP 1. Note that p is a diffeomorphism if and only if p has degree 1 (and the linear coefficient is non-zero); for all other p, tha map p itself is already non-injective. HW 2, # 5. Tori as quotients by lattices Let v, w R 2 be independent (hence a basis), and let Λ = vz + wz = {av + bw a, b Z}. We say that Λ R 2 is a lattice. The lattice Λ induces an equivalence relation on R 2 by putting x y iff x y Λ. Let M Λ = R 2 /Λ be the topological space obtained as the quotient of R 2 by this equivalence relation. (a) Fix a lattice Λ and let π : R 2 M Λ be the quotient map. Given p R 2 and r (0, 1), let Û r 2 p = B r (p) R 2, and show that π Û r p is a bijection onto its image Up r M Λ. Let ϕ r p = π 1 : U r Ûp r p Û p r. Show that the collection A = {(Up r, ϕ r p) p R 2, 0 < r < 1} 2 satisfies the conditions of the smooth manifold chart lemma, so M Λ is a smooth manifold with smooth structure generated by this atlas. Solution. As observed in class, the condition r < 1/2 needs to be replaced by r < r 0, where r 0 depends on Λ. Let r 0 be such that x Λ \ {0} implies x > 2r 0. Then given r (0, r 0 ) and p R 2, we observe that x, y B r (p) implies that x y 2r 0, and hence by the definition of r 0, we have x y / Λ, so x y. This implies that π(x) π(y), so π Û r p is 1-1. Now we check the conditions of the smooth manifold chart lemma. We just showed that ϕ r p is a bijection between U r p and the open set B r (p) R 2, which verifies (i). For (ii), first observe that U r p = U r p for all p p. Given U r p and U s q, let x Λ be such that p (q + x) is minimised. Then U r p U s q = U r p U s q+x = π(b r (p) B s (q + x)). We get ϕ r p(u r p U s q ) = B r (p) B s (q + x)

9 Math 7350 Selected HW solutions Page 9 of 30 and ϕ s q(u r p U s q ) = B r (p x) B s (q), both of which are open in R 2. Condition (iii) follows because the transition map ϕ r p (ϕ s q) 1 is a translation by x, where x is as in the previous paragraph. Condition (iv) follows by choosing p to have rational coefficients and r to be rational. Condition (v) follows since given any distinct points [p], [q] M/Λ, we can choose r such that B r (p ) B r (q ) = for all p p and q q. (b) Let Λ 1 and Λ 2 be any two lattices in R 2, and show that M Λ1 and M Λ2 are diffeomorphic (when equipped with the smooth structure from the previous part). Hint: Start by finding a smooth map F : R 2 R 2 such that F (Λ 1 ) = Λ 2. Solution. Let Λ 1 be generated by v, w R 2 and Λ 2 be generated by x, y R 2. Let F : R 2 R 2 be the linear map taking v x and w y. Then F is smooth and F (Λ 1 ) = Λ 2. Note that F is invertible and F 1 is also linear. Given p R 2 we have p + Λ 1 R 2 /Λ 1, and we define F : R 2 /Λ 1 R 2 /Λ 2 by F (p + Λ 1 ) = F (p) + Λ 2. This is well-defined because F is linear and F (Λ 1 ) = Λ 2. It is bijective because we can define its inverse by F 1 (q + Λ 2 ) = F 1 (q) + Λ 1. It only remains to show that both F and F 1 are smooth. With r (0, r 0 ) as in the previous part, we can write the coordinate representation ˆF relative to (Up r, ϕ r p) and (UF r (p), ϕr F (p)), where the first chart is in R 2 /Λ 1, and the second in R 2 /Λ 2. We get ˆF (p) = ϕ r F (p) F (ϕ r p) 1 (p) = ϕ r F (p)( F (p + Λ 1 )) = ϕ r F (p)(f (p) + Λ 2 ) = F (p), that is, the coordinate representation of F is the original linear 1 map F, which is smooth. The same argument shows that F is smooth, so F is a diffeomorphism.

10 Math 7350 Selected HW solutions Page 10 of 30 (c) Let Λ = Z 2 be the lattice generated by the standard basis vectors, and show that M Λ is diffeomorphic to the torus T 2 = S 1 S 1 with its standard smooth structure. Solution. Define a map p: R 2 T 2 by p(x, y) = (e 2πix, e 2πiy ). Then (x, y ) (x, y) + Λ implies that p(x, y ) = p(x, y), and so p passes to a map p: M Λ T 2. Moreover, p is a bijection, as we observed in the proof that T 2 is homeomorphic to the square with opposite edges identified. We must show that p and its inverse are smooth. S 1 is covered by four charts, each of which is the intersection of S 1 with a halfplane, and the corresponding coordinate maps are z Re(z) and z Im(z). Direct products of these charts give 16 charts that cover T 2, all with the same form of coordinate maps. Thus the coordinate representations of p have the form ˆp(x, y) = (cos 2πx, cos 2πy), and similarly with one or both cos replaced by sin. All these functions are smooth, and the coordinate representations of p 1 are the inverse trig functions on their appropriate domains, which are also smooth. HW 3, #1. Tangent space of a product manifold Let M 1,..., M k be smooth manifolds, and for each j, let π j : M 1 M k M j be the projection onto the M j factor. Prove that for any point p = (p 1,..., p k ) M 1 M k, the map is an isomorphism. α: T p (M 1 M k ) T p1 M 1 T pk M k v (d(π 1 ) p (v),..., d(π k ) p (v)) Solution. The map α is a direct sum of the linear maps d(π i ) p, hence it is linear. Moreover, because T p ( M i ) and T pi M i have dimension ni, in order to show that α is an isomorphism it suffices to show that it is onto. (This turns out to be easier than showing 1-1.)

11 Math 7350 Selected HW solutions Page 11 of 30 We prove surjectivity by exhibiting a linear map β : T pi M i T p ( M i ) such that α β is the identity on T pi M i. Given p i M i, let τ j : M j i M i be given by τ j (x) = (p 1,... p j 1, x, p j+1,... p k ). Define β by β(v 1,..., v k ) = d(τ 1 ) p1 (v 1 ) + + d(τ k ) pk (v k ). Then β is linear since each d(τ i ) pi is linear, and moreover we have α β(v 1,..., v k ) = j = j = j d(π j ) p (β(v 1,..., v k )) d(π j ) p d(τ i ) pi (v i ) i d(π j τ i ) pi (v i ) = i j where the last equality uses the fact that π j τ i is the identity if i = j and a constant function otherwise. Thus α β = Id i Tp i M i. v j, We remark that there is also the following alternate solution that uses local coordinates. Let (U i, ϕ i ) be charts for M i around p i. Then U = U i and ϕ = ϕ i gives a chart for M around p. Writing x 1 i,..., x n i i : U i R for the coordinate functions of ϕ i, we see that the coordinate functions of ϕ are x 1 1,..., x n 1 1, x 1 2,..., x n 2 2,..., x 1 k,..., x n k k, where x j i = xj i π i (note that we must compose with π i to get a function U R.) Thus a basis for T p ( M i ) is given by { } x j. p i To show that α is an isomorphism it suffices to show that ( ) d(π l ) p x j = p i x j if l = i and 0 otherwise, i since then α takes the basis for T p ( M i ) to a basis for T pi M i. We compute d(π l ) p ( x j p ) by observing that for any f C (M l ), we have i d(π l ) p ( x j i ) ( (f) = p x j i i,j ) (f π l ) = ĝ p x j (ˆp), i

12 Math 7350 Selected HW solutions Page 12 of 30 where g = f π l C (M) and ĝ = g ϕ 1 C (Û). (Recall that Û = ϕ(u) R n i.) But we also have ĝ x j i (ˆp) = (f π l ϕ 1 ) x j (ˆp) = i (f ϕ 1 l ˆπ l ) (ˆp) where ˆπ l : R n i R n l is projection onto the n l coordinates corresponding to ( x j l for ) j = 1,..., n l. The final quantity is equal to 0 if l i, and pi (f) if l = i. This proves that α maps a basis for x j i x j i T p ( M i ) onto a basis for T pi M i, hence it is an isomorphism. HW 3, # 2. (Lee, Problem 3-4). A trivial tangent bundle Show that T S 1 is diffeomorphic to S 1 R. Remark: You may find it interesting to consider whether or not the same is true for S 2. Be warned that we do not yet have the machinery in this course to answer this question. Solution. First recall that for any smooth manifold M, if (U, ϕ) is a chart on M then (Ũ, ϕ) is a chart on T M, where Ũ = p U T pu and ϕ: Ũ Û Rn is given by ϕ(v) = (π(v), v 1,..., v n ), where v j = v(x j ) so that v = v j x j π(v). (π : T M M is the canonical submersion.) The maps ϕ: Ũ Û Rn and ϕ Id: U R n Û Rn are both diffeomorphisms, thus G U : Ũ U Rn given by G U = (ϕ 1 Id) ϕ is a diffeomorphism. In light of the previous paragraph, it is natural to try to define a diffeomorphism G: T M M R n by fixing a smooth atlas A on M and then for each (U, ϕ) A, putting G(v) = G U (v) for v Ũ. The problem is that in general we have no reason to expect that G U U V = G V U V when (U, ϕ) and (V, ψ) are two overlapping charts in A.

13 Math 7350 Selected HW solutions Page 13 of 30 Writing τ : U R n U for projection to the first part of the ordered pair, we see that τ(g U (v)) = π(v) for every v Ũ, which does not depend on U. So the compatibility problems we may encounter have to do with the coordinates that different charts induce on the tangent spaces. The special thing about M = S 1 is that we can choose charts where these coordinates all agree. With S 1 C as the unit circle, let U = S 1 \ { 1} and let ϕ: U ( π, π) be defined by the condition that ϕ 1 (x) = e ix. Similarly, let V = S 1 \ {1} and let ψ : V (0, 2π) be defined by ψ 1 (x) = e ix. Let G U : Ũ U R and G V : Ṽ V R be as described above. (It is an easy exercise to check that ϕ, ψ are compatible with the usual smooth structure on S 1.) Note that given p U, every v T p S 1 can be expressed as v = v 1, x where v 1 = v(x) R. Thus G U (v) = (p, v 1 ) U R S 1 R. Writing x for the coordinate function relative to V, we similarly have v = ṽ 1 x for v T p S 1 when p V. The key is that the transition map between U and V is given by { x x (0, π), x(x) = x + 2π x ( π, 0). Thus for every p U V and v T p S 1, we have v(x) = v( x), hence v 1 = ṽ 1. In particular, G U = G V on Ũ Ṽ, so this gives a well-defined bijection G: T S 1 S 1 R. Note that the coordinate representations of G are the maps (ϕ Id) G U ϕ 1, but each of these is just the identity map on Û Rn, and hence is smooth. Similarly, the coordinate representation of G 1 is the identity, hence smooth, so G is a diffeomorphism. HW 3, # 4. Quotient manifolds Let G be a group and E a smooth manifold. A left action of G on E is a map G E E, often written as (g, p) g.p, that satisfies g 1.(g 2.p) = (g 1 g 2 ).p for all g 1, g 2 G, p E, e.p = p for all p E.

14 Math 7350 Selected HW solutions Page 14 of 30 Suppose we are given a left action of G on E such that for every g G, the map p g.p is a smooth map from E to itself. This induces an equivalence relation on E by putting x y iff there is g G such that g.x = y. We say that the action is free and proper (see p of Lee) if the following are true: (i) for every x E there is a neighbourhood U of x such that g.u U = for every g e; (ii) for every x, y E with x y there are neighbourhoods U of x and V of y such that g.u V = for every g G. Write M = E/G for the quotient space of E by the relation. (a) Prove that if the action is free and proper, then M is a topological manifold. (b) Let π : E M be the quotient map. Show that π is a covering map. (This means that for every x M there is a neighbourhood U x such that π 1 (U) = α A V α for some disjoint open sets V α E such that π Vα : V α U is a homeomorphism.) (c) Prove that if E is a smooth manifold, then M has a unique smooth structure such that π is a smooth covering map. ( Smooth covering map means that homeomorphism is replaced by diffeomorphism in the definition of covering map.) Solution. To show that M is a topological manifold, we first observe that quotients of second countable spaces are themselves second countable, so it suffices to check Hausdorff and locally Euclidean. The Hausdorff property will follow from (ii) in a moment; first we show that π is an open map. Indeed, given any U E we have π(u) = π(g.u), where G.U = g U g.u. Note that for every g G, the map p g.p is invertible with inverse q g 1.q, hence p g.p is

15 Math 7350 Selected HW solutions Page 15 of 30 a diffeomorphism from E to itself; thus if U is open, then g.u is open for every g G, so G.U is also open. Because G.U is saturated with respect to π, it follows from the definition of the quotient topology that π(u) = π(g.u) is open for every open U E, so π is an open map. Now given [x] [y] M, where x, y E and we write [x] = G.x = {g.x g G} for the equivalence class of x, then we have x y and so by (ii) there are neighbourhoods U x and V y such that g.u V = for every g G. Since π is an open map, the sets [U] = π(u) and [V ] = π(v ) are open in M; since [x] [U] and [y] [V ], to check Hausdorff it suffices to show that [U] [V ] =. Suppose [z] [U] [V ]. Then there are g, h G such that g.z U and h.z V. But then h 1 g.z h 1 g.u V, contradicting our choice of U, V. This shows that M is Hausdorff. To show that M is locally Euclidean, we start by proving that π is a covering map. Given any [x] M, by (i) there is a neighbourhood U x such that g.u U = for every g e. Note that [U] = π(g.u) = π( g G g.u) and that G.U is saturated, so π 1 [U] = g G g.u. To show that U is evenly covered it suffices to show that the union is disjoint and that π g.u is a homeomorphism onto its image for every g. First we observe that if z g.u h.u for some g h G, then h 1 z h 1 g.u U, but h 1 g e, contradicting (i). Thus the union is disjoint. Moreover, if π(y) = π(z) for some y, z g.u, then there is h G such that z = h.y, hence z hg.u g.u, so g 1 z g 1 hg.u U, so g 1 hg = e, so h = e. In particular, π g.u is 1-1. Since π is continuous and open, this shows that π g.u is a homeomorphism onto its image, thus U is evenly covered. Now we use the fact that π is a covering map to show that M is a topological manifold. Given [x] M, let U x be a neighbourhood as above, so that π U is a homeomorphism from U E to [U] = π(u) M. Because E is a topological manifold, there is a neighbourhood V x and a homeomorphism ϕ: V ˆV R n. Thus ϕ := ϕ π 1 U is a homeomorphism from [V ] [U] to an open subset of R n. This shows that M is locally Euclidean, hence a topological manifold (we already showed Hausdorff and second countable).

16 Math 7350 Selected HW solutions Page 16 of 30 Finally, we show that there is a unique smooth structure on M such that π is a smooth covering map. Observe that if there is such a smooth structure, then because each chart ϕ on E is a diffeomorphism onto its image, each of the maps ϕ π 1 U from the previous paragraph is also a diffeomorphism onto its image, so any smooth structure on M with π a smooth covering map must be compatible with the atlas defined in the previous paragraph. It only remains to show that the charts from this atlas are all smoothly compatible with each other, so that we have existence of such a smooth structure. Let ([U], ϕ) and ([V ], ψ) be any two such charts on M. That is, [U], [V ] M are evenly covered, with U, V E open; moreover, ϕ = ϕ π 1 U for a smooth chart ϕ: U Rn, and similarly for ψ. Note that we may assume without loss of generality that π U V maps U V homeomorphically to [U] [V ] (a priori we may have U V = even when [U] [V ], but we can replace V with g.v for some g G to get the desired property). Now we have ψ ϕ 1 = (ψ π 1 U V ) (π U V ϕ 1 ) = ψ ϕ 1, which is smooth since ϕ, ψ come from a smooth atlas for E. This shows that the atlas we defined on M is smooth, thus M has a smooth structure such that π is a smooth covering map. We showed above that this structure is uniquely determined by the smooth structure on E. HW 4, # 1. Let S, S be compact surfaces without boundary, and suppose that ρ: S S is a covering map with degree n. Prove that χ( S) = nχ(s). You may use without proof the following result: Given any open cover U of a surface S, there is a triangulation T of S such that every triangle in T is completely contained in some element U U.

17 Math 7350 Selected HW solutions Page 17 of 30 Solution. Because ρ is a covering map of degree n, every point x S has a neighbourhood U x x such that ρ 1 (U x ) is a disjoint union of open sets Vx 1,..., Vx n S with the property that ρ V i x is a homeomorphism from Vx i to U x. Let U = {U x x S}. By the result stated in the problem, there is a triangulation T of S such that every triangle is completely contained in some U x. Formally we defined T as a collection of triangles in R 2 together with an equivalence relation on the edges, such that the quotient space is the surface S. We may think of the triangles as being subsets of S (we identify a triangle with its image under the quotient map), and we will do this from now on; indeed, this is implicit in the formulation of the result in the previous paragraph (every triangle is completely contained in some U x ). Let T be a triangle in T, and let x be such that T U x. Then ρ 1 (T ) is a disjoint union of n triangles in S. Denote the collection of all such triangles in S by T. Note that the intersection of any two triangles T 1, T 2 T is of the form ρ 1 (ρ(t Vx i 1 ) ρ(t 2 )); since ρ(t 1 ) and ρ(t 2 ) are triangles in T and ρ V i x is a homeomorphism Vx i U x, we see that T 1 T 2 is a union of vertices and edges. Thus T is a triangulation of S. We showed above that the number of faces in T is n times the number of faces in T. The same is true for vertices and edges, because each edge and each vertex in T is contained in some U x, which is evenly covered by ρ. Thus we have χ( S) = #faces in T #edges in T + #vertices in T = n(#faces in T ) n(#edges in T ) + n(#vertices in T ) = nχ(s).

18 Math 7350 Selected HW solutions Page 18 of 30 HW 4, # 2. Let S k be the sphere with k handles. Give a necessary and sufficient condition on k, l N for the existence of a covering map ρ: S k S l. Hint: Use the previous problem to determine a natural necessary condition. Then given k, l satisfying this condition, describe a particular realization of S k as a surface in R 3 that is symmetric under rotation by 2π/(k 1) around the z-axis. Then for a suitable value of n, rotation by 2π/n around the z-axis will induce an equivalence relation on S k, whose quotient space is S l, and whose quotient map is the desired covering map ρ. Solution. From the previous problem we know that if there is a covering map S k S l then χ(s k ) = nχ(s l ) for some n N. Recall from lecture that χ(s k ) = 2 2k, so if a covering map exists then we have 2 2k = n(2 2l), or in other words, n = k 1 l 1 N. We claim that this necessary condition is also sufficient. It suffices to check the case when l k, since every surface covers itself (by the identity map). Note that if k = 0 or 1 then the condition can only be satisfied for l = k. Thus it suffices to consider the case when k > 1. Consider the realization of S k in R 3 sketched in the figure (for k = 5), where we start with the torus (S 1 ) as a surface of revolution and then add the remaining k 1 handles evenly spaced around the outside of the torus. We attach the handles symmetrically so that the surface is symmetric under rotation around the z-axis by 2π/(k 1).

19 Math 7350 Selected HW solutions Page 19 of 30 Now let n = k 1 and let G be the finite cyclic group with n elements. l 1 Let G act on S k by rotation; that is, if g is a generator for g then g m acts by rotation by 2πm/n around the z-axis. Note that this is a diffeomorphism from S k to itself and that this gives a smooth group action that is free and proper; a typical orbit of the action is shown below. The figure also shows how a fundamental domain for the quotient S k /G can be selected; cut S k into a wedge of angle 2π/n, and then glue the two boundary circles together to get S l. Note that because the quotient map is by a free and proper group action, we know from the previous assignment that it is a smooth covering map. HW 4, # 3. Consider the function f(x, y) = sin(4πx) cos(6πy) on the torus T 2 = R 2 /Z 2. (a) Prove that this is a Morse function (every critical point is nondegenerate) and calculate the number of minima, saddles, and maxima. (b) Describe the evolution of the sublevel sets f 1 (( ), c)) as c varies from the lowest minimum value to the highest maximum value.

20 Math 7350 Selected HW solutions Page 20 of 30 Solution. A straightforward computation shows that f = 4π cos(4πx) cos(6πy), x f = 6π sin(4πx) sin(6πy), y and since cos t, sin t never vanish simultaneously, we see that (x, y) R 2 is a critical point of f : R 2 R if and only if one of the following two conditions holds: cos(4πx) = sin(6πy) = 0, sin(4πx) = cos(6πy) = 0. The first condition is satisfied iff x = (2n + 1)/8 and y = m/6 for some n, m Z; the second is satisfied iff x = n/4 and y = (2m + 1)/12 for some n, m Z. Note that each critical point on the torus corresponds to a critical point in [0, 1) 2 (since this is a fundamental domain for the torus), and so the set of critical points is A B, where Another computation gives A = { 1 8, 3 8, 5 8, 7 8 } {0, 1 6, 1 3, 1 2, 2 3, 5 6 }, B = {0, 1 4, 1 2, 3 4 } { 1 12, 3 12, 5 12, 7 12, 9 12, }. 2 f x 2 = 16π2 sin(4πx) cos(6πy), 2 f x y = 24π2 cos(4πx) sin(6πy), 2 f y 2 = 36π2 sin(4πx) cos(6πy), so the Hessian determinant is ( ) 576π 2 sin 2 (4πx) cos 2 (6πy) cos 2 (4πx) sin 2 (6πy). If (x, y) A then the second term in brackets vanishes, and the first is positive, so the determinant is positive; thus every critical point in A is non-degenerate and is either a maximum or a minimum. Similarly, if (x, y) B then the first term vanishes and the second is positive, so the determinant is negative; thus every critical point in B is nondegenerate and is a saddle.

21 Math 7350 Selected HW solutions Page 21 of 30 We have shown that f is a Morse function and that it has 24 saddles, and 24 other critical points that are either minima or maxima (the points in A). Note that if (x, y) A then each of sin(4πx) and cos(6πy) is either ±1, so f(x, y) is either ±1; there are 12 elements of A with f = 1, which are minima, and 12 with f = 1, which are maxima. Now we consider the sublevel sets S c = f 1 ((, c)). For c < 1 this is empty; for c ( 1, 0) it is the union of 12 disjoint discs (geometrically these are ellipses). As c increases the discs grow, until at c = 0 they become rectangles which touch at the corners; these corners are the 24 saddle points. For c (0, 1) the sublevel set is a torus with 12 discs removed, and these discs are filled in as c increases past 1, so that S c = T 2 for c > 1. These sublevel sets are shown in the figure. c = 2 3 c = 1 3 c = 0 c = 1 3 c = 2 3 Notice that because there is more than one saddle with f = 0, passing from c = ε to c = ε does not correspond to gluing on a single pair of pants; rather, it corresponds to gluing on some kind of very complicated set of pants with lots of legs. More precisely, a regular pair of pants is the sphere with 3 discs removed. The very complicated set of pants with lots of legs that we encounter here (that is, S ε \ S ε ) is a torus with 24 discs removed. HW 5, #2. Lee, Problem 4-4. Let γ : R T 2 be the curve given by γ(t) = (e 2πit, e 2πiαt ), where α is any irrational number. Use Lemma 4.21 to show that the image set γ(r) is dense in T 2.

22 Math 7350 Selected HW solutions Page 22 of 30 Solution. Given ε > 0, Lemma 4.21 guarantees the existence of m, n Z such that mα n < ε. Let β = mα n and note that e 2πimα = e 2πi(mα n) = e 2πiβ. Given any open set V S 1 there is (a, b) R such that {e 2πix x (a, b)} V. Taking ε = b a and letting m, n, β be as above, we have β < ε and it follows that there is k Z such that kβ (a, b). In particular, we have e 2πikmα = e 2πikβ V. Now we turn our attention to the curve γ : R T 2. Recall that {U V U, V S 1 are open} is a basis for the topology on T 2. In particular, given any open U, V S 1 we can choose s R such that e 2πis U and then use the above argument to find k Z such that e 2πikmα e 2πisα V = {e 2πisα z z V }. Then e 2πi(s+km)α V, and we note that e 2πikm = 1, so γ(s + km) = (e 2πi(s+km), e 2πi(s+km)α ) = (e 2πis, e 2πisα e 2πikmα ) U V. This holds for arbitrary U, V, so γ(r) is dense in T 2. HW 6, #1. Lee, Problem 5-4. Let β : ( π, π) R 2 be the smooth curve given by β(t) = (sin 2t, sin t). (See Example 4.19.) Show that the image of this curve is not an embedded submanifold of R 2. Be careful: this is not the same as showing that β is not an embedding! Solution. Let S be the image of the curve. Let U be a small nbhd of 0 in R 2, so S U is open in S (with the subspace topology). Whenever U is small enough, (S U)\0 has four connected components. Thus S U cannot be homeomorphic to any open ball in R n, because an open ball with a point removed has two connected components if n = 1, and only one otherwise. Thus S with the subspace topology is not a topological mfd.

23 Math 7350 Selected HW solutions Page 23 of 30 HW 6, #4. Lee, Problem 5-18(a). Suppose M is a smooth manifold and S M is a smooth submanifold. Show that S is embedded if and only if every f C (S) has a smooth extension to a neighbourhood of S in M. Solution. ( ). Suppose S is embedded; we prove every f C (S) can be extended to F C (U) for some open U S. Since S is embedded it satisfies the local k-slice condition, so given any p S there is a slice chart (U p, ϕ p ) with p U p. Given f C (S) and p S, let f p C (U p ) be the function whose coordinate representation relative to (U p, ϕ p ) is ˆf p (x 1,..., x n ) = ˆf(x 1,..., x k ). Note that f p S Up = f S Up. Let U = p U p so that U is an open submanifold of M. Let {ψ p } p S be a partition of unity subordinate to the open cover (U p ) p S of U. Then the product ψ p f p gives a smooth function supported in U p, which we can take to be defined on U by setting it equal to zero outside of U p. Define F : U R by F = p S ψ pf p. Then F is smooth since the sum is finite at every point, and for all q S we have F (q) = p ψ p(q)f p (q) = p ψ pf(q) = f(q). ( ). Suppose every f C (S) has a smooth extension to a neighbourhood of S in M. We show that S is embedded by verifying the local k-slice condition. Given p S there is a nbhd V p in S (with respect to the topology making S an immersed submfd) such that ι: V M is an embedding. (This is a general property of immersions.) In particular, there is a chart U around p in which V U is a k-slice. Let f C (S) be a bump function supported in V, with f(p) > 0. By the hypothesis there is an open (in M) set W S and F C (W ) such that F S = f. Because F is continuous, W := {q W F (q) > 0} is an open set. Moreover, W S supp(f) V. Restricting the chart U to U W, we see that U W S U W V is a k-slice.

24 Math 7350 Selected HW solutions Page 24 of 30 HW 6, #5. Lee, Problem Show by giving a counterexample that the conclusion of Proposition 5.37 may be false if S is merely immersed. That is, give an example of a smooth immersed submanifold S M and p S for which T p S does not coincide with {v T p M vf = 0 whenever f C (M) and f S = 0}. Solution. Here is one example: let S R 2 be the figure eight from #1 of this homework assignment. Let S be the same set with the topology and smooth structure corresponding to β(t) = ( sin 2t, sin t). Then T 0 S R 2 is the line spanned by (2, 1), while T 0 S is the line spanned by ( 2, 1). Let v, ṽ be the derivations on C (R 2 ) corresponding to directional derivatives in these two directions. Note that both v, ṽ annihilate any f C (R 2 ) with f S = 0, but only one of them can be contained in T 0 S. Another example is the line with irrational slope on the torus. Let S be this submanifold and note that T p S is one-dimensional. But any smooth function vanishing on S must vanish on T 2 since S is dense, and so vf = 0 for every v T p T 2. HW 7, #1. Lee, Problem 5-6. Suppose M R n is an embedded m- dimensional submanifold, and let UM T R n be the set of all unit tangent vectors to M: UM := {(x, v) T R n x M, v T x M, v = 1}. It is called the unit tangent bundle of M. Prove that UM is an embedded (2m 1)-dimensional submanifold of T R n R n R n.

25 Math 7350 Selected HW solutions Page 25 of 30 Solution. Because M is embedded it satisfies the local m-slice condition: for every x M there is a chart (U, ϕ) around x with ϕ(u M) = {x ϕ(u) x m+1 = = x n = 0}. Let (Ũ, ϕ) be the associated chart for T R n R n R n, so Ũ = π 1 (U), and note that ϕ(ũ T M) = {(x, v) R n R n x m+1 = = x n = 0, v m+1 = = v n = 0}, so T M satisfies the local 2m-slice condition. This shows that T M is a 2m-dimensional embedded submanifold of R 2n. Now consider the function f : R 2n R defined by f(x, v) = v 2 = n i=1 (vi ) 2. Note that f is smooth, hence f restricts to a smooth function T M R. Moreover, UM = f 1 (1), so to show that UM is an embedded submanifold of T M (and hence also of R 2n ), it suffices to show that f T M has rank 1 at every (x, v) UM. To this end, it is enough to produce a curve γ : ( ε, ε) T M with γ(0) = (x, v) such that df (x,v) γ (0) 0. Let γ(t) = (x, (1 + t)v), then df (x,v) γ (0)(Id) = (f γ) (0) = d dt (1 + t)v 2 t=0 = 2. This shows that f T M has rank 1 at every (x, v) UM, hence UM is a codimension 1 embedded submanifold of T M, which completes the proof. HW 7, #3. Lee, Problem 7-1. Show that for any Lie group G, the multiplication map m: G G G is a smooth submersion. Hint: use local sections Solution. Given g G, define a map σ g : G G G by σ g (h) = (hg 1, g), so that m σ g (h) = hg 1 g = h. Thus σ g is a smooth section of m. Given any (h, g) G G, we have σ g (hg) = (hgg 1, g) = (h, g), so (h, g) is in the image of the smooth section σ g. This shows that m is a smooth submersion.

26 Math 7350 Selected HW solutions Page 26 of 30 HW 7, #5. Lee, Problem Considering S 2n+1 as the unit sphere in C n+1, define an action of S 1 on S 2n+1, called the Hopf action, by z (w 1,..., w n+1 ) = (zw 1,..., zw n+1 ). Show that this action is smooth and its orbits are disjoint unit circles in C n+1 whose union is S 2n+1. Solution. Given any z, the map (w 1,..., w n+1 ) (zw 1,..., zw n+1 ) is smooth because multiplication is smooth, and the orbits form a partition of S 2n+1 by basic properties of group actions. The only real task is to show that every orbit is a unit circle. To this end we observe that given (w 1,..., w n+1 ) S 2n+1 C n+1, the map F : C C n+1 given by F (z) = (zw 1,..., zw n+1 ) is linear and 1-1, so its image is a subspace of C n+1 with real dimension 2. The orbit of (w 1,..., w n+1 ) is the intersection of that image with S 2n+1 ; that is, it is the intersection of a plane with S 2n+1, which is a unit circle. HW 8, #2. Lee, Problem Prove that up to isomorphism, there are exactly one 1-dimensional Lie algebra and two 2-dimensional Lie algebras. Show that all three of these algebras are isomorphic to Lie subalgebras of gl(2, R). Solution. Let g be a one-dimensional Lie algebra and let q g be nonzero. Define a linear map f : g gl(2, R) by f(cq) = ( 0 c 0 0 ). Note that [q, q] = [q, q] = 0 by anti-symmetry and hence [cq, c q] = 0 for all c, c R by bi-linearity. Thus g is abelian and f is a Lie algebra homomorphism. The image of f is a 1-dimensional subspace of gl(2, R) in which all matrices commute, hence f is a Lie algebra isomorphism onto its image ( R ), so all 1-dimensional Lie algebras are abelian and are isomorphic to this one.

27 Math 7350 Selected HW solutions Page 27 of 30 Now suppose g is a two-dimensional abelian Lie algebra. Let v, w g be linearly independent and define f : g gl(2, R) by f(v) = ( ) and f(w) = ( ). Let h be the image of f, then h is a two-dimensional abelian Lie subalgebra of gl(2, R) and f : g h is a Lie algebra isomorphism. Thus every two-dimensional abelian Lie algebra is isomorphic to h, the Lie algebra of diagonal matrices in gl(2, R). Finally, suppose g is a two-dimensional non-abelian Lie algebra. Then there are v, w g such that [v, w] 0. As above, this implies that v, w are linearly independent. Thus they form a basis for g. Let u = [v, w] and note that one of the sets {u, v} and {u, w} is linearly independent (otherwise v, w are scalar multiples of each other). Without loss of generality assume u, v are linearly independent, hence a basis. Moreover, writing u = av + bw we have [u, v] = [av + bw, v] = b[w, v] = bu. Let x = v/b (note that b 0 since g is non-abelian), then [u, x] = b 1 [u, v] = u. Define f : g gl(2, R) by f(u) = ( ) and f(v) = ( ), and note that [f(u), f(v)] = f(u) = f([u, v]) so f is a Lie algebra homomorphism. In particular it is a Lie algebra isomorphism between g and h := f(g) since f is 1-1. Thus every two-dimensional non-abelian Lie algebra is isomorphic to h = {( 0 0 a b ) a, b R}. HW 8, #3. Lee, Problem Considering det: GL(n, R) R as a Lie group homomorphism, show that its induced Lie algebra homomorphism is Tr: gl(n, R) R. Hint: see Problem 7-4, which was #4 from the last assignment. Solution. Recall that gl(n, R) = M(n, R) with the commutator bracket. This is identified with T In GL(n, R) by identifying A gl(n, R) with the tangent vector at t = 0 to the curve γ A (t) = I n + ta, which is contained in GL(n, R) for sufficiently small values of t. Then T In GL(n, R) is identified with the space of left-invariant vector fields on GL(n, R) by

28 Math 7350 Selected HW solutions Page 28 of 30 associating to v T In GL(n, R) the left-invariant vector field (v L ) X = d(l X ) In (v) it generates (here X GL(n, R)). Similarly, the space of left-invariant vector fields on R is identified with T 1 R by associating v T 1 R with (v L ) x = d(l x ) In (v). In the canonical chart on R, where T x R is identified with R via the basis x, this gives (vl ) x = xv for x R and v R. Up to the identification of left-invariant vector fields with their representatives at the identity, the Lie algebra homomorphism induced by det is, by definition, the linear map T In GL(n, R) T 1 R given by d(det) In. It was shown on the last assignment that d(det) In (A) = Tr(A). Thus trace is the Lie algebra homomorphism induced by determinant. HW 9, #1. Let V be a finite-dimensional vector space. Recall that given two tensors ω T k V and η T l V, the tensor product ω η T k+l V is defined by ω η(v 1,..., v k, v k+1,..., v k+l ) = ω(v 1,..., v k )η(v k+1,..., v k+l ). Let det T 2 (R 2 ) be the 2-tensor defined by det(v, w) = det ( v 1 w 1 v 2 w 2 ) = v 1 w 2 v 2 w 1, where v = v i E i and w = w i E i. Recall that if (e 1, e 2 ) is the standard basis for (R 2 ), then det = e 1 e 2 e 2 e 1. Determine (with proof) whether or not there are 1-tensors (covectors) ω, η T 1 (R 2 ) = (R 2 ) such that det = ω η. Solution. If det = ω η then we would have 0 = det(e 1, E 1 ) = ω(e 1 )η(e 1 ), 0 = det(e 2, E 2 ) = ω(e 2 )η(e 2 ). Since det 0 we have ω 0 and η 0, hence either ω(e 1 ) = η(e 2 ) = 0 or ω(e 2 ) = η(e 1 ) = 0. But then we either have det(e 1, E 2 ) = 0 or det(e 2, E 1 ) = 0, a contradiction.

29 Math 7350 Selected HW solutions Page 29 of 30 HW 9, #2. Let (e 1, e 2, e 3 ) be the standard dual basis for (R 3 ). Show that e 1 e 2 e 3 is not equal to a sum of an alternating tensor and a symmetric tensor. Solution. Suppose ω, η T 3 (R 3 ) are alternating and symmetric, respectively. Then since (1, 2, 3) (2, 3, 1) is an even permutation we have ω(e 1, E 2, E 3 ) = ω(e 2, E 3, E 1 ), η(e 1, E 2, E 3 ) = η(e 2, E 3, E 1 ), and hence the same symmetry holds for ω+η. The result follows observing that (e 1 e 2 e 3 )(E 1, E 2, E 3 ) = 1 but (e 1 e 2 e 3 )(E 2, E 3, E 1 ) = 0. HW 9, #4. Let ω 1,..., ω k be covectors on a finite-dimensional vector space V. (a) Lee, Problem Show that ω 1,..., ω k are linearly dependent if and only if ω 1 ω k = 0. Solution. ( ). Without loss of generality suppose ω 1 = k i=2 a iω i for some a i R. Then ω 1 ω k = k a i ω i ω 2 ω k = 0 i=2 because each term has a repeated ω i. ( ). Suppose ω 1,..., ω k are linearly independent, then they extend to a basis ω 1,..., ω n for V. Let v 1,..., v n be the basis for V that is dual to this. (Formally, take the dual basis

30 Math 7350 Selected HW solutions Page 30 of 30 for (V ) corresponding to {ω i }, then use the canonical isomorphism between V and (V ).) Then (ω 1 ω n )(v 1,..., v n ) = 1, hence ω 1 ω n 0, hence ω 1 ω k 0. (b) Suppose ω 1,..., ω k are linearly independent, and so is the collection of covectors η 1,..., η k V. Prove that span(ω 1,..., ω k ) = span(η 1,..., η k ) if and only if there is some nonzero real number c such that ω 1 ω k = c η 1 η k. Solution. ( ). Since the spans are the same there are a i j R such that ω i = a i jη j for every i. Then we get ω 1 ω k = (a 1 j 1 η j 1 ) (a k j k η j k ) = A J η J where for a multiindex J = (j 1,..., j k ) we write A J = a 1 j 1 a k j k and η J = η j 1 e j k, and the sum is over all multi-indices (not just increasing ones). Note that the wedge product vanishes whenever J contains a repeated index, so this is actually a sum over all permutations, and if J is a permutation of {1,..., k} then the wedge product is a scalar multiple of η 1 η k. It follows that ω 1 ω k = cη 1 η k for some c R, and we observe that c 0 since ω 1,..., ω k are linearly independent hence the left hand side is not 0, by (a). ( ). Suppose c R\0 is such that ω 1 ω k = cη 1 η k. Then for every i we have ω 1 ω k η i = cη 1 η k η i = 0 where the last equality is because η i is repeated. By part (a) we conclude that {ω 1,..., ω k, η i } is linearly dependent. Since the ω j are linearly independent this implies that η i span{ω 1,..., ω k }. This holds for all i, hence span{η 1,..., η k } span{ω 1,..., ω k }. The reverse inclusion is similar.