DIFFERENTIAL GEOMETRY 1 PROBLEM SET 1 SOLUTIONS

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1 DIFFERENTIAL GEOMETRY PROBLEM SET SOLUTIONS Lee: -4,--5,-6,-7 Problem -4: If k is an integer between 0 and min m, n, show that the set of m n matrices whose rank is at least k is an open submanifold of M m n, R. Show that this is not true if at least k is replaced by equal to k. From basic linear algebra we know that if M is a matrix whose rank is l the an l l minor of M whose determinant is nonzero. So if we consider the map F : M m n, R R given by M M i an l l minor of M for l k det M i. Then since det and abs are continuous functions this is a sum of continuous functions and hence F is continuous. Moreover, by construction we know that F M = 0 iff M has rank strictly less than k. Therefore, F 0 = {M M m n, R rank M < k}. But R is Hausdorff and singletons are closed in a Hausdorff space so {0} is closed in R. Now since F is continuous we can immediately conclude that F 0 is closed in M m n, R. Therefore, M m n, R \ F 0 is open. But this is precisely the set {M M m n, R rank M k} So applying Lee s Example.7 we can conclude that the set of matrices in M m n, R whose rank is at least k is an open submanifold. Next we want to show that {M M m n, R rank M = k} is not an open submanifold of M m n, R. The simplest way to accomplish this is to show that this set is not open this is done by considering the following block matrix M = I k k This matrix has rank k, however for sufficiently small ɛ > 0 the matrix M = Ik k 0 0 ɛ is in a neighborhood of M and has rank k +. Thus {M M m n, R rank M = k} is not open in M m n, R and consequently cannot be an open submanifold. Date: January 3, 202.

2 2 Problem -5: Let N = 0,..., 0, be the north pole of S n R n+ and let S = N be the south pole. Define stereographic projection σ : S n \{N} R n by σ x,..., x n+ = x,...,x n and let σ x = σ x for x S n \ {S}. Part afor any x S n \ {N}, show that σ x is the point where the line through N and x intersects the linear subspace where x n+ = 0, identified with R n in the obvious way. Similarly, show that σ x is the point where the line through x and S intersects the same subspace. In R n+ we can express the line passing through N and x S n \ {N} by L N,x t = x N t + N. This line passes through the y n+ = 0 subspace when 0 = x N n+ t + N n+ = x n+ t +. This can be solved for t to find t =. So at this point we have L N,x x = n+ x,..., x n, 0 = σ x, 0. Similarly, the line passing through S and s S n+ \{S} is given by L S,x t = x S t+s. This line intersects the subspace defined by y n+ = 0 when 0 = L S,x t n+ = x n+ + t. Solving for t gives t =. +x n+ Thus L S,x +x = x x,..., n, 0 = σ x, 0. n+ +x n+ +x n+ Part b Show that σ is bijective and that σ u,..., u n = u 2 + 2u,, 2u n, u 2 By inspection we can see that σ is defined on all of R n and σ is defined on all of S n \{N}. So showing σ σ = Id R n and σ σ = Id S n \{N} is sufficient to establish that σ is a bijection and the σ given in the problem statement is its inverse. Proceeding by direct calculation we have σ σ 2u u = σ u 2 +,, 2u n u 2 +, u 2 u 2 + 2u,, 2u n u = 2 + u 2 + u 2 = 2u,..., 2u n u 2 + u 2 + = u = Id R n u

3 3 But this holds for all u R n and so σ σ = Id R n. Similarly, x σ σ x = σ x,, x n n+ x n+ 2x 2x,..., n, x 2 + +x n 2 2 = x 2 + +x n But x S n \ {N} and so x x n 2 = x n+ 2. So we have σ σ x = 2x,..., 2x n, xn+ 2 +2x n+ x n x n+ +x n+ 2 2x,..., 2x n, 2xn+ = 2 = Id S n \{N} x Since this holds for all x S n \ {N} we can conclude that σ σ = Id S n \{N}. Therefore σ is bijective and has inverse σ as defined above. Part c Compute the transition function s σ and verify that the atlas consisting of two charts S n \ {N}, σ and S n \ {S}, σ defines a smooth structure on S n First we consider σ σ. σ σ 2u u = σ u 2 +,..., 2u n u 2 +, u 2 u 2 + 2u,..., 2u n u = u 2 = u u 2 Inspection of this map reveals that it is differentiable. Next note that σ σ x = σ σ x = Id R n x = x x = x = Id R x

4 4 for all x R n whence σ x = σ x. Therefore σ σ u = σ σ u which, as before, is clearly smooth. = σ σ u = σ σ u u = u 2 = u u 2 Finally, note that S n = S n \ {N} S n \ {S}. Thus the atlas A = {S n \ {N}, σ, S n \ {S}, σ} defines a smooth structure on S n. Part d Show that the smooth structure defined in c is the same as the one defined in Example.20 Recall in example.20 the atlas was given by A = { U ± i, } ϕ± i where i n+ U + i = { x,..., x n+ S n+ x i > 0 } U i = { x,..., x n+ S n+ x i < 0 } ϕ ± i x,..., x n = x,..., ˆx i,..., x n+ ϕ ± i u,..., u n = u,..., u i, ± u 2, u i,..., u n To show that A as defined in c and A give the same smooth structure we need to show that A A is an atlas by Lee s Lemma.0. Since A and A independently define smooth structures on S n we know that A A will cover and that the transition functions from A and transition function from A are smooth. Thus it suffices to show that σ, ϕ ± i σ, ϕ ± i ϕ ± i σ, and ϕ ± i σ are smooth on their domain of definition. Applying the definitions of σ and ϕ ± i we find σ ϕ ± i which is smooth. Similarly, σ ϕ ± i { u u = u,...,u i,± u 2 if i=n u 2,...,u n u n { u ± u = u,...,u i,± otherwise u 2 if i=n u 2,...,u n +u n otherwise

5 which is again smooth. Finally, ϕ ± i σ u = ϕ ± i σ u = { { 2u 2u,..., 2u ˆi,...,2u n, u 2 2u 2u,..., 2u ˆi,...,2u n, u 2 if if i=n+ otherwise i=n+ otherwise Since these remaining transition functions are smooth we see that all transition function in A A are smooth and so A and A determine the same smooth structure on S n by Lee s Lemma.0. 5

6 6 Problem -6: By identifying R with C in the usual way, we can think of the unit circle S as a subset of the complex plane. An angle function on a subset U S is a continuous function θ : U R such that e iθp = p for all p U. Show that there exists an angle function θ on an open subset U S if and only if U S. For any such angle function, show that U, θ is a smooth coordinate chart on S with the standard smooth structure. Claim 0.. There exists an angle function θ on an open subset U S if and only if U S = Here we proceed by contradiction and suppose that there exists an angle function θ : S R. First note that if p, q S such that θ p = θ q then p = e iθp = e iθq = q and so θ is injective. Since S is compact and connect and θ is continuous we know that the image of θ is a compact connected subset of R. Consequently, a > b R such that θ S = [a, b]. But R is Hausdorff and so [a, b] is Hausdorff. Thus θ : S [a, b] is a continuous injective map from a compact space to a Hausdorff space and hence is a homeomorphism See Bredon s Topology and Geometry. On the other hand if any point in the interior of [a, b] is removed then the set has two connected components. On the other hand if any point is removed from S then the remaining piece is still connected. This contradicts S and [a, b] being homeomorphic. Therefore we can conclude that U S.. Suppose that U S is an open subset. Then p S \ U and consider the map arg : C \ 0 R defined to be the argument of the complex number z where the branch cut is taken to pass through the ray connecting 0 and p. From complex analysis we know that arg is a continuous function and hence restricts to a continuous function θ = arg U.

7 Claim 0.2. If U S is open and θ : U R is an angle function then A = { U ±, ϕ ±, U ± 2, ϕ ± 2 is an atlas for S where { } U ±, ϕ ±, U ± 2, ϕ ± 2 is the atlas on S corresponding to the standard smooth structure Lee example.2 and example.20. By Lee s Lemma.0 it suffices to show that on U U ± i the transition functions θ ϕ ± i and ϕ ± i θ are smooth where defined. To do this we recall that the angle function satisfies the property e iθp = p for all p U. So if we parameterize S by e it then we have e iθeit = e it and consequently θ e it = t+2k e it π for some k t Z. Furthermore, θ is continuous and so k : U Z must be continuous and consequently constant on each connected component of U. Since we can break U into a disjoint union of its connected components and consider each separately we may, without loss of generality, assume that U is connected. Thus θ e it = t + 2kπ. Moreover, as we saw above θ is injective and so θ x = e ix = cos x + i sin x. With these preliminary results in place we compute to find θ ϕ ± x = θ ± x 2, x = θ e iϕ± x 2,x = ϕ ± x 2, x + 2πk where ϕ x, y = arctan x y if x>0 arctan x+π y if x<0 and y 0 arctan x π y if x<0 and y<0 π 2 if x=0 and y>0 π 2 if x=0 and y<0 0 if x=0 and y=0 This is of course the argument of the complex number x + iy, or to put it another way the arctangent function with the quandrant accounted for. At any rate ϕ is smooth and so θ ϕ ± is smooth. Similarly, θ ϕ ± 2 x = θ x, ± x 2 = θ e iϕx,± x 2 = ϕ x, ± x 2 + 2πk which is smooth. On the other hand and which are smooth. ϕ ± θ t = ϕ ± cos x, sin x = cos x ϕ ± 2 θ t = ϕ ± 2 cos x, sin x = cos x 7, U, θ } The 2kπ isn t an issue due to the 2π periodicity of sin and cos

8 8 Problem -7: Complex projective n-space, denoted by CP n, is the set of all -dimensional complex-linear subspaces of C n+, with the quotient topology inherited from the natural projection π : C n+ \ {0} CP n. Show that CP n is a compact 2n-dimensional topological manifold, and show how to give it a smooth structure analgoous to the one we constructed for RP n. We identify C n+ with R 2n+2 via x + iy,..., x n+ + iy n+ x, y,..., x n+, y n+. Recall that a topological 2n-manifold is a locally Euclidean R 2n, second countable, Hausdorff, topological space. First, let [z : : z n+ ] denote the equivalence class of z,..., z n+ under π. i.e. [z] = [z : : z n+ ] = {w C n+ λ C \ {0} such that λw = z}. We now note that by construction of π we have π [z] = C z, the complex line passing through 0 and z with the origin excluded. So if U is open in C then π π U = {λz z U, λ C } which is nothing but the minimal collection of cones in C n+ \ {0} containing U and their reflections through 0. Consequently π π U is open and so by the definition of the quotient topology π U is open. Moreover, this holds for any open set U C n+ \ {0} and hence π is an open map. Recalling from topology that second countability is preserved under open quotients we can immediately conclude that CP n is second countable. Next we want to show that CP n is Hausdorff, to do this let [z] [w] CP n. Denote by L a the complex linear subspace of C n+ containing a and l a, b the minimal angle between L a and L b. Since [z] [w] we know L z L w. So let ɛ = l z, w and consider the sets U = 4 {ζ C n+ \ {0} l ζ, z < ɛ} and V = {ζ C n+ \ {0} l ζ, w < ɛ}. Then U and V are open in C n+ consequently their images are open in CP n since we showed that π was an open map. Moreover, [z] π U, [w] π V and π V π U = φ by construction and thus CP n is Hausdorff. 2 So to establish that CP n is a topological 2n-manifold it suffices to show that it is locally homeomorphic to R 2n. To this end consider the sets Ũi = {z,..., z n+ C n+ z i 0}. We can immediately see that Ũi is the complement of the closed set given by z i = 0 and hence is open. 2 An alternate approach is to recall that a topological space X is Hausdorff if = {x, x x X} is closed under the product topology in X. So we consider the set = {[z], [z] [z] CP n }, then π = { a, b µ, ν C µ such that a = z = νb }. However, this set is naturally identified with the set of all 2 n + complex rank matrices with nonzero columns. One can now write this set as the preimage of {0} R under a continuos function to show it is closed.

9 i Since π is an open map we have that U i = π Ũ = {[z : : z n ] CP n z i 0} is open in CP n under the quotient topology. Moreover, if [z] = [z : z n+ ] CP n then by definition of CP n we know that there is an i such that z i 0 and so [z] U i. Thus the U i constitute an open cover of CP n. Now define ϕ i : U i R 2n by [z : : z n+ ] Re z z i z z i,, Re Note that if [z] = [w] U i but z w then λ C such that z = λw and so zj z i and so ϕ i is well defined. Moreover, by inspection we see that ϕ i is continuous. Now consider ϕ i : R 2n CP n given by z i z i = λwj λw i 9 z i z, Re i+, z i z i = wj w i x, y,..., x n, y n [ x + iy : x i + iy i : : x i + iy i : : x n + iy n]. Then direct calculation reveals ϕ i ϕ i x = x and ϕ i ϕ i [z] = [z] where defined. Fur- reveals that it is continuous. Therefore ϕ i is a homeomorphism thermore, inspection of ϕ i from U i onto ϕ i U i. Thus CP n is locally Euclidean and has dimension 2n. Claim 0.3. CP n is compact. Let S n = {z C n+ z = } be the complex n-sphere. Then note any complex-linear subspace of C n+ intersects S n at exactly two points 0 z and z. Therefore, π S n = CP n. But S n is compact and π is continuous thus CP n is compact. Claim 0.4. A = {U i, ϕ i } i n+ is an atlas for CP n Given our above results it suffices to show that for i j, ϕ i ϕ j is smooth where defined. To see this we note that the i < j and i > j cases will be essentially symmetric calculations

10 0 and so without loss of generality we take i < j. Then ϕ i ϕ j x, y,..., x n, y n [ = ϕ i x + iy : : x j + iy j : : x j + iy j : x n + iy n] x + iy x + iy x i + iy i x i + = Re,, Re x i + iy i x i + iy i x i + iy i x i + x i+ + iy i+ x i+ + iy i+ Re,, x i + iy i x i + iy i x j + iy j x j + iy j Re, Re x i + iy i x i + iy i x i + iy i x i + iy i x j + iy j x j + iy j x n + iy n x n + iy n Re,, Re x i + iy i x i + iy i x i + iy i x i + iy i which is smooth. Remark 0.5. Note that if q : X Y is a quotient map and X is second countable then Y need not be second countable. Indeed, if we take X = n N [0, ] and take Y to be the quotient obtained by identifying all the left endpoints. Then Y is not first countable the distinguished and hence is not second countable. However, if q : X Y is an open quotient map and X is second countable then Y is second countable.