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UIC Math 549 fall 2006 Differentiable —Problems John Wood

The goal of problems 1 – 5 is stated in problem 5. They appear in Differential Topology by M. W. Hirsch and are taken from a 1940 paper by Heinz Hopf. The problems concern maps given in terms of a symmetric bilinear map β : Rn+1 × Rn+1 −→ R2n+1 where β(x, y) = z, x = (x0, . . . , xn), y = (y0, . . . , yn), and where z = (z0, . . . , z2n) is given by X zk = xiyj. i+j=k

1. Show if β(x, y) = 0 then x = 0 or y = 0.

2. Define g : Sn −→ S2n by g(x) = β(x, x)/|β(x, x)|. Show

(a) if g(x) = g(y) then β(x, x) = t2β(y, y) for some t 6= 0, (b) that β(x + ty, x − ty) = 0, and hence (c) x = ±y.

3. Define f : RP n −→ S2n by f([x]) = g(x). Show f is one-to-one. Conclude that f is a topological .

4. Show f is C∞.

5. Show the point p = (0, 1, 0,..., 0) ∈ S2n − im f and hence the stereographic projection ϕ from p gives an embedding ϕ ◦ f : RP n −→ R2n.

t t 6. Define f : Mn(R) −→ Mn(R) by f(X) = XX where X is an n × n-matrix and X is its transpose.

(a) f(X) is symmetric, (b) Df(X)M = XM t + MXt,

(c) If X ∈ O(n), that is, if f(X) = I, then Df(X) maps Mn(R) onto the space of 1 symmetric matrices. Hint: Given a symmetric S, take M = 2 SX.

7. Recall from §5, for a linear map `, ||`|| = sup{|`(x)|/|x| : x 6= 0}. Show ||λ ◦ `|| ≤ ||λ|| ||`||.

1 8. Let G`n(R) ⊂ Mn(R) be the open subset of invertible matrices; X ∈ G`n(R) iff det(X) 6= 0. Let f(X) = X−1. Show Df(X)M = −X−1MX−1.

9. Let ( x2y/(x2 + y2), if (x, y) 6= (0, 0); f(x, y) = 0, if (x, y) = (0, 0).

Show that Dvf(~0) exists for all v, but that Df(~0) does not exist since v 7→ Dvf(~0) is not linear. See Dieudonn´e8.4 problem 3(b) for an example where Dvf(~0) is linear in v, but still Df(~0) does not exist.

The next five problems are from M. Hirsch’s Differential Topology. They may require some thought or a suggested place to start. They are really exercises on the definitions.

10. An injective immersion of a compact Hausdorff is an embedding.

11. There is an immersion of the punctured S1 × S1 − {point} in R2. Suggestion: since a minus a point is homeomorphic to a disk minus a smaller disc in its interior, you can consider S1 × S1 minus a large homeomorph of the disk. It may help to regard S1 × S1 as a quotient space of the square.

12. Any product of can be embedded in Rm+1 where the product has dimension m. Suggestion: an inductive proof can be based on the statement, if M m embeds in Rn+1 with trivial , then M can be embedded in Rm+1+k with trivial normal bundle, and hence M × Sk embeds in Rm+1+k with trivial normal bundle.

13. If M is a compact C1 manifold, every C1 map M −→ R has at least two critical points.

14. Regarding the tangent bundle to Sn as

TSn = {(x, v): x ∈ Sn, v ∈ x⊥ ⊂ Rn+1},

construct a smooth map f,

TSn × R {(x, v, r): x ∈ Sn, v ∈ x⊥, r ∈ R}   f  y y Sn × Rn+1 {(x, w): x ∈ Sn, w ∈ Rn+1

2 The next 4 problems are from Glen Bredon’s, Topology and Geometry. Some manifolds are presented as inverse images of regular values. For more on problems 17 and 18 see John Milnor’s, Singular Points of Complex Hypersurfaces, Annals of Math. Studies 61.

15. Consider the real valued function f(x, y, z) = (2 − px2 + y2)2 + z2 on R3 − {(0, 0, z)}. Show that 1 is a regular value of f and identify the manifold M = f −1(1). [Solve f(~x) = 1 and grad f(~x) = ~0. To identify M, cylindrical coordinates may be helpful.]

16. Show the manifold M of problem 15 is not transverse to the plane

N = {(x, y, z) ∈ R3 : x = 1}.

Is M ∩ N a manifold? Describe M ∩ N. [Show grad f is normal to N at some point ~x

of M ∩ N, so that T~xN = T~xM.]

3 ~ 3 2 2 17. Let V = {(z1, z2, z3) ∈ C − {0} : p(z1, z2, z3) = 0} where p(z1, z2, z3) = z1 + z2 + z3. Show V is a 4-dimensional manifold by showing that 0 ∈ C is a regular value of p : C3 − {~0} −→ C. Show (0, 0, 0) is a singular point of p.

5 3 18. Let S = {(z1, z2, z3) ∈ C : g(z1, z2, z3) = 1} where g(z1, z2, z3) = z1z1 + z2z2 + z3z3. Show V ∩ S5 is a 3-manifold by showing that (0, 1) ∈ C × R is a regular value of (p, g). ⊥ Since, by problem 17, dp :(TzV ) −→ Tp(z) is onto for z ∈ V , it is enough to find 5 ~vz ∈ TzV with dg(vz) 6= 0 for z ∈ V ∩ S . Consider the curve γ : (1 − ε, 1 + ε) −→ V 2 3 3 0 defined by γ(t) = (t z1, t z2, t z3) and check (i) g ◦ γ(1) = 1 and (ii) (g ◦ γ) (1) 6= 0.

19. For 1-forms ω, the derivative satisfies

dω(X,Y ) = X(ω(Y )) − Y (ω(X)) − ω([X,Y ]) (∗)

for any vector fields X,Y . The Lie bracket is defined by

[X,Y ](f) = X(Y (f)) − Y (X(f)).

Verify this formula by computing each side for ω = f dg where f, g ∈ Ω0. Recall X(fg) = fX(g) + gX(f) and f dg(X) = fX(g). Since (∗) is independent of a choice of coordinate system, (∗) and the similar formula for p-forms show that the definition of d is invariant under change of coordinates.

20. Let i be the inclusion of the unit circle S1 into the plane R2. Define a 1-form ω by ω = i∗(x dy − y dx). This form ω is called dθ because, although θ is not a globally defined function on S1, local choices differ by an additive constant. For example, when x > 0, we may set θ = arctan(y/x).

3 (a) Verify that ω = d θ on the points of S1 where x > 0. (b) Show any α ∈ Ω1(S1) can be written as α = fω where f ∈ Ω0(S1). (c) Show the map I :Ω1(S1) −→ R is surjective where I is defined by

Z Z 2π I(fω) = fω = f(θ) dθ. S1 0

(d) If β = dg = g0(θ) dθ where g ∈ Ω0(S1), show I(β) = 0. R θ (e) Conversely if I(α) = 0, let g(θ) = 0 f(t) dt and show α = dg. (f) Deduce that H1(S1) = R. What is Hp(S1) for other values of p?

∗ # # # # 21. Let φ(x, y) = (ax + by, cx + dy) Then φ (e1 ∧ e2 ) is a multiple of e1 ∧ e2 . Find that multiple.

22. Let ω ∈ Ω1(M). Assume there exists a nonvanishing function f such that d(fω) = 0.

(a) If U ⊂ M is homeomorphic to a ball, find g ∈ Ω0(U) such that fω = dg. (b) Using ω = (1/f)dg, show ω ∧ dω = 0. (c) If ω = y dx − x dy + dz on R3, show ω ∧ dω 6= 0.

If ω = F # for a nonvanishing vector field F , then fF = grad g. F is orthogonal to the

level surfaces of g, f is called an integrating factor. For x ∈ U, Ex = {v : v ⊥ Fx} ⊂ TxM is the at x to a level of g. This is the local definition of a one foliation. A theorem of Frobenius says that ω ∧ dω = 0 is sufficient for such surfaces to exist.

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