ex08

MA40254 Differential and geometric analysis : Exercises 8

Hand in answers by 1:15pm on Wednesday 29 November for the Seminar of Thursday 30 November Homepage: http://moodle.bath.ac.uk/course/view.php?id=57709 ex08

0 (Warmup). Let $M$ be an $n$ -dimensional submanifold of $R^{s}$ . Prove (from the definition of $d$ on submanifolds) that if $α \in Ω^{k} (M)$ is exact, i.e., $α = d β$ , then $α$ is closed, i.e., $d α = 0$ .

Solution
Suppose $α = d β$ . We have seen in lectures that $d^{2} = 0$ so $d α = d^{2} β = 0$ , but let us recall how to prove this. Note that it suffices to prove $d α = 0$ locally on $M$ , i.e., on sufficiently small open neighbourhoods $U \cap M$ of each $p \in M$ . There are two ways to proceed.

First by definition, we may assume that on $U \cap M$ , $β = ι^{*} γ$ , where $γ \in Ω^{k} (U)$ and $ι : U \cap M \to U$ is the inclusion. Thus $d α = d^{2} (ι^{*} γ) = d (ι^{*} d γ) = ι^{*} (d^{2} γ) = 0$ on $U \cap M$ . Alternatively, we may assume that there is a parametrisation $φ : \tilde{U} \to U \cap M$ . Then $φ^{*} d α = φ^{*} d^{2} β = d φ^{*} d β = d^{2} φ^{*} β = 0$ , so $d α = 0$ on $U \cap M$ because $φ^{*} : Ω^{k} (U \cap M) \to Ω^{k} (\tilde{U})$ is a bijection with inverse $(φ^{- 1})^{*}$ .

1. Let $M \subseteq R^{s}$ be an $n$ -dimensional submanifold and $α \in Ω^{k} (M)$ for $k > 0$ .
- (i) Suppose that $M$ is diffeomorphic to $R^{n}$ and $α$ is closed; prove that $α$ is exact.
  
  Hint
  Let $φ : R^{n} \to M$ be a diffeomorphism, and consider the pullback $φ^{*} α \in Ω^{k} (R^{n})$ .
- (ii) Suppose $k = n$ ; show that $α$ is closed.
  
  Hint
  What is the dimension of ${Alt}^{n + 1} (T_{x} M)$ ?
- (iii) If $k = n$ , does $α$ have to be exact?
  
  Hint
  Look for examples on $M = S^{1}$ . A previous exercise may help.

2. Give an example of an $n$ -dimensional submanifold $M \subseteq R^{s}$ and an $α \in Ω^{k} (M)$ such that $α$ is not the pullback by the inclusion of any $β \in Ω^{k} (R^{s})$ .

Hint
The simplest examples have $n = s = 1$ and $k = 0$ .

3. Let $S^{2} = {z \in R^{3} : ‖ z ‖ = 1}$ , and $U = S^{2} ∖ {(0, 0, 1)}$ , and consider the parametrisation $φ : R^{2} \to U$ defined by

$(x_{1}, x_{2}) \mapsto \frac{1}{1 + ‖ x ‖^{2}} (2 x_{1}, 2 x_{2}, ‖ x ‖^{2} - 1) .$

Let $α = i^{*} d z_{3}$ where $i : S^{2} \to R^{3}$ is the inclusion and $z_{1}, z_{2}, z_{3}$ denote the coordinate functions on $R^{3}$ . Compute $φ^{*} α$ .

Hint
$φ^{*} i^{*} d z_{3} = (i \circ φ)^{*} d z_{3} = \dots$ .

MA40254 Differential and geometric analysis : Solutions 8

1.
- (i) Suppose $α$ is closed. Let $φ : R^{n} \to M$ be a diffeomorphism. Then $φ^{*} α \in Ω^{k} (R^{n})$ is closed, so exact by the Poincaré lemma, i.e, there is some $γ \in Ω^{k - 1} (R^{n})$ such that $d γ = φ^{*} α$ . If we let $β = (φ^{- 1})^{*} γ \in Ω^{k - 1} (M)$ then $d β = (φ^{- 1})^{*} d γ = (φ^{- 1})^{*} (φ^{*} α) = (φ \circ φ^{- 1})^{*} α = α$ .
- (ii) For each $p \in M$ , $d α_{p}$ is an element of ${Alt}^{n + 1} (T_{p} M)$ . Since $n + 1 > \dim T_{p} M$ , we have ${Alt}^{n + 1} (T_{p} M) = {0}$ , so $d α$ is zero everywhere.
- (iii) No. Consider $M = S^{1} \subseteq R^{2}$ , and let $α \in Ω^{1} (S^{1})$ be the pullback by $i : S^{1} \to R^{2} ∖ {0}$ of the non-exact 1-form
  
  $γ = \frac{- x_{2} d x_{1} + x_{1} d x_{2}}{x_{1}^{2} + x_{2}^{2}} \in Ω^{1} (R^{2} ∖ {0})$
  
  from Exercises 6 Q 4(iii). The same argument as there shows that $α$ is not exact either.

2. We can for instance take $M = (0, \infty) \subset R$ , and $k = 0$ . Then an element of $Ω^{k} (M)$ is just a smooth function $(0, \infty) \to R$ and pullback reduces to restriction. If we take it to be $x \mapsto \frac{1}{x}$ then that is not the restriction of any smooth function $R \to R$ .

3. We have $φ^{*} i^{*} d z_{3} = (i \circ φ)^{*} d z_{3} = d ((i \circ φ)^{*} z_{3})$ and $(i \circ φ)^{*} z_{3} = (x_{1}^{2} + x_{2}^{2} - 1) / (1 + x_{1}^{2} + x_{2}^{2})$ . The exterior derivative of this is

$\frac{4 x_{1} d x_{1} + 4 x_{2} d x_{2}}{(1 + x_{1}^{2} + x_{2}^{2})^{2}} .$