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0 (Warmup). Let \(M\) be an \(n\)-dimensional submanifold of \(\R ^s\). Prove (from the definition of \(d\) on submanifolds) that if \(\alpha \in \Omega ^k(M)\) is exact, i.e., \(\alpha =d\beta \), then \(\alpha \) is closed, i.e., \(d\alpha =0\).
SolutionSuppose \(\alpha =d\beta \). We have seen in lectures that \(d^2=0\) so \(d\alpha =d^2\beta =0\), but let us recall how to prove this. Note that it suffices to prove \(d\alpha =0\) 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 \(U\cap M\), \(\beta =\iota ^*\gamma \), where \(\gamma \in \Omega ^k(U)\) and \(\iota \colon U\cap M \to U\) is the inclusion. Thus \(d\alpha =d^2(\iota ^*\gamma )=d(\iota ^*d\gamma ) =\iota ^*(d^2\gamma )=0\) on \(U\cap M\). Alternatively, we may assume that there is a parametrisation \(\varphi \colon \wt U\to U\cap M\). Then \(\varphi ^*d\alpha =\varphi ^*d^2\beta = d\varphi ^*d\beta =d^2\varphi ^*\beta =0\), so \(d\alpha =0\) on \(U\cap M\) because \(\varphi ^*\colon \Omega ^k(U\cap M)\to \Omega ^k(\wt U)\) is a bijection with inverse \((\varphi ^{-1})^*\).
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1. Let \(M \subseteq \R ^s\) be an \(n\)-dimensional submanifold and \(\alpha \in \Omega ^k(M)\) for \(k > 0\).
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(i) Suppose that \(M\) is diffeomorphic to \(\R ^n\) and \(\alpha \) is closed; prove that \(\alpha \) is exact.
HintLet \(\varphi : \mathbb {R}^n \to M \) be a diffeomorphism, and consider the pullback \(\varphi ^*\alpha \in \Omega ^k(\mathbb {R}^n) \). -
(ii) Suppose \(k=n\); show that \(\alpha \) is closed.
HintWhat is the dimension of \(\mathrm {Alt}^{n+1}(T_xM)\)? -
(iii) If \(k=n\), does \(\alpha \) have to be exact?
HintLook for examples on \(M = S^1\). A previous exercise may help.
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3. Let \(S^2 = \{z \in \R ^3 : \norm {z} = 1\}\), and \(U = S^2 \setminus \{ (0,0,1) \}\), and consider the parametrisation \(\varphi \colon \R ^2\to U\) defined by
\[ (x_1, x_2) \mapsto \frac {1}{1+\norm {x}^2} \left (2x_1, \, 2x_2, \norm {x}^2-1 \right ).\]
Let \(\alpha = i^* dz_3\) where \(i\colon S^2\to \R ^3\) is the inclusion and \(z_1,z_2,z_3\) denote the coordinate functions on \(\R ^3\). Compute \(\varphi ^*\alpha \).
Hint\(\varphi ^*i^*dz_3=(i\circ \varphi )^*dz_3=\cdots \).
MA40254 Differential and geometric analysis : Solutions 8
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1.
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(i) Suppose \(\alpha \) is closed. Let \(\varphi : \R ^n \to M\) be a diffeomorphism. Then \(\varphi ^*\alpha \in \Omega ^k(\R ^n)\) is closed, so exact by the Poincaré lemma, i.e, there is some \(\gamma \in \Omega ^{k-1}(\R ^n)\) such that \(d\gamma = \varphi ^*\alpha \). If we let \(\beta = (\varphi ^{-1})^*\gamma \in \Omega ^{k-1}(M)\) then \(d\beta = (\varphi ^{-1})^*d\gamma = (\varphi ^{-1})^*(\varphi ^*\alpha ) = (\varphi \circ \varphi ^{-1})^*\alpha = \alpha \).
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(ii) For each \(p \in M\), \(d\alpha _p\) is an element of \(\mathrm {Alt}^{n+1}(T_pM)\). Since \(n+1 > \dim T_p M\), we have \(\mathrm {Alt}^{n+1}(T_p M) = \{0\}\), so \(d\alpha \) is zero everywhere.
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(iii) No. Consider \(M = S^1 \subseteq \R ^2\), and let \(\alpha \in \Omega ^1(S^1)\) be the pullback by \(i\colon S^1\to \R ^2 \setminus \{0\}\) of the non-exact 1-form
\[ \gamma = \dfrac {-x_2\,dx_1 + x_1\,dx_2}{x_1^{\,2} + x_2^{\,2}} \in \Omega ^1(\R ^2 \setminus \{0\}) \]
from Exercises 6 Q 4(iii). The same argument as there shows that \(\alpha \) is not exact either.
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2. We can for instance take \(M = (0,\infty ) \subset \R \), and \(k = 0\). Then an element of \(\Omega ^k(M)\) 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 \(\R \to \R \).
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3. We have \(\varphi ^*i^*dz_3=(i\circ \varphi )^*dz_3=d ((i\circ \varphi )^*z_3)\) and \((i\circ \varphi )^*z_3= (x_1^2+x_2^2-1)/(1+x_1^2+x_2^2)\). The exterior derivative of this is
\[ \frac {4x_1dx_1 + 4x_2dx_2}{(1+x_1^2+x_2^2)^2}. \]