ex10

0 (Warmup). Let \(M\) be an open subset of \(H^n\) with the standard orientation. Show that integration of differential \(n\)-forms on \(M\) (as in Section 4.3) defines an integration map \(\int _M\) on \(M\) (as in Section 4.5).

Solution
Multiple integration is linear by basic results of analysis, hence integration of forms defines a linear map \(\Omega ^n_c(M)\to \R \). Now if \(\varphi \colon \wt U\to U\sub M\) is an oriented parametrization (with \(U\) open in \(M\) and \(\wt U\) open in \(H^n\)), and \(\supp (\alpha )\sub U\), then \(\int _M\alpha = \int _U\alpha \) because of the way integration of forms is defined, and \(\int _U\alpha =\int _{\wt U} \varphi ^*\alpha \) by the change of variables formula for integration of forms. Thus both properties of integration maps are satisfied.

1. Let \(M \subseteq \R ^s\) be an oriented compact \(n\)-dimensional SMWB. Let \(\omega \in \Omega ^n(M)\) be any orientation form on \(M\) compatible with the chosen orientation. Show that \(\int _M \omega > 0\).

Hint
Use a partition of unity to write \(\omega \) as the sum of forms, each of which has support contained in the image of a parametrisation, and show that the integral of each term has positive integral.

2. Let \(M \subseteq \R ^s\) and \(N \subseteq \R ^\ell \) be oriented \(n\)-dimensional SMWBs, and let \(\varphi : M \to N\) be an orientation-preserving diffeomorphism. Show that for any \(\alpha \in \Omega ^n_c(N)\) we have

\[ \int _M \varphi ^* \alpha = \int _N \alpha . \]

Hint
Show that \(\Omega ^n_c(N) \to \mathbb {R}, \; \alpha \mapsto \int _M \varphi ^*\alpha \) is an integration map.

3. Let \(\omega \in \Omega ^2(S^2)\) be the orientation form defined by \(\omega _p = p \inner \Det \) for \(p \in S^2\). Consider the oriented manifold \(S^2\) with orientation defined by \(\omega \). Define parametrisations of \(S^2\) by

\[ \varphi : \R ^2 \to S^2 \setminus \{(0,0,1)\}, \; x \mapsto \frac {1}{1+\norm {x}^2}\left (2x_1, 2x_2, \norm {x}^2-1\right ) \]

and

\[ \psi : \R ^2 \to S^2 \setminus \{(0,0,-1)\}, \; y \mapsto \frac {1}{1+\norm {y}^2}\left (2y_1, 2y_2, -\norm {y}^2+1\right ) . \]
- (i) Express \(\varphi ^*\omega \in \Omega ^2(\R ^2)\) in terms of \(dx_1 \wedge dx_2\), and \(\psi ^* \omega \in \Omega ^2(\R ^2)\) in terms of \(dy_1 \wedge dy_2\).
  
  Hint
  The usual process would be to write \(\omega = z_1 dz_2 \wedge dz_3 - z_2 dz_1 \wedge dz_3 + z_3 dz_1 \wedge dz_2\) and then work out \(\varphi ^*\omega \) as \(\varphi ^*(z_1) d(\varphi ^*z_2) \wedge d(\varphi ^*z_3) + \cdots \). However, in this case it may be more convenient to organise the calculation as follows: if \(\varphi ^*\omega = f dx_1 \wedge dx_2\), then \(f(x) = \mathrm {Det}\left (\varphi (x), \frac {\partial \varphi }{\partial x_1}, \frac {\partial \varphi }{\partial x_2}\right )\).
- (ii) Is \(\varphi \) an oriented parametrisation? Is \(\psi \)? Is \(\varphi ^{-1}\circ \psi |_{\R ^2\setminus \{0\}}\) orientation-preserving?
  
  Hint
  You only need to examine the signs of the coefficients of \(dx_1 \wedge dx_2\) and \(dy_1 \wedge dy_2\) in the results from (i).
- (iii) Evaluate \(\int _{S^2} \omega \).
  
  Hint
  It is enough to evaluate \(\int _{\mathbb {R}^2} \psi ^* \omega \).

4. Let \(M\) be any orientable \(n\)-dimensional closed manifold (i.e., \(M\) is a compact SMWB with \(\partial M =\varnothing \)). Show that \(M\) admits a differential \(n\)-form \(\omega \) which is closed but not exact.

Hint
Use Q1 and Stokes’ Theorem.

5. Sketch a proof of the Poincaré Lemma. [Other examples of sketch proof questions can be found in past papers.]

MA40254 Differential and geometric analysis : Solutions 10

1. Let \(\varphi _i:i\in I\) be a family of oriented parametrisations \(\varphi _i : \wt U_i \to U_i\) of \(M\), such that the set of images covers \(M\). (Since \(M\) is compact, we can in fact take \(I\) to be finite.) Let \(\rho _i:i\in I\) be a partition of unity on \(M\) subordinate to this cover. Then

\[ \int _M \omega = \sum _i \int _M \rho _i\omega = \sum _i \int _{\wt U_i} \varphi ^*(\rho _i \omega ) . \]

To evaluate each term on the RHS, we write

\[ \varphi _i^*\omega = f dx_1 \wedge \cdots \wedge dx_n \]

for a function \(f : \wt U_i \to \R \). By definition of \(\varphi _i\) being oriented \(f\) takes positive values, so

\[ \int _{\wt U_i} \varphi ^*(\rho _i \omega ) = \int _{\wt U_i} \rho _i(\varphi (x)) f(x) dx_1\cdots dx_n \in \R \]

is non-negative, and positive unless \(\rho _i \equiv 0\). Since the \(\rho _i\) are certainly not all identically zero, the sum is positive.

2. Consider the linear map

\[ L : \Omega ^n_c(N) \to \mathbb {R}, \; \alpha \mapsto \int _M \varphi ^*\alpha . \]

If \(\psi : \wt U \to U\) is an oriented parametrisation of \(N\) and \(U' := \varphi ^{-1}(U) \subseteq M\), then \(\varphi ^{-1} \circ \psi : \wt U \to U'\) is an oriented parametrisation of \(M\). If \(\alpha \in \Omega ^n_c(N)\) has \(\supp \alpha \subseteq U\), then \(\supp \varphi ^*\alpha \subseteq U'\), so the characterising property of \(\int _M\) gives

\[ L(\alpha ) = \int _{\wt U} (\varphi ^{-1} \circ \psi )^*(\varphi ^*\alpha ) = \int _{\wt U} (\varphi \circ \varphi ^{-1} \circ \psi )^* \alpha = \int _{\wt U} \psi ^*\alpha . \]

Thus \(L\) is an integration map on \(N\), and hence \(L = \int _N\).

3.
- (i) To identify the function \(f : \R ^2 \to \R \) such that \(\varphi ^*\omega = f dx_1 \wedge dx_2\), we can compute \(f(x)\) as
  
  \[ \omega _{\varphi (x)}\left (\pd {\varphi }{x_1}, \pd {\varphi }{x_2}\right ) = \Det \left (\varphi (x), \pd {\varphi }{x_1}, \pd {\varphi }{x_2}\right ) . \]
  
  Now
  
  \[ \pd {\varphi }{x_1} = \pd {\left (\frac {1}{1+\norm {x}^2}\right )}{x_1} (1+\norm {x}^2) \varphi (x) + \frac {2}{1+\norm {x}^2}(1, 0, x_1) , \]
  
  and \(\pd {\varphi }{x_2}\) has an analogous expression. Because \(\Det \) is alternating, the \(\varphi (x)\) terms in \(\pd {\varphi }{x_i}\) do not contribute to
  \(\seteqnumber{0}{}{0}\)
  \begin{align*} &\Det \left (\varphi (x), \pd {\varphi }{x_1}, \pd {\varphi }{x_1}\right ) \; = \; \frac {4}{(1+\norm {x}^2)^3} \det \pmat {2x_1 & 1 & 0 \\ 2x_2 & 0 & 1 \\ \norm {x}^2-1 & x_1 & x_2} \\ &= \; \frac {4}{(1+\norm {x}^2)^3}(\norm {x}^2-1 - 2x_1^2 - 2x_2^2) \; = \; -\frac {4}{(1+\norm {x}^2)^2} \end{align*} Thus
  
  \[ \varphi ^*\omega = -\displaystyle \frac {4}{(1+\|x\|^2)^2}dx_1\wedge dx_2. \]
  
  A similar calculation shows
  
  \[ \psi ^*\omega = \frac {4 }{(1+ \norm {y}^2)^2} dy_1 \wedge dy_2. \]
- (ii) \(\varphi \) is not oriented, but \(\psi \) is, and therefore \(\varphi ^{-1}\circ \psi \) is orientation-reversing.
- (iii) Since \(\psi \) is an oriented parametrisation and the complement of its image in \(S^2\) is a lower-dimensional set,
  
  \[ \int _{S^2} \omega = \int _{\R ^2} \psi ^*\omega = \int _{\R ^2} \frac {4 dy_1 dy_2}{(1+ \norm {y}^2)^2} = \int _0^{2\pi } \int _0^\infty \frac {4 r dr d\theta }{(1+r^2)^2} = 4\pi , \]
  
  since the integral over \(\theta \) gives \(2\pi \) and the integral over \(r\) gives \(2\) (e.g. the integrand is the derivative of \(-2/(1+r^2)\).

4. Let \(\omega \) be an orientation form. Then \(d\omega =0\) and (using the orientation defined by \(\omega \)) \(\int _M\omega >0\) by previous exercises. We cannot have \(\omega =d\beta \) since \(\int _M d\beta =0\) by Stokes’ Theorem.

5. We do not give model sketch proofs. A good sketch, along the lines of the proof in the notes, would include the following main relevant points:
- • the decomposition of \(\Alt ^k(\R ^n)\), and hence \(\alpha \in \Omega ^k(U)\), with respect to one variable;
- • the existence of \(\gamma \) with \(\alpha '=\alpha -d\gamma \) independent of one variable (e.g. using the second fundamental theorem of calculus to define \(\gamma \), and the formula for \(\cL (d\alpha )\) when \(\cL (\alpha )=0\));
- • an argument (e.g. pullback and induction) that \(\alpha '\) is exact.