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0 (Warmup). Show that \(\alpha :=\sin (x_2)^3 d x_1 \wedge dx_2\) is exact.
SolutionThere are three easy ways and one hard way. The hard way is to integrate \(\sin (x_2)^3\) explicitly. The easy ways are: \(\alpha =d(x_1 \sin (x_2)^3 dx_2)\); \(\alpha =d(-(\int ^{x_2} \sin (t)^3 dt) dx_1)\) by the Second Fundamental Theorem of Calculus; or \(\alpha \) is exact by the Poincaré Lemma.
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1. Let \(U = \{ p \in \R ^4 : x_2(p) \neq 0 \}\), and
\[ \alpha = \frac {x_2 dx_1 - x_1 dx_2}{x_2^2} \wedge (x_1dx_3 + x_2^2 dx_4) \in \Omega ^2(U) . \]
Compute \(d\alpha \in \Omega ^3(U)\) and express it in standard form.
HintOne way to organise the calculation is to first show that \(d\left (\frac {x_2 dx_1 - x_1 dx_2}{x_2^2}\right ) = 0\).
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2. For each of the following differential 3-forms \(\alpha \), find a differential \(2\)-form \(\beta \) such that \(d\beta = \alpha \). (Note we abbreviate multi-index notation as \(dx_{ijk}=dx_i \wedge dx_j \wedge dx_k\).)
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(i) \(\alpha = x_3 x_4\, dx_{123} + x_3^2 \,dx_{124} + 2x_2x_3\,dx_{134} + x_1x_3 \,dx_{234}\; \in \; \Omega ^3(\R ^4)\).
HintFirst look for \(\gamma \in \Omega ^2(\mathbb {R}^4)\) of the form \(\gamma = fdx_{23} + gdx_{24} + h dx_{34}\) (for \(f, g\) and \(h\) functions \(\mathbb {R}^4 \to \mathbb {R} \)) such that \(\alpha - d\gamma \) has no \(dx_{123}, dx_{124}\) or \(dx_{134}\) component. -
(ii) \(\alpha = \log (x_1)\exp (x_2)\cos (x_3)^2 dx_{123} \; \in \; \Omega ^3(\R ^+ \times \R ^2)\).
HintWhich function is easiest to integrate?
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(i) Show that any \(\omega \in \Alt ^2 (\R ^3)\) can be written as \(\omega = \alpha \wedge \beta \) for some \(\alpha , \beta \in \R ^{3*}\).
HintIf \(\omega = \alpha \wedge \beta \), what can you say about \(\omega \wedge \alpha \)? What is the dimension of the subspace \(\{ \gamma : \omega \wedge \gamma = 0 \} \subseteq \mathbb {R}^{3*}\)? -
(ii) Show that \(\epsilon _1 \wedge \epsilon _2 + \epsilon _3 \wedge \epsilon _4 \in \Alt ^2(\R ^4)\) cannot be written in the form \(\alpha \wedge \beta \) for any \(\alpha , \beta \in \R ^{4*}\). (Here \(\epsilon _i\) is the standard dual basis of \(\R ^{4*}\) as usual.)
HintIf \(\epsilon _1 \wedge \epsilon _2 + \epsilon _3 \wedge \epsilon _4 = \alpha \wedge \beta \), consider the result of taking the wedge product of each side with itself.
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MA40254 Differential and geometric analysis : Solutions 7
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1. The first factor equals \(d\bigl (\frac {x_1}{x_2}\bigr )\), so its exterior derivative vanishes. Hence, using the Leibniz rule,
\(\seteqnumber{0}{}{0}\)\begin{align*} d\alpha \; &= \; -\frac {x_2 dx_1 - x_1 dx_2}{x_2^2} \wedge d(x_1dx_3 + x_2^2 dx_4) \\ &= \; -\frac {x_2 dx_1 - x_1 dx_2}{x_2^2} \wedge (dx_1 {\wedge } dx_3 + 2x_2 dx_2 {\wedge } dx_4) \\ &= \; -\frac {2x_2^2 dx_1 {\wedge } dx_2 {\wedge } dx_4 + x_1 dx_1 {\wedge } dx_2 {\wedge } dx_3}{x_2^2} \end{align*}
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2.
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(i) We can first look for, for instance, a \(\gamma \in \Omega ^2(\mathbb {R}^4)\) of the form \(\gamma = fdx_{23} + gdx_{24} + h dx_{34}\), for \(f, g\) and \(h\) functions \(\mathbb {R}^4 \to \mathbb {R} \), such that \(\alpha - d\gamma \) has no \(dx_{123}, dx_{124}\) or \(dx_{134}\) component. That means we should take \(f\) such that \(\pd {f}{x_1} = x_3x_4\), e.g., \(f =x_1x_3x_4\). Similarly we can take \(g = x_1x_3^2\) and \(h = 2x_1x_2x_3\). Evaluating \(\alpha - d\gamma \) we find that actually the \(dx_{234}\) terms cancel too. Thus we can take
\[ \beta = \gamma = x_1x_3x_4dx_{23} + x_1x_3^2dx_{24} + 2x_1x_2x_3 dx_{34} . \]
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(ii) \((-x_1 {+} x_1\!\log (x_1))\exp (x_2)\cos (x_3)^2dx_{23}\) and \(-\log (x_1)\exp (x_2)\cos (x_3)^2dx_{13}\) and
\[\tfrac {1}{4} \log (x_1)\exp (x_2)(\sin (2x_3)+2x_3)dx_{12} \]
are three possible choices for \(\beta \), the middle one being the easiest!
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3.
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(i) The linear map \((\R ^3)^* \to \Alt ^3 (\R ^3), \; \alpha \mapsto \omega \wedge \alpha \) has kernel of dimension at least 2. Pick linearly independent elements \(\alpha _1, \alpha _2\) in the kernel, and extend to a basis \(\alpha _1, \alpha _2, \alpha _3\) of \((\R ^3)^*\). We can then write
\[ \omega = \lambda _1 \alpha _2 \wedge \alpha _3 + \lambda _2 \alpha _3 \wedge \alpha _1 + \lambda _3 \alpha _1 \wedge \alpha _2 \]
for some coefficients \(\lambda _i \in \R \). Now
\[ 0 = \omega \wedge \alpha _1 = \lambda _1 \alpha _1 {\wedge } \alpha _2 {\wedge } \alpha _3 \]
implies that \(\lambda _1 = 0\), and similarly \(\lambda _2 = 0\). We can thus take \(\alpha = \lambda _3\alpha _1\) and \(\beta = \alpha _2\).
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(ii) Writing \((\gamma )^2\) for \(\gamma \wedge \gamma \), we have
\[ (\epsilon _1 \wedge \epsilon _2 + \epsilon _3 \wedge \epsilon _4)^2 = 2\epsilon _1 \wedge \epsilon _2 \wedge \epsilon _3 \wedge \epsilon _4 \in \Alt ^4 \R ^4 , \]
which is non-zero. On the other hand,
\[ (\alpha \wedge \beta )^2 = 0 \]
for any \(\alpha , \beta \in (\R ^4)^*\).
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4.
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(i) \(d(\alpha \wedge \beta )=d\alpha \wedge \beta +(-1)^k\alpha \wedge d\beta =0\) by assumption and the Leibniz rule.
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(ii) Suppose \(\beta =d\gamma \). Then \(d ((-1)^k \alpha \wedge \gamma ) =(-1)^kd\alpha \wedge \gamma + \alpha \wedge d\gamma =\alpha \wedge \beta \), by assumption and the Leibniz rule.
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5. I do not provide model sketch proofs as there is no single right answer, and I want to encourage you to develop your own. Instead, I indicate the most crucial ideas. For the proof in lectures of the IFT for \(f\colon U \to \R ^n\) with \(Df_x\) invertible, I would highlight the following.
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• Using the continuity of \(Df\) to find a domain \(U'\) on which \(Df\) is close to \(Df_x\) (hence \(f\) is a local diffeomorphism).
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• Using the Mean Value Inequality to establish the contraction mapping property on \(U'\).
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• Using the Contraction Mapping Theorem to show \(f(U')\) is open.
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• The estimate \(||y-z||\leq 2 ||f(y)-f(z)||\), which is used to establish both injectivity of \(f\) and differentiability of the inverse.
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