Homework 2 Problem 6

Problem 6

Is $\alpha $ $\left( 3a\,db\wedge dc\right)$ closed? Exact? What about $\beta $ $\left( \sin \theta \,d\theta \wedge d\phi \right)$ and $\gamma $ $\left( -\sin \theta \cos ^{2}\theta d\theta +\sin \theta d\phi \right)$?

Solution

For the closedness, we should check whether $d(.)=0$;

(1)
\begin{align} d\alpha =d(3a\,db\wedge dc)=\frac{\partial }{\partial a}(3a)da\wedge db\wedge dc+\frac{\partial }{\partial b}(3a)d^{2}b\wedge dc+\frac{\partial }{ \partial c}(3a)db\wedge d^{2}c=3da\wedge db\wedge dc \end{align}
(2)
\begin{align} d\beta = d\left( \sin \theta \,d\theta \wedge d\phi \right) =\frac{ \partial }{\partial \theta }(\sin \theta )d^{2}\theta \wedge d\phi +\frac{ \partial }{\partial \phi }(\sin \theta )d\theta \wedge d^{2}\phi =0 \end{align}
(3)
\begin{align} d\gamma = d\left( -\sin \theta \cos ^{2}\theta d\theta +\sin \theta d\phi\right) = \cos \theta d\theta \wedge d\phi \end{align}

Thus, only $\beta$ is found to be closed.

For the exactness, we should check if $\exists \eta$ $k-1$ form such that $(.)=d\eta$ where $(.)$ is a $k-form$;

If $\alpha$ is exact, then a candidate $1-form$, $\eta$ should have either $db$ or $dc$ in it. Let $\eta =f(a)db$, then

(4)
\begin{align} d\eta =\frac{\partial }{\partial a}(f(a))da\wedge db\neq \alpha. \end{align}

Let $\eta =f(a,c)db$, then

(5)
\begin{align} d\eta =\frac{\partial }{\partial a}(f(a,c))da\wedge db+\frac{\partial }{% \partial c}(f(a,c))dc\wedge db\neq \alpha . \end{align}

Similar argument is valid for $\eta =f(.)dc$. Therefore, $\alpha$ is not exact.

For $\beta$, let $\eta =-\cos \theta d\phi$, then

(6)
\begin{align} d\eta =\frac{\partial }{\partial \theta }(-\cos \theta )d\theta \wedge d\phi +\frac{\partial }{\partial \phi }(\cos \theta )d^{2}\phi =\sin \theta d\theta \wedge d\phi =\beta \end{align}

Hence, $\beta$ is exact.

For $\gamma$, $\eta =g(\theta ,\phi )$, so

(7)
\begin{align} d\eta =\frac{\partial }{\partial \theta }g(\theta ,\phi )d\theta +\frac{ \partial }{\partial \phi }g(\theta ,\phi )d\phi. \end{align}

If $\gamma$ is exact, then;

(8)
\begin{align} \frac{\partial }{\partial \theta }g(\theta ,\phi )=-\sin \theta \cos ^{2}\theta \end{align}
(9)
\begin{align} \frac{\partial }{\partial \phi }g(\theta ,\phi )=\sin \theta \end{align}

But, from the second equation, the function $g(\theta ,\phi )$ should be in the form of $g(\theta ,\phi )=(\phi + C) \sin \theta$ where $C$ is a constant.

If we plug that in to the first partial differential equation and solve the left hand side;

(10)
\begin{align} \frac{\partial }{\partial \theta }g(\theta ,\phi )=\frac{\partial }{\partial \theta }((\phi + C) \sin \theta) =(\phi + C) \cos \theta \end{align}

If $(\phi + C) \cos \theta =-\sin \theta \cos ^{2}\theta \Rightarrow \phi =-\sin\theta \cos \theta - C$ which results in $1$ dimensional manifold with coordinate $\theta$.

Therefore, this is a contradiction to the actual dimension of the $2-$torus and concludes that $\gamma$ is not exact.

Scott agrees that only $\beta$ is closed and only $\beta$ is exact. Note, however, that a form can't be exact if
it's not closed, so more work was done above than was absolutely necessary.

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